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KINEMATICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

KINEMATICS THEORY OF KINEMATICS Kinematics : The study of the motion of an object without taking into consideration cause of its motion is called kinematics. NOTE : The word ‘kinematics’ comes from the Greek word “Kinema” which means motion. The word ‘dynamics’ comes from the Greek word “dynamics” which means power. BASIC DEFINITIONS Distance and Displacement Suppose an insect is at a point A(x1, y1, z1) at t = t1. It reaches at point B(x2, y2, z2) at t = t2 through path ACB with respect to the frame shown in fig. The actual length of curved path ACB is the distance travelled by the insect in time t = t2 – t1. Y

C A rA

B rB X

Z

If we connect point A (initial position) and point B (final position) by a straight line, then the length of straight line AB gives the magnitude of displacement of insect in time interval t = t2 – t1. The direction of displacement is directed from A to B through the straight line AB from the concept of vector,    the position vector of A is r A = x1ˆi + y1ˆj+ z1kˆ and that of B is r B = x 2ˆi + y 2ˆj+ z 2 kˆ . According to addition law of vectors,    r A + AB = r B     AB = r B - r A  AB = ( x 2 - x1 ) ˆi + ( y 2 - y1 ) ˆj+ ( z 2 - z1 ) kˆ The magnitude of displacement is  2 2 2 | AB |= ( x 2 - x1 ) + ( y 2 - y1 ) + ( z 2 - z1 ) NOTE : Distance covered by the body is always equal to or greater than its magnitude of displacement. Example 1. A man walks 3m in east direction, then 4m in north direction. Find distance covered and the displacement covered by man. Sol. The distance covered by man is the length of path = 3m + 4m = 7m. N B Let the man starts from O and reaches finally at B (shown in figure).  OB represents the displacement of man. From figure, 4m  2 2 | OB |= ( OA) + ( AB) O 3m A W E  2 2 | OB |= (3m ) + (4 m ) = 5 m S

KINEMATICS

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KINEMATICS 4m 4 = 3m 3

and

tan q =



æ4 ö q = tan- 1ç ÷ è 3ø

æ ö - 1 4 The displacement is directed at an angle tan çè ÷ north of east. 3ø Average Speed and Average Velocity Suppose we wish to calculate the average speed and average velocity of the insect (in section (i)) between t = t1 and t = t2. From the path (shown in fig.) we see that at t = t1, the position of the insect is A(x1, y1, z1) and at t = t2, the position of the insect is B(x2, y2, z2). Y

C A rA

B rB X

Z

The average speed is defined as total distance travelled by a body in a particular time interval divided by the time interval. Thus, the average speed of the insect is The length of curve ACB v av = t 2 - t1 The average velocity is defined as total displacement of the body in a particular time interval divided by the time interval. Thus, the average velocity of the insect in the time interval t2 – t1 is   AB v av = t - t 2 1  rB - rA v av = t 2 - t1

 ( x - x1 ) ˆi + ( y2 - y1 ) ˆj+ ( z 2 - z1 ) kˆ v av = 2 t 2 - t1 Example 2. In one second a particle goes from point A to point B moving in a semicircle (fig.). Find the magnitude of average velocity. A  AB | vav |= Sol. Dt 1.0m  2.0 | vav |= m/s 1.0 B  | vav |= 2 m / s Example 3. A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h. If AB = 6BC the average speed in km/h between A and C is _______ Sol. AB = 40t1 ...(1) BC = 60t2 ...(2)

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KINEMATICS

total distance travelled time taken AB + BC Vav = t1 + t 2 From eqn. (1) and (2) A 40t1 + 60t 2 Vav = ...(3) t1 + t 2 According to question AB = 6BC 40t1 = 6 × 60t2 From eqn (1) and (2) t1 = 9t2 40  9t 2  60t 2 Vav  From eqn (3) from eqn (3) 9t 2  t 2 420t 2 Vav = Vav = 42 km / h  10t 2 Average speed =

B

C

Instantaneous Velocity Instantaneous velocity is defined as the average velocity over smaller and smaller interval of time.    Suppose position of a particle at t is r and at t + t is r + Dr . The average velocity of the particle for time   Dr interval t is v av = . Dt From our definition of instantaneous velocity, t should be smaller and smaller. Thus, instantaneous velocity is    Dr dr v = lim = Dt ® 0 Dt dt  Example 4. Let at any time t, the position vector of a particle is r = x ˆi + y ˆj+ z kˆ . Find the velocity of the particle.

Sol. x-component of velocity, v x =

dx dt

y-component of velocity, v y =

dy dt

vz =

dz dt

z-component of velocity,

Thus, velocity of particle  v = v x ˆi + v y ˆj+ v 2 kˆ  dx ˆi + dy ˆj+ dz kˆ v= dt dt dt

Average and Instantaneous Acceleration In general, when a body is moving, its velocity is not always the same. A body whose velocity is increasing is said to be accelerated. Average acceleration is defined as change in velocity divided by the time interval.  Let us consider the motion of a particle. Suppose that the particle has velocity v1 at t = t1 and at a later time

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KINEMATICS  t = t2 it has velocity v 2 . Thus, the average acceleration during time interval t = t2 – t1 is     v 2 - v1 Dv a av = = t 2 - t1 Dt

If the time interval approaches to zero, average acceleration is known as instantaneous acceleration. Mathematically,    Dv dv a = lim = Dt ® 0 Dt dt      Example 5. The velocity of a point depends on time t, as v = c + t b where c and b are constant vectors. Find the instantaneous acceleration at any time t. Sol. Acceleration at any time t,   dv d   a= = (c + t b ) dt dt    a = 0+ b = b

IMPORTANT FEATURES 1. If a body is moving continuously in a given direction on a straight line, then the magnitude of displacement is equal to distance. 2. Generally, the magnitude of displacement is less or equal to distance. 3. Many paths are possible between two points. For different paths between two points, distances are different but magnitudes of displacement are same. 4. The slope of distance-time graph is always greater or equal to zero. 5. The slope of displacement-time graph may be negative. 6. If a particle travels equal distances at speeds v1, v2, v3, ...... etc. respectively, then the average speed is harmonic mean of individual speeds. 2v1 v 2 7. If a particle moves a distance at speed v1 and comes back with speed v2, then v av = v1 + v 2  but v av = 0 8. If a particle moves in two equal intervals of time at different speeds v1 and v2 respectively, then v + v2 v av = 1 2 9. The average velocity between two points in a time interval can be obtained from a position versus time graph by calculating the slope of the straight line joining the co-ordinates of the two points. x2

x2

x1

x1

t1

t2 (a)

B A

( x2 – x1)

( t 2 – t1)

t1

t2 (b)

The graph [shown in fig.], describes the motion of a particle moving along x-axis (along a straight line). Suppose we wish to calculate the average velocity between t = t1 and t = t2. the slope of chord AB [shown in fig.(b)] gives the average velocity.

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KINEMATICS x 2 - x1 t 2 - t1 10. If a body moves with constant velocity, the instantaneous velocity is equal to average velocity. The instantaneous speed is equal to modulus of instantaneous velocity.

Mathematically,

v av = tan q =

11. x-component of displacement is y-component of displacement is z-component of displacement is

ò v dt Dy = ò v dt Dz = ò v dt Dx =

x

y

z

Thus, displacement of particle is  Dr = Dx ˆi + Dy ˆj+ Dz kˆ 12. If particle moves on a straight line, (along x-axis), then v = 13. 14. 15. 16. 17.

dx . dt

The area of velocity-time graph gives displacement. The area of speed-time graph gives distance. The slope of tangent at position-time graph at a particular instant gives instantaneous velocity at that instant. The slope of velocity-time graph gives acceleration. The area of acceleration-time graph in a particular time interval gives change in velocity in that time interval.

ONE, TWO AND THREE DIMENSIONAL MOTION One Dimensional Motion If only one of the three co-ordinates is required to specify the position of an object in space changing w.r.t. time, then the motion of the object is called one dimensional motion. Motion of a particle in a straight line can be described by only one component of its velocity and acceleration. For example, motion of a block in a straight line, motion of a train along a straight track, a man walking on a level and narrow road, an object falling under gravity, etc. Two Dimensional Motion If two of the three co-ordinates are required to spacify the position of an object on space changing w.r.t. time, then the motion of the object is called two dimensional motion. The motion of a particle through its vertical plane at some angle with horizontal. (¹ 90º) is an example of two dimensional (2–D) motion. This is a projectile motion. Similarly, a circular motion is an example of 2–D motion. A 2–D motion takes place in a plane and its velocity (or acceleration) can be described by two components in any two mutually perpendicular directions (vx and vy). Three Dimensional Motion If all the three co-ordinates are required to specify the position of an object in space changing w.r.t. time, then the motion of an object is called three dimensional motion. Such a motion is not restricted to a straight line or plane but takes place in space. In a 3–D motion velocity and acceleration of a particle can be resolved in three components (vx, vy, vz, ax, ay, az). A few examples of 3–D motion are a flying bird, a flying kite, a flying aeroplane, the random motion of gas molecules, etc. UNIFORMLY ACCELERATED MOTION A motion, in which change in velocity in each unit of time is constant, is called uniformly accelerated motion. So, for uniformly accelerated motion, acceleration is constant or approximately. So for uniformly accelerated motion  ( a = constant) , equations of motion are as under,,    ...(i) v = u+ a t

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KINEMATICS   1 s = u t + a t2 ...(ii) 2       and ...(iii) v . v = u . u + 2a . s  where u = initial velocity of particle  v = velocity at time t  s = displacement of particle at time t If motion is described in one dimension, so vector sign () need not to be used. Normally, vertical upward motion is taken negative and vertical downward motion is taken positive. Similarly, for horizontal rightward motion is taken positive and leftward motion is taken negative. Sign convention : Sign convention for (a) motion in vertical (b) motion in horizontal, is shown in Fig.

+ve

–ve (a)

+ve (b)

–ve

Example 6. At a distance L = 400m from the traffic light brakes are applied to a locomotive moving at a velocity v = 54 km/hr. Determine the position of the locomotive relative to the traffic light 1 min after the application of the brakes if its acceleration is –0.3 m/sec2. 5 Sol. u  54   15 m / s 18 a = –0.3 m/s2  v = u + at 0 = 15 – 0.3 t0 15 t0 = = 50 sec 0.3 After 50 second, locomotive comes in rest permanently.  v2 = u2 + 2as O2 = 152 – 2 × 0.3 S0

S0 = 

225 2250 = = 375m 0.6 6

the distance of the locomotive from traffic light = 400 – 375 = 25 metre

Example 7. A car moves in the x–y plane with acceleration (3iˆ + 4 ˆj) m / s 2 . (a) Assuming that the car is at rest at the origin at t = 0, derive expression for the velocity as function of time. (b) Find the equation of path of car and find the position vector as function of time. Sol. Here, ux = 0, uy = 0, uz = 0 2 2 ax = 3 m/s , ay = 4 m/s (a) vx = ux + axt or vx = 3t and vy = uy + ayt

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KINEMATICS or

vy = 4t  v = v x ˆi + v y ˆj  v = (3tiˆ + 4tjˆ)



1 2 a t 2 x 1 3 ´ 3t 2 = t 2 2 2 1 2 uyt + a yt 2 1 (4)t 2 = 2t 2 2 2 4 2´ x = x 3 3

x = uxt +

(b) or

x=

and

y=

or

y=



y=

æ 3 2ö ç x= t ÷ è 2 ø

4 x 3 Hence, the path is straight line. The position of car is  3 r = x ˆi + y ˆj = t 2 ˆi + 2t 2 ˆj 2



y=

Motion Under Gravity The most familiar example of motion with costant acceleration on a straight line is motion in a vertical direction near the surface of earth. If air resistance is neglected, the acceleration of such type of particle is gravitational acceleration which is nearly constant for a height negligible with respect to the radius of earth. The magnitude of gravitational acceleration near surface of earth is g = 9.8 m/s2 = 32 ft/s2. Discussion : Case-I : If particle is moving upwards : In this case applicable kinematics relations are : g v = u – gt ...(i) 1 h = ut - gt 2 ...(ii) u 2 v2 = u2 – 2gh ...(iii) Here, h is the vertical height of the particle in upward direction. NOTE : For maximum height attained by projectile h = hmax, v=0 2 2 i.e., (0) = u – 2ghmax



hmax =

u2 2g

Case-II : If particle is moving vertically downwards : In this case : v = u + gt v2 = u2 + 2gh

KINEMATICS

u

...(i) ...(ii)

g

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KINEMATICS 1 2 gt ...(iii) 2 Here, h is the vertical height of the particle in downward direction. h = ut +

Example 8. A particle is projected vertically upwards with velocity 40 m/s. Find the displacement and distance travelled by the particle in (a) 2s (b) 4s (c) 6s [Take g = 10 m/s2] Sol. Here, u is positive (upwards) and a is negative (downwards). So, first we will find t0, the time when velocity becomes zero.

t0 =

u 40 = = 4s a 10

(a) t < t0. Therefore, distance and displacement are equal. d = s = ut +

1 2 at 2

1 ´ 10´ 4 = 60 m 2 (b) t = t0. So, again distance and displacement are equal. 1 d = s = ut + at 2 2 1 d = 40´ 4- ´ 10´ 16 = 80 m 2 (c) t > t0. Hence, d > s 1 s = 40 ´ 6 - ´ 10´ 36 = 60 m 2 d = 40 ´ 2 -

While

d=

u2 1 2 + a (t - t 0 ) 2a 2

d=

(40)2 1 + ´ 10´ (6 - 4)2 = 100 m 2 ´ 10 2

Example 9. A ball is thrown upwards from the ground with an initial speed of u. The ball is at a height of 80 m at two times, the time interval being 6s. Find u. Take g = 10 m/s2. Sol. Here, u = u m/s, a = g = – 10 m/s2 and s = 80 m. 1 Substituting the value in s = ut + at 2 , we have 2 2 80 = ut – 5t or 5t2 – ut + 80 = 0 or and

t= t=

u+ u-

Now, it is given that

KINEMATICS

u 2 - 1600 10 u 2 - 1600 10

+ve

s = 80 m

–ve

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u

8

KINEMATICS

u+

u 2 - 1600 u 10

u 2 - 1600 =6 10

u 2 - 1600 or =6 or u 2 - 1600 = 30 5 or u2 – 1600 = 900  u2 = 2500 or u = ± 50 m/s Ignoring the negative sign, we have u = 50 m/s A

Example 10. A disc arranged in a vertical plane has two groves of same length directed 60º along the vertical chord AB and CD as shown in the fig. The same particles slide D down along AB and CD. The ratio of the time tAB/t CD is : (A) 1 : 2 (B) 1 : 2 (C) 2 : 1 (D) 2 : 1 B 1 2 SAB = g t AB Sol. 2 1 SCD = g cos 60º t 2CD 2 A C But SAB = SCD 60 60º º 60º 1 2 gcos g t g AB SAB 2 D = g  SCD 1 g cos 60º t 2 CD 2 B t AB t 2AB = 1 : 2 1 = 2 or  t CD t 2CD

C

Example 11. A stone is dropped from a height h. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height of 4h. Find the time when two stones cross each other. Sol. For second stone, v 2 = v02 - 2g (4h) 0 2 = v 20 - 8gh 

v 0  8gh

but they meth at height H in time t0.  Displacement of 1st stone is 1 h – H  gt 20 ...(1) 2 and that of second stone is 1 H = v0 t 0 - gt 02 ...(2) 2 After solving eqn (1) and (2) 1 1 h – v 0 t 0  g t 02  g t 02 or 2 2  h = v0t0

KINEMATICS

v=0

(1) (2)

4h

h v0

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KINEMATICS 

t0 =

h h = v0 8gh



t0 =

h 8g

Example 12. A rocket is launched at an angle 53º to the horizontal with an initial speed of 100 ms–1. It moves along its initial line of motion with an acceleration of 30 ms–2 for 3 seconds. At this time its engine fails & the rocket proceeds like a free body. Find : (i) the maximum altitude reached by the rocket (ii) total time of flight. Sol. S0 = ut + ½ at 2 = 100 × 3 + ½ × 30 × 9 = 300 + 135 = 435 m In OAB h sin 53º = S0 4 h  S0 sin 53º  435   87  4  348 m 5 v0 = u + at = 100 + 30 × 3 = 190 m/s (velocity at the time of switch off) After engin switch off 3 v0x  v0 sin 37º  190   114 m / s 5 4 v0y  v0 cos37º  190   152 m / s 5 ay = – 10 m/s2, ax = 0 (i) At maximum altitudes vy = 0 v 2y = v 20y + 2a y h 0

0 = v 20y + 2a y h 0 h0  –

A S0

v20y

u 53º

2a y O

152  152 h0   2  10 11552 h0 = h 0 = 1155.2 m  10 The maximum altitude reached by the rocket is

37º v0

37º h

Engin fail x

B

= h0 + h = (1155.2 + 348) m



1503.2 m

(ii)Total time of flight. 1 y = v0 y t + a y t 2 2 1 - h = v 0y t 0 + a y t 02 2 1155.2  152t 0 

KINEMATICS

1  10  t 20 2

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KINEMATICS - 1155.2 = 152t 0 - 5t 20 5 t 20 - 152 t 0 - 1155.2 = 0 t 02 - 30.4 t 0 - 231.04 = 0

t 0 = 35.54 sec.

NON-UNIFORMLY ACCELERATED MOTION When motion of a particle is not uniform i.e., acceleration of particle is not constant or acceleration is a function of time, then following relations hold for one dimensional motion : v=

(i)

ds dt

dv dv = v dt ds (iii) ds = vdt and (iv) dv = adt or vdv = ads Such problems can be solved either by differentiation or integration applying some boundary conditions. a=

(ii)

Example 13. A particle is moving with a velocity of v = (3 + 6t + 9t2) cm/s. Find out : (a) the acceleration of the particle at t = 3 s. (b) the displacement of the particle in the interval t = 5 s to t = 8 s. Sol. (a) Acceleration of particle a=

dv = ( 6 + 18t ) cm / s 2 dt

At t = 3 s, a = (6 + 18 × 3) cm/s2 a = 60 cm/s2 (b) Given, v = (3 + 6t + 9t2) cm/s

or

ds = (3+ 6t + 9t 2 ) dt ds = (3 + 6t + 9t2)dt



ò



s = ëé 3t + 3t 2 + 3t 3 ûù5 or s = 1287 cm

or

8

5

ds =

8

ò (3+ 6t + 9t ) dt 2

5

8

Example 14 : The motion of a particle along a straight line is described by the function x = (2t – 3)2 where x is in metres and t is in seconds. (a) Find the position, velocity and acceleration at t = 2 s. (b) Find the velocity of the particle at origin. Sol. (a) Position, x = (2t – 3)2 Velocity,

v

dx  4  2t  3 m / s dt

and acceleration, a =

dv = 8 m / s2 dt

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KINEMATICS At t = 2 s, x = (2 × 2 – 3)2 = 1.0 m v = 4(2 × 2 – 3) = 4 m/s and a = 8 m/s2 (b) At origin, x = 0 or (2t – 3) = 0  v=4×0=0 EQUATIONS OF MOTION FOR VARIABLE ACCELERATION Case-I : When acceleration a of the particle is a function of time : Since, acceleration of a particle is a function of time, i.e., a = f(t) dv = f (t) dt  dv = f(t)dt Integrating within the proper limits, we get



v = u+

t

ò0 f (t)dt

Case-II : When acceleration a of the particle is a function of distance : Since, acceleration of a particle is a function of distance, i.e., a = f(x) dv = f (x) dt



dv dx . = f (x) dx dt  vdv = f(x)dx Integrating with in proper limits, we get



x

v 2 = u 2 + 2ò f (x)dx x0

Case-III : When acceleration a of the particle is a function of velocity : Since, acceleration of a particle is a function of velocity, i.e., a = f(v) dv = f (v) dt dv dt =  f (v) Integrating within proper limits, we get



t=

v

dv

òu f (v)

Therefore, we get v as a function of t i.e., v(t). Again   

dv = f (v) dt dv dx . = f (v) dx dt dv v = f (v) dx vdv dx = f (v)

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KINEMATICS x

òx





v

vdv

v

vdv

v

vdv

dx =

òu f (v)

x - x0 =

òu f (v)

x = x0 +

òu f (v)

0

Therefore, we get x as a function of v i.e., x(v). GRAPHS IN ONE DIMENSIONAL MOTION The tabular forms of s–t and v–t graphs are given for one-dimensional motion with uniform velocity or with constant acceleration. Table-1 : Displacement-Time Graph S.No. Different Cases

s–t Graph

The main Features of Graph

s

1.

At rest

Slope = v = 0 t s

2.

s = ut

Uniform motion

s = 0 at t = 0 t

s

3.

1 s = 2 at 2

Uniformly accelerated motion with u = 0, s = 0 at t = 0

t

u = 0 i.e., slope of s–t graph at t = 0, should be zero.

s

4.

1 s = ut + 2 at2

Uniformly accelerated motion with u ¹ 0 but s = 0 at t = 0

t

Slope of s–t graph gradually goes on increasing.

s

5.

Uniformly accelerated motion with u ¹ 0 but s = s0 at t = 0

s0

2 s = s0 + ut + 1 2 at

s = s0 at t = 0

t

s

6.

Uniformly retarded motion t0

t

At t = t0, slope of s–t graph becomes zero.

Table-2 : Velocity-Time Graph S.No. Different Cases

v–t Graph v

1.

The main Features of Graph

v = constant a=0

Uniform motion t

KINEMATICS

(i) Slope of v–t graph = a = 0 (ii) v = constant

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KINEMATICS v

2.

v = at

Uniformly accelerated motion with u = 0, s = 0 at t = 0

t

(i) u = 0 i.e., v = 0 at t = 0 (ii) a or slope of v–t graph is constant.

t

At t = 0, v¹ 0 and slope of v–t graph is not zero.

v v = u + at

u

3.

Uniformly accelerated motion with u ¹ 0 but s = 0 at t = 0 v v = u + at

u

4.

Uniformly accelerated motion with u ¹ 0 but s = s0 at t = 0

v = u at t = 0 t

v u v = u – at

5.

Uniformly decelerated motion

Slope of v–t graph = –a (retardation) t

t0 v

6.

Non-uniformly accelerated motion

Slope of v–t graph increases with time

t

Acceleration-Time Graph (i) When a–t graph is a straight line parallel to time-axis, then acceleration = a = constant. a

Slope = 0

t (a)

(ii) When a–t graph is a straight line passing through origin, then acceleration of particle is increasing uniformly. a op Sl

e=

e +v

t (b)

(iii) When a–t graph is a straight line of negative slope, then acceleration is decreasing uniformly. a

Sl op e=

–v e

t (c)

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KINEMATICS v 20

Example 15. The fig. shows the v–t graph of a particle moving in straight line. Find the time when particle returns to the starting point. Sol. When the particle comes at initial position, total displacement is zero. Since, the area of v – t graph gives displacement. In this case area of v – t graph should be zero. 1 1  S   20  25   t  25 v 0 2 2 v0 or 0  250 –  t  25 2 500 or t – 25  v ...(1) 0 v 20  0 Also tan   from figure 25  20 t  25 v0 or 4  t – 25  v 0  4  t – 25 ...(2) From eqn (1) and (2) 500 125 t – 25   4  t – 25 t – 25

10

10 20

25 t

v 20

20

25

t

20

t(s) 40

v0

 t – 25 2  125 t – 25  125  5 5



t = (25 + 5 5 ) sec.



t = 36.2 sec

Example 16. From the velocity-time plot shown in figure, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. Sol. The distance fravelled by the particle during the first 20 second. 1 S1   v  t 2 1 S1   5  20 S1 = 50 m  2 The distance travelled by the partile during next 20 second is 1 S2   5  20 2 S2 = 50 m Since distance is a sclar quantity therefore total distance = S1 + S2 = 50 + 50

V 5m/s

–5m/s

S = 100 m

Example 17. A ball is dropped from a height of 80m on a floor. At each collision the ball losses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor, [Take g = 10 ms–2].

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KINEMATICS Sol. The time in first collision 1 2 gt (during downward motion) 2 1 80 = 0 + ´ 10 ´ t 2 ( u = 0) or 2 2´ 80 t= = 4s or 10 Final speed just before first collision v = 0 + 10 × 4 = 40 m/s h = ut +

æ 40 ö ÷m / s. It now loses half of initial speed after the collision i.e., when it first bounces its initial speed is 20çè = 2ø So, the time is loosing half of its speed. 0 = 20 – 10 × t´ (during upward motion) 20 t´= = 2s (final speed = 0) 10 In 2 s, it attains height 1 h´= 20´ 2 - ´ 10´ (2) 2 2 h´ = 40 – 20 = 20 m Now, it is dropped again from 20m with zero initial speed. Time taken in reaching the ground 1 20 = 0 + ´ 10(t) 2 2 t=2s Also final speed v´2 = 0 + 2 × 10 × 20 (from v2 = u2 + 2gh)  v´ = 20 m/s Thus, with the above data, we can draw the speed-time graph. Speed (m/s) 40 20 4

6

Time(s)

8

Since, velocity is a vector quantity so from the above graph, we can now draw the velocity-time graph. [Take downward motion positive and upward motion negative in case of v–t graph] Velocity (m/s) 40 20 4

6

Time(s)

8

–20

Example 18. The velocity-time graph of an object moving along a straight line is as shown in the fig. Calculate the distance covered by the object : 20 ms–1

A

B

A´ 2 t (s)

5

v O

KINEMATICS

0

B´

C 10

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KINEMATICS (a) between t = 0 to t = 5 s and (b) between t = 0 to t = 10 s. Sol. (a) Let x1 be the distance covered in the time interval between t = 0 to t = 5 s. Then, x1 = area of the trapezium OABB´ æ AB + OB´ ö 3+ 5 =ç ´ 20 = 80 m ÷´ AA´= è ø 2 2 (b) Let x2 be the distance covered in the time interval between t = 0 to t = 10 s. Then, x2 = area of the trapezium OABC. AB + OC 3 + 10 = ´ AA´= ´ 20 = 130 m 2 2 IMPORTANT FEATURES 1. For a particle having zero initial velocity if v  t, s  t2 and v2  s then acceleration of particle must be constant i.e., particle is moving rectilinearly with uniform acceleration. 2. For a particle having zero initial velocity if s  t, where  > 2, then particle’s acceleration increases with time. 3. For a particle having zero initial velocity if s  t, where  < 0, then particle’s acceleration decreases with time. 4. When a body is non-uniformly accelerated, then problem can be solved either be differentiation or integration (applying some boundary conditions). Differentiation : s – t  v – t  a – t Integration : a–tv–ts–t By boundary condition we mean that velocity or displacement at some time (usually at t = 0) should be known to us. Otherwise we cannot find constant of integration. dv 5. Equation a = v or vdv = ads is useful when acceleration displacement equation is known and velocity ds displacement equation is required.   6. When either u = 0 or u ­ ­ a , motion is only accelerated.     7. It can be observed when either u = 0, u ­ ­ a or u ­ ¯ a .   8. When u ­ ¯ a motion is first retarded (till the velocity becomes zero) and then accelerated in opposite direction. 9. Following are few points we may conclude in case of a one dimensional motion : æ ds ö (a) slope of displacement-time graph gives velocity çè as v = ÷ . dt ø æ dv ö ÷. (b) slope of velocity-time graph gives acceleration çè as a = dt ø (c) area under velocity-time graph gives displacement as (ds = vdt). (d) area under acceleration-time graph gives change in velocity (as dv = adt). (e) displacement-time graph in uniform motion is a straight line passing through origin, if displacement is zero at time t = 0 (as s = vt). (f) velocity-time graph is a straight line passing through origin in a uniformly accelerated motion if initial velocity u = 0 and a straight line not passing through origin if initial velocity u¹ 0 (as v = u + at). æ 1 2ö (g) displacement-time graph in uniformly accelerated or retarded motion is a parabola çè as s = ut ± at ÷ . 2 ø 10. Slopes of v–t or s–t graphs can never be infinite at any point, because infinite slope of v–t graph means infinite acceleration. Similarly, infinite slope of s–t graph means infinite velocity. Hence, the following graphs are not possible :

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KINEMATICS v

s

t

t

11. At one time, two values of velocity or displacement are not possible. Hence, the following graphs are not acceptable : v

s

v1 v2

s1 s2 t0

t

t0

t

RELATIVE MOTION If the velocities of two bodies are known w.r.t a common frame of reference, then the velocity of a body can be  measured w.r.t. the second body. Therefore, if the velocities of two bodies A and B w.r.t. the ground are v A  and v B then the relative velocity of A w.r.t. B is    v AB = v A - v B Similarly, we see that   v AB = - vBA Also, relative acceleration of A w.r.t. B is    and a AB = a A - a B   a AB = - a BA Example 19. Car A has an acceleration of 6 m/s2 due east and car B, 8 m/s2 due north. What is the acceleration of car B w.r.t. to car A ? Sol. It is a two dimensional motion.  aBA aB = 8 m/s 2 Therefore, a BA = acceleration of car B w.r.t. car A N   = aB - aA W E   a BA = (6) 2 - (8)2 = 10 m / s 2 S and  Thus, a BA

æ8 ö æ4ö –aA = 6 m/s 2 a = tan- 1ç ÷= tan- 1ç ÷ è 6ø è 3ø -1æ 4 ö is 10 m/s2 at an angle of α = tan çè ÷ from west towards north. 3ø

Discussion   (a) If a satellite is moving in equatorial plane with velocity v and a point on the surface of earth with velocity u relative to the centre of earth, the velocity of satellite relative to the surface of earth    vSE = v - u  (b) If a car is moving at equator on the earth’s surface with a velocity v CE relative to earth’s surface and a point on the surface of earth with velocity vE relative to its centre, then    v CE = vC - v E (c) If the car moves from west to east (the direction of motion of earth)

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KINEMATICS vC = vCE + vE and if the car moves from east to west (opposite to the motion to earth) vC = vCE – vE (d) For crossing the river in shortest time, the boat should sail perpendicular to the flow. If the width of river is d and v the velocity of boat in still water, then, d t= v The position of boat at the other bank is C (not B).    The displacement of the boat = OC = OB + BC

B

vr

C

vr v

O

(OB) 2 + (BC) 2

OC =

æ d ö2 OC = d + (v r t) = d + ç v r ÷ è vø (e) For crossing the river in shortest distance, the boat moves as such its horizontal component of velocity balances the speed of flow. A vr B OB = the shortest path = d vr v v sinθ = r vr = v sin  v 2

2

2

cos θ = 1­ sin 2θ



O

é æ v ö2 ù æ v 2 - v 2 ö r ÷ cos θ = ê 1­ç r ÷ ú = ç è ø v v ø ë û è d d t= = v cos q v v 2 - v 2 r

v

 (f)

t=

d 2

v - v 2r In this case, the magnitude of displacement = d. If boat crosses the river along the shortest path, then time is not least.

Example 20. A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hits the thief’s car ?    Sol. Let v p , v b and v t represent velocity of the police van, muzzle velocity of bullet and the velocity of the thief’s ’s car respectively. This is a one dimensional motion.  25 v p = 30km / h = m/s 3  v b = 150 m / s  160 v t = 192 km / h = m/s and 3 Since, bullet is fired from the moving police van, the effective velocity of the bullet will be    25 475 v´t = v b + v p = 150 + = m/s 3 3

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KINEMATICS  if the relative velocity of the bullet w.r.t the thief’s car is v bt , then    v bt = v´b - v t  475 160 315 v bt = = = 105 m / s 3 3 3

Example 21. A, B & C are three objects each moving wit constant velocity. A’s speed is 10m/sec in a  direction PQ . The velocity of B relative to A is 6 m/sec at an angle of cos–1(15/24) to PQ. The velocity  of C relative to B is 12 m/sec in a direction QP , then find the magnitude of the velocity of C.  Sol. v A = 10 ˆi  v B  6 cos  ˆi  6sin  ˆj 15 351 Here cos   ,  sin   24 24  And v CB = - 12 ˆi     v BA = v B - vA  or 6cos  i  6sin  j  v B  10 ˆi   v B  cos  ˆi  6sin  ˆj  10 ˆi  6

=

15 ˆ 351 ˆ i6 j  10 ˆi 24 24 351 ˆ  15 ˆ  4  10 i  4 j

55 ˆ 351 ˆ i j 4 4    v CB = vC - v B    55 351 ˆ v C = vCB + v B = - 12 ˆi + ˆi + j 4 4 7 351 ˆ = ˆi + j 4 4 

But 



2  7   351  vC       4   4 

=

49 + 351 = 4

2

400 20 = 4 4

Ans.

5 m/s

Example 22. A man crosses a river in a boat. If he crosses the river in minimum time he takes 10 min with a drift 120 m. If he crosses the river taking shortest path, he takes 12.5 min, find : (a) width of the river B (b) velocity of the boat with respect to water, (c) speed of the current. d v Sol. Let vr = velocity of river v = velocity of river in still water and A d = width of river For minimum time

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KINEMATICS Given,

tmin = 10 min

d = 10 v Drift in this case will be, x = vrt  120 = 10vr Shortest path is taken when vb is along AB. In this case,

or

Now,

vb =

v 2 - v 2r

125 =

d = vb

...(i)

...(ii)

vr

d v 2 - v 2r

Solving these three equations we get, v = 20 m/min, vr = 12 m/min and d = 200 m.

B

...(iii) v

vb

A Shortest path

Example 23. A man with some passengers in his boat, starts perpendicular to flow of river 200m wide and flowing with 2m/s. boat speed in still water is 4m/s. When he reaches half the width of river the passengers asked him they want to reach the just opposite end from where they have started. (a) Find the direction due to which he must row to reach the required end. (b) How many times more total time, it would take to that if he would have denied the passengers. Sol. (a) Event (1) — From A to B, Time taken by boat to recent from A to B is 100 100 t= = = 25sec. vy 4 Also, AD = vx t = 2t = 50 m Event (2) — From B to C, Actual velocity of boat should be along BC. This actual velocity is found by resultant of vrel and vr.    v rel = v b - vr x–comput of actual velocity is y vx = vr – vrel sin = 2 – 4sin C E vb and y – comput of actual velocity is B 200m vrel vy = vrel cos vrel vr The time taken to go from B to C is D A BE t0 = vy 100 100   v rel cos  4 cos  25 cos  EC = –vx t 0 = –(2 – 4 sin ) t 0 25   4 sin   2 cos  EC = AD = 50 m 

Also,

But

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KINEMATICS 

 (b)

25 cos  25 50   4 sin   2  cos  50 cos  = 100 sin  – 50 cos  = 2 sin  – 1 1 + cos  = 2 sin     2 cos 2  4sin cos 2 2 2 a 1 tan = 2 2   1  tan 1    2 2 EC   4 sin   2

 1   2 tan 1    2 If boat crosses the river with initial condition 200 200 t1 = = vy vrel

200 = 50sec. 4 If boat cross the river with final condition, t 2 = t AB + t BC t2 = t + t0 25 t 2  25  cos   1 tan   2 2  2 cos  2 5  cos   2 cos 2  1 2 4  2  1 5 8 3 = - 1= 5 5 25 125 200 t 2 = 25 + = 25 + = 3 3 3  5 t2 4 t2 200 4 =     t1 3 t1 3  50 3 t1 =

Example 24. To a person going west wards with a speed of 6 km/h rain appears to fall vertically downwards with a speed of 8 km/h. Find the actual direction of rain.  Sol. Let v M = velocity of man = 6 km/h  v = relative velocity of rain w.r.t. man = 8 km/h

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KINEMATICS  v R = actual velocity of rain

Vertical N

In this case

   v = v R + (- v M ) vM O W    E A v = vR - vM v vR B    or C vR = vM + v S  S  v R = (6) 2 + (8)2 = 10 km / h     The velocity of rain ( v R ) is given by the vector OC , the resultant of vectors OA and OB as shown in figure.  If  is the angle that v R makes with the vertical, then  BC | v M | 6 tan       0.75 OB | v | 8 or  = 36º 52´ (east of vertical) Example 25. Rain is falling vertically with a speed of 20 ms–1 relative to air. A person is running in the rain with a velocity of 5 ms–1 and a wind is also blowing with a speed of 15 ms–1 (both towards east). Find the angle with the vertical at which the person should hold his umbrella so that he may not get drenched.  Sol. v ra = - 20 kˆ z N  v m = 5 ˆi  E v a = 15 ˆi x W    v ra = v r - v a  S v r = - 20 kˆ + 15 ˆi    v rm = v r - v m = - 20kˆ +15iˆ - 5iˆ = - 20 kˆ + 10 ˆi 10 1 tan    20 2  1   tan1    2

vrm

vertically

Example 26. An aircraft flies at 400 km/h in still air. A wind of 200 2 km / h is blowing from the south. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if AB = 1000 km.  Sol. Given that vw = 200 2 km / h . vaw = 400 km/h and v a should be along AB or in north-east direction. Thus,    the direction of v aw should be such as the resultant of v w and v aw is along AB or in north-east direction. N B va A

KINEMATICS

45º

45º vw = 200 2 km/h C vaw = 400 km/h E

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KINEMATICS  Let v aw makes an angle  with AB as shown in fig. Applying sine law in triangle ABC, we get AC BC  sin 45º sin   BC  sin    sin 45º or  AC   200 2  1 1 sin       400  2 2   = 30º therefore, the pilot should steer in a direction at an angle of (45º + ) or 75º from north towards east.  | va | 400  Further, sin(180º 45º 30º ) sin 45º  sin105º km | v a |  (400) or sin 45º h  km  0.9659  km  cos15º  | va |   (400)  (400)    sin 45º   0.707  h h  km | va | 546.47 h  The time of journey from A to B is AB 1000 t   h  t = 1.83 h | va | 546.67

Example 27. A glass wind screen whose inclination with the vertical can be changed, is mounted on a cart as shown in figure. The cart moves uniformly along the horizontal path with a speed of 6 m/s. At what maximum angle  to the vertical can the wind screen be placed so that the rain drops falling vertically downwards with velocity 2 m/s, do not enter the cart ?  Sol. v c = 6 ˆi  v r = 2 ˆj    v rc = v r - vc vrc v 6 tan   c   3  vr 2 AE  AE   cos  ED  sin  tan    BE    cos  sin  3 2sin 2  / 2   2sin cos 2 2  cot  3  2 2sin 2 2

v=6m/s

vc

D

x

vr y

cos  



KINEMATICS

A cos sin

E

vr

D

B C

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KINEMATICS or or or 

 3 2  1 tan  2 3  1  tan 1 2 3

cot

 1   2 tan 1    3

MOTION IN TWO AND THREE DIMENSIONS INTRODUCTION A body is free to move in space. In this case, the initial position of body is taken as origin. Any convenient co-ordinate system is chosen. Let us suppose that at an instant t, the body is at point P(x, y, z).  The position vector of the body is r = x ˆi + y ˆj+ z kˆ . Thus, velocity   dr dx ˆi + dy ˆj+ dz kˆ v= = dt dt dt dt dx The velocity along x-axis is, v x = dt dv x and acceleration along x-axis is a x = . dt dy The velocity along y-axis is v y = dt dv y and the acceleration along y-axis is a y = dt dz dv z Similarly, v z = and a z = dt dt  The acceleration of the body a = a ˆi + a ˆj+ a kˆ . x

y

z

Discussion (a) If ax is constant, vx = ux + axt 1 x = uxt + axt2 2 v 2x = u 2x + 2a x x If ax is a variable,

ò v x dx ò dv x = ò a x dt x=

(b) If ay is constant 1 2 a t 2 y vy = uy + ayt y = u yt +

v 2y = u 2y + 2a y y

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KINEMATICS If ay is variable,

ò v ydt ò dv y = ò a y dt y=

(c) If az is constant, vz = uz + azt 1 z = uzt + azt2 2 2 2 v z = u z + 2a z z If az is variable,

ò vz dt ò dvz = ò a z dt z=

If the motion of the body takes place in x–y plane, then az = 0, vz = 0, uz = 0  Example 28. A bird flies in the x–y plane with a velocity v = t 2 ˆi + 3t ˆj . At t = 0, bird is at origin. Calculate position and acceleration of bird as function of time. Sol. We have given  v = t 2 ˆi + 3t ˆj

Here, Since,

vx = t2, vy = 3t and vz = 0 vx = t2

or

dx  t2 dt

or

0 dx  0 t dt

or Also,

x

t 2

t3 3 vy = 3t x

or

dy  3t dt

or

0 dy  0 3t dt

y

t

3t 2  2  Thus, position of bird is r  x ˆi  y ˆj  t3 3t 2 ˆ r  ˆi  j 3 2 vx = t2  y

 and

dv x d(t) 2   2t dt dt vy = 3t ax 

KINEMATICS

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KINEMATICS

dv y

dt dt dt  ay = 3 unit  Thus, acceleration of bird is a  a x ˆi  a y ˆj  a  2t ˆi  3 ˆj or

3

PROJECTILE MOTION We next consider a special case of two dimensional motion : A particle moves in a vertical plane with initial   velocity v 0 but its acceleration is always the free fall acceleration g , which is downward. Such a particle is referred to as a projectile (meaning that is projected or launched) and its motion is called projectile motion. A projectile might be a baseball in flight or a golf ball, but it is not air plane or a duck in flight. Or goal here is to analyze projectile motion using the tools for two-dimensional motion and making the assumption that air has no effect on the projectile. Now let us consider a projectile launched so that its initial velocity v0 makes an angle  with with the horizontal (shown in figure). For discussion of motion, we take origin at the point of projection. Horizontal direction as xaxis and vertical direction as y-axis is taken. y

g

v0

O

A

x

The initial velocity of projectile along x-axis is ux = u cos . The component of gravitational acceleration along x-axis is ax = g cos 90º = 0. The component of initial velocity along y-axis is uy = u sin . The acceleration along y-axis is ay = – g NOTE : In projectile motion, the horizontal and the vertical motions are independent of each other, that is, neither motion affects the other. Discussion (a) The instantaneous velocity of the projectile as function of time : Let projectile reaches at point (x, y) after time t [Fig.]. y  vx = ux = u cos  vy and vy = uy – gt = u sin  – gt g  u vx P(x,y) ˆ ˆ  v  vx i  vy j  O x A  v  u cos  ˆi  (u sin   gt) ˆj The instantaneous speed  | v | (u cos ) 2  (u sin   gt) 2 Also,

x = uxt = (u cos ) t = ut cos 

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KINEMATICS 1 y  u y t  gt 2 2 1 y   u sin  t  gt 2 2 The position of the projectile is  r  x ˆi  y ˆj

or

 1   r  ut cos  ˆi   ut sin   gt 2  ˆj  2 

(b) Trajectory of projectile : The y–x graph gives the path or trajectory of the projectile. From discussion of instantaneous velocity of projectile. x = ut cos  ...(i) 1 y  ut sin   gt 2 and ...(ii) 2 x t  ...(iii) u cos  Putting the value of t from Eq. (iii) into Eq. (ii).  x  1  x  y  u sin    g  u cos   2  u cos  

2

gx 2 y  x tan   2 or 2u cos 2  This is the required path or projectile.

Multiplying the eq. (iv) by 

x2 

...(iv)

2u 2 cos 2  to both sides, we get g

 2u 2 cos 2   2u 2 sin  cos  x    y g g  2

 u 2 sin  cos   Adding   to both sides, we get g  2

  2u 2 cos 2    u 2 sin  cos   u 2 sin 2   x    y       g g 2g  This is of the form, (x – a)2 = – c(y – b) ...(v) which is the equation of a parabola. Hence, the equation of the path of the projectile is a parabola. NOTE : The trajectory of a projectile will be parabolic when direction of velocity of projectile is different from direction of acceleration and its acceleration is constant both in magnitude and direction. (c) Time of flight : In Fig., the time taken by projectile to reach at point A from point O is known as time of flight. Here, OA = vxT, where T is time of flight. The total displacement along y-axis during motion of projectile from O to A is zero. So, y = 0

KINEMATICS

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KINEMATICS but

1 y  u y T  gT 2 2

or

1 0   u sin  T  gT 2 2

2u sin  ...(vi) g (d) Range of projectile : Distance OA is known as range of projectile [Fig.] The time taken to reach point A from point O is T



T

2u sin  g

 2u sin   The range R  u x T   u cos    g 

R  u

2sin  cos  g

u 2 sin 2 ...(vii) g Caution : This equation does not give the horizontal distance travelled by a projectile when the final height is not the launch height. Regarding range of projectile, two cases are discussed below : Case-I : Range will be maximum if sin 2 = 1 or  = 45º R

u2  (at = 45º) g Case-II : For given velue of v, the range of projectile will be same for angle  and 90º – though their times of flight and maximum heights are different. R max. 

R 90º  

u 2 sin 2  90º  g

R 90º  

u 2 sin 180º 2 g

u 2 sin 2  R g Thus, for the case as shown in fig, R30º = R60º

y

R 90º  

u 60º

O

u

30º

x

(e) Height attained by projectile : At the maximum height (at point B) the vertical component of velocity is zero. 

or

v 2y  u 2y  2gH 2

2

H

2

u sin  2g

KINEMATICS

vy = 0 B

(0) = (u sin ) – 2gH 2



y

ux H

O A

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x

29

KINEMATICS NOTE : • If air resists or opposes the projectile motion [Fig.], then y u

I II

O

0

0

>

x

0

I—Trajectory in vacuum II—Trajectory in presence of air resistance

• Time taken by projectile during upward motion < Time taken during downward motion. • The values of height attained and of range of a projectile decrease. • The projectile returns to the ground with less speed. At its trajectory its horizontal velocity also decreases. • Time of flight also decreases. • At which angle, the projectile returns to the ground, increases. Example 29. Prove that the maximum horizontal range is four times the maximum height attained by the projectile; when fired at an inclination so as to have maximum horizontal range. Sol. For  = 45º, the horizontal range is maximum and is given by R max 

u2 g

Maximum height attained

or

u 2 sin 2 45º u 2 R max H max    2g 4g 4 Rmax = 4 Hmax

Example 30. There are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection. Sol. There are two angles of projection  and 90º –  for which the horizontal range R is same. Now,

H1 

u 2 sin 2  2g

u 2 sin 2  90º   H2  and 2g 2 2 u cos  H2  2g 2 u u2 2 2 H  H  sin   cos   Therefore, 1 2 2g 2g Clearly, the sum of the heights for the two angles of projection is independent of the angle of projection.





Example 31. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60º with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec2. Sol. Here ax = 0 ay = –g ux = 100 sin60º = 50 3 uy = 100 cos60º = 50

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30

KINEMATICS 

 u 0 = u x ˆi + u y ˆj = 50 3 ˆi + 50ˆj  v y = u y + a y t = 50 – gt

y

v x = u x = 50 3 m / s   v = 50 3 ˆi + (50 - gt )ˆj   But u and v are perpendicular..    u v0 or or 

50

60º

 

x



3 ˆi  50 ˆj  50 3 ˆi   50  gt  ˆj  0

7500 + 2500 – 500 t = 0 10000 t=  500

t 0 = 20 second

Example 32. Fig. shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s. (a) At what angle 0 from the horizontal must a ball be fired to hit the ship ? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs? y

6 3º 27º

x R = 560 m

Sol. (a) Because the cannon and the ship are at the same height, the horizontal displacement is the range. The horizontal range is R

v 20 sin 2 g

...(i)

which gives us  gR  20  sin 1  2   v0 

 9.8  560  20  sin 1   (82)2  20 = sin–1(0.816) ...(ii) 1 0  (46.7º )  23º 2 and  = 90º – 0  = 90º – 23º = 67º The commandant of the fort can elevate the cannon to either of these two angles and (if only there were no intervening air!) hit the pirate ship. (b) We have seen that maximum range corresponds to an elevation angle 0 of 45º. Thus, from Eq. (i) with

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31

KINEMATICS 0 = 45º. v02 (82)2 sin 20  sin(2  45º ) g 9.8 R = 686 m  690 m As the pirate ship sails away, the two elevation angles at which the ship can be hit draw together, eventually merging at 0 = 45º when the ship is 690 m away. Beyond that distance the ship is safe. R

IMPORTANT FEATURES 1. As we have seen in the above derivations that ax = 0, i.e., motion of the projectile in horizontal direction is uniform. Hence, horizontal component of velocity u cos  does not change during its motion. 2. Motion in vertical direction is first retarded then accelerated in opposite direction. Because uy is upwards and ay is downwards. Hence, vertical component of its velocity first decreases from O to A and then increases from A to B. This can be shown as in fig. y

uy O

3.

A

u

ux

B

x

The co-ordinates and velocity components of the projectile at time t are x = sx = uxt = (u cos ) t 1 1 y  s y  u y t  a y t 2   u sin  t  gt 2 2 2 vx = ux = u cos  and vy = uy + ayt = u sin  – gt Therefore, speed of projectile at time ‘t’ is v  v 2x  v 2y and the angle made by its velocity vector with positive x-axis is

4.

 vy    tan 1    vx  Equation of trajectory of projectile x = (u cos )t x t  u cos  Substituting this value of ‘t’ in, 1 y   u sin  t  gt 2 , we get 2 gx 2 y  x tan   2 2u cos 2  gx 2 y  x tan   2 sec 2  2u gx 2 y  x tan   2 1  tan 2  2u These are the standard equations of trajectory of a projectile. The equation is quadratic in x. This is why the path of a projectile is a parabola. The above equation can also



KINEMATICS



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KINEMATICS be written in terms of range (R) of projectile as x  y  x 1   tan   R

5.

6.

Range R is given by

R

2  u cos  u sin  g

R

2  Horizontal component   Vertical component  g  of initial velocity   of initial velocity 

There are two unique times at which the projectile is at the same height h(< H) and the sum of these two times 1 2 equals the time of flight T. Since, h   u sin  t  gt is quadratic in time, so it has two unique roots ‘t1’ and 2 2u sin  2h ‘t2’ (say) such that sum of roots (t1 + t2) is and product (t1t2) is . The time lapse (t1 – t2) between g g these two events is (t1 – t2)2 = (t1 + t2)2 – 4t1 t2

7.

4u 2 sin 2  8h  g g If K´ is the kinetic energy at the point of launch then kinetic energy at the highest point is

8.

1 1 mv 2x  mu 2 cos 2  2 2 2 K´ = K cos  For complementary angles  and 90º – , if T and T90º – are the times of flight and R is the range, then

t1  t 2 

K´

T T90º  

e.g.,

T1º T89º 

2R  g



2R 90º  g



2R g

2R1º 2R 89º  g g

Example 33. A particle is projected in the X–Y plane. 2 sec after projection the velocity of the particle makes an angle 45º with the X-axis. 4 sec after projection, it moves horizontally. Find the velocity of projection. Sol. After 4 sec the particle reach at maximum hight. At maximum height it move horizontally. So that, T y =4 2 v0 2u sin   2u sin   T8  T   g g  O  x 80 u sin    40 ...(1) 2 u cos = v cos 45º v sin45 = 45 sin – 10 × 2

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KINEMATICS u sin   20 20   u cos  u cos  From eqn (1) and (2) u2 sin2 + u2 cos2 = 402 + 202 u2 = 1600 + 400 = 2000 1

ucos = 20

...(2)

u  20  100  10 20 20 5 m / s

PROJECTILE PROJECTED FROM SOME HEIGHT Projectile Projected in Horizontal Direction  Let a projectile is projected with velocity u. Take observation point O at a height h from ground [Fig.]. y O

u r

x

x y(x, y)

a y = –g vx

–h vy P

v A

Here, ux = u, uy = 0 and ax = 0, ay = –g (a) Let at time ‘t’, the co-ordinates of position of projectile is (x, y), then 1 2 x = ut and y  0  gt 2 Therefore, at time ‘t’ position vector  r  x ˆi  y ˆj  1 r  (ut) ˆi  gt 2 ˆj 2 2  2 2 2  1 2 | r |  x  y  (ut)    gt   2  y tan   and x (b) Let at time ‘t’ the horizontal and vertical velocities of projectile be vx and vy, so vx = u and vy = 0 + (–g)t = – gt  v  v ˆi  v ˆj  u ˆi  ( gt) ˆj x

and

y

v  v 2x  v 2y

v  u 2  ( gt) 2 v tan   x and vy (c) Let time taken by projectile from O to point A at ground is T, then

1  h  0  (  g)T 2 2

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KINEMATICS 2h g The horizontal distance in time T



T

2h g Therefore, the bomb dropped from an aeroplane moving with velocity u horizontally at height h, covers a PA  uT  u

horizontal distance u

2h on the ground. g

Projectile Projected Upwards at an Angle   Let projectile is projected upward at angle  with horizontal velocity u [Fig.].

u sin

u

vy = 0, vx = u x = u cos ay = –g A B u cos

O u cos –h

x

u sin

P

u D

C

ux = u cos  and uy = u sin  ax = 0 and ay = – g Now from 2nd equation of motion, 1  h   u sin   T    g T 2 2 or gT2 – (2u sin )T – 2h = 0 Solving this equation, we get horizontal distance covered in time T PC = (u cos )T

2u sin  g and horizontal distance covered in this time Time taken in covering path OAB 

u 2 sin 2 OB  g In such case for range PC to become maximum, ‘’ should be just less than 45º. NOTE : In such case on earth’s surface for maximum range, u sin α = . 2 2u + 2gh If h = 0, then α =

1  = 45º. 2

Projectile Projected Downward at an Angle   Let projectile is projected downward at an angle  with horizontal velocity u [Fig.].

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35

u y = –u sin

KINEMATICS y O

u x = u cos u

x a y = –g

–h P

A

ux = u cos  and uy = –u sin  ax = 0 and ay = –g From 2nd equation of motion, 1   g T 2 2 2 or gT + (2u sin )T – 2h = 0 To solve this euation, value of T can be evaluated. In this time the horizontal distance covered on the earth, PA = (u cos )T –h =  –u sin   T 

Example 34. A projectile is fired horizontal with a velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground (g = 9.8 m/s2) Sol. Here, ux = 98 m/s, ax = 0, uy = 0 and ay = g (a) At A, sy = 490 m. So, applying 1 sy  u y t  a y t2 2 1 490  0  (9.8)t 2  2  t = 10 s 1 BA  s x  u x t  a x t 2 (b) 2 or BA = (98) (10) + (0) or BA = 980 m (c) vx = ux = 98 m/s vy = uy + ayt = 0 + (9.8) (10) = 98 m/s 

v  v 2x  v 2y  (98)2  (98)2  98 2 m / s

and

tan  



 = 45º

vy vx



O

u = 98 m/s x y

B

A

vx

vy

98 1 98

Thus, the projectile hits the ground with a velocity 98 2 m / s at an angle of  = 45º with horizontal. Example 35. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find

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36

KINEMATICS (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed. Sol. (a) S1 = The distance moved by the train in first half minute. 1 S1 = u1t1 + a1t12 2 1 1 S1 = 0 × t1 + × 2 × (30)2 {  t1 = minute} 2 2 1 S1 = × 2 × 900  S1 = 900 m 2 The distance moved by the train after brakes applied. 1 S2 = u2t2 + a2t22 2 u2 = u1 + a1t1 u2 = 0 + 2 × 30 u2 = 60 m/s v2 = u2 – a2t2 0 = 60 – a2 × 60



a 2 = 1 m / sec2

v22 = u22 + 2 a2S2 0 = 60 × 60 – 2 × 1 × S2 S2 = 1800 m Total distance(s) = S1 + S2 S = (900 + 1800) m S = 2700 m



S = 2.7 km

(b) the maximum speed attained by the train is u 2 = 60 m / s (c) the position of the train at half of minimum speed. u22 = u12 + 2a1S’ (30)2 = 0 + 2 × 2 × S’ 900 = S' S' = 225 m  4

Example 36. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms–1 by a constant acceleration of 5 ms–2. It maintains a constant velocity of 20 ms–2 for 5 seconds and then is brought to rest again by a constant acceleration of –2 ms–2. Draw a velocity-time graph and find the distance covered by the car. Sol. v22 = u2 + 2aS (20)2 = 0 + 2 × 5 × S 400 S= S = 40 m  10 Car maintains a constant velocity of 20 m/s for 5 Second. S’ = v × t S '' = 100 m S’’ = 20 × 5 

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37

KINEMATICS The car comes to rest by a constant acceleration of –2 m/s2 (v’)2 = v2 + 2 × a’S’’ 0 = (20)2 – 2 × 2 S’’ 400 S '' = S'' = 100 m  4 The total distance covered by the car. = S + S’ + S’’ = 40 + 100 + 100 = 240 m Example 37. A ball rolls off the edge of a horizontal table top 4m high. If it strikes the floor at a point 5m horizontally away from the edge of the table, what was its speed at the instant it left the table ? Sol. Using h 

or

1 2 gt we have, 2 1 h AB  gt 2AC 2 2h AB t AC   g

Further,

v A

24  0.9 s 9.8 BC 5.0 v   5.55 m / s or t AC 0.9

BC = vtAC

4m C

B 5m

Example 38. A ball is projected at an angle of 30º above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle of 45º with the horizontal. Find the height of the tower and the speed with which it was projected. Sol. ux = v0 cos30º uy = v0 sin30º Vx = uX = v0 cos30º y vy = u y + a y t v0 vy = v0 sin30 – gt 30º x v0 H v y   10  5 2 v0 ax=0, ay= –g – 50 vy 2 But – tan 45º   v x v 0 cos 30 v  v 0 cos 30  0  50 2 1 3 v0    50 2 2 

v0 =

100 3 +1 100



=







3 –1



3 1

50





3 –1



3 –1

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KINEMATICS Also,



1 y = - H = u yt + a yt2 2 1  H  v0 sin 30  5   10  52 2





H  125 2 – 3 m

Example 39. A ball is thrown horizontally from a cliff such that it strikes ground after 5 sec. The line of sight from the point of projection to the point of hitting makes an angle of 37º with the horizontal. What is the initial velocity of projection. Sol. vx = v0cos37º vy = v0sin37º y ax = gsin37º = 6 m/s2 v0 O 37º 2 ay = –gcos37º = –8 m/s From O to A, displacement along y-axis is zero. 37º y = uyt + ½ ayt 2 A 0 = uy × 5 + ½ × –8 × 25 {  t = 5 sec.} 5uy = ½ × 8 × 25 100 uy = = 20 m / s 5 3 20  v0  5 100 v0 = m/s 3

37º

x

Example 40. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53º to horizontal. Its speed when it is at a height of 0.45 m from the point of projection is : (A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient Sol. According to conservation principle of machenic energy Ui + T i = Uf + T f 1 1 0 + mu 2 = mgh + mv 2 2 2 2 2 u = 2gh + v v = u 2 - 2gh

v  52  2  10  0.45 v  25  9



16



v  4 m/s u

Example 41. In the figure shown, the two projectiles are fired simultaneously. 20m/s What should be the initial speed of the left side projectile for the two 60º 45º projectile to hit in mid-air ? 10 m Sol. When two projectiles are projected from same height, then for collision, vertical component of velocities of both projectiles should be same

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KINEMATICS  

u sin 60º = 20 sin45º 20sin 45º u= sin 60º 20  2 u  2 3

u = 20

2 m/s 3

Example 42. The speed of a particle when it is at its greatest height is 2 / 5 times of its speed when it is at its half the maximum height. the angle of projection is _____ and the velocity vector angle at half the maximum height is ______. Sol. uy = usin u u cos ux = ucos vx = ucos ...(1) 2 2 vy = uy +2ayS H v 2y  u 2 sin 2   2g ...(2) 2 2 u cos   v 2x  v 2y 5 2 H u cos   u 2 cos 2   u 2 sin 2   2g 5 2 u cos  



2 5 2 u 5

u2  g

1

u 2 sin 2  2g

sin 2  2

2 sin   1 5 2 2 2  sin   cos2   1  5 2  cos  

2 sin 2   or , 5 5 or 5  5sin 2   sin 2   2 or 3 – 4 sin2  = 0 3 sin     60º  2 vy u 2 sin 2   gH tan    and vx u cos  cos 2  

u 2 sin 2  2g u cos 

1  sin 2  

{from eqn. (1) and (2)}



sin 2  sin   2 tan   cos 



tan  

u 2 sin 2   g tan  

sin 2  2 tan   cos 

KINEMATICS

sin 2  2  5 5

2

tan  tan 60º  2 2

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KINEMATICS

tan  

3 2



  tan 1

3 2

Example 43. A projectile is to be thrown horizontally from the top of a wall of height 1.7 m. Calculate the initial velocity of projection if its hits perpendicularly an incline of angle 37º which starts from the ground at the bottom of the wall. The line of greatest slope of incline lies in the plane of motion of projectile. Sol. ux = v0 uy = 0 ay = g = 10 m/s2 vx = v1 cos53º = 0.6 v1 vy = v1 sin53º = 0.8 v1  vx = u x = v0 or 0.6 v1 = v0 v 10v0 5v 0 v1 = 0 = = ...(1) v0 O x 0.6 6 3 1 2 vy = u y + a y t gt 2 C 0.8v1 = 10 t v1 h = 1.7 m A 37º 5v 0.8  0  10 t 3 53º 37º x 53º 6 60 grad B t= t  v0 = ...(2) y v1 0.8 8 In BAC 1 1.7 - gt 2 1.7 - 5t 2 AB 3 2 tan 37º =  = = AC 4 uxt v0 t 2 or 3v0t = 6.8 – 20t 8v 64 v02 3v 0  0  6.8  20  or 60 3600 2  72  64  2 24 2 64v 0 v0 + = 6.8 or    v 0  6.8 180  60 180 6.8  180 v 20  9 v0 = 3 m / s or   136 Example 44. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2 m/ sec. A dear running with a speed V in front at a distance of 4 5m moving perpendicular to the direction of motion of the elephant. If hunter can throw his spear with a speed of 10 m/sec, relative to the elephant, then at what angle  to it’s direction of motion must he thrown his spear horizontally for a successful hit. Find also the speed ‘V’ of the dear. 1 2 Sol. h = gt 2 2h 24 8 t   g 10 10

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KINEMATICS Assume horizontal plane at x–y–plane.  v rel  10 cos  ˆi  10sin  ˆj   v rel  u  2 ˆi    u  v rel  2 ˆi  10 cos   2 ˆi  10 sin  ˆj The deer is moving along y–axis. So, displacement of deer and displacement of spear along y–axis will be same in time t.  v t = u yt  v = 10 sin ...(1) Also, along x–axis :

ux t = 4 5 8 4 5 10 10 4  5   10 10cos   2  4 5 or 2 8  10 cos = 8 8 4 cos       = 37º 10 5 From eqn (1) or

10cos   2

v = 10 sin = 10 sin37º = 10 

3 = 5

4m 2m/s x

6 m/s

y A Example 45. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a P 1.25m plank P raised above the ground. At time t = 0 the plank starts moving along the + x direction with an acceleration 1.5 m/s2. At the same instant a stone is u x O 3.0m projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal. All the motions are in x – y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s2. Sol. ux = u cos uy = u sin ax = 0 u u =ucos 45º ay = –g 1.25 If the stone hitt the object after time t. So that virtical displacement of stone is 1.25 m u cos 1 2 y = uyt + a yt 2 1 2 therefore 1.25  (u sin  )t  gt 2 1 1.25  (u sin  )t  10  t 2 2 1.25 = (u sin) t – 5 t 2 (u sin)t = 1.25 + 5t 2 ...(1) Hotizontal displacement of stone is x = 3 + displacement of object A. Initial velocity of object is zero.

u sin

x

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KINEMATICS so displacement of object is 1 1 1.25  at 2   1.5  t 2  0.75 t 2 2 2 x = 3 + 0.75 t 2 (u cos)t = 3 + 0.75 t 2 Since velocity vector inclined at 45º with horizontals. u y u sin   gt tan(–45º )   ux u cos  u cos = – (u sin – gt) u cos = gt – u sin (u cos) t + (u sin) t = 10 t2 Add eqn (1) and (2) (u cos) t + (u sin) t = 4.25 + 5.75 t2 from eqn (3) and (4) 10 t2 = 4.25 + 5 .27 t2 4.25 t2 = 4.25 t2 = 1  From eqn (1) and (2) uy = u sin = 6.25 m/s uy = 6.25 m/s ux = 3.75 m/s

...(2)

...(3)

...(4)

t = 1 sec.

u = u 2x + u 2y u = (6.25) 2 + (3.75) 2



u = 7.29 m / s

IMPORTANT FEATURES Projectile motion is a two dimensional motion with constant acceleration (g). So, we can use    v  uat   1 a  u t  a t2 2 etc. in projectile motion as well. Here,   u  u cos  ˆi  u sin  ˆj and a   g ˆj y

Now, suppose we want to find velocity at time ‘t’.    v  u  a t v  u cos  ˆi  u sin  ˆj  gt ˆj  or v  u cos  ˆi   u sin   gt  ˆj Similarly, displacement at time ‘t’ will be,   1 S  u t  a t2 2  1 S  u cos  iˆ  u sin  ˆj t  gt 2 ˆj 2





KINEMATICS

u



O

g

x



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43

KINEMATICS  1   S  u cos  ˆi   ut sin   gt 2  ˆj  2 

PROJECTILE MOTION ON AN INCLINED PLANE One an inclined plane projectile is projected into two cases, one upwards and the other downwards. Up the Plane A projectile is projected up the inclined plane from the point O with an initial velocity u at an angle  with horizontal. The angle of inclination of the plane with horizontal is  [Fig.]. x A

u

9 o s( gc

y

y

90º

B

O

y´

x´

x´

0 º–

)

s =g

in

x

g sin(90º– ) = g cos

–

O y´

ux = u cos ( – ) and ax = – g sin  uy = u sin ( – ) and ay = – g cos  (i) Time of flight : During motion from point O to A, the displacement along y-axis is zero.  sy = 0 at t = T 

or

1 sy  u y t  a y t2 2 1 0  u sin      T  g cos  T 2 2

T



2u sin     g cos 

NOTE : Substituting  = 0, in the above expression, we get T =

2u sin α . g

Which is quite obvious because  = 0 is the situation shown in Fig. y u

O

x

g

(ii) Range : As shown in Fig, OA is the range of proectile. Horizontal component of initial velocity uH = u cos   OB = uHT (as aH = 0) =

 u cos θ 2u sin       2u 2 sin      cos  g cos 

KINEMATICS

g cos 

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44

KINEMATICS 2u 2 sin      cos  OB R  OA    cos  g cos 2  Using, 2 sin A cos B = sin (A + B) + sin (A – B) Range can also be written as, R

u2 sin  2     sin  g cos 2  

This range will be maximum when u2    R  2    1  sin   or    and max g cos 2  2 4 2

we see that for  = 0, range will be maximum for   R max

u2  1  sin 0º  g cos 2 0º

R max

u2  g

 or 45º. 4

Alternative method : For range, sx = R, t = T 1 sx  u x t  a x t 2 2 1 R  u cos      T  g sin  T 2 2 2u sin     Substituting the value of T  , in above equation for R. g cos 

or

R



u2 sin  2     sin  g cos 2  

Down the Plane A projectile is projected down the plane from the point O with an initial velocity u at an angle  with horizontal [Fig.]. The angle of inclination of plane with horizontal is . y

x

u O

( 90 os c g

90

º– º–

g )=

sin

g sin(90º– ) = g cos

A

ux = u cos ( + ), ax = g sin  uy = u sin ( + ), ay = –g cos  Proceeding in the similar manner, we get the following results : Therefore,

T

2u sin      g cos 

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KINEMATICS R

u2 sin  2     sin  g cos 2  

IMPORTANT FEATURES If two particles are projected at angles 1 and 2 respectively as shown in fig., then the relative motion of 1 with respect to 2 is a straight line at an angle. y

y

u2 u1

1

x

2

x

 u12y    tan 1  with positive x-axis.  u12x  where

u12x = u1x – u2x = u1 cos 1 – u2 cos 2 u12y = u1y – u2y = u1 sin 1 – u2 sin 2

Example 46. A particle is projected with a velocity of 20 m/s at an angle of 30º to an inclined plane of inclination 30º to the horizontal. The particle hits the inclined plane at an angle of 30º, during its journey. Find the (a) time of impact, (b) the height of the point of impact from the horizontal plane passing through the point of projection. Sol. The particle hits the plane at 30º (the angle of inclination of plane). It means particle hits the plane horizontally. (a)

t

u

T u sin   2 g

20sin  30º 30º  1.76 s 9.8 u 2 sin 2  H 2g t

(b)

30º 30º

2 20  sin 2 60º  H  15.3 m

2  9.8

Example 47. A particle is projected up an inclined plane with initial speed v = 20 m/s at an angle  = 30º with plane. Find the component of its velocity perpendicular to plane when it strikes the plane. Sol. Component of velocity perpendicular to plane remains the same (in opposite direction) i.e., u sin  = 20 sin 30º = 10 m/s Example 48. A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. Find the time after which it lands on the plane (g = 10 m/s2). Sol.

10m/s2 30º

ux = 10 cos30º = 5 3 m/s uy = 10 sin30º = 5 m/s arel =20 m/s2

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46

y 10sin30º 10 m/s2 v0= 10m/s KINEMATICS O 30º 0º o s3 c A 10

ax = arel sin30º = 10 m/s2 ay = – arel cos30º 3  10 3 m / s 2 2 when ball lands on inclined plane, y = 0 1 y = uyt + ay t 2 2 1 0  5 t   10 3 t 2 2 5 3 t =5 1 t= sec.  3  20 

x

30º

Example 49. A particle is projected from point P with velocity 5 2 m/s perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. Find the time of the flight of the particle. u0x = 5 2m / s u0y = 0 ax = –g sin45º ay = –g cos45º At point Q, vx = 0 {  the particle collied normally at point Q} The time taken by particle to go from P to Q is t0. vx = u x + a x t 0 = 5 2 – g sin45º t 0

P

Q 45º

y

x

x P

Q 45º

Sol.

y

g 45º

45º

5 2 5 2 2   1 sec. g sin 45º 10 t 0 = 1 sec.

t0 

Example 50. A ball is projected on smoot inclined plane in direction perpendicular to line of greatest slope with velocity of 8m/s. Find it’s speed after 1 sec. Sol. vy = uy + ayt vy = 0 + 10 sin37º × 1 3 vy = 10   1 7º 5 in3 gs vy = 6 m/s 37º vx = ux = 8 m/s x y 

8 m/s

º 37

v = v 2x + v 2y = 82 + 6 2 Ans. 10 m / s

Example 51. Two inclined planes OA and OB having inclination (with horizontal) 30º and 60º respectively, intersect each other at O as shown in fig. A particle is projected from point P with velocity u = 10 3 ms - 1

KINEMATICS

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47

u A

along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate (a) time of flight, (b) velocity with which particle strikes the plane OB, (c) vertical height h of P from O, (d) maximum height from O attained by the particle. Sol. (a) u x = 10 3 m / s

B

KINEMATICS

Q

P h 60º

30º O

uy = 0 ax = –g sin60º

B

u A

ay = –g cos60º

Q P

At point Q

h

vx = 0 The time taken by particle to go from P to Q is to vx = ux + axt

60º

30º O

0 = 10 3 - g sin 60º t 0 t0 = (b)

(c)

(d)

10 3 g sin 60º



t 0 = 2sec

vy = uy – ayt 0 vy = 0 – g cos60º × 2 1 v y  10   2  10 m / sec 2 h sin 30º = x 1 h =  x = 2h m 2 x 1 –2h = u y t + a y t 2 2 1 –2h  0  g sin 30º  (2) 2 2 1 1 2h   10   4  2 2 At maximum height vy = 0 v 2y = u 2y - 2gH 2

0 = (u cos30º) – 2gH 2

H=

h

x 60º

30º O

h= 5m uy = u cos30º X 30º Y h 60º

2

u cos 30º 2g

 3 10 3     2  H 2g 100  3  3 4 H 2  10



X

u

Y



30º 2

2

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48

KINEMATICS 25  9 45 H= =  20 4 Maximum height from O attained by the particle = H+h = 11.25 + 5 Ans. 16.25 m 

11.25 m

Example 52. A large heavy box is sliding without friction down a smooth plane of inclination . From a point P on the bottom of a box, a particle is projected. inside the box. The initial speed of the particle with respect to box is u and the direction of projection makes P an angle  with the bottom as shown in figure. (a) Find PQ if particle lands on Q. (b) If horizontal displacement of particle with respect to ground is zero. Find the velocity of box. Sol. ux = u cos uy = usin ax = –g sin ay = –g cos When the particle hitt the inclined plane then displacement in y-direction is zero. y x u =usin 1 y = u yt + a yt2 Q 2 s o uc s 1 P u = –gco ax = –gsin 0  (u sin  )t  g cos  t 2 – = 2 90 ay 1  g (u sin  )t   g cos   t 2 2 

Q

y

x

2u sin  ...(1) g cos  with respect to box ux = u cos ax = 0 1 x = uxt + axt2 2 x = u cost (2u sin  ) x  (u cos  ) (from eqn 1) g cos  u 2 sin 2 x g cos  According to question ux = –U ax = –gsin 1 x = uxt + axt2 2 2 2  2u sin   1  2u sin    u sin 2  U   g sin   g cos   g cos   2  g cos   t

KINEMATICS

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KINEMATICS u sin  cos  cos  u sin  cos  u cos   U  cos  u sin  cos  U  u cos   cos   cos  cos   cos  sin   U u  cos   u cos(   ) U cos 

u cos   U 

THINKING PROBLEMS 1.

Can you use the equations of kinematics to find the height attained by a body projected upwards with any velocity ?

2.

Can the relative velocity of two bodies be greater than the absolute velocity of either ?

3.

A boy sitting in a car moving with a constant velocity throws a ball straight up into the air. Will the ball fall behind him, in front of him or into his hand? What would happen if the car accelerated forward or went round a curve while the ball was in the air ?

4.

   v f  vi A student argues that the mean velocity during an interval of time can also be expressed as  v  2    r f  ri and this should always be equal to v f   . Is he right . t 2  t1

5.

Consider a collection of a large number of particles moving with the same speed v in random directions. Could you, by using simple logic, show that the magnitude of the relative velocity of a pair of particles averaged over all the pairs of the collection is greater than v ?

6.

The barrel of a gun and a target lie along the same horizontal. If the target is released and the gun is fired at the same time, the bullet will always hit the target whatever be the distance between the gun and the target. Is this true or false ?

7.

A black dot is made at the tip of an aerofoil of an aeroplane. What is the trajectory of the black dot as it appears to the pilot and to an observer on the ground?

8.

A body is dropped fron the window of a train. W ill the time of the free fall be equal if the train is stationary, moves with constant velocity, moves with constant acceleration?

9.

We can order events in time, such as past, present and future. Hence there is a sense of time. So is time a vector? If not, why not?

10. Average speed can mean the magnitude of the average velocity vector. Another meaning given to it is that the average speed is the total length of the path traversed divided by the elapsed time. Are these meanings different? if so, give an example. 11. Can a body have zero velocity and still be accelerating? can a body have a constant speed and still have a varying velocity? Can a body have a constant velocity and still have a varying speed? 12. Can an object have an eastward velocity while experiencing westward acceleration?

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KINEMATICS THINKING PROBLEMS SOLUTION 1. No, because the equations are applicable only so long as the acceleration is uniform. The acceleration due to gravity is uniform only near the surface of the earth. 2. Yes, e.g., when two bodies move in opposite directions, the relative velocity of each is greater than the individual velocity of either. 3. It will fall into his hand because of the ‘physical independence of vectors’ or the fact that the forward velocity of the ball in the air is the same as that of the car. But if the car accelerated, the ball would fall behind him. If the car went round a curve the ball would fall in front of him because it would move along the tangent to the curved path through a certain distance while the car moved along the curved path through the same distance. 4. No, he is not right. The correct definition of average velocity is the latter one. the first one can be used only when there is uniform acceleration. 5.

v rel  v 2  v2  2v 2 cos   2v sin  / 2. In other words between 0 and 60º the relative velocity is less than v, while between 60º and 180º it is greater than v. Hence the relative velocity averaged over all directions

v rel  v. 6. True, provided the target is within the range of the bullet, because the downward acceleration of both is the same. 7.

A circular path relative to the pilot and a helical path relative to the observer fixed to the earth.

8.

Same in all the three cases due to the physical independence of vectors.

9.

Time is not a vector though it may have a sense because in order to be a vector it must add and subtract by vector algebra rules and it does not.

10. The meanings are different. The point can be made clear by considering the motion of a particle along a circle. Let a particle describe half a circle of radius R in time .

 –R ˆi – R ˆi 2R ˆ  – i = average velocity vector  v    2R R |  v |    v = average speed    Obviously,  v   |  v  | , that is, average speed as the modulus of the average velocity vector is not the same as the average speed defined as the total length of the path traversed divided by time elapsed. 11. Yes, for example a particle in SHM at the extreme position has zero velocity but its acceleration is maximum. Yes, a particle in circular motion has constant speed but varying velocity. No, a body cannot have a constant velocity while having a varying speed. 12. Yes, a body can have eastward velocity and westward acceleration, e.g., SHM.

KINEMATICS

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KINEMATICS

ASSERTION & REASON it of (A) (B) (C) (D) (E) 1.

THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below the statements, mark the correct answer as – If both A and R are true and R is the correct explanation of A. If both A and R are true but R is not the correct explanation of A. If A is true but R is false. If both A and R are false. If A is false but R is true.

Assertion (A) : Reason (R) :

2.

Assertion (A) :

Horizontal component of the velocity of angular projectile remains unchange during its flight. At highest point the velocity reduces to zero. (A) (B) (C) (D) (E) Suppose a particle starts moving in a straight line with initial velocity +u and an acceleration –a, then velocity at displacement s comes out to be, v 2  u 2  2as . If we draw a graph between v 2 and s, it will be a straight line as shown in figure. 2

v

S0

Reason (R) :

S

v 2 versus s graph is a straight line passing through origin with positive intercept and negative slope. (A)

(B)

(C)

(D)

(E)

3.

Assertion (A) : Reason (R) :

A body can have acceleration even if its velocity is zero at a given instant of time. A body is momentarily at rest when it reverses its direction of motion. (A) (B) (C) (D) (E)

4.

Assertion (A) :

When a body is projected with an angle 45°, its range is maximum.

Reason (R) :

For maximisation of range sin 2 should be equal to one. (A) (B) (C) (D)

(E)

5.

Assertion (A) : Reason (R) :

Rocket in flight is not an illustration of projectile. Rocket takes flight due to combustion of fuel and does not move under the gravity effect alone. (A) (B) (C) (D) (E)

6.

Assertion (A) :

The slope of displacement-time graph of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Slope of displacement-time graph = Velocity of the body. (A) (B) (C) (D) (E)

Reason (R) :

7.

Assertion (A) : Reason (R) :

Displacement of a body may be zero when distance travelled by it is not zero. The displacement is the longest distance between initial and final position. (A) (B) (C) (D) (E)

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KINEMATICS 8.

Assertion (A) : Reason (R) :

9.

The relative velocity between any two bodies moving in opposite direction is equal to sum of the velocities of two bodies. Sometimes relative velocity between two bodies is equal to difference in velocities of the two. (A) (B) (C) (D) (E)

Assertion (A) : Reason (R) :

The displacement-time graph of a body moving with uniform acceleration is a straight line. The displacement is proportional to time for uniformly accelerated motion. (A) (B) (C) (D) (E)

10. Assertion (A) : Reason (R) :

A body falling freely may do so with constant velocity. The body falls freely, when acceleration of a body is equal to acceleration due to gravity. (A) (B) (C) (D) (E)

11. Assertion (A) : Reason (R) :

The speedometer of an automobile measure the average speed of the automobile. Average velocity is equal to total displacement per total time taken. (A) (B) (C) (D) (E)

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KINEMATICS

Level # 1 1.

A train starting from rest travels the first part of its journey with constant acceleration a, second part with constant velocity v and third part with constant retardation a, being brought to rest. The average speed for 7v the whole journey is . The train travels with constant velocity for .... of the total time 8 (A)

2.

3 4

(B)

7 8

(C)

5 6

(D)

9 7

A particle moving in a straight line with uniform acceleration is observed to be at a distance a from a fixed point initially. It is at distances b, c, d from the same point after n, 2n, 3n second. The acceleration of the particle is (A)

c  2b  a n2

(B)

cba 9 n2

(C)

c  2b  a 4 n2

(D)

c ba n2

3.

If a, b and c be the distances travelled by the body during xth, yth and zth second from start, then which of the following relations is true ? (A) a(y – z) + b(z – x) + c(x – y) = 0 (B) a(x – y) + b(y – z) + c(z – x) = 0 (C) a(z – x) + b(x – y) + c(y – z) = 0 (D) ax + by + cz = 0

4.

A body of mass 3 kg falls from the multi-storeyed building 100 m high and buries itself 2m deep in the sand. The time of penetration will be (A) 0.09 s (B) 0.9 s (C) 9 s (D) 10 s

5.

Two cars A and B, each having a speed of 30 km/hr are heading towards each other along a straight path. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km, heads directly towards car B, on reaching B, the bird directly flies back to A and so forth, then the total distance the bird travels till the cars meet is. (A) infinite (B) 30 km (C) 60 km (D) 120 km

6.

Two cars A and B, each having a speed of 30 km/hr are heading towards each other along a straight path. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km, heads directly towards car B, on reaching B, the bird directly flies back to A and so forth, the total no. of trips which the bird makes till the cars meet is (A) Four (B) Eight (C) Sixteen (D) Infinite

7.

Three particles start from the origin at the same time, one with a velocity v 1 along x-axis the second along the y-axis with a velocity v 2 and the third along x = y line. The velocity of the third so that the three may always lie on the same line. (A)

8.

v1 v 2 v1  v 2

(B)

2 v1 v 2 v1  v 2

(C)

3 v1 v 2 v1  v 2

(D) Zero

Choose the correct statement (A) A body starts from rest and moving with constant acceleration travels a distance y1 and the 3rd second y1 5 and y2 in 5th second. The ratio y  9 . 2 (B) A ball falls from the top of a tower in 8 second in 4 second, it till cover the first quarter of the distance starting from top. (C) The distance traveled by a freely falling stone released. With zero velocity in the last second of its motion to that traveled by it in the last second of its motion to that traveled by it in the last but one second is 7 : 5. The stone strike the ground with velocity 39.2 m/s. (D) None of these

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KINEMATICS 9.

A steam boat goes across a lake and comes back (i) on a quiet day when the water is still, and (ii) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey back. If the speed of the launch, on both days, was same, the time required for the complete journey on the rough day, as compared to that on the quiet day, will be. (A) Less (B) Same (C) More (D) Cannot be predicted

10.

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is (A) 33.3 km/hr

(B) 20 3 km/hr

(C) 25 2 km/hr

(D) 35 km/hr

11.

Starting from rest a particle moves in a straight line with acceleration a = (25 – t 2)1/2 m/s2 for 0  t  5 s, 3 a= m/s2 for t > 5 s. The velocity of particle at t = 7 s is. 8 (A) 11 m/s (B) 22 m/s (C) 33 m/s (D) 44 m/s

12.

A body is released from a great height and falls freely towards the earth. Exactly one sec later another body is released. What is the distance between the two bodies 2 sec after the release of the second body. (A) 4.9 m (B) 9.8 m (C) 24.5 m (D) 50 m

13.

The distance moved by a freely falling body (starting from rest) during 1st, 2nd, 3rd, .... nth s of its motion are proportional to. (A) Even numbers (B) Odd numbers (C) All integral numbers (D) Square of integral numbers

14.

A ball is thrown vertically upwards with a speed of 10 m/s from the top of a tower 200 m high and another is thrown vertically downwards with the same speed simultaneously. the time difference between them in reaching the ground in s(g = 10 m/s 2) is (A) 12 (B) 6 (C) 2 (D) 1

15.

Between two stations a train first accelerates uniformly, then moves with uniform speed and finally, retards uniformly. If the ratios of the time taken for acceleration, uniform speed and retarded motions are 1 : 8 : 1 and the maximum speed of the train is 60 km/h, the average speed of the train over the whole journey is. (A) 25 km/h (B) 54 km/h (C) 40 km/h (D) 50 km/h

16.

A ball is thrown vertically upwards. It was observed at a height h twice with a time interval  t. The initial velocity of the ball is. (A)

2

8 gh  g ( t )

2

(B)

 g t  8 gh     2 

2

(C)

1 2 2 2 8 gh  g ( t )

(D)

8 gh  4g2 ( t )2

17.

A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. a similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If a1 and a2 be the retardation offered to the bullet by wood and iron plates respectively then (A) a1 = 2a2 (B) a2 = 2a1 (C) a1 = a2 (D) Data insufficient

18.

The motion of a body falling from rest in a resisting medium is described by the equation a and b are constants. The velocity at any time t is a b (A) v t = b (1 – e–bt) (B) v t = e–bt a

19.

a (C) v t = b (1 + e–bt)

(D) v t =

dv = a – bv where dt

b bt e a

P . To o Mv increase the velocity of the vehicle from v 1 to v 2, the distance travelled by it (assuming no friction) is

A self-propelled vehicle of mass M whose engine delivers constant power P has an acceleration a =

(A) s =

3P 2 (v 2  v 12 ) M

KINEMATICS

(B) s =

M 2 ( v 2  v 12 ) 3P

(C) s =

M 3 ( v 2  v 13 ) 3P

(D) s =

3P 3 ( v 2  v 13 ) M

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55

KINEMATICS For an airplane to take-off it accelerates according to the graph shown and takes 12 s to take-off from the rest position. The distance travelled by the airplane is. (A) 21 m (B) 210 m (C) 2100 m (D) 120 m

A

5

B

m/s2

20.

6

t (in s)

12

21. A river is flowing from west to east at a speed of 5 meters per minute. A man on the south bank of the river, capable of swimming at 10 meters per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction. (A) due north (B) 30° east or north (C) 30° west of north (D) 60° east of north 22. A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is A (A) 1

(B) 3

(C) 4

(D)

41

23. In 1, 0, s, a particle goes from point A to point B, moving in a semicircle of radius 1.0 m (see Figure). The magnitude of the average velocity (A) 3.14 m/s (B) 2.0 m/s (C) 1.0 m/s (D) Zero

1.0 m

B

24. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height

d 2 . Neglecting subsequent motion and air resistance, its velocity v varies

with the height h above the ground as v

v d

(A)

h

(B)

v

d

v d

h

d

h

(C)

(D)

h

Acceleration 2 (m/s ) 10

25. A body starts from rest at time t = 0, the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be (A) 110 m/s (B) 55 m/s (C) 650 m/s (D) 550 m/s

Time 11 (Sec.)

26. The velocity is displacement graph of a particle moving along a straight line is shown v v0

x0 The most suitable acceleration-displacement graph will be

a

a

a

x

(A)

x

a x

(B)

(C)

x

(D)

x

KINEMATICS

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56

KINEMATICS

Multiple Choice Question 27. A particle is moving eastwards with a velocity of 5 m/s. In 10s the velocity changes to 5 m/s northwards. The average acceleration in this time is (A) Zero (C) 1

(B) 1

2 m s 2 towards north-east

(D)

2 m s 2 towards north-west

1 m s 2 towards north-west 2

28. A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. NO other information is available about its motion at intermediate times (0 < t < 1). If  denotes the instantaneous acceleration of the particle, then: (A)



(B)

 cannot exceed 2 at any point in its path.

cannot remain positive for all t in the interval

0  t 1.

 must be  4 at some point or points in its path. (D)  must change sign during the motion, but no other assertion can be made with the information given. (C)

29. The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt) where a, b(1) and the mass of B is also m. The coefficient of friction between A and floor is between B and floor is

1 and that

 2 ( 2  1 ) . Initially the left end of A

touches the left wall of B as shown in figure and both A and B moves with velocity  0 towards the right. All collisions between A ad B are elastic and contact time during each collision is very short. Find an expression for the period between two consecutive collisions.

3.

Consider the situation shown in figure the block B moves on a frictionless surface, while the coefficient of friction between A and the surface on which it moves is 0.2. Find the acceleration with which the masses move and also the tension in the strings. (Take = 10 m/s2)

4.

Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, upto what acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man = 65 kg. (g = 9.8 m/s2) The rear side of truck is open and a box of 40 kg. mass is placed 5 m away from the open end as shown in figure. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms-2 , find the time when box falls off the truck. (g = 9.8 m/s2)

5.

6.

A small disc A is placed on an inclined plane forming an angle  with the horizontal and is imparted an initial velocity v 0. Find how the velocity of the disc depends on the angle  if the friction cofficient   tan  and at the initial moment .  0   / 2

7.

A monkey of mass m clings to a rope slung over a fixed pulley. The opposite end of the rope is tied to a weight of mass M lying on a horizontal plate. The coefficient of friction between the weight and the plate is  . Find the acceleration of weight and the tension of the rope for three cases.(a) the monkey does not move with respect to the rope (b) the monkey moves upwards with respect to the rope with an acceleration b (c) the monkey moves downwards with respect to the rope with an acceleration b.

8.

–2

a = 2 ms

Each of the three plates has a mass of 10 kg. If the coefficients of

5m

M

m

18 N

static and kinetic friction at each surface of contact are  s = 0.3 and

 k  0.2 , respectively, determine the acceleration of each plate when the three horizontal forces are applied.

(Take g = 10 m/s2)

D C

15 N

100 N

B A

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43

DYNAMICS 9.

In the arrangement shown in figure, the rod of mass m held by two smooth walls, remians always perpendicular to the surface of the wedge of mass M. Assuming all the surfaces are frictionless, find the acceleration of the rod and that of the wedge.

m

10. A plank of mass M is placed on a rough horizontal surface and a constant horizontal force F is applied on it. A man of mass m runs on the plank. Find the acceleration of the man so that the plank does not move on the surface. Co-efficient of friction between the plank and the surface is  . Assume that the man does not slip on the plank.

M

F

11. A certain constant force starts acting on a body moving at a constant velocity v. After a time interval t , the velocity of the body is reduced by half and after the same time interval the velocity is again reduced by half. Determine the velocity of the body after a time interval 3  t from the moment when the constant force starts acting. M

12. A cart with a mass of M = 0.5 kg is connected by a string to a weight having a mass m = 0.2 kg. At the initial moment the cart moves to the right along a horizontal plane at a speed of v0 = 7 m/s. Find the magnitude and direction of the velocity of the cart, the place it will be at and the distance it covers in t = 5 seconds.

V0

m F(N)

13. A 20 kg block is originally at rest on a horizontal surface for which the coefficient of friction is 0.6. If a horizontal force F is applied such that it varies with time as shown. Determine the speed of the block in 10s.

F

200

0

5

14. For the system at rest shown in the figure, determine the accelerations of all the loads immediately after the lower thread keeping the system in equilibrium has been cut. Assume that the threads are weightless and inextensible, the strings are weightless, the mass of the pulley is negligibly small and there is no friction at the point of suspension.

t(s)

10

m1

m3

m2

m4

k 15. (a) In the arrangement shown in the figure the floor is smooth and the friction exists only between the blocks. The coefficient of static friction  s = 0.6 and coeffecient of kinetic friction

 k = 0.4, the

m1

m2 masses of the block are m 1 = 20 kg and m 2 = 30kg. Find the acceleration of each block if (1) F = 180 N (2) F = 200 N (b) Do the previous for F = 180 N if F is directed up as shown in the figure.

m1 F

k

F

m2

16. A body with zero initial slips from the top of an inclined plane forming an angle  with the horizontal.The coefficient of friction  between the body and the plane increases with the distance s from the top according to the law µ = bs. The body stops before it reaches the end of the plane. Determine the time t from the beginning of motion of the body to the moment when it comes to rest.

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44

DYNAMICS

Level # 3 1.

A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a force F(x) = – where K = 10–2 Nm 2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. (a) Find its velocity when it reaches x = 0.50 m. (b) Find the time at which it reaches x = 0.25 m (IIT 1998)

2.

3.

4.

A smooth semicircular wire-track of radius R is fixed in a vertical plane. One end 3R of a massless spring of natural length is attached to the lowest point O of the 4 wire-track. A small ring is held stationary at point P such that the spring makes an angle of 60° with the vertical. The spring constant K = mg/R. Consider the instant when the ring is released, and (a) draw the free body diagram of the ring. O (b) determine the tangential acceleration of the ring and the normal reaction. (IIT 1996)

Two masses m and 2m are connected by a massless string which passes over a light frictionless pulley as shown in the figure. The masses are initially held with equal lengths of strings on either side of the pulley. Find the velocity of the masses at the instant the lighter mass moves up a distance of 6.54 m. The string is suddenly cut at that instant. Calculate the time taken by each mass to reach the ground. (g = 981 cm/s2) (IIT 1977)

m

P

2m 13.08 m

Ground

m2

Two cubes of masses m 1 and m 2 be on two frictionless slopes of block A which rests on a horizontal table. The cubes are connected by a string which passes over a pulley as shown in the figure. To what horizontal acceleration  should the whole system (that is blocks and cubes) be subjected so that the cubes do not slide down the planes. what is the tension of the string in this situation? (IIT 1978)

A lift is going up. The total mass of the lift and the passengers is 1500 Kg. The variation in speed of the lift is as given in the graph. (a) What will be the tension in the rope pulling the lift at t equal to (i) 1 sec (ii) 6 sec (iii) 11 sec. (b) What is the height to which the lift takes the passengers? (c) What will be the average velocity and the average acceleration during the course of the entire motion? ( g = 9.8 m/s2) (IIT 1976)

2

60°

 ƒ m1

A 5.

K

2 x 

 v(m/s) 3.6

O

2

10

12 t (sec)

6.

Two balls A and B of masses 100 gm and 250 gm respectively and connected by a stretched string of negligible mass, and placed on a smooth table. When the balls are released simultaneously, the initial acceleration of ball B is 10 cm/s2 westward. What is the magnitude and direction of the initial acceleration of the ball A? (IIT 1975)

7.

A spring weighing machine kept inside a stationary elevator reads 50 kg when a man stands on it. What would happen to the scale reading if the elevator is moving upward with (a) constant velocity (b) constant acceleration ? (IIT 1972)

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45

DYNAMICS 8.

Two identical blocks A and B are placed on a rough inclined plane of inclination 45°. The coefficient of friction between block A and incline is 0.2 and that of between B and incline is 0.3. the initial

2m

9.

B

A

separation between the two blocks is 2m . The two blocks are released from rest, then find (a) the time after which front faces of both blocks come in same line and (b) the distance moved by each block for attaining above position. (IIT 2004)

B A

45°

In the figure masses m 1, m 2 and M are 20 kg, 5 kg and 50 kg P1 m1 respectively. The coefficient of friction between M and ground is zero. The coefficient of friction between m 1 and M and that between m 2 and ground is 0.3. The pulley and the string are massless. The string is perfectly horizontal between P1 and F M m 1 and also between P2 and m 2. The string is perfectly vertical P2 m2 between P1 and P2. An external horizontal force F is applies to the mass M. Take g = 10 m/s2. (a) Draw a free body diagram for mass M, clearly showing the forces. (b) Let the magnitude of the force of friction between m 1 and M be  1 and that between m 2 and ground be  2. For a particular F it is found that  1 = 2  2. Find  1 and  2. Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. (IIT 2000)

10. Block A of mass m block B of mass 2m are placed on a fixed triangular wedge by means of a massless, in extensible string and a frictionless pulley as shown in Figure. The wedge is inclined at 45° to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the system A and B is released from rest, find (a) The acceleration of A, (b) tension in the string, (c) the magnitude and direction of friction acting on A.

A m

2m 45°

45°

(IIT 1997, May)

11. Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and frictionless pulleys P1 and P2 as shown in the figure. Th masses move such that the portion of the string between P1 and P2 is parallel to the inclined plane and the portion of th string between and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37° with the horizontal. If the mass M1 moves downward with a uniform velocity, find: (a) the mass of M1 (b) tension in the horizontal portion of the string. 3 ) 5 12. In the Figure (a) and (b), AC, DG and GF are fixed inclined planes. BC = EF = x and AB = DE = y. A small block of mass M is released from the point A. It slides down AC and reaches C with a speed VC. The same block is released from rest from the point D. It slides down DGF and reaches the point F with speed VF . The coefficient of kinetic frictions between the block and both the surfaces AC and DGF are  . Calculate VC and VF . (IIT 1980)

B

(g = 9.8 m/s2, sin 37° =

P1 M2

P2

M1 M3

37°

(IIT 1981)

D

A

G

B

(a)

C

E

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(b)

46

DYNAMICS m

13. Two blocks connected by a massless sting slide down an inclined plane having an inclination of 37°. The masses of two blocks are m 1 = 4 kg m 2 = 2 kg respectively and the coefficient of friction of m 1 and m 2 with inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (a) the common acceleration of two masses and (b) the tension in the string (sin 37° = 0.6, g = 9.8 m/s2) (IIT 1979) 14. In the diagram shown, the blocks A, B and C weight 3 kg, 4 kg and 5 kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall while B and C are connected by a light flexible cord passing around a frictionless pulley. Find the force F necessary to drag C along the horizontal surface to the left at constant speed. Assume that the arrangement shown in the diagram, B on C and A on B, is maintained all through. (g = 9.8 m/s2) (IIT 1978)

1

m 2

37°

A B ƒ

C

15. An aeroplane requires for take off a speed of 80 km/h, the run on the ground being 100 metres. The mass of the plane is 10000 kg and the coefficient of friction between the plane and the ground is 0.2. Assume that the plane accelerates uniformly during the take off. What is the minimum force required by the engine of the plane for the take off? (g = 9.8 m/s2) (IIT 1977)

16. A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove and horizontal surface in contact 25 m/s2 2 is µ  . The disc has an acceleration of 25 m/s2 towards left. Find the 5 4 3 acceleration of the block with respect to disc. Given cos   , sin   . [2006] 5 5

Answer Key Assertion-Reason 1. (A)

2. (D)

3. (C)

4. (E)

5. (A)

6. (D)

7. (B)

9. (E)

10. (A)

11. (E)

12. (A)

13. (D)

14. (D)

15. (B)

8. (E)

Level # 1 Q. A ns. Q. A ns. Q. A ns. Q. A ns.

1 A 10 C 19 C 28 BD

2 C 11 A 20 D 29 B

3 C 12 A 21 A 30 C

4 A 13 A 22 B 31 B

5 A 14 B 23 C 32 D

6 C 15 C 24 D

7 A 16 A 25 A

8 D 17 D 26 A

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9 B 18 B 27 A

47

DYNAMICS

Fill in the Blanks / True-False / Match Table 1. 5 2. 0.5 p 3. (a) Earth (b) 4N, Book (c) No (e) 4N, hand, book, downward, (f) Second law 4. False 5. False 6. False 8. [(A — P, Q, R); (B — P, Q, R); (C — Q, S); (D — P, Q, R)] 9. [(A — P, S); (B — P, S); (C — Q); (D — R, S)]

(d) 4N, Earth, book, upward (g) Third law. 7. False

Passage Type Question Q. Ans. Q. Ans.

1 B 10 C

2 D 11 B

3 C 12 B

4 C 13 D

5 C 14 B

6 B 15 D

7 C 16 C

8 B

9 C

Level # 2

Section-A 1.

 7.47

 a   a g 

3. 2M 

2. (a) 7.35 m/s2, (b) 4.26 m/s (c) 1.54 m

4. (a) 37 N, (b) 55 N, (c) 36 m/s2, Upward 8. 2.02 m/s2, P = 15 N

7.

4 mg 17m  M

9. T = 174.1 N , f A = 1.27 m/s2 , downwards , f B = 1.08 m/s2 , f C = 0.9 m/s2

12. 1.272 m/s2 (down the plane) , 1.8 m/s2 (at 150 with horizontal ) , 149 N

Section-B 1. (a) 24.5 N , (b) 109 N

4.

2.

2(L  ) ( 2  1 )g

Net force on the man = 65 × 1 = 65 N a0 = Mg = 0.2 ×9.8 = 1.96 m/s2

6. v =

0 1  cos 

(b) a 

7. (a) a 

5.

2 5 = 4.34 s 0.53

( m  M )g mMg (1  ) , T m M Mm

m(g  b)  Mg 2m1m3 g , mM (m2 m3)(m1 m2)m2m3

8. aB = 0, aC = 4 m/s2 , aD = 0.2 m/s2

10.

3. a = 6 m/s2 T 2 = 48 N, T 1 = 32 N

F  ( M  m )g F  ( M  m )g  < a< + m m m m

(c) a 

m(g  b)  M g mM[g (1  )  b] , T mM mM

mg cos  sin  9.  m sin   M  sin    11.

mg cos  ,  m sin   M  sin   

7 v 4

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48

DYNAMICS 12. The cart will be at the same place and will have a speed v = 7 m/s, directed to right & total covered distance is 17.5 m. 14. a1 = a2 = a3 = 0 a 4 

13. 24.8 m/s

(ii) a1 = 6.08 m/s2 , a2 = 2.6 m/s2

m3  m 4  m1  m 2 g m4

15. (a) (i) a1 = a2 = 3.6 m/s2 16. t 

(b) a1 = 5.08 m/s2 , a2 = 3.4 m/s2

 gb cos 

Level # 3 1. (a) – 1 m/s (b) 1.48 sec

5 3    2. (a) a T   8  g  

3 (b) N    mg 8

3. Velocity = 6.54 m/s, Time taken by m is 2.8 second, Time taken by 2 m is

2 second. 3

 m1 sin   m2 sin   m1 m 2 sin   4.    m cos   m cos   g , T  m cos   m cos  2 1 2  1 

5. (a) (i) 17400 N (ii) 14700 N (iii) 1200 N (b) 36 m (c) Average velocity = 3 m/s, Average acceleration = 0 6. 25 cm/s2 Eastward. 7. (a) ramains same (b) increase 8. (a) t = 2 sec (b) SB = 7 2 m 9.

(c) SS = 8 2 m

(b)  1 = 30 N,  2 = 15 N, F = 60 N Tension = 18 N (Acceleration of M) = (Acceleration of m 1) = 0.6 m/s2 (m 1 is at rest w.r.t. M).

10. (a) zero

2 2    (b)  3  mg  

mg

(c)

3 2

11. (a) M1 = 4.2 kg

(b) 9.8 N

12. VC  VF  2 g y   x  13. (a) 1.31 m/s2

(b) 5.2 N

14. 78.4 N

15. 4.43 x 104 N

16. 10 m/s2

—X—X—X—X—

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49

GRAVITATION

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

GRAVITATION NEWTON’S LAW OF GRAVITATION Newton in 1687 proposed force law that we call Newton’s law of gravitation. According to him, “every particle of matter in the universe attracts every other particle with a gravitational force whose magnitude is directly proportional to the product of the masses of the particls and inversely proportional to the square of the distance between them”. Thus, m1m 2 r2 Here, m1 and m2 are the masses of the particles, r is the distance between them and G is the gravitational constant, with a value that is now known to be FG

G = 6.67 × 10–11 N–m2/kg2 G = 6.67 × 10–11 m3/kg–s2

1. 2. 3.

The direction of the force F is along the line joining the two particles. Regarding gravitational force, following points should be noted : The gravitational force between two particles is independent of the presence of other bodies or the properties of the intervening medium. Gravitational force is conservative force, therefore work done in displacing a body from one place to another is independent of path. It depends only on initial and final positions. The gravitational force obeys Newton’s third law i.e.,   F12   F21

C1 : Calculate the gravitational force of atraction between two spherical bodies, each of mass 1000 kg if the distance between their centre is 10 metre. (G = 6.67 × 10–11 Nm2 kg–2) Sol.

FG

m1m2 1000  1000  6.67  1011 2 r (10)2

F = 6.67 × 10–7 newton. Vector form of the law   Let us denote by F12 the force on m1 by m2. Similarly F21 denotes the force on m2 by m1. Then   F12   F 21 .

A m1 A

r12

F12 r1

B

F21

B That is, Newton’s law of gravitation obeys. Newton’s third law of motion m2 r2 (in strong form). The force on m1 is directed towards m2 and that on m2 is O directed towards m1. It is a central force.  Let be r12 be position vector of m1 with respect to m2. From vector addition law, we get,    r12  r 2  r1    r12  r1  r 2   Now F12 can be written as the product of magnitude and direction (unit vector). The direction of F12 is   r12   opposite to r12 . Hence  s the unit vector in the direction of F12 . | r12 |

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1

GRAVITATION

 Now

       r12    r 2  r1  F12  | F12 |     | F12 |      | r12 |   | r 2  r1 | 

  Gm1m 2 | F21 |  | F12 |  r2

  the distance between m1 and m2 may be written as r  | r1  r 2 | . This gives    Gm1m 2  r 2  r1  F12    2     | r1  r 2 |  | r 2  r1 | 

 Gm m   F12   1 23 r 2  r1 | r 2  r1 |



Thus,



    Gm m F12   1 2 3 r 2  r1  F21 | r 2  r1 |









This is the vector form of Newton’s law of gravitation. Principle of Superposition By using the principle of superposition, we can find the net (or result) gravitational force on any particle from others in a group of particles. This is a general principle that says a net effect is the sum of the individual effects. For n interacting particles, we can write the principle of superposition for gravitational force as       ...(i) F1,net  F12  F13  F14  F15  ......  F1, n  Here, F1,net is the net force on particle 1. We can express this equation more compactly as a vector sum. n   F1,net   F1i ...(ii) i2

For a particle which is from a real extended object the sum of Eq. (ii) becomes an integral and we have   F1   d F in which the integral is taken over the entire extended object and we drop the subscript “net”. y

Example 1 : Fig. shows an arrangement of three particles, particle 1 having mass m1 = 6.0 kg and particles 2 and 3 having mass m2 = m3 = 4.0 kg and with  distance a = 2.0 cm. What is the net gravitational force F1 that acts on particle 1 due to the other particles ?

m2 a m3

2a

x

m1

 Sol. The magnitude of the force F12 on particle 1 from particle 2 is F12 

Gm1m 2 a2

F12 

(6.67  1011 )(6.0)(4.0) (0.020)2

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GRAVITATION F12 = 4.0 × 10–6 N  Similarly, the magnitude of force F13 on particle 1 from particle 3 is

F13 

Gm1m 2 (2a)2

F13 

(6.67  1011 )(6.0)(4.0) (2  0.020) 2

F12 = 1.0 × 10–6 N

  the force F12 is directed in the positive direction of y and has only the y-component F12. Similarly, F13 is directed in the negative direction of x and has only the x-component – F13. y

Thus, the net force on particle 1 is

F12

F1, net  (F12 ) 2  ( F13 )2 6 2

F13

6 2

F1, net  (4.0  10 )  (1  10 )

x

m1

F1, net = 4.1 × 10–6 N C2 : Consider two particles of masses m and 4m separated by a distance of r. Find a point where the force exerted by these two particles on a third particle will be zero. Sol. Let the third particle of mass m´ be placed at a distance x from m on the line joining the particles. It must lie between the particles so that force may get balance. Then

G mm´ G 4 mm´  x2 (r  x)2

1 4 1 2    2 2 x (r  x) x rx r – x = 2x r = 3x x = r/3 The point is between the particles and at a distance of r/3 for m. C3 : Consider a particle moving in a circle of radius r due to gravitational force exerted by another particle of mass M at the centre. Calculate the speed of particle assuming the central particle at rest.

m M

r

Sol. Acceleration towards centre = v2/r GM m Force towards centre  r2 Using Newton’s second law,

GM mv 2 Mm v G 2 , r r r C4 : It is observed that two particles each of mass m are moving around a circle of radius R due to their mutual gravitational fore of attraction; then calculate the velocity of each particle. Sol. Here the force of attraction between them provides the necessary centripetal force. [as the two particles always ramain diametrically oppositive] hence.

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GRAVITATION V 2

2

mV m G R 4R 2

V

F O

m

Gm 4R

F

m

R V

Example 2: Three particles are located at the corners of an equilatural triangle of side a. Mass of each particle is m. They are all moving with the same speed in the same sense along the circumstance of the triangle. Find the speed of each. Sol. Figure shows forces acting on one particle, say A. Net force is vector sum of the two forces (superposition principle is obeyed). V m

Gm 2 Net force towards centre = 2 cos 30º × 2 a = From figure,

r cos 30º =

a

a r

3 Gm 2 a2

V

30º

m a/2

m

V

a 2

r 3 a  , 2 2

r

a 3

v2 3 v2  Acceleration towards centre  r a Using Newton’s second law of motion m 3 v2 3 Gm 2  , a a2

v

GM a

Example 3 : Three particles each of mass ‘m’ are placed on the three corners of a square of side L metre. Calculate the gravitational force on the force on the fourth particle of mass m placed at the corner D.

m A L B m

Gm 2 Sol. Force due to the particle at C = 2 along DC. L

L

m D L

L

C m

Gm 2 Force due to the particle at A = 2 along DA. L Gm 2 Force due to the particle at B = along DB. ( 2 L)2  Resultant force on the particle m at D  Gm 2 Gm 2 Gm 2    2 cos 45º  2 cos 45º  , along DB L  L ( 2 L)2 



Gm 2  1 2   , along DB. 2  L  2

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GRAVITATION Example 4 : A small mass and a thin uniform rod each of mass ‘m’ are positioned along the same straight line as shown. Find the force of gravitational attraction exerted by the rod on the small mass. Sol. The mass of considered element is m dm = dx 2L Gmdm Gm 2 dF = = dx  (L + x)2 2L(L + x)2 

F

Gm 2L

2 2L

2

1

 (L  x)

2

dx 

0

2L L m

m

dm m

F x

2L

Gm  1  2L  (L  x)  0

2

F

Gm  1 1   2L  3L L 

F

Gm 2 2 Gm 2   2L2 3 3L2

Example 5 : Figure shows two sphere of mass M each (S1 and S2). A light rigid rod has two lead balls each of mass m. The rigid rod is suspended from vertical wire. S1 pulls A and S2 pulls B so that the rod turns by . The twist in suspension wire is producing a restoring torque. If AS1 = BS2 = d, determine the torisonal constant of suspension fibre, S. Find its order if G = 6.67 × 10–11 Nm2 kg–2, m M ~ 102 kg L ~ 1 m and d ~ 1 mm. Sol. Torque due to gravitational forces on A and on B is  

S2

´ A d

d´ B

S1

GM m L . This must be balanced by the restoring d2

torque ´ = c.  Now,

GM m L d2 d  (L / 2) c 



Gmm L d2

GMmL2 c 2d 3

6.67  10 4  1011  1 c ~ 10 4119 ~ 103 Nm rad 1 3 2  (10) GRAVITATIONAL FIELD Can force be exerted by one body on another without contact? Some people think, yes ! A magnet can pull iron nail from distance, earth pulls an apple from distance and so on. This view is known as action-at-adistance point of view.

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GRAVITATION The question that haunted Newton also was : how is it that a body acts on another body from a distance ? He left the question unanswered but with the hope that there must be something that is mediating interaction; force cannot be exerted from distance. Today we know that this something is field. Faraday used to think in term of fields. Further investigations have given rise to certain particles mediating the interactions; we shall not go into such details but put here only the preliminary developments in the theory of gravitational field. Gravitational field is created by every mass around it and expends at speed of light. When this field reaches another mass, it exerts a foce on that mass. Thus one mass exerts force on the field and that field exerts force on this mass. The field plays intermediary role between two masses. Gravitational field intensity Let us put a test mass m0 at a point in a gravitational field created by some sources. The test mass experiences  a force F . This force is found to be proportional to the value of test mass m0.  F  m0  We can write equality sign by introducing proportionality constant, say g . Then   F  m0 g  Here g does not depend upon m0. Hence it is a vector property of the gravitational field. We call it ‘gravitational field intersity’. The strength of field at a point is described by force experienced by unit test mass placed at that point. The force per unit test mass is known as gravitational field intensity.  The value of F given above usually depends upon the location of observation the point and the mass distribution producting the field. Since the force is mutual, the test mass slightly disturbs the distribution of source masses  and g measured is not due to the given distribution but due to slightly different distribution of mases. This disturbance is negligible if m0  0. With this is mind the above definition is recast as    F g  lim  m0  0  m   0

 Unit and dimensions of g

S.I. unit — N kg–1 Dimensional formula : [LT–2] Gravitational Field intensity due to a point mass  Let us calculate as g at point due to a point mass M.

 For this we imagine a test mass m0 put at P. Then we write force using F using Newton’s law of gravitation.  Mm ˆ F  G 2 0 (r) Thus, r  GM g   2 rˆ . The r  The minus sign shows that is g is directed towards the source mass M.

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GRAVITATION Gravitational field intensity due to may point masses We can obtain net gravitational field intensity as the vector sum of field intensities due to each point mass.  If i-th source mass produces field intensity g i , then net field due to i = 1 to N point masses is given by     g  g1  g 2  .........  g N

 N  g   gi i 1

If the mass distribution be continuous, we have   g   dg where the integration covers the whole body. C5 : Three particles of equal mass are situated at the corners of an equilateral triangle of side ‘a’. How much is the gravitational field intensity at its centroid ? (2) m    Sol. Let us put a test mass m0 at the centroid. Forces acting on it are F1 , F2 and F3 . These are of equal magnitude and act at 120º with each other. Hence,    F1  F2  F3  0   g0

r F1

m

F2

r r

F3 m (3)

(1)

C6 : In the previous example, find the gravitational field intensity at mid point of one side of the triangle.    Sol. The figure shows the situation. F1 is force on test mass m0 at P, due to M at A, etc. F2 and F3 are equal and opposite. AM Hence   F2  F3  0  The only force remains is F1 , whose F1 

GMm 0  a 3  2   

2

M B

F2

F1 m0 P F3

M C

4 GM 3a 2 Thus the field intensity is 4 GM/3a2 in magnitude and directed towards the corner not jointed by the side containg observation point.



g

GRAVITATIONAL POTENTIAL Gravitational potentia at a point is defined as the work done by the field per unit test mass when the test mass is taken from given point to infinity or standard point. If W be the work done by the field when the test mass is taken to infinity the potential at the given point V

W m0 .

W is dependent upon the force exerted by the mass distribution and the mass distribution gets disturbed if m0 is not very small. Hence to get true gravitational potential m0 should be taken is very small. Thus

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GRAVITATION W m0  0 m 0

V  lim

S.I. Unit : J kg–1 or, m2s–2 Dimensional Formula [M0 L2 T–2] Gravitational potential due to a point mass Let M be a point mass producing gravitational field. We want to find potential at point P. A test mass m0 is taken from P towards infinite. At some position in this course, force on it is given by

Fx 

GM m0 GM m0 . If the displacement be dx, the work done by Fx is given by dW = Fxdx = – dx. 2 x x2

Now the total work done in moving it from p to W is given by integration. That is, 

W  GM m 0  r 

dx x2

W  GM m0  x 2 dx r 



 x 21   GM m 0 W   GM m 0    r  1  r W  GM V  It is a negative scalar. m0 r ,

Gravitational potential due to many particles If there be many particles producing the field, the potential at a point is the scalar sum of potentials due to all the N

point masses. that is, V   Vi i 1

If the mass distribution be continuous we resort to integration V   dV where integration covers the whole system. Example 6 : Three particles, each of mass m, are located at the corners of an equilateral triangle of side a. Calculate potential at the centroid. a Sol. The distance r of centroid from a corner of the triangle in given by r cos 30º = . m 2 a/2

ra/ 3

30º

a

The potential at O due to mass m at the corner is equal to –Gm/( a / 3 ). Hence total potential at O due to three masses is given by    GM  Gm V     3  3 3 a .  a   3 

O m

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m

8

GRAVITATION GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy of a mass m placed in the field of mass M will be calculated. Using the definition of potential at site of m, work per unit mass = potential = –

GM . r

now the work done by field in taking m to infinite distance away is given by W

GMm r

This work is called potential energy U, of m in the field of M. Thus U = potential × m (Potential energy = mass × potential) Work done is assembling a system : Work done by field in taking m to  is the same as work done by agent bringing m from  to the given point without acceleration. Hence W = potential × mass. In bringing single mass m1, when no other mass is there, no work is done by agent  W1 = 0 ...(1) The second mass m2 is brought is presence of m1. Hence W2 = (potential due to m1) × m2

Similarly,

 Gm1  W2 =   r  m2  12 

...(2)

 m1 m 3  W3 = – G  r  r  m3  13 23 

...(3)

Summation gives total work done.  W = W1 + W2 + W3

 mm mm mm  W   G 0  1 2  2 3  3 1  r12 r23 r31   This called mutual potential energy of the assembly of masses. If summation notation be used, we get W

where factor of

G N  2 i 1

 j i

mi .m j ri j

,

m j .mi m i .m j 1 is to take into account the double terms like r and r . 2 ji ij

C7. A particle of mass 1 kg is kept on the surface of a uniform sphere of mass 20 kg and radius 1.0 m. Find the work to be done against the gravitational force between them to take the particle away from the sphere. Sol. Potential at the surface of sphere

GM (6.67  1011 )(20)  J / kg R 1 V = – 1.334 × 10–9 J/kg V

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GRAVITATION i.e., 1.334 × 10–9 J work is obtained to bring a mass of 1 kg from infinity to the surface of sphere. Hence, the same amount of work will have to be done to take the particle away from the surface of sphere. Thus, W = 1.334 × 10–9 J Example 7 : Three particles each of mass m are placed at the corners of an equilateral triangle of side d. Calculate (a) the potential energy of the system, (b) work done on this system if the side of the triangle is changed from d to 2d. Sol. (a) As in case of two-particle system potential energy is given by (–Gm1m2/r), so UA = U12 + U23 + U31 So,

UA = –3

Gmm 3Gm 2 =– d d

C

(b) When d is changed to 2d,

UB  

m

60º

3Gm 2d

d

2

3Gm 2 So, work done = UB – UA = 2d

B

Am

m

Example 8 : Four masses (each of m) are placed at the vertices of a regular pyramid m (triangular base) of side ‘a’. Find the work done by the system while taking them appart so that they form the pyramid of side ‘2a’.

m a

Sol. Ui = Initial gravitation potential Gm 2 energy  6  a Uf = Final gravitational potential Gm 2 energy  6  2a wext = U = Uf – Ui



m

6Gm 2  6Gm 2   2a  a 

3Gm 2 6Gm 2 6Gm 2 6Gm 2 w ext       a a 2a 2a Binding Energy The minimum energy needed to dismantle a system into its constituents far apart is known as binding every of the system. Let E1 be energy of bound system and E2 the energy of dismantled system. Then E2 is greater than E1 and (E2 – E1) is the binding energy. Example 9 : Write the BE of the binary star. m v

v

d m

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GRAVITATION Sol. Let two identical stars move around common centre of mass at rest. They must have equal speeds and equal distance from centre of mass.

m2 The force on one star is F = G 2 , where d is separation between the two stars. This star has the speed v in d the circle of radius d/2, Newton’s second law of motion gives mv 2 Gm2  2 (d / 2) d Gm 2 2 mv  ...(1) 2d Using (1) we can write the kinetic energy K as 1 Gm 2 mv 2  ...(2) 2 4d The gravitational potential energy of the system is given by Gm 2 U ...(3) d The mechanical energy is bound state is K

Gm 2 Gm 2 3 Gm 2     4d d 4 d When the stars are far apart (dismantted), and at rest, E=0 The binding energy is now written as Eb = E´ – E E

...(4) ...(5)

 3 Gm2  Eb  0     4 d  3 Gm 2 . 4 d The binding energy of any bound system is positive. Eb 

Example 10 : A particle is projected from the surface of earth with an initial speed of 4.0 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km, and g = 9.8 m/s2. Sol. The maximum height attained by the particle is, h

v2

v2 R Substituting the values, we have 2g 

h

(4.0  103 )2 (4.0  103 )2 2  9.8  6.4  106

h = 9.35 × 105 m or

h  935 km

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GRAVITATION Example 11 : An object is projected vertically upward form the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. Calculate : (i) the initial speed of projection (ii) the speed at half the maximum height. Sol. (a)At maximum height, velocity becomes zero. Applying conservation principle of mechanical energy. Ui + Ti = Uf + Tf GmM 1  GmM   mv 20  R 2 9R 1 GmM GmM mv 20   or 2 R 9R GmM 8 8GmM   = R 9 9R  (b) or or

4 GM 3 R Ui + Ti = Uf + Tf  GmM 1  GmM 1  mv02   mv 2 R 2 5R 2 GmM 1 16 GM  GmM 1  m   mv 2 R 2 9 R 5R 2



v0 =

v=

2 3

2GM 5R

Example 12 : A projectile is thrown from the surface of the earth with a velocity v0. It loxses % of its mechanical energy in doing work against air friction. What height above the surface does it attain if (a) = 0 (b)   0. the radius of the earth is R and the value of gravitational field intensity is g at the surface. Sol. The initial energy of the projectile is the sum of kinetic energy

1 GmM mv20 and the potential energy  , i.e., 2 R

– mg R. Thus Ei 

1 mv 20  mg R 2

At a height h, the gravitational potential energy will be 

GMm GMm mgR   h h Rh  1 R 1    R R

The kinetic energy will be zero. Hence the mechanical energy at height h is mgR Ef   h   1   R According to question Ef = Ei –  % of Ei (a) For = 0, Ef = Ei .

v=0 v0

h R

(i)

(f)

 mg R 1  mv 20  mgR 1 h / R 2

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GRAVITATION

v02 1 1  2gR 1  h / R  2gR  2gR  v02  h  R  2  2gR  v0 

2 gR h  1 , 2 2 gR  v0 R

h

R  2Rg  ,  v 2  1

h

0

(b)

R n 1 2

ve where n  v 0

 mgR   1   1  mv02  mgR    h  100   2  1 R 1

h mgR h 2gR  1      R  R  2 2  1   mgR  mv 0 / 2 ,  1   mgR  v 0 100 100 









    2gR h  R  1  2gR  v 2 1     0    100  





Relationship between gravitational potential and field intensity  Let V be potential and g be gravitational field intensity at a point. If a test mass m0 be placed at this point, the potential energy U and force F, will be U = mass × potential U = m0V ...(1)   ...(2) F  m0 g    If m0 be displaced by dr , work done by gravitational force is m0 g . dr . This must be equal to the fall in potential energy, –dU. That is, –dU = dWcons   – d(m0 V) = m0 g . dr   – dV = g . dr ...(3) This is the relationship in differential form. In one dimension the above equation may be written as –dV = gx dx  dV gx  dx In two dimension, V is function of two variable. Then we use partial differentiations. We have y V y gx   x (x, y) V V x gy   y x

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GRAVITATION  means differentiate relative to x keeping other variables constant. x In three dimensions we have

Here

gy 

V V V ; gx   and g z  . y x z

       [In short g    ˆi  ˆj  kˆ   V   V ]  x y z  For a radial dependence V(r), we have gr = – dV/dr   Here gr is the component of g along r . Integral form : Integrating eqn. (3), we have V2



   dV   g . dr

V1

r2

r1

r2   V1  V2   g . dr r1

This is the integral for of the relationship. Example 13 : Consider a gravitational potential given by V =   z – components of gravitational feild intensity g .

Sol. Here

k     where r  x i  y j  z k . Write down x, y and r

 | r |  x 2  y2  z2

k



V(x, y, z) =

x 2  y2  z 2

gx  

 V   k    x x  x 2  y 2  z 2 

gx  k

d(x 2  y 2  z 2 )1/ 2  2  (x  y 2  z 2 ) 2 2 2 d(x  y  z ) x

k 2 (x  y 2  z 2 ) 3/ 2 . 2x 2  kx kx gx  2  3 2 2 3/ 2 (x  y  z ) r gx 

Similarly,



ky r3 kz gz   3 r   k ˆ ˆ   kr g   3 (x i  y ˆj  z k) r r3 gy  

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14

GRAVITATION Gravitation potential and field intensity at a point on the axis of a uniform ring The figure shows a uniform ring of mass M and radius R. On its axis is situated a point P, at a distance x from the centre. Let us imagine the ring to be made up of elements of masses like dM. Potential at P due to dM is given by dM

dV = –G

2

x R

2

dM

.

x2 + R2 R

Potential at P due to the whole ring is given by V



G x R

Here

P

x2  R 2

ring

2

x

dM

G

V

M

2

 dM   ring

GM x2  R2

x 2  R 2 is ‘slant’ distance of P from the ring.

Field intensity : As V = V(x), we use gx = 

dV . dx

Putting the expression for V, gx  

d   GM  dx  x 2  R 2 

g x  GM

d(x 2  R 2 )1/ 2 d(x 2  R 2 )  d(x 2  R 2 ) dx

 1  g x  GM   (x 2  R 2 )3 / 2  . 2x  2 

gx 

GMx ( x  R 2 )3 / 2 2

Plots : The graph of V(x) is shown in the figure. Its slope is

GMx which is zero at x = 0. (x  R 2 )3/ 2 2

V(x)

– GM R

x slope = zero

The plot of gx against x is as shown. We see that

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15

GRAVITATION (i)

x=0

gx = 0

(ii)

x > R

gx = –

GM x2

(iv)

x  

gx = 0

gx

x

Example 14 : A point P lies on the axis of a fixed ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be_______. Sol. Machenical energy consevation law Ui + Ti = Uf + Tf  GmM a2  a2

0

 GmM 1  mv 2 a 2

or

GmM  1  1 2 1    2 mv a  2

or

v2 



2GM  1  1   a  2

v

2GM  1  1   a  2

Example 15 : Find the gravitational field strength and potential at the centre of arc of linear mass density  subtending an angle 2at the centre. Sol.

2G  E sin R 2 2G 2 E sin R 2 E

2 R

E

2G sin  R

G m R  m = (2R) m = 2R G V 2 R v   2G   R Example 16 : Consider a uniform semicircular rod of mass m and length L. An equally massive particle is placed at the centre of curvature. Calculate the force acting on this particle. V

Sol. The length of the rod is half of perimetre of a circle.  r = L r = L/ ...(1) The mass per unit length of the rod is m/L.

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16

GRAVITATION Let us divide the rod into elementary masses, each of mass dm contained in angle d at position  (see figure). Then dm = m/Lr d. = /2 The force exerted by this ‘point mass’ on the mass m at centre  of currature is dF . It has two components –dF and dF sin . There is a symmetrical mass dm at – whose effect is to cancel out dF sin . The only contribution remains uncancelled is dF cos . Hence net force Fnet on the mass is givien by dm d

Fnet 



dF cos  =0

whole wire

dm Fnet  Gm 2 cos  r m cos  Fnet  Gm d L r

Fnet 

Gm 2 rL

m dF dF sin

–

/2



cos  d

 / 2

= – /2

Gm 2 [sin ]/ 2/ 2 , rL r = L/ Fnet 

Putting

dFcos

dF sin

dF

Fnet  2

Gm 2 rL

2 Gm 2 L2 This force is directed towards the mid point of the semicircular wire. We have

Fnet 

Gravitational potential and field intensity due to a uniform Spherical Shell Let us consider a uniform spherical shell of mass M. Its mass per unit area will be M/4R2 = . Let us consider a strip of width R d and radius R sin  (see figure). The whole shell is made up of such strips. Rd

M

z

d R sin x Uniform spherical shell

P

R

Potential at P due to the ring is given by dV = – G

2 R 2 sin  d z z  (x  R cos ) 2  (R sin ) 2 z2 = x2 + R2 – 2 xR cos  dV = – G

Now,

dM z

...(1)

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GRAVITATION Differentiating, we get 2 xR sin = 2z dz z dz sin d= xR From (1) and (2) GM z dz . dV = – 2z xR GM dz dV = – 2x R

...(2)

...(3)

Case I : P is outside the shell. In this case minimum value of z is x – R and maximum value is x + R. Therefore, Potential

x R

GM V 2x R

V



dz

xR

GM GM . 2R   2x R x

GM x The gravitational field intensity is obtained by using, dV d  GM  gx       dx dx  x  GM gx   2 x The shell is acting as if it were a point mass located at the centre.



V

Case II : Point P lies inside the shell. Here

Zmin = R – x



GM V 2x

V

GM R

x Z P

x R



dz

R

Rx



gx = 0

Potential at all the inside points is equal to that at the surface. V(x)

x=R – GM R

x

not differentiable

The plot V(x) is shown here. At x = R, the plot has a corner where it is not differentiable. The plot of g(x) has a x finite discontinuity at x = R.

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18

GRAVITATION g(x) gx = 0

discontinuity

x=R

x gx = – GM x2

C8. Two concentric spherical shells have masses m1 and m2 and radii r1 and r2(r2 > r1). What is the force exerted by this system on a particle of mass m3 if it is placed at a distance r (r1 < r < r2) from the centre. Sol. The outer shell will have no contribution in the gravitational field at point P. 

EP 

m2 m1

Gm1 r2

r1

Thus, force on mass m3 placed at P is,

m3

O

F = m3EP

r

P

r2

Gm1m3 r2   The field EP and the force F both are towards centre O. or

F

Gravitational potential and field intensity at a point due to homogenous solid sphere The point where potential and field are to be evaluated may be (i) outside the sphere (ii) inside the sphere. Case I : Outside point. We assume the sphere to be made up of concentric spherical shells, one of which is shown. Its potential at P is given by dV = – GdM/r dm Integration gives G dm V=– r  GM r For observation at outside points, sphere like a point mass located at its centre. V

gr  

dV d  GM  GM     2  dr dr  r  r

GM r2 The solid homogenous sphere acts as a point mass at its centre. Case II : Inside point : Potential at P due to the whole sphere may be treated as the sum of two potentials : The potential due to gr  

(i)

Sphere of radius OP, V1 and

(ii) Hollow sphere of inner radius OP = r and outer radius R, V2.

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19

GRAVITATION (i)

Potential (V1) due to sphere of radius OP. Here P lies at the surface of this sphere. Hence r

O

P 4 / 3r 3 V   G  1 r R M M  where  = density = V 4 / 3R 3 (ii) Potential (V2) due to hollow sphere : The hollow sphere is assumed to be made up of spherical shells. One such shell, shown in the figure is producing potential at P given by

dV2   G

dM x

dM

4x 2dx dV2  G x Due to the whole hollow sphere we have

P x dx

R

V2  4G  xdx r

4G 2 2 (R  r ) 2 Thus, using (i) and (ii) potential at P is given by V = V1 + V2 4 G r 3 4 2 2   3 V  G (R  r )  r 2   V2  

V

...(ii)

G 4r 2 G 4 2 2  (R  r ) 3 2

 r2 R 2  r2  V  G 4    2  3  3R 2  r 2  V   4 G  6   GM  3R 2  r 2  V  2 4 / 3R 3  3  GM (3R 2  r 2 ) 2R 3 This is as function of r. The gravitational field intensity along radial direction denoted by. It is given by dV  GM  gr      (0  2r)  3 dr  2R  GM gr   3 r R The graphs of V(r) and gr are shown below. Mark that | gr | is maximum at the surface of the sphere. V

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GRAVITATION V(r) r

r=R

outside sphere (hyperbola)

– 3GM 2R inside sphere (Parabola)

Table : Field and Potential Source

Potential (V) 

Point mass 

Ring

GM r



GM 2

x R



2

GM r2

GM x (x  R 2 )3/ 2 2

 GM  R , r  R   GM , r  R  r

rR  0,   GM  r 2 , r  R

 GM 2 2  2R 3 (3R  r ), r  R    GM , r  R  r

 GM  R 3 r, r  R   GM , r  R  r 2

Spherical Shell

Solid Sphere

Field intensity (gr)

C9 : A tunnel is made in the earth along its chord at a distance of R/2 from the centre. A mail is released from the mouth of the tunnel. there is no friction between the main and the tunnel wall. Calculate the normal force exerted by the tunnel on the mail if its mass is m. The mass of the earth is M and radius R.

N mg cos mg R/2

mg sin

Sol. Forces on the mail are (i) mg, towards the centre of the earth (ii) N, normal to tunnel Here g = GMr or, N = mg cos   GM  N = m  3 r cos  R

N=m N=

GM R R3 2

R   r cos    2

GM m 2R 2

Thus, N is independent of ; it depends on R.

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GRAVITATION C10 : A spherical mass of radius r = R/2 is taken out from a sphere of radius R and density . Calculate the force which this sphere having a cavity will exert on a mass m placed at a distance of x from its centre (x > R). Sol. For the whole sphere of radius R F1  G

Mm x2

4 / 3 R 3m x2 and for small sphere of radius r = R/2 M1m G4 / 3 (R / 2)3  m F2  G  (x  R / 2) 2 (x  R / 2)2 Resultant force = F1 – F2

R

F1  G



O

P r = R/2

x

1  4 1 R 3 Gm  2  2  3 2(2x  R)  x

KAPLER’S LAWS OF PLANETARY MOTION Before we actually put the kinematical laws of motion of planets given by Kepler, let us familiarise ourselves with an ellipse.

F2 F1

ellipse

Activity for an Idea of an ellipse : Take a drawing broad. Fix on it a graph paper. Fix two pins (F1, F2) on the paper. Take a thread whose length in larger than the distance between the two pins. Tie its ends to each of the pins. Take a pencil and pull the thread by it as in the figure. Move the pencil keeping the thread tight. You get a curve called an ellipse. The nails are two focii (plural world for focus) of the ellipse. Draw a line through F1 and F2. It cuts the ellipse at A and B. The length AB is called ‘major axis’ of the ellipse. Now draw a perpendicular bisector of the major axis. It is CD. It is called minor axis of the ellipse (because CD < AB as shown in the figure). The ratio OF1/OB or OF2/OA is known as eccentricity (e) of the ellipse. It is a measure of how much distorted the ellipse is releative to a circle (e = 0). Using the graph paper, you may find the area of the ellipse and compare with the standard formula, to judge how good your ellipse is : Area =  ab

A F2 C

2a D

F1 B 2b

The Kepler’s laws : Kepler (1571–1640) used an ellipse to describe his three laws of planentary motion. The three laws are :

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GRAVITATION (a) Law of orbits, (b) Law of areal velocity and (c) The law of periods. We shall now state and discuss these. The Law of orbits The closed path along which a planet moves around the sun is called an orbit. the first law gives its geometry and location of the sun relative to the orbit. Every plate moves around the sun in an elliptical orbit with the sun at one of the focii. P

In the figure sun is located at the focus F1, and F2 is vacant. The planet P is moving along the ellipse. Another planet will move in another ellipse but one focus of that ellipse will always be at the sun.

F2 S

F1

The Law of Areal velocity or sector velocity If we join a planet (P) to the sun (S), the line (SP) turns as the planet moves. It sweeps some area in given time. The area swept per unit time is known as areal velocity (or sector velocity) of the planet. The Kepler’s second law tells us that this areal velocity remains uniform. The line joining a planet to the sun sweeps equal and in equal time interval. P´

P

Q (Perihelion) Fast

S F1

F2

(Aphelion) slow

Q´

Let a planet move, from P to P´ in a time t, covering area SPP´. The planet goes from Q to Q´ in the same time interval t. Then, according to Kepler, Ar (SPP´) = Ar (SQQ´) If A be the area and t the time taken then, A = constant. t When the Earth, for example, is closest to the sun (at perihelion), its speed in larger than when it is farthest from the sun (apehelion).

The Law of Periods A planet takes some time to move once around the sun. This time is known as time period of the planet. If several planets are moving around the sun, they have their own periods. The time period of the earth for motion around the sun, for example, is one year. The law of periods connects this time period to size of ellipse. The law was formulated by Kepler’s in 1619. The square of period of revolution of planets around the sun are directly proportional to the cube of semimajor axes of respective orbits of the planets. Let T1 be the period of planet-1 moving in an ellipse of semimajor axis a1. Similarly T2, a2 are period and semimajor axis for planet-2, T3, a3 for planet-3, etc. Then

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GRAVITATION T12 T22 T32   = ....... = constant. a13 a 32 a 33 In short, T2  a3. Notice the Table giving the value of constant. Table : Law of Periods Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

Semi-major axis (a)

Period (T)

Ratio T2/a3

(× 1010 m)

(years)

(× 10–34 year2/metre3)

5.79 10.8 15.0 22.8 77.8 143 287 450

0.24 0.615 1 1.88 11.9 29.5 94 165

2.95 3.00 2.96 2.98 3.01 2.98 2.98 2.99

The Physical basis  The sun exerts a force of pull ( F ) on a planet that passes through the centre of mass of the planet. Newton discovered the law that gives the force : It is central and attractive (always directed towards the fixed point, the sun), and obeys inverse square law of distance (the force is inversely proportional to square of distance between the sun and the planet). The torque of this force about the sun is zero. Hence angular momentum of the planet about the sun is conserved.    If r be the position vector and v be the velocity of the planet, the angular momentum L of the orbiting planet is given by v    m L L  r  mv d   r F planet Now the area of triangle formed by v dt and r can be represented by geometrical meaning of cross product of two vectors. Thus L p   1 r dA  | r  v dt | 2 Using this we have    | L |  | r  v| m

  dA  | L |  2 m  dt 

 The direction of L is constant and magnitude is also constant. Thus,  dA | L |  = constant dt 2m

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GRAVITATION The law of areal velocity is, thus, consistent with conservation of angular momentum of the planet around the sun. In turn, it bases on the central nature of force between the planet and the sun. The other two laws are consistent with inverse square and attractive nature of gravitational force. Demonstrating these by calculations is straight forward but a bit cumbersome and not necessary at this level. Since a circle is also an ellipse with zero eccentricity, we have T2 = kr3 (Law of periods). 2

Also

 2  F  m r  m   r  T 2

4 2 m F .r kr 3 F

4 2 m kr 2

1 r2 Thus the period’s law is consistent with inverse square attractive force.

Thus

F

The value of ‘constant’ in Kepler’s law of periods for planetary motion The square of the period of revolution of a planet around the sun is proportional to the cube of semimajor axis of its orbit. T2  a3 T2 = ka3 ...(1) k is the constant of proportionality. To find k which is independent of the value of ‘a’, we take circle as a special case of an ellipse, where major axis equals minor axis Here we have P GmM s a S F= a2

mV 2 GmMs  a a2

v T

GMs a 2a v

T 2  4 2

 4 2  3 a2  a  GM s   GM s    a 

...(2)

From (1) and (2) k

4 2 GM s

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GRAVITATION Example 17 : A saturn year is 29.5 times the earth year. Find the distance of saturn from the sun if the Earth is at 1.50 × 108 km from the sun. Sol. Using Kepler’s law of periods, 2

 TS   rS    T   r  E E T  rS   S  T 

3

2 /3

. rE

E

 29.5 y  rS    1 y 

2/3

. 150  108 km .

We need log table for exact evaluation. However we have only three significant calculation and can use binomial method. (29.5)

2/3

= (27 + 2.5)

2/3

= 27

2/3

 2.5  . 1    27 

2/3

 2 2.5   (3 × 3 × 3)2/3  1  3  27  = 9.5 (Exact evaluation gives it as 9.5413)

Using this

rS = 14.25 × 108 km rS = 1.42 × 1012 m

To show that Kepler’s laws are consistent with Newton’s laws of gravitation Let a planet be moving in an elliptical orbit around the sun. Since the mass of the sun is very large than that of the planet, the sun will exert a gravitational force (attractive) on the planet.

m

F MS

R

G mM s ...(1) R2 Since the eccentricities are small and sun is more massive than the planets, the path can be assumed as approximately a circle. The centripetal force, exerted by the sun on the planet, is given by F

mv 2 m  2R  Fc     R R T 

2

...(2)

From (1) & (2)

GmM m  4 2 R 2    R2 R  T 2  T 2 4 2  R 3 GM Here, 42, G and M are constants, therefore, T2 = constant. R3

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26

GRAVITATION T2 = (const) R3 T2  R3 This is the law of periods. Areal velocity R 2 R 2   = constant. T  4 2 R 3   GM 

Thus the areal velocity in circular orbit is constant. C11 : The speed of a planet at perihelion is v0 and the distance from the sun is l. The major axis of elliptical path is 2a. Determine the speed at aphelion. If e be the eccentricity of the ellipse l = a(1 – e). Use it to expresses your answer.

ae

l

ae a

S v0

Sol. At perihelion the angular momentum about the sun-axis normal to plane of the orbit, is mv0 l where m is the mass of the planet. When it travels to aphelion let its speed be v. Then distance from the sun is (2a – l). The angular momentum about the above axis is now mv(2a – l). Since action line of the force exerted by the sun on the planet is passing through the sun, angular momentum about the sun is conserved. Hence mv0l = mv (2a – l)

v

v0 l v  2a  l  2a / l  1

Using l = a (1 – e),

v0  2   1  1 e   1 e v  1  e  . Example 18 : A body is launched from the earth’s surface at an angle  = 30º to the horizontal at a speed v

1.5 GM . Neglecting air resistance and earth’s rotation. Find the height to which the body will rise. R Here M is mass of earth and R the radius of earth. Sol. Let velocity at highest point be v and R + h = r Applying conservation of angular momentum between P and Q, we have Q v mvr = mv0R cos 30º v0 

3 v0 R ...(1) 2r Applying conservation of mechanical energy between P and Q, we have : 1 GMm 1 GMm mv02   mv 2  2 R 2 r substituting the value of v from equation (1), we get

or

v

=30 º

R

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R

v0

r

P

27

GRAVITATION

1 GMm 1  3v 20 R 2  GMm 2 mv0   m  2 R 2  4r 2  r or

2GM 3v02 R 2 2GM v    R 4r 2 r

or

1.5GM 2GM 3  1.5GM  R 2 2GM      R R 4  R  r2 r

2 0

or or

1 9 R2 2  .  2R 8R r 2 r –4r2 = 9R2 – 16Rr 4r2 – 16Rr + 9R2 = 0

or

r

  1.5 GM is given   v0  R  



or

16R  256R 2  144R 2 8

16R  10.58R = 3.323 R and 0.677 R 8 r  R or h = r – R = 2.323 R

r= but Hence

r = 3.323 R

ACCELERATION DUE TO GRAVITY AND ITS VARIATION The force on a particle near the earth can be written as  F = mass × gravitational field intensity Newton’s second law of motion gives   F = mass × a grav where agrav is the acceleration due to gravity. This gives gravitational field intensity mathematically equal to the acceleration due to gravity.   a grav  g (gravitational field intensity) Variation in acceleration due to gravity g with altitude and depth If we assuming that the earth is a homogenous sphere (which she is not exactly) the gravitational field intensity may be written as g=

GM (outside) r2

g=

GM r (inside) R3

where r = distance of observation point from the centre of the earth. The above expression shown that g decreases when we go away from the surface either inward (depth) outward (altitude).

|g|=g

For small height h above the surface

r=R

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r

28

GRAVITATION

g

g

Now,

GM  (R  h) 2

GM h  R 1    R

2

2

GM  h 1  2  R  R

2

GM  g 0 (surface value R2 2

h  g  g0 1    [which h M

v

GM r This is called orbital velocity of a satellite. It is independent of mass of satellite only if m ve, body again escapes but now the path is a hyperbola. Here, v0 = orbital  r  at A and 

ve = escape velocity at A. Note : 1. From case (i) to (iv) total energy of the body is negative. Hence, these are the closed orbits. for case (v) total energy is zero and for case (vi) total energy is positive. In these two cases orbits are open. 2.

A

v0

h R

If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.

C29 : Calculate the escape velocity from the moon. The mass of the moon = 7.4 × 1022 kg and radius of the moon = 1740 km. Sol. The escape velocity is

v

2GM R

v

2  6.67  10 11  7.4  1022 1740  103

v = 2.4 km/s C30 : Calculate the escape velocity on the surface of a planet whose mass is 1025 kg. and radius is 6.67 × 109 metre. Sol.

2  6.67  1011  1025 6.67  109

Ve 

2GM  R

Ve =

2  105 = 447.2 m/sec.

C31 : A body is thrown with a speed of 3ve where ve is escape speed from the surface of the Earth. Find its speed in interstellar space. Assume no other gravitational field than the earth, and neglect air. Sol. Conserving mechanical energy 

Now, 

GMm 1 1 2  m  3ve   0  mv 2 R 2 2

2 GM  v2e R

v 2e 1 1 2   3v e   v 2 2 2 2 2 2 2 v = (3ve) – ve 

v  2 2 ve Example 29 : A particle is fired vertically from the surface of the earth with a velocity ku e , where u e is the escape velocity and k < 1. Neglecting air resistance and assuming earth’s radius as Re. Calculate the height to which it will rise from the surface of the earth.

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44

GRAVITATION Sol. Applying conservation principle of mechanical energy, Ui + Ti = Uf + Tf here Tf = 0 GmM 1 GmM  mk 2 ve2   Re 2 Re  h  GmM 1  GmM  mk 2  2gR   or Re 2 Re  h or

 GM   v e  2gR e and g  2  Re  

GmM 1 GM GmM  mk 2 2 2 R e  Re 2 Re Re  h

By solving,

Re 1 k2

or

R e  R e  R ek 2 1 k2



Re  h  h

Re  Re 1 k2 Rek 2 h 1 k2

h





Example 30 : Two stars each of radius R are at rest at a distance where the force between them is negligible. The mass of one star is M and that of the other is 2M. Find their speeds when their centres are a distance d > 2R apart. Sol. Let the stars of mass M be moving with velocity V1 and 2M with V2. Then MV1 + 2 MV2 = (M + 2M)Vcm Since their initial velocities are zero, Vcm = 0  V1 + 2V2 = 0 V1 = – 2 V2 M M Negative sign indicates that they are moving in opposite directions. When separated by distance d, the potential energy is given by V V/2 U

GM (2M) d2

d

The kinetic energy is V1   1 1 MV 2  (2M) (V / 2) 2   V2   2  2 2 3 K  MV 2 4 The sum of two energies, K and U, must not change. Hence Ki + Ui = Kf + Uf K

00 

3 2GM 2 MV 2  4 d

8 GM V 3 d The speed of star of mass M is

8GM and that of mass 2 M is 3d

2 GM . 3d

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GRAVITATION Example 31 : A spaceship of mass 1000 kg. is situated on the surface of a planet whose distance from the sun is 2.28 × 108 km. (a) How much energy is the spaceship having ? (b) How much energy must be expended on it to launch it to go to infinity ? The mass of the sun is 2 × 1030 kg, planet is 6.4 × 1023 kg, radius of planet is 3395 km and G = 6.67 ×10–11 Nm2 kg–2. Neglect motion of the planet. Sol. The spaceship at rest has only potential energy. It is given by

M M  U  Gm   S  R r´ 

m r´ R

S

r

M

But r´ >> R hence we take r´  r. Thus U = – 6.67 × 10

–11

 6.4  1023  2  1030  × 10 ×  3395  103 2.28  108  103    3

 6.4  1017  2 U  6.67  10    1019  J 2.28  3.395  8

200   6.4 U  6.67  109   J  3.395 2.28 

U = – 6.67 × 89.5 × 109 J U = – 597.62 × 109 U = – 6.0 × 1011 J When the spaceship will go to infinity, its total energy will be zero. Ef = 0

M M  Ei  Gm   S  R r  Energy supplied to spaceship = Ef – Ei

M M   GM   S  R r  = 6.0 × 1011 J [as calculated in part (a)]

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GRAVITATION

THINKING PROBLEMS 1. If the force of gravity acts on all bodies in proportion to their masses, why does a heavy body not fall faster than a rigid light body? 2. Does the concentration of the earth’s mass near its centre change the variation of g (acceleration due to gravity) with height from its surface? 3. Why is it that we can learn more about the shape of the earth by studying the motion of an artificial satellite than by studying the motion of the moon? 4. One clock is based on an oscillating spring, the other on the pendulum. Both are taken to Mars. Will they keep the same time there as they keep on Earth? Will they agree with each other ? Explain. 5. The gravitational force exerted by the Sun on the Moon is almost twice as great as the gravitational force exerted by the Earth on the Moon. Why then doesn’t the Moon escape from the Earth? 6. Would you expect the total energy of the solar system to be constant? The total angular momentum? Explain. 7. Objects at rest on the earth’s surface move in circular paths with a period of 24 hours. Are they in ‘orbit’ in the sense that an earth satellite is in orbit? Explain. 8. The sun’s tide-raising power is only half as great as that of the moon. The direct pull of the sun on the earth, however, is about 175 times that of the moon. Why is it then that the moon causes larger tides ? 9. Does a rocket really need the escape valocity from the very beginning to escape from the earth? 10.Can a satellite move in a stable orbit in a plane not passing through the earth’s centre? Explain. 11. The total energy of the earth sun system is negative. How do you interpret the negative energy of a system? 12.As measured by a n observer on earth, would there be any difference in the periods of two satellites, each in a circular orbit near the earth in an equatorial plane, but one moving eastward and the other westward? 13.An artificial satellite in the presence of frictional forces will move into an orbit closer to the earth and may have increased kinetic energy. Explain this. 14.If a planet of given density were made larger, its force of attraction for an object on its surface would increase because of the planet’s greater mass but would decrease because of the greater distance from the object to the centre of the planet. Which effect predominates ? 15.A stone is dropped along the centre of a deep vertical mine shaft. Assume no air resistance but consider the earth’s rotation. Will the stone continue along the centre of the shaft? If not, decribe the motion. 16.Two air bubbles with radius r are present in water. Are these bubbles attracted or repelled? 17.Two setellites move along a circular orbit in the same direction at a small distance from each other. A container has to be thrown from the first satellite on to the second one. When will the container reach the second satellite earlier. If it is thrown in the direction of motion of the first satellite or in the opposite direction? The velocity of the container is small in comparison to that of the satellite. 18.When a train moves from west to east at high speed, does its weight increase or decrease? 19.Why is there ‘weightlessness’ in a satellite? 20.Is the velocity of escape of a particle at rest on the surface of the earth the same as that of a particle just orbiting the earth? Explain. 21.Two metal blocks ar shaped such that A can fit into B. When A is placed on B, the spring S is compressed. The system is now allowed to fall freely under gravity. It is found that A and B separate during the fall. explain.

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GRAVITATION

SOLUTION OF THINKING PROBLEMS 1. Though heavier bodies are acted upon by greater forces, the acceleration produced in them is the same as that produced in light bodies. Let F be the force on any body of mass m and g be the acceleration. Then Mm GM F = G 2 = mg g= 2 or R R an expression independent of mass. Hence, all bodies fall with the same repidity, provided there is no air resistance. 2. No. The acceleration due to gravity, that is, gravitational intensity at points outside the surface may be calculated by assuming the mass to be concentrated at the centre of the sphere. Hence, the distribution of mass over the earth does not affect the variation of g with height. 3. We can learn more about the shape of the earth by studying the motion of an artificial satellite than by studying the motion of the moon because the motion of the satellite is governed by the value of g at each of its positions. Since the moon is very far away, the variation of g near the surface of the earth is averaged out at the position of the moon. 4. The spring-operated clock will keep the same time but the pendulum clock will not keep the correct time as g on Mars is different from g on earth. They will not agree. 5. Because of the rotational equalibrium of the moon around the earth and also that of the earth + moon system around the sun. The attractive force on the moon due to the earth provides the necessary centripetal force for the rotational motion of the moon around the earth and the attractive force on the moon due to the sun provides necessary centripetal force for the rotational motion of the moon around the sun. 6. Yes. The gravitational forces are conservative. When a system is under the action of conservative forces, its total energy (kinetic + potential) is a constant and hence, the energy of the solar system is a constant. The gravitational forces on planets of the solar system are central forces directed towards the sun and hence, the moment of forces about the sun is zero. Thus, no torque acts on the planets and the angular momentum is a constant. 7. No, they are not in ‘orbit’ in the sense that an artificial satellite is, because the whole of the gravitational pull on the artificial satellites provides the necessary centripetal force, but for objects at rest on the earth, the whole of the gravitational pull (i.e., their weights) does not provide the necessary centripetal force; only a small friction provides the necessary centripetal force. In short, bodies in ‘orbit’ are weightless but bodies on the earth at rest are not weightless. 8. The tide-raising power of a body (sun or moon) is determined by the difference in the gravitational attraction between the water on the far and near sides, not by the total pull. The tide producting force varies inversely as the cube of the distance. Thus, the relatively short distance separating the earth and the moon makes the moon’s tide-raising force more than double that of the distance sun. 9. No. What is actually needed is the supply of total amount of energy required to take it beyond the gravitational field of the earth. 10.No. For stability of rotational motion there should not be any torque produced by the force acting on it about the centre of the orbit. When the orbit does not lie in a plane passing through the centre, the gravitational pull will have a moment (torque) about centre of the orbit. 11. The meaning of the total energy being negative is that the system (sun + earth) is a closed one, earth always being bound to the attracting solar centre and never escaping from it. 12.The westward satellite will appear to have a shorter period than the eastward satellite, owing to the rotation of the earth itself from west to east.

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GRAVITATION 1  GMm  mv 2     where m is the mass of the satellite, M = mass of the earth and r = radius  2 r  of the orbit. By the dynamics of circular motion. GM GMm mv 2 v2 =  or 2 r r r 1 GM 1 2 E or or  mv or –K (kinetic energy) 2 r 2 1 GM 1 GM  Ei (initial energy)   2 r and Ef (final energy)   2 r 0 By the principle of conservation of energy Ei = Ef + W (work done by frictional force) 1 GM 1 GM   W  2 r0 2 r

13.E (total energy) 

or  Also or 

1 1 1 GM     W (a positive quantity) 2  r r0  1 1 > or r < r0 r r0 –K0 = –K + W K – K0 = W (a positive quantity) K > K0

1 14.Mass-effect dominates over the distance-effect [mass µ R3. fprce pf attraction µ 2 ]. R 15.Due to the rotation of the earth from west to east the stone will deviate slightly to the east. 16.The two bubbles attract each other. A body immersed in an extended denser medium behaves like a negative mass so far as gravitational attraction is concerned. The two bubbles, therefore, behave like two negative masses, each of absolute value equal to the mass of an equal volume of water. So by the law gravitation  mm F  G 1 2 2 rˆ r    m1   m 2  rˆ  G m1m 2 rˆ F  G Here r2 r2 The negative force of interaction means force of attraction. 17.If the container is thrown in the direction opposite to the motion of the satellite, it will begin to move along an ellipse (say,1) lying inside the circular orbit of the satellites. The period of revolution will be slightly less than that of the satellites. Therefore, they can meet at the point of contract of the orbits (circular orbit of satellite and the elliptical orbit of the container) only after a greater number of revolutions. On the other hand, if the container is thrown in the direction of the satellite, the container begins to move along a ellipse like 2 in the figure. The velocity u should be such that during one revolution of the container, satellite B also makes one revolution and in addition covers the distance AB. So the container should be thrown in the direction of the first satellite.

18.The weight decreases. See Example 2. 19.The feeling of weight arises from the reaction of the ground on a man. In an orbiting satellite, the astronaut and the floor of the satellite both have the same acceleration (the centripetal acceleration towards the earth). Hence, the floor of the satellite offers no reaction to the astronaut and he feels weightless.

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GRAVITATION For circular motion of man GM  m mv 2  N  r2 r For a particle in circular motion GMm ' m ' v 2 GM v 2    r2 r r2 r 2 2 mv mv N  N0  r r 20.No, the velocity of escape arises from the energy required to take it from the present condition to infinity. This is equal to the difference between energy on the surface of the earth and infinity. For a particle at rest 1 GM mv 2e = m (only potential energy) 2 R 2GM ve  R For a particile in circular motion GM 1  mv 2c E (energy)   R 2 2 mvc GMm = But R R2 (from dynamics of circular motion) 1 GMm E  2 R 1 1 GMm mve2   2 2 R GM ve = or R 21.Bodies falling freely under gravity are in a state of weightlessness. In this state, the compressed spring will separate A and B.

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GRAVITATION

ASSERTION-REASON TYPE A statement of Statement-1 is given and a Corresponding statement of Statement-2 is given just below it of the statements, mark the correct answer as – (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 is NOT correct explanation of Statement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) If Statement-1 is false but Statement-2 is true. 1.

Statement-1 : Kepler’s laws for planetary motion are consequence of Newton’s laws. Statement-2 : Kepler’s laws can be derived by using Newton’s laws.

2.

Statement-1 : Three orbits are marked as 1, 2 and 3. These three orbits have same 2 3 semi-major axis although their shapes (eccentricities) are different. The three identical 1 satellites are orbiting in these three orbits, respectively. These three satellites have the same binding energy. Statement-2 : Total energy of a satellite depends on the semi-major axis of orbit according to the expression, E=

3.

 GMm . 2a

Statement-1 : A particle of mass m is projected towards the heavy planet as shown in figure. The angular momentum of the particle is conserved about point O. Statement-2 : Torque of gravitational force about point O is always zero.

v m M

d O

MATCH THE COLUMN 1.

An artificial satellite is in circular orbit around the earth. one of the rockets of the satellite is momentarily fired, the direction of firing of rocket is mentioned in Column I and corresponding change(s) are given in Column-II. Match the entries of ColumnI with the entries of Column II. Column I Column II (A) Towards the earth’s centre (P) Orbit changes and becomes elliptical (B) Away from the earth’s centre (Q) Orbit plane changes (C) At right angle to the plane of orbit (R) Semi-major axis of orbit increases (D) In forward direction (S) Energy of earth-satellite system increases.

2.

In Column-I, 4 artificial satellites of earth are shown, while in Column-II, some statements are given related to motion or other facts of satellites. Match the entries of Column-I with the entries of Column-II. Column I Column II Earth (A) (P) May be a geostationary satellite A

(B)

Earth

(Q) Not a geostationary satellite B

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GRAVITATION

Earth

(C)

(R) Not possible C

(D)

3.

4.

Earth

D

(S) If a rocket of a satellite fired in the direction to perpendicular to plane or orbit, then the orbit of the satellite changes.

Four identical satellites are orbiting in four elliptical orbits having same semimajor axis but different eccentricities. In Column-I some quantities associated with four orbits are given and in Column-II the words which can give the information about physical quantities mentioned in Column-I. Match the entries of Column-I with the entries of Column-II. Column I Column II (A) Total energy of all four orbits (P) Same (B) Speed of satellite in all four orbits (Q) Different (C) Velocity of satellite in all four orbits (R) Constant (D) Angular momentum of satellites about (S) Varying centre of earth in all four orbits.

3

4

1

E r1

r1

2

r2

r2

A particle is taken to a distance r (> R) from centre of the earth. R is radius of the earth. It is given velocity V which is perpendicular to r. With the given values of V in column-I you have to match the values of total energy of particle in column-II and the resultant part of particle in column-III. Here ‘G’ is the universal gravitational constant and ‘M’ is the mass of the earth. Column-I (Velocity) Column-II (total energy) Column-III (Path) (A)

V  Gm / r

P.

Negative

T.

Elliptical

(B)

V  2Gm / r

Q.

Positive

U.

Parabolic

(C)

V  2Gm / r

R.

Zero

V.

Hyperbolic

S.

Infinite

W.

Circular

(D)

Gm / r  V  2Gm / r

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GRAVITATION

LEVEL – 1 1.

Kepler’s law of are for motion of planets around the sun (A) the law of energy conservation (B) the law of momentum conservation (C) the law of charge conservation (D) the law of angular momentum conservation about the sun

2.

If the semi-major axis of orbit of a planet be doubled, the revolution period will become (A) 8 times

3.

2 times

(C)

3 times

(D) none

The major axis of an ellipse is l1 and minor axis l2. A planet moving along it has a period T. The area swept by the line joining the planet to the sun at one of the focii in a time t will be l1l2 (A) t T

4.

(B) 2

l1 l2 t (B) T

l12 t (C) T

l2 2 t (D) T

If the angular momentum of a planet of mass m around the sun is L, its areal velocity will be L (A) 2m

L (B) m

L (C) 2 m

L2 (D) 2m

5.

Within a uniform spherical shell (A) gravitational force is zero, but potential is –GM/R at every point. (B) any passenger will feel weightlessness relative to the sphere’s gravity. (C) the space is equipotential (D) all of these

6.

A particle is located in a uniform spherical shell at its centre. It experiences a force F. Now a small hole is made by removing a mass m from the shell. Now the force on the particle is F. Then (A) F = 0, F  0 (B) F  0, F = 0 (C) F = 0 = F (D) F  0  F

7.

Consider a uniform sphere with centre at C but heving a spherical cavity passing through C and surface of the sphere. There is a point A in the cavity. The direction of gravitational field intensity at A is along     (A) AC (B) CA (C) CC (D) C´C

8.

9.

A homogenous body has large extension such that at any point within the body the gravitational field intensity is zero. Now a cavity is made centred at some point A. Its radius is R. The gravitational field intensity at some point B outside the cavity is directed along   (A) AB (B) BA (C) Zero (D) None Consider a homogenous sphere of density . Pressure at some point A is P. Now a cavity of radius r < OA is made in it. The pressure at A is P´. Then (A) P = P´ (B) P > P´ (C) P < P´ (D) nothing can be said

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A C

C´

B A

r O

R A

53

GRAVITATION 10. The satellite is moving in an elliptical orbit about the earth as shown in figure : The minimum and maximum distance of satellite from earth are 3 units and 5 units, respectively. The distance of satellite from the earth when it is at P is equal to (A) 4 units (B) 3 units (C) 3.75 units (D) None of these

P

S

Q

E

m

11. A wreckage of mass m present in space, collides with a satellite of mass 10 m and orbital radius R as shown. As a result of collision, the wreckage sticks to satellite and satellite is transferred to an orbit whose minimum distance from the planet is R/2. Take mass of earth as M. The velocity of the wreckage with which it collides to the satellite is

GM R

(A)

58 GM R

(B)

63 GM R

(C)

v R R/2

59 GM R

(D)

12. A tunnel is dug along diameter inside the surface of earth and a particle is projected from the centre of tunnel. The minimum velocity of particle such that it escape out from the earth’s gravitation field is (Radius of earth = Re) (A)

3 g Re 2

(B)

2g R e

(C)

5 g Re 2

(D)

3g R e

13. A frictionless tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the centre of earth (where R is radius of earth). An object is released from one end of the tunnel. The correct graph, showing the variation of acceleration of particle with its distance r from centre of earth is a

a

(A)

r R/2

a

(B)

R

r R/2

R

a

(C)

r R/2

R

(D)

r R/2

14. A ring of mass m1 and radius R is fixed in space at some location. An external agent brings a point mass m2 from infinite to centre of the ring. Work done by the external agent will be (A) 

Gm1m 2 R

Gm1m 2 (B) R

G m12  m 22 (C) R

R

R m2

m1

Gm1m 2 (D) R  m  m  1 2

15. The work done to take a particle of mass m from surface of the earth to a height equal to 2R is (R is radius of earth) (A) 2 mgR (B) mgR/2 (C) 3 mgR (D) 2mgR/3 16. The radius of a planet is R1 and a satellite revolves around it in a circle of radius R2. The time period of revolution of satellite is T. Acceleration due to the gravitation of the platnet at its surface will be 4 2 R 32 R 32 4 2 R 13 R13 (A) 2 2 (B) (C) 2 2 (D) T R1 4 2 T 2 R12 T R2 4 2 T 2 R 22

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GRAVITATION 17. A planet of small mass m moves around sun of mass M along an elliptical orbit such that its minimum and maximum distance from sun are r and R respectively. Its period of revolution will be (r  R)3 (A) 2 6GM

(r  R)3 (B) 2 3GM

(r  R ) 3 (C)  2GM

(r  R)3 (D) 2 GM

18. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall, and the acceleration of the particle varies with x (distance of the particle from the centre) according to : Pressing Force

(A)

Pressing Force

x x=R/2 x=R

(B)

acceleration

x x=R/2 x=R

(C)

acceleration

x

(D)

x

x=R/2 x=R

x=R/2 x=R

19. Consider two satellites A and B of equal mass m, moving in same circular orbit about earth, but in opposite sense as shown in figure. The orbital radius is r. The satellites undergoe a collision which is perfectly inelastic. For this situation, mark out the correct statement(s). [Take mass of earth as M]. (A) The total energy of the two satellites plus earth system just before collision is –GMm/r. (B) The total energy of the two satellites plus earth system just after collision is –2GMm/r (C) The total energy of two satellites plus earth system just after collision is –GMm/2r. (D) The combined mass (two satellites) will fall towards the earth just after collision. 20. A solid sphere of uniform density and radius 4 unit is located with its centre at the origin O of co-ordinates. Two spheres of equal radii 1 unit, with their centres at A(–2, 0, 0) and B(2, 0, 0) respectively, are taken out of the solid leaving behind spherical cavities as shown in figure. Then (A) the gravitational field due to this object at the origin is zero. (B) the gravitational field at the point B(2, 0, 0) is zero (C) the gravitational potential is same at all points on the circle y2 + z2 = 36 (D) the gravitational potential is same at all points on the circle y2 + z2 = 4

r

A

B

Earth

y

x

m

A

O

B

z

21. Three planets of same density have radii R1, R2 and R3 such that R1 = 2R2 = 3 R3. The gravitational field at their respective surfacs are g1, g2 and g3 and escape velocities from their surfaces are v1, v2 and v3. Then (A) g1/g2 = 2 (B) g1/g3 = 3 (C) v1/v2 = 1/4 (D) v1/v2 = 3 22. Suppose the earth suddenly shrinks in size, still remaining spherical and mass unchanged (All gravitational forces pass through the centre of the earth). (A) The days will become shorter (B) The kinetic energy of rotation about its own axis will increase (C) The duration of the year will increase. (D) The magnitude of angular momentum about its axis will increase. 23. Two artificial satellites of the same mass are moving around the earth in circular orbits of different radii. In comparison to the satellite with lesser orbital radius, the other satellite with higher orbital radius will have : (A) greater kinetic energy (B) greater potential energy (C) greater total energy (D) greater magnitude of angular momentum, about the centre of the circular orbit

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55

GRAVITATION 24. Two objects of mass m and 4m are at rest at an infinite separation. they move towards each other under mutual attraction. If G is the universal gravitational constant. Then at separation r. (A) The sum of energy of the two objects is negative  10Gm  (B) Their relative velocity of approach is   r 

1/ 2

in magnitude

(C) The sum of kinetic energy of the objects is 4Gm2/r. (D) The sum of angular momenta of both the objects is zero about any point. 25. The earth is moving around the sun in an elliptical orbit. Point A is the closest and Earth point B is the farthest point in the orbit, as shown. In comparison to the situation B A when the earth passes through point B : Sun (A) total energy of the earth-sun system is greater when the earth passes through point A. (B) gravitational potential energy of the earth-sun system is greater when the earth passes through point A. (C) kinetic energy of the earth due to the motion around the sun is greater when it passes through the point A. (D) magnitude of angular momentum of the earth about the sun is greater when the earth passes through point A. 26. A person abandoned on a small spherical asteroid of mass m1 and radius R, sees a satellite orbiting the asteroid in a circular orbit of period T. The satellite appears stationary to a person on the asteroid. If the asteroid rotates with a period Ta, the radius of satellite is 1/ 3

1/ 3

 Gm1 Ta2  (A)   2  2 

 Gm1 Ta2  (B)   2  4 

1/ 3

 Gm1 T 2  (C)   2  4 

 Gm1 T 2  (D)   2  4 

3

27. If the distance between the earth and the sun were half its present value, the number of days in a year would have been [JEE, 06] (A) 64.5 (B) 129 (C) 182.5 (D) 730 28. An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) E0. Its potential energy is [JEE, 97] (A) –E0 (B) 1.5 E0 (C) 2 E0 (D) E0 29. In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure. The work done in different paths are W1, W2, W3 respectively then [JEE-(Scr), 03] (A) W1 = W2 = W3 (B) W1 > W2 > W3 (C) W1 = W2 > W3 (D) W1 < W2 < W3

(3)

B

C (2) (1)

A

30. A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then [JEE, 06] (A) TA > TB (if rA > rB) (B) TA > TB (if mA > mB) 2

 TA   rA  (C)       TB   rB 

3

(D) TA = TB

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56

GRAVITATION

PASSAGE TYPE Comprehension # 1 Changing from a circular to An elliptical orbit Let us identify the system as the spacecraft and the Earth but not the portion of the fuel in the spacecraft that we use to change the orbit. In a given orbit, the mechanical energy of the spacecraft-Earth system is given by GMm 2r This energy includes the kinetic energy of the spacecraft and the potential energy accosicated with the gravitational force between the spacecraft and Elliptical the Earth. If the rocket engines are fired, the thrut force moves the spacecraft Circular orbit orbit through a displacement. As a result, the mechanical energy of the spacecraft Earth system increass. The spacecraft has a new higher energy but is constrained to be in an orbit that includes the original strarting point. It can not be in a higher energy circular Earth orbit having a larger radius because this orbit would not contain the starting Rocket engine point. The only possibility is that the orbit is elliptical as shown in the figure. is fired here GMm E 2a Above equation gives the energy of the spacecraft - Earth system for an elliptical orbit where a is semimajor axis. Thus if we know the new energy of the orbit, we can find the semi-major axis of the elliptical orbit. Conversely, if we know the semi-major axis of an elliptical orbit we would like to achieve, we can calculate how much additional energy is required from the rocket engines. A spacecraft is moving in a circular orbit around the Earth (Radius 6400 Km), at a height of 300 km from the surface. To place the spacecraft in an elliptical orbit, the magnitude of the mechanical energy of the spacecraftEarth system is decreased by 10.0%. If the spacecraft-earth system had initial energy (–E0), then the total mechanical energy of the system after firing the rocket will be : (A) –1.1 E0 (B) –1.9 E0 (C) – E0 (D) None of these E

1.

2.

3.

4.

Semimajor axis of the new elliptical orbit is (A) 7437 Km (B) 6700 Km

(C) 7370 Km

maximum height of the spacecraft above the surface of the Earth will be : (A) 970 km (B) 1474 km (C) 300 km

(D) None of these

(D) 1774 km

Comprehension # 2 A triple star system consists of two stars, each of mass m in the same circular V orbit about central star with mass M = 2 × 1030 Kg. The two outer stars always lies at opposite ends of a diameter of their common circular orbit. The radius of M m m the circular orbit is r = 1011 m and the orbital period of each star is 1.6 × 107 S. 20 [Take = 2 = 10 and G = × 10–111 Nm2 kg–2] 3 V The mass m of the outer stars is (A) 16/15 × 1030 Kg (B) 11/8 × 1030 Kg (C) 15/16 × 1030 Kg (D) 8/11 × 1030 Kg

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57

GRAVITATION 5.

6.

The orbital velocity of each star is 5 5 10 × 103 ms–1 10 × 105 ms–1 (A) (B) 4 4 The total mechanical energy of the system is 1375 1375 (A)  × 1035 J (B)  × 1038 J 64 64

(C)

5 5 10 × 102 ms–1 (D) 10 × 104 ms–1 4 4

(C) 

1375 × 1034 J 64

(D) 

1375 × 1037 J 64

LEVEL – 2 1.

Ravi can throw a ball at a speed on earth which can cross a river of width 10 m. Ravi reaches on an imaginary planet whose mass density is twice of the earth. Find out the maximum possible radius of Planet so that if Ravi throws the ball at same speed it may escape from planet. Given radius of earth = 6.4 × 106 m.

2.

Two equal masses m and m are hung from a balance whose scale pans differ in vertical height by h. calculate the error in weighing, if any, in terms of density of earth .

3.

A sky lab of mass 2 × 103 kg is first launched from the surface of earth in a circular orbit of radius 2R (from the centre of earth) and then it is shifted from this circular orbit to another circular orbit of radius 3R. Calculate the minimum energy required (a) to place the lab in the first orbit, (b) to shift the lab from first orbit to the second orbit. Given, R = 6400 km and g = 10 m/s2.

4.

If a planet was suddenly stopped in its orbit supposed to be circular, show that it would fall onto the sun in a time





2 / 8 times the period of the planet’s revolution.

5.

A cord of length 64 m is used to connect a 100 kg astronaut to a space-ship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the space-ship is orbiting near earth surface. Also assume that the space-ship is orbiting near earth surface. Also assume that the space-ship and the astronaut fall on a straight line from the earth’s centre. The radius of the earth is 6400 km.

6.

The weight of a body is 81N at the surface of the earth. How much is the weight at a height equal to helf the radius of the earth ?

7.

A body weights w at a point half way from surface to centre of the earth. How much will it weight on the surface? Assume the Earth a homogenous sphere.

8.

A rocket is launched vertically upward with a speed of 5-6 km s–1. How high will it go without using any fuel? Neglect air resistance. G = 6.67 × 10–11 Nm2 kg–2, mass of the Earth = 6.0 × 1024 kg and mean radius of the Earth = 6.4 × 106 m.

9.

Calculate escape speed of molecules on the surface of moon where gm = 1/6 of gE. The density of moon is the same as that of the earth. The escape speed on the Earth is 11.2 km/s.

10. Consider a satellite of mass m is orbiting in a circle of radius r around a plant of mass M. Calculate its (a) kinetic energy (b) potential energy and (c) mechanical energy; take reference level of potential energy as zero at infinite radius.

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M

m

58

GRAVITATION 11. A light rod of length l connects two balls each of mass m. The system is free to rotate about pivot O. Two small spheres of masses M and 4M are placed respectively at normal distance x and y from the balls. the balls stay in equilibrium. For x = l/2, determine y. How much is the force exerted on the rod by pivot at O ?

O y

x

12. A binary star is a system of two stars separated by a distance l and moving around their centre of mass at rest (See figure) Determine the mechanical energy of the binary star shown. How much is its binding energy ?

l m

M

C

13. A smooth tunnel passes through a uniform solid sphere of radius R at a distance R/3 from the centre of the sphere. Neglect effect of tunnelling on the gravitational field. Determine the force exerted on a mail of mass m in the tunnel by tunnel wall. the mass of the sphere is M. Is this force gravitational or electromagnetic ? 44. A sensitive meter can measure changes in standard value of g in units of gals (1 gal = 1 cm s–2). the value of g at the surface of sphere of density  and radius R. A cavity is now made directly below the observation point. The change in the value of g is g. (See figure). Find an expresion for g.

h r R

15. A star is found to be moving in a circle with speed V. The radius of the circle is R. Determine the mass of a possible block hole at the centre of the circle. 16. What is the force exerted on a geostationary satellite of mass 200 kg ? Its distance from the centre is 4.24 × 107 m and radius of the earth is 6.4 × 106 m. 17. Three spherical shells, each of mass M and radius R, are in contact due to gravitational pull. Find the magnitude of the contact force between any two spheres. 18. The gravitational potential at a point (x, y) is (–4x + 6y) J kg–1. Determine the gravitational field intensity vector at this point. 19. A person is inside a statellite cabin moving round the earth. In the frame fixed to the cabin, draw forces acting on the person and show that if he stands on the platform of a spring balance in the satellite, the reading of the balance will be zero (w eightlessness). 20. An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite depends on velocity as F = av2 where a is constant. Calculate how long the satellite will stay in orbit before it falls onto the planet’s surface. 21. A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing energy due to small air resistance at the rate of C J/s. Find the time taken for the satellite to reach the earth. 22. A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at a distance 5R from the centre of the planet, the spaceship fires an instrument package with speed v0 as shown in the figure. The package has mass m, which is much smaller than the mass of the spaceship. For what angle  will the package just graze the surface of the planet ?

v0 R

m

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M

5R

59

GRAVITATION

LEVEL – 3 1.

Distance between the centres of two stars is 10 a. The masses of these stars are M and 16M and their radii a and 2a respectively. A body of mass m is fired at night from the surface of the larger star towards the smaller star. What should be its minimum initial speed to reach the surface of the smaller star ? Obtain the expression in terms of G, M and a. [JEE, 96]

2.

A cord of length 64 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. [REE, 98]

3.

A body is projected vertically upwards from the bottom of a crater of moon of depth R/100 where R is the radius of moon with a velocity equal to the escape velocity of the surface of moon. Calculate maximum height attained by the body from the surface of the moon. [JEE, 2003]

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60

GRAVITATION

Answer Key ASSERTION-REASON TYPE Q. Ans.

1 D

2 A

3 A

MATCH THE COLUMN 1.

[(A—PRS), (B—PRS), (C—PQRS), (D—P)]

2.

[(A—PS), (B—PS), (C—QS), (D—RS)]

3.

[(A—PR), (B—QS), (C—QS), (D—QR)]

4.

[(A—PW), (B—RW), (C—QV), (D—PT)]

LEVEL – 1

Q. 1 Ans. D Q. 11 Ans. B Q. 21 Ans. ABD

2 B 12 C 22 AB

3 A 13 D 23 BCD

4 A 14 A 24 BCD

5 D 15 D 25 C

6 A 16 A 26 B

7 D 17 C 27 B

8 A 18 BC 28 C

9 B 19 ABD 29 A

10 A 20 ACD 30 D

PASSAGE Q. Ans.

1 B

2 A

3 D

4 B

5 D

6 B

LEVEL – 2

1. R = 4 km

2. 8/3 Gmh

 2 GM e  3. (a) 9.6 × 1010 J; (b) 1.1 × 1010 J 5. T  m  r  2  r  

6. 36 N

7. 2 W

8. h = 2.13 × 106 m

10. (a)

1 GMm ; 2 r

(b) U  

GmM ; r

(c)  

1 GMm 2 r

9. 1.86 km/sec. 11. l´1

6GMm l2

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61

GRAVITATION

12. 

GMm(M  2m) GMm  M  2m    ;  2l (M  m) 2l Mm

13.

14. 

4Ger 3 3h

16. 44.66 N





18. 4 ˆi  6 ˆj N / kg

21. T 

GMm  1 1    2C  R r 

15.

Rv2 G

 19. N  0

GMm ; electromagnetic. 3R 2

20. t 

17.

GM 2 4R 2

m R ( n  1) a Gm

 8GM  1 1 22.   sin  5 1  5v 2 R    0

LEVEL – 3

1. vi min 

3 5GM 2 a

2. T = 3 × 10–2 N

3. h = 99 R

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62

CENTRE OF MASS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

CENTRE OF MASS

CENTRE OF MASS If we consider a system of many particles, each particle will have a complicated motion and it will be difficult to follow it. However, there is one point whose motion represents the translation motion of the whole system in a simple way. That point is located with respect to the system at a position decided by positions and masses of all the particles in the system. It is defined in such a way that it becomes possible to apply formulae of kinematics and dynamics in a form similar to those for single particle (See table). Item

Single Particle

Many - Particle System 



Velocity

 cm





Acceleration



a

a cm



Momentum Newton’s second



m 

Law of motion

m  cm 







Momentum Conservation



J ext = P



F = 0 ,  P = constant

p2 K= 2m

Kinetic energy





J = P

Impulse



Fext = M a cm

F = ma





Fext = 0,  P = constant

P2 K ;K= 2M

 pi2     i  1  2m i  N

NOTE : Concept of centre of mass loses its meaning at relativistically high speeds. H Centre of mass of two particles If we consider hydrochloric acid molecule is a system, it contains two Cl System atoms H and Cl. If we regard H as a single particle and Cl also as a single r1 r2 particle, this molecule is a system of two particles. Suppose we know the position vectors or hydrogen atom and chlorine atom, regarding them as as points, and their masses also. Using these, we want to define centre of mass of two particles. Since we want to write equations of motion of system of many particles in the form similar to those for a single particle this is possible only if position of centre of mass be defined in a particular way. And that way is 



rcm



m r  m Cl r2 = H 1 mH  mCl

We may compare this result with mean value of a quantity x given by —

x =

f1x1  f 2 x 2 f1  f 2

Here f are known as frequency of occurrence of variable x, and — x is usually called ‘weighted mean’ of x1 and x2. 

If expression for rcm ‘mass’ plays the role of ‘frequency’ or (statistical) weight.

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1

CENTRE OF MASS Definition For a two particle system, the centre of mass is the point whose position vector is the weighted mean of position vectors of the two particles with masses of particles playing the role of statistical frequency.

m1 C



In the figure rcm is the position vector of point C which is the CM of m1 and m2. Then 

m2

r1 r2



m1 r1  m 2 r2 rcm = m1  m 2



O

C1 : Properties 1. The centre of mass of two particles lies on the line joining them.

r1

To this, let r1' and r 2' be position vector of m1 and m2 relative to CM(C). 





m2

m1





rcm



Then, m1 ( rcm + r1' ) + m2( rcm + r1' )

2.

r1

r2



=

(m1 + m2) rcm



m1 r1' + m2 r1'









r2





Thus, r1' and r 2' are directed opposite to each other. This shows that r1' and r 2' are at 180º with each other but their tails lie at a common point. Thus, C lies in the line joining the particles. The centre of mass divides the distance between two particles in the inverse ratio of masses. As shown in the previous section 



m1 r1' = m2 r 2' 

| r1'| 

| r2' |

=

m2 m1

[taking modulus]

Thus, the centre mass is dividing the distance in inverse ratio. Thus, r

1 . Thus, the centre mass less closer m

to more massive particle. Note: Using m1r1 = m2( – r1 )

m1

r1

C

m2 > m1



 m2  r1 =  m  m    1 2  3.

The sum of first moment of masses of particles about centre of mass is zero. 

4.







The first moment of mass m1 about C is m1 r1' and that of m2 is m2 r 2' . The sum is m1 r1' + m2 r 2' . This sum is zero. When CM does not move,  rcm = 0







m1 r1 + m2 r2 = 0

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2

CENTRE OF MASS The relative displacement of 1 as seen by 2 is given by 





r1 – r2 = r12 Using these, we have 





m1 r1 + m2( r1 – r12 ) = 0 



(m2 + m1) r1 – m2 r12 = 0   m 2  r12  r1   m1  m2 



Similarly, 

m r r2 = 1 21 m1  m 2 

C2 : The masses of two particles are 300 g and 200 g. They are separated by 60 cm. Determine the distance of centre of mass from 200 g particle. Sol: This dist (r1) 300 g 200 g m2  m1 m2 We have =  r1' m  m r 1 1 2 300 = × 60 cm 200  300

m1

C

m2

That mass



3 × 60 cm 5 = 36 cm.

=

C3 : A system consisting of two masses connected by a massless rod lies along the x-axis. A 0.4 kg mass is at x = 2m while a 0.6 kg mass is at x = 7m. Calculate the x co-ordinate of the centre of mass. Sol.

XCM =

m1x1  m 2 x 2 (0.4  2)  (0.6  7) = =5m m1  m 2 (0.4  0.6)

C4 : Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the corners A, B, C and D respectively of a square ABCD of edge 1 m. If point A is taken as origin, edge AB is taken along x-axis and edge AD is taken along y-axis, the co-ordinates of centre of mass in SI is (a) (1, 1) (b) (5, 7) (c) (0.5, 0.7) (d) None of these Sol. XCM =

mA x A  mBx B  mC x C  mD x D mA  mB  mC  mD

=

1 0  2  1  3  1  4  0 1 2  3  4

=

23 1 = = 0.5 m 10 2

y

D (0, 1)

A (0, 0)

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C(1, 1)

B(1, 0)

3

x

CENTRE OF MASS m A y A  m B y B  mC yC  mD y D mA  mB  mC  mD

Similarly, yCM = =

1  0  2  0  3 1  4  1 1 2  3  4

=

7 = 0.7 m 10

C5 : A man of mass M stands at one end of a plank of length L which lies at rest on a frcitionless surface. The man walks to the other end of the plank. If the mass of plank is M/3, the distance that the mass moves relative to the ground is (a)

3L 4

(b)

L 4

(c)

4L 5

(d)

L 3

Sol: Since, external force on system is zero so, no change takes place in the centre of mass. XCM = XCM =

or

m1x1  m 2 x 2 m1  m 2 m1x1  m 2 x 2 m1  m 2

Here, xCM = 0  m1x1 + m2x2 = 0 or

Mx1 +

M x2 = 0 3



x1 = –

M x x2 = – 2 3M 3

.....(i)

xrel = x1 – x2 L = x1 – x2

But or or

L = –

x 2 4 – x2 = – x2 3 3

[From equation (i)]

3 x2 = – L 4



x 2 3L L = =– 3 4 3 3 Negative sign indicates that both moves in opposite directions.



x1 = –

Many particle system 







Let r1 , r2 ........... rN be position, vectors of masses m1, mi..........., mN in a system of N-particles. Then rcm is given by 





m1 r1  m 2 r2  ......m N rN rcm = m1  m 2  .........  m N



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4

CENTRE OF MASS In general, N



rcm =

i 1 N

mN

[discrete mass distribution]

m

m2

m1



 mi ri

Rod

mi

System System

i

Discrete mass distribution

i 1

Continuous mass distribution

If the system has continuous distribution of mass we have. 



rcm =

 dm r  dm

[continuous mass distribution]

where integration covers the whole body.

C1 C

System of systems If there be several systems having masses M1, M2 and having CM’s at 



C2

Rcm R1

R2

R 1, R 2 then CM of the system of systems is given by similar formula:

C3

R3

 

R cm =

M i R i M i

CENTRE OF MASS OF SOME RIGID BODIES We have several objects around us whose shapes and sizes remain the same. Such bodies are known as rigid bodies. The motion of such bodies can be described using their centres of mass. We take up the takes of locating centres of mass of a few rigid bodies like uniform rod, ring, disc, sphere, etc. Locating centre of mass of uniform rod Let a rod of length L be oriented along x-axis with ends at x = 0 and x = L. x=0 Let an elementary mass dm be located at position x. If  be mass per unit length, then dm = dx. The x-coordinate of centre of mass may be written as

x

dm

X

x=L

L

xcm

=

 dm x m

=

 ( dx)x 0

L

[x 2 / 2]L0 L = = L 2 This denotes mid point of the rod. Thus, centre of mass of uniform rod lies at the centre of the rod. Example 1. A non-uniform thin rod of length L is placed along X-axis as such it one of ends is at the origin. The linear mass density of rod is  = 0 x. The distance of centre of mass of rod from the origin is (a)

L 2

(b)

2L 3

(c)

L 4

(d)

L 5

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5

CENTRE OF MASS Sol: The mass of considered element is dm = dx = 0 xdx



XCM

=

L

L

 x dm

 x(

0

 dm

=

0

dm

x dx)

O

0

x

L



0

dx

x dx

0 L

=

 x3  0    3 0

L3 2 3 = L 2 = 2 L L 3 x   0 0   2 2  0 0

Locating centre of mass of circular ring Let a circular ring be placed with centre at origin in XY plane as in the figure. Let us choose an elementary mass dm at angular position . Then x = R cos y = R sin dm = µ Rd  [ = mass per unit length] The x-co-ordinate of centre of mass, xcm may be written as

Y dm d

=



2

=0 ( = 2)

 ( Rd).R cos  xcm =

X

0

 (2 R) =

R 2 [sin ]20  = 0 (2R) The y-co-ordinate of centre of mass, ycm may be written as

3 2

=

2

ycm =

 ( Rd).R sin  0

 (2 R)

R 2 [ cos ]20  = 0 = (2R) Thus, the centre of mass of uniform ring lies at the centre of the ring.

Y y=R

Locating centre of mass of a uniform circular disc Let  be the mass per unit area of the disc of radius R. Let it be dy composed of uniform rods of proper lengths. One such strip (rod) is located at position y. Its width is dy and length is 2

2

R

y X

2

R y .

Its whole mass is centered at position (0, y). Using the formula for system of systems, we have y = –R

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6

CENTRE OF MASS R

ycm =

R 2  y2 )}y / (R 2 )

 {dy(2

R

2 = R 2

R

y

R 2  y2 dy

R

For integration let us consider



2 2 d 3 2 2 3/2 – 1 d R  y 2 2 3/2 (R – y ) = (R – y ) dy 2 dy

=



3 ( R 2  y 2 )( 2y) 2

= –3 R 2  y 2 y

 (R 2  y 2 )3/ 2  d   = 3  

R 2  y 2 ydy

 (R 2  y 2 )3/ 2  d  =   3 



R 2  y 2 y dy

1 – (R2 – y2)3/2 = 3



R 2  y 2 y dy

or, Thus,

2  1   [(R 2  y 2 )3/ 2 ]R R 2  R  3  Thus, the centre of mass of a uniform disc lies at its centre.

ycm =

Locating centre of mass of a uniform sphere Let us imagine that a solid homogeneous sphere is made up of elementary discs, one of which is shown in the figure. Its thickness is dy and radius is R 2  y 2 . The mass in the volume of the disc is (R2 – y2)dy, where  is density. The centre of mass of the disc is located at (0, y). Using formula for CM of system of systems, we have yCM

R  4 3  2 2 =   (R  y ) ydy  /  3 R     R  

Y

2

2

R –y

y=R dy y y=0 y=0

y = –R

4 3  where  R   is the mass of the sphere. 3  Thus,

ycm

3 = 4R 3

R

 (R

2

ydy  y3dy)

R

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CENTRE OF MASS R R  y 4   3  2  y 2  = 4R 3 R  2    4   = 0.     R    R  Thus, the centre of mass of a homogeneous sphere lies at its centre.

m

Centre of mass of symmetric bodies If two equal masses be located in a line through C, at equal distances from C, their centre of mass lies at C. If this is true for all pairs of practices forming a system, the centre of mass lies at the centre of symmetry.

r r

C

m

CM-1 CM

CM

CM

CM

CM

CM

CM-2

NOTE : (i) There need not be a mass at the centre of mass. (ii) The centre of mass may lie outside the body. C5 : Which of the following has centre of mass not situated in the material of body ? (a) A rod bent in the form of a circle (b) Football (c) Handring (d) All of the above Sol: (a) If a rod is bent in the form of a circle, its centre of mass is at the centre of curvature. (b) Football is in teh shape of hollow sphere. So, its centre of mass is at the centre of sphere (c) The reason is same as (a). In all the above geometrical shape, no material is present at the centre of mass. Example 2. Find the position of centre of mass of the uniform lamina shown in figure. Sol: Here, A1 = area of complete circle = a2 A2 = area of small circle

Y

2

a 2 a =   = 4  2

O

a

X

(x1, y1) = co-ordinates of centre of mass of large circle = (0, 0)

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CENTRE OF MASS (x2, y2) = co-ordinates of centre of mass of small circle a  =  , 0 2 

Using

XCM

=

A1 x 1  A 2 x 2 A1  A 2

we get XCM

a 2  a  1     a 4 2 8 = a=– 2 = 6 a 3 a 2    4 4

and

= 0 as y1 and y2 both are zero.



yCM

 a  Therefore, co-ordinates of CM of the lamina shown in figure are   , 0  .  6 

4r 3 The position of COM of this system from point ‘C’ is m y  m2 y2 y cm  1 1 ...(i) m1  m 2 

2cm 8cm

Example 3. In the figure shown a hole of radius 2 cm is made in a semicircular disc of radius 6 at a distance 8cm from the centre C of the disc. The distance of the centre of mass of this system from point C is : (A) 4 cm (B) 8 cm (C) 6 cm (D) 12 cm Sol. The centre of mass of semicircular disc is

C

y m2 8cm

m1

m1  r12   (6 ) 2 m1  36 3

m 2  r22     2

2

m2  4 From equation (i)

4  6  4  8 3 36  3  4 

36  3  y cm 

y cm  y cm

36 3  8  32 36 3  4 

72 2  8 701.89  = = 8 cm 87.73 9 2  1

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9

CENTRE OF MASS IMPORTANT FEATURES C6 : Centre of mass of some well known rigid bodies are given below: (a) Centre of mass of a uniform rectangular, square or circular plate lies at its centre.

CM CM

CM

(b) Centre of mass of a uniform semicircular ring lies at a distance of 2R from its centre, on the axis of symmetry where R is the radius  of the ring.

h=

R

CM

2R 

O

(c) Centre of mass of a uniform semicircular disc of radius R lies at a distance of h =

4R from the center on the axis of symmetry as 3

R

shown in figure.

CM

4R 

O

(d) Centre of mass of a hemispherical shell of radius R lies at a distance of h =

R from its centre on the axis of symmetry as shown in 2

R

figure.

CM

R 

O

(e) Centre of mass of a solid hemisphere of radius R lies at a distance of h=

3R from its centre on the axis of symmetry.. 8

R

CM

3R 

O

MOTION OF SYSTEM OF PARTICLES If the particles of a system are displaced, the CM may or may not be displaced. But the displacement of CM is possible only lay displacing the particles. Displacement of centre of mass 



If  ri be the displacement of ith particle (mass = mi), then the displacement of centre of mass,  rcm may be written as 

mi  ri  rcm = m i where summation  runs over all the particle in the system. 

C7 : Two particles of masses 10 g and 20 g are displaced such that centre of mass does not move. If 10 g particle is displaced by 0.02 cm, what is the displacement of 20 g particle ?

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10

CENTRE OF MASS

Sol: Using

mi x i mi , we have 10 × 0.02 + 20 xm = 0

xcm =

x = –0.01 m The negative sign shows that the direction of displacement of this particle is opposite to that of the given displacement. The velocity of centre of mass: 





If the particles of a system have velocities 1 ,  2 , .........  N , the velocity of centre of mass of the system is defined by Y



mi  i = mi



 cm

where summation includes all the particles within the system. C8 : Consider the particles confined in the system. Write X-components of the velocity of centre of mass of the system. Sol: We have

cmx =

3ms–1

System 1.4 ms–1 1.2 kg 37º 1.0 ms–1

 mi  i x

0.50 kg 37º

1.5 kg

 mi

2ms–1 60º 7 kg

–1

1.5ms

5 kg

37º –1 1.0 kg 2ms

2ms–1

=

1.2  0  0.50  3  1.0  ( 1.5 cos 37º )  1.0  2 cos 37º 1.5  1  cos 37º –1 ms 1.2  0.50  1.0  1.0  1.5

=

0  1.5  1.2  1.6  0.3 5.2

5

2.2 –1 ms = 0.45 ms–1 5.2 Note that 7 kg and 5 kg particles are outside the system.

3

37º

=

4

C9 : Two blocks of masses m1 and m2 are lying on a table when m2 is given a velocity 0 to the right. Write the velocity of the centre of mass. m1

Sol: Here



 cm =

 0

m1  0  m 2   m1  m 2

Rest

m2 0

m2  = m  m 0 1 2

Momentum and centre of mass motion 

The momentum of a system of particles ( P ) is defined as the vector sum of momenta of all the particles enclosed. 



However,



m i  = P

Thus,



mi i P  cm = =  mi M 

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CENTRE OF MASS 



P = MVcm





Total momentum = Total mass × Vcm . C10 : The mass of a system is 200 kg. Its particles are moving in unknown ways but centre of mass is moving at 10 ms–1. What is the magnitude of the momentum of the system ? 



P = MVcm

Sol: Using



| P | = 200 kg × 10 ms–1 = 2000 kg ms–1 The Acceleration of centre of mass The rate of change of velocity of centre of mass is its acceleration. 



If various particles have acceleration a i , the acceleration a cm is given by 



a cm

mi a i =  mi

Here, summation runs over all the particles in the system. C11 : Two blocks in an Atwood machine have masses m and 2m. Calculate the acceleration of centre of mass of the two blocks when they are released, by using the formula for individual acceleration. m 2  m1 a = m  m g. 2 1

Sol: In the figure m is accelerating upward while 2m downward.  a1 = acceleration of m

=

2m  m g, up 2m  m 

= –

g 3

m

 a 2 = acceleration of 2m

2m

 2m  m g = g, downward = 2m  m 3





Using, a cm



m1a1  m 2 a 2 = , we have m1  m 2 



a cm

=





m( g / 3)  2m(g / 3) g = m  2m 9 

Thus, the acceleration of CM is

g . 9

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CENTRE OF MASS Dynamics of system of particles Now we consider forces acting on various particles in a system. 



Let F1 be the net force acting on a particle of mass m1 producing acceleration a1 in it. Then according to Newton’s second law of motion, 



F1 = m1 a1 





Now F1 may arise due to internal force F1int and force exerted to by particles outside the system, F1ext . That is 





F1 = F1ext + F1int

Similarly we can write the following: 







m1 a1 = F1 = F1int + F1ext 



fji fji



m2 a2 = F2 = F2int + F2ext





 



 



mi a1 = Fnet = Fint + Fext Now the sum of internal forces is zero as 



Fij + Fji = 0

[Newton’s third law of motion]



 M acm = Fext This shows that 1. Net force on a system of particles is equal to net external force. 2. Sum of internal forces on particles of a system is zero. 3. CM cannot be accelerated by internal forces. 4. External forces equal mass multiplied by acceleration of CM. 5. Action point of external force is immaterial in producing acceleration in centre of mass. C12 : If a projectile exploids in air in different parts, the path of the centre of mass remains unchanged. This is because during explosion no external force (except gravity) acts on the centre of mass. The situation is as shown in figure.

U

B

Path of COM is ABC, even though the different parts travel in different A directions after explosion,.

Explosion Path of COM

C

C13 : Suppose a system consists of more than one particle (or bodies). Net external force on the system in a particular direction is zero. Initially the centre of mass of the system is at rest, then obviously the centre of mass will not move along that particular direction even though some particles (or bodies) of the system may move along that direction. The following example will illustrate the above theory. C14 : Can a car accelerate on a smooth horizontal road ? Sol: No, for accelerating the car (i.e. accelerating the centre of mass of the car), external horizontal force is needed. This force can come only from friction. Here the road is smooth. Hence no horizontal external force acts on the car. It cannot accelerate.

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CENTRE OF MASS C15 : A uniform rod of mass m and length L is tied to a vertical shaft. It rotates in horizontal plane about the vertical axis at angular velocity . How much horizontal force does the shaft exert on the rod ?

L

L Sol: Let T be the force applied horizontally on the rod. The acceleration of CM is 2   , as it is moving in a circle 2 L of radius . 2

Using



2

acm =  (L/2)

Fext = M acm

T cm

L T = M   2 2

Thus, the horizontal force exerted is

1 M2 L. 2



C16 : A chain of irregular shape and total mass m is tied to a string whose other and is connected to the shaft. The shaft make it rotate at angular velocity  as in the figure. Calculate the distance of centre of mass of the chain from the centre of circle in which the CM is moving. Sol: Here, acm = 2r 





r

m



Fext = T + mg

Using,





Fext = M a cm Fextc = mac Fext= ma = 0

T

Here acm is horizontal vector.. 2

T sin + 0 = m r T cos – mg = 0



.....(i) .....(ii)

acm

CM

From these, tan =

r g

g tan. 2 This radius is the required distance.

m C

2

mg

r=

C17 : A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The centre of mass (a) of the box remains constant (b) of the box pulls the ball system remains constant (c) of the ball remains constant (d) of the ball relative to the box remains constant Sol: Since, no external force is present on teh system so, the centre of mass of system will not be changed.

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CENTRE OF MASS C18 : A loaded spring gun of mass M fires a shot of mass m with a velocity  at an angle of elevation . The gun is initially at rest on a horizontal frictionless surface. After firing, the centre of mass of the gun-shot system (a) moves with a velocity  (b) moves with a velocity

m M

m cos in the horizontal direction M

(c) remains at rest

(M  m) in the horizontal direction (M  m) Sol: If we consider a system consisting of gun and mass m, then no external force is present on the system. So, the acceleration of centre of mass of the system will be zero. Since, initially the system is in rest so, centre of mass will always remain at rest. (d) moves with a velocity

C19 : Two blocks A and B are connected by a massless string (shown in figure). A force of 30 N is applied on block B. The distance travelled by centre of mass in 2 s starting from rest is

B

A 10 kg

(a) 1 m (b) 2 m (c) 3 m (d) non of these Sol: The acceleration of centre of mass is

20 kg

F = 30N

Smooth

F aCM = m  m A B

= 

s=

30 = 1 m/s2 10  20

1 1 aCMt2 = × 1 × 22 = 2m 2 2

C20 : Two bodies A and B of masses m1 and m2 respectively are connected by a massless spring of force constant k. A constant force F starts acting on the body A at t = 0. Then k F m2 m1 F (a) at every instant, the acceleration of centre of mass is m  m 1 2 A B (b) at t = 0, acceleration of B is zero but that of A is maximum (c) the acceleration of A decreases continuously (d) all of the above Sol: (a) Since, external force on the system is F F  aCM = m  m 1 2 (b) At t = 0, elongation in spring is zero. So, force on m2 is zero, but force on m1 is only in forward direction, i.e., F. 

F a1 = m at t = 0 1

This is maximum acceleration of m1.

Hence, (d) is correct.

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CENTRE OF MASS C21 : In the given figure, two bodies of masses m1 and m2 are connected by massless spring of force constant k and are placed on a smooth surface (shown in figure), then: k (a) the acceleration of centre of mass must be zero at every instant m1 F (b) the acceleration of centre of mass may be zero at every instant (c) the system always remains in rest (d) none of the above Sol: The resultant force on the system is zero. So, the centre of mass of system has no acceleration.

m2

F

Example 4. Two planks each of mass m and length L are connected by a frictionless, massless hinge as shown in the figure. Initially the system is at rest on a level frictionless surface. The vertical plank falls anticlockwise and finally comes to rest on the top of the horizontal plank. Find the displacement of the hinge till the two planks come in contact. Sol. In horizontal direction centre of mass does not move. xic = xf c 0 m  m 

L 2

2m

L  2m   x 0  2  = 2m

A B

L L or = – x0 2 4

L 2 O

L 2 C

 x0

2L  L L L L = – = = 2 4 4 4

L – x0 2

Example 5. Two masses, nm and m, start simultaneously from the intersection of two straight lines with velocities v and nv respectively. It is observed that the path of their centre of mass is a straight line bisecting the angle between the given straight lines. Find the magnitude of the velocity of centre of inertia. [here  = angle between the lines]    m1 v1  m1 v 2 v cm  Sol. m1  m 2  nm v ˆi  m nv cos  ˆi  nmv sin  ˆj v cm  m  nm 2

v cm 

 nmv  mnv cos    nmv sin  m 1  n  2

v cm 



nmv 1  cos    sin 

x0

nv m

cm O

v nm

y 2

x

2

m 1  n  nv 1  cos 2   2 cos   sin 2  1  n 

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CENTRE OF MASS  nv   2  2 cos   n  1

  nv   2  2 cos 2   n  1 2   nv   2 cos   n  1 2 2nv cos v cm 

 2

n 1

Example 6. Two blocks of equal masses m are released from the top of a smooth fixed wedge as shown in the figure. Find the magnitude of the acceleration of the centre of mass of the two blocks. Sol. 

 mg sin 30º ˆi  mg sin 60º ˆj a cm  mm  g sin 30º ˆi  g sin 60º ˆj a cm  2  g 3g ˆ acm  ˆi  j 4 4

m

m

m 60º

30º

90º m

mgsin30º mgcos30º 60º

30º x

y

acm = g/2 Example 7. An open water tight railway wagon of mass 5 × 103 kg coasts at an initial velocity 1.2 m/s without friction on a railway track. Rain drops fall vertically downwards into the wagon. The velocity of the wagon after it has collected 103 kg of water will be (A) 0.5 m/s (B) 2 m/s (C) 1 m/s (D) 1.5 m/s Sol. Applying conservation principle of momentum, mivi = mfvf 5 × 103 × 1.2 = (5 × 103 + 103) vf 

5  103  1.2 vf  1m/s 6  103

Example 8. A parallel beam of particles of mass m moving with velocity v impinges on a wall at an angle  to its normal. The number of particles per unit volume in the beam is n. If the collision of particles with the wall is elastic, then the pressure exerted by this beam on the wall is : (A) 2 mn v2 cos (B) 2 mn v2 cos2 (C) 2 mn v cos (D) 2 mn v cos2 Sol. The change in velocity is v = 2v cos The no. of particles in volume dv = sdx is dN = ndv = nsdx  dm = mdN = mn sdx  

dm dx  mns  mnsv dt dt dp dm F  v dt dt F = (2v cos )mnsv

vsin s v

v

www.physicsashok.in vcos n

vcos

17

CENTRE OF MASS F = 2mn v2s cos    F.s p 2  Pressure, s Fs cos  F cos   s2 s 2 2 p = 2mn v cos  p

THE C-FRAME In many cases when we examine only the relative motion of particles within a system, but not the motion of this system as a whole, it is mot advisable to resort tot he reference frame in which the centre of inertia is at rest. Then we can significantly simplify both the analysis of phenomena and the calculations. The reference frame rigidly fixed to the centre of inertia of a given system of particles and translating with respect to inertial frames is referred to as the frame of the centre of inertia, or, briefly, the C frame. The distinctive feature of the C frame is that the total momentum of the system of particles is equal to zero; this immediately follows from equation. In other words, any system of particles as a whole is at rest in its C frame. The C frame of a closed system of particles is inertial, while that of a non-closed system is non-inertial in the general case. Example 9. In the diagram shown, no friction at any contact surface. Initially, the spring has no deformation. 2M What will be the maximum deformation in the spring ? Consider all the strings to be sufficiency large. Consider the spring constant to be K. (A) 4F / 3K (B) 8F / 3K (C) F / 3K (D) none Sol. At the time of maximum separation, blocks come in rest in C-frame. In C-frame, T = 0 v=0 Wpeseudo =WS + W1F + W2F = T = 0 ...(i) 2M 2F Workdone by pseudo frame in C-frame is zero. W1F = Fx1C W2F = 2Fx2C 1 0  kx 2max  Fx1C  2Fx 2C  0 2

F M

v=0 M

F

...(ii)

1 2 kx max  Fx1C  2Fx 2C 2 xmax = x1C + x2C From the concept of centre of mass Mx1C = 2Mx2C x1C = 2x2C  x1C + 2x2C = xmax x  3x2C = xmax

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CENTRE OF MASS

x 2C 

x max 3

2x max 3 Putting the values in equation (2) x1C 

1 2  2x  x  kx max  F  max   2F  max   3   3  2 1 2F 2F kx max   2 3 3

x max 

8F 3K

Ans.

MOMENTUM CONSERVATION AND CENTRE OF MASS MOTION The momentum of a system of particles is the vector sum of the momenta of all its particles. This is related to the velocity of centre of mass as 



P = M Vcm Hence momentum of a system is possible due to centre of mass motion. 

The change in momentum P of a body occurs only if an external force be present 

dP  = Fext dt 

What does happen if Fext be zero ? The Law of Conservation of momentum We know that 



Fext = M a am







Fext = 0  acm = 0  Vcm = constant 



M Vcm = constant 

 P = constant. If the external force on a system be zero, its momentum remains constant. This is the law of conservation of total momentum of a system of particles. Conservation of component of momentum 



If external force is not zero, then a component of P taken perpendicular to the Fext is conserved. C22 : A block is placed on a smooth triangular block and released. (a) Is external force on the system zero ? (b) Which component of total momentum is conserved ?

Fext x = 0 W1

acm

X  Px = const

W2

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19

CENTRE OF MASS Sol: The external forces on the system are pull of the earth and normal reaction of table - both are vertical. Since the block slides down, the centre of mass is coming down from rest into motion. Hence external forces are not balanced. But the x-component of forces is zero. Hence Px is conserved. (a) No. (b) Horizontal component (X- component in the figure). Example 10. Two balls of equal masses are projected upward simultaneously, one from the ground with speed 50 m/s and other from a 40 m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass. Sol. The initial position of centre of mass is yic =

ucm =

m  0  m  40 = 20m 2m

y 30 m/s B m

m  50  m  30 = 40 m/s 2m

acm = –g

h = 40m



2 2 = u cm + 2acmH vcm



0 2 = 402 – 2 × 10 × H



H =

50 m/s A m

1600 = 80 m 20

 yfc – yic = 80  yfc = 80 + 20 = 100 m C23 : A body at rest break up into three parts. If two parts having equal masses fly off perpendicularly to each other with a velocity of 18 m/sec, then calculate the velocity of the third part which has a mass 3 times the mass of each part. mv1 Sol: 6 2 m/sec at an angle 135º from the either ball.

mv1 2

3mv = mv1 2 v=

2v1 3

=

2  18 =6 2 3

mv1 3 mv

C24 : A body which is moving with a velocity of 10 m/sec towards east breaks up into three equal parts. If two of them move in the north and south direction (i.e., moves up and down) respectively with a velocity of 10 m/sec, then calculate the velocity of third part. mv1 Sol: Momentum along X-axis before and after explosion is conserved. Mv = 0 + 0 + mv2 cos

mv2

 www.physicsashok.in

mv1

20 Mv

CENTRE OF MASS Mv = mv2 cos .....(i) Momentum along Y-axis before and after explosion is conserved. mv1 – mv1 + mv2 sin = 0  mv2 sin = 0 or,  = 0º From (i) v2 =

Mv 3m  10 = = 30 m/sec. m cos  m 1

Example 11. A hemisphere of radius R and of mass 4m is free to slide with its base on a smooth horizontal table. A particle of mass m is placed on the top of the hemisphere. Find the angular velocity of the particle relative to hemisphere at an angular displacement  when velocity of hemisphere has become v. N Sol. Actual velocity of ball is v1 is horizontal direction.  v1 – (–v) = vrel cos  v  v1 + v = vrel cos  ...(i) Applying conservation principle of momentum in horizontal direction. vrel sin Pix = Pfx 0 = mv1 – 4mv  v1 = 4v ...(ii) Applying mechanic energy conservation principle Loss in P.E. = gain in K.E.



1 2 m v12   v rel sin   2 After solving equation (i), (ii) and (iii) mg  r  r cos  



2

1  4mv 2 2

vrel cos

...(iii)

5v cos  v 5v  rel  R R cos 

v rel 

 rel



C25 : Disintegration of stationary nucleus U238 Initially U238 is in state of rest, so its momentum is zero. Therefore, after disintegration, momentum will be zero. Hence, U238

92



Th234 + 2He4

92

U238

90 

-particle

or M234 v 234 + M v  = 0 M  4  v or v 234 = – M v  = – 234  234

92



and

U

234

4

2

H

234 M K = 234 = 4 M K Th

Disintegration of stationary kayon Final momentum





P  P– = 0

K0

Before disintegration (at rest)

P–

P+

www.physicsashok.in – + 0 + K   + –

After 21 disintegration

CENTRE OF MASS 



So,

P  P– = 0

or

P = – P– or | P | = |– P– |



K 

Thus,

K 





m  1 = m = 1 



[ m+ = m–]

C26 : Recoil force In n balls reach of mass m are fired from a cannon time t second, then F=

recoil force F = rate of change of momentum = reaction force = upthrust =

nmv t

nmv Cannon

nm N t

C27 : A man of mass m1 is standing on platform of mass m2 kept on a smooth horizontal surface. The man starts moving on the platform with a velocity r relative to the platform. Find the recoil velocity of platform. Sol: Absolute velocity of man = r – , where  = recoil velocity of platform.

vr – v v

Taking the platform and the man as a system, net external force on the system in horizontal direction is zero. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Hence 0 = m1(r – ) – m2 

m1 r  = m m 1 2

Example12. A gun (mass = M) fires a bullet (mass = m) with speed r relative to barrel of the gun which is inclined at an angle of 60º with horizontal. The gun is placed over a smooth horizontal surface. v Find the recoil speed of gun.

60º

M

vsin60º r vr cos60º–v

Sol: Let the recoil speed of gun is . Taking gun + bullet as the system. Net external force on the system in horizontal direction is zero. Initially the system was at rest. Therefore, applying the principle of conservation of linear momentum in horizontal direction, we get Mv – m(r cos60º – v) + 0   =

n r cos 60º Mm

or  =

n r 2(M  m)

C28 : A bullet of mass 20 gram is fired from a gun of mass 6 kg, if the speed of the bullet is 600 m/sec, then calculate the speed of the recoil of the gun. Sol: As the momentum of the bullet and the gun must be equal and opposite therefore,

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22

CENTRE OF MASS MV = mv

or 6V =

20 × 600 1000

 V = 2 m/sec. C29 : A machine gun can fire bullets of 50 grams at a speed of 2000 m/sec; the man holding the gun can exert an average force of 200 newtons against the gun. Calculate the maximum number of bullets which he can fire per minute. Sol: We have, Force = mnu where m = mass of bullet, u = speed of bullet, n = number of bullets per unit time. 

F = mnu

or, 200 =

50 × 2000 × n 1000

Hence, number of bullets fired per minute = 60 × 2 = 120. M

C30 : A plate of mass M is held at rest by firing bullets from below. Each bullet has a mass m, velocity u(up) just before hitting and stops after hitting the plate for a moment and falls. Determine the number of bullets striking the plate per unit time. 



Sol: The change in momentum of a bullet due to hitting the plate is p = 0 – mu . If N bullets be hitting the plate 

per unit time, the rate of change of momentum of the bullets is –N mu . This is also the force exerted by the 

plate during collision. Using Newton’s third law of motion, an opposite force, N mu , will act on the plate due to the colliding bullets. (rest) u

p

Considering the equilibrium of forces acting on the plate,

m





m

N mu + Mg = 0  Nmu = Mg N=

The change in bullet's momentum

Just after the hitting

Going to hit

Mg mu

Nm u

This is the number of bullets striking the plate per unit time. Mg

C31 : A projectile is launched at an angle  with velocity V0. At the highest point of its path, it explodes into two parts of masses in the ratio 1 : 3. The lighter mass retraces its path. Find the velocity of the heavier part just after the explosion.

V0 V0 0

Sol: The figure shows the mass m moving with horizontal velocity of V0 cos0 at the top just before explosion. The lighter mass (m0) formed after explosion retraces its path, its velocity at the top is just reversed due to explosion. Thus m0 moves leftward at V0 cos0. At this moment, 3 m0 is moving rightward with the requested velocity, V. Conservation of momentum along horizontal X-axis, gives

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23

CENTRE OF MASS mV0cos0 = m0(–V0 cos0) + 3m0V m m = – V0 cos0 + 3. V 4 4



V0cos0 = –

m

V0cos 0

V0cos0

V m0

V0 3 cos0 + V 4 4

X

Just before explosion

3m0

= m/4 =3m/4 X Just before explosion

5 V cos0 3 0 This is the velocity of the heavier part just after the explosion.



V =

C32 : Motion of a ball under gravity (i)

(ii)

If the ball is the system, the force of gravity will be external and so the linear momentum o the ball will change. This is not violation of law of conservation of linear momentum as linear momentum is conserved only if Fext = 0. If ball and the earth is the system, all the acting forces will become internal and so, v 





p S = p B + p E = constant

m

Ball

Now as initially both ball and the earth are at rest 



p B + pE = 0

Earth

M

From this it is clear that 



(a) p E = – p B , i.e., if ball acquires some momentum, the earth also acquires equal and opposite momentum. 



V



 (b) As p = mv + MV = 0, i.e., V = –(m/M)  v i.e., the velocity of earth is opposite to that of ball in direction and much smaller in magnitude (as M > > m). So, if ball moves away from the earth, the earth also moves away from the ball and if ball moves towards the earth, the earth also moves towards the ball but the speed of the earth is much lesser than that of the ball (as its mass is much greater than that of ball).

KE m p2   (c) Now as K = and | p E | = | p B |, K = i.e., KE of ball and earth will not be equal and as M >>m M 2m B the, KE of earth will be much lesser than that of the ball. (d) Initially KE of the system is zero (as both are at rest) and after some time KE of the system is not zero(as both are in motion). So, KE of the system is not constant but changes. However, (KE + PE), i.e., mechanical energy of the system is conserved. When the ball and earth approach each other KE increases and PE decreases and when they recede from each other KE decreases and PE increases.

Lecture 5 C33 : Motion of two masses connected by a spring Consider two blocks. resting on a frictionless surface and connected by a massless spring as shown in figure. If the spring is stretched (or compressed) and then released from rest. Then as

v2

v1 m2

m1

Fext = 0 





pS = p1 + p 2 = constant

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CENTRE OF MASS However, initially both the blocks were at rest so, 



p1 + p 2 = 0

From this it is clear that: 



(i)

p 2 = – p1 i.e., at any instant the two blocks will have momentum equal in magnitude but opposite in direction (Though they have different values of momentum at different positions).

(ii)

As p = mv , m1 v1 + m2 v 2 = 0, i.e., v 2 = –(m1/m2) v1 i.e., the two blocks always move in opposite directions with lighter block moving faster.













K1 m2 p2 and | p1 | = | p2 | , K = m , i.e., the KE of two blocks will not be equal but in the inverse 2m 2 1 ratio of their masses and so lighter block will have greater KE. (iv) Initially KE of the blocks is zero (as both are at rest) and after sometime KE of the blocks is not zero (as both are in motion). So, KE is not constant but changes. here during motion of blocks KE is converted into elastic potential energy of the spring and vice-versa such that KE + PE = ME = constant

(iii)

Now as K =

C34 : (a) Can kinetic energy of a system be changed without changing its momentum ? (b) Can momentum of a system be changed without changing its kinetic energy ? Sol. (a) Yes; In explosion of a bomb or inelastic collision between two bodies as force is internal, momentum is conserved while kinetic energy changes. (b) Yes; If a force acts perpendicular to motion, work done will be zero and so kinetic energy will remain constant. However, the force will change the direction of motion and so, the momentum, e.g., in case of uniform circular motion of a body KE is constant while momentum changes. C35 : A meteorite burns in the atmosphere before it reaches earth’s surface. What happens to its momentum ? Sol. The momentum of meteorite is transferred to air molecules by air drag and so, the momentum of meteorite plus air molecules (system) always remains constant. C36 : Can a sail boat be propelled by air-blow. at the sails from a fan attached to the boat ? Sol. As sails and fan both are attached to the boat, force due to the air-blown on the sails by the fan is an internal force. Now as by an internal force momentum of the system cannot be changed, so the system (boat + fan + sails) cannot be propelled by blowing air at the sails from a fan attached to the boat. C37 : A bird is held in a light container which is completely closed. Can we tell when the bird is resting or flying by weighing the container ? Sol. According to the law of conservation of linear momentum we know that the linear momentum of a system of constant mass remains unchanged in the absence of any external force. Now treating the bird, container and the air in the container as the system of constant mass we find that the force which the bird exerts in flying is an internal force. But as by an internal force the momentum of a system cannot be changed, so the weight of the system will not change, i.e., we cannot tell when the bird is flying or sitting by weighing the container. Note : However, if in the above problem the container is of wire gauge, the momentum of the system will not be conserved (why ?) and due to this the weight of the system will be lesser when the bird is flying as compared to the weight of the same system when bird is resting, i.e., we will be able to tell when the bird is resting or flying by weighing the container.

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CENTRE OF MASS Example 13. Two trolleys A and B are free to move on a level frictionless track, and are initially stationary. A man on trolley A throws a bag of mass 10 kg with a horizontal velocity of 4 m/s with respect to himself on to trolley B of mass 100 kg. The combined mass of trolley A (excluding bag) and the man is 140 km. Find the ratio of velocities of trolleys A and B, just after the bag lands on trolley B. Sol. Applying conservation principle of momentum for trolly A and bag mv – m1v1 = 0 v A

or mv = m1v1 or 10v = 140 v1 or v = 14v1 also, v1 + v = 4 or v1 + 14v1 = 4 

v1 =

v1

m1 = 140 kg

B m2 = 100 kg

v2

.....(i)

4 m/s , 15

v = 14v1 =

56 m/s 15

Applying conservation principle of momentum for bag and trolley B, mv = (m + m2)v2 56 10  56 mv 15  v2 = = = 110 15  11 m  m2 v1 4 11  v = × 15 × 11 = 15  56 14 2

Example 14. A bob of mass m attached with a string of length l tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass m gently stuck to if. Find the angle from the vertical to which it rises. Sol. The velocity of bob just before collision is u = 2g Apply conservation principle of momentum mu = 2mv0 

v0 =

u = 2

m

g =1 2

Applying conservation principle of energy,

v0

2m

1 (2m)v 20 = 2mgH 2



g v 20  H= = = 2  2g 2g 4

But

 = (1 – cos) , 4

or

cos =

3 4

,

or

1 = 1 – cos 4

3  = cos–1   4

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CENTRE OF MASS Example 15. A block of mass m starts from rest and slides down a frictionless semicircular track from a height h as shown. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass m. If the block h and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is : (A) h/4 (B) h/2 (C) h (D) independent of h Sol.

v 0  2gh before collision. Apply conservation principle of momentum, mv0 = (m + m)u

v0 2 Applying conservation principle of momentum, 

u

1  2m u 2  2mgh´ 2



u2 h´ 2g v02 h´ = 4g h´ =

2gh 4g

h´ =

h 2

Question No. 16 to 19 A small ball B of mass m is suspended with light inelastic string of length L from a block A of same mass m which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle  from equilibrium position & then released. Example 16. The displacement of block when ball reaches the equilibrium position is

A L

L B

u=0

L sin  (B) L sin (C) L (D) None of these 2 Sol. If ball and blocks are taken as system, then in horizontal direction no external force is present. Hence, centre of mass remains in rest in horizontal direction.  m(L sin  – x) – mx = 0 or mL sin  – mx – mx = 0

(A)



x

L sin  2

Example 17. Tension in string when it is vertical, is (A) mg (B) mg(2 – cos) Sol. Loss in P.E. = gain in K.E. mg(L – L cos ) =

B

(C) mg(3 – 2 cos )

(D) None of these

1 1 mv12  mv 22 2 2

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CENTRE OF MASS Applying momentum conservation principle in horizontal direction. Pix = Pfx 0 = mv1 – mv2  v1 = v2 = v  mg(L – L cos ) = mv2 

v  gl (1  cos ) m(2v 2 ) T  mg  l

4mg l 1  cos  l  T = mg + 4mg (1 – cos ) Hence (D) is correct. T  mg 



Example 18. maximum velocity of block during subsequent motion of the system after release of ball is (A) [gl (1 – cos)]1/2 (B) [2gl (1 – cos)]1/2 (C) [gl cos]1/2 (D) informations are insufficient to decide v max  gl 1  cos 

Sol.

(From sol. of previous problem.) Example 19. The displacement of centre of mass of A + B system till the string becomes vertical is (A) zero

(B) y cm 

Sol.

(C)

L 1  sin  2

(D) None of these

m1 y1  m 2 y 2 m1  m 2

ycm 



L 1  cos  2

m1y1  m 2 y 2 m1  m 2

0  mL 1  cos  mm L  1  cos  2

y cm  ycm



in downward direction.

IMPULSE AND IMPULSIVE FORCE A great force acting for a very short interval of time on a body is called an impulsive force. There are numerous examples of impulsive forces. When we kick a ball, the force on the ball is an impulsive force. The force experienced by the coil of a moving coil galvanometer due to sudden flow of current is an impulsive force. According to Newton’s second law,  m(v  u)  F = ma = t 



where F is the average force during the short interval t and a is the average acceleration produced. The

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CENTRE OF MASS impulse of an impulsive force is the great change in momentum produced by it. 

Thus the impulse of an impulsive force = change in momentum = F t. In fact an impulsive force is variable force. The plot of an impulsive force is a curve which is zero in the beginning and the end of the interval showing a peak value at some intermediate instant.

A (a) F A

The total impulse of the force during the interval for which it acts is given by the summation of the impulse from instant to instant. At the instant t the elementary impulse is F.dt.

(b) O

O

t

t



total impulse

=

 Fdt 0

= area OAO = change in momentum. When a cricket ball is hit by a bat, it exerts an impulsive force on the ball and the impulse of the force on the balls is so great that it moves in the opposite direction with tremendous speed. If the same ball is to be stopped in the same interval t by a fielder he must apply the same average impulsive force of figure (a). The force being very large he (the fielder) will feel severe pain, particularly in the bare hands. If the lowers his hand so as to lengthen the duration of the force, the same change of momentum will need a comparatively smaller force (figure b) and so he will not feel the pinch of catching the ball. For the same reason when a man jumps from a high wall on the side concrete road he hurts himself more seriously than when he jumps from the same wall on a heap of sand. C38 : A tennis ball of mass 200 gm is dropped on the floor from a height of 1.2 m. It rebounds to a height of 0.9m. If the ball was in contact with the floor for 0.01s, what was the average impulsive force and what was its impulse. Sol. Let 0 be the velocity of the ball immediately before reaching the ground.  02 = 2 × 9.8 × 1.2

Let n be the velocity after bouncing n2 = 2 × 9.8 × 0.9 Taking upward as the positive direction 

[From  = 0 + at)

2  9.8  .9 = – 2  9.8  1.2 + a × 0.01 where a is the average acceleration (upward) during the interval. or 



a =



4.2  4.85 9.05 = = 905 ms–2 .01 .01 

F = ma =

200 × 905 = 181 newton. 1000



Impulse F .t = 181 × .01 = 1.81 kms–1. C39 : A croquet ball (0.5 kg) is struck by a mallet, receiving the impulse shown in graph. What is the ball’s velocity

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29

CENTRE OF MASS just after the force has become zero ?

There are in all 42 × 2 = 84 squares (approximately) within OAA.  Area of OAA = 84 × 0.1 = 8.4 newton second = impulse experienced = change in momentum = 0.5 × v – 0.5 × 0 = 0.5 v 

 =

Force in newton

Sol: Each square represents 100 × 1 × 10–3 = .1 Ns

A

1500

1000

500 A 4 8 Time in milli second

0

8.4 = 16.8 ms–1 0.5

C40 : A truck of mass 2 × 103 kg travelling at 4 m/s is brought to rest in 2 s when it strikes a wall. What force (assume constant) is exerted by the wall ? Sol: Using impulse = change in linear momentum We have F.t = mvf – mvi = m(vf – vi) or F(2) = 2 × 103 [0 – (–4)] or 2F = 8 × 103 or F = 4 × 103 N

F u

^

^

+ve

^

C41 : A ball of mass m, travelling with velocity 2 i  3 j receives an impulse – 3m i . What is the velocity of the ball immediately afterwards ? Sol: Using





^





j = m( v f  v i ) ^

^

– 3m i = m[ v f – ( 2 i  3 j )] or or



^

^

^

vf = – 3 i + ( 2 i  3 j ) 

^

^

vf =  i  3 j

C42 : A bullet of mass 10–3 kg strikes an obstacle and moves at 60º to its original direction. If its speed also changes from 20 m/s to 10 m/s. Find the magnitude of impulse acting on the bullet. Sol: Mass of the bullet m = 10–3 kg Consider components parallel to J1 J1 = 10–3[–10 cos60º – (–20)] or J1 = 15 × 10–3 N-s

J1

20 m/s

Similarly, parallel to J2, we have

10 cos60º

120º 60º 10 m/s J2

10 sin60º

J2 = 10–3[10 sin60º – 0] = 5 3 × 10–3 N-s The magnitude of resultant impulse is given by

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30

CENTRE OF MASS

or

J =

J12  J 22 = 10–3

J =

3 × 10–2 N-s

(15) 2  (5 3)2

Example 20. A force exerts an impulse I on a particle changing its speed from u to 2u. the applied force and the initial velocity are oppositely directed along the same line. The work done by the force is (A) Sol.

3 Iu 2

(B)

1 Iu 2

(C) I u

I = P I = 2mu – (–mu) I = 3mu ...(i) According to work energy theorem W = T W = Tf – Ti W

1 1 2 2 m  2u   m   u  2 2

W

1 1 m 4u 2  mu 2 2 2

+

I u

m

m

2u

 

3mu 2 W 2 From equation (i) and (ii) W

(D) 2 I u

1 Iu 2

...(ii)

Ans.

Example 21. A man whose mass is m kg jumps vertically into air from a sitting position in which his centre of mass is at a height h1 from the ground. When his feet are just about to leave the ground his centre of mass is h2 from the ground and finally rises to h3 when he is at the top of the jump. (a) What is the upward force exerted by the ground on him treating it as a constant ? (b) Find work done by normal reaction from ground. Sol. a =

F m

[Here F is force applied by ground on man]

 F  mg   V02 = 02 + 2   (h – h )  m  2 1 After leaving the surface,

0 2 = V02 – 2g(h3 – h2)  V02 = 2g(h3 – h2)  F  mg or 2  (h – h ) = 2g(h3 – h2) m  2 1

Here F is impulsive for a so mg may be neglected.

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CENTRE OF MASS 

2F(h 2  h1 ) = 2g(h3 – h2) m

 F=

mg(h 3  h 2 ) (h 2  h1 )

(b) Since, contact point does not move. Hence, workdone by normal reaction is zero. Example 22. In the figure shown, each tiny ball has mass m, and the string has length L. One of the ball is imparted a velocity u, in the position shown, in which the 60º initial distance between the balls is L / 3 . The motion of ball occurs on smooth u horizontal plane. Find the impulse of the tension in the string when it becomes taut.

Sol. 

L L  sin 60º 3 sin  3

L 3

or or

1   = 30º 2 3 2 Tdt = mv1 and –Tdt = mv1 – mucos30º –mv1 = mv1 – mucos30º 2v1 = u cos 30º



v1 



Tdt  mv1 

sin  

60º

Tdt



3u 4

v1 Tdt u cos30º u cos30º 30º u v1 Tout position

3 mu 4

Example 23. A platform of mass m and a counter weigth of mass (m + M) are connected by a light cord which passes over a smooth pulley. A man of mass M is standing on the platform which is at rest. If the man leaps vertically upwards with velocity u, Find the distance through which the platform will descend. Show that when the man meets the platform again both are in their original positions. Sol. FdT = Mu Fdt – Tdt = mv0 Tdt = (m + M)v0  Fdt – (m + M)v0 = mv0 or Mu = (2m + M)v0 Mu v0   2m  M The acceleration of platform is

 m  M g  mg a m Mm

Mg 2m  M v2 = v02 – 2aS a



upward upward

Fdt m v0

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Tdt v0

m+M

32

CENTRE OF MASS  2Mg  0  v02   S  2m  M  0

or

S0



 2m  M  v20   2m  M   

S0 

2Mg

2Mg

Mu    2m  M  

2

Mu 2 2g  M  2m

Example 24. A mass 2m rests on a horizontal table. It is attached to a light inextensible string which passes over a smooth pulley and carries a mass m at the other end. If the mass m is raised vertically through a distance h and is then dropped, what is the speed with which the mass 2m begins to rise ?

(a)

2gh 5

(c)

gh 3

(b)

h

2gh 3 2m

(d)

2gh

Sol: In the figure it is clear that v =

.....(i)

2gh

Using impulse = change in momentum For mass 2m J = 2mv1 .....(ii) For mass m J = mn – mv1 .....(iii) Solving equations (i), (ii) and (iii), we get

J

J v

Just when the string jerks

v=

v1

v1

Just after string jerks

2gh 3

COLLISION A cricket ball is struck by a bat an there is large change ein momentum of the ball and bat in a very short time of contact. We say that collision between the ball and the bat has taken place. Similarly an -particle moving towards gold nucleus in Rutherford like experiment undergoes large change in momentum in short time. We say that collision has taken place. The feature we observe is ‘short time’ and ‘large change in momentum’. The rate of change of momentum will be very high. The force will be high but short lived. A force of high magnitude but acting for short period of time is known as an impulsive force. Hence in collision, impulsive forces act on colliding bodies. Collision is the process in which bodies exchange energy and momentum in a very short period of time.

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33

CENTRE OF MASS If a collides with B, the force exerted on A is equal and opposite to that exerted by A on B, i.e., 



FAB = – FBA .

Since these forces are impulsive, external forces may be ignored as they F BA produce very small change in momentum of a colliding body in the very short period. We can ignore this change and may conserve momentum of colliding bodies: Total momentum just before the collision equals the total momentum just after the collision.

FBA mg < < FAB mg (external force)

Demonstration Three identical steel balls are suspended by threads of equal length. ‘A’ is moved to left and released. It goes to collide with B. But A stops, and C is thrown out. Why ?

A

B

C

A

B

A

C

C

B

Steel is highly elastic. Hence collision is elastic. A and B collide horizontally. The velocity of A goes to B and that of B (zero) goes to A. A is stopped, B collides with C. Velocity of C goes to B and that of B goes to C. Thus B is stopped and C is thrown out. This happens quickly. DIRECT CENTRAL IMPACT As an introduction to impact, we consider the collinear motion of two spheres of masses m1 and m2, figure a, travelling with velocities v1 and v2. If v1 is greater than v2, collision occurs with the contact forces directed along the line of centres. This condition is called direct central impact. Following initial contact, a short period of increasing deformation takes place until the contact area between the spheres ceases to increases. At this instant, both spheres figure b, are moving with the same velocity v0. During the remainder of contact, a period of restoration occurs during which the contact area decreases to zero. The final condition is shown in the c-part of the figure where the spheres now have new velocities v1 and v2, where of necessity v1 is less than v2 . All velocities are arbitrarily assumed positive to the right, so that this scalar notation a velocity to the left would carry a negative sign. If the impact is not overly severe and if the spheres are highly elastic, they will regain their original shape following the restoration. With a more severe impact and with less elastic bodies, a permanent deformation may result. v1 m1

>

v2 m2

(a) Before impact

v1

v0 m1

m2


> m2, then m1 – m2 = m1 and m1 + m2 = m1  1 ~ u1 and 2 = 2u1 i.e., when a massive moving ball collides with a light ball at rest then after collision the massive ball continues to move with the same velocity but light ball starts moving with twice the velocity of massive ball.

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36

CENTRE OF MASS (c) If m1 < < m2, then m1 – m2 = –m2 and m1 + m2 = m2  v1 = –u1 and v2 = 0 i.e., when a light moving ball collides with a massive ball at rest then the light ball rebounds with the same velocity and massive ball continues to be at rest. (2)

If m1 >> m2 equation (v) and (vi) become v1 = u1 and v2 = 2u1 – u2

(3)

If m2 1

body is maximum when m1 m2 is (B) After the collision, the momentum of

(Q) < 1

second body is maximum when m1 m2 is (C) After the collision, K.E. of second body

(R) = 1

is maximum when m1 m2 is (D) For elastic collision, coefficient of restitution is

(S) 0

1.

A ball kept in a closed box moves in the box making collisions with the walls. the box is kept on a smooth surface. The centre of mass: (A) of the box remains constant (B) of the box plus the ball system remains constant (C) of the ball remains constant (D) of the ball relative to the box remains constant

2.

In the given figure, two bodies of masses m 1 and m 2 are connected by massless spring of force constant k and are placed on a smooth surface (shown in figure), then: F (A) the acceleration of centre of mass must be zero at every instant (B) the acceleration of centre of mass may be zero at every instant (C) the system always remains in rest (D) None of the above

m1

k

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m2

F

50

CENTRE OF MASS 3.

4.

m1 

A

Two blocks of mass m 1 and m 2 are connected by light inextensible string passing over a smooth fixed pulley of negligible mass. The acceleration of the centre of mass of the system when blocks move under gravity is

 m2  m1  g (A)   m1  m2  5.

m2

In the given figure the mass m 2 starts with velocity v 0 and moves with constant velocity on the surface. During motion the normal reaction between the horizontal surface and fixed triangular block m 1 is N. Then during motion: (A) N = (m 1 + m 2) g (B) N = m 1 g (C) N < (m 1 + m 2) g (D) N > (m 1 + m 2) g

2

 m1  m2  g (B)   m2  m1 

2

 m2  m1  (C)   g  m1  m2 

 m1  m2  (D)   g  m2  m1 

A block of mass M is tied to one end of a massless rope. The other end of the rope is in the hands of a man of mass 2M as shown in the figure. the block and the man are resting on a rough wedge of mass M as shown in the figure. The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is massless and frictionless. What is the displacement of the wedge when the block meets the pulley. (Man does not leave his position during the pull) (A) 0.5 m

(B) 1 m

(C) Zero

(D)

M 2m

M

2 3m

6.

The figure shows the positions and velocities of two particles. If the particles v1= 5 m/s v2= 3 m/s move under the mutual attraction of each other, then the position of centre of 1 kg 1 kg mass at t = 1 s is x=0 (A) x = 5 m (B) x = 6 m (C) x = 3 m (D) x = 2 m x = 10 m

7.

A block of mass m moving with a velocity v 0 collides with a stationary block of mass M at the back of which a spring f stiffness k is attached, as shown in the figure. Choose the correct alternative(s)

v0 k m

Smooth

M

(A) The velocity of the centre of mass is v0 2 (B) The initial kinetic energy of the system in the centre of mass frame is (C) The maximum compression in the spring is v0

1  mM  2   v0 . 4 M m

mM 1 m  M  k .

(D) When the spring is in the state of maximum compression, the kinetic energy in the centre of mass frame is zero. 8.

A hemisphere and a solid cone have a common base. The centre of mass of the common structure coincides with the centre of the common base. If R is the radius of hemisphere and h is height of the cone, then (A)

9.

h  3 R

(B)

h 1  R 3

(C)

h 3 R

(D)

h 1  R 3

Masses of 1 g, 2 g, 3 g, ........ 100 g are suspended from the 1 cm, 2 cm, 3 cm ......... 100 cm marks of a light metre scale. The system will be supported in equilibrium at (A) 60 cm (B) 67 cm (C) 55 cm (D) 72 cm

10. A drum major’s baton consists of two masses 2 M and M separated by a thin light rod of length L. The baton



is thrown into the air. Neglecting air drag, if R is the position vector of the centre of mass of the baton then the equation of motion for the centre of mass is

  d 2R 2g  (A) dt 2 3

  d 2R g  (B) dt 2 3

 d 2R  g (C) dt 2

  d 2R 4g  (D) dt 2 3

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51

CENTRE OF MASS 11. A rectangular block of length  and breadth b is held with one corner resting on a frictionless table and is gently released, thus falling in a complex tumbling motion. The trajectory of the centre of mass of the block is (A) a circuit track of radius

 b or depending on the manner it is released (or placed before releasing). 2 2

(B) a straight line. (C) an ellipse. (D) unknown, as data provided is insufficient to arrive at any conclusion. 12. A cannon and a supply of cannon balls are inside a sealed rail road car at rest. The cannon fires to the right, the car recoils to the left. The balls remain in the car after hitting the farther wall. Assume friction to be absent. If L is the length of the rail road car, D is the distance traversed by the rail road car relative to the ground, m is the total mass of cannon balls fired and M is the total mass of remaining system (cannon + car) then (A) D > L

(C) D 

(B) D < L

ML M m

(D) D 

mL M m

13. Four point masses A, B, C and D are placed at points which are coplanar but non-collinear. If P is the centre of mass of the system comprising A, B, C and D, then (A) P must lie within the quadrilateral ABCD. (B) P must lie on a line joining any two of the points A, B, C and D. (C) P may or may not coincide with one of the point masses. (D) P lies inside or on the edge of at least one of the triangles formed by taking any three of the points from A, B, C and D. L

14. The figure shows a small prism of mass M sliding on the bigger prism of mass 5M from the top to the bottom. Assuming all surfaces to be frictionless and the system (big prism + small prism) to be at rest initially, the distance moved by the prism combination is

3 (A) L to the left 2

3 (B) L to right 2

2 (C) L to right 3

M 5M 

2 (D) L to left 3

5L

15. A man of mass m, the light rope and the balloon of mass M are at rest as shown. The man starts climbing the rope and reaches the top. Assume the crate in which the man is standing to be light.

mL . M m mL (B) The balloon ascends by a distance M m ML (C) The balloon descends by a distance M m (A) The balloon descends by a distance

(D) The balloon ascends by a distance

ML M m

16. A large tray of mass M has in it a cubical block of ice of mass m and edge length L. The ice starts melting. The shift in the height of centre of mass of system (ice + tray) is

1

mL

(A) 2 m  M ; upward  

1

ML

1

mL

1

ML

(B) 2 m  M ; upward (C) 2 m  M ; downward (D) 2 m  M ; downward      

17. An ideal spring is connected between two blocks of mass M and m. Blocks-spring system can move over a smooth horizontal table along a m straight line along the length of the spring as shown in figure. The blocks are brought nearer to compress the spring and then released. In the subsequent motion: (A) Initially they move in opposite directions with velocities inversely proportional to their masses (B) the ratio of their velocities remains constants. (C) Linear momentum and energy of the system remains conserved. (D) The two blocks will oscillate about their centre of mass, which remains stationary.

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M

52

CENTRE OF MASS 18. All the particles of a body are situated at a distance R from the origin. The distance of the centre of mass of the body from the origin is: (A) = R (B)  R (C) > R (D)  R. 19. Select the correct alternative(s) : (A) It is possible to have a collision in which whole the kinetic energy is lost. (B) Only external forces can cause the centre of mass of a system to accelerate. (C) Centre of mass of a solid body necessarily lie within the body. (D) A ball is dropped form a height h to the ground. If the coefficient of restitution is e, the height to which ball goes up after it rebounds second time is e2 h. 20. Which of the following are not correct about centre of mass? (A) It depends on frame of reference. (B) In centre of mass frame, momentum of a system is always zero. (C) Internal forces may affect the motion of centre of mass. (D) Centre of mass and centre of gravity are synonymous. 21.

A spring is compressed between two blocks of masses m 1 and m 2 placed on a horizontal frictionless surface as shown in figure. When the blocks are released, they have initial velocity of v 1 and v 2 as shown in figure. The blocks travel distances x 1 and x 2 respectively before coming to rest. The ratio x 1 / x 2 is m1 m2 m1 (A) (B) (C) m2 m1 m2

(D)

m2 m1

22.

A spherical ball falls from a height h on a floor of coefficient of restitution e. Find the total distance covered by the ball, before coming to rest. (A) h (1 + e2)/(1–e2) (B) h (1 – e2)/(1+ e2) (C) h (1 + e)/(1–e) (D) h (1 – e)/(1+e)

23.

A ball of mass m approaches a moving wall of infinite mass with speed v along the normal to the wall.The speed of the wall is u towards the ball. The speed of the ball after an elastic collision with the wall is (A) u + v away from the wall (B) 2u + v away from the wall (C) u - v away from the wall (D) v- 2u away from the wall

24.

Two particle of mass m 1 and m 2 in projectile motion have velocities V 1 and V 2 respectively at time t = 0. They collide at time t 0 their velocities become V1 and V2 and time 2t 0 while still moving in air . The value of     | (m1v 1  m 2 v 2 )  (m1v 1  m 2 v 2 ) | is 1 (A) 0 (B) (m 1 + m 2) gt 0 (C) 2(m 1 + m 2) gt 0 (D) (m 1 + m 2) gt 0 2

25.

A particle strikes a horizontal smooth floor with a velocity u making an angle  with the floor and rebounds with velocity  making an angle  with the floor. The coefficient of resitution between the particle and the floor is e. (A) The impulse delivered by the floor to the body is mu(1-e) sin  (B) tan  = tan (  /e) (C) v = u 1  (1  e 2 ) sin 2 

(D) none of these

26.

A man of mass 50 kg. is standing on a platform of mass 100 kg. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed 15m/sec relative to the platform, then calculate with what velocity relative to the ice the platform will recoil? (A) 5 m/sec (B) 10 m/sec. (C) 15 m/sec (D) none of these

27.

Two balls shown in figure are identicial, the first moving with a speed v towards right and second staying at rest. The wall at the extreme right is fixed and smooth. Assuming all collision to be elastic, which of the following statements are correct (A) the speed of the first ball is reduced to zero finally after all collision (B) only two collision are possible (C) the speed of the balls remain unchanged after all collisions have taken place (D) none of these

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53

CENTRE OF MASS L

28.

A simple pendulum is suspended from a peg on a vertical wall pendulum is pulled away from the wall to a horizontal position and released. The ball hits the wall, the coefficient of resitution being 2 . What is the aminimum number of collisions after 5 which the amplitude of ossicillation becomes less than 600 ? (A) 2 (B) 3 (C) 4 (D) none of these

29.

A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular directions with a speed v. What will the speed of the third (lighter) fragment ? (A) v

30.

(B) 2 v

(C) 2 2 v

(D) 3 2 v

The bob A of a pendulum released from a height h hits head-on another bob B of the same mass of an identical pendulum initially at rest. What is the result of this collision? Assume the collision to be elastic (A) Bob A comes to rest at B and bob B moves to the left attaining a maximum height h h (B) Bobs A and B both moves to the left, each attaining a maximum height 2

A h B

(C) Bob B moves to the left and bob A moves to the right , each attaining a maximum height

h . 2

(D) Both bobs come to rest 31.

A ball of mass m moving with a veloctity v undergoes an oblique elastic collision with another ball of the same mass m but at rest. After the collision, if the two balls move with the same speeds, the angle between their directions of motion will be (A) 300 (B) 600 (C) 900 (D) 1200

32.

A boy throws a tennis ball vertically downwards. He wants the ball to rebound from the floor and just touch the ceiling of the room which is at the height of 4 m. If the coefficient of restitution is 0.8, with what velocity should the ball strike the floor ? (A) 13.4 m/s (B) 11.2 m/s (C) 10.3 m/s (D) none of these

33.

An empty box is put on the pan of a physical balance and the scale is adjusted to zero by counterpoising. A stream of small identical beads each of mass m are then dropped into the box from a height h at a constant rate of n beads per second. If the collision between the beads and the box is completely inelastic, find the reading of the scale, t second after the beads begin to fill the box : (A) nm

34.

2h t g

nmt gh

 2 h  (C) nmg  t   g  

(D) nmght

A pendulum consists of a wooden bob of mass m and length  . A bullet of mass m 1 is fired towards the v1 pendulum with a velocity v 1. The bullet comes out of the bob with speed and the bob just completes 3 motion along a vertical circle. The velocity v 1 is : 3 m  (A) 2  m  5g   1

35.

(B)

(B)

m 5g  m1

(C)

3 m    g 2  m1 

(D)

3  m1    5g  2 m 

A body of mass M at rest explodes into three pieces, two of which of mass (M/4) each are thrown off in perpendicular directions with velocities of 3 m/s and 4m/s respectively. The third piece will be thrown off with a velocity of : (A) 1.5 m/s (B) 2 m/s (C) 2.5 m/s (D) 3 m/s

36. An isolated particle of mass m is moving in a horizontal plane (x – y) along the x-axis at a certain height above

the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later the smaller fragment is at y = + 15 cm. The larger fragment at this instant is at [JEE, 97] (A) y = – 5cm (B) y = + 20 cm (C) y = + 5 cm (D) y = – 20 cm

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54

CENTRE OF MASS 37. A set of n-identical cubical blocks lie at rest parallel to each other along a line on a smooth horizontal surface.

The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed V towards the next one at time t = 0. All collisions are completely inelastic, then (A) the last block starts moving at t = n(n – 1)L/(2v) (B) the last block starts moving at t = (n – 1)L/v (C) the centre of mass of the system will have a final speed v/n (D) the centre of mass of the system will have a final speed v. [JEE, 95] 38. Two particles of masses m1 and m2 in projectile motion have velocities

  v1 and v 2 respectively at time t = 0.

  They collide at time t0. Their velocities become v1´ and v´2 at time 2t0 while still moving in air. The value of  m v ´1  m v ´2  m v 1  m v 2  [JEE(Scr), 01] 2 1 2  1  is





(A) zero





(B) (m1 + m2)gt0

(C) 2(m1 + m2)gt0

(D) ½(m1 + m2)gt0

39. Two block of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless

horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block The velocity of the centre of mass is : [JEE (Scr), 02] (A) 30 m/s (B) 20 m/s (C) 10 m/s (D) 5 m/s 40. Look at the drawing given in the figure which has been drawn with ink of uniform line-

thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. the mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are : outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (a, –a). The y-coordinate of the centre of mass of the ink in this drawing is [JEE, 09] (A)

a 10

(B)

a 8

(C)

a 12

41. A disk of radius a/4 having a uniformly distributed charge 6C is placed

(D)

y

x

a 3 y

in the x – y plane with its centre at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis x from x = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The electric flux through this cubical surface is [JEE, 09] 2C (A)  0

2C (B)  0

10C (C)  0

12C (D)  0

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55

CENTRE OF MASS

Passage Type Problems PASSAGE – 1 In a game of billiards ball A is given an initial velocity V0 of magnitude V0 = 3 m/s along line DA parallel to the axis of the table. It hits ball B and then ball C, which are both at rest knowing that A and C hit the sides of the table squarely at points A’ and C’, respectively that B hits the side obliquely at B’, and assuming frictionless surfaces and perfectly elastic impacts. The motion of billiard balls are are assumed to be particle moving freely in a horizontal plane rather than the rolling and sliding spheres they actually are. 1.

Mark Correct option: (A) The momentum of balls system remains conserved before hitting the wall of table. (B) The angular momentum of balls system about point O remains conserved before striking with the walls of table. (C) Mechanical energy of system will remains constant. (D) All the above.

2.

VA , VB and VC represent the velocities of balls A and B with the balls hit the sides of the table. Then (B) VA  VBy

(A) VA  VB 3.

(D) VA  3  VC

(C) VA  3  VB

(D) VA  VB

Which of following is correct relation: (A) VBx  3  VC

4.

(C) 3  VB  VC

(B) VA  3  VC

Which of following equation is correct? (A) 20VC2  78 VB  72  0

(B) VC2  VB2  72  0

(C) 20VC2  78 VC  72  0

(D) 20 VC  78 VB  72 VA  0

PASSAGE – 2 Suppose a body of mass m 0 is placed on a smooth horizontal surface at rest. The mass of the body is decreasing exponentially with disintegration constant  . Assuming that the mass is ejected backwards with a relative velocity 5.

The mass of body at an instant t is



(A) m0 e t 6.

(B) m0 1  e



(C) m0 1   t 

(D) None

(B)

0

(C) m0  0  e t

(D) Zero

The acceleration of the body is (A)

8.

 t

The thrust force on the body is (A) m0 0

7.

0 .

0 

(B)

0 

(C)

0 t2

(D) Zero

The velocity - time graph is

(A)

v

(B) t

v

(C) t

v

(D)

v

t

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t

56

CENTRE OF MASS

9.

PASSAGE – 3 A multistage rocket consists of two stages. The first stage has a total mass of 12000 kg, of which 8000 kg is fuel. The total mass of second stage is 1000 kg, of which 600 kg is fuel. Assume that the relative speed V0 of ejected material is constant and neglect any effect of gravity. (The latter effect is small during the firing period if the rate of fuel consumption is large). Suppose the entire fuel supply carried by the two stage rocket is utilized in a single stage rocket of the same total mass of 13000 kg. In terms of V0, what is the speed of the rocket, starting from rest, when its fuel is exhausted? (A) 0.956 V0 (B) 1.87 V0 (C) 1.08 V0 (D) 3.74 V0

10. For the two stage rocket, what is the sped when the fuel of the first stage is exhausted if the first stage carries the second stage with it to this point? (A) 0.956 V0 (B) 9.56 V0 (C) 3.74 km/s (D) 1.08 V0 11. What is final speed of the second stage? (A) 1.08 V0 (B) 1.87 V0

(C) 9.56 V0

(D) 3.74 km/s.

12. What value of V0 is required to give the second stage of the above rocket a speed of 7 km/s. (A) 1080 km/s (B) 14 km/s (C) 374 km/s (D) 3.74 km/s PASSAGE – 4 A satellite travels through interstellar space. In do so, it picks up interstellar dust and its mass increases at a rate

dm   AV where A is satellite surface area, V its velocity and  the (constant) dust density. At dt

t = 0, the velocity is V0 and the mass is m 0. Assume that there are no external forces. 13. The acceleration of satellite as function of velocity is (A)

A 2 V m

(B) 

A 2 V m

(D) 

(C) g

 AV m

14. The velocity of satellite is

 1  At   (A)   m   V0

1



(B)  V0 



m    At 



(C)  V0 



m    At 

(D) None.

Level – 2

1.

Find the centre of mass of the system shown in figure.

2.

A uniform wire is bent into the form of a rectangle with length L and width W. If two of the sides coincide with the +x and +y axes, where is its center of mass ?

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57

CENTRE OF MASS 3.

The uniform solid sphere shown in the figure has a spherical hole in it. Find the position of its center of mass.

4.

A rigid body consists of a 3-kg mass connected to a 2-kg mass by a massless rod. The 3-kg mass is located at r1 = 2i + 5j m, and the 2-kg mass at r2 = 4i + 2j m. Find the length of the rod the coordinates of the center of mass.

5.

A straight rod of length L has one of its ends at the origin and the other at x = L. If the mass per unit length of the rod is given by Ax where A is a constant, where is its center of mass ?

6.

Consider the system composed of a 1-kg body and a 2-kg initially at rest at a center-to-center distance of 1m. All numerical values quoted in this exercise are to be considered exact, and the 2-kg body is to the right of the 1-kg body. (A) How far is the system’s center of mass from the center of the 1-kg body ? (B) Beginning at t = 0s, a net rightward force of 2 N acts on the 2-kg body. What is its resultant acceleration ? (C) How far does the 2-kg body move between t = 0s and t = 1s ? (D) How far is the center of mass from the 1-kg bosy at t = 1s ? (e) How far did the center of mass move between t = 0s and t = 1s ? (f) What is the acceleration of the center of mass, beginning at t = 0s? (g) Suppose that all the mass in both objects were concentrated at the center of mass, and the net rightward force of 2 N acted on this concentrated mass. What would its acceleration be ? (h) State the general theorem which is illustrated by the results of parts (f) and (g) . m

7.

A particle of mass m moving with a speed u strikes a smooth horizontal surface at an angle  . The particle rebounds at an angle  with a speed v. Determine an expression for v and  if the coefficient of restitution is e.

m





v

Smooth Surface m

8.

u

In the figure shown, a ball of mass m collides perpendicularly on a smooth stationary wedge of mass M. If the coefficient of restitution of collision is e (< 1) then determine the velocity of the wedge after collision.

v0



9.

A projectile is fired from a gun at an angle of 450 with the horizontal and with a muzzle speed of 1500 ft/sec. At the highest point in its flight the projectile explodes into two fragments of equal mass. One fragment, whose initial speed is zero, falls vertically. How far from the gun does the other fragment land, assuming a level terrain ?

10.

A machine gun fires 50-gm bullets at a speed of 1000 meters/sec. The gunner, holding the machine gun in his hands, can exert an average force of 180 nt against the gun. Determine the maximum number of bullets he can fire per minute ?

11.

A block of mass m 1 = 100 kg is at rest on a very long frictionless table, one end of which is terminated in a wall. Another block of mass m 2 is placed between the first block and the wall and set in motion to the left with constant speed v 2. as in fig. Assuming that all collisions are completely elastic, find the value of m 2 for which both blocks move with the same velocity after m 2 has collided once with m 1 and once with the wall. The wall has infinite mass effectively.

12.

A bullet of mass 10 gm strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance 12 cm. Assuming that bullet remains embedded in the pendulum, calculate the initial speed.

13.

An electron, mass m, collides head-on with an atom, mass M, initially at rest. As a result of the collision a characterstics amount of energy E is stored internally in the atom. What is the minimum initial velocity v 0 that the electron must have ?

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58

CENTRE OF MASS 14.

Mass m 1 collides head-on with m 2, initially at rest, in a completely inelastic collision. (A) What is the kinetic energy of the system before collision ? (B) What is the kinetic energy of the system after collision ? (C) What fraction of the original kinetic energy was converted into heat ? (D) Let v cm be the velocity of the center of mass of the system. View the collision from a primed reference frame moving with the center of mass so that v'1i = v 1i - v cm, v'2i = - v cm. Repeat parts (A), (B) and verted to heat the same in each case ? Explain.

15.

Two small blocks A and B of masses m 1= 0.5kg and m 2 = 1 kg respectively, each carryin positive charge of q = 40mC are kept stationary on a smooth horizontal floor. W hen the blocks are released, due to electrostatic repulsion, a moves towards left while

A

B

+

A 10 cm

+ 10 cm

10 cm

B towards right. After moving 10 cm, A comes into contact with a massless spring of force constant K = 6750 Nm-1 while after moving 20 cm, B collides nelastically with a rigid wall as shown in fig. Calculate (i) velocity of A when it comes in contact with the spring and (ii) maximum compression of the spring. 16.

A sphere of mass m falls with velocity v on a smooth inclined wedge of mass M and angle

 which rests on 2 mv cot  . m  M cos ec 2 

a smooth horizontal plane. Show that the velocity of the wedge immediately after collision is The bodies are all perfectly elastic.

B

A 17.

18.

Show that when two identical spheres A and B with a coefficient of restitution e = 1 collide while moving with velocity v A and v B which are perpendicular to each other they will rebound with velocities v´A and v´B which are also perpendicular to each other.



90°–  vB

vA

C

A 700 g ball B is hanging from an inextensible cord attached to a support C. A 350 g ball A strikes B with a velocity v 0 at an angle of 60° with the vertical. Assuming perfectly elastic impact (e = 1) and no friction, determine the velocity of each ball

60°

immediately after impact. Check that no energy is lost in the impact.

B

v0

A

19.

/2 Three spheres, each of mass m, can slide freely on a frictionless, horizontal surface. Spheres A and B are attached to an inextensible, C B v0 inelastic cord of length l and are at rest in the position shown when sphere B is struck squarely by sphere C which is moving to the right with a velocity v 0; knowing that the cord is slack when sphere B is  struck by sphere C and assuming perfectly elastic impact between B and C, determineA (A) the velocity of each sphere immediately after the cord becomes taut. (B) the fraction of the initial kinetic energy of the system which is dissipated when the cord becomes taut.

20.

A 1 kg block B is moving with a velocity v 0 of magnitude v 0 = 2 m/s as it hits the 0.5 kg sphere A, which is at rest and hanging from a cord attached at O. Knowing that µk = 0.6 between the block and the horizontal surface and e=0.8 between the block and the sphere, determine after impact (A) the maximum height h reached by the sphere, (B) the distance x traveled by the block.

O

A

h

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v0

x

59

CENTRE OF MASS 21.

A mass M = 1kg lies on a smooth horizontal base of a rough inclined plane at an angle 370 with the horizontal as shown in figure. A bullet of mass m=0.1 kg. is fired horizontally with a velocity u = 110m/s and gets embeded in it almost immediately. The impulse imparted carries the combined mass up the incline and finally lands on the horizontal level with the horizontal base. If the length of the incline is l = 1.8m and the incline offers a coefficient of kinetic friction 0.5 to the sliding of the mass, find the horizontal distance from the base of the incline to the point of landing of the combined mass. Assume the contact of the incline with the horizontal plane is smooth and mass is not jerked when starts up on incline plane. m

22.

A small ball of mass m is connected by an inextensible massless string of length l with an another ball of mass M = 4 m. They are released with zero tension in the string from a height h as shown in figure. Find the time when the string becomes taut for the first time after the mass M collides with the ground. Take all collisions to be elastic.

 M

F

23. The figure shows the force versus time graph for a particle. (i) (ii)

Find the change in momentum p of the particle. Find the average force acting on the particle.

100 N

O

0.2

0.4 t(s)

24.

While waiting in his car at a stoplight, an 80-kg man and his car are suddenly accelerated to a speed of 5 m/s at the result of a rear-end collision. Assuming the time taken to be 0.3 s, find (A) the impulse on the man and (B) the average force exerted on him by the back of the seat of his car.

25.

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in the fig. The horizontal speed of the water is 80 cm/s and 30 cm 3 of the water hits the plate each second. Assume that the water moves parallel to the plate after striking it. One cubic centimeter of water has a mass of one gram.

26.

Three identical particles A, B and C lie on a smooth horizontal table, Light inextensible strings which are just taut connect AB and BC and  ABC is 1350. An impulse J is applied to the particle C in the direction BC. find the initial speed of each particle. The mass of each particle is m.

Level – 3 1.

A cylinderical solide of mass 10-2 kg and cross-sectional area 10-4 m 2 is moving a parallel to its axis (the X-axis) with a uniform speed of 103 m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with stationary dust particles of uniform density 10-3 kg/m 3. When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimesions of the cylinder remain partically unchanged, and that the dust sticks only to the front face of the cylinder, find the x-coordinate of the cylinder at t = 150 s. [JEE 1993]

2.

A small sphere of radius R is held against the inner surface of larger sphere of radius 6R (as shown in the fig.). The masses of large and small spheres are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the large sphere, when the smaller sphere reaches the other extreme position. [JEE 1996]

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60

CENTRE OF MASS 3.

A large open top container of negligible mass and uniform cross-sectional area A has a small hole of croos -sectional are A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density  and mass m 0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate (i) the acceleration of the container, and (ii) its velocity when 75% of the liquid has drained out. [JEE 1997]

4.

A particle of mass m 1 is moving in a circular path of radius R with a constant speed v 2 is located at point (2R,0) at time t = 0 and a man starts moving with a velocity v 1 along the positive y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. the man as a function of time. [JEE 2003]

5.

Two point masses m 1 and m 2 are connected by a spring of natural length l 0. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity v 0 along positive x-axis. When the system reaches the origin the strings breaks (t = 0). The position of the point mass m 1 is given by x1 = v 0t - A (1-cos t) where A and  are constants. Find the position of the second block as a function of time. Also find the relation between A and l 0.

A cart is moving along +x direction with a velocity of 4m/s. A person in the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in y – z plane making an angle of 30º with the vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine (a) the speed of the combined mass immediately after the collision with respect to an observer on the ground. (b) the length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass. [JEE, 97] 6.

7.

Two blocks of mass 2kg and M are at rest on an inclined plane and are separated by a m M distance of 6.0 m as shown. The coefficient of friction between each of the blocks and 6.0 2kg the inclined plane is 0.25. The 2kg block is given a velocity of 10.0 m/s up the inclined plane. It collidies with M, comes back and has a velocity of 1.0 m/s when it reaches its initial position. The other block M after the collision moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M. [Take sin  tan = 0.05 and g = 10 m/s2] [JEE, 99]

8.

A car P is moving with a uniform speed of 5(31/2)m/s towards a carriage of mass 9 kg at rest kept on the rails at a point B as shown in fig. The height AC is 120 m. Cannon balls of 1 kg are fired from the car with an initial velocity 100 m/s at an angle 30º with the horizontal. The first cannon ball hits

9.

P

C

the stationary carriage after a time t0 and sticks to it. Determine t0. At t0, the A B second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to the carriage. What will be the horizontal velocity of the carriage just after the second impact ? [JEE, 01] There is a rectangular plate of mass M kg of dimensions (a × b). The plate is held b in horizontal position by striking n small balls each of mass m per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding a elastically with velocity v. What is v ? [JEE, 06] It is given n = 100, M = 3 kg, m = 0.01 kg; b = 2m; a = 1m; g = 10 m/s2

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61

CENTRE OF MASS

Answer Key Assertion-Reason Type Q. Ans.

1 B

2 C

3 D

4 A

5 A

6 B

Match the Column 1. 2. 3. 4. 5.

[(A — Q), (B — P), (C — R), (D — S)] [(A — R), (B — P), (C — S), (D —Q)] [(A — Q), (B — P), (C — R), (D — S)] [(A — Q), (B — P), (C — S), (D — R)] [(A — P), (B — Q), (C — R), (D —R)]

Level – 1 Q. A ns. Q. A ns. Q. A ns. Q. A ns. Q. A ns.

1 B 10 C 19 AB 28 C 37 AC

2 A 11 B 20 A CD 29 C 38 C

3 C 12 BD 21 B 30 A 39 C

4 D 13 CD 22 A 31 A 40 A

5 A 14 C 23 B 32 B 41 A

6 B 15 A 24 C 33 C

7 CD 16 C 25 C 34 A

8 A 17 A BCD 26 A 35 C

9 B 18 B 27 C 36 A

Passage Type Problems Q. Ans. Q. Ans.

1 D 10 A

2 B 11 B

3 A 12 D

4 C 13 B

5 A 14 A

6 C

7 A

8 A

9 C

Level – 2 1. 0.167 L

2.

L W , 2 2

3.

a 3b R  a3 3

4. 3.71 m

5.

2 L 3

6. (a) 2/3 m, (b) 1 m/s2 (righwards) (c) 0.5 m (d) 1 m (e) 1/3 m (f) 2/3 m/s2 rightwards (g) 2/3 m/s2 (right wards) (h) 2/3 m/s2 (right wards

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62

CENTRE OF MASS





7.   tan 1 e tan  , v  u cos 2   e 2 sin2 

10. 216 bulltes/min.

14. (a)

8.

11. m 1/3

1  e mv 0 sin 

9. 1.1 × 105 ft

M  m sin2 

 (M  m)  13. 2 E  Mm  

12. 310 m/s

1/ 2

2 m2 1 m1v 1i 1 1 1 2 2 2 m1v 12i (b) (c) m  m (d) (m1 v i  v 1i v cm  v cm )  m 2 v cm , zero, 10%, no 2 (m1  m2 ) 2 2 2 1 2

15. (i)12m/s (ii) 10cm.

17. 25m/s,

36.9°, 43.3 m/s

53°.1

18. v´A = 2.73 m/s; v´B = 3.78 m/s 19. (a) VA’ = V0 (1 – e)/2 , VB’ = V0 (1 + e)/2 (c) VC’ = V0 (1 + e)n – 1/2n – 1 (d) 0.881 V0

(b) VB’ = V0(1 – e2 ) 4, VC’ = V0 (1 + e)2/4

 20. (a) 0.294 m (b) 0.0544 m

24. 1330 N

21. 8.787 m.

25. 0.024 N

26.

22.

2gh

23. (i) 20 N-S

(ii) 50 N

3 2 10 J, J, J 7 m 7m 7m

Level – 3 1. 105m

2. (L + 2R, 0)

3. (i) g/50 (ii)

 v2  ˆ  v2  ˆ 4. m   v 2 sin R t  i   v 2 cos R t  v 1  j      

 m1  m1 5. x 2 = v 2t + m A(1 – cos  t),  0   m  1 A 2  2 

6. 2.5 m, 0, 319 m

8. 12 sec., 15.75 m/s

7. 0.84, 15.12 kg

m0 g 2A

9. 10 m/s

—X—X—X—X—

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63

FLUID MECHANICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

FLUID MECHANICS DEFINITION OF A FLUID The substances which flow are called fluids. Both liquids and gases come in this category. The science of fluids at rest is called fluid statics, while that of moving fluids hydrodynamics. Fluid statics involves hydrostatics pressure, floatation, Pascal’s law and Archimedes’ principle. Hydrodynamics involves continuity equation, Bernoullis’s principle and Torricelli’s theorem. DENSITY AND PRESSURE With fluids, we are more interested in the extended substance rather than small piece of the substance and in properties that can vary from point to point in that substances. It is more useful to speak of density and pressure. Density Density () of any substance is defined as the mass per unit volume or Mass = Volume or  = m/V Regarding density following points are worth noting : C1: Density of a body means the ratio of mass of a body to the volume of the body while density of a substance means the ratio of mass of substance to the volume occupied by the substance. Thus, for a solid body, Density of body = Density of substance While for a hollow body, density of body is lesser than that of substance because volume of hollow body is greater than volume of substance. C2: In case of a liquid, sometimes an another term relative density (RD) is defined.

Density of substance Density of water at 4º C RD is a pure ratio. So, it has no units. It is also some times referred as specific gravity. RD =

Note : Density of water at 4ºC in CGS is 1 g/cm3. Therefore, numerically the RD and density of substance (in CGS) are equal. In SI units the density of water at 4ºC is 1000 kg/m3. C3: If two liquids of densities 1 and 2 having masses m1 and m2 are mixed together, then the density of mixture will be =

=

If m1 = m2,

=

(m1 + m 2 ) Total mass = (V1 + V2 ) Total volume

(m1 + m 2 )  m1 m 2       1 2 212 = Harmonic mean 1  2

C4: If two liquids of densities 1 and 2 having volumes V1 and V2 are mixed, then the density of the mixture is, =

(m1 + m 2 ) Total mass = (V1 + V2 ) Total volume

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1

FLUID MECHANICS =

1V1  2 V2 V1  V2

1  2 = Arithmetic mean 2 C5: If the temperature of a liquid is increased, the mass remains the same while the volume is increased, then

If V1 = V2,

=

1  density of the liquid decreases     . Thus, V

´ V   V´ ´ V V    V  V V  V ´ 1   1   Here,  = thermal coefficient of volume expansion and  = rise in temperature or

   1   1   C6: With increase in pressure due to decrease in volume, density will increase, i.e., 

´

´ V V V     V´ V  V V   P / B V

´ 1   1  P / B Here, P = charge in preseure and B = bulk modulus of elasticity of the liquid or

Therefore,

´

  P    1  1  P / B  B 

Example 1. A king orders for a gold crown having a mass of 4 kg. When it arrives from the goldsmith, the volume of the crown is found to be 100 cm3. What is your opinion about the crown if the density of pure gold is 8 × 103 kg/m3 ? Sol. The density of crown mass 4  volume 100  106 crown = 4 × 104 kg/m3 which is just half that of pure gold. There are two possibilities about this deviation : (a) The crown contains impurities of light metals decreasing its mass for a given volume and so the density. (b) The crown contains cavities increasing its volume for a given mass and so decreasing the density.

crown =

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2

FLUID MECHANICS C7:. Relative density of an oil is 0.8. Find the obsolute density of oil in CGS and SI units. Sol. Density of oil (in CGS) = (RD) g/cm3 = 0.8 g/cm3 = 800 kg/m3 Example 2. When equal volumes of two metals are mixed togethers, the specific gravity of alloy is 4. When equal masses of the same two metals are mixed together, the specific gravity of the alloy is 3. Calculate the specific gravity of each metal. Sol. In case of two metals  m  m2   1  V1  V2  So, when equal volumes are mixed, V1 = V2 = V and m1 = V1 and m2 = V2, then 

V1  V2 VV

1  2  4 ...(1) 2 When equal masses are mixed, m1 = m2 = m and V1 = (m/1) and V2 = (m/2), then mm  m / 1  m / 2

or

or

212 3 1  2

...(2)

Solving Eqs. (1) and (2) for 1 and 2, we find specific gravities of metals are 2 and 6. CONCEPT OF PRESSURE The word ‘pressure’ is linked with ‘pressing’. Two surfaces in contact exert on each other ‘normal contact force’. With this force they press each other. w

w

A´

A´ N

w N

.... .... A

N

w

.... .... N

A´

In the figure an anvil of weight W is placed on a horizontal surface making contact in area A or A´. The free body diagram shows that contact force N is equal to W, because anvil is not accelerated along vertical. N=W In the two cases surface have different intensity of pressing. How much intense is the pressing ? This is described by a term pressure. In the figure same contact force N is distributed over large area A´ in one case and smaller area A is other. We define pressure as the magnitude of normal contact force per unit area. Thus, pressure for area A is  |N| P A and that for area A´ is

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3

FLUID MECHANICS  |N| P´ A´ We define presure at a point by taking a small area A around the point. The magnitude of normal contact   force over this area is |  N | . Then pressure at the point is limiting value of |  N | / A | , where A tends to zero. That is,  | N| P  lim A  0 A dN P dA SI. Unit : The unit of pressure is just unit of force divided by unit of area; Nm–2 in SI. We cal 1 Nm–2 = 1 Pa (pascal).

Dimensional Formula : [MLT–2/L2] = [ML–1 T–2] Example 3. A sample of metal gasket is placed between flat narrow ends of two conical diamond anvils and compression of 50 kN is applied. The diameter of the end faces is 5.0 × 10–3 m., Calculate how much pressure is existing on the sample. (Metal Gasket)

50 kN

50 kN

Diamond Cone

Sol. As anvils are unaccelerated, the contact force N is equal to the applied force F. Thus N = F = 50 kilo newton. A N N

F

A acceleration = 0

This force is distributed over the contact area. The gasket is slightly protruding outside hence the whole end area of anvils are in contact. Thus, area A of contact is given by A = D2/4 3.141 A= × (5.0 × 10–3)2 m2 4 Pressure existing at the face of sample is given by P

N A

4  50  103 N (3.141)  (5.0  10 3 ) 2 m 2 P = 2.5 × 1011 Pa. P

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4

FLUID MECHANICS PRESSURE DUE TO FLUID COLUMN (AT REST) One of the interesting things is the pressure at the surface of the earth exerted by air surrounding it. This is known as atmospheric pressure at the surface. How does it arise ? It is due to collision of gas z (up) molecules (see kinetic theory of gases). We can, however, find its value by considering statics of a fluid column at rest. (P+dP)A We know that the magnitude of normal contact force per unit area of contact surface is equal to pressure. dz PA Hence we calculate the contact force by assuming z g Adz vertical acceleration of air column to be zero. This is z=0 known as hydrostatic condition in an inertial frame. acceleration = 0 The figure shows an air column of cross-sectional area A, extending vertically along Z-axis. At a height z we take an elementary height dz. The volume of this elements is A dz. multiply it by density of air at this level, , and gravitational field intensity g. We get pull due to the Earth. Let P be pressure at z and P + dP at z + dz. The free body diagram of the fluid element is shown. We have PA = (P + dP) A + Adz g 0 = dP + dz g dP  g dz This is the fundamental equation of lapse rate of pressure in fluid of density . There are several implications of this equation. Z C8: If we move along z = constant in a homogeneous fluid (horizontal level) we get dP = 0 or P = constant. Thus pressure at points in the same horizontal level is the same in a homogeneous fluid at rest in an inertial frame A B C9: If we move along a path passing through heterogeneous fluid, pressure in the 1 same horizontal level is not the same. In the figure pressure at A is not equal to g 2 pressure at B, because the path ACB is passing from A to C and C to B through liquids of different densities (hetergeneous). C C10: In the same liquid, the pressure decreases with height because dP/dZ is negative. In other words, pressure increases with depth in a fluid. Z-axis P0 g

Z0

Z0 – h Z=0

To see the above fact, let P0 be pressure at the surface of a liquid of density . The pressure at depth h may be found using the equation dP/dz = –g. We have, dP = – g dz.

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FLUID MECHANICS Let us take z-coordinate at the top as z0. At z = z0 (top of liquid) the pressure is due to air. Let it be denoted by P0. Then at depth h, the z-coordinate will be h less than z0, i.e., z0 – h. Hence integrate L.H.S. from P to P0 and R.H.S. from z0 – h to z0. That is, P0

z0

 dP   g 

dz

z0  h

P

P0 – P = – .g. [z0 – (z0 – h)] P = P0 + gh This formula can be more easily derived as below. P0 A (force due to pressure)

A

Ah g(weight of column)

PA(force due to pressure)

Consider the vertical equilibrium of fluid column of height h. acting on if forces the area as shown in the figure. P0A + A h g = PA P = P0 + hg As we go in the depth, pressure increases. Here P0 is atmospheric pressure. Using this formula we cannot know P without knowing P0. A barometer does the job of measuring atmospheric pressure P0. C11: A U-tube of uniform cross section (see Fig.) is partially filled with a liquid I. Another liquid II which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are the same, while the level of liquid I has risen by 2 cm. If the specific gravity of liquid I is 1.1, the specific gravity of liquid II must be (A) 1.12 (B) 1.1 (C) 1.05 (D) 1.0 Sol. [B] The level of liquid is same in both the limbs. Pressure in limbs I at B = Pressure of limb II at A h A

P2 II

h

II

I

P1 I

B

h1g = h2g 1 = 2

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FLUID MECHANICS Variation of pressure at two points when fluid is in horizontal acceleration : When a fluid is subjected to a constant acceleration, different elements of the fluid orient themselves so as to obtain a new equilibrium position with respect to the vessel. Since the fluid particles are relatively at rest with respect to one another, therefore, the laws of fluid statics area applicable but in the modified form to take into account the effect of acceleration. l Consider a horizontally accelerating container with liquid in it. Take two points A and B at a horizontal distance l (along the line of acceleration a). hA hB Take a horizontal cylinder within the two areas S containing points A and a A B B. Let pressure at A is PA & at B is PB. S

The forces along the line AB are PA S towards right and PB S towards left. Under the action of these forces the element is accelerating. thus, PA S – PB S = ma or (PA – PB) S = (S)l a or P1 – P2 = la i.e., two points in same horizontal level do not have equal pressure for a fluid accelerating horizontally. If atmospheric pressure is P0 then, PA = P0 + hA g and PB = P0 + hB g, as there is no vertical acceleration. hA and hB are the depths of points A & B from free surface.  h1 g – h2 g = l a or

h1  h 2 a  l g

or

tan  

a g

where is the inclination of the free surface with the horizontal. Example 4. A vessel containing water is given a constant acceleration a towards the right along a straight horizontal path. Which of the following diagrams in Fig. represents the surface of the liquid ?

(A)

a

a

(B)

(C)

a

(D) none of these

Sol. [C] A (dx)

B

D

C

Let us consider a small dotted segment of thickness dx for observation. Since this segment is accelerated towards right, a net force is acting in this segment towards right from the liquid towards the left of ABCD. According to Newton’s third Law, the segment ABCD will also apply a force on the previous section creating a pressure on it which makes the liquid rise.

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FLUID MECHANICS Variation of pressure at two points when fluid is in vertical acceleration: Consider point A and B inside liquid with a vertical separation h where the liquid is accelerating upward. Consider a vertical cylinder of length h with boundaries at A & B. S is the surface area of flat boundaries. Let PA & PB are the pressures at A & B respectively. Thus, forces acting on the cylinder are PA S upward PB S downward and W = (S)hg downward. Under the action of these forces liquid cylinder is accelerating upward with acceleration ‘a’. Thus, PA S – PB S – mg = ma or (PA – PB) =  (g + a) h Special cases : (i) If a is (–)ve i.e. the vessel is accelerating downward then, (PA – PB) = (g – a) h (ii) If a is zero then pressure is equal everywhere inside the liquid. (iii) If a is greater than g then fluid occupies the upper part of the container.

a

B h A

Example 5. A bucket contains water filled upto a height = 15 cm. The bucket is tied to a rope which is passed over a frictionless light pulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket + water). The water pressure above atmosphere pressure at the bottom is (A) 0.5 kPa (B) 1 kPa (C) 5 kPa (D) None of these Sol. The acceleration of bucket is 2mg  mg g a  downward 3m 3  P = P0 + (g – a) h 10   P  P0  1000  10    0.15   3 20  0.15 3 P – P0 = 103 N/m2 = 1 kPa P  P0  1000 

Hence option (B) is correct.

Variation of pressure when fluid is under both horizontal and vertical acceleration : When a fluid is accelerating in arbitrary direction then pressure varies both horizontally and vertical separately for horizontal and vertical component of acceleration respectively.  Here, a  a ˆi  a ˆj and a = a cos  & a = a sin  x

y

x

y

Now, difference in pressure at two points in same horizontal level at a separation l is given by

ay

a ax

P  l a x  l a cos 

And similarly for two points in same vertical line with a separation h, the pressure difference is P  (g  a y )h  (g  a sin )h Inclination of the free surface with respect to the horizontal is given by ax a cos  tan   tan   ; g  ay g  a sin 

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FLUID MECHANICS Example 6. A trolley containing a liquid of density slides down a smooth inclined plane of angle with the horizontal. Determine the angle of inclination of the free surface with the horizontal. Sol. The acceleration of the trolley along the inclined plane is a = g sin  ax = a cos  = g sin  cos  ay = a sin  = g sin2  The inclination of the free surface is ax g sin  cos  tan     tan  g  a y g(1  sin 2 )

= ax ay

a=gsin

or = Note : That the free surface becomes parallel to the inclined plane. It is because the net effective gravity inside the trolley is perpendicular to the inclined plane. Example 7. A fluid container is containing a liquid of density is accelerating upward with acceleration a long the inclined place of inclination  as shown. Then the angle of inclination of free surface is : a  1  (A) tan    g cos  

1  a  g sin   (B) tan    g cos  

1  a  g sin   (C) tan    g(1  cos  )  Sol.  N cos  = mg cos  and N sin  = ma + mg sin 



tan  

a

1  a  g sin   (D) tan    g(1  cos  ) 

N cos

a  g sin  g cos 

N

N sin

ma+mg sin mg cos

 a  g sin     tan 1    g cos   Hence option (B) is correct.



Variation of pressure when a fluid is rotating : When a liquid container is rotating with constant angular velocity, liquid particles orient themselves to a new equilibrium position at which no relative motion exists between different parts of the liquid. Consider the container is rotated about its axis with constant angular speed . Consider a small fluid element which is under the action of two forces, horizontal centrifugal reaction due to rotation and down-ward force due to gravity. The slope of the free surface at the position of element is

dy 2r  tan   dr g

or

 2r dy  dr g

2 rm

mg

y O

r

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FLUID MECHANICS

2 2 r Integrating we get, y = y0 + 2g where y0 is the integration constant which depends upon the initial values. Let R is the radius of the cylinder and h0 is the height of the liquid before the start of rotation then

2R 2 r = R, y = ymax = y0 + 2g

at

at r = 0, y = ymin = y0 Here original height of the liquid is the mean of maximum and minimum heights.

y max  y min  2 R 2   y0 2 4g

So,

h0 

or

2R 2 y0  h 0  4g



y max  h 0 

2R 2 4g

y min  h 0 

2R 2 4g

2R 2 2r 2  and 4g 2g Note : The above equation shows that the free surface of liquid is that of a paraboloid. y  h0 

Example 8. A cylinder of radius R = 1 m and height H = 3 m, two-third filled with water, is rotated about its vertical axis, as shown in figure. Determine the speed of rotation when (i) The water just starts spilling over the brim, (ii) The point at the centre of the base is just exposed. Sol. (i) We know that ymax = H = 3 ; H = 2/3 H = 2 m; R = 1 m; g = 10 ms–2.



Thus

H h=(2/3)H

R

4g(H  h) 4(10)(3  2)  2 R 12

  40 rad s 1

y=H

(ii) We know that

y Here,

r 2  y0 2g

R

y = H = 3m; y0 = 0; r = R = 1m g = 10 ms–2



2gH 2(10)(3)  ; 2 R 12

  60 rad s 1

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FLUID MECHANICS Example 9. Figure shows a three arm tube in which a liquid is filled upto levels of height l. It is now rotated at an angular frequency  about an axis passing through arm B. The angular frequency at which level of liquid in arm B becomes zero. (A)

2g 3l

(B)

g l

(C)

3g l

(D)

A

B

C

l

3g 2l

l l

l

3g l  3l  P  P0  g    P0   2 2

Sol.

l

 P  P0  S   S l  2  2 3g l S S l 2 2   2 2 3 l 2 g 2 2



2 

3g l



3g l

Hence option (C) is correct.

a = 6 cm

Example 10. Length of a horizontal arm of a U-tube is l = 21 cm and ends of both of the vertical arms are open to surroundings of pressure 10500 Nm–2. A liquid of density  = 103 kg m–3 is poured into the tube such that liquid just fills horizontal part of the tube. Now, one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity 0 = 10 radians/sec. l = 21cm If length of each vertical arm be a = 6cm, calculate the length of air column in the sealed arm. Sol. When tube is rotated, liquid starts to flow radially outward and air in sealed arm is compressed. Let the shift of liquid be x as shown in figure. Let cross-sectional area of tube be S. Initial volume of air, V0 = Sa and initial pressure P0 = 10500 Nm–2 Final volume, V = S (a – x) a–x  Final pressure, P 

P0 V0 P .a  0 V (a  x)

A

x

B (l – x) P0 a x or pressure at B, P2 = P + xg = + xg (a  x) Centripetal force required for circular motion of vertical column of height x of liquid is provided by reaction of the tube while that to horizontal length (l – x) is provided by excess pressure at B. Force exerted by pressure difference is

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FLUID MECHANICS  P0 x   xg  S F1 = (PB – PA) S = (P2 – P0) S =   (a  x)  Mass of horizontal arm AB of liquid is, m = S(l – x)

Radius of circular path traced by its centre of mass is r = x +  Centripetal force,

l  x  l  x   2  2

F2 = m02r

  l  x  2  P0 x  xg  S  0   F2 = F1 or {S (l – x)}   2   (a  x)  or x = 0.01 m = 1 cm  Length of air column in sealed arm = (a – x) = 5 cm.

But

BAROMETER We can measure atmospheric pressure P0 using a liquid column in a tube. This arrangement is shown in the figure and is called barometer. (There are several other modifications). Mercury is filled in a tube of length about 1m. Its mouth is covered (by thumb, say) and inverted into a vessel containing mercury. It is held in position by a stand and mouth is now uncovered. Z

P1A

z0 h

h

Z0 – h

A gh

P0 A

It is observed that some mercury comes out in the vessel. A vacant space is created at the top (closed) end. This space is historically called Torricelli’s vacuum, although same mercurry vapour is present in it. If pressure at the top of mercurry column be P1, we have, for equilibrium of column, P1A + A 0gh = P0 A. Here P1 is mercurry vapour pressure which we generally ignore. Then P0 = 0gh The atmospheric pressure is, thus, measured if h be measured. The normal density (0) of mercury is 13.6 g cm–3 and g = 981 cms–1. For a given place and temperature, these data are fixed. Hence directly h can be used to show pressure. One atmospheric pressure is a unit of pressure equal to 76 cm of Hg. Example 11. A mercury barometer reads h0 as pressure of air in a closed lift at rest. If the lift accelerates up at g/2, what will be : (a) atmospheric pressure in the lift; and (b) the reading of the barometer ? Sol. (a) The air enclosed in the lift will have the same temperature and volume. Its pressure will be the same as P0 = h0g. Here is the density of mercury in barometer.

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FLUID MECHANICS (We shall see in the kinetic theory of gases that pressure is arising due to collision of molecules.) (b) When the lift accelerates, mercury column also accelerates. Let h be the length of mercury column now, and a (= g/2) be the acceleration. Then, using Newton’s second law of motion, P0A – h Ag = (hA)a = hAg/2 h Ag a P0 = 3/2 hg Now we compare it with h0. P0 = gh0 P0 A  h0 = 3/2h h = 2/3 h0 (reading) The reading of barometer falls, although pressure in the lift remains the same. Gauge Pressure : The pressure in excess of atmospheric pressure is called gauge pressure. P – P0 = gh is the gauge pressure. It is measured by a manometer. P0

h g P

manometer

If the pressure be lower than atmospheric, we define vaccum pressure as the deficiency in pressure. Variation of pressure in air with height The pressure decreases as we go up in the atmosphere as the temperature and density of air at different heights is not the same. A rough estimate shows that at constant temperature, pressure falls exponentially with height. Other pressure units : 1.013 × 105 Pa = 1 atm. 105 Pa = 1 bar 1.013 bar = 1 atm. 1 mm of Hg = 1 torr = 133 Pa. Example 12. Find the absolute and gauge pressures of the gas Gas 1.5 cm above the surface of the oil in the tank shown in the figure. 1.0 cm 3 Take the density of oil as 800 kg/m , density of mercury as Oil 2.0 cm 13.6 × 103 kg/m3, atmospheric pressure as 1.0 × 105 A B 2 2 N/m and g = 10 m/s . Sol. As the points A and B are at the same horizontal level in mercury, the pressure at these points is the same. Now, if P is the absolute pressure of the gas and P0 is the atmospheric pressure, then pressure at A is

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FLUID MECHANICS PA = P + (1.5 + 2.0) × 800 × 10 PA = P + 0.28 × 105 N/m2 Similarly pressure at B is PB = P0 + (1.0 + 2.0) × 13.6 × 103 × 10 PB = P0 + 4.08 × 105 N/m2 Using PA = PB´ Gauge pressure we get P – P0 = 3.8 × 105 N/m2 and Absolute Pressure P = (1.0 × 105 + 3.8 × 105) Pa P = 4.8 × 105 Pa. PASCAL’S LAW (i) The pressure in a fluid at rest is the same at the same height (or depth relative to top surface). h A

B

Let two points A and B at the same depth h from the top surface. We consider a narrow tube of fluid AB between points A and B. Let  be its cross sectional area. Force at end A is PA  and that at end B is PB. Since the tube is not accelerating horizontally,

PA

PB

these forces must be balanced. Thus, PA  = PB  PA = PB. This proves the Pascal’s observation. (ii) The change in pressure at any point in a closed liquid is totally transmitted to all the points of the liquid. We can show this by using P = P0 + gh. Let P0 be changed by p. Then, the pressure at depth h is given by P´ = (P0 + p) +  gh.  P´ – P = p. This does not contain h. Hence the change in pressure at the top is transmitted to all the points totally. (iii) Pressure at a point does not depend upon orientation of the surface (area) used to define it. To show this we take a small prism of liquid around the observation point O. Let N1, N2 and N3 be normal forces acting on faces of the prism having area A1, A2 and A3. Now

N1 N2 N3 P1 = A , P2 = A and P3 = A . 1 2 3

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FLUID MECHANICS Since the prism is at rest,    N1  N 2  N3  0 Resolving forces, N2 = N1 sin  N3 = N1 cos  Also A1 cos  = A3 A1 sin  = A2 N1  P1 = A 1 N2 N1 sin  N1 P2 = A = A sin  = A 2 1 1

N1 N2

...(1)

A2 O A3

A1 N3

...(2)

N1 cos  N1 N3 P3 = A = A cos  = A 1 1 3

Thus, P1 = P2 = P3. This also distinguishes pressure from normal stress. Although they are dimensionally the same, normal stress changes by changing the orientation of surface whereas pressure remains the same. Pressure is a scalar quantity. APPLICATIONS OF PASCAL’S LAW Since the change in pressure at any point in a liquid is transmitted to all points of the liquid, we can obtain larger force by using larger contact area. This is the principle of hydraulic lift or hydraulic brakes. Hydraulic Machines Figure shows a horizontal cylinder with a piston and three vertical tubes. Liquid has the same level in all the three tubes. This shows that pressure is the same at their bases. Now we push the piston and wait. We find that the level is raised but by the same height in the all the tubes. Thus the change in pressure has been transmitted to all the points without any decrease. (This is Pascal’s law of transmission of pressure).

This is the principle of hydraulic machines. Example : Hydraulic lift Figure shows two pistons of areas A1 and A2. They are connected by a liquid. If we apply a force F1 a change in pressure F1/A1 is produced. This, according to Pascal, is transmitted to all the points of the liquid, and hence also at the base of right piston (of area A2). Corresponding to the change in pressure, there is produced a force F2, acting over area A2, such that F2 = (change is pressure) × A2

 F1  F2 =  A  A2.  1

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FLUID MECHANICS F1 A1 A2 F2

As A2 > A1, the force F2 is larger than the applied force F1. Mechanical advantage of a machine is defined as the ratio of load (F2) to effort (F1). There (A2/A1) is equal to (F2/F1), the mechanical advantage of hydraulic lift. the large force F2 may be used to press a drum of rotating wheel, acting as brake. This is the principle of a hydraulic brake. The force F2 may be used to press big bundles into small for typing ‘Brahma press’ is such a device. Example 13. The area of cross-section of the wider tube shown in figure is 800 cm2. If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is : (A) 10 cm (B) 6 cm (C) 15 cm (D) 2 cm Sol. Pressure in horizontal pipe at both sides are same. mg  gH  P0 + gh + gH  P0  A mg gh   A mg m 12 h    gA A 1000  800  104 

h

12 3 m  100  15 cm 80 20

12 kg

Hence option (C) is correct.

FORCES ACTING ON SOLID BOUNDARIES (a) Whenever a fluid comes in contact with solid boundaries it exerts a force on it. (b) This force on a particular bounary may be obtained by integrating the pressure over the entire area of the boundary. (c) The variation of liquid pressure acting at base, vertical wall and inclined wall are shown in the figure. Force at the base : As the pressure is uniform at the flat base, therefore, force acting at the base is F = P × (area of the base) F = gh (lb) 

P = gh

F  g (hlb)

where is the density of liquid, h is the height of the liquid column, l & b are the length and breadth of the base.

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FLUID MECHANICS C12: A beaker of circular cross-section of radius 4 cm is filled with mercury upto a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmmospheric pressure = 105 N/m2. Density of mercury = 13600 kg/m3. Take g= 10 m/s2. Sol. The pressure at the surface = atmospheric pressure = 105 N/m2. The pressure at the bottom. 105 N/m2 + hg kg   m   = 105 N/m2 + (0.1 m)  13600 3   10 2  m s 5 2 2 = 10 N/m + 13600 N/m = 1.136 × 105 N/m2. The force exerted by the mercury on the bottom = (1.136 × 105 N/m2) × (3.14 × 0.04 m × 0.04 m) = 571 N. Example 14. The vessel shown in the adjoining fig. is filled with water of density w = 103 kgm–3. 5 cm (i) Find the net force acting at the base of the vessel. (ii) What is the weight of water in the vessel ? Sol. (i) Pressure at the base of the vessel is 1 cm P = w gh = (103) (10) (6 × 10–2) = 600 Pa A2 = 50 cm  F = A = (600) (50 × 10–4) = 3 N (ii) Weight of water in the tube is W = wg (A1h1 + A2h2) or W = (103) (10)[(10 × 10–4) (5 × 10–2) + (50 × 10–4) (1 × 10–2)] W=1N

= 10 cm2

Example 15. A U–tube having horizontal arm of length 20 cm, has uniform cross-sectional area = 1 cm2. It is filled with water of volume 60 cc. What volume of a liquid of density 4 g/cc should be poured from one side into the U–tube so that no water is left in the horizontal arm of the tube ? (A) 60 cc (B) 45 cc (C) 50 cc (D) 35 cc Sol. P0 + l gh0 = P0 +  gh 4 h0 = 60 60 h0   15 cm 4  Volume of liquid pushed = A (h0 + 20) = 1 (15 + 20) = 35 cm3. Hence option (D) is correct.

Force on the vertical wall Pressure acting on the vertical wall is not uniform but increases linearly with depth. Let us take an element dy of the boundary at a depth y from the free surface.

h0

h

y Py = gy h

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Ph = gh

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FLUID MECHANICS  Pressure at depth y, Py = gy Force acting on the element dy of breadth b is dF = Py (bdy) = gb ydy  Total force on the wall, 1 gbh 2 0 2 Thus, total force acting per unit width of the vertical wall h

F   dF   gbydy 

F 1  gh 2 b 2 The point of application (or centre of force) of the total force from the free surface is

he

h

h

h

0

0

0

2

 ydF   ygbydy   y dy   dF  gbydy  ydy h

h

h

0

0

0

1 3 h 2 he  3  h 1 2 3 h 2 2 h 3 i.e., the total force acts at a depth of 2/3 h from the free surface. 

he 

Force on an inclined wall : We know pressure always acts normal to a surface, and it depends upon the depth y of the point from the free surface. The horizontal force Fx acts on the vertical projection of the inclined wall

h h/sin

1 gbh2 P = gh 2 And the vertical force Fy acts due to the weight of the liquid supported by the wall,

Fx =

Fy =

1 1 gb (h) (h cot ) = gbh2 cot  2 2



F  Fx2  Fy2 

or

F

or

1 gbh 2 1  cot 2  2

1 1 gbh 2 . 2 sin 

F 1 h2  g b 2 sin 

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FLUID MECHANICS Example 16. Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let F2 be the force applied by the liquid on the bottom of this new vessel. (Neglect atmosphere pressure) Then : (A) F1 = F2 Sol.

(B) F1 = F2/ 2

(C) F1 =

 F2

(D) F1 = F2

2

V = R h = R H H = h F1 = gh R2 F2 = gH R2 = g R2h = F1 Hence option (D) is correct.

Example 17. A liquid of mass 1 kg is filled in a flask as shown in figure. The force exerted by the flask on the liquid is (g = 10 m/s2) [Neglect atmospheric pressure] : (A) 10 N (B) greater than 10 N (C) less than 10 N (D) zero Sol. Fy Fx Fx

mg

Force diagram of liquid

ghA

Fy + gh A = mg = 10 N

Hence option (A) is correct.

Example 18. A heavy hollow cone of radius R and height h is placed on a horizontal table surface, with its flat base on the table. The whole volume inside the cone is filled with water of density . the circular rim of the cone’s base has a watertight seal with the table’s surface and the top apex of the cone has a small hole. Neglecting atmospheric pressure find the total upward force exerted by water on the cone is (A) (2/3)R2hg (B) (1/3)R2hg (C) R2hg (D) None Sol. By force diagram of liquid : Fy + gh R2 = mg Fy  Fy = mg – gh R2 Fx

1 Fy = R2h g – gh R2 3

Fy = –

Fx gh R2 mg

2 R2h g 3

Hence upward force exter on come by water is

2 R2hg. 3

Hence option (A) is correct.

BUOYANCY (a) If a body is partially or wholly immersed in a fluid, it experiences an upward force due to fluid surrounding it. The phenomenon of force exerted by fluid on the body is called buoyancy and the force is called buoyant force.

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FLUID MECHANICS (b) When a solid body is immersed in a liquid, it displaces the liquid, partially or completely of its volume. While doing so it pushes or applies a downward force on the liquid, hence, in accordance to newton’s third law, the liquid also exerts an upward force (equal to the weight of the displaced liquid) Archimedes´ principle It states that when a body is partically or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid. If V be the volume of displaced fluid of density  then, buoyant force, FB = Vg NOTE : 1. Buoyant force arises because the pressure in the fluid is not uniform but increases with depth. 2. Buoyancy force acts through the centroid of the displaced fluid. 3. Buoyant force may be different from the weight of the displaced fluid in non-inertial frame of reference. Flotation (a) When the force of buoyancy is equal to the weight of the dipped (partially or completely) solid, the solid will remain in equilibrium. This is called flotation. (b) When the overall density of the solid is smaller than that of fluid, the solid floats with a part of it in the fluid. The fraction dipped is such, that the weight of the displaced fluid equals the weight of the solid. (c) APPARENT WEIGHT : A solid of volume Vs is dipped inside a liquid, where it displaces the liquid by a volume Vl. If s and l be the density of solid and liquid respectively then, apparent weight of the solid is, Wa = Vs s g – Vl l g or

Wa  (Vs s  Vl  l )g

For completely submerged solid, Vl = Vs and we have Wa = Vs (s – l)g (a) Stability of floating body depends on the effective point of application of the buoyant force in relation to the point of application of its weight. (b) The weight of a body acts through its centre of gravity G and the buoyant force acts through the centre of mass/buoyancy B of the displaced liquid. (A) STABILITY OF COMPLETELY SUBMERGED BODY (a) Neutral equilibrium : When a homogeneous cylinder is completely immersed in a liquid G and B coincide with each other. Then all positions are equilibrium position.

W G

B FB

G W

(b) Unstable equilibirum : When an additional weight is placed on top of a homogeneous cylinder so that G lies above B. Then by a slight disturbance the cylinder topples. (c) Stable equilibrium : When an additional weight is placed at the bottom of a homogeneous cylinder so that G lies below B. A slight disturbance in the cylinder produces restoring couple which tries to bring the cylinder in earlier position.

FB

B

G

B W

B

FB

FB B

G W

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B G W

20

FLUID MECHANICS (B) STABILITY OF PARTIALLY SUBMERGED BODY (a) For partically submerged body under certain condition may be stable even if G lies above B. (b) It is because a slight rotation in the body may cause a sufficient shift in the position of centre of buoyancy (as shape of the displaced liquid changes) which may provide a restoring couple. (c) The body remains in stable equilibrium as long as B remains below the meta centre M (point of enter section of shifted line to original line of action of buoyant force)

W G B FB

M

W G

FB B

C13: A body floats in a liquid contained in a beaker. The whole system as shown in Figure falls freely under gravity. The upthrust on the body is (A) zero (B) equal to the weight of the liquid displaced (C) equal to the weight of the body in air (D) equal to the weight of the immersed portion of the body Sol. [A] The whole system falls freely under gravity Upthrust = weight of fluid displaced = (mass of fluid displaced) × g For a freely falling body, g = 0  Upthrust = 0. C14: The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation : (A) the balance A will read more than 2kg (B) the balance B will read more than 5 kg (C) the balance A will read less than 2 kg and B will read more than 5 kg (D) the balance A and B will read 2 kg and 5 kg respectively Sol. [BC] When the block of mass m is arranged as shown in the figure, an upthrust FT will act on the mass which will decrease the reading on A. but according to Newton’s second law, to each and every action, there is equal and opposite reaction. So FT will act on the liquid of the beaker which will increase the reading in B.

A

m B

A

FT

FT

B

Example 19. A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm cm3). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in gm/cm3 is (A) 3.3 (B) 6.4 (C) 7.2 (D) 12.8

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21

FLUID MECHANICS Sol. [C] Wt. of sphere = Uptrust due to Hg + Upthrust due to oil V V Vdg  d Hg g  d oil  g 2 2 d Hg  d oil 13.6  0.8 d   7.2 g / cm3 2 2

Oil

Mercury

Example 20. In English the phrase ‘tip of the iceberg’ is used to mean a small visible fraction of something that is mostly hidden. For areal iceberg what is this fraction if the density of sea water is 1.03 g/cc and that of ice is 0.92 g/cc ? Sol. In case of floatation, weight = upthrust, i.e., mg = Ving, i.e., V = Vin [as  = m/V]   Vout = V – Vin = V  1   

or

Vin =

 V 

or

fout 

Vout     0.92  0.11  1    1       1.03  1.03 V

so

fout = 0.106 or 10.6% Example 21. A cork of density 0.5 gcm–3 floats on a calm swimming pool. The fraction of the cork’s volume which is under water is (A) 0% (B) 25 % (C) 10 % (D) 50 % Sol. mg = Vg V0g = Vg 0.5 V0g = 1 × Vg

V0 2 Fraction = 50 % V

Hence option (D) is correct.

Example 22. A small wooden ball of density  is immersed in water of density to depth h and then released. The height H above the surface of water up to which the ball will jump out of water is (A)

h 

  (B)   1 h  

(C) h

(D) zero

Sol. When ball is insides water : The acceleration of ball inside water is



a0 

Vg  mg m

a0 

    Vg  Vg       1 g   g V     

The velocity of ball at the surface is v02 = 2a0 h

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FLUID MECHANICS  or

v2 = v02 – 2 gH 02 = v02 – 2gH



H



      H h    1 h      

v02 2a 0 h a 0 h   2g 2g g Hence option (B) is correct.

Example 23. A sphere of radius R and made of material of relative density has a concentric cavity of radius r. It just floats when placed in a tank full of water. The value of the ratio R/r will be    (A)     1

1/3

   1 (B)    

1/3

   1 (C)    

1/ 3

   1 (D)     1

1/ 3

Sol. Since, sphere just floats,  mg = Vg or

4 4 4   0  R 3  r 3  g   R 3g 3  3 3

But

0  



 (R3 – r3) = R3

or

 r3   1  3   1  R 

or

1 r3 1  3  R

or

R      r    1

r3 1  R3 

or

1

or

3    1 r      R3

1/ 3

Hence option (A) is correct.

Example 24. A cubical block of wood 10 cm on a side and of density 0.5 g/cm3 floats in a jar of water. Oil of density 0.8 g/cm3 is poured on the water until the top of the oil layer is 4 cm below the top of the block. (a) How deep is the oil layer ? (b) What is the gauge pressure at the lower face of the block ? Sol. (a) Let the oil layer be x cm deep. Then weight of block = weight of oil displaced + weight of water displaced. or (103)(0.5) = 102 (x) (0.8) g + 102 (6 – x) 1 4 cm x cm or x = 5 cm. Oil (10–4–x)cm (b) Gauge pressure at the lower face of the block = 5 × 0.8 × 98 + 1 × 1 × 980 water = 4900 dynes/cm2.

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FLUID MECHANICS Example 25. A cone of radius R and height H, is hanging inside a liquid of density  by means of a string as shown in the figure. The force, due to the liquid acting on the slant surface of the cone is (neglect atmosphere pressure) (A) gHR2 (B) HR2 (C) 4/3 gHR2 (D) 2/3 gHR2 Sol.  Fy = PR2 = Vg = Buoyancy force

R

1 R2hg 3

or

Fy + ghR2 = 



1 Fy = R2hg – ghR2 3

Fy = –

H

T Fy Fx

2 R2hg 3

Fx P R2 mg

2 R2hg in downward direction. 3 Hence option (D) is correct.

Hence Fy =

Example 26. A dumbbell is placed in water of density . It is observed that by attaching a mass m to the rod, the dumbbell floats with the rod horizontal on the surface of water and each sphere exactly half submerged as shon in the figure. The volume of the mass m is negligible. The value of length l is

d(V  3M) 2(V  2M) Sol. For equilibrium, (A)

(B)

d(V  2M) 2(V  3M)

Mg + 2Mg + mg = 

(C)

V V g+ g 2 2

d(V  2M) 2(V  3M)

2M,V Water d

d(V  2M) 2(V  3M)

(D)

d(V  2M) 2(V  3M)

...(i)

Taking torque about mass m,  V   V    g  Mg  d  l     g  2Mg l 2 2 From (i) and (ii)

l

l M,V

...(ii)

Hence option (B) is correct. F

B

b Example 27. A rod of length 6 m has a mass of 12 kg. It is hinged at one end W at a distance of 3 m below the water surface. C 3m (i) What weight must be attached to the other end so that 5 m of the rod is O O´ submerged ? H A 12kgf N (ii) Find the magnitud and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5. Sol. Mass per unit length of the rod is 2 kg. Therefore, mass of the submerged portion of the rod is 10 kg and its 10 3 volume = m 500

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FLUID MECHANICS (using the simple formule, volume =

mass and density = specific gravity × 1000 kg m–3). Therefore, density

buoyant force, 10 × 1000 = 20 kgf 500 Let N and H be the vertical downward and horizontal reactions of the hinge on the rod. Considering horizontal and vertical translational equilibrium of the rod, N + 12 + W = 20 (where W is the weight to be attached and H = 0). N + W = 8 and H = 0 Considering the rotational equilibrium of the rod about A –20 × g × 2.5 cos  + 12 × g × 3 cos  + W × 6 cos  = 0 6W = 50 g – 36 g = 14 g

Fb =

W

7 7 g  kgf 3 3

N  8

7 17  kgf 3 3

Example 28. A wooden plank of length 1 m and uniform cross-section is hinged at F one end to the bottom of a tank as shown in figure. The tank is filled with water C up to a height of 0.5 m. The specific gravity of the plank is 0.5. B l A mg Find the angle  that the plank makes with the vertical in the equilibrium position. O (Exclude the case  = 0). Sol. The forces acting on the plank are shown in the figure. The height of water level is l = 0.5 m. The length of the plank is 1.0 m = 2l. The weight of the plank acts through the centre B of the plank. We have OB = l. The buoyant force F acts through the point A which is the middle point of the dipped part OC of the plank. OC l OA   We have 2 2 cos  Let the mass per unit length of the plank be . Its weight mg = 2lg.  l  The mass of the part OC of the plank =   . cos   The mass of water displaced =

1 l 2l   . 0.5 cos  cos 

2l g . cos  Now, for equilibrium, the torque of mg about O should balance the torque of F about O. So, mg (OB) sin  = F (OA) sin   2l    l  (2 l ) l   or  cos    2 cos   or cos2 = 1/2

The buoyant force F is, therefore, F =

or

cos =

1 , 2

or

 = 45º

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FLUID MECHANICS Example 29. A solid rod of length L cross-sectional area A and density  is suspended freely in a wide tank filled with a liquid of density 0 to a height h = L/ (where > 1). The lower end of the rod touches the base of the tank. (i) Determine the angle of inclination of the rod with the horizontal in the equilibrium position. (ii) Determine the normal reaction between the rod and the base of the tank. Sol. In the figure shown, let us assume L = length of the rod l = immersed length of the rod  = angle of inclination of rod with the horizontal. (i) If the rod is in equilibrium, then the net torque about the point A is zero, i.e.  A  (ALg) 2

or

L 0  l2 

Now,

sin  

B

L l cos   (0 Al g) cos   0 2 2

h L 1 0   l l  

FB h=L/ W

A

0

N

(ii) Using Newton’s second law N = W – FB = ALg – 0Alg or

  l N  ALg 1  0   L 

or

   N  ALg 1  0    

FLUID DYNAMICS Blowing air, water flowing in a river, liquid flowing in a tube, blood flowing in vein are examples of fluid motion. The study of fluid in motion is known as fluid dynamics. Types of fluid flow A fluid element is a small volume of fluid containing a large number of molecules. During motion, this fluid element may be observed moving in various fastion as decided by the forces acting on it. These define various types of fluid flow. Compressible flow is defined as that is which the density of fluid changes during the flow. Incompressible flow is that in which fluid density does not change during the flow. The flow of liquids is generally incompressible and that of gases is compressible (at low speeds). A fluid element may be rotating while flowing. This is known as rotational flow. Irrotational flow is that is which a fluid element does not rotate while moving. If we look at various points of space in which fluid has irrotational flow, velocity vectors of fluid elements at various points of space (at a given time) have their line integral in a closed path equal to zero :    v . dl  0 (irrotational flow)

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FLUID MECHANICS During flow heat exchange or work exchange may take place between fluid and its surrounding. In such a flow, we define isothermal flow as that in which temperature of fluid element remains the same during the flow. Adiabatic flow is that in which no heat exchange between fluid element and surrounding is allowed. Real fluid have fractional motion. The name of fluid friction is viscosity. If viscous forces are considerable the flow is called ‘viscous flow’. If the, viscous force has negligible effect, the flow is idealised as ‘non-viscous flow’.

Streamline flow

Turbulent flow

uncertain type of motion

Figure shows water tap turned on slowly. Water flow in upper part looks smooth while in lower part (larger speed) the flow is not smooth (it is zig-zag). The reason for such a look is that flow at low speed is streamline while at large speed it is turbulent. Steady or Streamline flow The flow of fluid in which every fluid element crosses a given point of space with the same velocity as its predecessor, is known as steady flow. Thus, at a very given point of space we have a vector of velocity. We have a velocity field, just like electrostatic fleid or gravitational field. The path followed by all the particles of fluid passing through V a point is the same. This path is ‘streamline’. hence we A also call the steady flow a streamline flow. A streamline is V´ analogous to electric lines of force. Velocity at a point is Streamlines tangent to streamline passing through that point. B The magnitude of velocity is proportional to the density of streamlines (per unit area normal to velocity vector). In the figure same number of streamlines are spaced more closely at A than at B and hence velocity at A is larger then that at B : V > V´ Two streamlines never cross each other because that would mean two directions of velocity at the crossing point which is not possible in a steady flow. v2 The line integral of velocity in a closed path is zero is steady flow. The velocity field here is an example of a conservative A2 field. v1 We can always imagine a pipe of flow constituted by streamlines. Fluid in this pipe flows in a disciplined way, Flow pipe A1 never crossing its wall. A flow pipe or flow tube is a bundle of streamlines. The figure shows a flow tube. Its cross-sectional area is non-uniform. In a uniform velocity field, streamlines are parallel and equispaced. We have in that case cross-sectional area as the same every where for a flow tube and fluid elements have equal velocity at all sections at a time.

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FLUID MECHANICS In the case of non-uniform steady flow, larger area means rarefied streamlines and hence low speed of particles. In a flow tube the same number (N) of streamlines are crossing different areas (A). Hence, by definition of streamlines, N A N Vk A VA = kN = constant. The product of flow tube area and fluid velocity at each cross-section of a flow pipe remains the same. This is the constant of equation of continuity (describing continuous flow) : A1V2 = A2V2 = ........... We can show that equation of continuity is based a conservation of fluid mass during flow. The product Av  is, in fact, volume of fluid per unit time passing through surface normal to V . If we multiply it by density , we get the rate of mass flow. If mass is neither created nor destroyed, this rate must be the same. Hence AV= constant. For incompressible flow, is also constant. Then AV = constant. Narrower the flow tube, faster the flow. V

C15: From a tube of inner radius 4 cm, water is flowing at an average speed of 2m/sec. Calculate the rate of flow. Sol. Q = Rate of flow Q = volume of liquid flowing per sec. = area of cross-section × velocity  Q = A × v = r2v =

22 2 (4) × 2 × 100 c.c./sec. = 10053.09 c.c./sec. 7

C16: The rate of flow in a tube of inner radius 4 cm is 36 litre/hour. Calculate the speed of the water in the tube. Sol. Q = A v = r2v or

36  1000 22 2  (4) v 60  60 7 v = 0.198 cm/sec.

C17: A horizontal pipe having a constriction is shown in the figure. The radius at M and N are respectively 8 cm and 4 cm. Calculate the velocity at N if velocity of water at M is 16 cm/sec. Sol. We have A1v1 = A2v2 M N r2 =8cm v 2 2 r1 v1 = r2 v2 r1 =4cm 1 v 2

2

r  v1   2  v 2  64 cm / sec.  r1  BERNOULLI’S THEOREM Deniel Bernoulli (1700 – 1782) studied flow of fluids using energy method. If we apply conservation of energy of steady flow, we get an equation known as Bernoulli’s equation (representing Bernoulli’s theorem). It relates velocity of flow, elevation of flowing fluid and pressure.

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FLUID MECHANICS The sum of kinetic energy per unit volume, gravitational potential energy per unit volume and pressue, remains constant at all the points of a stream line. This is Bernoulli’s theorem.

P

v

Stream line

Y Reference land (y = 0

If a fluid of density be flowing with speed V along a streamline, and P be the pressure and y be height relative to a reference line, then kinetic energy per unit volume = 1/2 V2 gravitational potential energy per unit volume = gy. The sum is gy + 1/2 V2 + p. According to Bernoulli’s theorem p + gy + 1/2V2 = constant. This is Bernoulli’s equation. To prove it, we consider a flow pipe and fluid contained in the pipe between sections a and b (see fig.) The pressure p1 causes a force p1 A1 at section-a and pressure p2 causes a force p2A2 at section-b. Due to velocity the section-a moves by v1dt and section-b by v2dt. The fluid we considered as (i) in the figure, acquires the position (f) in time dt. Work done on the fluid is W = p1A1V1dt – p2V2V2dt. p2 A2 (i) This must be equal to change in mechanical energy, E. Now E = U + K. b v dt 2 where U is the change in gravitational energy, and K is (f) a´ the charge in kinetic energy of the fluid lying, between section-a and section-b. a´ During motion section-a goes to a´ and b to b´. the fluid p1 A1 a v1 dt between a´ and b has the same kinetic energy and potential energy in the states i and f. Hence K = (Ka´b + Kbb´ ) – (Kaa´ + Ka´b) K = Kbb´ – Ka´a 1 1 (A2V2dt ) V22 – (A1V1dt )V12 2 2 U = (Ua´b + Ubb´ ) – (Uaa´ + Ua´b) U = Ubb´ – Uaa´ U = (A2V2dt ) gy2 – (A1V1dt)gy1 Therefore using E = W, we have

K =

Now, 

1 1 (A2V2dt ) V22 – (A1V1dt )V12 + (A2V2dt ) gy2 – (A1V1dt)gy1 2 2 = p1A1V1dt – p2A2V2dt A1V1 = A2V2 1 1 V22 – V12 + gy2 – gy1 = p1 – p2 2 2

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FLUID MECHANICS Arranging subscripts-1 on RHS and subscripts-2 on LHS, 1 1 V22 + gy2 + p2 = V12 + gy1 + p1 2 2 The sum of pressure (p), kinetic energy per unit volume (1/2 V2) and gravitational potential energy per unit volume (gy) has the same value at section-1 as at section-2. This is true for any other section also. Area A1 or A2 of the flow tube are small enough so that unique velocity can be taken. Hence the result is true when pipe becomes very narrow, i.e., a stream line. The Bernoulli’s equation

1 V2 = constant 2 must be used for ideal flow only. By this we mean incompressible, non-viscous, irrotational and streamline flow. In the special case of hydrostatics, (we take v = 0). This gives p + gy = constant. As y decreases, p increases, i.e. hydrostatic pressure increases with depth.

p + gy +

C18: The work done in pushing 10 cubic metre of water in a pipe of diameter 8 cm is 16000 joule. Calculate the difference in the pressure at the two ends of the pipe. Sol. We have, work done W = (P1 – P2)V 

P1  P2 

W 16000   1600 N / m 2 V 10

Example 30. Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 N/m2 between A and B where the areas of cross-section are 30 cm2 and 15 cm2 respectively. Find the rate of flow of water through the tube.

A

B

Sol. Let the velocity at A = vA and that at B = vB. v B 30 cm2  2 By the equation of continuity, v A 15 cm 2 By Bernoulli’s equation, PA +

1 1 vA2 = PB + vB2 2 2 1 1 3 (2vA)2 – vA2 = vA2 2 2 2

or

PA – PB =

or

600

or

v A  0.4 m 2 / s 2  0.63 m / s

N 3 kg    1000 3  v2A 2 m 2 m 

The rate of flow = (30 cm2) (0.63 m/s) = 1890 cm3/s

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FLUID MECHANICS Example 31. A liquid of density 0.83 g/cm3 is contained in a vessel under a total internal pressure of 3 atm. Neglecting viscosity, calculate the speed with which the liquid will escape from a small hole made in the side wall of the vessel.

Here

1 1 P1  v12  P2  v 22 2 2 P1 = 3 atm P2 = 1 atm v1 = 0



v2 

2(P1  P2 ) 

v2 

2  2  1.013  105  22.1 m / s 0.83  103

Sol.

P1

v2 P2

v1 = 0

Example 32. A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of crosssection of the tube at the points P and Q at heights of 2 metre and 5 metre P are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at 2m point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q. Take g = 9.8 m/s2. Sol. Given : A1 = 4 × 10–3 m2, A2 = 8 × 10–3 m2, h1 = 2 m, h2 = 5 m, v1 = 1 m/s and  = 103 kg/m3 From continutiy equation, we have A1v1 = A2v2 or

A  v 2   1  v1  A2 

or

 4  103  v2   1 m / s  8  103 

A2

Q 5m

v2

A1 v 1 h1

2 h2

1 m/s 2 Applying Bernoulli’s equation at section 1 and 2, v2 

1 1 P1  v12  gh1  P2  v 22  gh 2 2 2 1 P1  P2  g(h 2  h1 )  (v 22  v12 ) ...(1) 2 Work done for unit volume by the pressure as the fluid flows from P to Q W1 = P1 – P2 W1 = g(h2 – h1) + (v22 – v12) [from eq. (1)]

or (i)

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FLUID MECHANICS  1  1  W1  (103 )(9.8)(5  2)  (103 )   1  J/m3 = {29400 – 375} J/m3  4  2  or W1 = 29025 J/m3 Ans. (ii) Wor done per unit volume by the gravity as fluid flows from P to Q. W2 = – g(h2 – h1) = – (103)(9.8)(5 – 2) J/m3 or W2 = – 29400 J/m3 Ans. Example 33. A siphon tube is used to remove liquid from a container as shown in fig. In order to operate the siphon tube, it must initially be filled with the liquid. (i) Determine the speed of the liquid through the siphon. (ii) Determine the pressure at the point C. Sol. (i) Applying bernoulli’s equation at points A and D, we get

C H

Datum A

h B y

1 1 pA + vA2 + gyA = pD + vD2 + gyD 2 2 Assuming datum for potential energy at the free surface, we have yA = 0; yD = – (h + y); pA = pD = patm vA2  0 ; vD = v

patm + 0 + 0 = patm +

1 2 v = g[–(h + y)] 2

or

v  2g(h  y) (ii) Applying Bernoulli’s equation at A and C, we get

Here,

1 1 vA2 + gyA = pC + vC2 + gyC 2 2 yC = + H ; vC = v (according to the continuity equation)



patm + 0 + 0 = pC +

pA +

or or

1 2 v + gh 2 patm = pC + g (h + y) + gh pC = patm – g (h + H + y)

Example 34. The figure shows a siphon in action. The liquid flowing through the siphon has a density of 1.5 g/cc. Calculate the pressure difference between (a) points A and D, and (b) points B and C. Sol. (a) At A and D the pressure is atmospheric, therefore, PA – PD = 0 (b) PA – PB = 1.5 × 980 × 20 dynes/cm2 PD – PC = 1.5 × 980 (200 – 20) As PA = PD ; PA – PC PA = 1.5 × 980 × (200 – 20) PA = 26400 dynes/cm2.

B

C

20 cm A

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200 cm

D

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FLUID MECHANICS Example 35. A liquid flows out of a broad vessel through a narrow vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the pipe and the length of the pipe is h ? Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure. Applying Bernoulli’s theorem to points 1, 2 and 3

1

x

2

h

p0 1 2 p2 1  v1  gH   v 22  g(h  x)  2  2 p0 1 2  v3  0  2 By continuity equation v1 A1 = A2v2 = A2v3 Since A1 >> A2, v1 is negligible and v2 = v3 = v (say) 



p0 p 1 p 1  gH  2  v 2  g(h  x)  0  v 2   2  2



v  2gH

and

p0 p  gH  2  gH  g(h  x)  

...(i)

p0 = p2 + g (h – x) p2 = p0 – g (h – x) ...(ii) Thus pressure varies with distance from the upper end of the pipe according to equation (ii) and velocity is a constant and is given by (i). Applications of Bernoulli’s theorm A variety of phenomena involving ideal fluid flow, or approximately ideal flow, could be explained using Bernoulli’s theorem. 1.

Speed of efflux : Torricelli’s law : If we puncture an open tank filled with water, (or any liquid) water (with liquid) will flow out of the pin hole. This hole of very small radius is called on orifice. The speed of outflow is called ‘speed of efflux’. Torricelli observed that speed of efflux at the orifice is equal to the speed of freely falling stone released from the top of liquid surface. This is known as y=h Torricelli’s theorem. If h be the depth of the orifice from free surface, the speed of efflux will be

2gh .

y–axis

Let us consider a streamline joining orifice to the top of the liquid. Let the reference level for gravitational potential energy be at the level of orifice. Pressure at the top of liquid is atmospheric pressure P0 and also at the point where orifice is made. The speed at the top is negligible as area here is very large. Hence p0 +  gh + 0 = P0 +

h

y=0 Reference level

1 2 v . 2

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FLUID MECHANICS Solving it we get

v  2gh . In certain situations the tank is not open. The pressure above the liquid top is not equal to pressure at the orifice. In that case 1 2 v . 2 In rockets, the inside pressure is much larger and the gh - term is ignored.

P1 +  gh + 0 = P0 + 0 +

C19: A tank contains water upto a height H. There is a very narrow hole at a depth h below the level of water in the tank. At what distance the stream of water coming out from the orifice in the wall will strike the floor ? Sol. Water coming out from the orifice has a horizontal velocity

v  2gh Let in time t the water strikes the floor. Then 1 H  h  0  gt2 2



t

2(H  h) g

Therefore horizontal distance travelled x  vt  2gh 

2(H  h)  2 (H  h)h g

C20: In the above problem at what depth the hole should be so that the stream of water coming out from it may strike the floor at maximum distance from the tank. Sol. We have,

x  2 (H  h)h

 x2 = 4(H – h)h or x2 = H2 – H2 + 4(H – h)h Clearly x will be maximum, if (H – 2h)2 = 0 Hence,

h

H 2

C21: In the above problem calculate the maximum distance from the tank where the stream of water will strike the floor. Sol. We have, x2 = H2 – (H – 2h)2 Clearly x is maximum, when (H – 2h)2 = 0 Hence maximum value of x = H. C22: A tank contains water upto height H. What should be the position of two different holes in its wall so that water stream coming out from them may strike the ground at the same point ?

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FLUID MECHANICS Sol. The height of second hole from the bottom must be equal to the depth of the 1st hole from the level of the surface. For same range we have x1 = x2 h h 1



2 (H  h)h  2 (H  h1 )h1

or

H – h1 = h

H h

x1 = x2

C23: There are two narrow holes in the wall of a vessel filled with water at a distance of 50 cm. The area of cross-section of each hole is 0.2 sq. cm; calculate the distance of the point of intersection of their streams if 140 cm3 of water is filled in the vessel in every second. Sol. Let M and N be the two holes.

v1  2gh , v 2  2g(H  h) If x is the distance of the point of intersection then, x = v1t1 = v2t2 It can be shown that 2 1  V2 2 2g  x   H  120 cm 2  2g 2 V 2 

h M H=50cm

V1 V2

Example 36. An open tank is filled with two immiscible liquids of densities 1 and 2 as in the figure. Determine the speed of efflux. Sol. As Torricelli’s theorem is applicable to open tank, single density system, we shall use Bernoulli’s theorem for lower liquid. Pressure at the junction of liquids is P0 + 1 gh1 and presure at the orifice is P0. Hence

x

1

h1

2

h2

v

1 (P0 + 1gh1) + 2gh2 + 0 = P0 + O + 2V2 2

V

2(1gh1  2 gh 2 ) 2

  V  2gh 2  2  1  gh1  2  NOTE : That it depends upon (1/2), whereas in single liquid system it does not depend upon the type of liquid. Example 37. A cylindrical tank having cross-sectional area A = 0.5 m2 is filled with two liquids of density 1 = 900 kg m–3 and 2 = 600 kg m–3, to a ( 2) h height h = 60 cm each as shown in figure. A small hole having area a = 5 cm2 is made in right vertical wall at a height y = 20 cm from the bottom. Calculate h ( 1) (i) velocity of efflux, y F (ii) horizontal force F to keep the cylinder in static equilibrium, if it is placed on a smooth horizontal plane and (iii) minimum and maximum values of F to keep the cylinder in static equilibrium, if coefficient of friction between the cylinder and the plane is µ = 0.01. (g = 10 ms–2)

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FLUID MECHANICS Sol. Since area of hole is very small in comparison to base area A of the cylinder, therefore, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be v and atmospheric pressure P0. Consider two points A (inside the cylinder) and B (just outside the hole) in the same horizontal line as shown in figure. Pressure at A, PA = P0 + h2g + (h – y)1g Pressure at B, PB = P0 2 According to Bernoulli’s theorem, pressure energy at A PA = pressure energy at B + kinetic energy at B 1 B A 1 2  PA = PB + 1v 2 –1  v = 4 ms Ans. When cylinder is on smooth horizontal plane, force F required to keep cylinder stationary equals horizontal thrust exerted by water jet. But that is equal to mass flowing per second x change in velocity of this mass.  F = (av ) (v – 0) = av2 mg or F = 7.2 N Ans. Total mass of the liquid in the cylinder is m = Ah1 + Ah2 = 450 kg Fmax 7.2 N Limiting friction = µmg = 45 N  F < Limiting friction, therefore, minimum force requied is zero. Ans. µN Consider free body diagram for maximum value of force. N Considering vertical forces, N = mg Now considering horizontal forces, Fmax = F + µN or Fmax = 52.2 N Ans. Position of Liquid level in tank Let A be the area of the top of liquid and y be its position relative to the hole. Then in a time dt, the level will fall by | dy |, expelling a volume A | dy | through the hole. If x be area of the hole andy V be the velocity of efflux, volume expelled in time dt is  Vdt. Hence A | dy | =  V dt ...(1) using Torriceli’s theorem (for  0  – <  – 0 1     w v'  1 or v’ > v. So the ball rises.  v 11. No. There is no buoyant force on the solid because the liquid is not in contact at the bottom and so it exerts no upward thrust on it. 12.The level will remain parallel to the plane during the journey down the plane. The component of the force mg down the plane, which is m sin  produces acceleration down the plane. The surface is subjected to the force mg cos  perpendicular to the plane. A liquid surface keeps itself at right angles to the force to which it is subjected. So the level rests parallel to the plane during the downward journey of the bucket. 13.A floating body float s in stable equilibrium when its centre of gravity is well below the centre of buoyancy, that is, the centre of gravity of the displaced liquid. When loaded by enough weight the centre of gravity goes well below the centre of buoyancy. 14.It is found that a liquid will flow faster and more smoothly from a sealed can when two holes are punctured in the can then when one hole is made. Explain. 15.When they are close to each other, the velocity of the water between them increases, resulting in a fall of pressure there according to Bernoulli’s theorem. The pressures from the sides push them together. 16.When the parachute opens out, the pressure of the air above drops and so an upward thrust is called into play to balance the weight of the parachutist. 17.The viscous force on a body depends on its velocity. The greater the velocity the greater is the viscous force. When a body falls from a sufficient height, it acquires enough velocity to produce a viscous force that balances its weight. The resultant force on the body being zero, the body moves with uniform velocity, called terminal velocity. 18.While taking off an airplane needs maximum dynamic lift, which is associated with an unsymmetrical set of streamlines relatively close together on the upper side and relatively far apart below. The unsymmetric distribution of streamlines can be obtained better by moving into the wind. While landing, it should move with the wind when the streamlines will be uniformly distributed on the two sides of the airplane and so no lift will arise and the plane will land under gravity. 1 2 p1 1 2 p 2 v1   v 2  19.Yes, because 2  2  1 p  p 2  p1   v12  v22  2 At high altitudes the density is considerably low and so the pressure difference will depend on the altitude of the airplane. 20.The formula ‘thrust = pressure difference × area’ holds for fluids at rest. This does not hold for fluid in motion. p 1 2 p0 1 2 By Bernoulli’s theorem (a theorem applicable to fluids in streamline motion)  V   v and AV =  2  2





av where A = area of the chamber, a = area of the orifice. Since A >> a, v >> V, hence V2 may be neglected in comparison to v2. 

v2  2

 p  p0  

Now thrust on the rocket  v 

dM But the mass (dM) of gas flowing out in time dt is (a v  dt). dt

dM  av dt

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FLUID MECHANICS 2  p  p0   2a  p  p0   21.Near the edge the velocity is zero and the pressure is the normal atmospheric pressure. At the centre, the velocity is maximum and so there is maximum drop of pressure at the centre. Thus, the pressure difference between the central region and the surrounding air being the maximum, the destructive power of a tornado is maximum at the centre. 22.Since the stream has a high velocity, the pressure inside the stream is below the atmospheric pressure. The ball is supported from the bottom by the thrust of the stream and kept in position by the lateral pressure difference. 23.The pressure of the air streaming over the roof is less than that of the stationary air below the roof. This is according to Bernoulii’s theorem. The surplus pressure of the stationary air of the room acts normally on the two inclined halves of the roof. When the roof is secured at the edges A and B, the thrusts on the two halves will have turning effects about the edges and so the roof is likely to tear along the ridge. On the other hend if the roof is secured rigidly along the ridge, the thrusts will then lift the roof against gravity and it will be carried away by the wind. 24.When air is blown through the pipe, the air between the discs is set in motion, resulting in a heavy drop of pressure. The surplus pressure from below pushes the disc B and knocks it against the upper disc. 25.When it is submerged more by a short distance, the air inside the balloon is compressed resulting in a reduction of the buoyant force on it. Thus, downward force (weight of the sinker attached + weight of air in the balloon) being in excess of the buoyant force, it sinks to the bottom. 26.Let V’ be the total volume of the block of wood and W be the weight of the block. Then by the law of flotation (V’ – V) × tg = W where t = density of water at tºC. Since wood has a negligible coefficient of expansion V’ may be considered constant.  V = V’ – W/tg As the temperature is gradually increased from 0ºC, t increases and so V increases up to 4ºC, when the density of water becomes maximum. Above 4ºC, t decreases and so V decreases continuously. 27.Reading on the scale = weight of liquid and beaker + buoyant force = W = Vtg, where W = weight of liquid + beaker V = volume of the solid at tºC and t = density of liquid at tºC 0 g  Reading W  V0 1   t   1  ' t where V0 = volume of the solid at 0ºC,  = coefficient of expansion of the solid, ’ = coefficient of expansion of the liquid  Reading = W + V00 {1 – (‘ – )t}g As t is increased, {1 – (‘ – )t} decreases and so the reading also decreases because W is a constant. 28.In air Wg – Fg = Wb – Fb where Wg = weight of glass bulb, Fg = buoyant force on the bulb, Wb = weight of brass weights, Fb = buoyant force on the brass weights Since the volume of the bulb is greater than the brass weights Fb < Fg or Wb – Wg + Fg < Fg or Wb < Wg Thus when air is removed the bulb will become heavier than the brass weights. 29.Let H be the height of the liquid in the beaker and H’ be the height of the liquid after all the ice melts. Let A be the area of cross-section of the beaker. 

thrust = av2 = a ×

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FLUID MECHANICS Let V be the volume of the ice block and V’ the volume of it inside the liquid. Then V ice Vice × g = V’ × 1.2 × 1000 × g  V ' 1200 Vice Volume of liquid in the beaker = AH – V’ = AH  1200 When all the ice melts, it is converted into water of the same mass but of volume, say, V’’. V’’ × 1000 = Vice  V '' Vice 1000 Therefore, total volume of liquid after ice melts Vice  Vice    AH    AH '  1200  1000 1   1 A  H ' H   Vice     ve   1000 1200   H’ > H. So the level of the liquid will rise. 30.Let h1 and h2 be the heights above the horizontal line. Let p1 be the pressure of the entrapped air. Then p1 – p0 = gh2 and also and p1 – p0 = gh1  gh1 = gh2  h1 = h2 So the entrapped air will not affect the levelling test. 31.Steam is highly compressible. When a steam boiler blows up, steam expands violently causing considerable damage while water being incompressible expands nominally with release of pressure. So it does not cause much damage. 32.Since they are of the same radius, the buoyant force on each of them is the same. W W  Fb  a  g where W is the weight of the hollow sphere W' W ' Fb  a' and g where W’ = weight of solid sphere  a  a ' W  1    Fb  W '  1     g  g But W’ > W a a' 1  1  a  a'  g g Hence the solid sphere arrives at the ground earlier. 33.No. Raindrops of different sizes reach the earth with different speeds. The terminal velocity acquired by a drop is proportional to the square of its radius. At the final stage, upward viscous force = downward driving force 4 3 6rv  r     g  v  r2  3 34.For flotation Weight of the body = weight of the liquid displaced Vg = V’’g where V = volume of the body, V’ = volume of the submerged portion of the body.  V’/V = /’ 

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FLUID MECHANICS Thus, the ratio of the submerged volume to the total volume is independent of the acceleration due to gravity. Hence in free space and on the surface of Jupiter the cork will neither sink more nor rise more. 35.The pressure difference on the two sides of a tennis ball due to the spin imparted to it is greater than it is in the case of a baseball because the former is roughter. Hence, it is easier to make the flight a tennis ball curve. 36.Due to the high speed of the train, the pressure of the air between the train and the man may fall considerably and the pressure difference may be sufficiently high to push the man towards the speeding train. 37.To pass on to the stable equilibrium position. 38.p = p0 (atmospheric pressure). In general p = p0 +  (g + a)h. Here a = –g. Therefore p = p0. 39.Initially, the ship will move to the right because the in-flowing water (flowing to the left) will exert a force 2pA to the right, where p is the pressure at a depth h of the hole, and A is its area. As soon as the stream of water reaches the opposite wall, this wall will be acted upon by the leftward force F = Av2, where v is the velocity of the stream with respect to the ship. As a result, the motion will begin to retard. 40.Siphonic action will start and water will flow to the reservoir. 41.By Bernoulli’s theorem the velocity of flow is v = 2gh where h is the height of water above the hole. The force on the plank = v2S = 2ghS. Once the hole is covered, the velocity of flow is zero and so the force on the plank = ghS which is just half as great. 42.In the beginning when the stream has not yet reached the pan, equilibrium will be disturbed. The pan will swing upwards since the water flowing out of the vessel will not longer exert force on its bottom and hence on the pan. That is, the reading of the balance will decrease. The moment the stream reaches the pan, equilibrium will be restored. On each element gravity imparts a downward impulse mg

2h  m 2gh (impulse = force × time) and on reaching the pan the same element g

of the liquid experiences on upward impulse m 2gh (impulse = change in momentum). Thus, each element experiences equal and opposite impulse. since the stream is continuous, the upward and downward impulses of elements cancel out and o the equilibrium is restored. The moment when the stream stops flowing, the pan will swing down, since the last elements of the liquid falling on the pan act on it with a force that exceeds the weight of the elements. 43.When air is blown over the paper there is a drop in pressure and the atmospheric pressure from below balances the weight of the paper. 44.Due to wind on the two sides of the flag, the difference of pressure produces ‘folds’ in the flag in compliance with Bernoulli’s theorem. 45.This is in compliance with the equation of continuity. V (rate of volume flow) a1v1 = a2v2 i.e., a 1/v. At the lower end the velocity is greater and so the are of cross-section is less. SURFACE TENSION 1. No, there will not be a fountain. The curvature of the free liquid surface at the top of the tube will change till the upward force 2rT cos balances the gravitational pull of the column of liquid of height h/2. 2. When the soil is not harrowed, there are large capillaries in it. Water in the soil rises up the capillary holes to the surface from where it evaporates continuously. Thus the soil continuously loses water. When harrowed, all these capillaries are destroyed, and so capillary suction stops altogether. Thus, water is retained longer in the soil. 3. Because of surface tension of water. Due to surface tension the surface of a small volume of water tends to occupy the minimum area. The area of a given volume of water has the minimum value for a spherical shape. 4. The end of a glass rod becomes round on being heated because of the surface tension of molten glass.

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FLUID MECHANICS

5. The grease melts, and capillary forces carry it to the surface of the cold fabric placed under the clothing, where it is soaked. 6. The end in the shadow is colder than the end in the sun. The capillary forces drive the water in this direction. 7. No, the water will not flow out. When the tube is taken out, a convex meniscus is formed and a force due to surface tension is called into play in the upward direction which keeps the water in the tube. Thus the length of water column remaining in the tube will be l + h. 8. In a capillary tube the height of the liquid that can be retained is given by r2hg = 2rT cos where  is the angle at which the liquid meets the solid. hgr = 2T cos (a) Since H > h, surface tension cannot sustain the column of water and so water flows out. (b) Since H = h, the water does not flow out. The meniscus is convex. (c) The water does not flow out. The meniscus is convex and is less curved than in the second case. (d) The water does not flow out. The meniscus is flat. (e) The water does not flow out. The meniscus is concave. 9. Water cannot remove grease stains. This is because water does not wet a greasy spot. The molecules of a detergent are hair pin shaped, experiencing different amounts of forces at the ends due to water molecules and grease molecules. The molecules are as if pinned to small globes of greasy dirt and form an interface between water and greasy dirt globes. The greasy dirt is thus dislodged from the clothes and is carried away with running water. 10.The excess pressure inside a soap bubble is inversely proportional to its radius. Hence, the pressure inside A is greater than the pressure inside B. Air will flow from A and B. Therefore, A will become smaller and B will become larger. 1. 2. 3. 4. 5. 6. 7.

ELASTICITY Pressure is the external force per unit area, while stress is the internal force called into play from within a strained body acting transversely per unit area of the body. No. It depends on the radius of the wire as well. Steel. Elasticity is measured by the stress per unit strain. For the same amount of strain, much greater stress is produced in steel than in rubber. The steel wire has a greater force constant. The one with greater area will have a smaller force constant. Only solids. Liquids and gases have only bulk modulus. Solids.

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94

FLUID MECHANICS

ASSERTION-REASON 1.

Statement-1 : Imagine holding two identical bricks under water. Brick A is completely submerged just below the surface of water, while Brick B is at a greater depth. The magnitude of force exerted by the person (on the brick) to hold brick B in place is the same as magnitude of force exerted by the person on the brick) the hold brick A in place. Statement-2 : The magnitude of buoyant force on a brick comletely submerged in water is equal to magnitude of weight of water it displaces and does not depend on depth of brick in water.

2.

Statement-1 : Upto elastic limit of a stress-strain curve the steel wire tends to regain its original shape when stress is removed. Statement-2 : Within elastic limit the wire follows Hook’s law.

3.

Statement-1 : When an ideal fluid flows through a horizontal tube of variable cross-section, the pressure becomes different as different points. Statement-2 : Raindrops falling from a great height reach the ground with a relatively small velocity. This phenomena involves the viscosity of air.

4.

Statement-1 : When a drop of ink falls on a newspaper, it spreads on it. Statement-2 : Adhesive force between ink and paper is greater than cohesive force between ink molecules.

5.

Statement-1 : In the diagram shown, a cube is floating in water in tilted position, in shown situation, the cube is in stable position i.e., cube won’t sink. Statement-2 : The point of application of gravitational force and buoyancy force are passing through the same line in above described case.

C

MATCH THE COLUMN 1.

2.

Bernoulli’s equation can be written in the folliwing different forms (column-I). Column-II lists certain units each of which pertains to one of the possible forms of the equation. Match the unit associated with each of the equations : Column-I Column-II 2 v p  (A) + z = constant (P) Total energy per unit mass 2g g (B)

V 2 + P + gz = constant 2

(Q)

Total energy per unit weight

(C)

V2 P  + gz = constant 2 

(R)

Total energy per unit volume

In Column-I, a uniform bar of uniform cross-section area under the application of forces is shown in the figure and in Column-II, some effects/phenomena are given. Match the entries of Column-I with the entries of Column-II. Column-I Column-II (A) F (P) Uniform stresses developed in the rod F

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95

FLUID MECHANICS (B) (C) (D) 3.

4.

F Smooth

F F Smooth

F

Match the following : Column-I (A) Splitting of bigger drop into small droplets (B) Formation of bigger drop from small droplets (C) Spraying of liquid

(Q)

Non-uniform stresses developed in the rod

(R)

Compressive stresses developed

(S)

Tensile stresses developed

(P) (Q) (R) (S)

Column-II Temperature increases Temperature decreases Surface energy increases Surface energy decreases

Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column-II. [JEE, 09] Column-I Column-II Y (A) The force exerted by X on Y (P) Block Y of mass M left has a magnitude Mg. on a fixed inclined plane X P X, slides on it with a constant velocity. (B) The gravitational potential energy (Q) Two ring magnets Y and Z, P of X is continuously increasing. each of mass M, are kept Z in frictionless vertrical plastic Y stand so that they repel each other. Y rests on the base X X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity. (C) Mechanical energy of the system (R) A pulley Y of mass m0 is fixed to X + Y is continuously decreasing. a table through a clamp X. A P Y block of mass M hangs from a X string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity. (D) The torque of the weight of Y about (S) A sphere Y of mass M is put in a point P is zero. nonviscous liquid X kept in a Y container at rest. The sphere is X released and it moves down in P the liquid. (T) A sphere Y of mass M is falling with its terminal velocity in a Y viscous liquid X kept in a container. P

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X

96

FLUID MECHANICS

LEVEL – 1 1.

The force due to pressure in a liquid at a given depth below the free surface (A) is always exerted downward of orienting the surface (B) is the same in all directions of orienting the surface (C) equals the total weight of liquid above that depth (D) depends upon the amount of liquid below that depth S

2.

In a U-tube experiment, a column PQ of water is balanced by a column RS of (some liquid see figure). The relative density of the liquid is (A) h2/h1 (C) 2 h1/h2

(B) (h1 – h2)/h1 (D) 2h2/(h1 + h2)

P h1

h2 R

Q

3.

The pressure at the bottom of a liquid tank is not proportional to the (A) height of the liquid above the bottom (B) acceleration due to gravity (C) density of the liquid (D) area of the liquid surface at the top

4.

A sample of a metal weights 210 g in air, 180 g in water and 120 g in an unknown liquid. Then (A) the density of the metal is four times the density of the unknown liquid (B) the density of the metal is 3 g/cm3 (C) the density of the metal is 7 g/cm3 (D) the metal will float on water

5.

A beaker containing water is put on the platform of a spring balance. A piece of iron suspended from a string is immersed in the water without touching the sides or the bottom of the beaker. How will the reading on the spring balance be affected, assuming that no water flows out of the beaker ? (A) The balance will indicate less weight (B) The balance will indicate more weight (C) The reading will not change (D) The reading will be less or more depending the upon position of piece of iron in the liquid

6.

Which is based on an application of Bernoulli’s equation for fluid flow ? (A) Capillary rise (B) Dynamic lift of an aero plane (C) Viscosity meter (D) Hydraulic press

7.

A steel ball of mass m falls in a viscous liquid with a terminal velocity 4 cm s–1. Another steel ball of mass 64 m will fall through the same liquid with a terminal velocity of (A) 4 cm s–1 (B) 16 cm s–1 (C) 8 cm s–1 (D) 64 cm s–1 B

8.

In the apparatus shown, initially the stop-cock A is closed and B is open. Now B is connected to a vacuum pump so that pressure is reduced and the liquid rises up to some point C above A. B is closed at this position of liquid and A is opened. Then (A) liquid in the tube will fall till its level is same as outside (B) liquid will flow out of A (C) liquid will rise further (D) liquid will not flow out of A

C

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A

97

FLUID MECHANICS 9.

In a hydraulic press, f and F are the forces acting on the small piston and the large piston having diameters d and D respectively. Then f/F is (B) 2 D2/d2

(A) 2 d/D

(C) d2/D2

(D)

D/d

10. Two capillary tubes of the same radius and lengths l1 and l2 are fitted horizontally side by side to the bottom of a vessel containing water. The length of a single tube of the some radius that can replace the two tubes such that the rate of steady flow through this tube equals the combined rate of flow through the two tubes, is 2l1 l2 (A) l  l 1 2

l1 l2 (C) l  l 1 2

(B) l1 + l2

(D)

l1  l2 2

11. Two capillary tubes of the same length and radii r1 and r2 are fitted horizontally side by side to the bottom of a vessel containing water. The radius of a single tube that can replace the two tubes such that the rate of steady flow through this tube equals the combined rate of flow through the two tubes, is (A)

(B) (r12 + r22)1/2

r1r2

(D) (r14 + r24)1/4

(C) r1 + r2

12. A capillary tube is dipped in a beaker containing a liquid. The angle of contact of the capillary with the liquid is 90º. The liquid in the tube will (A) fall (B) rise (C) neither rise nor fall (D) may rise or fall depending on the density f the liquid 13. An oil drop is placed on the surface of water. It will (A) partly be as a spherical droplets and partly as thin film (B) remain on the surface as a sphere (C) spread as a thin layer (D) remain on the surface as distorted drop 14. If the work done in blowing a bubble of volume V is W, then the work done in blowing a bubble of volume 2V will be (A) (4)1/3 W (B) 2 W (C) 8 W (D) (2)1/3 W 15. Which one of the following curves shows correctly the variation of velocity v with time t for a small spherical body falling vertically in a long column of viscous liquid ? v

v

(A)

v

(B)

(C) t

t

v

(D) t

t

16. A cylindrical vessel filled with water is released on an inclined surface of angle  as shown in figure. The friction coefficient of surface with vessel is µ (< tan ). Then the constant angle made by the surface of water with the incline will be : (A) tan–1 µ

(B)  – tan–1 µ

(C)  + tan–1 µ

Fixed

(D) cot–1 µ

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98

FLUID MECHANICS 17. A cylindrical container of radius R and height h is completely filled with a liquid. Two horizontal L shaped pipes of small cross-section area a are connected to the cylinder as shown in the figure. Now the two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions. Then the torque due to ejected liquid on the system is : (A) 4 aghR (B) 8 aghR (C) 2 aghR (D) none of these 18. In the figure shown water is filled in a symmetrical container. Four pistons of equal area A are used at the four opening to keep the water in equilibrium. Now an additional force F is applied at each piston. The increase in the pressure at the centre of the container due to this addition is (A) F/A (B) 2F/A (C) 4F/A (D) zero

(A) V0

(B) V0/2

h/2

h/2

R

R

F2

F1

19. A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by gH gH (A) (B) 2L L gH 2gH (C) (D) 2L L 20. A square box of water has a small hole located in one of the bottom corner. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed v0, as shown in figure (1). When the box is still half empty, it is tilted by 45º so that the hole is at the lowest point. Now the water will flow out with a speed of

2R

F3

F4

open to atmosphere H

L a

V0 V0

Fig. (1) Initial

(C) V0 / 2

Later-on

(D) V0 / 4 2

21. A mosquito with 8 legs stands on water surface and each leg makes depression of radius a. If the surface tension and angle of contact are T and zero respectively then the weight of mosquito is : (A) 8T . a (B) 16 Ta (C) Ta/8 (D) Ta/16  22. In determination of young modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is 1m. the curve between extension and stress is depicted then young modulus of wire will be : (A) 2 × 109 N/m2 (C) 2 × 1010 N/m2

(B) 1 × 109 N/m2 (D) 1 × 1010 N/m2

4 mm Extension

2 mm 4000 KN/m2

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8000 KN/m2

Stress (KN/m2)

99

FLUID MECHANICS Area ‘A’

23. A block of mass M area of cross-section A & length l is placed on smooth horizontal floor. A force F is applied on the block as shown. If y is young modulus of material, then total extension in the block will be : (A) Fl /Ay (B) Fl / 2Ay (C) Fl / 3Ay (D) cannot extend

F l

24. A uniform rod of mass m and length l is rotating with constant angular velocity  about an axis which passes through its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its young’s modulus is Y. Neglect gravity. The stream at the mid point of the rod is : (A)

m 2 l 8AY

(B)

3m  2 l 8AY

(C)

3m  2 l 4AY

(D)

m 2 l 4AY

25. A ball of mass 10 kg and density 1 gm/cm3 is attached to the base of a container having a liquid of density 1.1 gm/cm3, with the help of a spring as shown in the figure. The container is going up with an acceleration 2 m/s2. If the spring constant of the spring is 200 N/m, the elongation in the spring is (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm 26. In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densitieis 2 and 3. The height h for the equilibrium of cylinder must be (A)

3R 2

(B) R

3 2

(C) R 2

2 m/s 2

h

2

(D) R

R

3

R

3 4

27. A conical flask of mass 10 kg and base area 103 cm2 is floating in liquid of specific gravity 1.2 as shown in the figure. The force that liquid exerts on curved surface of conical flask is (g = 10 m/s2) (A) 20 N in downward direction (B) 40 N in downward direction (C) 20 N in upward direction (D) 40 N in upward direction 28. In the figure shown, forces of equal magnitude are applied to the two ends of a uniform rod. Consider A as the cross-section area of the rod. For this situation, mark out the incorrect F statement(s). (A) The rod is in compressive stress. (B) The numerical value of stress developed in the rod is equal to F/A. (C) The stress is defined as internal force developed at any cross-section per unit area. (D) None of the above.

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10 cm

F

100

FLUID MECHANICS 29. For the block shown in the figure, the Poisson’s ratio is . The body is under state of compression. F In this situation (A) the volume of the stressed body decreases a (B) the decrease in length is more, than compensating for increase in area (C) the volume of the stressed body increase (D) both (A) and (B) are correct.

c F b

30. In above question, if the fractional change in length of block due to application of force F is 0.001 and corresponding change in volume is 0.0005, find the value of Poisson’s ratio for the material of the blocks? (A) 1/2 (B) 1/10 (C) 1/4 (D) Not possible to determine the value of Poisson’s ratio from the given data 31. The space between two larger horizontal metal plates, 6 cm apart, is filled with a liquid of viscosity 0.8 N-s/m2. A thin plate of surface area 0.01 m2 is moved parallel to the length of the plate such that plate is at a distance of 2 cm from one of the plates and 4 cm from the other. If the plate moves with a constant speed of 1 m/s, then (A) the layer of the fluid which is having maximum velocity is lying mid-way between the plates. (B) the layers of the fluid which is in contract with the moving plate is having the maximum velocity. (C) the layer of the fluid which is in contact with the moving plate and is on the side of farther plate is moving with the maximum velocity. (D) the layer of the fluid which is in contact with the moving plate and is on the side of nearer plate is moving with the maximum velocity. 32. The coefficient of restitution for collision of two bodies (A) depends on Young’s modulius of both the bodies (B) does not depend on Young’s modulius of both the bodies (C) depends on Young’s modulus of the lighter body (D) may or may not depend upon Young’s modulus 33. A heavy block of mass 150 kg hangs with the help of three vertical wires of equal length and equal cross-section area as shown in the figure. Wire II is attached to the midpoint (centre of mass) of block. Take Y2 = 2Y1. For this arrangement mark out the correct statement(s) (A) The wire I and III should have same Young’s modulus. (B) Tension in I and III would be always equal. (C) Tension in I and III would be different. (D) Tension in Ii is 75 g.

I

II

x

III

x

Y1

Y2

Y3

150 kg

34. A metal wire of length L, cross-section area A and Young’s modulus Y is stretched by a variable force F. F is varying in such a way that F is always slightly greater than the elastic forces of resistance in the wire. When the elongation in the wire is l, upto this instant

YAl 2 (A) the work done by F is 2L

YAl 2 (B) the work done by F is L

YAl 2 (C) the elastic potential energy stored in wire is 2L

(D) no energy is lost during elongation

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101

FLUID MECHANICS 35. A solid floats in a liquid is in partially dipped position, (A) the solid exerts a force equal to its weight on the liquid (B) the solid doesn’t exert any force on the liquid (C) the solid exerts a force equal to buoyancy force on the liquid (D) the liquid exerts a force of buoyancy on the solid which is equal to the weight of solid 36. All the four wires in the options below are made up of the same material. Which of these will have the largest extension, when the same tension is applied ? (A) Length 200 cm and diameter 2 mm (B) Length 100 cm and diameter 0.5 mm (C) Length 300 cm and diameter 1 mm (D) Length 50 cm and diamter 0.5 mm 37. A uniform plank of length L and Young’s modulus Y is pushed over a smooth horizontal surface by a constant smooth F0 horizontal force F0. The area of cross-section of the plank is A. The compressive strain (L/L) of the plank in the direction of the force is : 3F0 F0 F0 2F0 (A) (B) (C) (D) AY AY 2AY AY 38. Two wires of same material and length but cross-sectional area in the ratio 1 : 3 are used to suspend the same loads. The extensions in them will be in the ratio : (A) 1 : 3 (B) 3 : 1 (C) 4 : 1 (D) 1 : 4 39. A wire elongates by l mm when a load W newton is hang from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire (in mm) will be : (A) zero (B) l (C) l / 2 (D) 2l 40. A work of 2 × 10–2 J is done on a wire of length 50 cm and area of cross-section 0.5 mm2. If the Young’s modulus of the material of the wire is 2 × 1010 N/m2, then the wire must be : (A) elongated to 50.1414 cm (B) stretched by 0.707 mm (C) contracted by 2.0 mm (D) none of these 41. The value of Poisson’s ratio for all practical bodies lies between : (A) 0 to 0.5 (B) –1 to 1.5 (C) 0.5 to 1 42. Two bodies of masses 1 kg and 2 kg are connected by a metal wire shown in figure. A force of 10 N is applied on the body of mass 2 kg. The breaking stress of metal wire is 2 × 109 N/m2. What should be minimum radius of the wire used, if it is not to break ? (A) 0.23 × 10–4 m (B) 4 × 10–4 m (C) 5 × 10–4 m

(D) –1 to 1

2 kg

1 kg

F = 10N

Smooth surface

(D) 5.2 × 10–4 m

43. Two wires, ne mode of coper and other of steal are joined end to end. (as shown in figure). The area of cross-section Cooper of copper wire is twice that of steel wire. F They are placed under compressive force of magnitudes F. The ratio of their lengths such that change in lengths of both wires are same is : (YS = 2 × 1011 N/m2 and YC = 1.1 × 1011 N/m2) (A) 2.1 (B) 1.1 (C) 1.2 (D) 2

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Steel

F

102

FLUID MECHANICS y 44. The curve in figure represents potential energy (U) in between two atoms in a diatomic molecules as a function of distance x between atoms. The U atoms are : A (A) attracted when x lies between A and B and repelled when x lies between B and C (B) attracted when x lies between B and C and repelled when x lies between A and B (C) attracted when they reach B (D) repelled when they reach B

B

C x

a0 = 2m/s2

45. one end of a steel wire is fixed to ceiling of an elevator moving up with an acceleration 2 m/s2 and a load of 10 kg hangs from other end. Area of crosssection of the wire is 2 cm2. The longitudinal strain in the wire is : (Take g = 10 m/s2 and Y = 2 × 1011 N/m2) (A) 4 × 1011

(B) 3 × 10–6

(C) 8 × 10–6

(D) 2 × 10–6

46. Two bodies of masses 1 kg and 2 kg are connected by a steel wire of crosssection 2 cm2 going over a smooth pulley (as shown in figure). The longitudinal strain in the wire is : (Take g = 10 m/s2, Y = 2 × 1011 N/m2) (A) 3.3 × 10–7

(B) 3.3 × 10–6

(C) 2 × 10–6

1 kg

(D) 4 × 10–6

2 kg

47. Water from a tap emerges vertically downwads with an initla speed of 1.0 ms–1. The cross-sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is [JEE, 98] –4 2 –5 2 –5 2 (A) 5.0 × 10 m (B) 1.0 × 10 m (C) 5.0 × 10 m (D) 2.0 × 10–5 m2 48. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per section from both holes are the same. then, R is equal to :[JEE,2000] (A) L / 2

(B) 2L

(C) L

(D) L/2

49. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density  where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is [JEE, 01] 2 (A) Mg (B) Mg – vg (C) Mg + R hg (D) g (V + R2h) 50. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there. After some time the coin falls into the water. Then [JEE, 02] (A) l decreases and h increases (B) l increases and h decreases (C) both l and h increase (D) both l and h decrease

h 2r

coin l h

51. Water is filled in a container upto height 3m. A small hole of area ‘a’ is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If a/A = 0.1 then v2 is (where v is the velocity of water coming out of the hole) [JEE, 05] (A) 48 (B) 51 (C) 50 (D) 51.5

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103

FLUID MECHANICS

PASSAGE

1.

2.

Comprehension – 1 Two cylindrical tanks of radii r and 2r with their bases at the same level contain a liquid of density  to heights H and 3H, respectively as shown in figure. The tanks are linked through a pipe of very small cross-sectional area A. Due to pressure difference liquid starts flowing from narrower vessel to broader vessel to equalize the pressure. Based on above information, answer the following questions : The final common level of liquid in both vessels is (A) 2H (B) 7H/5 (C) 3H/2

5H  r 2 gA

3H H

A

(D) 5H/2

5r 2 (B) A

r 2 H  (C) A g

H g

8r 2 H  (D) 5A g

The work done by gravity in equalising these levels is (A) r2 × gH2

4.

2r

The time taken for the liquid levels to become equal from initial levels is (A)

3.

r

(B)

41 × r2 ×gH2 10

(C)

8 × r2 × gH2 5

Comprehension – 2 A wooden cylinder of diameter 4r, height h and density /3 is kept on a hole of diameter 2r of a tank, filled with water of density as shown in the figure. The height of the base of cylinder from the base of tank is H. If level of liquid starts decreasing slowly when the level of liquid is at a height h1 above the cylinder, the block just starts moving up. Then, value of h1 is (A) 2h/3 (B) 5h/4 (C) 5h/3

(D)

h1

15 × r2 ×gH2 2

/3

4r h2

h

H 2r

(D) 5h/2 [JEE, 06]

5.

Let the cylinder is prevented from moving up, by applying a force and water level is further decreases. Then, height of water level (h2 in figure) for which the cylinder remains in original position without application of force is [JEE, 06] (A) h/3 (B) 4h/9 (C) 2h/3 (D) h

6.

If height h2 of water level is further decreased, then (A) cylinder will not move up and remains at its original potition. (B) for h2 = h/3, cylinder again starts moving up (C) for h2 = h/4, cylinder again starts moving up (D) for h2 = h/5 cylinder again starts moving up

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[JEE, 06]

104

FLUID MECHANICS

LEVEL – 2 1.

The solid ball in the figure hangs from a spring balance B1 and submerged in a liquid contained in a beaker placed on a spring balance B2. The mass of the beaker is 1.0 kg and that of the liquid is 2.5 kg. Balance B1 reads 3.5 and B2 reads 8.5 kg. The volume of ball is 0.005 m3. (a) Write the buoyant force. (b) Find the density of the liquid. (c) What will the balances B1 and B2 read if the ball is pulled up out of the liquid ?

B1

B2

2.

A uniform rod of length L m, specific gravity 0.5 and mass m is hinged at one end at a distance of 1/2 L m below water surface. (a) What mass M must be attached to the other end of the rod so that 5/6 L m of the rod gets submerged? (b) Find the magnitude and direction of the force exerted by the hinge on the rod.

3.

A U-tube, placed in a vertical plane, is partially filled with liquid X. A liquid Y, immiscible with X, is next poured into one side until X rises on the other side by 25 cm. If the density of Y relative to X is 0.8, by what distance the Y-level will stand higher than the X-level ?

4.

A vessel contains a liquid X (density 0.8 g/cm3) over another liquid Y (density 13.6 g/cm3), not mixing together. A homogeneous sphere floats with half its volume immersed in upper and the other half in lower liquid. Find the density of the material of the sphere.

5.

6.

A liquid flows out of two small hles X and Y in the wall of a tank. The two streams strike the ground at the same point. If the hole X is at a height h above the ground and the level of water stands at a height H above the ground, find the height of Y.

M

X Y

Y H

Two copper (density ) balls A and B are released in a liquid of viscosity . They are connected by a narrow light rod. How does the tension in the rod vary with time ?

X h

A

B

r 2r

7.

Water rises to a height of 4 cm in a capillary tube. If the area of cross-section of the tube is reduced to 1/16 of the former value, find how high will the level rise.

8.

A soap bubble A of radius r1 and another soap bubble B of radius r2 (> r1) are brought together so that the combined bubble has a common interface. Find the radius of the interface.

9.

Find the work that must be done to get n small equal size spherical drops from a bigger spherical drop (radius = R) of a liquid having surface tension S. A mercury drop breaks into 8 equal drops. The surface energy increases by a fact or n. Find n.

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FLUID MECHANICS 10. Calculate the work done against surface tension in blowing a soap bubble from a radius of 10 cm to 20 cm if the surface tension of soap solution is 2.5 × 10–2 Jm–2. 11. A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and a light slider supports a weight of W = 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface of the liquid forming the film ? 12. A tube T of radius r is connected to a bubble of a liquid having surface tension S. The radius of the bubble is R at time t = 0. Show that it will shrink to zero in time t = 2lR4/Sr4, where  is the co-efficient of viscosity of air.

l T

R

13. Length of horizontal arm of a uniform cross-section U-tube is l = 21 cm and ends of both of the vertical arms are open to surrounding of pressure 10500 N/m2. A liquid of density  = 103 kg/m3 is poured into the tube such that liquid just fills the horizontal part of the tube. Now one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity 0 = 10 rad/sec. If length of each vertical arm be a = 6 cm. Calculate the length of air column in the sealed arm. [g = 10 m/sec2] 14. A simple accelerometer (ann instrument for measuring acceleration) can be made in the form of a tube filled with a liquid and bent as shown. During motion, the level of the liquid in the left arm will be at a height h1, and in the ring arm at a height h2. Determine the acceleration a of a carriage in which the accelerometer is installed, assuming that the diameter of the tube is much smaller than h1 and h2. 15. A hemispherical tank of diameter 4m is filled with water. A very small hole is punched at 1m above the bottom as shown in figure. Find x (distance at which water strikes the surface).

6 cm

21 cm

h1

h2 /4

/4

................................................. ................................. ............................. ......................... . ................... . . . ....... . . 1m x

16. Two arms of a tube have unequal diameters d1 = 1.0 mm and d2 = 1.0 cm. If water (surface poured into the tube held in the vertical position, find the difference of level in the tube. Assure the angle of contact to be zero. T = 0.07 N/m. 17. A block mass 0.5 kg is attached to free end of 80 cm long aluminium wire of 85º diamter 0.7 mm and suspended vertically. It is now given circular motion into a horizontal plane at a rate such that the wire makes an angle of 85º with the vertical. Find the increase in length of the wire. (sin 85º = 0.9962, cos 85º = 0.0872, Young’s modulus of aluminium = 7 × 1010 Nm–2) 18. A metal wire of cross section 4 mm2 and length 5 m is suspended vertically from a rigid support. A bob of mass 100 kg is now attached at the lower end. If the is bob gets snapped, calculate the thermal energy generated in the wire. The Young’s modulus is 210 PGa and density is 7800 kg/m3.

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FLUID MECHANICS 19. The graph of stress  against strain () for a material is shown in the figure. The arrow shows the case of increasing stress. How much is the elastic energy density in the material when the strain is : (a) 0.1 (b) 0.3 and (c) 0.4 ?

(GPa) 2.0 1.5 1.0 0.5 0.1 0.2 0.3 0.4

20. An elastic band of length l is attached to a block of mass M which is held by a person at rest so that the band is unstretched. The person suddenly releases the block. (a) Find the maximum elastic energy stored in the band as the block falls. l (b) If the block finally comes at rest, how much is thermal energy generated in the bank ? The Young’s modulus of the band is Y, cross-sectional area is  and mass is negligible relative to that of the block. 21. A column of mercury of 10 cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will be column of mercury be displaced if the tube is held vertically ? 22. A sinker of weight w0 has an apparent weight w1 when weighted in a liquid at a temperature t1 and w2 when weight in the same liquid at temperature t2. The coefficient of cubical expansion of the material of sinker is  . What is the coefficient of volume expansion of the liquid. 23. A point mass m is suspended at the end of a massless wire of length l and cross section A. If Y is the Young’s modulus for the wire, obtain the frequecy of oscillation for the simple harmonic motion along the vertical line. 24. A cube of wood supporting 200 gm mass just floats in water. When the mass is removed, the cube ruses by 2 cm. What is the size of the cube ? 25. A beaker containing water is placed on the pan of balance which shows a reading of M gms. A lump of sugar of mass m gms and volume Vcc. in now suspended by a thread in such a way that it is completely immersed in water without touching the beaker and without any overflow of water. What will be the reading of the balance just when the lump of sugar is immersed ? How will the reading change as the time passes on? 26. A boat floating in a water tank is carrying a number of large stones. If the stones are unloaded into water, what will happen to the water level ? 27. Two idendical cylindrical vessels with their bases at the same level each contain a liquid of density  (rho). The height of the liquid in one vessel is h1 and in other is h2. The area of either base is A. What is the work done by gravity in equalizing the levels when the two vessels are connected. 28. A wooden plank of length 1 m and uniform cross-section in hinged at one end to the bottom of a tank as shown in fig. The tank is filled with water upto a height 0.5 m. The specific gravity of the plank is 0.5. Find the anlge  that the plank makes with the vertical in the equilibrium position. (Exclude the case  = 0º) 29. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal and containing air at the same pressure P. When the tube is held at an angle of 60º with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at 30ºC).

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FLUID MECHANICS

LEVEL – 3 1.

(a)

(i) (ii) (b)

(i) (ii) (iii) 2.

(i) (ii) (iii) 3.

4.

(i)

A container of large uniform cross-sectional area A resting on a horizontal surface, holds two immiscible, non-viscous & incompressible liquids of densities d & 2d, each of height H/2 as shown in figure. The lower density loquid is open to the atmosphere having pressure P0. A homogeneous solid cylinder of length L(L < H/2) cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with the length L/4 in the denser liquid. Determine : The density D of the solid and The total pressure at the bottom of the container. d H/2 The cylinder is removed and the original arrangement is restored. A tiny hole of area s(s ). [JEE, 99]

A uniform solid cylinder of density 0.8 gm/cm3 floats in equilibrium in a combination of two non mixing liquids A and B with its axis vertical. The densities h air 3 3 of the liquids A and B are 0.7 gm/cm and 1.2 g/cm , respectively. The height hA A of liquid A is hA = 1.2 cm. The length of the part of the cylinder immersed in hB liquid B is hB = 0.8 cm. B (a) Find the total force exerted by liquid A on the cylinder. (b) Find h, the length of the part of the cylinder in air. (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released. [JEE, 02] 7.

8.

9.

Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger D=8mm is 8 mm and the diameter of the nozzle is 2 mm. The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range Ground of water stream on the ground. Take g = 10 m/s2. [JEE, 04]

d=2mm 1.25 m

A solid sphere of radius R is floating in a liquid of density  with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing simple harmonic motion. Find the frequency of these oscillations. [JEE, 04]

10. A horizontal pipe line carries water in a streamline flow. At a point along the pipe where the cross-sectional are is 10 cm2, the water velocity is 1 ms–1 & the pressure is 2000 Pa. The pressure of water at another point where the cross sectional area is 5 cm2, is __________ pa. [Density of water = 103 kg m–3] [JEE, 04] 11. A U tube is rotated about one of it’s limbs with an angular velocity  . Find the difference in height H of the liquid (density ) level, where diameter of the tube d

1)

Equation of Linear Motion and Rotational Motion Linear Motion

Rotational Motion

(a) If acceleration is 0, v = constant and s = vt. (b) If acceleration a = constant, (i) s =

(u  v) t 2

(a) If acceleration is 0,  = constant and  = t. (b) If acceleration a = constant then (i)

vu t (iii) v = u + at

=

(1   2 ) t 2

 2  1 t (iii) 2 = 1 +  t

(ii) a =

(ii)  =

(iv) s = ut + (1/2) at2

(iv)  = 1t +

1 2 t 2 (v) 22 = 12 + 2 

(v) v2 = u2 + 2as 1 a (2n – 1) 2 (c) If acceleration is not constant, the above equation will not be applicable. In this case

(vi) Snth = u +

dx (i) v = dt (iii) vdv = ads

dv d 2 x (ii) a = = 2 dt dt

 2 (c) If acceleration is not constant, the above equation will not be applicable. In this case

(vi) nth = 1 + (2n – 1)

d (i)  = dt (iii) d = d 

d d 2 (ii)  = = 2 dt dt

C6: A flywheel of radius 0.5 metre starts from rest with angular acceleration of 0.34 radian/sec2. What will be its angular velocity after 100 sec ? Sol. Here 2 = 1 + t 2 = 0 + 0.34 × 100 = 34 radian/sec. C7: A circular disc of radius 2 m is revolving at 240 revolutions per minute a torque is applied which slows it at constant rate of  radian/sec2. In what time the disc will completely stop ? Sol. Here 2 = 1 + t or 0 = 2n – t 

t

2n 2  240   8 sec.    60

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C8: In the above question, calculate how many revolutions the disc makes before it comes to rest. Sol. If  is the angular displacement then, 22 = 12 – 2 or 0 = 12 – 2 2





12 4 2 n 2 4 2  240       32 2 2 2    60 

No of revolutions = /2 = 16

Example 1: Two gears G1 and G2 have radii and R > r. They are coupled and angular velocity of G1 varies at uniform rate of . Determine the angle turned by G2 in time t if the motion starts from rest. Sol: Since the contact point of gears is not sliding, the velocity is common. Hence



1 =

v (magnitude) r

2 =

v (magnitude) R

2 =

1r (magnitude) R

T R

G1 G2

2 = v R

V 1 = v r r

R

Differentiating this r (magnitudes) R Now we can use for gear –2

2 = 1

 = Z 0t + Then

= 0+ =

1  t2 2 Z

1  r  2  t 2 R 1  r  2  t 2 R

[The Gear-1 turns by

r  1 2 t while Gear-2 turns   –times this value.] 2 R

Example 2. A disk 8 cm in radius rotates about its central axis at a constant rate of 1200 rev/min. Determine (a) its angular speed, (b) the speed at point 3 cm from its centre, (c) the radial acceleration of a point on the rim, and (d) and total distance a point on the rim moves in 2.00 s. Sol. (a)

(b)

1200  2  40  rad / s 60 = 40 × 3.14 rad/s = 126 rad/s v = r  = 0.03 × 126 = 3.78 m/s 

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ar = R2 = 0.08 × (126)2 = 1.26 km/s2

(c) (d)

 = t = 126 × 2 = 252 radian S R S = R = 0.08 × 252 = 10.1 m 

 

Dynamics of rotation The study of rotational motion with explanation of its cause is known as rotational dynamcis. According to rotational dynamics, “torque” is responsible for changing angular velocity. Torque is also known as moment of force.



MOMENT OF FORCE If a body is free to rotate about a point O at rest in an inertial frame S, and a force F is applied at A, if produces rotation about O (see figure). We find by experiment that rotational effect is directly proportional to r, R and sin. We define a quantity  = rF sin  We define torque as a vector by 



O r

F

 F

  r



 = r × F. Which is combined effect of force and its action point. We call it moment of force F about O or torque about O. 

The torque of a force F about a point is defined as the cross product of the position vector of action point of the force and the force vector. 





In the figure, torque about O is  = r × F . Note how the direction is obtained using right hand rule for cross  product and vector  is shown at O. S.I. unit of torque is Nm and [ML2T–2] is its dimensional formula. Moment arm : Moment arm of a torque is the perpendicular (r) dropped on action line of force from the point about which torque is calculated. 





In the figure r is moment arm. We can write r as vector sum of r

r|| F

r moment arm

r





and r || which is parallel to F . Then 













= ( r + r ||) × F 



= ( r × F ) + ( r || × F ) 



= r × F Giving

Y

F



|  | = r Fsin90º = r1F. (2, 2 3)

C9: A force of 20 N acts at (2, 2 3 ) along the line y = 3 x. Calculate the torque of this force about the origin.

y=

Z

3x

X

(0, 0)

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Sol: The line of action of F passes through origin. Hence 

r = 0 





 = r × F = 0. Y

C10: In the previous example, calculate the torque of the force about ponit (1, 0)m.

F (20N)

Sol: We have 



3x (2, 2 3) r r

y= 



= r × F

Z





60º

O

(1, 0)

X

= r × F 

 |  | = r F From figure 

| r | = (1m) sin60º 3 m 2 = 20 N

= F

  = 10 3 Nm (inward along - Z axis) In vector from ^

= (10 3 Nm)(– k ).  Example 3. A force of F  2 ˆi  3 ˆj N is applied to an object that is pivoted about a fixed axle aligned along the z.  coordinate axis. if the force is applied at the point r  4 ˆi  5 ˆj m, find (a) the magnitude of the net torque about the z axis and (b) the direction of the torque vector .    Sol.  r F    4 ˆi  5 ˆj  2 ˆi  3 ˆj    12 kˆ  10 kˆ  2 kˆ



 



Torque about a point due to several forces If several forces act on a body at various points, their rotational effect about a point is described by net torque. Net torque or resultant torque is the vector sum of torques due to all the forces about the same point. We  denote it by  (gamma).

     = 1 + 2 + .......... N 





   where 1 , 2 , .......... N are torques about the same point due to forces F1 , F2 , ........ FN . If these be acting 

 at r1 ,  r2 , ...... rN relative to the point about which torque is requested.

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

 1

  = r1 × F1 



2 =  r2 × F2 | | |    N = rN × FN In the special case, when the forces are concurrent, we can obtain net torque as torque due to net force.

Torque about an axis 



Let several forces F1 , F2 ...... be acting on a body. We want to find torque about Z-axis. Then we take any 

point O on this axis. The action points have positions r1 ,  r2 ........... relative to O. Give the total of this torque  about O is  : Z

      = ( r1 × F1 ) + ( r2 × F2 ) + ............... Its Z-component,Z, is called torque about Z-axis. Torque about an axis is defined as the component of torque vector along the axis where the torque is calculated about any point on the axis.

torque about Z-axis Z

F2

F1

In the figure, z = cos It may be positive, negative or zero. It is a scalar quantity.

O

Couple Moment Two forces of the same magnitude, opposite directions and acting along parallel lines constitute a couple. The torque due to couple of forces is called ‘couple moment’. 

F r2





r2

d

In the figure, F and – F constitude a couple. Their torque about O is given by       0 = r2 × F + r1 × (– F )

–F r1

r1



= (r2 × F ) + r1 × (– F ) 



= d× F [see figure]  |  0| = d × F (one of the forces × separation) “Couple moment” is independent of the point about which it is calculated. It is a “Free vector”. F

Anticlockwise and clockwise torque The torque which tends to rotate the body clockwise is known as clockwise torque.

C F



In the figure grav is clockwise about O. An anticlockwise torque is that which tends to rotate the body anticlockwise.

r

grav

O mg



In the figure F is anticlockwise torque about O.

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C11: The more is the value of r, the more will be torque and easier to rotate the body. (i) The handle of screw driver is taken thick. (ii) In villages the handle of flour-mill is placed near the circumference. (iii) The handle of handpump is kept-long. (iv) The rinch used for opening the tap, is kept-long. C12:

Work done by torque =



2

1

 d

= Torque × angular displacement.   Example 4. Given that, r  2 ˆi  3 ˆj and F  2iˆ  6 kˆ . The magnitude of torque will be —

(A) 405 N–m (B) 410 N–m Sol. We know that,      rF    2 ˆi  3 ˆj  2 ˆi  6 kˆ    12  ˆj  6  kˆ  18 ˆi    12 ˆj  6 kˆ  18 ˆi  Now |  |  (12) 2  ( 6) 2  (18) 2  |  |  144  36  324  504



(C)

504 N–m

(D)

510 N–m

      

Hence (C) is correct.

Example 5. An automobile engine develops 100 kilo-watt, when rotating at a speed of 1800 rev./min. The torque developed by it will be – (A) 60 N–m (B) 531 N–m (C) 5.31 N–m (D) 6.0 N–m Sol. The power delivered by the torque  exerted on rotating body is given by P  P = 100 kW = 100,000 Watt

P =  or Here



 1800    2  60  rad / sec.  60 

105   513 N.m 60  3.14

Hence (B) is correct.

Equilibrium of rigid bodies Several forces and torques acting on a rigid body may be ‘balanced’. Then the body is said to be in equilibrium or meahanical equilibrium’. It is classified into two parts: rotational equilibrium and translational equilibrium. Translational Equilibrium: A body is said to be in translational equilibrium if the vector sum of all the forces be zero. The forces are said 

to be balanced and Newton’s first law of motion is obeyed if  Fi = 0.

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Rotational Equilibrium: If the vector sum of all the torques acting on a body about a point be zero, the body is said to be in rotational equilibrium about that point. If all the forces acting on the body lie in a plane, we can classify torque into clockwise and anticlockwise. For rotational equilibrium, we get Sum of clockwise torques = Sum of anticlockwise torques Principle of moments For a body in rotational equilibrium under coplanar forces clockwise torque is balanced by anticlockwise torque. In vector form, the vector sum of all the torques is zero. i = 0 Rotational equilibrium is rotational version of the Newton’s first law of motion in which forces acting on a particle are balanced producing translational equilibrium. Mechanical Equilibrium Mechanical equilibrium refers to a system under rotational as well as translational equilibrium. In this case: 

or,

 i = 0

or,

 Fi = 0



[rotational equilibrium] [translational equilbrium]

Action point of resultant force. The action point of resultant force is that point about which torque due to the forces is zero. Centre of gravity (CG.) is the action point of resultant of gravitational forces acting on all the particles forming the body. In the figure, let C be the centre of gravity of the body. Let dm be an elementary

C.G.



C

mass at position . Torque d about C is given by 

r



dc = r × ( dm g )

dmg

If the total torque is evaluated, we have 

c =





 r  dm g

whole body By definition, this torque is zero. 

If g is uniform over the whole body, then 

c





=  ( r  dm g)    =   dm r   g = 0  

[by definition of CG]



Now  dm r is the first moment of mass about C. If C happens to be centre of mass also, the first moment of mass of the whole body about it is zero.

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PHYSICS

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 dm r = 0

ROTATIONAL MECHANICS SHEET

[C is centre of mass]

CM

whole body

CG



Thus, for uniform g , centre of gravity coincides with centre of mass. 

If g varies over points of the body CG and CM do not coincide as indicated for a long rod shown in the figure. C13: (i) When a rigid body is rotating about a fixed axis and a force is applied on it at some point then we are concerned with the component of torque of this force about the axis of rotation not with the net torque. (ii) The component of torque about axis of rotation is independent of the choice of the origin O, so long as it is chosen on the axis of rotation, i.e., we may choose point O anywhere on the line AB. (iii) Component of torque along axis of rotation AB is zero if 



(a) F || AB

(b) F intersects AB at some point



(iv) If F is perpendicular to AB, but does not intersect it, then component of torque about line AB = magnitude 



of force F × perpendicular distance of F from the line AB (called the lever arm or moment arm) of this torque. 

(v)



If there are more than one force F1 , F2 ........... acting on a body, the total torque will be 













 = r1 × F1 + r2 × F2 + .......... But if the forces act on the same particle, one can add the forces and than take the torque of the resultant force, or 



 = r × ( F1 + F2 + ......) (vi) In general, if many forces are acting on a body the net torque is different about different points but if the 



body is in translational equilibrium (i.e. F1 + F2 + ...... = 0) the net torque about different points is equal. This can be shown as follows: 







Let the force F1 + F2 ,.......... are acting on a rigid body at position vectors r1 , r2 ,.......... etc. The position vectors are about the origin O. Then 









 0 = r1 × F1 + r2 × F2 + .........

.....(i) 

Now, suppose the position vector of an another point P is rp and now we want to find the torque about this point P, then, 













p = ( r1 – rp ) × F1 + ( r2 – rp ) × F2 + ........ 













= ( r1 × F1 + r2 × F2 + .....) – rp ( F1 + F2 + ......) From equations (i) and (ii) we can see that 





.....(ii)



 0 = p if F1 + F2 + ..... = 0 or the net torque of all the forces about any point is same if the body is in translational equilibrium.

C14: It is correct that whenever the resultant force acting on a body is zero, the body is in static equilibrium. Sol. No; zero resultant torque in also necessary.

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C15: A meter rod is acted upon by two forces, 2N and 4N, normal to its length as in the figure. Determine the distance of the point from CM at which their resultant force is acting. Sol: By definition, action point of resultant force is that about which the net torque becomes zero. Case (a): In this case the two torques will cancel each other only if the point lies lower than the action point of 1N force. Let its distance from CM be x cm. Then 4 × (10 – x) = 2 × (50 – x) 40 – 4x = 100 – 2x x = –30 cm

2N 4N 10cm

CM

4N

PA 10 – x CM x

C16: A rectangular block having uniform density has centre of gravity at C. It is placed on horizontal plane. Draw the normal contact force and its action point.



2 x(50 – x)

4 x(10 – x) Ncm

x

P

2N P 4N

x

10cm CM

\ \\\\ \\\\ \\\\ \ \ \ \ C \\\\ \\\\ \\\\ \\\\ \ \ \ \ \\\\

C

The plane is now tilted by an angle  (sec figure (b). Still the block is at rest under frictional force. Draw the normal force and its action point.

CM (b)

Negative sign shows that the action point P is 30 cm below the centre of mass. Case (b): In this case let the point P be at distance x-above CM. Equating moment about P. 4 × (x – 10) = 2 × (50 – x) 4x – 40 = 100 – 2x 6x = 140 x = cm = 36.7 cm The point P lies 23.3 cm above the CM.



10cm

(a)

2N 50 – x

2N 4N

(a)





Sol: N , mg and f and must be concurrent forces for they are balanced. N

Action point of mg

N Action point of mg Action point of N mg (a)

\\\ Action point \\\\ \ \ \ of N \\\\ \\\\ \ \ \ mg \\\ \\\\ \ \ \ \\\ \\\\

\\\ Action point of friction



(b)

[The largest angle  is that for which mg passes through edge. Above this angle, equilibriumis disturbed]. Example 6. A horizontal force F = mg/3 is applied on the upper surface of a uniform cube of mass m and side a which is resting on a rough horizontal surface having µS = 1/2. The distance between lines of action of mg and normal reaction N is : (A) a/2 (B) a/3 (C) a/4 (D) None

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Sol. In this case, torque of mg is balanced by torque due to normal reaction and applied force F. 

a a  N   x   Fa  mg 2  2

or

a  mga mga N   x   2  3 2

or

a  mga mga mga N   x    2  2 3 6

But

N = mg



a mga mg  mgx  2 6

or

x

a a 3a  a a    2 6 6 3

Hence (B) is correct.



l mg cos  = N´ l sin  2 mg cot 2 Using these,

Smooth vertical wall P l Q



Rough

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\



passes through intersection of N and mg . Considering horizontal and vertical forces, we have f = N .....(i) N = mg .....(ii) Considering torque about a horizontal axis passing through Q,

 N=

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 7: A uniform rod of length l lies at rest. Draw forces acting on it. Considering its rotational and translational equilibrium, determine the force of friction acting at Q as a function of angle . How does it varry by decreasing angle  ? What factor will decide minimum value of  without slipping ? Sol: Figure shows how forces can be shown to be concurrent. Forces on Q are N and f which adjust to produce the resultant Fc that P

N´ Fc Fc

N

N mg f

mg

Q

.....(iii) cot 

 mg  f=   cot  2  As  is decreased, cot increases. Hence frictional force f increases.

Using law of limiting friction f  SN

2 s

No Slippapge  min

90º



.....(iv)

1 mgcots mg 2  cot 2S

Thus,

1 tan 2 s

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PHYSICS  The factor

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 1  min  tan  2s  –1

1 comes from the cube’s centre of gravity and S from nature of 2

contact surfaces. These control min. You may notice that a person walking on ice slips for larger steps (small , limiting friction crossed). Small steps (large 0) ensure a friction less than limiting (no slipping). Example 8. A uniform board of weight W and width 2L hangs from a light, horizontal beam, hinged at the wall and supported by a cable (fig.) Determine (a) the tension in the cable and (b) the components of the reaction force exerted by the wall on the beam, in terms of W, L and . Sol. Taking torque about O, 2L = (T sin ) 3L  × 2 L = 3TL sin  Fy 2 = 3 T sin  ...(1) O Also, for translatory equilibrium, Fx L Fx = T cos  ...(2) Fy + T sin  =  ...(3) From equation (1)

 

L 2L

T

w

2L

2 3sin  From equation (1), (2) and (3) T

and

Fx 

2 cot  3

Fy 

 3

MOMENT OF INERTIA Moment of inertia takes same role in rotatory motion as mass in translatory motion. Smaller the moment of inertia about an axis, easier to produce rotation about that axis; larger the moment of inertia, harder to produce rotation. This means that moment of inertia about an axis is a measure of rotational inertia of the body about that axis. This is its physical meaning. Radius of gyration: To gyrate means to rotate. The radius of gyration about an axis is defined as that distance from the axis where total mass can be centred so as to give the same moment of inertia as the actual mass distribution.

I = miri2

CM

M

I . It can be shown that M it has dimensions of length and depends on geometry of the system. From figure, radius of gyration K is given by K =

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The radius of gyration of a homogenous sphere about its diameter, for ex, may be written as K=

2 MR 2 5 = M

2 R 5

Theorems on Moment of Inertia We shall discuss two important theorems on MI that help it. Theorem of perpendicular axes for laminear bodies and Theorems of parallel axes. Theorem of perpendicular axes Let Z axis be normal to a lamina. Let us draw XY axes intersecting Z-axis at (0, 0, 0). The points of lamina have z = 0. We shall denote a point in lamina only by (x, y) a for simplicity without a loss. Let mi be mass of i-th particle at (xi, yi, 0). Its distance ri from Z-axis is given by ri2 = x i2 + y i2 .

Its moment of inertia about Z-axis is mi ri2. For all the N-particles we have IZ = mi ri2 = mi( x i2  yi2 ) = mi x i2 + mi y i2 = Iy + Ix Statement: If thre axes x, y and z be drosen such that two of them lie in the plane of a lamina, the sum of moment of inertia about these two axes is equal to the moment of inertia about the third axis. This is known as theorem of perpendicular axes. C17: A uniform square plate has moment of inertia Ix and Iz about X and Z axis shown in the figure. Show that Iy = Ix = Iz/2. Also show that rotation of XY axes about t-axis by 45º or by  does not change the result. Sol: Using the theorem of perpendicular axes, Iz = Ix + Iy .....(i) By symmetry of mass distribution Ix = Iy .....(ii)  Iz = 2Ix = 2Iy  Ix = Iy = Iz/2. If the axes be rotated about Z-axis by , even then Ix = Iy Ix+ Iy Iz 

I x´´  I y´´ 

y

CM z

x

Iz 2

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Parallel Parallel axes theorem (Steiner’s theorem) axis through CM Let a body have a moment of inertia Iz about Z-axis and Icm about an axis passing through CM of the body and parallel to the Z-axis. Let d ri mi R be the separation between the two parallel axes and M be the mass of ri d M the whole body.  CM Then Iz = Icm + Md2. This is known as parallel axes theorem. d The moment of inertia of a body about an axis is equal to the sum of its moment of inertia about an axis parallel to the given axis and passing through CM, and the product of mass of the body with the square of distance between the parallel axes. Proof: Let i-th particle have mass m. Drop perpendiculars on Z-axis and CM-axis from this particle. The MI about zaxis is given by (see figure). Iz = miri .....(i) as from Z-axis the distance of mi is ri. The summation runs over all the particlar in the body. Now, construct vectors by arrows as in the figure. i

i





 i

r = d + Ri

We have  i







r . ri = ri2 = d2 + 2 d . R i + R i2 Using (ii) in (i) 

.....(ii)



Iz = mi(d2 + R i2 + 2 d . R i ) 

= (mi)d2 + mi R i2 + 2.mi R i 



= Md2 + Icm + 2 d .mi R i

.....(iii)









Let ri' be position vector of i-th particle from CM. Let i + R i = ri' . By definition of CM, the sum of first moment of mass about CM is zero. 



mi ri' = 0



mi i + mi R i = 0







Now mi i is a vector along CM-axis while mi R i is normal to this axis. But their sum is zero. We know that 







a b & ab =0







a =0= b 

 mi R i = 0 From (iii) & (iv), we have I = Md2 + Icm This proves the theorem.

.....(iv)

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Some Calculation of Moment of Inertia 1. Moment of inertia of a uniform rod about an axis passing through its CM and normal to its length: Let us choose an elementary mass dM at position x on x-axis. Its distance is also x from axis. Its moment of inertia dIz about Z-axis is given by

dM x=– L 2

x dx

x=L 2

X

M  dIz = dMx2 =  dx  x2 L  The total MI will be obtained by inetragion over the whole rod (from x = –L/2 to x = L/2). Thus M Iz = L

1/ 2

L/ 2

3 M x  x dx =    L  3  1/ 2 L / 2

2

M  L3  L3   = 3L  8    8     

ML2 12 This is the MI of rod about CM-axis, normal to rod. =

About one of its ends: If Z-axis passes normally through one end, we may use parallel axes theorem.

Z

2

Here,

Icm =

ML 12

CM L d= 2

L 2 Using parallel axes theorem Iz = Icm + Md2

d =

2

ML2 ML2 L IL = M   + = 12 3 2 2.

Rectangular Plate: Axis through CM, in the plane of the plate and parallel to an edge: Let M be the mass of the plate. M . Let us consider the plate is made up of thin bars perpendicular to the axis one lb of which is shown in the figure and has mass dM. The moment of inertia of this bar about the given axis is given by X

Then mass per unit area is

1 dM l2 12 Integrating over all the bars, the total moment of inertia is given by

dIx =

I = =

dM

Y CM

b

l

1 2 l . dM 12  1 Ml2 12

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Similarly, the moment of inertia about an axis (Y) through CM in the plane of the plate normal to breadth b can be written as 1 Mb2 12

I =

Moment of inertia about an axis normal to the plate & passing through CM: It can be obtained using perpendicular axis theorem. In the figure



Ix =

1 Mb2 12

Iy =

1 Ml2 12

Iz = Ix + Iy =

1 1 Mb2 + Ml2 12 12

1 M(l2 + b2) 12 If the plate is a square, its sides are equal then, l=b=a

=

Iz = 3.

1 Ma2 6

Rectangular bar: A rectangular bar may be made up of rectangular plates of mass dM. Then Z

dM(l 2  b 2 ) Iz = 12 Integrating, we get Iz = 4.

M(l 2  b 2 ) 12

b

dM

M

h

[Effect of h contained in M]

Ring or Hoop: An element of mass dM lies at a distance R from Z-axis.  dIz = dMR2 Integrating over the whole ring

Z R

2

Iz = R  dM = MR2 5.

dM

MI or Disc about its axis: Let M be mass of uniform disc of radius R. Then mass per unit area is M/R2. Let the disc be made up of elementary rings of radius r and width dr. The mass of this ring is obtained: Area = 2rdr dM = (mass per unit area) × area  M  =  2  × 2rdr  R  Now we may write the MI of the ring about Z-axis as dIz = dM.r2

dr 2r

Z

dr

r R

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2M 3 r dr R2 Integrating over the whole disc we get

=

2M Iz = R2

4 2M  R  1 0 r dr = R 2  4  = 2 MR2 R

3

Diametrical axis: Let X and Y be mutually perpendicular diametrical axes. Then Ix = Iy [by symmetry] Also, Iz = Ix + Iy [perpendicular axes theorem] Now Iz = mR2

6.



1 mR2 = 2Ix = 2Iy 2



Ix = Iy =

Y

Z X

1 mR2 4

Uniform solid cylinder: MI about its axis: We may imagine the cylinder to be made up of large no of elementary discs, one of the them has mass dM. Its MI about z axis is given by 1 dMR2 2 Integrating over the whole cylinder,

Z

dIz =

R M

1 2 1 Iz = R  dM = MR2 2 2 MI about perpendicular bisector axis: It is equal to

Iz = 7.

dM

MR 2 Ml 2 + 4 12

Homogenous sphere, diametrical axis: Let us consider a homogenous sphere of mass M and radius R. Its moment of inertia about diametrical axis shown in the figure may be calculated as follows. Let us consider the sphere to be made up of disc like slices one of which is located at position z. Its thickness is dz and radius is R 2  z 2 . The MI of this elementary disc is Z

1 dIz = × Mass × (radius)2 2

=

1 × (R2 – z2)dz × (R2 – z2) 2

where  = density =

Z=R 2

2

R –z

dz z=0

R

M z = –R

4 R 2 3

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 2 2 (R – z )dz 2 Integrating over the whole sphere

dIz =

 Iz = 2  = 2

R

 (R

4

 2R 2 z 2  z 4 ) dz

R

R R  4R  2 2 R dz  2R z dz  z 4dz       R R R 

=

 2

 5 4 5 2 5  2R  3 R  3 R   

=

 2

 30  20  6  5  R 15  

8 5 R 15 Putting the value of , we get

=

Iz =

2 mR2. 5

Moment of Inertia for simple geometrical objects Some bodies of simple geometry are thin uniform rod, uniform rectangular plate, circular ring or hoop, uniform circular disc, uniform solid cylinder, homogeneous sphere, hollow sphere, right circular cone etc. We shall give you values of moment of inertia of these about frequently used axes. Table MI of some regular bodies Body Position of axis

Moment of inertia

Z

Y

Laminar body X

Iz = Ix + Iy (perpendicular axis theorem)

laminar body

Z CM

IZ = ICM + Md2 (Parallel Axes theorem)

Any body d

Body

M

Position of axis

Moment of inertia

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ROTATIONAL MOTION 90º CM l Total mass = m

Uniform rod

ml 2 Iz = 12

Z

Iz =

l

ml 2 3

Z

ml 2 2 Iz = sin  3



Z 

Iz =

ml 2 2 sin  12

CM

Y

CM

b

l [The axis is the plane of plate]

Uniform rectangular plate

X

CM

b

Iy =

l

ml 2 12

Y

Z

X

mb 2 Iz = 12

[Axis Z is normal to plate]

Iz =

m 2 (l + b2) 12

Z COM

Ring or Hoop

R

Iz = mR2

[Z-axis : through CM and normal to plane of hoop]

CM R

I=

1 mR2 2

[Axis as diameter]

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ROTATIONAL MECHANICS SHEET

Z COM R

Uniform Circular Disc

Iz =

1 mR2 2

I=

1 mR2 4

I=

1 mR2 2

[Z-axis: through CM and normal to disc]

[Axis as diameter]

l R CM

Right Circular Cylinder

[Axis through CM along the length]

l R CM [Axis through CM normal to length]

CM

Right Circular Cone

R

mR 2 ml 2 I= + 4 12

I=

3 mR2 10

I=

2 mR2 3

I=

2 mR2 5

CM

Spherical Shell [Diametrical Axis]

Solid homogenous sphere R [Dimetrical axis]

R2

R2

Party Hollow sphere

5 5 2  R 2  R1  I = m R3  R3  5  2 1 

[Dimetrical axis]

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C18: Theorem of parallel axes is applicable for any type of rigid body whether it is a two dimensional or three dimensional, while the theorem of perpendicular axes is applicable for laminar type or two dimensional bodies only. C19: In theorem of perpendicular axs, the point of intersection of three axes (x, y and z) may be any point on the plane of body (it may even lie outside the body). This point may or may not be the centre of mass of the body. C20: Moment of inertia of a part of a rigid body (symmetrically cut from the whole mass) is the same as that of the whole body. e.g., in figure (a) moment of inertia of the section shown (a part of a circular disc) about an axis perpendicular to its plane and passing through point O is is also is

R O

1 MR2 as the moment of inertia of the complete disc 2

M

(a)

(b)

1 MR2. This can be shown as in figure. Suppose the given section 2

1 th part of the disc, then mass of the disc will be nM. n 1 Idisc = (nM)R2 2 1 1 Isection = Idisc = MR2 n 2

C21: Calculate the M.O.I. of a circular disc about an axis passing through its centre and perpendicular to its plane; mass of the disc is 2 kg and its radius is equal to 50 cm. Sol. Here

I=

1 MR2 2 2

1  50  I  2  0.25 kgm 2   100  2

Example 9. In the above question calculate its M.O.I. (1) about a tangent in its own plane; and (2) about a tangent  r to its plane. Sol. (1) From the theorem of parallel axis, M.O.I. about tangent in its own plane. I = M.O.I. about a parallel diameter + MR2 1 I  MR 2  MR 2 4 2 5 5  50  2 I  MR 2   2     0.625 kgm  4 4 100 (2) M.O.I. about tangent  r to its plane. = M.O.I. about  r axis + MR2 

1 3 MR 2  MR 2  MR 2 2 2



3  50   2  0.75 kgm 2  100  2

2

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ROTATIONAL MECHANICS SHEET Y

X

C22: The figure represents a thin circular disc of mass 4 kg and radius 0.60 metre. Calculate its moment of inertia about an axis XX´, passing through A and  r to its plane. A

O

Sol. From theorem of parallel axis, M.O.I. about XX´. = M.O.I. about YY´ + Mr2

0.2m

=

1 MR2 + Mr2 2

=

1 × 4 × (0.6)2 + 4 × (0.2)2 = 0.88 kgm2. 2

Y´

X´

C23: Calculate the M.O.I. of a uniform cylinder of mass 6 kg, redius 0.4 metre and length 1 metre about an axis passing through its centre and  r to its length.

 L2 R 2  2 I  M    0.74 kgm 12 4  

Sol.

C24: There is a uniform circular disc of mass 10 kg and radius 2 metre. Calculate the radius of gyration if it is rotating about an axis passing through its centre and  r to its plane. Sol. Here 

I=

1 MR2 2

R 2 = = 1.414 m 2 2

K=

C25: A solid sphere of mass 6 kg and radius 2 metre is rotating about its diameter; then calculate the radius of gyration. Sol. Here 

I=

1 MR2 2

K

2 2 R  2  1.26 metre 5 5

C26: For points masses each of mass m are attached to a massless string. Calculate the M.O.I. of the system about XX´. Sol. M.O.I about XX´ = mL2 + 0 + mL2 + m(2L)2 = 6 mL2. C27: Three point masses each of mass m are kept as shown in the figure, the string is massless. Calculate the M.O.I. of the system about an axis passing through (1) AB, (2) BC and (3) about an axis passing  rly through A. Sol. (1) M.O.I about AB = 0 + 0 + m(6)2 = 36 m (2) 64 m (3) 164 m

X m

m L

m L

m L

X´

A m 10m

8m B

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m

6m

m C

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Example 10. Three bodies have equal masses m. Body A is solid cylinder of radius R, body B is a square lamina of side R, and body C is a solid sphere of radius R. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina) (A) A (B) B (C) C (D) D Sol. For A For B

MR 2 IA  2 IB 

m 2 MR 2 R  R2  12 6





2 mR 2 5 where IA > IC > IB

For C

IC 

Hence, (B) is correct. A

Example 11. A point mass mA is connected to a point mass mB by a massless rod of length l as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axies BB and AA, which is parallel to each other and perpendicular mA I BB to the rod is I = 3. The distance of the centre of mass of the system from the mass AA A A is (A) (3/4)l (B) (2/3)l (C) (1/2)l (D) (1/4)l 2 Sol. IAA = mBl IBB = mAl2

B

l

mB

B

IBB 3 I AA mA 3 mB

But



mA = 3mB

mAxA = mB(l – xA) 3mBxA = mB (l – xA) 3xA = l – xA 4xA = l xA 

l 4

Hence (D) is correct.

Example 12. Three uniform rods each of mass m and length L metre are connected to form an equilateral triangle. Calculate the M.O.I. of the system about an axis passing through one of the vertex and  r to plane of the triangle. Sol. M.O.I. of the rod AB, and AC about the given axis are each equal to

1 mL2. and M.O.I. of the rod BC about 3

the given axis (from the theorem of parallel axis) is  3  1 1  mL2  mx 2  mL2  m  L 12 12  2 

2

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1 3mL2 10 2 2  mL   mL 2 4 12 Hence, M.O.I. of the whole system about given axis

A

1 1 10 18 3 I  mL2  mL2  mL2  mL2  mL2 3 3 12 12 2

m, L B

x

L, m

D m, L

C

Y

C28: The figure below represents two spheres of mass M and radius R. They are kept in contact as shown. Calculate M.O.I. about axis YY´. Sol.

I

7 7 mR 2  mR 2 5 5

I

14 mR 2 5

Y´

Example 13. Four spheres each of mass M and radius r are kept with their centres on the corners of a squre of side L. Calculate M.O.I. of the system about an axis passing through one side of the square. Sol. M.O.I. about AB 2 2 2  2   Mr 2  Mr 2   Mr 2  ML2    Mr 2  ML2  5  5  5 5

8 2  Mr 2  2ML2  M 4r 2  5L2 5 5



M A

L

L

M

BM

L D

L

C

M



C29: Two point masses are kept at distance of 9 metre on a thin uniform rod of negligible mass. The rod remains horizontal if it is pivoted at A. Calculate M.O.I. about a vertical line XX´ through the pivot A. Sol. Since the rod remains horizontal about A, hence X 1 × g(9 – y) = 2g × y 1kg 2kg A  y = 3 metre y 9–y  I = 2(3)2 + 1(6)2 X´ I = 54 kgm2 Example 14. The M.O.I. of a thin uniform rod of mass M and length L about an axis passing through its centre is I1. This rod is bent in form of a ring, and if I2 is the M.O.I. of the ring formed about its centre then compare I1 and I2.

For the ring,

1 ML2 12 2r = L



r

Sol. For thin rod,

I1 

L 2

L I2  Mr 2 M    2 



2

I1 1 4 2  2  ML2   I 2 12 ML2 3

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Example 15: Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of the triangle. Sol: Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC (i.e., D) is

A

CM C

B

ml 2 12 From theorem of parallel axes, moment of inertia of this rod about the asked axis is I2 = I1 + mr2

A

I1 =

CM r 30º

2

ml 2 ml 2  l   m   2 3  12 6

B

C

D

 Moment of inertia of all the three rods is

 ml 2  ml 2 I = 3I2 = 3  6  = 2   Example 16. Three identical thin rods each of mass m & length l are placed along x, y & z-axis respectively they are placed such that, one end of each rod is at origin ‘O’. Then moment of inertia of this system about z-axis is

ml2 2m l 2 (A) (B) 3 3 Sol. Since the one end of each rod is at origin ‘O’

(C) m l

2

ml2 (D) 4

ml2  3 According to perpendicular theorm. Iz = Ix + Iy Ix  I y 

Iz 

m l 2 m l 2 2m l 2   3 3 3

Hence (B) is correct.

Example 17: Find the moment of inertia of a solid sphere of mass M and radius R about an axis XX shown in figure. Sol: From theorem of parallel axis, IXX = ICM + Mr2 =

2 MR2 + MR2 5

=

7 MR2 5

x

x

x

CM

x r=R

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Example 18:Consider a uniform rod of mass m and length 2l with two particles of mass m each at its ends. Let AB be a line perpendicular to the length of the rod and passing through its centre. Find the moment of inertia of the system about AB.

A l

l

m

Sol:

m

IAB = Irod + Iboth particles

B

2

=

m(2l ) + 2(ml2) 12 y B

7 = ml2 3

Example 19: Find the moment of inertia of the rod AB about an axis YY as shown in figure. Mass of the rod is m and length is l. Sol: Mass per unit length of the rod =

A

m l

y

m Mass of an element PQ of the rod, dm =   dx  l  Perpendicular distance of this elemental mass about yy is r = x sin.  Moment of inertia of this small element of the rod (can be assumed as a point mass) about yy is, m  m dI = (dm)r2 =  dx  (x sin)2 = sin2 x2 dx l  l   Moment of inertia of the complete rod, x l

y B

r

1

m 2 I =  dI = sin2  x dx l x 0 0

=

P x

A

ml 2 sin2 3

Note: (i) I = 0 if  = 0

Q

y

(ii) I =

 ml 2 if  = or 90º 2 3

Example 20. Two rods of equal mass m and length l lie along the x-axis and y-axis with their centres origin. What is the moment of inertia of both about the line x = y : (A)

ml2 3

(B)

ml2 4

(C)

ml2 12

(D)

ml2 6

y

2

Sol.

Ix  I y 

ml sin 2  12

2

I

ml ml  , 24 24

I

ml2 12

45º

2

I

2m l 24

2

x

Hence (C) is correct.

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Example 21. A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by I = 2x2 – 12x + 27 the x-coordinate of centre of mass is (A) x = 2 (B) x = 0 (C) x = 1 (D) x = 3 2 Sol. I = 2x – 12x + 27 dI  4x  12 dx For minimum moment of inertia

or

dI 0 dx  4x – 12 = 0  x=3 But M.I. about an axis passing through centre is minimum.

Hence (D) is correct.

Example 22. The figure shows a uniform rod lying along the x-axis. The locus of all the points lying on the xy-plane, about which the moment of inertia of the rod is same as that about O is (A) an ellipse (B) a circle (C) a parabola (D) a straight line

y

O

x

ml 2 I0  3

Sol.

2

Here

 But

l  r   x    y2  2

P (x, y) r y

m l2 IP   mr 2 12 I0 = IP

l/2 x – l/2

or

2   m l2 m l2 l   m  x    y 2  3 12 2  

or

2   m l2 m l2 l   m  x    y 2  3 12 2  

or

2   3m l 2 l  m  x    y 2  12 2  

l2 l2 2  x   xl  y 2 or 4 4 or x2 + y2 – xl = 0 l2 l2 x 2  y 2  xl    0 or 4 4 2 l l2  2 x   y  or   2 4 This is an equation of circle.

Hence (B) is correct.

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Y

Question No. 23 to 26 The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90º. The apex lies at the origin and the base is parallel to X-axis. Example 23. The moment of inertia of the plate about the z-axis is

ML2 (A) 12

ML2 (B) 24

ML2 (C) 6

(D) None of these tan 45º 

Sol. 

A

l/2 45º

D

l/2 45º

B

h

In ACD, tan 45º 

y l  x 2h

  

l x 2l /2 y=x dA = 2y dx = 2xdx dm = 6 dA = 2 x dx



2 2y  dI  dm  dm(x) 2



M

X

h l/2

l 2

h

l

y

y

dx

C

12

 4x 2  dI  2xdx   2xdx x 2   12  dI 

2x 3 dx  2x 3  dx 3

8 dI  x 3  dx 3





l /2

0

8 3 8  x4  x  dx     3 3 4 8 l4  l4    3 64 24

Here

 



M area of plate M 4M  2 1 l l  l 2 2

I

4M l 4 M l 2   l 2 24 6

I

ML2 6

( l = L )

Hence (C) is correct.

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Example 24. The moment of inertia of the plate about the x-axis is

ML2 (A) 8

ML2 (B) 32

ML2 (C) 24

ML2 (D) 6

dI = (dm) x2 dI = (2x  dx) x2 dI = 2 x3 dx

Sol.



I  2

l /2

0

x 3 dx

 x4   l4  I  2    2   4  64  l 4  l 4 l 4 4M I  2     64 32 32 l 2 Ml2 I 8

Hence (A) is correct.

Example 25. The moment of inertia of the plate about its base parallel to the x-axis is

ML2 (A) 8

ML2 (B) 36 l  dI  dm   x  2 

Sol.



l/2

ML2 I 24

(D) None of these

2

l  dm   x  x0 2 

I

ML2 (C) 24

2

Hence (C) is correct.

Example 26. The moment of inertia of the plate about the y-axis is

ML2 (A) 6

ML2 (B) 8

ML2 (C) 24

(D) None of these

dm 4dmx 2 2 dI   2y  12 12

Sol. 

I

l /2

0

I

dmx 2 3

Ml 2 8

Hence (B) is correct.

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TORQUE AND NEWTON’S SECOND LAW In section 2.4, we have discussed about the turning effect of a force, torque. We know that for a body to be in totational equilibrium, the sum of all the torque acting on it must be zero. Now what happens if net torque is not zero. The case is similar if net forces acting on a body is not zero, the body wil accelerate according to Newton’s second law. In rotational motion also the law holds good, but require some modification as when net torque on a body about a given axis of rotation is not zero, body will have angular acceleration. The magnitude of angular acceleration can be obtained by Newton’s Second law in rotational motion. In translational motion we use

F = ma

...(1)

In rotational motion we use

=I

...(2)

Let hand side is the net torque acting on the body and on right hand side I is the moment of inertia of the body about the given axis and is the angular acceleration of the body. This corresponds to Newton’s second law for translational motion, a F, where torque has taken the place of force and correspondingly the angular acceleration takes the place of the linear acceleration a. In the linear case, the acceleration is not only proportional to the net force but is also inversely proportional to the inertia of the body, which we call its mass m, thus we can write a = F/m. In case of rotational motion moment of inertia plays the role of mass. As we have discussed that the rotational inertia of an object depends not only on its mass, but also on how that mass is distributed. For example, a large diameter cylinder will have greater rotational inertia than one of equal mass but smaller diameter (longer than previous). The former will be harder to start rotating, and harder to stop as its moment of inertia is larger. When the mass is concentrated farther from the axis of rotation, the rotational inertia is greater. This is the reason why in rotational motion the mass of a body can not be considered as concentrated at its centre of mass. For understanding the application of Newton’s second law in rotational motion, consider the rod shown in figure, pivoted at an end about which it can rotate. If two forces F1 and F2 are applied on it as shown from opposite directions, tend to rotate the rod. The respective torque of these forces are given as Clockwise torque due to force F1 is

1 = F1 sin  . r1

Anticlockwise torque due to force F2 is

2 = F2 . r2

F1 r1 r2 90º

Let 1 > 2, the rod rotates in clockwise direction with an angular acceleration , which can be shown from equation (2), as net torque is in clockwise direction, we have

 M l2  F1r1 sin   F2 r2     3 

...(3)

M l2 As moment of inertia of the rod of mass M and pivoted at one of its end is given as . Above equation (3) 3 will given initial angular acceleration of the rod.

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APPLICATION OF  = I For problems based on pure rotation : Solution technique : Step (i) Draw the free body diagram (if required) Step (ii) Identify, the axis of rotation and angular acceleration w.r.t. axis of rotation. Step (iii) Calculate net torque w.r.t. axis of rotation and then us  = I with + ve/ – ve sign convention. Example 27. A rod of length L and mass M is free to rotate about a vertical axis passing through its one end. A constant force F starts acting in horizontal plane. Initially the  direction of F is perpendicular to length of the rod. Calculate angular velocity of rod as a function of angular displacement  at any time t. Sol. Let = angular displacement of rod at time t = angular velocity at this time. = angular acceleration.  = FL cos 



L F

L cos

ML2 = 3

3F cos  ML

= 

F t=t

d 3F cos   d ML

3F

  d  ML  cos  d When

 2 3Fsin   C 2 ML = 0 ; = 0 ;







C=0

6Fsin  ML

KINETIC ENERGY OF ROTATING BODY A rigid body is made up of large number of particles. The kinetic energy of a rotating body is the sum of kinetic energy of its particles. Let  be the angular velocity. Then kinetic energy of i-th particle at distance ri from rotation-axis is 1 mi vi2 2 where ri = vi. The sum of such terms over the whole body gives K

K =  =  =

1 m (r )2 2 i i



1 m 2 r 2 2 i i

mi ri

1 (mi ri2 )2 2

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The quantity mi ri2 is the moment of inertia of the body about rotation axis, denoted by I. Then 1 2 I 2 Multiplying and dividing by I,

K =

K =

1 I2 2 2 I

L2 2I Here L is angular momentum about the axis. 

K =

C30: Whether a particle is in translational motion, rotational motion or in both it merely depends on the reference point with respect to which the motion of the particel is desribed. For example: Suppose a particle P of mass m is moving in a straight line as shown in figure (a), (b) and (c). v cos P P

v

v

90º r B

A

(a)

(b)

P



v

r v sin

C (c)

Refer figure (a): With respect to point A, the particle is in pure translation motion. Hence, Kinetic energy of the particle can be written as 1 mv2 2 Refer figure (b): With respect to point B, the particle is in pure rotational motion. Hence, the kinetic energy of the particle can be written as

K.E. =

1 1 v K.E. = l2 = (mr2)   2 2 r

2

1 mv2 2 Refer figure (c): With respect to point C, the particle can be assumed to be in rotational as well as translational motion. Hence, the kinetic energy of the particle can be written as

=

K.E. =

1 1 m(v cos)2 + I2 2 2

1 1  v sin   = m(v cos)2 + (mr2)   2 2  r 

2

1 2 mv 2 Thus, in all the three cases, the kinetic energy of the particle comes out to be the same.

=

C31: If a particle is moving in a circle it is in pure rotational motion about the centre of the circle, while for a moment it may be in pure translational motion about some other point.

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C32: If a particle P is moving in a circle, its angular velocity about centre of the circle (c) is two times the angular velocity about any point on the circumference of the circle (0) P O or c = 20 P This is because PCP = 2POP (by property of a circle) C c =

PCP t pp ,

0 =

POP t pp

From these relations we can see that c = 20. C33: If a rigid body is rotating about a fixed axis with angular speed , all the particles in rigid body rotate same angle in same interval of time, i.e., their angular speed is same (). They rotate in different circles of different radii. The planes of these circles are different. Linear speed of a particle situated at a distance r from the rotational axis is v = r or vr d . Here  is the angle between the dt line joining any two points (say A and B) on the rigid body and any reference line (dotted) as shown in figure. For example AB is a rod of length 4 m. End A is resting against a vertical wall OY and B is moving towards right with constant speed vB = 10 m/s. To find the angular speed of rod at  = 30º, we can proceed as under. OB = x = AB cos  x = 4 cos



r

C34: Angular velocity of a rigid body () is

or

or

 d  dx = –4 sin   dt  dt   d  (dx / dt)   =– dt 4 sin    10 = – = –5 rad/s 4 sin 30º

B A

 Y A

 O

[



vB = 10 m/s B

X

dx = vB = 10 m/s] dt

 d   0 . Here negative sign implies that  decreases as t increases   dt 

C35: A circular disc of mass 2 kg and radius 1 metre is to be rotated about its centre in a horizontal plane with angular acceleration of 4 rad/sec2. Calculate the torque required. Sol.   I 1 MR 2   2 1 a = × 2 × (1)2 × 4 = 4 kgm2/sec2. 2 

C36: A constant torque of 40 Nm is applied to a wheel pivoted on a fixed axis. At what rate power is being furnished to the wheel when it is rotating at 120 revolution per minute. 120 Sol. P =  .  = t . (2n) = 40 × 2 × = 502.6 Watt. 60

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C37: A fly wheel of moment of inertia 7.5 kgm2 is rotating at 240 revolution per minute; calculate its K.E.  = 2n = 2 ×

Sol.  K.E.

240 =8 60

=

1 2 1 I = [7.5] [8]2 2 2

=

1 × 7.5 × 64 × p2 = 2366 joule. 2

Example 28. A uniform rod of mass m and length l hinged at its end is released from rest when it is in the horizontal position. The normal reaction at the hinge when the rod becomes vertical is : Mg 3Mg (B) 2 2 Sol. Loss in P.E. = gain in K.E.

(A)

(C)

5Mg 2

(D) 2 Mg

l 1  I 2 2 2

or

mg

or

mg l 

or

3g  2 l

m l2 2  3

F

mg

l F  mg  m  2 2

or

l F  mg  m  2 2 F  mg  m

l 3g 5 mg  2 l 2

Hence (C) is correct.

Problems with pure rotation and pure translational of two or more connected objects. Solution techinque : Step (i) Draw the free body diagram. Step (ii) Assume linear acceleration of translatory object & angular acceleration of rotatory object.    Step (iii) Use F  m a for translation motion and   I  for rotational motion. Step (iv) Relate linear and angular acceleration and solve the obtained equation and get the answer. Example 29. Consider a pulley fixed at its centre of mass by a clamp. A light rope is wound over it and the free end is tied to a block of mass m. Find the linear acceleration of block. Sol. For rotational motion of pulley : TR =

MR 2 .  2

...(1)

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For translational motion of object : Mg – T = ma and a = Ra

...(2) ...(3)

MR 2  a  MaR .     R 2 2 Solving (2) and (4) TR 

mg 

a

ROTATIONAL MECHANICS SHEET

...(4) T a

Ma  ma 2 Mg

mg  m  M / 2

Example 30. a uniform rod of length l, hinged at the lower end is free to rotate in the vertical plane. If the rod is held vertically in the beginning and then released, the angular acceleration of the rod when it makes an angle of 45º with the horizontal (I = ml2/3) (A)

3g 2 2l

Sol. Loss in

(B)

6g 2l

(C)

2g l

(D)

2g l

P.E. = gain in K.E.

l 1 1  cos   I 2 2 2 Differenting both sides,

or

mg

l d d mg sin   I 2 dt dt

or

l mg sin   I  2



l mg sin 45º 2  m l2 3





3g 2 2l

Hence (A) is correct.

Example 31. A person pulls along a rope wound up around a pulley with a constant force F for a time enterval of t seconds. If a and b are the radii of the inner and the outer circumference (a < b), then find the ratio of work done by the person in the two cases shown in the figure is W1/W2. Sol.

 

W1 = F a  Fa = I 1 Fa 1  I 1   1 t 2 2

F Case I

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 1 Fa  2 W1  Fa  t  2 I  F2 a 2 2 W1  t 2I F2a 2 2 t Similarly, W2  2I W1 a 2   W2 b2



Example 32. A uniform cylinder of mass m can rotate freely about its own axis which is horizontal. A particle of mass m0 hengs from the end of a light string would round the cylinder which does not slip over it. When the system is allowed to move, the acceleration of the descending mass will be 2m 0g m0g 2m 0 g m 0g (A) m  2m (B) m  m (C) m  m (D) 2m  m 0 0 0 0 Sol. m0g – T = m0a Tr = I  mr 2 Tr   or 2 mr  ma T  or 2 2 m0 ma m0 g   m0 a or 2 m  m0 g   m0   a or  2 m0 g 2m0 g a  m m  2m0  Hence (A) is correct. m0  2 l/4

Example 33. For the pivoted slender rod of length l as shown in figure, the angular velocity as the bar reaches the vertical position after being released in the horizontal position is

g (B) l Sol. Loss in P.E. = gain in K.E. (A)

or or or 

mg

24g 19l

(C)

24g 7l

(D)

l

l/4

4g l

l 1  I 2 4 2

2 l 1  m l 2  l   2 mg    m     4  4 2  12  2 2 l 1 ml ml  2 mg     4 2  12 16 



24 g 7l

Hence (C) is correct.

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Problem based on situation in which same object is performing translation and rotational motion. This type of motion of an object can be considered as superposition of two types of motion taking place simultaneously. (A) translatory motion of centre of mass (B) rotational motion w.r.t. axis passing through centre of mass Solution technique : Step (i) Draw the free body diagram of object. T

T a mg

mg

Step (ii) Assume linear acceleration of centre of mass and angular acceleration w.r.t. an axis passing through the centre of mass. Step (iii) Use F = ma for translating motion of centre of mass and  = I for rotational motion w.r.t an axis passing through centre of mass. Step (iv) Relate a and  and solve the equations. Example 34. A cylinder of mass M is suspended through a two strings wrapped around it an shown in figure. find the linear acceleration of centre of mass of the cylinder.

T

T Mg

a

Sol.

a

mg mg mg For translatory motion of centre of mass Mg – T = Ma ...(1) For Rotational motion of cylinder w.r.t. an axis passing through the centre of mass MR 2  2 Relation between a &  a = R solving (1), (2) and (3) we get TR 

Mg 

...(2) ...(3)

MR  a  .    Ma 2  R

Mg = 3 Ma/2 a = 2g/3 Example 35. Two discs A and B touch each other as in figure. A rope lightly wound on A is pulled down at 2 m/s2. Find the friction force between A and B if slipping is absent.

R 1 kg 2

2R

2 kg

m/s2

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mR 2 2 mR a T  f   2 R ma 1  2 1 T  f    2 2 T–f=1

ROTATIONAL MECHANICS SHEET

TR  fR 

Sol.



f

f

...(1)

2

Also,

m´ 2R  2  4  R2 a f 2R  ´ 2 2 2R F=a=2N LECTURE – 7

ROLLING MOTION (a) Rolling motion of an object can be consider as superposition of two motions taking place simultaneously. (i) Translatory motion of the centre of mass (ii) Rotational motion of body w.r.t. an axis passing through the centre of mass. (b) In pure translational motion, all points move with the same linear velocity Vcm. In rotational motion about centre of mass all points move with the same angular velocity about the central axis. v

v+ R v

v v2 +

2 2

R

v

R v

v

v– R

R

R

Perfect Rolling : (a) If an object is in perfect rolling on a surface than point of contact of body with surface must be relatively at rest w.r.t. surface. v for perfect rolling A vAB = 0 B surface net velocity of A = net velocity of B. (b) Perfect rolling on ground : (i) The relative instantaneous speed of the point of contact during rolling is zero. (ii) For perfect rolling motion, work done against friction is zero. (iii) In case of perfect rolling the rotating object remain’s relatively at rest w.r.t. surface on which it is rolling therefore the force of friction will be static in nature. (iv) Due to static nature of friction in perfect rolling its value lives in between O and µN. Where µ is coefficient of friction between rolling object and surface. (v) For perfectly rolling v – R = 0 where v is velocity of centre of mass and is angular velocity. v

v

v– R VG = 0

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C38: If sphere is in perfect rolling on platform or on plank

ROTATIONAL MECHANICS SHEET sphere

v = R = v1

v

Solution technique : (i) Draw the free body diagram of different components of system. (ii) Assume linear and angular acceleration of different components of system.    (iii) User F  m a and   I  for translational and rotational motion respectively.. (iv) Relate linear and angular acceleration. (v) Finally solve the equations. Note : Friction is not necessary for perfect rolling.

v1

Example 36. The spool shown in figure is placed on rough horizontal surface and has inner radius r and outer radius R. The angle  between the applied force and the horizontal can be varied. The critical angle () for which the spool does not roll and remains stationary is given by 1  r  (A)   cos   R

1  2r  (B)   cos   R

Sol. For equilibrium F cos  = f and (F)r = fR f r  F R From equation (1) and (2) f r cos    F R  r   cos 1     R

or

1 (C)   cos

r R

F R r

1  r  (D)   sin   R

...(1)

...(2)

Hence (A) is correct.

Example 37. A round object of mass M and radius R rolls without slipping down an inclined plane set on angle  to horizontal. Find – (a) Linear acceleration of centre of mass. (b) Minimum coefficient of friction required for perfect rolling. assuming M.I. w.r.t. an axis passing through centre of mass is MK2. Sol. For translatory motion R mg sin  – fr = ma ...(1) a f r For rotatory motion m g fr R = MK2 () ...(2) sin mg cos and a – R = 0 ...(3) mg solve equation (1), (2) and (3) a

g sin  1  K2 / R 2

fr 

and

mg sin   R2  1  K 2 

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

fr  µ mg cos 

Where

 R2 5  K 2  2 



µ

(for perfect rolling)

2 tan  7

Imperfect Rolling (i) When point of contact of Rolling object is not relatively at rest. w.r.t. surface on which it in rolling then it is called imperfect Rolling. (ii) Nature of friction in this case is kinetic & it should opposing the motion of point of contact (strictly). (iii) Direction of linear acceleration and angular acceleration should be taken according to direction of friction (strictly). (iv) Nature of friction is kinetic therefore its value is µKN. Where µK is kinetic friction coefficient between surface and object. Transformation of Imperfect Rolling into perfect Rolling N v0 >

0R

v= R

v0 0

µN

imperfect Rolling

perfect Rolling

N v0
Ti (Because angular momentum remains conserved) But w = T = Tf – Ti > 0 Hence, workdone is positive. Hence (B) is correct. Example 69. A small object is attached to a light string which passes through a hollow tube. The tube is held by one hand and the string by the other. The object is stet into rotation in a circle of radius r1. The string is then pulled down, shortening the radius of the circle to r2. The ratio of the new kinetic energy to original kinetic energy is r1 (A) r 2

(B) 1

Sol. By conservation principle of angular momentum, mv1r1 = mv2r2  v1r1 = v2r2 ...(1) 1 mv 22  v  2  r  2 2   2   1 But ratio = 1  r2  mv12  v1  2

 r1  (C)   r 

2

2

 r2  (D)   r 

2

1

Hence (C) is correct.

Example 70. A small bead of mass m moving with velocity v gets threaded on a stationary

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semicircular ring of mass m and radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system ? (A) v/R (B) 2v/R (C) v/2R (D) 3v/R Sol. Applying conservation principle of angular momentum, mvR = (mR2 + mR2) 



vR v  2 2R 2R

R O

v m

Hence, (C) is correct.

Example 71. A child with mass m is standing at the edge of a disc with moment of inertial I, radius R, and initial angular velocity . See figure given below. The child jumps off the edge of the disc with v tangential velocity v with respect to the ground. Then new angular velocity of the disc is (A)

I 2  mv2 I

I  mvR (C) I

(B)

(I  mR 2 ) 2  mv 2 I

 I  mR    mvR (D) 2

I

Sol. Applying conservation principle of angular momentum, (I + mR2) = I´ + mv1R ...(1) But v1 – R´ = v (By relative concept) ...(2) From equation (1) and (2) (I  mR 2 )  mvR ´ Hence (D) is correct. I Example 72. Two men, each of mass 75 kg, stand on the rim of a horizontal large disc, diametrically opposite to each other. The disc has a mass 450 kg and is free to rotate about its axis. Each man simultaneously start along the rim clockwise with the same speed and reaches their original starting points on the disc. Find the angle turned through by the disc with respect to the ground. Sol. vrel = v1 + R ad vrel = v2 + R  v1 = vrel – R and v2 = vrel – R v1  Li = Lf vrel  or

 

vrel

MR 2  O = mv1R + mv2R – 2 MR 2  = mR (vrel – R + vrel – R) 2 = 2mR (vrel – R) 2 = (v1 + R) t

v2

...(1)

2 2 2   v1  R v rel  R  R v rel

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  t   

2 v rel

ROTATIONAL MECHANICS SHEET

...(2)

From equation (1) MR  2mR  2mR v rel 2

or

M    2m   2m v rel 2







 

 

2m v rel M / 2  2m 2 2 2m v rel   v rel v rel M / 2  2m

4m 75  4  75   4   M  225  150  375  2m 2 

4 radian 5

Example 73. A uniform disc of mass m and radius R rotates about a fixed vertical axis passing through its centre with angular velocity . A particle of same mass m and having velocity 2R towards centre of the disc collides with the disc moving horizontally and sticks to its rim. Find (a) the angular velocity of the disc. (b) the impulse on the particle due to disc. (c) the impulse on the disc due to hinge. Sol. (a) Applying conservation principle of angular momentum about centre of disc.

 mR 2  mR 2   mR 2  ´ 2  2 

 (b)

´

O

mR 2   3 3 2  mR 2 2

Fydt = Jy = 2 mR Fx dt  J x  mR´



mR 3

J  J 2x  J 2y  mR 22   mR

=

1 9

37 3

37 mR 3

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Example 74. A block of mass m is attached to a pulley disc of equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 5 m/s. Its velocity when the string becomes taut will be (A) 3 m/s (B) 2.5 m/s (C) 5/3 m/s (D) 10/3 m/s Sol. For block, –Tdt = mv – mv0 For disc, rTdt = I I Iv  2 r r

or

Tdt 

or

mv0  mv 

or

v0  v 



v

mv 2

v 3  v 2 2

2 2 10 v0   5  m/s 3 3 3

Hence (D) is correct.

Example 75. A uniform rod of length l is given an impulse at right angles to its length as shown. Find the distance of instantaneous centre of rotation from the centre of the rod.

cm x impulse

Sol. Let instantaneous axis of rotation is passing through point O. vcm – y = 0  vcm = y \

Here

v y  cm  Fdt = mvcm

vcm O

C y

x

Fdt

ml2  xFdt = I= 12 or

m l 2 Fdt  12 x

or

m v cm 

m l 2 12 x

vcm l2 y   12 x Example 76. A solid sphere of mass m and radius R is placed on a smooth horizontal surface. A sudden below is given horizontally to the sphere at a height h = 4R/5 above the centre line. If I is the impulse of the blow then find (a) the minimum time after which the highest point B will touch the ground. (b) the displacement of the centre of mass during this internal.

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Sol. (a) Here Fdt = mvcm and hFdt = Icm  hI 

or

4R 2 I  mR 2 4 5

 or

2I mR  = t  = t



t

or

Fdt = mvcm I = mvcm



v cm 



S  v cm t 

(b)

Fdt

2 mR 2 5

or



ROTATIONAL MECHANICS SHEET



 mR   2I

I m

I mR R   m 2I 2

Angular momentum conservation in collision If we consider a ball colliding with a bat, and neglected the force exerted by hand during collision, there is no external torque and bodies are freely colliding. If we pivot a body and then consider another body colliding with it, the pivot exerts an impulsive external force, in general. The external force acting on the system during collision is not negligible. Case (a): Free collision of free extended bodies: We use two conservation laws in this case. (i) Total momentum is conserved. (ii) Angular momentum about any point is cosnerved. 

Case (b) Pivoted bodies: In this case, external force ( Fext ) on the system arises from pivot (O) and is considerable. (i) Total momentum is not conserved. (ii) Angular momentum is conserved about the pivot.

O r N N

ROTATIONAL COLLISIONAND ANGUALR MOMENTUM We have discussed head on and oblique collisions. In this section we will discuss the different cases of collision of two bodies in which during or after collision rotational motion of the body is also taken into account.

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ROTATIONAL MECHANICS SHEET

A

A

L

L w

m

u

B

v

B

We first discuss teh simples case of a collsion which is shown in figure. A rod AB of mass M and length L is hinged at the point A, is hanging vertically. A small ball of mass m moving with a speed u horizontally strikes the end B of the rod elastically. Let us consider that after collision the rod will start rotating with angular speed  and the ball continues to move forward with a less speed v. Sot the ball may rebound but we need not to consider this case, as if b all will rebound, the result will give the velocity v negative. The values of v and w in this problem can be obtained in two ways using energy and angular momentum conservation or using impulse equations. Note about hinge at point A: Students should note that when ball will strike the rod, an external impulse will be developed at the hinge which will prevent and rod to move forward, as rod as only rotate, can not translate. Due to this external linear momentum of the system can not be conserved but angular momentum can be conserved about the hinge. As stated in above paragraph, here we cannot use linear momentum conservation but as so external torque is present we use angular momentum conservation about the hinge as

 ML2   muL = mvL +   3 

.....(i)

Here mul is the angular momentum of the ball before striking the rod and it is the only angular momentum before collision as rod is at rest. After collision aas ball will move with a speed v in same direction, its angular momentum is mul and that of rod is I as it starts rotation with initial angular velocity . As collision is elastic we use kinetic energy conservation before and after collison is 2 1 1 1  ML  2 2 mu = mv +  3  .....(ii) 2 2 2   Now using the above two equations, we get the values of v and . The other way of solving this problems is by breaking equation (i). It can be broken in two parts if required in some problems using impulse equations. Again consider the initial case when ball strikes the rod, a rnormal force is developed between the surface of ball and that of rod due to the push of ball against rod, the situtation is shown in figure. This interaction for F acts for a short duration (say dt) when the ball and rod are in contact. It retards the ball and its torque accelerates the rod in anticlockwise direction. We can use the impuse equatons for ball (linear) and rod (angular). For motion of ball, we have the linear impulse equation is mu = Fdt = mv .....(iii) For motion of rod, we have the angular impulse equation as 0 + Fdt = I

or

 ML2  FLdt =  3    

A

L

m

Fdt

u Fdt

B

.....(iv)

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If we use the above two equations along with equation (ii), we can solve the problem. Here if eqauation (iii) and (iv) are merged, it results equation (i). This type of working might be of more utility in solving the problem instead of directly using energy conservations. The previous case might also be of inelastic or partical elastic collision. If in previous problem with same initial conditions the collision is partical elastic and the coefficient of restitution is given as e, the angular momentum conservation equation (i) remains same as no external torque is acting but now we cannot use enrgy conservation as collision is not perfectly elastic. Here we use the definition of coefficient of restitution that it is the ratio of velocity of separtion after collision on the velocity of approach before collision. In this case it is used as L  v u or L = v eu .....(v) Solving the equations (i) and (v), we get the results v and . If this collision were perfectly inelastic, we use e = 0, which comes from equation (v), v = L, as no separation occurs in perfectly inelastic collision. Here it is important to be noted that the ball and rod will be separate even if inelastic collision takes place because ball is in translational motion and rod is in rotational motion. Figure explains the situation.

e=

A

A

L

m

L

L 

u

B e=0

A

A

v

B

m

u

 B B

v = L

Now we consider one more when ball sticks to the end of the rod and will rotate along with the rod. It is a specific type of inelastic collision as ball sticks to rod. Here again we can neither use energy conservation nor the coefficient of restitution equation. Only angular momentum conservation equation is sufficient for solving the solving the problem as one of previous varialbes reduces, the linear velocity of the ball. Consider figure. Ball and rod will start together in rotation with an initial angular velocity w. Thus we write the angular momentum conservation equation as muL = I [H = I = M.I. of rod plus ball]

 ML2 2 muL =  3  mL    

.....(vi)

Heremoment of inertia of the system taken combined that of rod pulsball as both are in rotational motion with angular velocity w after collision. Example (vi) will give us the angular velocity  of the system. A Now we discuss another case when rod is not hinged at end end. A Consider the collision shown in figure. Here rod AB is placed on a L L smooth surface which is free to move on it and ball moving shown in v  figure. d m As no external force is acting on the system, here we can also conserve u v linear momentum along with angular momentum. When the ball strikes B B the rod, the instantaneous axis of rotation is taken at the centre of the rod hence the equation of angular momentum conservation is written abut the centre of the rod. If after collision, ball moves with speed v1 and centre of rod moves with some angular velocity, let it be  , then according to angular momentum consevation we have. 2

1

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mud = mv1d l

 ML2  or mud = mv1d +  12   ....(vii)   From linear momentum conservation, we have mu = mv1 + Mv2 .....(viii) Equations (vi), (vii) and (viii) gives the unknown parameters after collisions v1, v2 and  if this collision is not elastic, we use coefficient of restitution instead of energy convervation equation (iv), as (v 2  d)  v1 .....(ix) u Here v2 + d is the linear speed of point P after collision as rod moves translationally with velocity v2 and rotates with angular velocity  and v1 is the final velocity of the ball after collision. For inelastic collisions, e can be taken as zero.

e=

Example 77. A rod of length R and mass M is free to rotation about ta horizontal axis passing through hinge P as in figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the rod hits the block B of mass m and stops. Find the ratio of masses such that the block B completes the circle. Neglect any friction.

M P

R m B

Sol. Minimum velocity required by block ‘m’ to complete the motion in 5gR conserving mechanical energy 1 2 R I  Mg  2 2

MgR I Cons. angular momentum wrt P before & after collision. 

I  m.R 5gR I.

MgR  mR 5gR I

P R.M

m

5gR

MgRI = m2R2 5 gR puting

I

ML2 MR 2  3 3

(since L = R)

M  15 m

Example 78: A rod of mass M is lying on a smooth horizontal table. A small disc sliding at speed V0 hits the rod at one end normal to its length. The disc comes to a stop just after the collision. (a) Write the velocity of the centre of mass of the rod just after the collision. (b) Write the angular velocity of the rod about its CM just after collision. (c) How much distance does the rod move in one rotation about CM ? Sol: Since there is no external force on the rod + disc system, momentum is conserved. Let V be the velocity of CM of the rod.

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(a) Conserving momentum, mv0 + M × 0 = m × 0 + MV V=

mv 0 M

M

.....(i)

 m

(b) Angular momentum conservation about a point of space initially coinciding with the centre of the rod: mv0

v

L = I 2

ML2   12 6mv 0 .....(ii) ML (c) As there is no friction, the rod slides with constant velocity after the collision. Let it move by s in time t. In one rotation angle turned is 2. Hence 2 = t .....(iii)

 =



s = =

v2  mv 0 L 2 × = M  6mv0  3    ML 

Example 79. A rod AC of length L and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving with velocity v strikes rod at point B which is at a distance L/4 from mid point making angle 37º with the rod. The collision is elastic. After collision find A B C (a) the angular velocity of the rod. 37º (b) the distance which centre of the rod will travel in the time in which it makes l/4 half rotation. (c) the impulse of the impact force. Sol. (a) The ball has v´, component of its velocity perpendicular to the length of rod immediately after the collision u is velocity of COM of the rod and  is angular velocity of the rod, just after collision. The ball strikes the rod with speed v cos 53º in perpendicular direction and its component along the length of the rod after the collision is unchanged. u Using for the point of collision. Velocity of separation = Velocity of approach 3v   l    u   v´  5  4

...(1)

Conserving linear momentum (of rod + particle), in the direction  to the rod. 3 mv .  mu  mv´ ...(2) 5 Conserving angular moment about point ‘D’ as shown in the figure

V´ D

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ROTATIONAL MECHANICS SHEET

 l ml 2  0  0   mu   4 12   u

l 3

...(3)

u

By solving

24 v 72 v , W 55 55 l  

(b) Time taken to rotate by  angle t =

l/4

Ndt

l 3 Using angular impulse-angular momentum equation.

In the same time, distance travelled = u2 . t =

 N . dt .

l m l 2 72 v  . 4 4 55 l

24mv 55 or using impulse – momentum equation on Rod

 N . dt 

 Ndt  mu 

24m 55

m is the collision elastic ? M Sol: For elastic collision initial kinetic energy should be equal to the KE just after the collision. That is, K i = Kf

Example 80: In the previous example for what ratio

1 1 1 mv 20 = 0 + MV2 + I2 2 2 2 1 1 ML2 2 2 = MV +  2 2 12 2

1 1 ML2  6mv 0   mv 0  = M.  +    M  2 2 12  ML 



m2 m2 m = +3 M M 1 =4



2

m M

=

m M

1 4

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Example 81: A small disc and a thin uniform rod of length L, whose mass is  times greater than the mas of disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity v, after which it elastically collides with the end of the rod. Find the velocity of the disc angular velocity of the rod after the collision. At what value of h will the velocity of the disc after the collision be equal to zero ? Reverse the direction ? Sol: The situation is shown in figure. If mass of disc is m, then mass of rod hm. if v1 and v2 be the velocities of the disc and rod after collision, using linear momentum conservation, we have mv = mv1 + hmv2 or v = v1 + v2 ....(x) If after collision, rod starts rotating with angular speed w, using consrvation of angular momentum we have

or

L mv = 2

 mL2  L    + mv1 12 2  

v

1 L + v1 6

=

6(v  v1 ) .....(xi) L As collision is elastic, using kinetic energy conservation, we get or



=

1 2 mv = 2

or

2

v

1 1 1 mv12 + v 22 + I2 2 2 2

3 (v  v1 )2 2 = v + (v - v1) +   2 2 1

Solving for v1, we get

4 v 4 Here we can see that v1 is zero if h = 4 and v1 will be negative when  > 4. v1 =

Example 82. On a smooth table two particles of mass m each, travelling with a velocity v0 in opposite directions, strike the ends of a rigid massless rod of length l, kept perpendicular to their velocity. The particles stick to the rod after the collision. Find the tension in rod during subsequent motion. l l mv 0  mv 0  I Sol. 2 2 ml 2 mv0 l    2 ml 2 mv0 l   or 2 2v  0  l 2  l 2  l   2v0  T  m   m        2 2 l  2mv02 T  l

v0 m l v0 m

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Example 83. A solid uniform sphere of radius R and mass M rolls without slipping with angular velocity w0 when it encounters a step of height 0.4 R. Find the angular velocity immediately after inelastic impact with the rough step. Sol. Applying conservatio principle of angular momentum about point of impact, mvcm (R – h) + Icm 0 = I´

0

h

2 7 mR 0  R  h   mR 2 0  mR 2´ 5 5

or

7 7 mR 2 0  mRh 0  mR 2´ 5 5

or

7 7 R 0  h 0  R´ 5 5



´

5  7  5h  0   7R  5 

´

5  7R  5h  0   7R  5 

´

0  7R  2R  7R

´

5 0 7

CONCEPT OF TOPPLING Rolling is generally a difficult concept to grasp. The primary question that rises in the students’ mind is how can a point in the body have zero velocity and yet the body moves forward. Let us consider a simple example as shown in figure.

F Fig. : Force acting on a flat body

Assume that a force F acts on a flat body as shown above. How will the body move ? Obviously the body will move forward translationally. The situation is shown in figure.

v v Fig. : Motion of the body

All points in the body will move with the same velocity and acceleration. Therefore, the body will move translationally. Now assume that the body becomes less wide. F

What will the situation be ? This is shown in figure.

Fig. : Force acting on a body with less width

In this case as well, the body is likely to move translationally. However as the body becomes less and less wide, something else happens. Let us now consider the situation in figure. In this case as the force F increases, the frictional force also increases. At some point

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the body starts toppling. At this instance the body topples about point A. the motion in this situation is shown in figure.

Fig. : Force on a very thin body

F

As the force increases, the body starts toppling about point A. The body now has fallen down. Let us step back and consider the situation in greater depth.

• • •

Obviously this kind of motion is not translational. The body has moved forward A as a whole. We can say that though because the centre of mass of the body has Fig. : Body topples moved, yet the motion has not been translational. What can we say about the motion ? We can observe the following points : Different points in the body move with different velocities and accelerations. Whereas the topmost point has the greatest velocity, the point A has zero velocity. The point of contact A is at rest, even while the body moves. The body has moved forward even when one point has zero velocity. We can therefore conclude that the body can move forward even if a single point in the body has zero velocity. v each one of these points is shown below : Different points in the body move with different velocities and accelerations. Whereas the topmost point has the greatest velocity, point A has zero velocity. This is shown in figure. The point of contact A is at rest, even while the body moves. This is shown in figure.

v1 v2 Fig. : Different points have different velocities

The body has moved forward even when one point has zero velocity and zero acceleration. This is shown in figure.

0 A

F

Fig. : Point A has zero velocity. Velocity profile of other points

B

F C Fig. : Point has moved

The centre of mass was initially at B. It has then moved forward to point C. Therefore the body has moved forward. • • •

We call this kind of motion toppling. Now let us look at when exactly a body will topple : A body will topple when the base is smaller. A body will topple when the force exerted is higher from the ground level. A body resting on a surface, which has friction, is more likely to topple.

Look at the situation below. Which body is more likely to topple ? F

F B

C

Fig. : Which body will topple ?

Which body will topple ?

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Obviously body B, because body C is wider then body B. This illustrates the first point that for a body to topple, the base has to be smaller. Consider the next situation now : F B

F

C

Fig. : Which body will topple ?

In this case, the width is the same.

Which body will topple ? In the second case, the height at which the force is applied is lower. Therefore the body as shown first is likely to topple. If the surface has no friction, the body will be more likely to move by pure translation than by toppling. Look at the situation below. F C

F C

Fig. : Which body will topple ?

Which body will topple ? Obviously the the square, because it has a less wide base. Why? Explanations follow. Similarly, we can see that a pentagon is more likely to topple than a square. Similarly an octagon will be more likely to move by toppling than a pentagon. Therefore, we have seen that as the body becomes less and less wide, it tends to move more and more by toppling.

F

Fig. : An octagon is more likely to move by toppling

The question, however, still remains unanswered.

F

How and when will a body move by translation and when will it move B b by pure translation ? Let us consider the situation shown in figure a A Assume that the width is a and height is b. Let us draw the force diagram, just as Fig. : When will the body topple ? the body is about to topple. Assume that (µ) is the coefficient of friction. We know from the earlier discussions that the body will topple when the normal force shifts to the rightmost point. For the body to topple about A, taking moments about A, (why do we take point A ? Because maximum number of force i.e. friction and normal force pass through A. It is also correct to take torque about centre of mass, though this will be lengthier). Total torque = Fb –

F B

mga 2

If torque > 0 then the body topples.

mg a

b N A

Fig. : Force diagram on the body

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PHYSICS There

• •

ROTATIONAL MOTION

F>

ROTATIONAL MECHANICS SHEET

mga 2b

What this means is that mga If F < then body will move in pure translation and will not topple. 2b mga If F > , the body will topple. 2b Let us study the concept to toppling in a little more mathematical detail. While we outline every scenari, try to visualize physically the motion of the body. We have already established that for force F >

mga , the body will start moving by topping. 2b

Let us now bring into this picture the friction, width and height of the force and study toppling. Consider first a square body of width a. Assume that the force is applied at the top of the square. Assume also that the coefficient of friction is µ. The situation is shown in figure.

F N mg

Using the equations described above, the body will topple for force

The body will topple when F >

f Coefficient of friction = µ Fig. : Condition for toppling a square

mga F> (since a = b). 2b mg or F > 0.5 mg 2

If the coefficient of friction = 0.2 what will happen to the square ? Will it topple or translate ? For the body to just topple or translate, the force equations are same. The force equation for just toppling or just translation is F–f=0



F=f

N – mg = 0



N = mg;

...(1) f  µmg

If µ = 0.2, Maximum frictional force = 0.2 mg Therefore using equation (1) the force required to move the body by translation is F = 0.2 mg. The Force required to topple the body is F = 0.5 mg. Therefore transalation happens first. The body will translate and not topple. What happens if the coefficient of friction is 0.6 ? In such a case, the force required to topple is F = 0.5 mg. This remains unchanged. However the force required to move the body by translation is 0.6 mg. In this case toppling will take place first. We can now summarize the scenario below : For

µ > 0.5, the body will move by toppling.

For

µ < 0.5, the body will move by translation.

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This off course assumes that the force is exerted at the top. What happens if the force is exerted at any other point ? In such a case, the coefficient of friction will have to be higher to move the body by toppling. Visualize this. It is very difficult for a square to topple. it requires a very rough surface (µ > 0.5). What is the value of frictional forces when µ < 0.5 and µ > 0.5 ? The figure explains it all. F = µmg

F < µmg N mg

N

mg

A

A

f = µmg Coefficient of friction = µ < 0.5 Moves by translation Friction = µmg

f < µmg Coefficient of friction = µ > 0.5 Moves by translation Friction < µmg

Note the interesting fact in the second case. The body topples and moves. but during this time f < µmg. We all know that when a body translates frictional force = µmg. The interesting thing to note is that a body a shown above can move in other methods — not necessarily by just translation. It can topple. And what happens when the body topples ? The body moves, but yet frictional force < µmg. This is a very important point. Students need to visualize this scenario very clearly in their minds. When the body topples, frictional force is less than µmg. Also it requires less effort to move a body. This is a classic property of toppling. • • • •

Let us summarize the characteristics of toppling that we have already discussed earlier : All points have different velocities and accelerations. The lowermost point will have zero velocity and acceleration when the body just starts toppling. Frictional force is less than the maximum value of friction. It requires less effort or force to move the body by toppling.

If this is the case, why doesn’t a square body not topple so easily ? The reason is very simple. The coefficient of frictions needs to be higher than 0.5 for a square body to topple (assuming that the force acts at the topmost point). This does not happen in reality because if µ = 0.5, the surface must really be very rough. Most surfaces are smoother than this. Hence a square body prefers to move by translation and not by toppling. F mg

N A

Fig. : Force required to topple a hexagon

Again we can take torque about A. The side of the hexagon is a. Therefore the height will be a tan 60. You can easily see that this is the case, by looking at figure. 60º a

a tan 60

But now let us consider a hexagon. a hexagon. Again assume that the coefficient of friction is µ and that the force is being exerted at the top. Assume the width of the sides to be a.

Fig. : Height is a tan 60 www.physicsashok.in 83

PHYSICS The torque about A is,

ROTATIONAL MOTION

mga =0 2

F a tan 60 –

Therefore the body will topple for F >

ROTATIONAL MECHANICS SHEET

mga 2 tan 60

Now tan 60 ~ 1.732. Therefore the body will topple for F > 0.21 mg You can now see the huge difference. for µ < 0.21, the body will translate and for µ > 0.21, the body will topple. The situation is again reproduced below F

F mg

mg

N

N A

A f = µmg

f = µmg

Coefficient of friction = µ < 0.21 Moves by translation Friction = mµg

Coefficient of friction = µ > 0.21 Moves by translation Friction < mµg

Fig. : Motion for toppling a hexagon body

Again the same conditions that were discussed before, regarding toppling hold true. When a body just topples, the coefficient of friction is less than µmg. The point of constact is a rest (i.e. point A). Different points have different accelerations. These are typical characteristics of toppling and you should now be clear about these. Note that we can make an interesting observation. The value of µ required to make a square topple is 0.5, whereas the value of µ required to make a hexagon topple is 0.21. this obviously bears out the fact that it is easier to topple a hexagon than a square. Similarly you will find that the minimum value of µ required to make a body topple will decrease if the body is an octagon. It will reduce still further for a decagon. This means that a decagon will move by toppling more easily than by pure translation. The force required to move it by toppling is less. What can we say about an infinite-sided polygon ? Extending the same logic, you will find that for values of µ < 0, the body will topple. What does this mean ? The body always topples. Even if a small force is applied, the body will topple. Therefore note the interesting observation. Circular bodies move predominantly by toppling. Again let us reproduce the properties of toppling here. • All points have different velocities and accelerations. • The lowermost point will have zero velocity and acceleration when the body just starts toppling. • Frictional force is less than the maximum value of friction. • It requires very little effort to move a body by toppling. These are precisely the properties of toppling a circular body as well. The only difference is that since the side of the circle is 0, virtually no force is required to topple the body. The body moves by toppling and F < µN. Now let us introduce a very interesting property. Toppling is Rolling. They are one and the same. Visualize toppling motion and you can visualize rolling motion. If you look at the properties of toppling described above, these are exactly the properties of rolling.

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Toppling is therefore the same as rolling. To visualize rolling motion always think of toppling motion. It is easier to visualize toppling. How does a body move when it is toppled ? This is shown below F

C

F B

F A

C

A B C Fig. : Visualization of toppling

To start with, A is at the edge. During the first round of toppling, A is at rest and B falls down. Now the body topples about point B. Point C touches the gound. In the next round, the body topples about point C. An interesting point to observe here is that the body topples about different points. First it is point A, which is at rest, and the body topples about A. Then the body topples about point B and B is at rest. Therefore different points have zero velocities and accelerations. This is exactly the situation with rolling as well. In rolling the point of contact has zero motion. This point however keeps on changing. It is very important for the students to be able to visualize toppling and understand rolling in this contaxt. • • • • •

To summarize Toppling is the same as rolling. During toppling or rolling F < µmg. It requires negligible force to topple or roll a disc. Point of contact has zero motion with respect to the surface. Different points have zero motion at different points of time.

F > 0.21 mg

Another question that the student may ask is what happens in the case of a hexagon if the value of the force F > 0.21 mg. In such a case, the body will difinitely topple, but the point of contact will also move. This will not have zero acceleration. This is shown in figure.

A Acceleration

f = µmg

The point of contact will move. The body will topple, but in a different way. the point of contact moves. We call this kind of motion sliding or skidding. Notice that when a body slides or skids,

Coefficient of friction = 0.21 Fig. : Value of force is higher than the force required to topple.

Frictional force – µmg and not less than µmg. F

What is the minimum force required for the body to topple ? The body will topple when the normal force passes through the edge. The force diagram at this stage is shown in figure.

Mass=1 kg

mg N f

taking torque about point A we have

A Fig. : Force diagram when body is just going to topple.

mg (1/2)= 4F F = (1/8)Mg = 0.125 mg The force required for the body to move by translation is 0.2 mg, but the body topples when the force is 0.125

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ROTATIONAL MOTION

mg. Therefore, it is obvious that the body will topple before it translates. Example 84. A uniform cube of side ‘b’ and mass M rest on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point, at a height 3b/4 above the base. What should be the coefficient of friction (µ) between cube and table so that b is will tip about an edge before it starts slipping ? (A) µ > 2/3 (B) µ > 1/3 (C) µ > 3/2 (D) None Sol.

b 3b F 2 4 F = f < µ mg From equation (1) and (2) F < µ mg 2mg  µg 3 2 µ or 3 mg

F 3b/4

...(1) ...(2)

Hence (A) is correct.

Example 85. A uniform cylinder rests on a cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm in height, which of the following is the minimum acceleration of the cart needed to cause the cylinder to tip over ? (A) 2 m/s2 (B) 4 m/s2 (C) 5 m/s2 (D) the cylinder would slide before it begins to tip over. Sol. Taking torque about O, ma0 × 5 > mg × 2 ma0 or 

a0 

2g 4

a 0 min

2  10   4 m / s2 5

O

a0 mg

Hence (B) is correct.

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THINKING PROBLEMS 1. Can the mass of a body be taken to be concentrated at its centre of mass for the purpose of calculating its rotational inertia? 2. If two circular discs of the same weight and thickness are made from metals of different densities, which disc will have the larger moment of inertia about its central axis? 3. About what axis would a uniform cube have its minimum rotational inertia? 4. Consider four bodies : a ring, a cube, a disc and a sphere. All the bodies have the same mass. The ring, disc and sphere have the same diameter, equal to the length of the cube on each edge. All rotate about their axes through their respective centres of mass. Which one has the largest moment of inertia? Which is the smallest? 5. A person can distinguish between a raw egg and a hard boiled one by spinning each one on a table. Explain. 6. A wooden sphere rolls down two different inclined planes of the same height but different inclines. Will it reach the bottom with the same speed in each case ? Will it take longer to roll down one incline than the other? If so, which one and why ? 7. A man, with two equal masses held at arm’s length, stands on a rotating table. If he moves the masses and puts them over his shoulders, does this change his speed of rotation? Explain. 8. A student stands on a platform that can rotate only about a vertical axis. In his hand he holds the axle of a rimloaded bicycle wheel with its axis vertical. The wheel is spinning about the vertical axis with an angular speed 0, but the student and the platform are at rest. The student turns the wheel through 180º, that is, the axle of the wheel is held downward by the student. What happens ? 9. A diver can turn several somersaults before striking water. Explain. 10.Show that in the course of diving, the rotational kinetic energy of a diver, turning somersaults before striking water, increases. What is the source of this increased energy? 11. If the polar ice caps melt and spread uniformly, how will the length of the day be affected ? 12.A circular turntable rotates at constant angular velocity about a vertical axis. There is no friction and no driving torque. An icepan containing ice also rotates along with it. The ice melts but none of the water escapes from the pan. Is the velocity now greater, the same as, or less than the original velocity? Give reasons for your answer. 13.In order to get a billiard ball to roll without sliding from the start, the cue must hit the ball not at the centre but exactly at a height of 2/5 R above the centre, where R is the radius of the sphere. Explain. 14.A student stands on a turntable with a rim-loaded bicycle wheel. he holds the shaft of the wheel vertical and in the beginning, there is no motion of the turntable. Now the student is asked to rotate the wheel in a clockwise direction relative to him. What happens to the motion of turntable + student? 15.The melting of the polar ice caps is supposed to be a possible cause of the variation of the earth’s time of rotation. Explain.

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SOLUTION OF THINKING PROBLEMS 1.

No.

I1 =

2.

Now

 

1 2 1 mr1 I 2 = mr22 and 2 2 I1 r12 = I 2 r22 m   r12 t 1   r22t  2

r12 2  r22 1 I1  2  I 2 1

or

I

1 

Thus, the one with greater density will have less rotational inertial. 3.

About a diagonal, because the mass is more concentrated about a diagonal.

4.

I (ring) = (1/4)ma 2 where a is the length of each edge of the cube, I (cube) = (1/6)ma 2, I (disc) = (1/8)ma 2, I (sphere) = (1/10) ma 2. I1 : I2 : I3 : I4 = 1/4 : 1/6 : 1/8 : 1/10 = 30 : 20 : 15 : 15. Thus, the ring has the greatest moment of inertia and the sphere the smallest.

5.

The hard boiled egg will continue to sping if they are momentarily stopped and then let go.

6.

The acceleration down an inclined plane is given by a = g sin /(1 + k2/r 2) where k is the radius of gyration. h (height of the plane) = s sin  where s is the length of the plane.

1 2 1 at = g sin a t 2 / (1 + k 2 / r 2 ) 2 2

Now

s=



t 2 sin 2 a = constant



t  1/ sin 

Thus, the greater the time of descent the less the inclination to the horizontal. Now

v 2  2as  2

g sin  h 2 2 1  k / r sin 



v = a constant

Thus, the sphere will reach the bottom with the same speed, whatever be the inclination. 7.

The only external force acting on the system (man + table) is gravity and reaction of the ground, and those exert no torque about the axis of rotation. Hence, the angular momentum of the system is conserved about this axis. When the man puts the masses on his shoulders, the moment of inertia of the system about the axis decreases and so the angular speed of the system increases.

8.

The platfor starts rotating with almost double the initial angular momentum of the wheel. To justify this, let us consider ‘wheel + student + platform’ asour system. The initial angular momentum of the system is I00, where I0 is the moment of inertia of the wheel about the vertical axis. Since there is no external torque on the system about the vertical axis, the angular momentum about the vertical axis must be conserved about this axis. Let I be the moment of inertia of ‘student + platform’ about the vertical axis. The angular momentum of the wheel about the vertical axis is now –I00. Let  be the angular velocity of the platform. Then by the principle of conservation of angular momentum I00 = –I00 + I or I = 2I00    2 0 on account of negligible m.i. of the platform.

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Let us consider the diver as a system and his limbs as its components. When he leaves the diving board, he has a certain angular speed 0 about the horizontal axis through his centre of mass. Now there are no external forces acting on him except gravity, and gravity exerts no torque about his centre of mass. His angular momentum remains constant, and I00 = I. When he pulls in his limbs, his moment of inertia decreases and consequently, his angular velocity increases.

10. Since the diver is free from external torque about the horizontal axis through his centre of mass, his angular momentum remains constant. Therefore. I00 = I or I 02w 02 = I 2w 2 ....(i) When he pulls in his limbs, his moment of inertia about the horizontal axis decreases.

1 1 > I I0



I < I0

or

1 2 1 I   I 0 02 2 2

or

or

1 1  I 2 2  . I 02 02 I I0

The source of this increased kinetic energy is the diver himself, who does work when he pulls the parts of his body together. His body energy is converted into his kinetic energy. 11. If the polar ice caps were to melt, the water formed would spread on the surface of the earth and so its moment of inertia would increase. By the law of conservation of angular momentum, the speed of rotation would be reduced, that is, the length of the day would increase. 12. Due to the accumulation of mass near the edge, the moment of inertia of the system increases and so the angular velocity of the system decreases. 13. Let the ball be struck by a horizontal force F at a height d above the centre. Then the torque exerted by it on the ball about its centre of F.d. By the formula  = I, we have F.d =

2 MR2 ×  5

Let a be the acceleration of the centre of mass produced by this force. Then F = Ma. For rolling without sliding, v = R and a = .R;  F = M R 

MR  d 

2 MR 2   5



d=

2 R 5

14. The system (turntable + student) will turn in the opposite direction. Since the entire system is free from external torques, the angular momentum of the system (wheel + turntable + student) remain constant at zero. If I = moment of inertia of turntable + student and I0 = moment of inertia about the vertical axis of rotation of the table and 0 = angular velocity of rotation (relative to the earth), then I00 + I= 0 or

 

I 0 0 I

15. A liquid of given mass has less inertia for rotational motion than the same mass of solid because a liquid cannot sustain shearing force. It always tends to move translation-wise. Hence, the melting of the ice caps would result in the reduction of the rotational inertia of the earth and hence, the angular speed of rotation of the earth would increase or time of rotation decrease. If water were to spread uniformly, the result would be as in Example 11.

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ASSERTION & REASON A statement of Statement-1 is given and a Corresponding statement of Statement-2 is given just below it of the statements, mark the correct answer as – (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 is NOT correct explanation of Statement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) If Statement-1 is false but Statement-2 is true. 1. 2.

Statement-1 : Statement-2 : Statement-1 :

3.

Statement-2 : Statement-1 :

4.

Statement-2 : Statement-1 : Statement-2 :

5.

Statement-1 : Statement-2 :

6.

Statement-1 : Statement-2 :

In pure rolling motion, net work done by friction is zero. Sum of translational work done by friction and rotational work done by friction is zero. A uniform solid cylinder rolling with angular velocity  along a plane surface strikes a vertical rigid wall. Angular velocity of cylinder when it begins to roll up a wall is less than the initial Angular velocity (). After striking the vertical wall angular velocity If bodies slide down an inclined plane without rolling then all bodies reach the bottom simultaneously. Acceleration of all bodies are equal and independent of the shape. The force of friction in the case of a disc rolling without slipping down an inclined plane is zero. When the disc rolls without slipping, friction is required because for rolling condition velocity of point of contact is zero. As the radius of earth reduces by 50% without any change in mass, length of a day reduces. Angular momentum conservation provides drop in time as  decreases to 25% of the original. If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. The linear momentum of an isolated system remains constant. [JEE, 07]

MATCH THE COLUMN 1.

A circular body of mass M and radius R, initially spinning about its centre of mass with

0 is gently placed

on a rough horizontal surface. The moment of inertia of body about its C.M. is  CM  MK 2 . If the coefficient of friction between the body and the surface is

 then:

Column I (A) Translational work done by the friction (B) Rotational work done by the friction (C) Larger the moment of inertia of body, the time required for rolling motion (D) Larger the moment of inertia of the body, work done by the friction

Column II (P) –ve (Q) +ve (R) smaller (S) longer (T) greater

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In the following column - I mass of each object is m and circular of radius R. Column - II represents moment of inertia.

Column I

Column II

(A) Full ring

(P) mR 2

(B) Half ring

(Q)

mR 2 2 mR 2 (R) 4  mR 2 (S) 2

(C) Quarter ring (D) Arc making an angle 3.



at the centre

Suppose a force Of is applied at the top most point of a rigid body of radius R and mass M.

Column I

4.

Column II

(A) Force of friction will be zero for (B) Force of friction will be forward for (C) Force of friction will be backward for

(P) Solid sphere (Q) Zero (R) Ring

(D) If

(S) No body

a  R then force of friction

A rigid body of mass M and radius R rolling without slipping on the inclined plane, then the magnitude of force of friction.

Column I

5.

Column II

(A) For ring

(P)

Mg Sin  2.5

(B) For solid sphere

(Q)

Mg Sin  3

(C) For solid cylinder

(R)

Mg Sin  3.5

(D) For hollow sphere

(S)

Mg Sin  2

A rigid body is rolling without slipping on the horizontal surface

Column I

Column II

(A) Velocity at point A i.e. VA

(P)

(B) Velocity at point B i.e. VB

(Q) Zero

(C) Velocity at point C i.e. VC

(R) V

(D) Velocity at point D is i.e. VD

(S) 2 V

V 2

LEVEL – 1 1.

A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given, kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll with the same KE as soon as released towards a wall which is at the same distance from the ring and the cylinder. Then: (A) The cylinder will reach the wall first. (B) The ring will reach the wall first. (C) Both will reach the wall simultaneously. (D) None of the above.

2.

A solid homogeneous sphere is moving on a rough horizontal surface, partially rolling and partly sliding. During this kind of motion of the sphere: (A) total KE is conserved. (B) angular momentum of the sphere about the point of contact is conserved. (C) only the rotational KE about centre of mass is conserved. (D) angular momentum about the centre of mass is conserved.

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ROTATIONAL MECHANICS 3.

The angular momentum of a projectile (in X-Y plane) about origin (point of projection) at time t has: (A) only Z component. (B) only X and Y components. (C) All the three X, Y and Z components. (D) only Y-component.

4.

A solid sphere of radius R and mass M is pulled by a force F acting at the top of the sphere as shown in figure. Friction coefficient is sufficient enough to provide rolling without slipping. Work done by force F when the centre of mass moves a distance S is : (A) FS

5.

7.

v

(B) v

3FS 2

Rough

(C)

2

(D)

2v

v 2

(B) a smooth inclined surface. (D) a rough inclined surface.

A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. At what height from the ground should it be hit so that the shell does not slip on the surface?

2 3.R

(B)

5 4.R

(C)

5 3.R

(D)

3 2.R A

What is the moment of inertia of a triangular plate ABC of mass M and side BC = a about an axis passing through A, shown in figure and perpendicular to the plane of the plate?

Ma 2 (A) 6 9.

(D)

A sphere cannot roll on: (A) a smooth horizontal surface. (C) a rough horizontal surface.

(A) 8.

(C) Zero

A body of mass M slides down an inclined plane and reaches the bottom with velocity v. If a ring of same mass rolls down the same inclined plane, what will be its velocity on reaching the bottom? (A)

6.

(B) 2FS

F

3Ma 2 (B) 4

Ma 2 (C) 24

Ma 2 (D) 12

45°

a

B

45°

C

A thin circular ring of mass ‘M and radius r is rotating about its axis with a constant angular velocity w, Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity. (1983)

w  M  2m 

wM (A) M  m  

(B)

(C)

 M  2m 

wM  M  2m 

(D)

w  M  2m  M

10. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. the rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of (1995) (A) 0.42 m from mass 0.3 kg (B) 0.70 m from mass 0.7 kg (C) 0.98 m from mass 0.3 kg (D) 0.98 m from mass 0.7 kg 11. A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The angular speed of the block after it hits O is(1999) (A) 3V

 4a 

(B) 3V

 2a 

(C)

3V



2a



a M

(D) Zero

V O

12. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities

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92

ROTATIONAL MECHANICS that are conserved as the beads slide down, are (A) angular velocity and total energy (kinetic and potential) (B) Total angular momentum and total energy (C) angular velocity and moment of inertia about the axis of rotation. (D) total angular momentum and moment of inertia about the axis of rotation. 13. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is (A)

1 MR 2 2

(B)

1 MR 2 4

(C)

1 MR 2 8

(D)

(2000)

(2001)

2MR 2

14. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The direction of the frictional force acting on the cylinder are (2002) (A) up the incline while ascending and down the incline descending. (B) up the incline while ascending as well as descending. (C) down the incline while ascending and up the incline while descending (D) down the incline while ascending as well as descending. 15. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity

0 . When the

tortoise move along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform

  t  will vary with time t as

(t)

(A)

(t)

0

(B)

(2002) (t)

0

(C)

(t)

0

(D)

16. Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse M J = MV is imparted to the body at one of its ends, what would be its angular velocity? (2003) (A)

V L

(B)

2V L

(C)

V 3L

(D)

0

L M J = MV

V 4L

17. A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (2003) (A) centre of the circle (B) on the circumference of the circle. (C) inside the circle (D) outside the circle. 18. A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity

0 . A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be (2004) (A) 2 K

(B)

K 2

(C) K

(D)

K 4

19. A disc is rolling without slipping with angular velocity  . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is (2004) (A) VQ  VC  VP

(B) VP  VC  VQ

(C) VP  VC , VQ  VC 2

(D) VP  VC  VQ

C

P

Q

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93

ROTATIONAL MECHANICS 20. A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect? (2005) (A)  = mg [where  is the friction force] (B) F = N [where N is the normal force] (C) F will not produce torque (D) N will not produce torque

21. A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? (2005)



(A) L (angular momentum) is conserved about the centre



(B) only direction of angular momentum L is conserved (C) It spirals towards the centre (D) Its acceleration is towards the centre. 22. Select the correct statement(s): (A) A particle performs uniform circular motion with an angular momentum L. If the frequency of particle’s motion is doubled and its KE is halved, the angular momentum becomes one-fourth. (B) If a particle moves in X-Y plane, the resultant angular momentum about origin has only Z-component. (C) A particle is moving along a straight line parallel to X-axis with constant velocity. Its angular momentum about the origin decreases with time. (D) Action of angular impulse is to change the angular momentum. 23. A ring, a cube, a cylinder, a prism and a sphere all having equal masses, equal heights and equal maximum width: (A) The cube has the largest moment of inertia about an axis perpendicular to cross-section and passing through the centre of mass. (B) The ring has the largest moment of inertia about the axis mentioned above. (C) The cylinder has the smallest moment of inertia. (D) The sphere has the smallest moment of inertia. 24. Select the correct alternative(s): (A) The mass of a body can be taken to be concentrated at its centre of mass for the purpose of calculating its rotational inertia. (B) Two circular discs of the same mass and thickness are made from metals of different densities, the one with greater density will have less rotational inertia about its central axis. (C) About an axis passing through diagonally opposite ends of a uniform cube, have its minimum rotational inertia. (D) None of these. 25. Select the correct statement(s): (A) A very small particle rests on the top of a hemisphere of radius 20 cm. The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, is 3.2 m/s. (B) A solid cylinder at rest at the top of an inclined plane of height 2.7m rolls down without slipping. If the same cylinder has to slide down a frictionless inclined plane and acquires the same velocity as that acquired by the centre of mass of the rolling cylinder at the bottom of the inclined plane, the height of the inclined plane should be 1.8m. (C) Three thin rods each of mass M and length L are welded so as to form an equilateral triangle. The moment of inertia of the triangle about an axis perpendicular to the plane of triangle and passing through of its vertices is (D)

3 ML2 . 2

None of these.

26. Two boys of equal masses are sliding freely on smooth horizontal surface with the same speed v on parallel straight paths in opposite directions. The paths are separated by a perpendicular distance d. One of the boys carries a light pole of length d, held firmly at one end. The other boy grasps the other end of the pole just when they are crossing each other. Which of the following statements is/are true for the subsequent motion?

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94

ROTATIONAL MECHANICS (A) Each boy moves in a circular path of radius

d 2 with a constant speed v..

(B) The tension in the pole remains constant. (C) They come to rest when the pole has rotated through 90° to lie along the direction of original motion. (D) The tension in the pole varies with time. 27. A uniform rod AB of length  is free to rotate about a horizontal axis A passing through A as shown in figure. The rod is released from rest from horizontal position. If the rod gets broken at mid-point C when it becomes C vertical, just after breaking of the rod: (A) Angular velocity of upper part starts to decrease while that of B lower part remains constant. (B) Angular velocity of upper part starts to decrease while that of lower part starts to increase. (C) Angular velocity of both the parts is identical. (D) Angular velocity of lower part becomes equal to zero. (E) lower part falls vertically. 28. If 1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and  2 is the moment of inertia of the ring formed by bending the rod, then: (A)  2  1 4  2

(B)  2  1

2

(D) 1  2   2 3

(C)  2 1  0.3

29. If the polar ice caps melt and spread uniformly: (A) The moment of inertial of the earth about centroidal axis would increase. (B) The moment of inertia of the earth about centroidal axis would decrease. (C) Length of the day would increase. (D) The speed of rotation would be reduced. 30. A sphere is rolling down a plane of inclination  to the horizontal. The acceleration of its centre, down the plane is: (A) g sin 

(B) less than g sin 

(D) greater than  g sin   2

(C) greater than g sin 

2m 2

31. The moment of inertia of the pulley system shown in figure is 4 kg m . The radii of bigger and smaller pulleys are 2m and 1m, respectively. The angular acceleration of the pulley system is: (A) 2.1 rad/s2 (B) 4.2 rad/s2 2 (C) 1.2 rad/s (D) 0.6 rad/s2.

1m

5 kg

4 kg

32. Two masses 1 kg and 2 kg are connected by an inextensible light thread passing over a pulley of mass 2 kg and radius 20 cm. There is no slipping anywhere: (A) Tensions at the two sides of the pulley will be different. (B) Accelerations of the two blocks will be (C) Ratio of tensions on the two sides will be

56

g 4.

(D) Ratio of tensions on the two sides will be 1

2.

33. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of tis ends with a uniform angular velocity  . The force exerted by the liquid at the other end is : (A) M  2 L 2

(B) M  2 L

(C) M  2 L 4

(D) M  2 L2 2

34. Let  be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle  with AB. The moment of inertia of the plate about the axis CD is then equal to: (A) 

(B)  sin 2 

(C)  cos 2 

(D)  cos

2

 2 

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95

ROTATIONAL MECHANICS 35. A rod of mass M kg and length L metre is bent in the form of an equilateral triangle as shown in the figure. The moment of inertia of triangle about a vertical axis to perpendicular to the plane of triangle and passing through the centre (in units of kg m 2) is

ML2 (A) 12

ML2 (B) 54

ML2 (C) 162

O

ML2 (D) 108

36. A cord is wound round the circumference of a solid cylinder of radius R and mass M. The axis of the cylinder is horizontal. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the cylinder will be (A)

2mg M  2m

(B)

2gh R

4mgh  M  2m  R 2

(C)

(D)

37. A disc of moment of inertia  is rotating about its axis, which is initially along the vertical direction, making ‘n’ revolution per minute. The axis gradually tilts becomes horizontal and the disc rotates about the horizontal axis with the same revolution per minute. The total time, taken in doing so is t seconds then the torque acting on the body is (A) Zero

(B)

2 n 60 t

(C)

2 2 n 60 t

(D) None of the above.

38. From the circular disc of radius R and mass M, a concentric circular disc of small radius r is cut and removed, the mass of which is m. the moment of inertia of the angular disc remaining, about its axis perpendicular to the plane and passing through the centre of mass will be (A)

1 M  R2  r 2  2

(B)

1  M  m  R2  r 2  2

(C)

1  M  m  R2  r 2  2

(D)

1  MR 2  mr 2  2

39. Three discs each of mass M and radius R are placed in contact with each other as shown in the figure here. Then the M  of the system about an XX’ is: (A)

MR 2 4

(B)

11MR 2 4

(C)

2 MR 2 3

(D) 7MR 2

40. In which case, it is easier to rotate?

(A) Case I

(B) Case II

(C) same

(D) data is insufficient

41. A small mass m is attached to the rim of a circular disc of mass M and radius R. The disc rotates with angular velocity  about an axis passing through the centre O of the disc and perpendicular to its plane. Then the angular momentum of the system will be (A)

1 MR 2 2

(B) MR 2

 M  2m  2 R  2  

(C) 

(D)  M  m  R 

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96

ROTATIONAL MECHANICS 42. When a force F acts on a side of the hexagonal body will (A) topple (B) translate (C) both (D) None

43. A particle of mass m is moving along a straight line y = x + 4 with constant velocity v, then angular momentum of the particle about the origin is (A) Zero (B) constant (C) decreasing continuously (D) increasing continuously 44. A cylinder can maintain its rolling motion on a (A) rough horizontal surface (C) rough inclined surface

(B) smooth horizontal surface (D) smooth inclined surface

45. Choose the correct statement(s) for a particle moving along a circular path of constant radius (A) centripetal acceleration is always perpendicular to velocity vector. (B) centripetal acceleration is always perpendicular to angular velocity vector. (C) angular acceleration vector is always perpendicular to angular velocity vector. (D) angular velocity vector is always perpendicular to the linear velocity vector. 46. A solid sphere is executing rolling with slipping motion on a rough horizontal surface (A) The frictional force will always perform a negative work on the sphere. (B) The work done by friction cannot be zero. (C) The angular momentum of the sphere is conserved about its centre of mass. (D) The angular momentum of the sphere is conserved about any point on the horizontal surface. 47. A solid sphere of mass M and radius R is pulled horizontally on a rough surface as shown in the figure. Choose the incorrect alternatives

F M. 2F (B) The acceleration of the centre of mass is . 3M (A) The acceleration of the centre of mass is

F

CM

(C) The frictional force on the sphere acts forward. (D) The magnitude of the frictional force is F 3 .

Rough Surface

48. A constant force F is applied at the top of ring as shown in figure. Mass and the radius of the ring are M and R respectively. Angular momentum about the point of contact at time ‘t’ (A) is constant (B) increases linearly with time (C) is 2 FRt (D) 2FRt

F M

R

49. A disc of radius R rolls on a horizontal surface with linear velocity V and angular velocity  . There is a point P on circumference of disc at angle  , which has a vertical velocity. Here  is equal to

V R 1 V (C)   cos R (A)

  sin 1

 V  sin 1 2 R 1 V (D)   cos R (B)

50. Which of the following statement(s) is/are correct for a spherical body rolling without slipping on a rough horizontal surface at rest (A) the acceleration of the point of contact with the ground is zero. (B) the speed of some of the point(s) is/are zero. (C) frictional force may or may not be zero (D) work done by friction may or may not be zero.

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ROTATIONAL MECHANICS 51. A 2kg mass attached to a string of length 1m moves in a horizontal circle as a conical pendulum. The string makes an angle  = 30° with the vertical. Select the correct alternative(s) (g = 10 m/s2). (A) The horizontal component of angular momentum of mass about the point of support P is approximately 2.9 kg m 2/s. (B) The vertical component of angular momentum of mass about the point of support P is approximately 1.7 kg m 2/s. (C)

The magnitude of

dL  ( angular momentum of mass about point of support P) is approximately dt L

kgm 2 . s2 dL (D)   will not hold good in this case. dt 10



52. A particle of mass m is travelling with a constant velocity v  V0 along the line y = b, z = 0. Let dA be the area swept out by the position vector of (from origin) particle in time dt and L is the magnitude of angular momentum of particle about origin at any time t. Then (A) L = constant

(B) L  constant

(C)

dA 2 L  dt m

(D)

dA L  dt 2m

53. A uniform rod kept on the ground falls from its vertical position. Its foot does not slip on the ground. (A) No part of the rod can have acceleration greater than g in any position (B) At any one position of rod, different points on it have different acceleration. (C) Any one particular point on the rod has different acceleration at different positions of the rod. (D) The maximum acceleration of any point on the rod, at any position is 1.5 g. 54. A thin uniform rod of mass m and length  is free to rotate about its upper end. when it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately, after impact (A) the angular momentum of rod is jl .

(C) the kinetic energy of rod is 55. The torque

3J 2 . 2m

(B) the angular velocity of rod is

3J . ml

(D) the linear velocity of mid point of rod is

3J . 2m

     on a body about a given point is found to be equal to A  L where A is a constant vector and

 L is the angular momentum of the body about this point. It follows that:

dL  is perpendicular to L at all instant of time. dt   (B) the component of L in the direction of A does not change with time.  (C) the magnitude of L does not change with time.  (D) L does not change with time. (A)

56. Two particles A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is (A) 3 V (B) V (C) 1.5 V (D) Zero (JEE, 1982) 57. When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts (A) in the backward direction on the front wheel and in the forward direction on the rear wheel. (B) in the forward direction on the front wheel and in the backward direction on the rear wheel. (C) in the backward direction on both the front and the rear wheels. (D) in the forward direction on both the front the rear wheels. (JEE, 1990)

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ROTATIONAL MECHANICS 58. Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. the ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (JEE, 1998) (A) q/2m (B) q/m (C) 2q/m (D) q/m 59. Let  be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle  with AB. The moment of inertia of the plate about the axis CD is then equal to (A) 

(B)  sin 2 

(C)  cos 2 

(D)  cos

2

 2 

(JEE, 1998)

60. Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes as angle  with AB. The moment of inertia of the plate about the axis CD is then equal to [JEE, 98] (A) I (B) I sin2  (C) I cos2  (D) I cos2 (/2)

     61. The torque  on a body about a given point is found to be equal to A  L where A is a constant vector and L is the angular momentum of the body about that point. From this it follows that [JEE, 98]   (A) dL / dt is perpendicular to L at all instants of time   (B) the components of L in the direction of A does not change with time   (C) the magnitude of L does not change with time (D) L does not change with time 62. A smooth sphere A is moving on a frictionless horizontal plane with angular speed  and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angularspeeds are A and B, respectively. Then [JEE, 99] (A) A < B (B) A = B (C) A =  (D) B =  63. A disc of mass M and radius R is rolling with angular speed  on a horizontal as shown. The magnitude of angular momentum of the disc about the origin O is : [JEE, 99] (A) (1/2)MR2

(B) MR2

(C) (3/2)MR2

(D) 2MR2

y

M O

x

64. A cubical block of side L rest on a rough horizontal surface with coefficient of friction F µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is : [JEE, (Scr) 2000] (A) infinitesimal (B) mg/4 (C) mg/2 (D) mg(1 – µ) 65. A thin wire of length L and uniform linear mass density  is bent into a circular loop with x centre at O as shown. The moment of inertia of the loop about the axis XX´ is : [JEE, (Scr) 2000] (A) L3/82 (B) L3/162 (C) 5L3/162 (D) 3L3/82

L

B x´

60º O

66. Two particles each of mass M are connected by a massless rod of length l. The rod is lying on the smooth surface. If one of the particle is given an impulse MV as shown in the figure then angular velocity of the rod would be : [JEE, (Scr) 03] (A) v/l (B) 2v/l (C) v/2l (D) None

Mv M

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ROTATIONAL MECHANICS 67. A disc has mass 9m. A hole of radius R/3 is cut from it as shown in the figure. The R/3 moment of inertia of remaining part about an axis passing through the centre ‘O’ of the 2R/3 Q disc and perpendicular to the plane of the disc is : [JEE, (Scr) 05] 2

(A) 8 mR

40 (C) mR2 9

2

(B) 4 mR

R

37 (D) mR2 9

68. A particle moves in circular path with decreasing speed. Which of the following is correct   (A) L is constant (B) only direction of L is constant [JEE, (Scr) 05]  (C) acceleration a is towards the centre (D) it will move in a spiral and finally reach the centre 69. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickkness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be [JEE, 06] (A) 2R / 15

(B) R 2 /15

(C) 4R / 15

(D) R/4

70. A solid cylinder of mass m and radius r is rolling on a rough inclined plane of inclination . The coefficient of friction between the cylinder and incline is µ. Then [JEE, 06] (A) frictional force is always µmg cos  (B) friction is a dissipative force (C) by decreasing , frictional force decreases (D) friction opposes translation and supports rotation 71. A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies hA of the ball at A, B and C, respectively. Then (A) hA > hC ; KB > KC (C) hA = hC ; KB = KC

C

A B

(B) hA > hC ; KC > KA (D) hA < hC ; KB > KC

hC

[JEE, 06]

72. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v2/(4g) with respect to the initial position. The object is [JEE, 07] (A) ring (B) solid sphere (C) hollow sphere (D) disc

v

FILL IN THE BLANKS 1.

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height

3a 4 above the base.

The minimum value of F for which the cube begins to tip about the edge is ........... (Assume that the cube does not slide). (1984) 2.

According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal area in equal intervals of time. This law is a consequence of the conservation of ................ (1985)

3.

A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity

0 about an axis perpendicular to the rod and passing through the midpoint

of the rod (see figure). There are no external forces.

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100

ROTATIONAL MECHANICS

When the beads reach the ends of the rod, the angular velocity of the system is ..............

(1988)

4.

A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given by x = A cos (  t). There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is ................ (1988)

5.

A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn where A is a constant, r is the instantaneous radius of the circle and n = ....................... (1993)

6.

The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is 4.4 x 1015 m 2/s. The area enclosed by Earth’s orbit approximately ........... m 2. (1997)

7.

A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of 30° with the horizontal. If the coefficients of static and kinetic friction are each equal to  and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is ........... and its direction is .............. (wire up or down) the inclined plane. (1997)

8.

A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. the knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is ........... and on B is ................. (1997)

B

A 9.

A symmetric lamina of mass M consists of a square shape with a semicircular section over of the edge of the square as shown in Figure. The side of the square is 2a. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is 1.6 Ma2. The moment of inertia of the lamina about the tangent AB in the plane of the lamina is ....................

2a O

(1997)

TRUE & FALSE 10. A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of the paper and (a) passing through A, (b) passing B, by the application of the same force, F, at C (midpoint of AB) as shown in the figure. The angular acceleration in both the cases will be the same. (1985)

A

C

B

F

11. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity  . Another disc of the same dimensions but of mass

2 5 .

M 4 is placed gently on the first disc coaxially. The angular velocity of the system now is (1986)

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101

ROTATIONAL MECHANICS 12. A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given the same kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll as soon as released towards a wall which is at the same distance from the ring and the cylinder. The rolling friction in both cases is negligible. The cylinder will reach the wall first. (1989)

PASSAGE PASSAGE – 1 Moment of inertia is a physical term which oppose the change in rotational motion. Moment of inertia depends on distribution of mass, shape of the body as well as distance from the rotational axis. Moment of linear momentum is called angular momentum. If no external torque act on the system then angular momentum of the system remains conserved. Geometrical meaning of angular momentum relates to the areal velocity. 1.

Mass M is distributed over the rod of length L. If linear mass density (  ) linearly increases with length as  = Kx. The M.  . of the rod about one end perpendicular to rod i.e. (YY’) (A)

2.

ML2 3

(B)

(C)

2 ML2 3

(D)

KL4 4

Four holes of radius R are cut from a thin square plate of side 4R and mass M. The moment of inertia of the remaining portion about z-axis. (A)

Y

X

4     MR 2 3 4

 MR 2 12

(B) 

4     MR 2 3 6

(D) 

 8 10  6 3

(C)  3.

ML2 12

 2  MR 

A particle of mass m is moving along the line y = 3x + 5 with speed V. The magnitude of angular momentum about origin is (A)

5 mV 12

(B)

5 mV 2

(C)

1 mV 2

(D)

1 mV 3

4.

A hollow sphere is rolling without slipping total energy during rolling is 60 J, then (A) translational K.E. = 36 J (B) rotational K.E. = 36 J (C) translational K.E. = 24 J (D) rotational K.E. = 24 J

5.

Acceleration of block of mass m 1 is (given moment of inertia of pulley is  and string does not slip over the pulley).

 m1  m2  g

 m1  m2  g (A)   m1  m2 

(B)  m  m     1 2 2 

    m1  m2  2  g R  (C)  m1  m2

    m1  m2  2  g R   (D)     m1  m2  2  R  



R 

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102

ROTATIONAL MECHANICS PASSAGE – 2 Angular velocity is defined as the rate of change of angular displacement. The angular velocity is a measure of the degree of rotation of body. For a rigid body  is constant. All angular variable (such as angular displacement, angular velocity and angular acceleration) are directed along the axis of rotation and perpendicular to plane. Rotating rigid object has two acceleration, one centripetal and other tangential both accelerations being normal to each other. 6.

A disc of radius R rolls on a horizontal ground with linear acceleration “a” and angular acceleration  as shown in figure. The magnitude of acceleration of point P at an instant, when its linear velocity is V and angular velocity is  , will be:

 a  ra 

(A)

2

  r 2 

2

(B)

ar R

(C)

r 2 a 2  r 2 4

(D)

r

7.

A disc is rotating clockwise at 20 radian/sec. Its centre has velocity 30 m/s in the forward direction. It has radius of 20 m. Then (A) velocity of topmost point is 430 m/s in forward direction. (B) velocity of lowermost point is 370 m/s in backward direction. (C) velocity of topmost point is 400 m/s (forward). (D) velocity of lower most point is 50 m/s (forward).

8.

The topmost and bottom most points have velocities V1 and V2 in the same direction. The radius of sphere is R. Then the correct option are (A) Angular velocity of sphere about cm is (B) The Linear velocity at point P is VP 

(C) The Linear velocity at point Q is



V1  V2 clockwise 2R

3V1  V2 . 4

 3 3 V1  V2 1   2 2 . 

 V1  V2  .  2 

(D) Velocity of centre of mass is  9.

Three identical rods, each of mass m and length  are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is (A)

l 2

(B) l

3 2

(C)

l 2

(B) MR 2

3 2 (C) MR  2



60°

(D) l

10. A Disc of mass M and radius R is rolling with angular speed  on a horizontal plane as shown in figure, The magnitude of angular momentum of the disc about O is

MR 2  (A) 2



(D) 2MR 2

3



Y  M O

V X

PASSAGE – 3 In figure 1, the winch is mounted n an axle, and the 6-sided nut is welded ot the winch. By turning the nut with a wrench, a person can rotate the winch. For instance, turning the nut clockwise lifts the block off the ground, because more and more rope gets wrapped around the winch. Three students agree that using a longer wrench makes it easier to turn the winch. But they disagree about why. All three students are talking about the case where the winch is used, over a 10-second time interval, to lift the block one meter off the ground.

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103

ROTATIONAL MECHANICS Student : 1 By using a longer wrench, the person decreases the average force he must exert on the wrench, in order to lift the block one meter in ten seconds.

nut

winch

person grips wrench here

block Figure : 1

11.

Figure : 2

Student : 2 Using a longer wrench reduces the work done by the person as he uses the winch to lift the block one meter in ten second. Student : 3 Using a longer wrench reduces the power that the person must exert to lift the block one meter in ten seconds. Student 1 is: (A) Correct, because the torque that the wrench must exert to lift the block doesn’t depend on the wrench’s length (B) Correct, because using a longer wrench decreases the torque it must exert on the winch. (C) Incorrect, because the torque that the wrench must exert to lift the block doesn’t depend on the wrench’s length. (D) Incorrect, because using a longer wrench decreases the torque it must exert on the wrench.

(A)

(B)

length

(C)

length

(D&&)

length

force

If several wrenches all apply the same torque to a nut, which graph best expresses the relationship between the force the person must apply to the wrench, and the length of the wrench?

force

13.

force

Which of the following is true about student 2 and 3? (A) Students 2 and 3 are both correct. (B) Student 2 is correct, but student 3 is incorrect. (C) Student 3 is correct, but student 2 is incorrect. (D) Students 2 and 3 are both incorrect.

force

12.

length

PASSAGE – 4 When a force Of is applied on a block of mass m resting on a horizontal surface then there are two possibilities, either the block moves by translation or it moves by toppling. If the surface is smooth then the block always translates but on a rough surface it topples only when the torque of the applied force Of is greater than the torque of mg about a point in contact with the ground. When the force Of is applied the body may topple about A or it may translate.

F m

h

a

A

Answer the following question.

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104

ROTATIONAL MECHANICS 14.

When the block topples about A, the normal force – (A) passes through centre of mass (B) is / zero (C) shifts to the right and passes through right most edge containing A. (D) is zero if the surface is smooth.

15.

The block will move in pure translation if – (A) F 

16.

mga h

(B) F 

mga 2h

(C) F 

mga 2h

(D) None

mga 2h

(C) F 

mga 2h

(D) None

The block will topple about A if – (A) F 

mga h

(B) F 

17.

If the block be a cube of edge a and  = 0.2 then – (A) The body will translate (B) The body will topple (C) The body may translate or topple (D) None

18.

If the block is a cube of edge a and  = 0.6 then– (A) The body will translate (B) The body will topple (C) The body first translates and then topples (D) None PASSAGE – 5

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparated an initial angular velocity 2 using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an angular velocity  by a spring having the same spring constant and compressed by a distance x2. Both the discs rotate in the clockwise direction. 19. The ratio x1/x2 is [JEE, 07] (A) 2

(B) 1/2

(C)

2

(D) 1/ 2

20. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is [JEE, 07] (A) 2I/(3t) (B) 9I/(2t) (C) 9I/(4t) (D) 3I/(2t) 21. The loss of kinetic energy during the above process is (A) I2/2 (B) I2/3 (C) I2/4

[JEE, 07] (D) I2/6

LEVEL – 2 1.

A uniform solid sphere of mass 1 kg and radius 10 cm is kept stationary on a rough inclined plane by fixing a highly dense particle at B. Inclination of plane is 37° with horizontal and AB is the diameter of the sphere which is parallel to the plane, as shown in figure. Calculate:

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105

ROTATIONAL MECHANICS (i) (ii)

mass of the particle fixed at B, and minimum required coefficient of friction between sphere and plane to keep sphere in equilibrium.

2.

A ball of radius R = 20 cm has mass m = 0.75 kg and moment of inertia (about its diameter)   0.0125 kg m 2 . The ball rolls without sliding over a rough horizontal floor with velocity v 0 = 10 ms–1 towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is e = 0.7, calculate velocity v of the ball long after the collision. (g = 10 ms–2).

3.

AB is a horizontal diameter of a ball of mass m = 0.4 kg and radius R = 0.10 m. At time t = 0, a sharp impulse is applied at B at angle of 45° with the horizontal, as shown in Figure so that the ball immediately starts to moves with velocity v 0 = 10 ms–1. (i) Calculate the impulse If coefficient of kinetic friction between the floor and the ball is  = 0.1, calculate (ii) velocity of ball when it stops sliding. (iii) time t at that instant, (iv) horizontal distance ravelled by the ball upto that instant. (v) angular displacement of the ball about horizontal diameter perpendicular to AB, upto that instant, and (vi) energy lost due to friction.

4.

A solid ball of diameter d = 11 cm is rotating about its one of the horizontal diameters with angular velocity

0  120 rad sec . It is released from a height so that it falls h = 1.8 m freely and then collides with the horizontal floor. Co-efficient restitution is

e  5 6 and co-efficient of friction between the ball and the ground

is   0.2 . Calculate fraction of energy lost during collision and the distance between the points where the ball strikes the floor for the first and second time. 5.

A heavy plank of mass 102.5 kg is placed over two cylindrical rollers of radii R = 10 cm and r = 5 cm. Mass of rollers is 40 kg and 20 kg respectively. Plank is pulled towards right by applying a horizontal force Of = 25 N as shown in Figure. During first second of motion the plank gets displaced by 10 cm.

If plank remains horizontal and slipping doesnot take place, calculate magnitude and direction of force of friction acting between (i) plank and bigger roller. (ii) plank and smaller roller. (iii) bigger roller and floor, and (iv) smaller roller and floor. 6.

A wheel of radius R = 10 cm and moment of inertia I = 0.05 kg-m 2 is rotating about a fixed horizontal axis O with angular velocity

0 = 10 rad/sec. A uniform rigid rod of mass m = 3 kg and length l = 50 cm is hinged

at one end A such that it can rotate about end A in a vertical plane. End B of the rod is tied with a thread as shown in figure such that the rod is horizontal and is just in contact with the surface of rotating wheel. Horizontal distance between axis of rotation O if cylinder and A is equal to a = 30 cm.

If the wheel stops rotating after one second after the thread has burnt, calculate co-efficient of friction (g = 10 ms–2)

 between the rod and the surface of the wheel.

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106

ROTATIONAL MECHANICS 7.

Two heavy and light cylindrical rollers of diameters D and d respectively rest on a horizontal plane as shown in figure. The larger roller has a string wound round it to which a horizontal force P can be applied as shown. Assuming that the coefficient of friction  ahs the same value for all surfaces of contact, determine the limits of one.

8.

P D

d

 so that the larger roller can be pulled over the smaller

A block of mass m height 2h and width 2b rests on a flat car which moves horizontally with constant acceleration ‘a’ as shown in figure.

Determine: (a) the value of the acceleration at which slipping of the block on the car starts, if the coefficient of friction is µ. (b) the value of the acceleration at which block topples about A, assuming sufficient friction to prevent slipping and (c) the shortest distance in which it can be stopped from a speed of 72 km/hr with constant deceleration so that the block is not disturbed. The following data are given: b = 0.6 m, h = 0.9 m; µ = 0.5 and g = 9.8 m/s2. 9.

A block of and M = 4 kg of height ‘h’ and breath ‘b’ is placed on a rough plank of same mass M. A light inextensible string is connected to the upper end of the block and passed through a light smooth pulley as shown in the figure.

A mass m = 1 kg is hung to the other end of the string. (a) What should be the minimum value of coefficient of friction between the block and the plank so hat, there is no slipping between the block and the wedge? (b) Find the minimum value of b/h so that the block does not topple over the plank, friction is absent between the plank and the ground. 10. A homogeneous cylinder and a homogeneous sphere of equal mass m = 20 kg and equal radii R are connected together by a light frame and are free to roll without slipping down the plane inclined at 30° with the horizontal. Determine the force in the frame.

Assume that the bearings are frictionless. Take g = 10 m/s2)

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107

ROTATIONAL MECHANICS 11. A uniform ring of mass m, radius a and centre C lies at rest on a smooth horizontal table. The plane of the ring is horizontal. A point P on the circumference is struck horizontally and it begins to move in a direction at 60° to PC. If the magnitude of impulse is mv 7 , find the initial speed of point P.. 12. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through D and perpendicular to face ABCD. A bullet of mass m and speed v is shot at a height of

4a as shown in figure. 3 The bullet becomes embedded in the cube. Find the minimum value of v r equired to topple the cube. Assume m 0, along positive x-axis for x < 0 and zero for x = 0. Thus, the particle will oscillate about stable equilibrium position, x = 0. The force in this case is called the restoring force. Of these, if n = 1, i.e., F = – kx the motion is said to be SHM. Example 24. Describe the motion of a particle acted upon by a force (i) F = –2(x – 2)3

(ii) F = –2(x – 2)2

(iii) F = – 2(x – 2)

Sol. (i) F = –2(x – 2)3 F=0 at x=2 Force is along negative x-direction for x > 2 and it is along positive x-direction for x < 2. Thus, the motion of the particle is oscillatory (but not simple harmonic) about x = 2. (ii) F = 0 for x = 2, but force is always along negative x-direction for any value of x except at x = 2. Thus, the motion of the particle is rectilinear along negative x-direction. (iii) Let, us take x – 2 = X, then the given force can be written as, F=–2X This is the equation of SHM. Hence, the particle oscillates simple harmonically about X = 0 or x = 2. Example 25: A body of mass 1kg is executing simple harmonic motion which is given by y = 6.0 cos(100t + 4) cm. What is the (i) amplitude of displacement, (ii) Angular frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy ? Sol. The given equation of SHM is y = 6.0 cos (100t + /4) cm. Comparing it with the standard equation of SHM, y = A sin (t + ), we have (i) Amplitude A = 0.6 cm (ii) Angular frequency  = 100 sec–1. (iii) Initial phase  = /4 (iv) Velocity v =  (A 2  y 2 ) = 100

(36  y 2 ) cm/sec

(v) Acceleration = –2y = –(100)2y = – 104y (vi) Kinetic energy =

1 1 mv2 = m2(A2 – y2) 2 2

The kinetic energy of a particle in SHM is maximum, when it passes its mean position i.e. at y = 0 1 1 2 (K.E.)max = mv max = mA22 2 2

=

1 × 1 × 104 × (0.06)2 = 18 joules 2

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28

SIMPLE HARMONIC MOTION Example 26: A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. Calculate the velocity and the kinetic energy of the particle at the moment when its displacement is 0.6 metre. Sol. We know that, at a displacement y from mean position particle’s velocity is given as v =  (A 2  y 2 ) or

v

2 (A 2  y 2 ) T

v

2  3.14 [(1.0) 2  (0.6)2 ]  3.2 m/sec. (11/ 7)

Kinetic energy at this displacement is given by K=

1 mv2 2

K=

1 × 0.8 × (3.2)2 = 4.1 joule 2

Example 27: A person normally weighing 60 kg stands on a platform which oscillates up and down harmonically at a frequency 2.0 sec–1 and an amplitude 5.0 cm. If a machine on the platform gives the person’s weight against time, deduce the maximum and minimum reading it will show, take g = 10 m/sec2. Sol. As shown in figure, platform is executing SHM with amplitude and angular frequency given us A = 5.0 cm and

 = 2n = 4 rad/sec

[as n = 2 sec–1]

+A weighing machine –A

Here weighing machine will show weight more then that of man when it is below its equailibrium position when the acceleration of platform is in upward direction. In this situation the free body diagram of man relative to N platform is shown in figure. Here ma is a pseudo force on man in downward direction relative to platform (or weighing machine). As weighing machine will read the normal reaction on it thus for equilibrium of an relative to platform, we have

N = mg + ma

or

N = mg + m(2y)

[as | a | = 2y]

mg + ma

Where y is the displacement of platform from its mean position. We wish to find the maximum weight shown by the weighing machine, which is possible when platform is at its lowest extreme position as shown in figure, thus maximum reading of weighing machine will be N = mg + m2A N = 60 × 10 + 60 × (4)2 × 0.05

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29

SIMPLE HARMONIC MOTION N = 600 + 480 N = 1080 Nt = 108 kg t Similarly the machine will show minimum reading when it is at its upper extreme position when pseudo force on man will be in upward direction, thus minimum reading of weighing machine will be N = mg – m2A N = 600 – 480 = 120 Nt = 12 kg t Example 28: A 20 g particle is oscillating simple harmonically with a period of 2 second and maximum kinetic energy 2 J. The total mechanical energy of the particle is zero. Write the following — (a) Amplitude of oscillation (b) Potential energy as a function of displacement x relative to mean position. Sol. (a)

m = 20 g = 20 × 10–3 kg T = 2s Kmax = 2 J 1 m2 A2 2

Using

Kmax =

we have

1  2  2   20  103    . A 2  2 2

2

2 3

5  10  4 2 A

 A2

10 m 

(b) Etot = EOSC + U0 0 = 2 + U0 U0 = –2 J 

U(x) =

1 m2 x2 – 2 2 2

1 20  2  2 U(x)     x 2 2 1000  2 

2 2 u(x)  x 2 (S.I. units) 100 Example 29: A body is executing SHM under the action of force whose maximum magnitude is 50N. Find the magnitude of force acting on the particle at the time when its energy is half kinetic and half potential. Sol. Total energy E = 1/2 mA22 Up = Uk 

1 1 mA 2 2 sin 2 t  mA 2 2 cos 2 t 2 2

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30

SIMPLE HARMONIC MOTION or

tan2t = 1



 4 a = –2A sin t



F = ma = 2Asin



t 

 4

(in the sence of magnitude)

50  25 2 N 2 Example 30: A point particle of mass 0.1 kg is executing SHM with amplitude of 0.1 m. When the particle passes through the mean position, its K.E. is 8 × 10–3 J. Obtain the equation of motion of this particle if the initial phase of oscillation is 45º. 

F

1 mv 2max  8  10 3 2

Sol. or or or 

1 mA 2 2  8  103 2 1  0.1  (0.1) 2  2  8  10 3 2 2 = 16  = 4 rad/s



  x = A sin  t   4



  x = 0.1 sin  4t   4

Example 31: In case of simple harmonic motion (a) what fraction of total energy is kinetic and what fraction is potential when displacement is one half of the amplitude, (b) at what displacement the kinetic and potential energies are equal? Sol. In case of simple harmonic motion 1 1 1 K  m 2 A 2  y 2 , U  m 2 y 2 and E  m 2 A 2 2 2 2 2   K y A  1 3   f K   1  2   1    as y  given  (a) So  E  A   4 4 2  





2

U y 2  (A / 2)  1 fP   2     E A 4  A  (b) According to given problem 1 1 K = U, i.e., m2(A2 – y2) = = m2y2 2 2 A y  0.7 A i.e. 2y2 = A2 or 2

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31

SIMPLE HARMONIC MOTION Example 32: A particle starts oscillating simple harmonically from its equilibrium position then, the ratio of kinetic energy and potential energy of the particle at the time T/12 is : (T = time period) (A) 2 : 1 (B) 3 : 1 (C) 4 : 1 (D) 1 : 4 Sol: x = A sin t v

dx = A  cos t dt



Uk 

1 1 mv 2  mA 2 2 cos 2 t 2 2

and

UP 

1 mA 2 2 sin 2 t 2



U k cos 2 t   cot 2 t 2 U P sin t Uk 2 T  cot 2  UP T 12 Uk   cot 2  cot 2 30º UP 6 Uk  3 :1 UP

Hence option (B) is correct.

Example 33: A body performs simple harmonic oscillations along the straight line ABCDE with C as the midpoint of AE. Its kinetic energies at B and D are each one fourth of its maximum value. If AE = 2R, the distance between B and D is (A)

Uk 

Sol:

A

R (B) 2

3R 2

(C)

3R

B

(D)

C

D

E

2R

1 m 2 (A 2  x 2 ) 2

or

1 1 1  mA 2 2  m 2 (A 2  x 2 ) 4 2 2

or

A2  A2  x 2 4

or

x 2  A2 



x





A 2 3A 2  4 4

3A 3  R 1 2

Hence option (A) is correct.

Example 34: The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9J and its amplitude is 0.01 m, its time period would be (A) /10 sec (B) /20 sec (C) /50 sec (D) /100 sec

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32

SIMPLE HARMONIC MOTION 1 mv 2max  9  5  4 2

Sol: or

mv 2max  8

or

2 v 2max  8

or

v max = 2 m/s = A





or 

v max 2   200 A 0.01

2  200 T 2  T  200 100

Hence (D) option is correct.

Example 35: A spring mass system preforms SHM. If the mass is doubled keeping amplitude same, then the total energy of SHM will become : (A) double (B) half (C) unchanged (D) 4 times 1 mA 2 2 2 1  k E  mA 2    m 2 E

Sol:

1 2 A k (Independent of mass) Hence option (C) is correct. 2 C36: A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A. Its speed as it passes through the equilibrium position is V. If extended 2A and released, the speed of the mass passing through the equilibrium position will be (A) 2V (B) 4V (C) V/2 (D) V/4 Sol:  v = A  v´ = 2A  v´ = 2v Hence option (A) is correct. E

C37: A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean position is 50 J. Find, (i) the maximum kinetic energy (ii) the minimum potential energy (iii) the potential energy at extreme positions. Sol. At mean position, potential energy is minimum and kinetic energy is maximum. Hence, Umin = 50 J and

(at mean position)

Kmax = E – Umin = 200 – 50 Kmax = 150 J

(at mean position)

At extreme positions, kinetic energy is zero and potential energy is maximum 

Umax = E Umax = 200 J

(at extreme position)

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33

SIMPLE HARMONIC MOTION C38:

The potential energy of a particle oscillating on x-axis is given as U = 20 + (x – 2) 2

Here U is in joules and x in metres. Total mechanical energy of the particle is 36 J. (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle. Sol. (a)

F=–

dU = – 2(x – 2) dx

By assuming x – 2 = X, we have F = – 2X F–X

Since,

The motion of the particle is simple harmonic (b) The mean position of the particle is X = 0 or x – 2 = 0, which gives x = 2 m (c) Maximum kinetic energy of the particle is, Kmax = E – Umin = 36 – 20 = 16 J. Note : Umin is 20 J at mean position or at x = 2 m. Example 36. If a particle moves in a potential energy field U = U0 – ax + bx2, where a and b are positive constant, obtain an expression for the force acting on it as a function of position. At what point does the force vanish? Is this a point of stable equilibrium ? Calculate the force constant and frequency of the particle. dU  a  2bx dx F = 0, at x = a/2b F

Sol.

d2U  2b  0 dx 2 i.e., x 

a is a point of minimum potential energy. Hence, the equilibrium is stable. 2b k = 2b

f

and

1 k 1 2b  2 m 2 m

Example 37: A particle of mass m moves in a one-dimensional potential energy U(x) = –ax2 + bx4, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to (A) 

a 2b

(B) 2

a m

(C)

2a m

(D)

a 2m

U = –ax2 + bx4

Sol: 

F

dU  2ax  4bx 3 dx

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34

SIMPLE HARMONIC MOTION At mean position F = 0  2ax = 4bx3 2a a  4b 2b

or

x2 



x0  



F = – 2a(x0 + x) + 4b(x0 + x)3

a 2b

F = – 2ax0 – 2ax + 4bx03

 x 1   x  0

3

 3x  F = – 2ax0 – 2ax + 4bx03  1  x   0 12bx 30 x F = –2ax0 – 2ax + 4bx0 + x0 3

F = –2ax0 – 2ax + 4bx03 + 12bx02x

F  2a

a a a a  2ax  4b  12b x 2b 2b 2b 2b

a a  2ax  2a  6ax 2b 2b ma´ = + 4ax F  2a

  

4a x m a´  – x a´ = 2x a´

4a a 2 m m Example 38: A particle of mass m moves in the potential energy U shown above. The period of the motion when the particle has total energy E is 

U(x) 2 U= 1 2 kx , x > 0

U=mgx. x > 0 x

(A) 2 m / k  4 2E / mg 2 (C)  m / k  2 2E / mg 2 Sol: For x 0 U = mgx 

F

But

E



v0 



T2 



dv   mg dx

1 mv 20 2

2E , m

it is speed at lowest point.

2v 0 2 2E  g g m

T1 m 2 2E  T2    2 k g m Hence option (C) is correct. T

PASSAGE (Q. 39 to 41) The graphs is figure show that a quantity y varies with displacement d in a system undergoing simple harmonic motion. y

y

(I)

y

(II) O

d

Example 39: The total energy of the system (A) I (B) II Sol: In SHM, total energy remains constant. Example 40: The time (A) I (B) II Sol: x = A sin t Hence option (D) is correct.

y

(III) O

d

(IV) O

d

O

(C) III

(D) IV

(C) III

(D) IV

Example 41: The unbalanced force acting on the sytem (A) I (B) II (C) III 2 Sol: a = – x  F = ma = – m2x Hence option (D) is correct.

d

(D) None

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36

SIMPLE HARMONIC MOTION Energy method for SHM problems : 1. Write the equilibrium equation. 2. Take displaced position and write the potential plus kinetic energies : E = K + U 3.

Set

dE = 0 and obtain dynamical equation. dt

Example 42: The friction coefficient between the two blocks shown in figure is µ and the horizontal plane is smooth. If the system is slightly displaced and released, (a) find the time period

m k M

(b) find the magnitude of the frictional force between the blocks when the displacement from the mean position is x, (c) find the maximum amplitude if the upper block does not slip relative to the lower block during oscillations. Sol. (a) For small amplitude, the two blocks oscillate together. The angular frequency is of a single block spring system :

  [k /(M  m)] and so the time period is : T  2 [(M  m) / k] (b) When the blocks are at distance x from the mean position, their acceleration is : kx Mm The resultant force on the upper block is, therefore

a = – 2x = –

mkx Mm The force is prompted by the frictional of the lower block. Hence the magnitude of the frictional force is

mf = –

mk | x | . Mm (c) the maximum frictional force required for simple harmonic motion of the upper block is mKA/(M + m) at the extreme position. But the maximum frictional force can only be µ mg.

Therefore

mkA = µmg Mm

or

A=

µ(M  m)g k

Example 43: A 2 kg mass is attached to a spring of force constant 600 N/m and rests on a smooth horizontal surface. A second mass of 1 kg slides along the surface toward the first at 6 m/s. (a) Find the amplitude of oscillation if the masses make a perfectly inelastic collision and remain together on the spring. What is the period of oscillation? (b) Find the amplitude and period of oscillation if the collision is perfectly elastic. (c) For each case, write down the position x as a function of time t for the mass attached to the spring, assuming that the collision occurs at time t = 0. What is the impulse given to the 2 kg mass in each case ? Sol. (a) From conservation of linear momentum, 1 × 6 = (1 + 2)v  v = 2 m/s

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37

SIMPLE HARMONIC MOTION Now from conservation of mechanical energy, 1 1 mv 2  kA 2 2 2

1kg

 m  3  A v    2 or  k  600 

6m/s

2kg

A = 0.141 m = 14.1 cm

m 3  2  0.44 s k 600 (b) For perfectly elastic collision, T  2

 m  m1   2m1  v´2   2 v2   v1   m1  m 2   m1  m 2 

or

 2  1 v´2  0   6  4 m/s  1  2 

 m  ´  2  1/ 2 A  v 2   600   4  k

or

A = 0.23 m = 23 cm

T  2

m 2  2  0.36 s k 600

2 2   14.28 rad / s T 0.44 and amplitude A = 14.1 cm  x = A sin t = (14.1 cm) sin (14.28 t) In the second case,

(c) In the first case,



2 2   17.45 rad / s T 0.36 and amplitude A = 23 cm  x = 23 sin (17.45 t) cm Impulse, J = P = 2 × 2 = 4 N-s in the first case and 4 × 2 = 8 N-s in the second case. Example 44: A body A of mass m1 = 1.00 kg and a body B of mass m2 = 4.10 kg are interconnected by a spring as shown in figure. The body A performs free vertical harmonic oscillations with amplitude a = 1.6 cm and frequency  = 25 rad/s. Neglecting the mass of the spring, find the maximum and minimum values of force that this system exerts on the bearing surface. 

Sol.

A

B

Nmax : k = 2m1 = (25)2 (1.0) = 6.25 N/m Acceleration of centre of mass in extreme position, 2

a CM 

m1a m A (upwards)  1 m1  m 2 m1  m 2 m1 2 A  (m1  m 2 ) (m1  m 2 )

Now

N max  (m1  m 2 )g 



Nmax = (m1 + m2)g + m12A Nmax = (1.0 + 4.10)9.8 + (1.0)(25)2(1.6 × 10–2) = 59.98 newton

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m1

a=

2

A

m2

38

SIMPLE HARMONIC MOTION Example 45: In the figure shown, the block A collides with the block B and after collision they stick together. Calculate the amplitude of resultant vibration. u Sol. Common velocity after collision = 2

M B

u

M A

2

Hence,

1 1  u 2 K.E. = (2M)    Mu  2 2 4

It is also the total energy of vibration because the string is unstretched at this moment, hence if A is the amplitude, then 1 1 M KA 2  Mu 2 A u. or 2 4 2K Example 46: A body of mass m falls from a height h on to the pan of a spring balance. The mass of the pan is M and spring is massless. The force constant of the spring is k. The body gets struck to the pan and starts performing harmonic oscillations in the vertical direction. Calculate the period, amplitude and energy of these oscillations. Sol. Let v be the velocity immediately after impact.

Then

m 2gh = (M + m)v

m 2gh ...(1) Mm The period of oscillation of a spring depends on the mass attached and its force constant and it is given by

or

 here

v

T  2

m k

T  2

Mm k

x´0 l0



or

l

k Mm

m 1 M

x´´ 0 2

x0 3

The body vibrates about the equilibrium position. Let the spring be compressed by x´0 under the weight of the pan. Then Mg = kx´0. Let the spring be compressed by x´´0 from the previous position under the weight (M + m)g. Then (M + m)g = k(x´´0 + x´0) This positive (position 2) is the equilibrium position of the system (pan + body). Let x0 be the maximum compression of the spring relative to position 1. This is the lower extreme position (position 3) of the system. Applying Conservation of energy between positions 1 and 3 1 1 (M  m)gl  k x´20  (M  m)v 2 2 2 1  (M  m)g(l  x 0 )  k(x´0  x 0 ) 2  0 2

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39

SIMPLE HARMONIC MOTION

m 2 gh 1  (M  m)g x 0  k x´0 x 0  k x 02 Mm 2   Mg x 0  mg x 0  k

Mg 1 x 0  k x 20 k 2

2m 2 gh kx  2mg x 0  0 Mm 2 0

mg mg 2kh  1 k k Mm The negative value is not acceptable x0 



x0 

mg mg 2kh  1 k k (M  m)g

Since the amplitude of SHM is the distance between the mean position and the extreme position, here a, amplitude  x 0  x´´0 

and

E

mg 2kh 1 k (M  m)g

1 (M  m)a 2 2 2

1 m 2g 2  2kh  k E  (M  m) 2  1  2 k  (M  m)g  M  m 1 m2g 2  2kh  E 1  2 k  (M  m)g 

Example 47: The spring shown in figure is kept in a stretched position x0. Assuming the horizontal surface to be frictionless, calculate the frequency of oscillations after the system is released.

k M

m

Sol. As there is no external force acting on the system of two blocks plus the spring, the centre of mass of the system will remain at rest. The mean positions of two simple harmonic motion or two blocks occur when the spring becomes unstretched. If the mass m moves towards left through a distance X area the mass M moves towards left through a distance x before acquires natural length, x + X = x0 ...(1) where x and X will represent the amplitudes of two blocks m and M respectively. Because the centre of mass of the system must not move during the motion, we can wtire mx = MX ...(2) From (1) and (2)

Mx 0 mx 0 and X= ...(3) Mm Mm x and X also give the amplitudes of m and M during oscillations. At the mean positions the two blocks have x=

kinetic energies

1 1 1 m2x2 and M2x2. Their sum must be kx02. 2 2 2

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40

SIMPLE HARMONIC MOTION Now, from (3)

X0 =

Mm X M

1 1 kx02 = 2(mx2 + Mx2) 2 2 Using (3), angular frequency



K(M  m) Mm

and the frequency

n

 K(M  m)  2 Mm

Example 48: Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring-mass system is constrained to move inside A B a rigid smooth pipe bent in the form of a circle as shown in fig. The pipe is fixed in y1 y2 a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. /6 /6 Q Each spring has a natural length of 0.06  m and force constant 0.1 N/m. Initially, P O both the balls are displaced by an angle  = /6 radian with respect to diameter PQ of the circle and released from rest. (a) Calculate the frequency of oscillation of the ball B. (b) What is the total energy of the system ? (c) Find the speed of the ball A when A and B are at the two ends of the diameter PQ. Sol. (a) As here two masses A and B are connected by two springs, this problem is equivalent to the oscillation of a reduced mass m by a spring of effective force constant keff given by m

and

m1m 2 0.1  0.1   0.05 kg m1  m 2 0.1  0.1

keff = k1 + k2 = 0.1 + 0.1 = 0.2 N/m

1 k eq 1 0.20 1   Hz 2 m 2 0.05  (b) As here one spring is compressed while the other is stretched by same amount (say y) and balls are at rest at A and B, so

So

f

E

1 1 k1y 2  k 2 y 2  ky 2 2 2

[as k1 = k2]

But from above figure y = y1 + y2 = R1 + R2 = 2R y = 2 × 0.06 × (/6) = 0.02 m So E = (0.1)(0.02)2 = 42 × 10–5 J (c) As at P and Q springs are unstretched so the whole energy becomes kinetic of the balls A and B, i.e.

Here So

1 1 m1v12  m 2 v 22  E  4 2  105 J 2 2 m1 = m2 = 0.1 kg and v1 = v2 = v 2 2 –5 0.1 v = 4 × 10 , i.e., v = 2 × 10–2 m/s

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41

SIMPLE HARMONIC MOTION Example 49: Figure shows a block P of mass M resting on a horizontal smooth floor at a distance l from a rigid wall. Block is pushed toward right by a distance 3l/2 and released, when block passes P k from its mean position another block of mass m1 is placed on it m which sticks to it due to friction. Find the value of m1 so that the combined block just collides with the left wall. l Sol. When block P is released from rest from a distance 3l/2 toward right from mean position, this will be the amplitude of oscillation, so velocity of block when passing from its mean position is given as

v  A 

3l 2

k m

[as  =

k ] m

If mass m1 is added to it and just after if velocity of combined block becomes v1, from momentum conservation we have mv = (m + m1)v1 or

v1 

 3l m  m  m1   2

k m 

If this is the velocity of combined block at mean position, it must be given as [now 1 =

v1 = A1

k m  m1 ]

Where A1 and 1 are the new amplitude and angular frequency of SHM of the block. It is given that combined block just reaches the left wall thus the new amplitude of oscillation must be l so we have

or

m 3l . (m  m1 ) 2

k k  l1 m m  m1

3 m 1 2 m  m1

or

9m = 4m + 4m1

or

m1 

5 m 4

Example 50: For the arrangement shown in figure, the spring is initially compressed by 3cm. When the spring is released the block collides with 4 m = 1 kg k = 10 N/m the wall and rebounds to compress the spring again. (a) If the coefficient of restitution is 0.7, find the maximum compression in the 4 cm spring after collision. (b) If the time starts at the instant when spring is released, find the minimum time after which the block becomes stationary. Sol. (a) Velocity of the block just before collision, 1 1 1 mv 20  kx 2  kx 02 2 2 2

here,

k 2 x  x2 m 0 x0 = 0.03 m, x = 0.01 m, k = 104 N/m, m = 1 kg



v0  2 2 m / s

or

v0 





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42

SIMPLE HARMONIC MOTION After collision, v = ev0 = (0.7) 2 2 = 2 m/s Maximum compression in the spring is 1 2 1 2 1 kx m  kx  mv 2 2 2 2 m 2 1(2) 2 2 x m  x  v  (0.01)   2.23 cm or k 104 (b) In the cse of spring-mass system, since the time period is independent of the amplitude of oscillation, therefore 2

T

T0   1    2 sin 1      3  2  m k

T

 1  1      2 sin  3    

sin –1(1/3)

+3 x O –1

/2

/2

–3

v

sin –1(1/3)

Example 51: The system shown in the figure can move on a smooth surface. the spring is initially compressed by 6 cm and then released. Find k=800N/m (a) time period 6 kg 3kg (b) amplitude of 3 kg block (c) maximum momentum of 6 kg block.

µ k m1m 2 3 6 µ   2 kg m1  m 2 9 T  2

Sol. (a) Here 

T  2



T

(b) Here But and or or  or or 

2 1  2  800 20

 sec ond 10

x0 = 6 cm x10 + x20 = x0 m1x10 = m2x20 3x10 = 6x20 x10 = 2x20 x10 + x20 = x0

x10 

x10  x0 2

3 x10  x 0 2 2 2 x10  x 0   6  4 cm 3 3

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43

SIMPLE HARMONIC MOTION (c) Applying conservation principle of momentum 3 × v1 – 6 v2 = 0  3v1 = 6v2  v1 = 2v2 Applying mechanical energy conservation 

1 1 1  3  v12   6  v 22  kx 20 2 2 2

or

3 2 1 v1  3v 22   800  (0.06) 2 2 2

or

3 (2v 2 ) 2  3v 22  400  36  104 2 3  4v 22  3v 22  1.44 2 9 v22 = 1.44

or or

1.2  0.4 m / s 3 P2max = 6 × 0.4 = 2.4 kg m/s v2 

or 

Example 52: A 2 Kg block moving with 10 m/s strings a spring of constant 2 N/m attached to 2 Kg block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be (A)

2 sec

(B)

T  2

Sol: Here

µ

1 sec 2

10 m/s 2 kg

(C) 1 sec

2 kg

(D)

1 sec 2

µ k

22  reduced mass = 1 2 2

1  2 s. 2 Initially, system is at mean position. Again blocks separate when spring comes in natural length (i.e. mean position). Hence, t = T/2 = 1 second. 

T  2

Example 53: Two blocks A(2kg) and B(3kg) rest up on a smooth horizontal surface are connected by a spring of stiffness 120 N/m. Initially the spring is undeformed. A is imparted a velocity 3kg 2kg of 2m/s along the line of the spring away from B. Find the displacement of 2m/s B A A t seconds later. Sol.

22 4   0.8 m / s 23 5 In C-frame, the motion is SHM. v cm 

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44

SIMPLE HARMONIC MOTION But But 

m1x1C = m2x2C x0 = x0OC + x0OC

x1OC 

m2 x 0 m1  m 2

x 2OC 

and

m1x 0 m1  m 2

Applying conservation principle of mechanical energy in C-frame, 1 2 1 2 µrrel  kx 0 2 2

or or 



1 23 1   4   120 x 02 2 5 2 64  x 20 5  120 2 x0  10 2 3 m2 x 0 10  6 x1OC   m1  m 2 5 50 x1OC 

12  0.12 m 100

k µ

Also,



here

µ

m1m 2 3 2 6   m1  m 2 5 5





120  10 rad / s 1.2

 or or 

x1C = x1OC sin t x1 – xC = 0.12 sin 10t x1 = xC + 0.12 sin 10t x1 = 0.8t + 0.12 sin 10t

Example 54. Two blocks of masses m1 and m2 are connected by a spring of stiffness k. Another block of mass m1, sliding at V0 without friction, hits the set up elastically as shown in the figure.

m1

m1

k

m2

V0

Plot elastic energy of the spring as a function of time t after the collision. Sol. Due to one dimensional elastic collision of identical bodies, velocity is transferred totally to m1. m1

m2 V0

Let x1 and x2 be positions of m1 and m2 at a time t. Then extension in the spring (with natural length = l) x = x2 – x1 – l ...(1)

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45

SIMPLE HARMONIC MOTION The equation of motion of m1 is writen as d 2 x1 d 2 x1 m1 2  kx m1 2  kx or dt dt From these d2 x d 2 x 2 d 2 x1 (x  x  l )   2 1 dt 2 2 dt 2 dt  k   k k k    x x     x m2   m1  m1 m 2 

where

d2x   2 x 2 dt

...(2)

 1 1    k    m 2 m1 

...(3)

Solving these, x = A sin (t + ) t = 0, x = 0, sin (t + ) sin = 0  = 0,  ...(1) dx 2 dx 2  0,  V0 At t = 0, dt dt dx  0  V0  or A cos  = – V0 dt =  ...(2) From (i) and (ii) V02 2 2 sin  + cos  = 0 + 2 2 A V0 A  V0 V x sin(t  )   0 sin t    1 U(x)  kx 2 Now we can write 2 At

The plot of

2 0

 kV  2 U(t)   sin t  2CO 2 

U(t) kv02 2 2

Umax

t

Circular Representation of SHM We have discussed that an oscillation can be regarded as ShM if it satisfies the basic requirements to be SHM. Every SHM can be best represented as a projection of a particle in circular motion on its diameter. In fact the motion of projection of a uniform circular motion on its diameter satisfies both the conditions required to be SHM. Let us analyze it from the figure shown.

B 2´ 1´

y t=0 x

x´

x´ C

y

P´ O

2 1 P x t=0 A

y´ D y´

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46

SIMPLE HARMONIC MOTION A particle P is executing a uniform circular motion on the circle with a uniform angular velocity  as shown. The particle is shown at its initial position A at t = 0. It starts in anticlockwise direction with uniform angular velocity . If we follow the motion of the projection of particle on its vertical diameter in figure then we can see, along withP, its projection P´ starts from mid point (centre of circle) in upward direction and the respective position of P´ for position of P at point 1 and 2 are 1´ and 2´ as shown in figure. When P reaches the topmost position, P´ will also be at this point and when P starts tracing the second quadrant of circle, P´ starts coming down towards point O. Thus we can see, as point P traces its circular path ABCD, its projection on diameter YY´ follows oscillatory motion along OBODO .... and so on. This oscillatory motion of P´ can be taken as SHM as it follows the conditions to be a SHM. We’ll prove it in next section. Here motion of particle is in a straight line with amplitude equal to the radius of circle. As shown in figure, the displacement of P´ from point O as a function of time can be given as y y = R sin t ...(1) P´ t=t R y As P´ is going up, its velocity can be given as t

x´

dy v= =Rcos t dt

...(2)

Its acceleration is

x A t=0

y´

dv a= = – R2 sin t dt From equation (1), we have

a = – 2y

...(3)

Here is a constant thus the acceleration of P´ is directly proportional to the displacement from its mean position O and negative sign in equation (3) shows that direction of acceleration is opposite to y that is towards mean position O. Hence the motion of projection P´ can be regarded as SHM. Equation of SHM Equation of an oscillation is the mathematical expression giving the displacement of oscillating particle from its mean position as a function of time. B If a particle is executing SHM with amplitude A, it can be regarded as the P t=t projection of a circular motion of redius A as shown in figure, if circular motion P´ R y of point P is at a constant angular velocity then this is termed as angular P t C frequency of the point P´ which is in SHM. A t=0 P Let us consider at t = 0, P was at point A and starts in anticlockwise direction as shown in figure. Now in time t, point P traversed by an angle t (as shown) and P´Dreaches to a displacement y as shown, can be given as y = A sin t ...(4) This equation-(4) in this case is called as equation of SHM of point P´ which is in SHM. Here t which is the angular displacement of point P (in circular motion) is called phase angle of point P´ in SHM. In previous article we’ve discussed the SHM of point P´ which was at its mean position at t = 0. But it is not necessary that particle starts its SHM from its mean position. It can start from any point on its path, thus equation-(4) can not be accepted as a general equation of SHM, this being the equation of those all SHMs where particle starts (at t = 0) their SHM from their mean position.

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47

SIMPLE HARMONIC MOTION Now consider figure where a particle P´ starts its SHM with a point having initial phase angle anticlockwise on the circle of point P of which it is the projection. Let as t = 0, point P was at an angular displacement from its reference point (point A). Thus the point from which P´ will start its SHM is shown in figure. At this position (projec-

Y B y X´ C

t=t P

P´´ P´

t

P t=0 X A

tion of P at t = 0 on YY´) is called as initial phase of point P´ in SHM, as is the initial angular displacement of particle P which is in circular motion. Now after time t, the angular displacement of P is ( D Y´ +t) which is called the instantaneous phase of point P´ at time t = t and at this instant the displacement of point P´ from mean position is y = A sin (t + ) ...(5) This equation-(5) is called as general equation of SHM of a particle which starts its SHM with initial phase . Here initial phase of SHM implies it is the initial angular displacement of the particle which is in circular motion of which the projection is executing SHM. As equation-(4) gives the equation of SHM of those particles which starts their SHM from mean position. Similarly we can define an equation of SHM of those particles which start their SHM from their extreme position by substituting  = /2 in equation (5). As if a particle starts from its extreme position, we can take its initial phase /2 thus its equation of SHM from equation (5) can be given as y = A cos t ...(6) In all type of problems in which a body or a particle execute SHM, we assume that this is the projection of an another perticle who is in circular motion and its projection is executing SHM (that body or particle which is in SHM). Valocity and acceleration of a particle in SHM Equation (5) gives the general expression for displacement from mean position of a particle executing SHM as a function of time. Thus velocity of this particle as a function of time can be given as v=

dy = A cos (t + ) dt

...(7)

To convert it in displacement function, we can write. v = A 1  sin 2 (t   ) y2 v = A 1  2 A

v =  A 2  y 2

[as y = A sin t] ...(8)

Equation (8) gives the velocity of a particle in SHM with amplitude A, and angular frequency as a function of its displacement from mean position. From equation-(8) we can state that in SHM, particle’s velocity is maximum when y = 0 i.e. at its mean position and is given as at

y = 0,

vmax = A

At extreme positions of particle when y = ± A, its velocity is zero where it returns towards its mean position. From equations-(5) and (8) we can plot the graphs of displacement and velocity as a function of time as shown in figure.

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48

SIMPLE HARMONIC MOTION y A sin O

T

t

2T

T= 2 y = A sin ( t + ) (a) +A

y

7T/4

T/2 3T/4

O T/4

T

–A

5T/4

t

2T

T= 2

y = A sin t (b)

y

T/4 T/2

O

3T/4

T

t

2T

y = A cos t (c)

Similarly acceleration of particle in SHM can be given as a=

or

dv = – A2 sin(t + ) dt

a = – 2y

...(9)

[as y = A sin (t + )] ...(10)

From equation-(10) we can see that at mean position (y = 0) when velocity of particle is maximum, its acceleration is zero and at extremities where y = ± A, accelration of particle is maximum and its magnitude is given as amax = 2A

(towards mean position)

It also shows that as particle moves away from mean position, its acceleration continuously increases till it reaches its extreme position (at amplitude) when its velocity becomes zero and it returns. Equation-(10) can be rewritten in differential from as

d2 y + 2y = 0 dt 2

...(11)

This equation-(11) is called “Basic Differential Equation” of motion of a particle in SHM. Every expression of displacement y as a function of time which satisfies this equation can also be regarded as an equation of SHM. C39: Prove that yAeit is an equation of SHM. Sol. According to given equation in problem, differentiating with respect to time we get,

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49

SIMPLE HARMONIC MOTION dy = i A eit dt

Differentiating again with respect to time, we get

d2 y = – 2Aeit = – 2y dt 2 Thus we have

[as y = Aeit]

d2 y + 2y = 0 dt 2

This is the basic differential equation of SHM hence y = Aeit is an equation of SHM. Example 55: The displacement of a body executing SHM is given by x = A sin (2t + /3). The first time from t = 0 when the velocity is maximum is (A) 1/12 sec. (B) 0.16 sec (C) 0.25 sec (D) 0.5 sec v

Sol: or or Hence

 or

dx    2A cos  2t    dt 3

± 2A = 2A cos (2t +

 ) 3

t = t0

  cos  2t    1  3     2 3 3  2  6  = t0   2 t 0 6



t=0 /3

Phasor diagram

t0 

 1  sec. 12 12

Hence (A) option is correct

Phase analysis of a particle in SHM We’ve already discussed about phase angle of a particle in SHM. It is actually the angular displacement of that particle who is in circular motion and whose projection is in SHM. At a general time t, the instantaneous phase of a particle in SHM can be written as (t + ) if is its initial phase (already discussed). Phase difference in two SHM Case I : When two SHMs are of same agnular frequency Figure shows two particles P´ and Q´ in SHM with same angular frequency . P and Q are the corresponding particles in circular motion for SHM of P´ and Q´. Let P and Q both starts their circular motion at the same time at t = 0 then at the same instant P´ and Q´ starts their SHM in upward direction as shown. As frequency of both are equal, both will reach their extreme position (topmost point) at the same time and will again reach their mean position simultaneously at time t = T/2. [T = Time period of SHM = 2/] and move in downward direction together or we can state that the oscillations of P´ and Q´ are exactly parallel and at every instant the phase of both P´ and Q´ are equal, thus phase difference in these two SHMs is zero. These SHMs are called same phase SHMs.

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50

SIMPLE HARMONIC MOTION

P t=t P´

Q t=t Q´

t

P A

t=0

Q

t A

ref. point

t=0

ref. point

Now consider figure, where we assume if P´ starts its SHM at t = 0 but Q´ will start at time t1. In this duriation from t = 0 to t = t1, P´ will move ahead in phase by t1 radians while Q´ was at rest. Now Q´ starts at time t1 and move with same angular frequency . It can never catch P´ as both are oscillating at same angule frequency. Thus here Q´ will always lag in phase by t1 then P´ or we say P´ is leading in phase by t1 then Q´ ans as  of both are constant their phase difference will also remains constant so keep it in mind that in two SHMs of same angular frequency, if they have some phase difference, it always remains constant. t = t1

P´ t1

P A t=0

Q´

Q t = t1

Now we consider a spacial case when time lag between the starting of two SHMs is T/2 i.e. half of oscillation period. Consider figure, here if we assume, particle P´ starts at t = 0 and Q´ at t = T/2 when P´ completes its half oscillation. Here we can see that the phase difference in the two SHM is by which Q´ is lagging. Here when Q´ starts its oscillation in upward direction P´ moves in downward direction. As angular velocity of P and Q are same, both complete their quarter revolution in same time. Thus when P´ reaches its bottom extreme position, Q´ will reach its upper extreme position and then after Q´ starts moving downward, P´ starts moving upward and both of these will reach their mean position simultaneously but in opposite directions, P´ has completed its one oscillation where as Q´ is at half of its oscillation due to a phase lag of .

t = T/2 P´

P t=0

Q´

Q t = T/2

P´

Thus if we observe both oscillations simultaneously we can see that oscillations of the two particls P´ and Q´ are exactly antiparallel i.e. when P´ goes up, Q´ comes down and at all instants of time their displacements from mean position are equal but in opposite direction if there amplitudes are equal. Such SHMs are called opposite phase SHMs. Case II : When two SHMs are of different angular frequency We’ve discussed that when two SHMs are of same angular frequency, their phase difference does not change with time. Consider two particles in SHM as shown in figure. Their corresponding particles for circular motion are A and B respectively as shown. If both A´ and B´ starts their SHM from mean position at t = 0 with angular frequencies 1 and 2, then we say at t = 0 their phase difference is zero but after time t, their respective phase

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51

SIMPLE HARMONIC MOTION are 1t and 2t. Thus after time t, the phase difference in the two SHMs is  = (1 – 2)t thus equation above shows that if angular frequencies of the two SHMs are different their phase difference continuously changes with time. t=t A´ A´

2 1

1t

A t=0

B´ B´

t=t t 2 B t=0

Same Phase SHMs As discussed above two particles execute SHM in such a way that their oscillations are exactly parallel to each other or their phase difference during oscillation is zero, they are said to be in same phase we’ve seen that this happens when both SHMs are started at same time with same angular frequency. This can happen also when the time lag between starting of the two SHM is T or an integral multiple of time period of SHM. Because if one starts at t = 0 and other starts at t = T, in this duration first particle will complete its first oscillation and is going to start its second oscillations and the second particle will start in synchronization with the first. Hence the two oscillation will still be parallel or in same phase. There phase diference in the two SHMs will be 2. Not only this even if the time lag in starting of the two SHMs is 2T, 3T ..... nT or the phase difference in the two SHMs is 4, 6, 8 ...... 2n, then also these SHMs can be treated in same phase. Thus phase difference in two SHMs of same phase is  = 2, 4, 6 .......... 2n Opposite Phase SHMs As discussed in previous article, two SHMs are said to be in opposite phase when their oscillations are antiparallel this happens when two SHMs of same angular frequency start with a time lag of T/2 and phase difference among the two SHMs is . By analyzing the situation it can also be stated that the same thing also happens when the time lag in starting of the two SHMs is 3T/2, 5T/2 ........ (2n + 1)T/2 or the phase difference between the two is 3, 5 ..........(2n + 1). Thus phase difference in two SHMs of opposite phase is  = , 3, 5 ........., (2n + 1) To understand the concept of phase and phase difference in SHM, we take few examples. Example 56: Two particles are in SHM in a straight line about same equilibrium position. Amplitude A and time period T of both the particles are equal. At time t = 0, one particle is at displacement y1 = +A and the other at y2 = – A/2, and they are approaching towards each other. After what time they cross each other ? (A) T/3 (B) T/4 (C) 5T/6 (D) T/6 Sol: From phasor diagram, t = t0    t 0  6 2     3 4 2 2t 0      6 2 6 6 3 2 2 T t0    6 6  2 / T 6

t0

t 0 

or 

t = t0

t0

C 6 –A 2 3

t0 –

6

t0

Hence option (D) is correct.

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52

SIMPLE HARMONIC MOTION Example 57. Two particles execute SHM with same amplitude A and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance A/2 from mean position. Find phase difference in the two SHMs. Sol. Figure shows that two respective particles P´ and Q´ in SHM along with their corresponding particles in circular motion. Let P´ moves in upward direction when crossing Q´ at A/2 as shown in figure(a), at this instant phase of P´ is  1  1 = sin–1   = 2 6

P´

...(1)

P N2

Q N2

Q´ 2

1

A

Similarly as shown in figure(b) we can take particle Q´ is moving in downward direction (opposite P´) at A/2, this implies its circular motion particle is in second quadrant thus its phase angle is  1  5 2 =  – sin–1   =  – = ...(2) 2 6 6 As both are oscillating at same angular frequency their phase diff. remains constant which can be given from equation-(1) and (2), as

 = 2 – 1 =

5  2  – = 6 6 3

Example 58: A particle executes SHM with amplitude A and angular frequency . At an instant when particle is at a distance A/5 from mean position and moving away from it. Find the time after which it will come back to this position agin and also find the time after which it will pass through mean position. B Sol. Figure shows the circular motion representation for the particle P´ given in prolbem. The initial situation of particle is shown in figure. As P moves, its projection P´ will P go up and then come back to its initial position when P reaches to the corresponding P A/5 position in second quadrant as shown. In this process P traversed an angular C A displacement  with angular veliocity , thus time taken in the process is D  1 2cos 1 (1/ 5) 1 t   [2 cos (1/ 5)]     Now when P reaches point C, P´ will reach its mean position. Thus time taken by P from its initial position to point C is

(   )   sin 1 (1/ 5) t    The same time P´ will take from A/5 position to mean position through its extreme position. Example 59: Two particles executing SHM with same angular frequency and amplitudes A and 2A on same straight line with same mean position cross each other in opposite direction at a distance A/3 from mean position. Find the phase difference in the two SHMs.

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53

SIMPLE HARMONIC MOTION Sol. Figure shows the two corresponding particles of circular motion for the two mentioned particles in SHM. Let particle P is going up and particle Q is going down. From the figure shown, the respective phase differences of particles P´ and Q´ are

and

–1 

 1 1 = sin  3 

[Phase angle of P´]

 1 2 =  – sin–1   6

[Phase angle of Q´]

Q

Q´

2A

2

P´ N3

1

Thus phase difference in the two SHMs is  1  1  = 2 – 1 =  – sin–1   – sin–1  3  6

C40: A particle stars its SHM from mean position at t = 0. If its time period is T and amplitude A. Find the distance travelled by the particle in the time from t = 0 to t = 5T/4. Sol. We know in one complete oscillation i.e. in period T, a particle covers a distance 4A and in first one quarter of its period it goes from its mean position to its extreme position as its starts from mean position thus the distance travelled by the particle in time 5T/4 is 5A. Example 60: A spring block pendulum is shown in figure. The system is hanging in equlibrium. A bullet of mass m/2 moving at a speed u hits the block from downward direction and gets embedded in it. Find the amplitude of oscillation of the block now. Also find the time taken by the block to reach its upper extreme position after hit by bullet.

Nature length of spring m

h

Sol. If block is in equilibrium then spring must be at some stretch if it is h, we have mg = kh If a bullet of mass m/2 gets embedded in the block, due to this inelastic impact its new mass becomes 3m/2 and now the new mean position of the block will be say at a dept. h1 from old mean position then, we must have 3 mg = k(h + h1) 2 3m g = k(h + h1) 2 mg h1  (as mg = kh) 2k Just after impact due to inelastic collision if the velocity of block becomes v, we have according to momentum conservation

or

v m u

m/2

m 3m u v 2 2 u v 3

Now the block executes SHM and at t = 0 block is at a distance h1 

mg above its mean position and having 2k

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54

SIMPLE HARMONIC MOTION a velocity u/3. If amplitude of oscillation is A, we have u  mg    A2    2k  3

2

2 u 2 2k  2  mg    A      or 9 3m  2k   k [As for this spring block system  = ] (3m / 2) 2

mu 2  mg    6k  2k  Now time taken by particle to reach the topmost point can be obtained by circular motion representation as shown in figure. This figure shows the position of block P and its corresponding circular motion particle P0 at the t = 0. Block P will reach its upper extreme position when particle P0 will travers the angle  and reach the topmost point. As P0 moves at constant angular velocity , it will take a time given as A

t

P h1

P1 A

 cos 1 (h1 / A) 3m  3mg    cos 1   2kA   2k k 3m / 2

Example 61: Figure(a) shows a spring block system, hanging in equilibrium. The block of system is pulled down by a distance x and imparted a velocity v in downward direction as shown in figure(b). Find the time it will take to reach its mean position. Also find the maximum distance to which it will move before returning back towards mean position.

k equilibrium position

Sol. As shown in figure (b), when the block is pulled down by a distance x and thrown downward, it will start executing SHM. It will go further to a distance A (amplitude) from mean position before returning back which can be found by using the velocity of block v at a displacement x from its mean position as

m

x m

(a)

(b)

v   A2  x 2

or

v2 

k 2 A  x2 m





[as for spring block system  =

k ] m

mv 2  x2 k Now time of motion of bob can be obtained by circular motion representation of the respective SHM. Corresponding circular motion representation for this SHM is shown in figure at t = 0. At t = 0, block P is at a distance x from its mean position in downward direction and it is moving downward so we consider its corresponding circular motion particle in III quadrant as shown in reference angular velocity we consider P will reach its mean position when particle P0 reaches position A by traversing an angle . Shown in figure. Thus it will take a time given as

or

A

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55

SIMPLE HARMONIC MOTION B

1    sin  x / A  t   k m

C

A A

x P

P0

  v   D m x  t   sin 1 or k   mv 2  x2   k   The maximum distance to which block will move from its initial position is A – x as it gas upto its lower extreme at a distance equal to its amplitude A from to mean position. Example 62: Figure shows a block of mass m resting on a smooth horizontal ground attached to one end of a spring of force constant k in natural length. If another block of same mass and moving with a velocity u toward right is placed on the block which stick to it due to friction. Find the time it will take to reach its extreme position. Also find the amplitude of oscillations of the combined mass 2m.

u

m k m smooth

Sol. When second mass sticks to the lower mass, due to such an inelastic collision, the velocity of combined block is reduced by half that is u/2 to conserve momentum. Now at mean position the velocity of block can be written as u  A 2

or

u k A 2 2m

(if A is the amplitude of oscillation) [As here for combined block new angular frequency of SHM is  =

k

2m ]

u 2m m u 2 k 2k As oscillation starts from mean position, in reaching its extreme position, particle has to cover a phase of /2 radians, thus time taken by particle to reach its extreme position is or

A

t

/2  k   2 2m

Example 63: In previous example if block is pulled toward riht by a distance x0 and released, when the block passes through a point at a displacement x0/2 from mean position, another block of same mass is gently placed on it which sticks to it due to friction. Find the new amplitude of oscillation and find the time now it will take in reaching its mean position and extreme position on left side. Sol. When block was released at x0 from mean position, this will be the amplitude of oscillation and when it is passing through the position of half amplitude x0/2, its velocity can be given as

x  v   A  0  2 2

v

2

or

v

k m

x 02 

x 02 4

3 k x0 2 m

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56

SIMPLE HARMONIC MOTION When another block of same mass is added to it, due to momentum conservation its velocity becomes half and  of oscillation will, also change from

k to m

k . Again using the formula for velocity of SHM at a 2m

distance x0/2 from mean position, we get

v x  v´   A´2   0   2 2 or

3 k x0  2 m

2

2

k x  A´2   0  [if A´ is the new amplitude of oscillation]  2 2m

5 3 2 x 02 5 2 A´ x0 A´  x 0   x0 or 8 4 8 2 2 To find time of motion in SHM we use circular motion representation of the respective SHM. The figure shows the corresponding circular motion. If at t = 0, the second mass is added to the oscillating block, it was at a position x0/2 from mean position and moving towards it, and after adding the mass the new amplitude of 2

oscillation changes to A´ and  changes from

k to m

k . Figure shows the corresponding position of 2m

particle in circular motion at t = 0 in II quadrant. When this particle P0 will reach the position C after traversing the angle , particle P in SHM will reach its mean position and similarly when P0 will reach position D, P will reach the extreme position on other side. thus the time taken by P to reach mean position from a position of x0/2 from mean position is given as  x  B sin 1  0  P0  2A´  t=0 P t1   A0 v´ ´ k x 0 /2 A A´ C 2m  2 sin 1  D  2  5  2m t1   sin 1   or k k  5 2m Similarly time taken by P to reach the left end extreme position is t2 

/2  ´

2m k

  2 1   sin    5    2

Example 64: Figure shows a block P of mass m resting on a smooth horizontal surface, attached to a spring of force constant k which is P k rigidly fixed on the wall on left side, shown in figure. At a distance l to m the right of block there is a rigid wall. If block is pushed toward lift so that spring is compressed by a distance 5l/3 and released, it will start l its oscillations. If collision of block with the wall is considered to be perfectly elastic. Find the time period of oscillations of the block. Sol. As shown in figure, as the block is released from rest at a distance 5l/3 from its mean position, this will be the amplitude of oscillation. But on other side of mean position block can move only upto a distance l from mean

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57

SIMPLE HARMONIC MOTION position and then it will return from this point with equal velocity due to elastic collision. Consider figure. If no right wall is present during oscillation block P will executing complte SHM on right side of mean position also up to its amplitude 5l/3. Thus we can observe the block P at a point X at a distance l from mean position (where in our case wall is present), if block passes this position at speed v (which is the speed with which block hits the wall in our cse), after reaching its extreme position Y, it will return and during return path it will cross the position X with the same speed v (as displacement from mean position is same). v

X v

P m

l Y

5l/3

(a)

5l/3

Thus we can state that in our case of given problem, block P is executing SHM but it skips a part XYX on right half of its motion due to electric collision of the block with the wall. If T0 is the time period of this oscillation we look at figure (b), which shows the corresponding circular motion representation. Here during oscillation of point P, particle P0 covers its circular motion along ECDAF, and from F it instantly jumps to E (due to elastic collision of P with wall at X) and again carry out ECDAF and so on, thus we can find the time of this total motion as

t

  2   2sin 1 (3/ 5)   k m

B Y P0 E

P X

l

C

 1  3   ...(1)    2 sin  5     We can also find the time period of this motion by subtracting the time of FYE from the total time period as t

m k

 3 2  2 cos 1   m  5  or t t k  these equation-(1) and (2) will result same numerical value.

5l/3

F A

5l/3 D

(b)

 1  3    2  2 cos  5    

...(2)

Example 65: Figure shows a spring block system hanging in equilibrium. If a velocity v0 k is imparted to the block in downward direction. find the amplitude of SHM of the Natural length block and the time after which it will reach a point at half of the amplitude of block. h Sol. Initially in equilibrium if block is at a depth h below the natural length of spring then m P we have Mean position mg = kh If at mean position block is imparted a velocity v0, this would be the maximum velocity of block during its oscillation. If its amplitude of oscillation is A, then it is given as v0 = A

[where  =

k ] m

v0 m  v0  k Now to find the time taken by block to reach its half of amplitude point we or

A

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58

SIMPLE HARMONIC MOTION consider the corresponding circular motion of the SHM as shown in figure. Here block P will reach to half of its amplitude when particle P0 will reach point E shown in figure at an angular displacement  from mean position relative to point A, thus time taken by it is 1  sin 1 2  m t    6 k km

B

A/2

C

Angular Harmonic Motion : If the angular acceleration of a system be directly proportional in magnitude to the angular displacement and be directed opposite to angular displacement relative to fixed position, its motion is said to be angular simple harmonic.

v0 P

E A A P0

D

Z Z

In the figure rotation of the body about z-axis is such that angular acceleration z is given by z = – 2 where 2 is a positive constant. The motion is then angular SHM. As  z 

d 2 , dt 2

d 2   2 the differential equation may be written as 2 dt Multiplying both the sides by the moment of inertia of the body about the z-axis (I), we get z = – I2  where z is the z-component of net torque  acting on the body.. Table : SHM Features Quantity

Linear SHM

Equation

m

Angular frequency

  k/m

  C/I

Displacement

x = A sin(t + )

 = 0sin(t + )

Phase Phase constant

t +  

t +  

Amplitude

A

0

Total energy of oscillation

1/2mA2 = 1/2kA2

1/2I202 = 1/2C02

Potential energy

1/2m2x2

1/2I22

Kinetic Energy

1/2I2(A2 – x2)

1/2I2(02 – 2)

d2x + kx = 0 dt 2

Angular SHM I

d 2 + C = 0 dt 2

Simple Pendulum (mathematical pendiulum) A heavy point mass suspended from a rigid support using inextensible, elastic and massless thread and free to oscillate without friction is called a simple pendulum. All these conditions are ideal and cannot be realised completely in particle. Hence such a pendulum is also called mathematical pendulum. In laboratory a thread, a bob, a split cork and stand are arranged to get an approximate simple pendulum.

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59

SIMPLE HARMONIC MOTION Splitcork

l

L= l + h + r h r

rigid

friction Zero massless, inextensible, elastic cord heavy point mass

Motion of simple pendulum Let the thread be deflected by an angle  about z-axis as in the figure. The torque abotu z-axis is given by

Z-axis

l sin l

z = –mgl sin  For small angle, sin   

(see table)

=0

Table : Some values of sin  (radian)

(degree) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.01745 0.03490 0.05235 0.06981 0.08727 0.10472 0.12217 0.13963 0.15708 0.17453 0.19199 0.20944 0.22689 0.24435 0.26180 (Values reounded at 5th place)

mg

sin  0.01745 0.03490 0.05234 0.06976 0.08716 0.10453 0.12187 0.13917 0.15643 0.17365 0.19081 0.20791 0.22495 0.24192 0.25882

Then,

z = –mgl

(angular SHM)

Now,

d 2 d 2 2 = ml dt 2 dt 2 d 2  = ml2 2 = – mgl dt 2 d  g     2 l dt Comparing this with standard equation, we get

z = I

2 

g l

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60

SIMPLE HARMONIC MOTION Thus, the angular frequency  of the simple pendulum is given by

g l The time period of the pendulum is given by 

l 2  T = 2 g  This expression contains four informations known as laws of simple pendulum.

T=

Laws of simple pendulum : (i) T  l (law of length)

1 (ii) T  g (law of acceleration due to gravity) (iii) T doesn’t depend upon mass (law of mass) (iv) T doesn’t depend upon amplitude (law of amplitude) Second’s Pendulum : The simple pendulum having time period of 2s is called second’s pendulum. A pendulum clock generally uses this pendulum. Simple pendulum in accelerated frame : Case I : Let the simple pendulum be located in a lift. The acceleration D the lift is a0.

 a 0 upward : In this case ma0 and mg are downward giving effective-g geff = (g + a0) a0 ma0



T  2

l g eff

m(g+a0)

mg

 2

l g  a0

Time period has decreased. A pendulum clock in such a lift will be running fast. Case II :

 a directed downward : In this case pseudoforce ma0 acts upward. The net downward force is mg – ma0. This gives effective-g equal to ‘g – a0’. Then

T  2

l g  a0

ma0 mg

a0< g m(g+a0)

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61

SIMPLE HARMONIC MOTION Case III :

   a 0 is horizontal : In this case ma 0 and mg are at right angles. They give as effective - g

geff  g2  a20 Hence the period is

a0

ma0

l

T  2

2

g  a 02 mgeff = m g2 + a02

mg

Simple pendulum in a fluid (non-visicous) :  In a non-visicous fluid, the vertical buoyant force (m/)g acts opposite to mg . The net downward force is mg – (m/)g. m = Volume displaced Fb = m g

  The effective-g is  1   g .  

m = mass of bob = density of bob = fluid density


Sol. Let the string make and angle  with the downward vertical. Let t = 0 at  = . Then  =  cos t

g l At the wall  = – – =  cos t where



t

= =–

   1 cos 1     

The period is double of this value. Hence T2

  l cos 1    g  

Example 68: Figure shows two identical simple pendulums of length l. One is tilled at an angle and imparted an initial velocity v1 toward mean position and at the same time other one is projected away from mean position at a velocity v2 at an initial angular displacement B. Find the phase difference in oscillations of these two pendulums.

l

l A

v1

v2

B

Sol. It is given that first pendulum bob is given a velocity v1 at a displacement l from mean position, using the formula for velocity we can find the amplitude of its oscillations as

V1   A12  (l )2 g we have l

As for simple pendulum = V12 

[If A1 is the amplitude of SHM of this bob]

g [(A12 )  (l ) 2 ] l

v12l A1  l   g 2

2

...(1)

Similarly if A2 is the amplitude of SHM of second pendulum bob we have

V2   A 22  (l) 2 or

v 22l A2  l   g 2 2

...(2)

Now we represent the two SHMs by circular motion representation as shown in figure.

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64

SIMPLE HARMONIC MOTION A0

B t=0 B0 A2 l 2

I SHM (a)

t=0 A l

A1

1

II SHM (b)

In first pendulum at t = 0 the bob is thrown from a displacement l from mean position with a velocity v1 toward mean position. As it is moving toward mean position, in figure(a), we consider the corresponding circular motion particle A0 of the bob A is second quadrant, as the reference direction of , we consider anticlockwise. As shown in figure initial phase of bob A is given as

 l  1    sin 1    A1 

...(3)

Similarly for second pendulum bob B, its corresponding circular motion particle B0 at t = 0 is considered as shown in figure (b) its initial base is given as

 l  2  sin 1    A2 

...(4)

As both the pendulums are identical, their angular frequency for SHM must be same, so their phase difference will not change with time, hence their phase difference can be given as

 l   l     2  1  sin 1    sin 1      A2   A1  Example 69: In previous question if second pendulum bob is thrown at velocity v2 at an angle  from mean position but on other side of mean position. Find the phase difference in the two SHMs now as shown in figure.

l

v1

l v2

B

A

Sol. In this case still amplitudes of the two SHMs will remain same and are given by equation (1) and (2) but when we represent the two SHMs on their corresponding circular motions, the position of the particle in circular motion in second pendulum is now different as shown in figure. As shown in figure (a) and (b) the initial phases of the A0 two pendulum bobs are t=0 A  l  l A1 2 1 =  – sin–1  A  A 2 1  1 l and

 l 2 =  – sin–1    A2 

B

B0

II SHM (a)

II SHM (b)

As  for both SHM are same, their phase difference remains constant so it is given as  l   l –1 –1  = 2 – 1 = sin   + sin  A   1  A2 

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65

SIMPLE HARMONIC MOTION Example 70: Two pendulums have time periods T and 5T/4. They starts SHM at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase : (A) 5 (B) 4 (C) 11 (D) 9 Sol: n1T1 = n2T2 or

n1T  n 2

or

n1 5  n2 4



n1 = 5

5T 4

Hence, option (A) is correct.

C47: A simple pendulum having a bob of mass m swings with an angular amplitude of 40º. Show that the tension in the string is greater than mg cos 20º when its angular displacement is 20º. Sol. The tension is the string at an angle 20º is

mV 2 T = mg cos 20º + L or 

mV 2 0 T – mg cos 20º = L T > mg cos 20º.

COMPOUND PENDULUM This is made whenever a rigid body is hanging freely from a horizontally pivoted axis, as shown in figure. A body of mass m is pivoted at point O through a horizontal A axis AA´. The body is hanging freely under gravity and in equilibrium position its centre of mass C is vertically below the suspension point O, at a distance l from O. If the body is slightly tilted from its equilibrium position by an angle , mg will exert a restoring torque on it in opposite direction to restore the equilirbium position. Thus restoring torque on body in dotted position after tilting is R = – mg. l sin  R = –mgl If its angular acceleration is , we have I = – mgl or

=–

mgl  I

A´ O l l c c mg

[–ve sign for restoring nature] [for small , sin   ] [Here I is the momentum of inertia of body about axis AA´] ...(1)

Comparing equation (1) with standard differential equation of angular SHM we get

mgl I Thus its time period of oscillation is 

T

2 I  2  mgl

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66

SIMPLE HARMONIC MOTION EQUIVALENT LENGTH OF A SIMPLE PENDULUM We know that the time period of oscillations of a simple pendulum is given by 2

l g

...(3)

Similarly we’ve also discussed that the time period of a compound pendulum is given as

2

I Mglc

...(4)

Where I is the moment of inertia of the rigid body about the axis of rotation and lc is the distance of centre of gravity of body from the suspension point (Axis of rotation). if we consider a simple pendulum of length leq which has a bob of same mass M as that of rigid body of compound pendulum and has the time period same as that of the compound pendulum then this length of simple pendulum leq is called “Equivalent length of simple pendulum for the given compound pendulum”. For the time periods of the two pendulums to be equal, we have

2

l I  2 g mglc

...(5)

If k be the radius of gyration of the rigid body about an axis passing through its centre of mass then the moment of inertia of the rigid body about point of suspension is given as I = Mk2 + Mlc2

...(6)

Now from equation-(5), we have leq g

or



leq  lc 

Mk 2  Mlc2 mglc

k2 lc

...(7)

Equation-(7) gives the equivalent length of simple pendulum for the given compound pendulum. One important point to be noted here is if in equation-(7) we replace lc by k2/lc, we get k2 k2 leq   lc k 2 / lc leq 

k2  lc lc

...(8)

Which is same as that of equation-(7). Thus we can say that if the same rigid body, which is suspended from a point, situated at a distance lc from centre of gravity of body, we suspend if from a point at a distance k2/lc from centre of gravity of the body and oscillate like a compound pendulum, its equivalent length of simple pendulum remains same or the time period of oscillation of body remains same. Consider figure, a rigid body is suspended at a point O through a horizontal axis AA´. Here C is the centre of gravity of the body. If it oscillates then the time period of small oscillations can be given as

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67

SIMPLE HARMONIC MOTION k2 lc

lc  T  2

...(9)

g

As discussed in last section we can say that if the same body is suspended either from any point on circular arcs PQ or RS [as shown in figure(b)] of radius lc with centre at C (circle M1) or any point on the circle of radius k2/lc with centre at C (circle M2), the time period of small oscillations of the body will remain same. B

A´ P

O1 2 lc

A

Q M2

lc k2 /l c C

M1

c 3

R

S 4 B´

(a)

(b)

Here we can develop one property of the body when it is used as a compound pendulum. This is, when a straight line is drawn passing through the centre of gravity of body as in figure (b) a line BB´ is drawn. There axis four points on this line (as here points 1, 2, 3 and 4) about which if body is suspended, the time period of small oscillation of body remains same. This we can also prove graphically as if we plot the time period of oscillation, the curve looks like as shown in figure. T As shown in firure, if the body is suspended from C, from equation-(9), we can see that if lc = 0, time period becomes , and at lc = k i.e. if the body is suspended from a point at a 2 3 4 T1 1 distance equal to radius of gyration of body from C, the time period of oscillation is minimum and at all other suspension points the time period is higher and if we draw a horizontal Tmin line in graph at time period T1 which is more than minimum x period, it cuts the graph at four points as shown, which verifies k k C the statement we´ve discussed earlier. lc = k

lc = 0

lc = k

From equation-(9) we can also find the value of lc at which this time period has a minimum value by equating dT dlc = 0, as

dT 2  . dlc g

or

 k2  1  2   0 k 2  lc  2 lc  lc 1

lc = ± k

Which also verifies the experimental result obtained by graph shown in figure.

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68

SIMPLE HARMONIC MOTION Example 71: A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is (A) 2m (B) 4m (C) 1.5 m (D) 3m I T  2  mgl

Sol:

or

T  2

or

2

3mr 2 2mgr

3r 2m

l 3r  2 g 2g

3 3 r   1  1.5 m Hence option (C) is correct. 2 2 C48: Calculate the time period of a metre scale if it is free to rotate about a horizontal axis passing through its one end. Sol. Total length of the metre scale = 1 metre



l

Here,

K = (radius of gyration) =

1 12 l = distance of the C.G. from one end = 0.5 m T = 2

C49:

K2  l 2 = 1.6 sec (nearly). lg

Calculate the time period of small oscillation of a uniform metre stick, if it is suspended through 10 cm mark.

Sol. For a metre stick, K2 =

L 1  12 12

 L = 1 metre

K2 1   0.208 l 12  0.4

Hence,

T  2

K2 / l  l 0.208  0.4  2  1.56 sec (nearly) g 9.8

C50: A uniform circular disc of mass m and radius R is suspended from a small hole in the disc. Calculate the minimum possible time period from small oscillations. T  2

Sol.

k2 / l  l ; g

T will be minimum when (k2/l + l) is minimum, when k / l  l  0 ; or l  k  R 2 C51: Calculate the length of a simple pendulum whose time period is equal to that of a particular physical pendulum. Sol. We have or

T  2 L

L I  2 g mgl

I ml

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69

SIMPLE HARMONIC MOTION Example 72: Consider a pulley of moment of inertia I supporting a block of mass m, tied to an ideal spring of stiffness k. The block is slightly pulled downward and released. Determine its period of oscillation.

R

Sol. In equilibrium

m

mg = T . T = kx0 Also,  mg = kx0 ...(i) Let the block be displaced downward by x and moving with velocity dx/dt. Then U

I

k

1 k(x  x 0 )  mgx 2 2

1  dx  1 K  m    I 2 2  dx  2

2

1  dx  1 I  dx  K  m     2  dt  2 R 2  dt 

or

2

2

1 1  dx   I  E  k(x  x 0 )2  mgx     m  2  .  2 2  dt   R  Differentiating it and setting to zero, dE 0 dt

dx dx  dx   I  d2x  mg  m  k(x + x0) =0   dt dt  dt   R 2  dt 2 d2x  I  k(x + x0) mg  2  m  2   0 ...(ii) dt R Using (i) and (ii), we get d2 x k  x 2 dt m  I / R2 Comparing it with standard equation of SHM, d2x   2 x , 2 dt k m  I / R2 2  T  2   m  I / R2 K Example 73: An L-shaped bar of mass M is pivoted at one of its end so that it can L freely rotate in a vertical plane, as shown in the figure. (a) Find the value of 0 at equilibrium. (b) If it is slightly displaced from its equilibrium position, find the frequency of oscillation. B Sol. (a) Taking B as the origin, the co-ordinates of its c.m. are ML A L xC  2 2  M M 4 M/2 L c.m.  2 2 yC L ML B M/2 L xC and yC  2 2  M 4

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A

0

L

C

C

70

SIMPLE HARMONIC MOTION tan 0 

L/4 1  3L / 4 3

 1 0  tan 1    3

(b) The frequency of oscillation for a compound pendulum is

1 mgd 2 1 where d = distance of the COM from the point of suspension. I = moment of inertia about the point of suspension. f

2

A 3L/4 0

2

L  3L   L  d       10  4   4 4

L/4 B

2 2  M L  M L  M   L I           L2     2  3  2  12  2    2

L 1 Mg 4 10 f 2 ML2 3



or

f

2

C

 ML2  3 

1 g (2.37) 2 L

Example 74: A system of two identical rods (L–shaped) of mass m and length l are resting on a peg P as shown in the figure. If the system is displaced in its plane by a small angle , find the period of oscillations : (A) 2

2l 3g

(B) 2

T  2

Sol:

2 2l 3g

L/4

(C) 2

2l 3g

l

(D) 3

P

l

l 3g

I mgL

ml 2 ml 2 2mgl 2 I   Here 3 3 3 L sin 45º  From figure : l/2

l/2

l/2 º 45 L

45º G

l



L



2ml 2 2 2l T  2  2 l 3g 3 mg 2 2

2 2

Hence option (C) is correct.

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SIMPLE HARMONIC MOTION Example 75: A solid cylinder is attached to a horizontal massless spring so that it can roll without slipping along a horizontal surface. The spring constant k is 3 N/m. If the system is released from rest at a point in which the spring is stretched by 0.25 m, find (a) the translational kinetic energy and (b) rotational kinetic energy of the cylinder as it passes the equilibrium position. (c) Show that under these conditions the centre of mass of the cylinder executes SHM with time period

T  2 (3M / 2k) Sol. If at any position the stretch in the spring is x and the velocity of centre of mass of the cylinder is V, then as U

1 2 kx , 2

KT 

1 MV 2 2

1 2 1 I  MV 2 2 4 So total mechanical energy of the system E = KT + KR + U 1 1 1 i.e. E = MV2 + MV2 + kx2 2 4 2 3 1 i.e. E = MV2 + kx2 4 2 According to given problem V = 0 if x = 0.25

and

KR 

[as L =

1 Mr2 and  = V/r] 2

...(1)

2

so

1 3  1 E   (3)     J  4 2 32

Now at equilibrium position U = 0 [as x = 0]. 3 3 1 MV 2  J, MV 2  J i.e. 4 32 8 So at equilibrium position : (a) translation KE = (1/2) MV2 = (1/16) J (b) Rotational KE = (1/4)MV2 = (1/32)J (c) As in SHM energy is conserved, i.e., dE/dt = 0. From eqn. (1), we have

so

3 dV 1 dx M . 2V  k . 2x 0 4 dt 2 dt

or

3 d2x 1 M 2   kx 4 dt 2

 dx dV d 2 x  as  V and  2  dt dt   dt

2k d2x 2    2 x with 2 3M dt This is the standard equation of SHM with time period T = (2/).

or

So here

T  2

3M 2k

Example 76: A cylinder of radius r and mass m rests on a curved path of radius R as shown in Fig. Show that the cylinder can oscillate about the bottom position when displaced and left to itself. Find the period of oscillation. Assume that the cylinder rolls without slipping.

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SIMPLE HARMONIC MOTION Sol. Restoring torque acting on cylinder after a small displacement , about an axis through point of constant O with the curved path,  = – mg sin  r  mgr  (as for small , sin  = ) Angular acceleration of cylinder,

d2 dt 2 r = (R – r) (R  r)   r 

R A r

d 2  (R  r) d 2  2  Consequently dt r dt 2 Moment of inertia about contact point, I

m

A

3mr 2 2

3 2 (R  r) d 2 mr 2 r dt 2 d 2 g  0 2 3 dt (R  r) 2

–mgr 

or

Hence



3 (R  r) 6(R  r) T  2 2  g g

g 3 (R  r) 2

or

Example 77. Two identical cylinders C1 and C2 are placed with their axes horizontal and in the same horizontal plane. They are rotated uniformly about their once axes in clockwise and anticlockwise direction respectively. A uniform plank AB of mass M is kept resting of these rotating cylinders.

2L A

B C1

mg

C2

If the plank is displaced slightly from its equilibrium position the demostrate that it performs simple harmonic oscillations. Calculate the time period of these oscillations. The axes of the cylinders are separated by a distance 2L and the co-efficient of friction between the plank and the wheels is µ. Sol. Let the plank be displaced to the right (x-axis) through a small distance x and released. The normal forces executed by the two cylinders not equal. (L+x)

A

(L–x) G´

G

B

x C1

X mg

C2

Let them be N1 and N2. The frictional forces are µN1 and µN2 to the right and left respectively. Considering the vertical forces, N1 + N2 = mg

...(1)

Also, considering the rotational equilibrium of the plank about C. N1(L + x) = N2 (L – x)

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SIMPLE HARMONIC MOTION (N1 – N2) L = – (N1 + N2)x

...(2)

Hence resultant force acting on the plank along x-axis : Fx = µN1 – µN2 = µ(N1 – N2) Fx = –µMgx/L

[using (1) and (2)]

This is the restoring force, acting towards the centre. Comparing with standard equation of SHM Fx = M2x Here time-period of SHM is T

2 L  2 .  mg

TORISIONAL PENDULUM In a Torisional Pendulum an object is suspended from a wire with rigidity coefficient C. If such a wire is twisted by an angle , due to its elasticity it exerts a restoring torque  = C on the twisted object attached to it. A general torisional pendulum is shown in figure. Here a disc D of radius R and mass M is attached to a stiff wire whose other end is suspended from ceiling as shown. From the equilibrium position of this disc if it is twisted by an angle  as shown, the wire applies a restoring torque on it, which is given as R = –C

[–ve sign, for restoring nature]

If during restoring motion the angular acceleration of disc is , we can write  = I where I is the moment of inertia of disc about its central axis, thus we have I = –C

or

=–

C  I

...(1)

M D

R

Equation-(1) resembles with the basic differential equation of SHM in angular form thus we can state the angular frequency of this SHM is   C/I

...(2)

2 I  2  C In the above cases of some pendulums we’ve discussed, how to find the angular frequency and time period of a body in SHM. Now we take some similar examples of physical situations in which an object is in SHM. Thus the period of SHM is T 

C52: A circular disc of mass 1.5 kg and radius 15 cm is suspended with the help of a wire, whose one end is fixed to the centre of the disc. When a torque of 15 × 10–2 Nm/rad is applied, then the disc begins to oscillate. Calculate the time period for small displacement. Sol. M.O.I. of the dis, I  and

1 Mr 2 2

T  2

I Mr 2 (1.5)(0.15) 2  2  2  2.107 sec. C 2C 2  15  102

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SIMPLE HARMONIC MOTION Example 78: Figure shows a torisional pendulum consists of a uniform disc D of mass M and radius R attached to a this rod of torisional constant C. Find the amplitude and the energy of small torisional oscillations of the disc, if initially the disc was imparted angular speed 0. Sol. We know for a torisional pendulum angular frequency of small oscillations is given as   C/I Where I = 1/2MR2, thus



D R

2C MR 2

At it is given that from mean position the disc is imparted as\an angular speed 0, if the angular amplitude of oscillations of disc is , we have 0 = 

2C MR 2 Thus angular amplitude is given as or



MR 2  0 2C For angular SHM the total oscillation energy is given as 1 2 2 I  [If Umin = 0] 2 Thus here oscillation energy of the disc is given as ET 

2 11   2C   MR 2  ET   MR 2      MR 2   2C 0  2 2

or

ET 

1 MR 2 02 4

SUPERPOSITION OF PERPENDICULAR SHM’s : Let us consider two SHM’s acting on the same particle along perpendicular directions. The resulting motion of the particle will be obtained by eliminating t from the expressions for x and y. y Let

x = A1 sin t

...(1)

y = A2 sin(t + )

...(2)

P(x, y)

From (2) x

y/A2= sin t cos  + cos t sin  x2 y x 1  = cos  + sin  A12 A 2 A1

[using (1)]

2

 y   x x2  2  cos   1   A   A 2  sin  A1 2 1

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SIMPLE HARMONIC MOTION y2 x 2 2xy x2 2 2  2 cos   cos   sin   2 sin 2  2 A 2 A1 A1A 2 A1 y2 x 2 2xy  2 cos   sin 2  2 A 2 A1 A1A 2 This is an equation to an ellipse. Special cases : Case (i)  = 0. In this case we have y2 x 2 2xy  2 0 2 A 2 A1 A1A 2 2

 y x  A  A   0 2 1



y

A2 x A1

A2 –A1 A1

The path is a straight line segment (see figure) –A2

A tan   2 A1 Y

This motion is SHM with amplitude A12  A22

tan = – A2 A1

A2

Case (ii)  = . In this case, we have

y

A1 X

–A1

y2 x 2 2xy  2 0 2 A 2 A1 A1A 2

–A2

A2 x A1

The path is a straight line segment inclined at and  where the amplitude of oscillation is A12  A22 .

Y A2

Case (iii)  = /2. In this case, we have x2 x 2  1 A 22 A12 This is an ellipse. Case (iv)  = /2, A1 = A2. In this case, we have

(x, y) A1

–A1

X –A2

y

x2 + y2 = A2 This is a circle.

A x

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SIMPLE HARMONIC MOTION Table : Superposition of SHMs Superposition

Superposition features

Same line

x = x1 + x2 = A sin (t + ); A=

A12  A 22  2A1A 2 cos 

A 2 sin  tan  = A  A cos  1 2

Perpendicular lines

y2 x 2 2xy  2 cos   sin 2  2 A 2 A1 A1A 2

Lissajous figure : This figure is the locus of particle in motion under action of mutually perpendicular SHM’s having different frequencies. It is a closed path known as Lissajous figure. Example 79: Find the amplitude and initial phase of a particle in SHM, whose motion equation is given as y = A sin t + B cos t Sol. Here in the given equation we can write A = R cos  ...(12) and B = R sin  ...(13) Thus the given equation transforms to y = R sin(t + ) ...(14) Equation-(14) is a general equation of SHM and here R is the amplitude of given SHM and is the initial phase of the oscillating particle at t = 0. Here R is given by squaring and additing equation-(12) & (13) R = A 2  B2 Initial phse can be given by dividing - (13) & (12) tan =B/A  = tan–1 (B/A) Note : Equations y = Aei(t + ) and y = A sin t + B cos t, we can also represent the general equation of SHM. Example 80. Two simple harmonic motions are represented by the following equations y1 = 10 sin(/4) (12t + 1) y2 = 5 (sin 3t +

3 cos 3t)

Here t is in seconds. Find out the ratio of their amplituds. What are the time periods of the two motions ? Sol. Given equations are y1 = 10 sin(/4) (12t + 1) ...(1) y2 = 5(sin 3t + 3 cos 3t) ...(2) We recast these in the form of standard equation of SHM which is y = Asin (t + ) ...(3) Equation (1) may be written as y1 = 10sin[(12t/4) + (/4)]

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77

SIMPLE HARMONIC MOTION y1 = 10sin[(3t + /4)] Comparing eqn. (4) with eqn. (3) we have Amplitude of first SHM = A1 = 10 cm s–1 and 1 = 3  Time period of first motion T1 = 2/1 = 2/3 = (2/3) s Eq. (2) may also be written as

...(4)

y2 = 5sin 3t + 5 3 cos 3t Let us put 5 = A2 cos 

...(5)

5 3 = A2 sin  Then y2 = A2cos  sin 3t + A2sin  cos3t y2 = A2 sin(3t + ) Squaring (5) and (6) and adding, we have and

...(6) ...(7)

A2  [(5)2  (5 3)2 ]  10 cm i.e. amplitude of second SHM = 10 cm and time period of second SHM = T2 2/2 = 2/3 = 2/3 s Thus, the ratio of amplitudes A1 : A2 = 1 : 1 and periodic times are T1 = T2 = (2/3) s Example 81: The resulting amplitude A´ and the phase of the vibrations  A cos 2 ________ respectively.

S = A cos(t) +

Sol. 

 A 3   A   t   + cos(t + ) + cos  t   = A´ cos (t + ) are _______ and 2 2 4 8

ax  A 

A 3A  4 4

y A a2 = 2

A A 4A  A 3A    2 8 8 8

and

ay 



A  a 2x  a 2y 

9A 2 9A 2  16 64 2

A  a 2x  a 2y 

36A  9A 64

x a1= A

A a3 = 4 2

A a4 = 8

45A 3 5 A  8 8 ay 1 tan    ax 2  1   tan 1    2

A= and

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SIMPLE HARMONIC MOTION Example 82: A particle is subjected to two mutually perpendicular simple harmonic motions such that its x and y coordinates are given by   x = 2 sin t ; y = 2 sin  t   4 The parth of the particle will be : (A) an ellipse (B) a straight line (C) a parabola (D) a circle x 2

Sol: Hence

sin t 



cos t 



y  2sin t cos

or

y

or

y

   2 cos t sin 4 4

2 2 sin t  cos t 2 2

y 2

or

4  x2 2

x 4  x2  2 2 2

x 4  x2  2 2

2 y  x  4  x2

or

2y2 + x2 – 2 2 xy = 4 – x2

or

2y2 + 2x2 – 2 2 xy = 4

or x2 + y2 – 2 xy = 2 This is an equation of ellipse.

Hence option (A) is correct.

Example 83: The amplitude of the vibrating particle due to superposition of two SHMs,   y1 = sin  t   and y2 = sin t is : 3 (A) 1

(B) a x  1cos

Sol:

(C)

2

 3 1  3 2

a y  a1 sin

y

 3 3  1  3 2 2 2



 3  3  A  a 2x  ay2        2  2 



(D) 2

3

9 3   3 4 4

a1 = 1

/3

x a2 = 1

2

Hence option (C) is correct.

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SIMPLE HARMONIC MOTION Example 84: Two simple harmonic motions y1 = A sin t and y2 = A cos t are supre imposed on a particle of mass m. The total mechanical energy of the particle is : 1 m2A2 (B) m2A2 2 y1 = A sin t y2 = A sin (t + /2) ax = A ay = A

(A) Sol:

 a = Resultant amplitude

=

(C)

1 m2A2 4

(D) zero

y a2 = A 2 x

a a

2 y

 A2  A 2  2 A



90º

x

2 2

E = 1/2ma 

a1 = A

E = 1/2 m2a22

E  mA 2 2

or

Example 85: Vertical displacement of a plank with a body of mass m on it is varying according to law y = sin t + 3 cos t. The minimum value of  for which the mass just breaks off the plank and the moment it occurs first after t = 0 are given by : (y is positive vertically upwards) (A)

g 2 , 2 6

 g

(B)

g 2 , 2 3

or

amax = 2A g = 2A

or

g = 2 A12  A22

or

2 g = 2 1  3

or

g = 22







A = sin t + 3 cos t

or or

2 = sin t0 + 3 cos t0 2 = 2sin(t0 + )

But

tan  

or

  2  2 sin  t 0    3

or

t 0 

or

t0 

Sol:

g 2

 g

(C)

g  , 2 3

2 g

(D)

2g ,

2 3g

and if occurs at extreme at upper extreme position.

3  3 1



  60º 

or

t 0 

 3

   3 2

     2 3 6

  2  6 6 g

Hence option (A) is correct.

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SIMPLE HARMONIC MOTION

THINKING PROBLEMS 1.

Which of the following functions are (a) aperiodic (b) periodic but not simple harmonic (c) simple harmonic : (i) sin 2t (ii) 1 + cos 2t (iii) a sin t + b cos t (iv) sin t + sin 2t + cos 2t (v) sin3 t (v) log (1 + t) (vii) exp (–t) ? 2. Can every oscillatory motion be treated as simple harmonic motion in the limit of small amplitude ? 3. What would happen to the motion of an oscillating system if the sign of the force term in equation F = – kx is changed ? 4. (a) Can a motion be periodic but not oscillatory ? (b) Can a motion be oscillatory but not simple harmonic ? If your answer is yes, give an example and if not, explain why ? 5. Can a body have acceleration without having velocity ? 6. Determine whether or not the following quantitie scan be in the same direction for a simple harmonic motion : (a) displacement and velocity, (b) velocity and acceleration, (c) displacement and acceleration. 7. (a) Can we ever construct a simple pendulum strictly according to its definition ? (b) Is the motion of a simple pendulum linear SHM or angular SHM ? 8. A girl is swinging in a sitting position. How will the period of swing be affected if : (a)The girl stands up while swinging ? (b) Another girl of same mass comes and sits next to her ? 9. A hollow metal sphere is filled with water and a small hole is made at its bottom. It is hanging by a long thread and is made to oscillate. How will the period of oscillation change if water is allowed to flow through the hole till the sphere is empty ? 10. The resultant of two simple harmonic motions at right angles and of the same frequency is always a circular motion. True or false? Explain.

SOLUTION OF THINKING PROBLEMS (a) Functions (vi) log (1 + t) and (vii) exp (–t) increase (or decrease) continuously with time and can never repeat themself so are aperiodic. (b) Function (ii) (1 + cos 2t), (iv) sin t + sin 2t + cos 2t and (v) sin3 t are periodic [i.e., f(t + T) = f(t)] with periodicity (/), (2/) and (2/) respectively but not simple harmonic as for these functions (d2y/dt2) is not  – y. (c) Functions (i) sin 2t and (iii) a sin t + b cos t, i.e., (a2 + b2)1/2 sin [t + tan–1 (b/a)] are simple harmonic [with time period (/) and (2/) respectively] as for these (d2y/dt2) – y. 2. No, not at all. In the limit of small amplitudes only those oscillatory motions can be treated as simple harmonic for which the restoring force (or torque) becomes linear. For example, the oscillatory motion of a simple or spring pendulum or motion of atoms in a molecule becomes simple harmonic in the limit of small amplitude as restoring force (or torque) becomes linear while in case of oscillatory motion of a ball between two inclined planes or two perfectly elastic walls, the motion does not become simple harmonic even for vanishingy small amplitude. 3. If the sign is changed in the force equation, acceleration will not be opposite to displacement and hence the particle will not oscillate, but will accelerate in the direction of displaement. So the motion will become accelerated translatory. However, equations of motion cannot be applied to analyse the motion as acceleration (= 2y) is not constant. Mathematical analysis shows that in this situation both velocity and displacement will increase exponentially with time. 1.

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SIMPLE HARMONIC MOTION 4.

Yes; Uniform circular motion, (b) Yes; When a ball is thrown from a height on a perfectly elastic plane surface the motion is oscillatory but not simple harmonic as the restoring force F = mg = constt. and not F  – x] 5. Yes; In SHM at extreme position velocity is zero but acceleration is maximum = 2A. 6. (a) Yes; when the particle is moving from equilibrium position to extreme position, (b) Yes; when the particle is moving from extreme to equilibrium position, (c) No; as in SHM displacement is always opposite to acceleration.] 7. (a) No, (b) Angular SHM. 8. (a) decreases, (b) unchanged 9. Time period will first increase, reaches a maximum and then will decrease. 10. False. The general motion is elliptical. A circle is a particular case of ellipse when its major axis is equal to its minor axis. In general, the motion is described by

When

d 2 y 2 2xy  2 cos   sin 2  2 a b ab a = b and  = /2, x2 + y2 = a2

ASSERTION-REASON TYPE A statement of Statement-1 is given and a Corresponding statement of Statement-2 is given just below it of the statements, mark the correct answer as – (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 is NOT correct explanation of Statement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) If Statement-1 is false but Statement-2 is true. 1.

Statement-1 : The force acting on a particle moving along x–axis is F = –kx + v0t, where k is a constant. Statement-2 : To an observer moving along x-axis with constant velocity v0, it represents SHM.

2.

Statement-1 : A particle executing simple harmonic motion comes to rest at the extreme positions. Statement-2 : The resultant force on the particle is zero at these positions.

3.

Statement-1 : Soldiers are asked to break steps while crossing the bridges. Statement-2 : The frequency of marching may be equal to the natural frequency of bridge and lead to resonance which can break the bridge.

4.

Statement-1 : The function Y = cos2 t + sin t does not represent a simple harmonic motion. Statement-2 : Sum of two harmonic function may not be harmonic motion.

5.

Statement-1 : A block of mass m is attached to a spring inside a trolley of mass 2m as shown in figure, all surfaces are smooth. Spring is stretched and released. Both trolley and block oscillate simple harmonically with same time period and same amplitude. Statement-2 : In absence of external force, centre of mass of system of particles does not accelerate.

2m

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m

82

SIMPLE HARMONIC MOTION

MATCH THE COLUMN 1.

2.

3.

V(m/s)

A simple harmonic oscillator consists of a block attached to a spring 2 with k = 200 N/m. The block slides on a frictionless horizontal surface, with equilibrium point x = 0. A graph of the block’s velocity v as a 0 function of time t is shown. Correctly match the required information in the column I with the values given in the column II. (use 2 = 10) –2 Column-I Column-II (A) The block’s mass in kg (P) –0.20 (B) The block’s displacement at t = 0 in metres (Q) –200 2 (C) The block’s acceleration at t = 0.10 s in m/s (R) 0.20 (D) The block’s maximum kinetic energy in Joule (S) 4.0 Match the following Column-I

0.10

0.20 t(s)

Column-II (P)

(B) y = A sin 1t + A sin(2t + ) (C) Time period of a pendulum of infinite length.

(Q) SHM for equal frequencies and amplitude (R) Superposition may not be a SHM always.

(D) Maximum value of time period of an oscillating pendulum.

(S)

Match the following Column-I (A) A constant force acting along the line of SHM affects (B) A constant torque acting along the arc of angular SHM affects. (C) A particle falling on the block executing SHM when the later crosses the mean position affects (D) A particle executing SHM when kept on a uniformly accelerated car affects.

T = 2

R (R is radius of the earth) g

(A) Linear combination of two SHM’s

amplitude will be 2 A for 1 = 2 and phase difference of /2.

(P)

Column-II The time period

(Q)

The frequency

(R)

the mean position

(S)

The amplitude

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SIMPLE HARMONIC MOTION

LEVEL – 1 1.

Two particles P and Q describe SHM of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is a 2 . The phase difference between the particle is : (A) zero (B) /2 (C) /6 (D) /3

2.

A rod of length l is in motion such that its ends A and B are moving along x-axis and y d B P  2 rad/s always. P is a fixed point on the rod. y-axis respectively. it is given that dt Let M be the projection of P on x-axis. For the time interval in which  changes from 0 to /2, choose the correct statement, M (A) The acceleration of M is always directed towards right (B) M executes SHM (C) M moves with constant speed (D) M moves with constant acceleration

3.

The coefficient of friction between block of mass m and 2m is m = 2 tan . There is no friction between block of mass 2m and inclined plane. The maximum amplitude of two block system for which there is no relative motion between both the blocks. (A) g sin 

4.

k m

(B)

mg sin  k

(C)

3mg sin  k

l A

x

k m 2m

(D) None of these

Graph shows the x(t) curves for three experiments involving a particular spring-block system oscillating in SHM. The kinetic energy of the system is maximum at t = 4 sec. for the situation : x 1 2 0

4se c.

t(in sec)

3

(A) 1

(B) 2

(C) 3

(D) Same in all y

5.

A particle of mass m = 2 kg executes SHM in xy-plane between points A and B  underaction of force F  F ˆi  F ˆj . Minimum time taken by particle to move x

A(2, 2) x

y

from A to B is 1 sec. At t = 0 the particle is at x = 2 and y = 2. Then Fx as function of time t is B(–2, –2) (A) –42 sin t (B) –42 cos t (C) 42 cos t (D) None of these 6.

The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by v2 = 108 – 9x2 (all quantities are in cgs units) : (A) the motion is uniformly accelerated along the straight line (B) the magnitude of the acceleration at a distance 3cm from the point is 27 cm/sec2 (C) the motion is simple harmonic about the given fixed point. (D) the maximum displacement from the fixed point is 4 cm.

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SIMPLE HARMONIC MOTION 7.

The time period of an ideal simple pendulum is given by : T = 2 l / g The time period of actual simple pendulum (T´) is slightly different due to small damping (friction). Then (A) T´ > T (B) T´ < T (C) T = T´ (D) None

8.

A small bob of a simple pendulum contains water. Its periodic time is T. During oscillation, the temperature of surrounding is lowered so that water gets freezed. The new time period is T´. Then : (A) T > T´ (B) T < T´ (C) T = T´ (D) None

9.

Two simple pendula A and B are shown in the figure. The phase difference between A and B is : (A)  (B) /2 (C) 0 (D) /4

l

l

A

B

10. A body executes simple harmonic motion under the action of a force F1 with a time period (4/5) second. If the force is changed to F2 it executes SHM with a time period (3/5) second. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period in second is : (A) 12/25 (B) 24/25 (C) 35/24 (D) 25/12 11. What should be the displacement of a simple pendulum whose amplitude is A, at which potential energy is 1/4th of the total energy ? (A) A / 2 (B) A/2 (C) A/4 (D) A / 2 2 12. The potential energy U of a particle is given by U = {20 + (x – 4)2} J. Total mechanical energy of the particle is 36 J. Select the correct alternative(s) (A) The particle oscillates about point x = 4 m (B) The amplitude of the particle is 4 m (C) The kinetic energy of the particle at x = 2 m is 12 J (D) The motion of the particle is periodic but not simple harmonic 13. A rod of mass M and length L is hinged at its centre of mass so that it can rotate in a vertical plane. Two springs each of stiffness k are connected at its ends, as shown in the figure. The time period of SHM is (A) 2

M 6k

(B) 2

M 3k

(C) 2

ML k

k

L, M Hinge k

(D) 

M 6k

14. A particle moves along the x-axis according to the equation x = A sin2 t (A) The particle oscillates about the origin (B) The particle oscillates about the point x = A (C) The particle oscillates with a period T = / (D) The particle oscillates with amplitude A/2 15. In the arrangement shown in figure the pulleys are smooth and massless. The spring k1 and k2 are massless. The time period of oscillation of the mass m is

m (A) 2 k  k 1 2 (C) 2

m(k1  k 2 ) 2k1k 2

2m (B) 2 k  k 1 2 (D) 2

m(k1  k 2 ) k1k 2

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k2

k1

m

85

SIMPLE HARMONIC MOTION   16. The equation of SHM of a particle oscillating along the x-axis is given by x = 3 sin  t   cm. The acceleration 6 of the particle at t = 1 s is (A) –1.5 2 cms–2 (B) 2.6 2 cms–2 (C) –2.6 2 cms–2 (D) 1.5 2 cms–2

17. For the system shown in the figure, initially the spring is compressed by a distance a from its natural length and when released, it moves to a distance b from its equilibrium position. The dicrease in amplitude for one half cycle (–a to +b) is (A)

µmg k

(B)

2µmg k

(C)

k m

µ a

µmg 2k

b

(D) none of these

18. The motion of a particle is given x = A sin t + B cos t. The motion of the particle is (A) not simple harmonic (B) simple harmonic with amplitude A + B (C) simple harmonic with amplitude (A + B)/2

(D) simple harmonic with amplitude A 2  B2

19. Two masses m1 and m2 are suspended together by a massless spring of force constant K. When the masses are in equilibrium is removed without disturbing the system. The amplitude of oscillation is (A) m2g/K (B) m1g/K (C) (m1+m2)g/K (D) (m1–m2)g/K 20. Two bodies M and N of equal masses are suspended from two seperate massless springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that on N is : (A) k1/k2

(B)

k1 / k 2

(C) k2/k1

(D)

k 2 / k1

21. A particle executing SHM while moving from one extremity is found at distance x1, x2, x3 from the centre at the ends of three successive seconds, the period is (A)  (B) /2 (C) 2/ (D) 1/ –1 where = cos (x1 + x3)/2x2 22. One end of a spring of force constant k is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the x0 body. The spring is then compressed by 2x0 and released. The time taken m to strike the wall is (A)

1 k  6 m

(B)

m k

(C)

2 m 3 k

(D)

 k 3 m

23. In the above problem the velocity with which the body strikes the other wall is (A) x 0

m k

(B)

3k x m 0

(C)

k x m 0

(D) 2x 0

m k

 24. A particle moves in the X–Y plane according to the equation : r  2 ˆi  4 ˆj sin t. The motion of the particle is: (A) parabolic (B) circular (C) straight line (D) None of these





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86

SIMPLE HARMONIC MOTION 25. The speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is given by V = (108 – 9x2)1/2. All quantities are in SI units. (A) The motion is uniformly accelerated along the straight line. (B) The magnitude of acceleration at a distance 3 m from the fixed point is 27 m/s2. (C) The magnitude of acceleration at a distance 3 m from the fixed point is 9 m/s2. (D) The maximum displacement of the particle from the fixed point is 4 m. 26. The potential energy U of a particle is given by U = (2.5 X2 + 100) Joule. If the mass of the particle is 0.2 kg, then : (A) the motion of the particle is SHM. (B) the mean position is X = 0. (C) angular frequency of the oscillation is 5 rad/s

(D) the time period of oscillation is 1.26 sec.

27. In SHM : (A) displacement and velocity may be in the same direction. (B) displacement and velocity can never be in the same direction. (C) velocity and acceleration may be in the same direction. (D) displacement and acceleration can never be in the same direction.  28. A particle moves in the X–Y plane according to the equation r  3 ˆi  ˆj cos 5 t . The motion of the particle is: (A) Along a straight line (B) along an ellipse (C) periodic (D) along a parabola





29. The potential energy of a particle of mass 2 kg, moving along the x–axis is given by U(x) = 16(x2 – 2x) J, where x is in metres. Its speed at x = 1 m is 2 ms–1 : (A) The motion of the particle is uniformly accelerated (B) The motion of the particle is oscillatory from x = 0.5 m to x = 1.5 m. (C) The motion of the particle is simple harmonic (D) The period of oscillation of the particle is /2 s. 30. If a SHM is given by y = (sin t + cos t) m, which of the following statements is/are true ? (A) The amplitude is 1 m (B) The amplitude is 2 m. (C) Particle starts its motion from y = 1 m. (D) Particle starts its motion from y = 0 m. 31. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45º, then : (A) the resultant amplitude is (1 + 2 ) a. (B) the phase of the resultant motion relative to the first is 90º. (C) the energy associated with the resulting motion is (3 + 2 2 ) times the energy associated with any single motion. (D) the resulting motion is not simple harmonic. 32. A linear harmonic oscillator of force constant 2 × 103 N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its : (A) maximum potential energy is 100 J. (B) maximum kinetic energy is 100 J. (C) maximum potential energy is 160 J. (D) minimum potential energy is zero.

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87

SIMPLE HARMONIC MOTION 33. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-sub merged in a liquid of density  at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is : 1  k  Ag  (A)   2  M 

1/ 2

1  k  Ag  (B)   2  M 

1/ 2

1  k  gL2  (C) 2   M   

1/ 2

1  k  Ag  (D) 2  Ag 

1/ 2

34. A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x) = k|x|3 where k is a positive constant. If the amplitude of oscillations is a, then its time period T is (A) proportional to 1/ a

(B) independent of a

(C) proportional to

(D) proportional to a3/2

a

[JEE, 98]

35. A particle free to move along the x-axis has potential energy given by U(x) = k[1 – exp(–x2)] for – < x < +, where k is a positive constant of appropriate dimensions. The (A) at point away from the origin, the particle is in unstable equilibrium. (B) for any finite nonzero value of x, there is a force directed away from the origin. (C) if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin. (D) for small displacements from x = 0, the motion is simple harmonic. [JEE, 99] 36. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45º, then [JEE, 99] (A) the resultant amplitude is (1  2) a (B) the phase of the resultant motion relative to the first is 90º (C) the energy associated with the resulting motion is (3  2 2) times the energy associated with any single motion. (D) the resulting motion is not simple harmonic. 37. The period of oscillation of simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination  is given by [JEE, 2000] (A) 2

L g cos 

(B) 2

L g sin 

(C) 2

L g

(D) 2

L g tan 

38. A particle executes simple harmonic motion between x = –A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then [JEE(Scr), 01] (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2 39. A particle is executing SHM according to y = a cos t. Then which of the graphs represents variations of potential energy : [JEE(Scr), 03] P.E.

(A) (I) & (III)

I

II

P.E. III

i

(B) (II) & (IV)

(C) (I) & (IV)

IV

x

(D) (II) & (III)

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88

SIMPLE HARMONIC MOTION 40. A block P of mass m is placed on a frictionless horizontal surface. Another block Q k of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. µS is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q is [JEE, 04] (A) kA (B) kA/2 (C) zero (D) µSmg

µS

Q P

41. A simple pendulum has time period T1. When the period of suspension moves vertically up according to the equation y = kt2 where k = 1 m/s2 and t is time then the time period of the pendulum is T2 then (T1/T2)2 is [JEE(Scr), 05] (A) 5/6 (B) 11/10 (C) 6/5 (D) 5/4 42. Function x = A sin2t + B cos2t + C sin t cos t represents SHM (A) for any value of A, B and C (expect C = 0) (C) if A = B; C = 0

[JEE, 06]

(B) if A = –B; C = 2B, amplitude = | B 2 | (D) if A = B; C = 2B, amplitude = | B |

44. The x–t graph of a particle undergoing simple harmonic motion is shown below. the acceleration of the particle at t = 4/3 s is (A)

2 2 3 / 32  cm/s

(B) –2/32 cm/s2

(C) 2/32 cm/s2

x(cm)

43. A student performs an experiment for determination of g = (42l/T2)l = 1 m and he commits an error of l. For the takes the time on n oscillations with the stop watch of least count T and he commits a human error of 0.1 sec. For which of the following data, the measurement of g will be most accurate ? [JEE, 06] l T n Amplitude of oscillation (A) 5 mm 0.2 sec 10 5 mm (B) 5 mm 0.2 sec 20 5 mm (C) 5 mm 0.1 sec 20 1 mm (D) 1 mm 0.1 sec 50 1 mm 1

t(s) 0

4

8

12

–1

(D) – 3 / 32 2cm/s2

[JEE, 09]

45. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle  in one direction and released. the frequency of oscillation is [JEE, 09] (A)

1 2k 2 M

(B)

1 k 2 M

(C)

1 6k 2 M

(D)

1 24k 2 M

46. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is [JEE, 09] k 1A k2A k1 k2 (A) k (B) k 2 1 M P k1A k 2A (C) k  k (D) k  k 1 2 1 2

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89

SIMPLE HARMONIC MOTION

PASSAGE

1.

2.

Passage – 1 A block of mass m is suspended from one end of light spring as shown. The origin O is considered at distance equal to natural length of spring from ceiling and vertical downward direction as +ve Y axis. When the sytem is in equilibrium a bullet of mass m/3 moving in vertical upward direction with velocity v0 strikes the block and embeds into it. As a result the block (with bullet embedded into it) moves up and starts oscillating. Based on above information, answer the following equations : Mark the correct statement(s). (A) The block-bullet system performs SHM about y = mg/k. (B) The block-bullet system performs oscillatory motion but not SHM about y = mg/k. (C) The block-bullet system performs SHM about y = 4mg/3k. (D) The block-bullet system perform oscillatory motion but not SHM about y = 4mg/3k. The amplitude of oscillation would be 2

(A) 3.

4.

5.

6.

mv 20  4mg     3k  12k

(B)

mv20  mg    12k  3k 

2

(C)

mv20  mg    6k  k 

2

(D)

l

k

O m m/3

Y v0

mv02  4mg    6k  3k 

2

The time taken by block-bullet system to move from y = mg/k (initial equilibrium position) to y = 0 (natural length of spring) is [A represents the amplitude of motion] (A)

4m 3k

 1  mg  1  4mg   cos  3kA   cos  3kA    

(B)

(C)

4m 6k

 1  4mg  1  mg   sin  3kA   sin  3kA    

(D) None of the above

3k 4m

 1  mg  1  4mg    cos  3kA   cos  3kA    

Passage – 2 A platform is executing SHM in a vertical direction, with an amplitude of 5 cm and a frequency of 10/ vibrations/sec. A block is placed on the platform at the lowest point of its path. [Take g = 10 m/s2] Answer the following questions based on above information : At what point will the block leave the platform ? (A) 2.5 cm from mean position when acceleration is acting down and velocity is in upward direction. (B) 2.5 cm from mean position when platform is moving up. (C) 2.5 cm above mean position when platform is moving down. (D) 2.5 cm below the mean position. Mark the correct statement(s). (A) Normal contact force between the platform and block is constant. (B) As platform approaches mean position from bottom, the normal contact force between the block and platform increases. (C) As platform moves up away from mean position, the normal contact force between the block and platform decreases. (D) Both (B) and (C) are correct. At what point, the block returns to the platform ? (A) 1.3 cm above equilibrium position (B) 1.3 cm below equilibrium position (C) 4.3 cm above equilibrium position (D) 4.3 cm below equilibrium position

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90

SIMPLE HARMONIC MOTION

LEVEL – 2 1.

2.

A ring of mass m can freely slide on a smooth vertical rod. The ring is symmetrically attached with two springs, as shown in the fig., each of stiffness k. Each spring makes an angle  with the horizontal. If the ring is slightly displaced vertically, determine its time period.

k

m rod

k

A

A block B of mass m = 0.5 kg is attached with upper end of a vertical spring of force constant K = 1000 Nm–1 as shown in fig. Another identical block A falls from a height h = 49.5 cm on the block B and gets stuck with it. The combined body starts to perform vertical oscillations. Calculate amplitude of these oscillations. (g = 10 ms–2)

h B

3.

One end of an ideal spring is fixed to a wall at origin O and axis of spring is parallel to x-axis. A block of mass m = 1 kg is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in Figure is x = 10 + 3 sin(10t), where t is in second and x in cm. Another block of mass M = 3 kg, moving towards the origin with velocity 3 kg 1 kg 30 cm/sec collides with the block performing SHM at t = 0 and gets x stuck to it. Calculate O (i) new amplitude of oscillations, (ii) new equation for position of the combined body and (iii) loss of energy during collision. Neglect friction. 4.

Two identical blocks A and B of mass m = 3 kg are attached with ends of an ideal spring of force constant K = 2000 Nm–1 and rest over a smooth horizontal floor. Another identical block C moving with velocity v0 = 0.6 ms–1 as shown in fig. 100 strikes the block A and gets stuck to it. Calculate for subsequent motion m m m K (i) velocity of centre of mass of the system, V0 (ii) frequency of oscillations of the system, (iii) oscillation energy of the system, and (iv) maximum compression of the spring. K2

5.

6.

In the arrangement shown in fig, pulleys are small and light and springs are ideal. K1, K2, K3 and K4 are force constants of the springs. Calculate period of small vertical oscillations of block of mass m. Fig., shows a particle of mass m = 100 gm, attached with four identical springs, each of length l = 10 cm. Initial tension in each spring is F0 = 25 newton. Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure.

In the arrangement shown in Fig., body B is a solid cylinder of radius R = 10 cm with mass M = 4 kg. It can rotate without friction about a fixed horizontal axis O A block A of mass m = 2 kg suspended by an inextensible thread is wrapped around the cylinder. A horizontal light spring of force constant K = 100 Nm–1 fixed at one end keeps the system in static equilibrium. Calculate (i) initial elongation in the spring and (ii) period of small vertical oscillations of the block. (g = 10 ms–2)

K4 m

K1

K3 B

p m

A

C

7.

D R O B A

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91

SIMPLE HARMONIC MOTION 8.

9.

A rigid rod of mass m with a ball of mass M attached to the free end is restrained to oscillate in a vertical plane as shown in the figure. Find the natural frequency of oscillation.

3/4L

M

L/4

m k

Find the period of small oscillations in a vertical plane performed by a ball of mass m fixed at the midle of a horizontally stretched string of length l. The tension of the string is assumed to be constant and equal to F >> mg.

10. A particle moves along the x-axis according to the law x = a cos t. Find the distance that the particle covers during the time interval t = 0 to t. 11. Find the amplitude, frequency and epoch of the simple harmonic motion represented by x = 3 sin t + 4 cos t. 12. A helical spring elongates 10 cm when subjected to a tension of 5 N. Find the mass which should be attached to the bottom of the spring so that when pulled down and released the mass will vibrate twice per second. Find also its maximum velocity when the amplitude of vibration is 1 cm. 13. Find the frequency of small oscillations of the arrangement illustrated in figure. The radius of the pulley is R, its moment of inertia relative to the axis of rotation is I, the mass of the body is m, and the spring stiffness is k. The mass of the thread and the spring are negligible, the thread does not slip over the pulley and there is no friction in the axis of the pulley. 14. A plank with a body of mass m on it executes simple harmonic motion of cyclic frequency  = 11 rad s–1 between the levels 1 and 2 separated by a distance 2a a m as shown in fig. Find : (a) the force that the body exerts on the plank as it moves from level 1 to 2 when a a = 4 cm; (b) the minimum value of a when the body starts falling behind the plank; (c) the amplitude of oscillation at which the body jumps up to a height h = 50 cm relative to level 1.

R m

2 O

1

15. (a) A block of mass m is tied to one end of a string which passes over a smooth fixed pulley A and under a light smooth movable pulley B. The other end of the string is attached to the lower end of a spring of spring constant k2. Find the period of small oscillations of mass m about its equilibrium position. A

k2

k2 B A

m

B k1 Figure (a)

k1 m Figure (b)

(b) A block of mass m is attached to one end of a light inextensible string passing over a smooth light pulley B and under another smooth light pulley A as shown in the figure. The other end of a string is fixed to a ceiling. A and B are held by springs of spring constants k1 and k2. Find angular frequency of small oscillation of the system.

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92

SIMPLE HARMONIC MOTION m

D

16. Find the angular frequency of motion of block m for small motion of rod BD when we neglect the inertial effects of rod BD. Spring constant are k1 and k2. Neglect friction forces.

b

k1

k2

C B

17. Find the angular frequency for small oscillation of block of mass m in the arrangement shown in the figure. Neglect the mass of the rod. k1 m k2 l/2

l/2

k3

18. A rod AB of mass M is attached as shown below to a spring of constant k. A small block of mass m is placed on the rod at its free end A. (i) if end A is moved down through a small distance d and released, determine the period of vibration. (ii) determine the largest allowable value of d if the block m is to remain at all times in contact with the rod.

k m A

19. A rod of mass m and length l is hinged at its upper end and carries a block of mass M at its lower end. Two springs having the spring constants k1 and k2 respectively are attached to the rod at the distances b and c from the hinge as shown in the figure. Find the angular frequency of vibration for small oscillation of the system. 20. A uniform broad of length L and weight W is balanced on a fixed semi-circular cylinder of radius R as shown in the figure. If the plank is tilted slightly from its equilibrium position, determine the period of its oscillations.

B l

b

b

C

k1 k2

l m M l

r

21. A plank with a body of mass m placed on it start moving straight up according to the law y = a(1 – cos t), where y is the displacement from the initial position,  = 11 rad/s. Find : (a) The time dependence of the force that the body exerts on the plank. (b) The minimum amplitude of oscillation of the plank at which the body starts falling behind the plank. 22. A particle of mass m free to move in the x–y plane is subjected to a force whose components are Fx = –kx and Fy = –ky, where k is a constant. The particle is released when t = 0 at the point (2, 3). Prove that the subsequent motion is simple harmonic along the straight line 2y – 3x = 0. 23. In the shown arrangement, both the springs are in their natural lengths. The k2 coefficient of friction between m2 and m1 is µ. There is no friction between m1 m2 k1 and the surface. If the blocks are displaced slightly, they together perform m1 simple harmonic motion. Obtaion. (a) Frequency of such oscillations. (b) The condition if the frictional force on block m2 is to act in the direction of its displacement from mean position. (c) If the condition obtained in (b) is met, what can be maximum amplitude of their oscillations ?

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SIMPLE HARMONIC MOTION

LEVEL – 3 1.

State whether true or false “Two simple harmonic motions are represented by the equations x1 = 5sin [2t + /4] and x2 = 5 2 (sin 2t + cos2t) their amplitudes are in the ratio 1 : 2” [REE,96]

2.

A block is kept on a horizontal table. the table is undergoing simple harmonic motion of frequency 3 Hz in a horizontal plane. The coefficient of static friction between block and the table surface is 0.72. Find the maximum amplitude of the table at which the block does not slip on the surface. [REE,96]

3.

A bob of mass M is attached to the lower end of a vertical string of length L and cross sectional area A. The Young’s modulus of the material of the string is Y. If the bob executes SHM in the vertical direction, find the frequency of these oscillations. [REE, 2000]

4.

A diatomic molecule has atoms of masses m1 and m2. The potential energy of the molecule for the interatomic separation r is given by V(r) = –A + B(r – r0)2, where r0 is the equilibrium separation, and A and B are positive constants. The atoms are compressed towards each other from their equilibrium positions and released. What is the vibrational frequency of the molecule ? [REE, 01]

5.

Two masses m1 and m2 connected by a light spring of natural length l0 is compressed completely and tied by a string. This system while moving with a velocity v0 along +ve x–axis pass through the origin at t = 0. At this position the string snaps. Position of mass m1 at time t is given by the equation. x1(t) = v0t – A(1 – cos t) Calculate : [JEE, 03] (a) Position of the particle m2 as a function of time. (b) l0 in terms of A.

6.

A small body attached to one end of a vertically hanging spring is performing SHM about it’s mean position with angular frequency  and amplitude a. If at a height y* from the mean position the body gets detached from the spring, calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it’s subsequent motion after detachment. (a2 > g). [JEE, 05]

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y0 m

94

SIMPLE HARMONIC MOTION

Answer Key ASSERTION-REASON TYPE Q. Ans.

1 A

2 C

3 A

4 B

5 D

MATCH THE COLUMN 1. 2. 3.

[(A—R), (B—P), (C—Q), (D—S)] [(A—QR), (B—QRS), (C—P), (D—P)] [(A—RS), (B—RS), (C—PQS), (D—RS)]

LEVEL – 1 Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans.

1 B 11 C 21 C 31 AC 41 C

2 B 12 A BC 22 C 32 BC 42 ABD

3 C 13 A 23 B 33 B 43 D

4 A 14 CD 24 C 34 A 44 D

5 B 15 D 25 B 35 D 45 C

6 BC 16 D 26 ABC 36 AC 46 D

7 A 17 B 27 A CD 37 A

8 B 18 D 28 AC 38 A

9 A 19 B 29 BCD 39 A

10 A 20 D 30 B 40 B

PASSAGE Q. Ans.

1 C

2 B

3 A

4 A

5 C

6 D

LEVEL – 2 1. T  2

m 2k

2. 5 cm

4. (i) 0.2 m/sec.; (ii)

6. 0.02  sec.

3. (i) 3 cm; (ii) x = 10 + 3sin(5t + )cm or x = 10 – 3sin(5t) cm; (iii) 0.135 Joule

5 10 Hz ; (iii) 0.09 Joule; (iv) 3 10 mm. 

7. (i) 20 cm ; (ii) 0.4  sec.

8. f 

1 1 1 1 5. T  4 m  k  k  k  k   1 2 3 4

1 3k 2 27M  7m

9. T  

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ml F

95

SIMPLE HARMONIC MOTION 10. S = a[n + 1 – cos (t –

n n )] when n is even; S = [n + cos (t – )] when n is odd. 2 2

11. A = 5, f = /2 and epoch = 53º

12. m = 310 g, vmax = 12.5 cm/sec.

13. f 

1 k 2 m  1/ R 2

 a 2  g 14. (a) N  mg  1  g cos t  ; (b) a min  2 ; (c) a = 0.2 m.    15. (a) T  2

m(k1  k 2 ) k1k 2 1  ; (b) k1k 2 2 m(k1  k 2 )

4k 2 k 3  1 k1  17.    m k 2  4k 3  19.  

(k1b 2  k 2 c2 )  (M / m / 2)gl (M  m / 3)l 2

16.  

k 1k 2 C 2 m k1C 2  k 2 (a  b) 2





gl 2 (m  M / 3) l 2 (m  M / 3) 18. (i) T  2 ; (ii) d max  kb 2 kb 2

20. T  2L

3 gr 4

21. (a) N = m(gta2 cost); (b) 8 km.

k1 m1 µ(m1  m 2 )m 2g k1  k 2 1 23. (a) 2 m  m ; (b) k  m ; (c) m k  m k 2 2 1 2 2 1 1 2

LEVEL – 3

1. True

2. 2cm

3.

1 YA 2 ML

 m1  m1 5. (a) x 2  v0 t  m A(1  cos t) ; (b) l0   m  1 A  2  2

m1m 2 1 4. f  2 2B(m  m ) 1 2 6. y 

mg g  2 a k 

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96

OPTICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

OPTICS

Review of Concepts (a) (B)

Due to reflection, none of frequency, wavelength and speed of light change. Law of reflection : (i) Incident ray, reflected ray and normal on inident point are coplanar. (ii) The angle of incidence is equal to angle of reflection.

Some important points : In case of plane mirror : (i) For real object, image is virtual. (ii) For virtual object, image is real. (iii) Image size = Object size. (iv) The converging point of incident beam behaves as object. (v) If incident beam on optical instrument (mirror, lens etc) is converging in nature, object is virtual. (vi) If incident beam on the optical instrument is diverging in nature, the object is real. (vii) The converging point of reflected or refracted beam from an optical instrument behaves a image. (viii) If reflected beam or refracted beam from an optical instrument is converging in nature, image is real.

n P

P

Real Image

n

Virtual Object

n

(ix) If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual.

  P’

P n Real Object

 

Virtual Object

(x) For solving the problem, the reference frame is chosen in which optical instrument (mirror, lens, etc.) is in rest. (xi) The formation of image and size of image is independent of size of mirror.

(xii) Visual region and intensity of image depend on size of mirror.

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1

OPTICS (xiii) If the plane mirror is rotated through an angle  , the reflected ray and image is rotated through an angle 2  in the same sense. (xiv) If mirror is cut into a number of pieces, then the focal length does not change. (xv) The minimum height of mirror required to see the full image of a man of height h is h/2. v sin

Rest

Rest

v sin

Object



v

Image

Object

v cos Image

v cos

(xvi)

(xvii)

Rest vm

vm Object

v

Image

2 vm - v

vm

v

2 vm

(xix)

(xviii)

Object

Image

Object In rest

Image

2 vm + v

(xx) (C) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirrors are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle. 360  where  = angle between mirrors.

Here, n 

(i) If

360 is even number, the number of image is n–1. 

360 is odd number and object is placed on bisector of angle between mirrors, then  number of images is n–1.

(ii) If

360 is odd and object is not situated on bisector of angle between mirrors, then the  number of images is euqal to n.

(iii) If

(D)

n

Law of reflecteion in vector form : Let eˆ1 = unit vecotr along incident ray..

nˆ

Let eˆ 2 = unit vector along reflected ray

ˆ 1

ˆ 2

nˆ = unit vector along normal on point of incidence Then, eˆ 2  eˆ1  2 eˆ1 . nˆ nˆ

(e)

Spherical mirrors : (i) It is easy to solve the problems in geometrical optics by the help of co-ordinate sign convention.

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OPTICS y

y

x x’

x’ y’

y

x x’ y’

(ii) The mirror formula is

y

y

x x’ y’

x x’

x

y’

y’

1 1 1    u 

Also, R = 2  These formulae are only aplicable for paraxial rays. (iii) All distances are measured from optical centre. It means optical centre is taken as origin. (iv) The sign convention are only applicable in given values. (v) The transverse magnification is image size     object size u Sun 1. If object and image both are real,  is negative. 2. If object and image both are virtual,  is negative D F  3. If object is real but image is virtual;  is positive.  4. If object is virtual but image is real,  is positive. d 5. Image of star; moon or distant object is formed at focus of mirror. If y = the ddistance of sun or moon from earth. D = diameter of moon or sun’s disc.  = focal length of the mirror d = diameter of the image  = the angle subtended by sun or moon’s disc D d  Then tan  =  = y  Here,  is in radian.

Laws of Refraction 1.

(a)The incident ray, the refracted ray and normal on incidence point are coplanar.

1 1 (B) 1 sin 1   2 sin 2    cons tan t .

2 2

nˆ ˆ 1

1

(C) Snell’s law in vector form: Let,

eˆ1 = unit vector along incident ray

ˆ 2

2

eˆ 2 = unit vecotr along refracted.

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OPTICS nˆ = unit vector along normal on incidence point. Then 1 eˆ1  nˆ   2 eˆ 2  nˆ .

Some important points : (i) The value of absolute retractive index  is always greater or equal to one. (ii) The value of refractive index depends upon material of medium, colour of light and temperature of medium. (iii) When temperature increases, refractive index decreases. (iv) Optical path is defined as product of geometrical path and refractive index. i.e., optical path =  x (v) For a given time, optical path remains constant. i.e.,

1 x1   2 x 2   cons tan t



1



1 c1   2 c 2  2 c1  1 c 2



i.e.,

dx 1 dx 2  2 dt dt



1 c

(vi) The frequency of light does not depend upon medium. c1   1 , c2   2  1 c 2  2 1      2 c1 1  (a) When observer is in rarer medium and object is in denser medium: real depth  Then apparent depth (B) When object is in air and observer is in denser medium: Apparent Real depth P’ apparent position depth  real position

2.

Air Observer

Denser medium () P

Object

t  1 (C) The shift of object due to slab is x  t 1     (i) This formula is ony applicable when observer is in rarer medium. (ii) The object shiftness does not depend upon the position of object. (iii) Object shiftness takes place in the direction of incidence ray.

P

Q

P’

Object shiftness =x

(D) The equivalent rerfractive index of a combination of a number of slabs for normal incidence 

is

 ti t  i i

Here,

 t i = t 1 + t 2 + ....... 

1 2

t1 t2

ti t t  1  2   i 1  2

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OPTICS (e) The apparent depth due to a number of media is 

i

ti . i

 r

(f) The lateral shifting due to a slab is d = t sec r sin (i – r).

t d

3.

Rarer 1

(a) Cricital angle : When a ray passes from denser medium (  2 ) 90°

to rarer medium ( 1 ), then for 90° angle of refraction, the corresponding angle of incidence is critical angle. Mathematically,

Rarer medium (1)

(B) (i) When angle of incidence is lesser than critical angle, refraction takes place. the corresponding deviation is     sin 1 2 sin i   i for i = C   1  (ii) When angle of incidence is greater than critical angle, total internal reflection takes place. the corresponding deviation is

    2i

4.

Denser 2

c

 sin C  1 2

when i > C

r

i

c

i cosec 2 (f) For limiting angle of prism, i = i’ = 90°, the limiting angle of prism = 2C where C is critical angle. If angle of prism exceeds the limiting values, then the rays are totally reflected. (g)   i graph for prism: (h) For minimum deviation,

(i) i = i’ and r = r’

 A sin  m  2   (ii)   A sin 2

n

n’

i r

r’

B

i’

C

 m i

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OPTICS In the case of minimum deviation, ray is passing through prism symmetrically. (i) For maximum deviation max  . i = 90° or i’ = 90° (j) For thin prism,     1 A (k) Angular dispersion, D     r  A





(l) Angular deviation,  y   y  1 A     r  (m) dispersive power =      1  y

   

    r   (n)  y   2  

B

2

1

(o) For dispersion without deviation,   y  0 (p) For deviation without dispersion,  D  0 Refractive surface formula,

x’

A

O

x

C

 2  1  2  1    u r Here,  = image distance, u = object distance, r = radius of curvature of spherical surface (a) For plane surface, r =  (B) Transverse magnification,

m

Im age size 1   object size  2u

(C) Refractive surface formula is only applicable for paraxial ray.

LENS 1.

1 1 1    u 

Lens formula :

(a) Lens formula is only applicable for thin lens. (B) r = 2  formula is not applicable for lens. (C) m 

image size   object size u

(D) Magnification formula is only applicable when object is perpendicualr to optical axis. (e) Lens formula and the magnification formula is only applicable when medium on both sides of lenses are same.

f(+ve)

(f)

f(+ve) (i)

f(-ve) (ii)

f(-ve)

f(-ve) (iii)

(iv)

f(+ve)

(v)

(g) This lens formula is applicable for converging as well diverging lens. Thin lens maker’s formula :

(vi) 1

2

1

1   2  1   1 1        1   r1 r2 

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OPTICS 2.

(a) Thin lens formula is only applicable for paraxial ray. (B) This formula is only applicable when medium on both sides of lens are same. (C) Intensity is proportional to square of aperture. (D) When lens is placed in a medium whose refractive index is greater than that of lens. i.e.,

1   2 . Then converging lens behaves as diverging lens and vice versa. (e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other. (f) If a lens is cut along the diameter, focal length does not change.

(g) If lens is cut by a vertical, it converts into two lenses of different focal lengths.

+

1 1 1    1  2

i.e.,

f1 f

f2

1 2 3 + + + + + + 4

(h) If a lens is made of a number of layers of different refractive index (shown in figure). Then number of images of an object formed by the lens is equal to number of different media.

5 6

(i) The minimum distance between real object and image in is 4  .

f1

(j) The equivalent focal elngth of co-axial combination of two lenses is given by 1 1 1 d    F 1  2 1  2

f2 d m 2). (A)  

x m1  m 2

(B) m 1 m 2 = 1

(C)  

D2  x 2 4D

(D) all the above

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OPTICS 26. A liquid of refractive index 1.33 is placed between two identical plano-convex lenses, with refractive index 1.50. Two possible arrangement P and Q are shown. The system is (A) divergent in P, convergent in Q. (B) convergent in P, divergent in Q. (C) convergent in both (D) divergent in both.

Q

P

27. A lens of refractive index  is put in a liquid of refractive index  . If the focal length of the lens in air is

 , its focal length in liquid will be (A)

     (B)    1

    1   

   1 (C)     

(D)

  

  

28. A convergent lens is placed inside a cell filled with a liquid. The lens has a focal length +20 cm when in air and its material has a refractive index 1.50. If the liquid has a refractive index 1.60, the focal length of the system (A) –160 cm (B) – 24 cm (D) –80 cm (D) + 80 cm 29. A double convex lens, made of glass of refractive index 1.5, has focal length 6 cm. The radius of curvature of one surface is double than that of other surface. The small radius of curvature has value (A) 4.5 cm (B) 6 cm (C) 4 cm (D) 9 cm 30. If the distance between a projector and screen is increased by 1%, then illumination on the screen decreases by (A) 1 % (B) 2 % (C) 3 % (D) 4 %

31. A lens forms a sharp image of a real object on a screen. On inserting a parallel slide between the lens and the screen with its thickness along the principal axis of the lens it is found necessary to shift the screen parallel to itself ‘d’ away from the lens for getting image sharply focused on it. If the refractive index of the glass relative to air is  , the thickness of slab is (A)

d 

(B)  d

(C)

d  1

(D)   1

d 

32. A thin convex lens in used to form a real image of a bright point object. The apeture of the lens is small. A graph, shown is obtained by plotting a suitable parameter Y against another suitable parameter x. If

 = the focal length of the lens u = object distance v = image distance and Real Positive Convention is used then (A) (uV)  x; (u + V)  y (B) (u + V)  x; (uV)  y 1 u 1 (C) u  x; (D)  y  x;  y u v v

Y O

 X

-1

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OPTICS 33. Which of the following best represents object distance u vs image distance v graph for a convex lens. y y y y

(A)

(B)

(C)



(D)







34. Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are  r,  y and  v respectively and those for the flint glass are  ’r,  ’y and  ’v respectively. The ratio A’/A for which there is no net angular dispersion.





2 y  1 (A)

 

y  1

 

 y  1  y (C)   1  y y

 y   y (B) 2   y y

2  y .  y (D)    y y

35. A point object is placed at distance of 0.3 m from a convex lens of focal length 0.2 m cut into two equal halves, each of which is displaced by 0.0005 m, as shown in figure. If C1 and C2 be their optical centres then, (A) an image is formed at a distance of 0.6 m from C1 or C2 along principal axis.

C1 O

(B)

two images are formed, one at a distance of 0.6 m and other at a distance of 1.2 m from C1 or C2 along principal axis.

(C) (D)

an image is formed at a distance of 0.12 m from C1 or C2 along principal axis. two images are formed at a distance of 0.6 m from C1 or C2 along principal axis

C2

at a separation of 0.003 m.

36.

A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if (1983) A B 

(A) sin   37.

8 9

(B)

2 8 sin   3 9

(C) sin  

2 3

A ray of light from a denser medium strike a rarer medium at an angle of incidence i (see Figure). The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r’ The critical angle is r

i

r'

(A) sin

1

 tan r 

(B) sin

1

 tan i 

(C) sin

1

 tan r 

(D) tan

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1

 sin i  14

OPTICS 38.

Two coherent monochromatic light beams of intensities  and 4  are superposed. The maximum and minimum possible intensities in the resulting beam are (A) 5  and  (B) 5  and 3  (C) 9  and  (D) 9  and 3 

39.

An isosceles prism of angle 120° has a refractive index 1.44. The parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite faces

120°

(A) are parallel to each other (B) are diverging (C) make an angle 2 [sin–1 (0.72) – 30°] with each other (D) make an angle 2 sin–1 (0.72) with each other 40.

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing (A) a concave mirror of suitable focal length (B) a convex mirror of suitable focal length (C) a convex lens of focal length less than 0.25 m (D) a concave lens of suitable focal length

41.

A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (A) convergent lens of focal length 3.5 R (B) convergent lens of focal length 3.0 R (C) divergent lens of focal length 3.5 R (D) divergent lens of focal length 3.0 R

42.

A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 and L2 having refractive indices

1 and  2 respectively   2  1  1 . the lens will

diverge a parallel beam of light if it is filled with (A) air and placed in air (B) air and immersed in L1 (C) L1 and immersed in L2 (D) L2 and immersed in L1 43.

A diverging beam of light from a point source Is having divergence angle  , falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is

(A) zero

44.

(B)

1

1  n

(C) sin 



(D) 2sin

1

1   n

A ray of light passes through four transparent media with refractive indices

1 , 2 , 3 and 4 as

shown in the figure. the surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

(A)

1  2

(B)

 2  3

(C)

3  4

(D)

 4  1

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OPTICS 45.

A given ray of light suffers minimum deviation in an equilateral prism p, Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer

P

(A) greater deviation (C) same deviation as before 46.

R

(B) no deviation (D) total internal reflection

Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams. (A) R1

47.

Q

R2

(B) R

(C) R

R

(D) R

Two plane mirrors A and B are aligned parallel to each other, as shown in the figure A light ray is incident at an angle 30° at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is

(A) 28

(B) 30

(C) 32

(D) 34

48.

The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 2 0 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image. (A) 1.25 cm (B) 2.5 cm (C) 1.05 cm (D) 2 cm

49.

A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of

(A)  4 3 sin i 50.

(B)

1 sin i

 g would be

(C)

43

(D) 1

A beam of white light is incident on glass air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are (A) Violet, Indigo, Blue (B) All colors except green (C) Yellow, Orange, Red (D) White light

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16

OPTICS 51.

An equilateral prism is placed on a horizontal surface. A ray pQ is incident onto it. For minimum deviation R S

Q P

(A) PQ is horizontal (C) RS is horizontal

(B) QR is horizontal (D) Any one will be horizontal

52.

A source emits sound of frequency 600 Hz inside water. The frequency heard in air will be equal to (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s) (A) 3000 Hz (B) 120 Hz (C) 600 Hz (D) 6000 Hz

53.

A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface is (A) 6 cm (B) 4 cm (C) 12 cm (D) 9 cm

54.

A convex lens is in contact with concave lens. the magnitude of the ratio of their focal length is

2 3.

Their equivalent focal length is 30 cm. What are their individual focal lengths? (A) –15, 10 (B) –10, 15 (C) 75, 50 (D) –75, 50 55.

A container is filled with water

   1.33 upto a height

of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. Focal length of the mirror is (A) 15 cm (B) 20 cm (C) –18, 31 cm (D) 10 cm

Multiple Choice Question with ONE or MORE THAN ONE correct answer: 56.

A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination is (A) –1.5 dioptres (B) –6.5 dioptres (C) +6.5 diopres (D) +6.67 dioptres

57.

A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen (A) half the image will disappear (B) complete image will be formed (C) intensity of the image will increase (D) intensity of the image will decrease.

58.

A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to 12

u f  (A) b    f  59.

12

 f  (B) b   u f 

u f  (C) b    f 

 f  (D) b   u f 

2

A beam of light consisting of red, green and blue colours is incident on a right angled prism, figure. The refractive indices of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will

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17

OPTICS

45°

(A) separate part of the red colour from the green and blue colours (B) separate part of the blue colour from the red and green colours (C) separate all the three colours from one another. (D) not separate even partially any colour from the other two colours. 60.

A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P2 is (A) 5.33° (B) 4° (C) 3° (D) 2.6°

61.

Two thin convex lenses of focal lengths f 1 and f 2 are separated by a horizontal distance d (where

d  f1  d  f 2 ) and their centres are displaced by a vertical separation  as shown in Figure.

Taking the origin of coordinates O, at the center of the first lens the x and y coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by: (A) x 

f1 f 2 ,y f1  f 2

(B) x 

f1  f 2  d   ,y f1  f 2  d f1  f 2

(C) x 

f1 f 2  d  f1  d    f1  d  ,y f1  f 2  d f1  f 2  d

(D) x 

f1 f 2  d  f1  d  ,y0 f1  f 2  d

62.

Which of the following form(s) a virtual and erect image for all positions of the object? (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror.

63.

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence of 45°. The ray just undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible value(s) of n from the following: (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.6

64.

A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be. (A) real, and will remain at C (B) real, and located at a point between C and  (C) virtual, and located at a point between C and O (D) real, and located at a point between C and O

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18

OPTICS

Fill in the blanks: 1.

A light wave of frequency 5 x 1014 Hz enters a medium of refractive index 1.5, In the medium the velocity of the light wave is .................. and its wavelength is ................ (2 Marks)

2.

A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then d is equal to .......... cm.

3.

A monochromatic beam of light of wavelength 6000Å in vacuum enters a medium of refractive index 1.5. In the medium its wavelength is ..., its frequency is ............. (1985)

4.

In young’s double-slit experiment, the two slits act as coherent sources of equal amplitude ‘A’ and of wavelength ‘  ’. In another experiment with the same set-up the two slits are sources of equal amplitude ‘A’ and wavelength ‘  ’, but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is ................. (1986)

5.

A thin lens of refractive index 1.5 has 7a focal length of 15 cm in air. when the lens is placed in a medium of refractive index

4 , its focal length will become ........... cm. (1987) 3

6.

A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at a distance of 9 meters and 25 meters respectively from the source. The ratio of amplitudes of the waves at P and Q is .................... (1989)

7.

A slab of a material of refractive index 2 shown in Figure, has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in the figure.

n1=1.0

A

P

C n2=2.0 C

n3 =

4 3

O

15 cm B

D 20 cm

An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from the left is ............... (1991)

8.

A thin rod of length

f is placed along the optic axis of a concave mirror of focal length f such that itss 3

image which is real elongated, just touches the rod. The magnification is ............. 9.

A ray of light undergoes deviation of 30° when incident on an equilateral prism of refractive index The angle made by the ray inside the prism with the base of the prism is .............

10.

(1991)

2.

(1992)

A light of wavelength 6000Å in air, enters a medium with refractive index 1.5. Inside the medium its frequency is ........ Hz and its wavelength is ............... Å. (1997)

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OPTICS 11.

Two thin lenses, when in contact, produce a combination of power +10 diopters. When they are 0.25 m apart, the power reduces to +6 diopters. The focal length of the lenses are ...... m and .... m. (1997)

12.

A ray of light is incident normally on one of the faces of a prism of apex angle 30° and refractive index

2 . The angle of deviation of the ray is .......... degrees.

(1997)

True / False : 13.

The intensity of light at a distance ‘r’ from the axis of a long cylindrical source is inversely proportional to ‘r’. (1981)

14.

A convex lens of focal length 1 meter and a concave lens of focal length 0.25 meter are kept 0.75 meter apart. A parallel beam of light first passes through the convex lens, then through the concave lens and moves to a focus 0.5 m away from the concave lens. (1983)

15.

A beam of white light passing through a hollow prism give no spectrum.

16.

A parallel beam of white light fall on a combination of a concave and a convex lens, both of the same material. Their focal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the other side of the lens system, one sees coloured patterns with violet colour at the outer edge. (1988)

17.

Match List I and List II and select the correct answer using the codes given below the lists: The arrangement shows different lenses made of substance of refractive index 1.5 and kept in air. R1 = 30 cm, R2 = 60 cm. Match the focal lengths

(1983)

Table Match

Table I

I.

II.

III.

IV.

R1

R1

R1

R1

Table II R2

R2

R2

R2

(A) I-A, II-B, III-D, IV-C (C) I-D, II-C, III-A, IV-B

A.

–120 cm

B.

+40 cm

C.

–40 cm

D.

+120 cm

(B) I-C, II-A, III-B, IV-D (D) I-B, II-D, III-C, IV-A

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20

OPTICS 18.

Table I

Table II

I.

An object is placed at focus before A. Magnification is –  a convex mirror II. An object is placed at the centre of B. Magnification is +0.5 curvature before a concave mirror III. An object is placed at focus before C. Magnification is +1 a concave mirror. IV. An object is placed at centre of curvature D. Magnification is – 1 before a convex mirror. (A) I-B, II-D, III-A, IV-E (B) I-A, II-D, III-C, IV-B (C) I-C, II-B, III-A, IV-E (D) I-B, II-E, III-D, IV-C

19.

Match the followings:

Table I

Table II

A.

Magnification m = +1

(i)

Convex mirror

B.

Magnification m  

2 3

(ii)

Plane mirror

C.

Magnification m  

3 2

(iii)

Concave mirror

(A) A  (ii) B  (iii) C  (i) (C) A  (ii) B  (i) C  (iii)

20.

(B) A  (i) B  (ii) C  (iii) (D) A  (iii) B  (ii) C  (i)

For a concave mirror of focal length 20 cm, match the followings:

Table I Objective distance A. 10 cm B. 30 cm C. 40 cm D. 50 cm (A) A  II, B  I, C  III, D  IV (C) A  I, B  IV, C  III, D  II

Table II Nature of image (i) Magnified, inverted and real (ii) Equal size, inverted and real (iii) Smaller, inverted and real (iv) Magnified, erect and virtual (B) A  IV, B  I, C  II, D  III (D) A  IV, B  I, C  III, D  II.

PASSAGE TYPE QUESTIONS THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision, the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grownup person.

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21

OPTICS A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can ciliary muscles are most strained in this position. For an average grown-up person, minimum distance of object should be around 25 cm. A person suffering for eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle lens becomes object for eye-lens and whose image is formed on retina. The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens equal to the numerical value of the power of lens with sign. For example, power of lens required is +3 D (converging lens of focal length

100 cm), then number of lens will be 3

+3. For all the calculations required you can use the lens formula and lens maker’s formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 1.

Minimum focal length of eye-lens of a normal person is (A) 25 cm

2.

(B) 2.5 cm

25 cm 9

(C)

(D)

25 cm 11

(D)

25 cm 11

Maximum focal-length of eye lens of normal person is (A) 25 cm

(B) 2.5 cm

25 cm 9

(C)

3.

A near-sighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be (A) + 1 (B) – 1 (C) + 3 (D) – 3

4.

A far-sighted man cannot see object only upto a distance of 100 cm from his eyes. The number of the spectacles lens that will make his range of clear vision equal to an average grown up person is (A) + 1 (B) – 1 (C) + 3 (D) – 3

5.

A person who can see objects clearly from distance 10 cm to  , then we can say that the person is (A) normal sighted person (B) near-sighted person (C) far-sighted person (D) a person with exceptional eyes having no eye defect. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ' f ' on angle of incidence

' ' as shown in figure. is given by f  R

R sec  2

  Pole (P)

Principal axis

C

F f

where R is radius of curvature of mirror and  is the angle of incidence. The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays. As a result of above dependence different rays are brought to focus at different points and the image of a point object is not a point.

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OPTICS 6.

If f p and f m represent the focal length of paraxial and marginal rays respectively, then correct relationship is: (A) f p  f m

7.

(B) f p  f m

(C) f p  f m

(D) None

If angle of incidence is 60°, then focal length of this rays is: (A) R

(B)

R 2

(C) 2R

(D) 0

8.

The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60° is: (A) 180° (B) 90° (C) Can’t be determined (D) None

9.

For paraxial rays, focal length approximately is: (A) R

10.

(B)

R 2

(C) 2R

(D) None

Which of the following statements are correct regarding spherical aberration: (A) It can be completely eliminated (B) it can’t be completely eliminated but is can’t be minimised by allowing either paraxial or marginal rays to hit the mirror (C) It is reduced by taking large aperture mirrors (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Rainbow is formed during rainy season due to refraction and total internal reflection of rays falling on suspended water droplets. When rays of the sun fall on rain drops, the rain drops disperse the light and deviate the different colours by refraction and total internal reflection to the eye of the observer. A person observing the drops will see different colours of the spectrum at different angles. The rainbow which results from single total internal reflection is called primary rainbow and secondary rainbow is formed due to two total internal reflections suffered by rays falling on water drops.

Rays from sun

A V

C R

Re d

D l et vio

11.

12.

Secondary rainbow

R Re d

v io le t

B

V

Primary rainbow

Figure shows formation of rainbow due to four drops A, B, C and D. The light surffers only one total linternal reflection in drops C and D forming primary rainbow. Secondary rainbow is formed by drops A and B where light suffers two total linternal reflections. Rainbow is an arc of: (A) Circle (B) Ellipse (C) Parabola (D) Can’t be determined The visibility of the rainbow is due to: (A) All rays (B) Rays undergoing maximum deviation (C) Rays undergoing minimum deviation (D) None

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OPTICS 13.

14.

15.

In primary rainbow, the colour of outer edge is: (A) Blue (B) Violet (C) Red

(D) None

In secondary rainbow, the colour of inner edge is: (A) Red (B) Violet (C) Indigo

(D) None

The necessary condition for the observer to see rainbow is: (A) Sun, observer’s eye and the centre of the rainbow arc lie on the same line (B) Sun, observer’s eye and the centre of the rainbow arc lie on the different line (C) From any position provided sun is at the back of the observer (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The laws governing the behavior of the rays namely rectilinear propagation, laws of reflection and refraction can be summarised in one fundamental also known as Fermat’s principle. According to this principle a ray of light travels from one point to another such that the time taken is at a stationary value (maximum or minimum). If c is the velocity of light in a vacuum, the velocity in a medium of refractive index n is

c nl , hence time taken to travel a distance l is . If the light passes through a number of n c

1 1  nl and  n dl . If refractive index varies continuously. Now,, nl c c

media, the total time taken is 

is the total optical path, so that Fermat’s principle states that the path of a ray is such that the optical path in at a stationary value. This principle is obviously in agreement with the fact that the ray are straight lines in a homogenous isotropic medium. It is found that it also agrees with the classical laws of reflection and refraction. 16.

If refractive index of a slab varies as

  1  x 2 where x is measured from one end, then optical path

length of a slab of thickness 1 m is: (A) 17.

4 m 3

(B)

3 m 4

(C) 1 m

(D) None

The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is: A

B

P (A) Maximum 18.

(B) Minimum

(C) Constant

(D) None

The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is A

(A) Maximum

(B) Minimum

B

(C) Constant

(D) None

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OPTICS 19.

The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is A

B (A) Maximum 20.

(B) Minimum

(C) Constant

(D) None

The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is

A B A and B are focii of ellipse (A) Maximum

(B) Minimum

(C) Constant

(D) None

THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE One hard and stormy night you find yourself lost in the forest when you come upon a small hut. Entering it you see a crooked old woman in the corner hunched over a crystal ball. You are about to make a hasty exit when you hear the howl of wolves outside. Taking another look at the gypsy you decide to take your chances with the wolves, but the door is jammed shut. Resigned to bad situation your approach her slowly, wondering just what is the focal length of that nifty crystal ball. 21.

If the crystal ball is 20 cm in diameter with R.I. = 1.5, the gypsy lady is 1.2 m from the central of ball, where is the image of the gypsy in focus as you walk towards her? (A) 6.9 cm from the crystal ball (B) 7.9 cm from the crystal ball (C) 8.9 cm from the crystal ball (D) None

22.

The image of old lady is: (A) real, inverted an enlarged (C) erect, virtual and magnified

23.

(B) erect, virtual and small (D) real, inverted and diminished

The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no longer get an image of her. At what object distance will there be no change of the gypsy formed? (A) 10 cm (B) 5 cm (C) 15 cm (D) None

THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The table below contains some physical properties of common optical materials. The refractive index of a material is a measure of the amount by which light is bent upon entering the material. The transmittance range is the range of wavelengths over which the material is transparent.

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25

OPTICS

Material

Lithium fluoride Calcium fluoride Sodium chloride Quartz Potassium bromide Flint glass* Cesium iodide

Physical Properties of Optical Materials Refractive Transmittance Useful range index for light for prisms range (m) of 0.589 m (m) 1.39 0.12-6 2.7-5.5

Chemical resistance Poor

1.43

0.12-12

5-9.4

Good

1.54

0.3-17

8—16

Poor

1.54 1.56

0.20-3.3 0.3-29

0.20-2.7 15—28

Excellent Poor

1.66 0.35—2.2 0.35-2 1.79 0.3—7.0 15-55 *Flint glass is lead oxide doped quartz.

Excellent Poor

24.

According to the table, which material(s) will transmit light at 25  m – (A) Potassium bromide only (B) Potassium bromide and cesium iodide (C) Lithium fluoride and cesium iodide (D) Lithium fluoride and flint glass

25.

A scientist hypothesizes that any material with poor chemical resistance would have a transmittance range wider than 10  m . The properties of which of the following materials contradicts this hypothesis– (A) Lithium fluoride (B) Flint glass (C) Cesium iodide (D) Quartz

26.

When light travels from one medium to another, total internal reflection can occur if the first medium has a higher refractive index than the second. Total internal reflection could occur if light were travelling from– (A) Lithium fluoride of flint glass (B) potassium bromide to cesium iodide (C) quartz to potassium bromide (D) flint glass to calcium fluoride

27.

Based on the information in the table, how is the transmittance range related to the useful prism range– (A) The transmittance range is always narrower than the useful prism range (B) The transmittance range is narrower than or equal tot he useful prism range (C) The tranmittance range increases as the useful prism range decreases (D) The tranmittance range is wider than and includes within it the useful prism range

28.

The addition of lead oxide to pure quartz has the effect of– (A) decreasing the transmittance range and the refractive index (B) decreasing the transmittance range and increasing the refractive index (C) increasing the transmittance range and the useful prism range (D) increasing the transmittance range and decreasing the useful prism range. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE A periscope viewing system is to be used to observe the behavior of primates in a large environmentally controlled room on the upper floor of a large research facility. The periscope, like those used on

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26

OPTICS submarines, is essentially a large, folded-path, low power telescope (using prisms to fold the light path). A sketch of the preliminary design appears below. Like all Newtonian telescopes, it uses a relatively long focal length objective lens to form a real image in front of the eyepiece lens (of shorter focal length). The observer looks through the eyepiece lens to see the final image, in the same manner that one would use a magnifying glass.

The distance between the lenses is approximately equal to the sum of their focal lengths. The eyepiece, in this design, can be moved forward or back in order to focus on the primates as they move closer to or further away from the objective lens. 29.

The total tube length of the three sections is to be 4 m. The objective lens available has a focal length of 3 m. What should the focal length of the eyepiece lens be? (A) 0.75 m (B) 1 m (C) 1.33 m (D) 7 m

30.

A visitor seeing the sketch points out an important flaw that will require a design change. what is the flaw? (A) The focal length of the eyepiece lens is too short. (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. (D) The prisms cannot be used in this way.

31.

A visitor seeing the sketch points out an important flaw that will require a design change. What is the flaw? (A) The focal length of the eyepiece lens is too short. (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. (D) The prisms cannot be used in this way.

32.

What will be the approximate magnification of this periscope? (A) 0.67x (B) 1x (C) 3x (D) 300x

33.

The prisms (45–45–90° prisms) turn the light path through 90° by “total internal reflection” from the inside hypotenuse faces of the prisms when the incident angle is 45° as in the sketch. Can one use crown glass with an index of refraction of 1.52 for the prism? (A) yes, because the critical angle for crown glass is 47° (B) yes, because the critical angle for crown glass is 41°. (C) No, because the critical angle for crown glass is exactly 47° (D) No, because the critical angle for crown glass is exactly 41°.

34.

Describe the properties of the image that one sees with this preliminary design (A) real, inverted, magnified (B) real, upright, magnified (C) virtual, upright, same size as object. (D) virtual, inverted, magnified

35.

The telescope is focused on a primate rather far away on the farside of the large habitat. As the primate moves rather closer to the telescope, what must the observer do to see the primate clearly? (A) No change, the image remains clear. (B) Move the eyepiece away from the objective. (C) Move the eyepiece closer to the objective. (D) Use an inverting eyepiece because the image flips.

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27

OPTICS THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the normal human eye, light from an object is refracted by the cornea-lens system at the front of the eye and produces a real image on the retina at the rear of the eye. For a given eye, its lens-to-retina distance is fixed at about 2.5 cm. Most of the focusing of an image is done by the cornea, which has a fixed curvature that is convex with respect to incoming light. The importance of the lens is that its radius of curvature ccan be changed, allowing the lens to fine-tune the focus. The lens is surrounded by the ciliary muscle. Contraction of the muscle decreases tension on the lens. This allows the natural elasticity of the lens to produce an increase in the radius of curvature. when the muscle relaxes, the lens flattens out, decreasing tis radius of curvature. Unfortunately, the lens losses elasticity with age and the ability to alter curvature decreases. The range over which clear vision is possible is bounded by the far point and the near point. In normal vision the far point is infinity and the near point depends on the radius of curvature of the lens. For normal eyes the average near point for reading is 25 cm.

AGE, years 10 20 30 40 50 60

NEAR POINT, cm 7 10 14 22 40 200

In the myopic (nearsighted) eye, the lens-to-retina length is too long and/or the radius of curvature of the cornea is too great. This causes rays from an object at infinity to focus at a point in front of the retina. The far point is closer than normal .A corrective lens will put a virtual image of a distant object at the position of the actual far point of the eye. In the hyperopic (farsighted) eye, the lens-to-retina length is too short and/or the radius of curvature of the cornea is not great enough. This causes rays from an object at infinity to focus at a point behind the retina. The near point is farther away than normal. A corrective lens will put a virtual image of the close object at the position of the actual near point. The relation among the object (o) and image (i) distances from the eye and the focal length (f) of the lens is given by the lens-distance rule :

1 o 1 i  1 f .

When using this equation, all distances are given in centimeters. The power of corrective lenses is usually given in units called diopters. Power, in diopters, is the reciprocal of the focal length in meters : Pdiopter  1 f meter . By convection – I. Converging lenses have positive focal lengths, and diverging lenses have negative focal lengths. II. Real images have positive distances from the lens, and virtual images have negative distances from the lens. 36.

The lens system of the myopic eye is best described as – (A) producing too much convergence. (B) producing too little convergence. (C) producing too much divergence. (D) producing too little divergence.

37.

An optometrist examined John’s eyes. The farthest object he can clearly focus on with his right eye is 50 cm away. What is the power of the contact lens required to correct the vision in his right eye – (A) –0.50 diopters (B) –2.0 diopters (C) +2.0 diopters (D) +5.0 diopters

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OPTICS 38.

In a mildly hyperopic eye, the focal length of the eye’s natural lens can be corrected by – (A) contracting the muscle and increasing the radius of curvature. (B) contracting the ciliary muscle and decreasing the radius of curvature (C) relaxing the ciliary muscle and increasing the radius of curvature. (D) relaxing the ciliary muscles and decreasing the radius of curvature.

39.

Jane must wear a contact lens with a power of +3.00 diopters in one eye to be able to clearly focus on an object 2.5 cm in front of the eye. Based on the vision in this eye, which of the following is the most likely age range for Jane – (A) Less than 40 years old (B) From 40 to 49 years old (C) From 50 to 59 years old (D) 60 years or older

40.

George wears eyeglasses that sit 2.0 cm in front of his eyes. His incorrect far point is 50 cm. What is the focal lengths of his eyeglasses – (A) –50 cm (B) +50 cm (C) –48 cm (D) +48 cm

41.

In a surgical procedure called radial keratotomy, (RK), a laser is used to flatten the cornea by placing as series of hairline cuts around the perimeter of the cornea. Which statement is most accurate – (A) RK corrects myopia by decreasing the focal length of the eye. (B) RK corrects myopia by increasing the focal length of the eye. (C) RK corrects hyperopia by decreasing the focal length of the eye. (D) RK corrects hyperopia by increasing the focal length of the eye. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Student are given a variety of lenses and optics equipment, such as lens holders, lighted object sources. Optical benches, meter sticks and tapes, image screens, and several examples of commercial optical equipment, such as microscopes and telescopes. They are to work in an open-ended optics lab in order to learn the general principles of lenses and the optical devices that can be constructed using lenses.

42.

A student is given a short focal length converging lens and long focal length converging lens. One lens is placed in a holder. A lighted object is placed 18 cm in front of the lens and it is found that a clear image can be focused on a screen placed 36 cm behind the lens. what is the focal length of this lens? (A) 8 cm (B) 12 cm (C) 27 cm (D) 46 cm

43.

What magnification is produced by the above lens when the object is 18 cm in front of the lens and the image is 36 cm behind the lens? (A) 2x (B) 3x (C) 4x (D) 6x

44.

A lighted object is placed 6 cm in front of the second lens, which has a focal length of +24 cm. Where is the image and which kind of image is it? (A) 8 cm in front of the lens: a virtual image. (B) 8 cm behind the lens: a real image (C) 16 cm in front of the lens: a real image (D) 16 cm behind the lens; a virtual image.

45.

The 24 cm focal length lens is used as the objective of a simple refracting telescope and a third converging lens of focal length +8 cm is used as the eyepiece. What is the magnification of this simple refractor? (A) 0.6x (B) 3x (C) 4x (D) 6x

46.

A commercial microscope is examined by the student. The objective is marked 20x and the eyepiece is marked 10x. what power objective should replace the above objective so that the microscope’s magnification will be 400x (A) 5x (B) 10x (C) 40x (D) 100x

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OPTICS 47.

A lighted object is placed 24 cm in front of a +12 cm focal lengths lens. The image formed by this lens is the object for a second lens of +24 cm focal length. The second lens is placed 72 cm behind the first lens. where is the final image with respect to the second lens? (A) 24 cm in front of # 2 (B) 24 cm behind # 2 (C) 36 cm in front of # 2 (D) 48 cm behind # 2

48.

A lens of focal length +24 cm is used to view an object placed 12 cm in front of the lens. The object is 5 cm tall. How tall is the image? (A) 2.5 cm (B) 3.3 cm (C) 7.5 cm (D) 10 cm

49.

A diverging lens of focal length –24 cm is now used with the object 12 cm in front of the lens. How tall is the image if the object is 5 cm tall? (A) 2.5 cm (B) 3.3 cm (C) 8 cm (D) 10 cm

50.

A near sighted student cannot see objects clearly unless they are as close as 80 cm (his “far-point”). The image that he sees through his new contact lens is a virtual image because he looks through the lens to see the image. what focal length lenses does he need in order to see very distance objects, such as the starts? (A) –20 cm (B) –30 cm (C) –4 cm (D) – 25 cm THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The phenomenon of refraction has long intrigued scientists and was actually used to corroborate one of the major mysteries of early science: the determination of the speed of light. The refractive index of a transparent material irrelated to a number of the physical properties of light. In terms of velocity, the refractive index represents the ratio of the velocity of light in a vacuum to its velocity in the material. From this ratio, it can be seen that light is retarded when it passes through most types of matter. It is worth noting that prisms break up white light into the seven “colors of the rainbow” because each color has a slightly different velocity in the medium. Snell’s law allows one to follow the behavior of light in terms of its path when moving from a material of one refractive index to another with the same, or different refractive index. It is given by:

n1 sin 1  n2 sin  2 , where “I” refers to the first medium through which the ray passes, “2” refers to the second medium, and the angles refer to the angle of incidence in the first medium of refraction in the second

1  and the angle

 2  .

A ship went out on a search for a sunken treasure chest. In order to locate the chest, they shone a beam of light down into the water using a high intensity white light source as shown in Figure. The refractive index for sea water is 1.33 while that for air is 1.00. 51.

From the information in the passage, how would you expect the speed of light in air to compare with the speed of light in a vacuum (which is given by “c”)? (A) It would be the same (=c) (B) It would be greater than c. (C) It would be less than c. (D) This cannot be determined from the information given.

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OPTICS 52.

Using the information in the passage, what must the approximate value of chest as shown in Figure? (A) 15.2° (B) 30.4°

53.

(C) 45.6°

2 be such that it hit the

(D) 63.4°

How does the refractive index in water light compare with that of red light given that violet light travels more slowly in water than red light? (A) nviolet  nred

(B) nviolet  nred

(C) nviolet  nred

(D) This depends on the relative speeds of the different colors in a vacuum.

54.

Total internal reflection first occurs when a beam of light travels from one medium to another medium which has a smaller refractive index at such an angle of incidence that the angle of refraction is 90°. This angle of incidence is called the critical angle. What is the value of the sine of this angle when the ray moves from water towards air? (A) 2 (B) 0.75 (C) 0.50 (D) 0

55.

What would happen to the critical angle, in the previous question, if the beam of light was travelling from water to a substance with a greater refractive index than air, but a lower refractive index than water? (A) It would increase (B) It would decrease (C) It would remain the same (D) Total internal reflection would not be possible.

56.

Which of the following would you expect to remain constant when light travels from one medium to another and the media differ in their refractive indices? (A) Velocity (B) Frequency (C) Wavelength (D) Intensity. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The invention of the compound microscope by Jansen in the late 1500’s truly revolutionized the world of science, particularly the field of cellular and molecular biology. The discovery of the cell as the fundamental unit of living organisms and the insight into the bacterial world are two of the contributions of this instrument to science. It is unseemly that such a relatively simplistic apparantus took generations to be developed. Its main component are two convex lenses: one acts as the main magnifying lens and is referred to as the objective, and another lens called the eyepiece. The two lenses act independently of each other when bending light rays. The actual lens set-up depicted in Figure.

Light from the object (O) first passes thought he objective and an enlarged, inverted first image is formed. The eyepiece then magnifies this image. Usually the magnification of the eyepiece is fixed (either x 10 or x 10) and three rotating objective lenses are used : x 10, x 40 and x 60. The most recent development in microscope technology is the electron microscope which uses a beam of electrons instead of light. Photographic film must be used otherwise no image would be formed on the retina. This microscope has a resolution about a hundred times that of the light microscope. 57.

Based on the passage, what type of image would have to be produced by the objective magnification? (A) Either virtual or real (B) Virtual (C) Real (D) It depends on the focal length of the lens.

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OPTICS 58.

Where would the first image have to be produced by the objective relative to the eyepiece such that a second, enlarged image would be generated on the same side of the eyepiece as the first image (first image distance = d1)? (A) di  Fe

(B) di  Fe

(C) 2  Fe  d i  Fe

(D) di  2  Fe

59.

Two compound microscopes A and B were compared. Both had objectives and eyepieces with the same magnification but A gave an overall magnification that was greater than that of B. Which of the following is a plausible explanation? (A) The distance between objective and eyepiece in A is greater than the corresponding distance in B. (B) The distance between objective and eyepiece in A is less than the corresponding distance is B. (C) The eyepiece and objective positions were reversed in A. (D) The eyepiece and objective positions were reversed in B.

60.

A student attempted to make a compound microscope. However, when she tried to view an object through the apparatus, no image was seen. Which of the following could explain the mishap? I. The object distance = focal length of objective. II. The object distance for eyepiece lens as her eyepiece. III. The student used a diverging lens as her eyepiece. IV. The student used a converging lens as her objective (A) I, II, III and IV (B) I, II, III (C) I, II, IV (D) II, III, IV

61.

The magnification of the eyepiece of a compound microscope is x15. The image height is 25 mm and the magnification of the objective is x40. What is the object height? (A) 1.67 mm (B) 0.60 mm (C) 0.38 mm (D) 0.04 mm

62.

What is the refractive power of an objective lens with a focal length of 0.50 cm? (A) 0.2 diopters (B) 2.0 diopters (C) 20 diopters (D) 200

THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Magnification by a lens of an object at distance 10 cm from it is –2. Now a second lens is placed exactly at the same position where first was kept, without changing the distance object and lens. The magnification by this second lens is – 3. 63.

Now both the lenses are kept in contact at the same place. What will be the new magnification. (A) 

64.

13 5

(B) 

12 7

(C) 

6 11

(D) 

5 7

What is the focal length of the combination when both lenses are in contact. (A)

60 cm 17

(B)

5 cm 17

(C)

12 cm 7

(D)

13 cm 9

THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the case of convex lens, when object is moved from f to 2f, its image is real, inverted and magnified. It moves from f to infinity on other side. 65.

Focal length of a convex lens is 10 cm. When the object is moved from 15 cm to 25 cm, the magnitude of linear magnifications. (A) will increase (B) will decrease (C) will first increase then decrease (D) will first decrease than increase.

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OPTICS 66.

Image of object AB shown in figure will be like: B 2F

A

F

A' (A)

F

A'

(B)

2F

F

2F

B' A'

(C)

F

A'

(D)

2F

B'

F

2F

B'

B'

THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Figure shows a simplified model of the eye that is based on the assumption that all of the refraction of entering light occurs at the cornea. The cornea is a converging lens located at the outer surface of the eye with fixed focal length approximately equal to 2 cm. Parallel light rays coming from a very distant object are refracted by the cornea to produce a focused image on the retina. The retina then transmits electrical impulse along the optic nerve to the brain. cornea retina

Two common defects of vision are myopia and hyperopia. Myopia, sometimes referred to as nearsightedness, occurs when the cornea focuses the image of a distance object in front of the retina. Hyperopia, sometimes referred to as farsightedness, occurs when the cornea focuses the image of a nearby object behind the retina. Both of these problems can be corrected by introducing another lens in front of the eye so that the two lens system produces a focused image on the retina. If an object is so far away from the lens system that its distance may be taken as infinite, then the following relationship holds:

1 1 1   , where f is the focal length of the cornea, f , is the focal length of the correcting c l f c fl  x v

lens, x is the distance from the correcting lens to the cornea, and v is the image distance measured from the cornea. (Note : The index of refraction is 1.0 for air and 1.5 for glass). 67.

How far away should the retina be from the cornea for normal vision? (A) 0.5 cm (B) 1.0 cm (C) 2.0 cm

(D) 4.0 cm

68.

For a distant object, the image produced by the cornea is: (A) real and inverted (B) real and upright (C) virtual and inverted (D) virtual and upright.

69.

What kind of lens would be suitable to correct myopia and hyperopia respectively? (Note : Assume that the correcting lens is at the focal point of the cornea so that x  f c .) (A) Converging, converging (C) Diverging, diverging

70.

(B) Converging, diverging (D) Diverging, converging

The focal length of a woman’s cornea is 1.8 cm, and she wears a correcting lens with a focal length of 16.5 cm at a distance x = 1.5 cm from her cornea. What is the image distance v measured from the cornea for a distant object? (A) 1.0 cm (B) 1.5 cm (C) 2.0 cm (D) 2.5 cm

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OPTICS 71.

In the case of contact lens, the cornea and the correcting lens are actually touching and act together as a single lens. If the focal length of both the cornea and the contact lens are doubled, then the image distance v for a distant object would: (A) be 1/4 the old value (B) be 1/2 the old value. (C) be the same as the old value (D) be twice the old value.

Level # 2 A

1.

In what direction should a beam of light be sent from point A (Figure) contained in a mirror box for it to fall onto point B after being reflected once from all four walls? Point A and B are in one plane perpendicular to the walls of the box (i.e., in the plane of the drawing).

2.

A concave mirror has the form of a hemisphere with a radius R = 55 cm. A thin layer of an unknown transparent liquid is poured into this mirror, and it was found that the given optical system produces, with the source in a certain position two real image, one of which (formed by direct reflection) coincide with source and the other is at a distance of  = 30 cm from it. Find the refractive index  of the liquid.

3.

A point source of light S is placed on the major optical axis of concave mirror at a distance of 60 cm. At what distance from the concave mirror should a flat mirror be placed for the rays to converge again at the point S having been reflected from the concave mirror and then from the flat one? Will the position of the point where the rays meet change if they are first reflected from the flat mirror? The radius of the concave mirror is 80 cm.

4.

A pile 4 m high driven into the bottom of a lake is 1 m above the water. Determine the length of the shadow of the pile on the bottom of the lake if the sun rays make an angle of 45° with the water surface. The refractive index of water is 4/3.

5.

In figure, a fish water watches a fish through a 3.0 cm thick glass wall of a fish tank. The watcher is in level with the fish; the index of refraction of the glass is 8/5 and that of the water is 4/3.

8.0 cm

3.0 cm

B

6.8 cm

Observer Water

Wall

(a) (B)

6.

To the fish, how far away does the watcher appear to be? To the watcher, how far away does the fish appear to be?

A hollow sphere of glass of refractive index  has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite the center. The inner cavity is concentric with external surface and the thickness of the glass is every where equal where equal to the radius of the

  1 R inner surface. Prove that the mark will appear nearer than it really is, by a distance 3   1 , where R is the radius of the inner surface.

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OPTICS 7.

8.

Y

A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to the x-axis and width parallel to the y-axis. A ray of light is travelling along y-axis at origin. the refractive 0 index  of the medium varies as   , where  0 and 1  x r  r (> 1) are constants. The refractive index of air is 1. (a) Determine the x-coordinate of the point A, where the ray intersects the upper surface of the slab-air boundary. (B) Write down the refractive index of the medium at A. (C) Indicate the subsequent path of the ray in air.

A d medium X

O

A man of height 2.0 m is standing on level road where because of temperature variation the refractive index of air is varying as   1 ay , where y is height from road. If a = 2.0 x 10–6 m –1. Then find distant point that he can see on the road.

9.

A portion of straight glass rod of diameter 4 cm and refractive index 1.5 is bent into an arc of circle of radius R. A parallel beam of light is incident on it as shown in the figure. Find the smallest value of R which permits all the light to pass around the arc.

R

Observer

10. A glass sphere has a radius of 5.0 cm and a refractive index of 1.6. A paperweight is constructed by slicing through the sphere on a plane that is 2.0 cm from the centre of the sphere and perpendicular to a radius of the sphere that passes through the center of the circle formed by the intersection of the plane and the sphere. The paperweight is placed on a table and viewed directly above by an observer who is 8.0 cm from the table top as shown in figure. when viewed through the paperweight, how far away does the tabletop appear to the observer?

8.0 cm 3.0 cm 5. 0

cm

11. A ray of light is incident on a composite slab at a angle of incidence i as shown in the figure. Find the lateral shift x of the ray when it comes out from the otherside.

A

12. A prism of apex angle A is made up of a material of refractive index  . The refractive indices of the mediums on the left and right sides are  1 and  2 respectively. A ray of light is incident from the side of medium of refractive index  1 at an angle i and comes out from the other side as shown in the figure. Find the angle of deviation.

i

1



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2

35

OPTICS 13. A hemisphere of radius a/2 and made up of a material of variable refractive index is placed with its base centre O at the origin as shown in the figure. a . ax A ray of light is incident at the point O at an angle  with the normal in the xy plane and comes out through a point P on its curved surface. Find the coordinate of the point P if   0 .

The refractive index of the material of the hemisphere varies as  

14. A ray of light is incident on the sphere of radius R and refractive index  as shown in the figure. The incident ray is parallel to a horizontal diameter and the distance between the incident ray and the horizontal diameter is b. Find the angle of deviation  suffered by the ray..

b R

15. An intense beam parallel to the principal axis is incident on a convex lens. Multiple extra images F 1, F 2, ...... are formed Principal due to feeble internal reflections, called flare spots as shown axis F1 in the figure. The radius of curvature of the lens is 30 cm and 60 cm and the refractive index is 1.5. Find the position of the first flare spot.

u

F2 F0

16. The image of the object shown in the figure is formed at the bottom of the tray filled with water. From the details given in the figure, calculate the value of h.  = 30 cm /4 O

36 cm

1m

85 cm

h

Y

17. In the given figure there are two thin lenses of same focal length  arranged with their principal axes inclined at an angle  . The separation between the optical centers of the lenses is 2  . A point object lies on the principal axis of the O convex lens at a large distance to the left of convex lens. (a) Find the coordinates of the final image formed by the system of lenses taking O as the origin of coordinate axes, and (B) Draw the ray diagram.



X

2

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OPTICS 18. (a) A prism has refracting angle equal to  /2. It is given that  is the angle of minimum deviation and  is the deviation of the ray entering at grazing incidence. Prove that sin  = sin2  and cos  =  cos  (b) A ray of light passes through a prism in a principal plane the deviation being equal to angle of 2 incidence which is equal to 2  . It is given that  is the angle of prism. Show that cos  

1





8 2  1

where  is the refractive index of the material of prism. 19. A thin flat glass plate is placed in front of a convex mirror. At what distance b from the plate should a point source of light S be placed so that its image produced by the rays reflected from the front surface of the plate coincides with the image formed by the rays reflected from the mirror? The focal length of the mirror is  = 20 cm and the distance from the plate to the mirror a = 5 cm. How can the coincidence of the images be established by direct observation?

S

a

20. A concave mirror forms the real image of a point source lying on the optical axis at a distance of 50 cm from the mirror. The focal length of the mirror is 25 cm. The mirror is cut in two and its halves are drawn a distance of 1 cm apart in a direction perpendicular to the optical axis. How will the images formed by the halves of the mirror be arranged?

b

1 cm

21. A glass hemisphere of radius 10 cm and  = 1.5 is silvered over tis curved surface. There is an air bubble in the glass 5 cms from the plane surface along the axis. Find the position of the images of this bubble seen by observer looking along the axis into the flat surface of the hemisphere. 22. The height of a candle flame is 5 cm. A lens produces an image of this flame 15 cm high on a screen. Without touching the lens, the candle is moved over a distance of  = 1.5 cm away from the lens, and a sharp image of the flame 10 cm high is obtained again after shifting the screen. Determine the main focal length of the lens. 23. A thin converging lens of focal length  is moved between a candle and a screen. The distance between the candle and the screen is d (> 4  ). Show that for two different positions of the lens, two different images can be obtained on the screen. If the ratio of dimensions of the image is  , find the value of (  + 1/  ). 24. Three convergent thin lenses of focal lengths 4a, a and 4a respectively are placed in order along the axis so that the distance between consecutive lenses is 4a. Prove that this combination simply inverts every small object on the axis without change of magnitude or position. 25. A converging bundle of light rays in the shape in the shape of a cone with the vertex angle of 40° falls on a circular diaphragm of 20 cm diameter. A lens with a focal power of 5 diopters is fixed in the diaphragm. What will the new cone angle be? 26. A ray of light is incident on the spherical surface of radius of curvature R as shown in the figure. Therefractive index on the right side of spherical surface is  . The medium ont he left side of the spherical surface is air.. The distance of the incident ray from the axis of the spherical surface is b. After refraction the ray intersects the axis at a point. F. Find the distance  of the point F from the pole O.

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OPTICS 10 cm 27. Consider an arrangement of two equibi convex lenses of focal length in air 10 cm. The refractive index of the glass of which the lenses are g between the made is  g = 3/2 and the refractive index of water filling the space two lenses is  w = 4/3. A small object O is placed on the axis O AIR at a distance of 10 cm from the first lens in air as shown in the figure. The distance of separation between the two lenses is 10 cm. Find the position and magnification of the final image.

10 cm g AIR

A

28. A thin convex lens of focal length 1m is cut into three parts A, B and C along the diameter. The thickness of the middle layer C is 1 cm. The middle layer is now removed and the two parts A nad B are put together to form a composite lens. Then the part C is also placed infront of this composite lens symmetrically as shown in the figure. A paraxial beam of light is incident along tyhe axis of the part C. Find the distance between the two images formed.

C

1 cm

B A

C

B 29. An equi biconvex lens of focal length 10 cm in AIR and made up of material of refractive index 3/2 is polished on one side. Another identical lens (not polished) is placed infront of the polished lens at a distance of 10 cm as shown in the figure. The space between the two lenses is filled with a liquid of refractive index 4/3. An object O is placed infront of the unpolished lens at a distance of 10 cm. Find the final position of the image.

10 cm

10 cm

B 60° 30. Consider an equilateral prism ABC as shown in the figure. A ray of light is incident on the face AB and gets transmitted into the prism. Then total internal reflection takes place at the face BC and the ray comes out of prism through the face AC. The total angle of deviation is 120°. Find the refractive index  of the material of the prism.

60° A

60° C

Level # 3 1.

An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1 cm). [IIT 1980]

2.

The x-y plane is the boundary between two transparent media. Medium –1 with z  0 has refractive index

2 and medium –2 with z  0. 0, has a refractive index 3 . A ray of light in medium –1 given by  the vector A  6 3 ˆi  8 3 ˆj  10 kˆ is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium – 2. [IIT 1999]

3.

An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no parallux between the images formed by two mirrors. What is the radius of curvature of the convex mirror? [IIT 1973]

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OPTICS 4.

A rectangular block of glass is placed on a printed page lying on a horizontal surface. Find the value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block. [IIT 1979]

5.

A glass lens has focal length 5 cm in air. What will be its focal length in water. (Refractive index of glass is 1.51 and that of water is 1.33). [IIT 1977]

6.

A ray of light is travelling form diamond to glass. Calculate the minimum angle of incidence of the ray as the diamond glass interface such that no light is refracted into glass. W hat will happen if the angle of incidence exceeds the angle? (refractive index of glass is 1.51 and that of diamond is 2.47)[IIT 1977]

7. 8.

What is the velocity of light in glass of refractive index 1.5? (Velocity of light in air = 3 x 1010 cm/sec.) [IIT 1976] Photographs of the ground are taken from an aircraft flying at an altitude of 2000 meters by a camera with a lens of focal length 50 cm. the size of the film in the camera is 18 cm x 18 cm. What area of the ground can be photographed by this camera at any one time? [IIT 1976]

9.

A rectangular glass block of thickens 10 cm and refractive index 1.5 is placed over a small coin. A 4 to a height of 10 cm and is placed over the glass block. 3 (a) Find the apparent position of the object when it is viewed at near normal incidence. (b) Draw a neat ray diagram. (c) If the eye is slowly moved away from the normal at a certain position the object is found to disappear due to total internal reflection. At which surface does this happen and why? [IIT 1975]

beaker filled with water of refractive index

10. A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1.0 m (see figure). The point of incidence is the origin A (0, 0). The medium has a variable index of refraction n (y) given by n(y) = [Ky3/2 + 1] ½ –3/2 where K = 1.0 (meter) The refractive index of air is 1.0. (a) Obtain a relation between the slope of the trajectory of the ray at a point B (x, y) in the medium and the incident angle at that point. (b) Obtain an equation for the trajectory y (x) of the ray in the medium. (c) Determine the coordinates (x 1, y1) of the point P, where the ray intersects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently. [IIT 1995]

11. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. [IIT 1999]

12. A light ray is incident on an irregular shaped slab of refractive index 2 at an angle of 45° with the normal on the incline face as shown in the figure. the ray finally emerges from the curved surface in the medium of the refractive index  = 1.514 and passes through point E. If the radius of curved surface is equal to 0.4 m, find the distance OE correct up to two decimal places. [IIT 2004]

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39

OPTICS 13. A point object O is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of the convex mirror? [IIT 1976] 14. An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm as shown in figure. A plane mirror is placed inclined at 45° to the lens axis 10 cm to the right of the lens (see figure). Find the position and size of the image formed by the lens and mirror combination. trace the rays forming the image. [IIT 1972]

15. An object is placed at 20 cm left of the convex lens of focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens find the magnification and the nature of the final image. Draw the ray diagram and locate the position of the final image. [IIT 1974]

16. An object is approaching at thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of 0.4 m from the lens. [IIT 2004]

3 is placed on a horizontal 2 plane mirror as shown in the figure. The space between the lens

17. A thin biconvex lens of refractive index

4 . 3 It is found that when a point object is placed 15 cm above the lens on its principle axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid. [IIT 2001]

and the mirror is then filled with water of refractive index

18. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A’B’ is the image after refraction from the lens and reflection from the mirror, find the distance of A’B’ from the pole of the mirror and obtain its magnification. Also locate position of A’ and B’ with respect to the optic axis RS. [IIT 2000]

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40

OPTICS 3 and 2 of focal length 0.3 m in air is sealed into an opening at one

19. A thin equiconvex lens of glass of refractive index  

4  end of a tank filled with water     . On the opposite side 3  of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. [IIT 1997, May]

20. A thin plano-convex lens of focal length  is split in to two halves: one of the halves is shifted along the optical axis (see figure). The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half-lenses is 2. Find the focal-length of the lens and separation between the two halves. Draw the ray diagram for image formation. [IIT 1996] 21. A plano convex lens has a thickness of 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be 25/8 cm. Find the focal length of the lens. [IIT 1984] 22. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface. (a) Where should a pin be placed on the optic axis such that its image is formed at the same place? 4 (b) If the concave part is filled with water of refractive index , 3 find the distance through which the pin should be moved so that the image of the pin again coincide with the pin. [IIT 1981]

23. Find the focal length of the lens shown in the figure. The radii of curvature of both the surfaces are equal to R. [IIT 2003]

24. The refractive indices of the crown glass for blue and red lights are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and 1.73 respectively. An isosceles prism of angle 6° is made of crown glass. A beam of white light is incident at a small angle of this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light. Determine the angle of the flint glass prism. Calculate the net dispersion of the combined system. [IIT 2001]

25. A prism of refracting angle 30° is coated with a thin film of transparent material of  is refractive index 2.2 on face AC of the prism. A light of wavelength 5500 A incident on face AB such that angle of incidence is 60°, find (a) the angle of emergence, [Given refractive index of the material of the prism is

3 ].

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41

OPTICS (b) the minimum value of thickness of the coated film on the face AC for which the light emerging from the face has maximum intensity. [IIT 2003] 26. A right angle prism (45° – 90° – 45°) of refractive index n has a plate of refractive index n1 (n1 < n) cemented to its diagonal face. The assembly is in air. A ray is incident on AB (see figure), (a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. (b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated. [IIT 1996] 27. A right angled prism is to be made by selecting a proper material and the angles A and B (B  A), as shown in figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. (a) What should be the minimum refractive index n for this to be possible? (b)

5 is it possible to achieve this with the angle B 3 equal to 30 degrees ? [IIT 1987]

For n =

28. Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 and n2 (n2>n1) at an angle of incidence  as shown in the figure. The angle  is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown in the figure), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases. (i) n3 < n1 and (ii) n3 > n1. [IIT 1986] 4 ) is refracted by a spherical air bubble of 3 radius 2 mm situated in water. Assuming the light rays to be paraxial, (a) find the position of the image due to refraction at the first surface and the position of the final image. (b) draw a ray diagram showing the positions of both the images. [IIT 1988]

29. A parallel beam of light travelling in water (refractive index =

30. Light is incident at an angle  on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of  . [IIT 1992] 31. The radius of curvature of the convex face of a plano convex lens is 12 cm and its refractive index is 1.5. (i) Find the focal length of this lens. (ii) The plane surface of the lens is now silvered. At what distance from the lens will parallel rays incident on the convex face converge. (iii) Sketch the ray diagram to locate the image, when a point object is placed on the axis, 20 cm from the lens (polished). (iv) Calculate the image distance when the object is placed as in (iiii). [IIT 1979] 32. A ray of light is incident at an angle of 60° on one face of prism which has an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism. [IIT 1978]

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42

OPTICS 33. A pin is placed 10 cm in front of a convex lens of focal length 20 cm., made of a material of refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature are 22 cm. Determine the position of the final image. Is the image real as virtual? [IIT 1978] 34. The refractive index of the material of a prism of refracting angle 45° is 1.6 for a certain monochromatic ray. What should be minimum angle of incidence of this ray on the prism so that no total internal reflection takes place as the ray comes out of the prism. [IIT 1976] 35. A prism of refractive index n1 and another prism of refractive index n2 are stuck together without a gap as shown in the figure. The angles of the prisms are as shown n1 and n2 depend on  , the wavelength of light, according to n1  1.20 

10.8  10 4 2

and

n2  1.45 

1.80  10 4 2

where,  is in nm. (a) Calculate the wavelength  0 for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light of wavelength  0, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum. [IIT 1998] 36. A projector lens has a focal length 10 cm. It throws an image of a 2 cm x 1 cm slide on a screen 5 metre from the lens. Find : (a) the size of the picture on the screen and (b) ratio of illuminations of the slide and of the picture on the screen. [IIT 1975] 37. A ray of light incident normally on one of the faces of a right angled isosceles prism is found to be totally reflected as shown in the figure. What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water, trace the path of the emergent rays for the same incident ray, indicating the values 4  of all the angles.      . 3 

[IIT 1973]

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43

OPTICS

Answer Key Assertion & Reasion

Assertion & Reasion Que.

1

2

3

4

5

6

7

8

9

10

Ans.

A

A

A

A

B

E

C

A

C

A

Que.

11

12

13

14

15

16

Ans.

B

D

D

B

B

D

Level # 1

Objective Type Que.

1

2

3

4

5

6

7

8

9

10

Ans.

B

C

A

A

C

C

B

D

D

C

Que.

11

12

13

14

15

16

17

18

19

20

Ans.

D

A

B

C

A

D

B

CD

C

C

Que.

21

22

23

24

25

26

27

28

29

30

Ans.

B

B

B

A

D

C

A

A

A

B

Que.

31

32

33

34

35

36

37

38

39

40

Ans.

C

C

C

A

D

A

A

C

D

C

Que.

41

42

43

44

45

46

47

48

49

50

Ans.

A

D

B

D

C

C

B

B

B

C

Que.

51

52

53

54

55

56

57

58

59

60

Ans.

B

C

A

A

B

A

BD

D

A

C

Que.

61

62

63

64

Ans.

C

BC

CD

D

Fill in the Blanks / True–False / Match Table 1. 2 x 108 m/s, 0.4 x 10–6 m

2. d = +15 cm

5. 60 cm

6.

9. Zero 15°

10. 5 x 1014 Hz, 4000Å

25 9

3. 4000Å, 5 x 1014 Hz

4. 2

7. 35 cm

8. 1.3

11. 0.125 m, 0.5 m

1 2 .

Que.

13

14

15

16

17

18

19

20

Ans.

T

F

T

T

D

A

C

B

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44

OPTICS

Passage Type Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D

B

B

C

D

B

D

D

B

A

A

C

C

A

A

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

A

B

A

B

C

A

D

B

B

A

D

D

B

B

B

Que.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

C

C

B

D

B

A

B

C

C

C

B

B

A

A

B

Que.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

C

D

D

B

C

A

B

C

B

A

B

C

A

A

B

Que.

61

62

63

64

65

66

67

68

69

70

71

Ans.

D

D

C

A

B

C

C

A

D

C

D

Level # 2. 2. 1.6

7.

3. 90 cm., Yes.

4. 2.88 m

2  1   2   d   2  1   x  r 0 0 (a) A  r     (c) Ray will become parallel to y-axis.

8. –2 Km.

9. R  12 cm .

1  1 12.   i  A  sin  2

10. 7.42 cm

    sin 2 i  cos A sin i sin A 1  

A

(b)

5. (a) 13.3 cm (b) 14.975 cm 0  1 2 2  d    2 2  0     0  1   r   

  cos i t 1  x  1 11.   12  sin 2  

13.

   cos i  1   sin i  t  2   2 2  i   sin i  

a a  15   , 8 8 

 b2 b2  1  b 2    2 sin    1  14. 15. The first spot is at 12 cm on left side from the optical R  R R2    centre.   2 cos   1   , 0 16. –20 cm 17.     19. b = 15 cm.  cos   1    20. At a distance of 15 cm. from the mirror 2 cm. from each other.

21. First Image at a distance of 3.33 cm from flat surface and the second at infinity. 22. 9 cm     1 2     R 1  d  2 2  23.   2   2 25. 81°40’ 26. b b    2  2  1  2   R R  27. 25 cm from the second lens on the right side magnification m = –2. 28. 0.5 cm 29. 6 cm back side of unpolished lens.

30.  

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7 3

45

OPTICS

Level # 3 3 ˆi  4 ˆj  5 kˆ

1. –7.67 cm from the mirror.

2.

5. 18.84 cm

6.  = sin–1 0.6115 = 38° No ray is refracted into glass.

7. 2 x 1010 cm per second.

8. 720 m x 720 m

9. (a) 14.16 cm below the water surface. 10. (a) Slop

dy  cot i dx

2 x  (b) y  K    4

3. 25 cm

5 2

4.

2

(b) No glass water interface. 4

(c) (4m, 1m)

(d) Ray emerges parallel to the positive x-axis.

4 12. OE = 6.06 m 13. 25 cm. 3 14. At a perpendicular distance of 20 cms from the lens axis 8 cm is size oriented parallel to lens axis.

11. m 

15. At the position of the object magnification = –1.

16. 0.09 m/s, 0.3 per second.

17. 1.6

18. Distance of A’B’ from pole of mirror a15 cm, magnification = –1.5. Distance of A’ above RS is 0.3 cm, Distance of B’ below RS is 1.5 cm. 19. 0.9 m from the lens (0.1 m behind the mirror)

20. (a)  = 0.4 m, d = 0.6 m.

21. 75 cm

22. (a) 15 cm

(b) 1.15 cm towards the lens

3 R 23.    3 1

24. 4°, 0.04°

25. (a) 0

(b) 1250Å

1 27. (a) sin B

(b) No.

1  1  2 2   n  n1  n1  26. (a) sin    2 

1 (b) sin

1.352

 73

2

29. (a) Image due to first surface at a distance of 6 mm before the first surface final image at a distance of 1mm before the first surface. 30.

2

31. (a) 24 cm

33. 17 cm infront of lens, Real. 36. (a) 100 cm x 50 cm

(b) 12 cm

(d) 80 cm.

34. sin 1 0.176   10.1

 slide  2401 (b)  picture

32.   3 . 35. (a)  0  600 nm

(b) i  sin1 ¾

37.   2 , Angle of refraction in water r = sin 1 ¾ .

—X—X—X—X—

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46

ELECTROSTATICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

ELECTROSTATICS ELECTROSTATICS ELECTRIC CHARGE Charge is the property associated with matter due to which is produces and experiences electrical and magnetic effects. The study of electrical effects of charge at rest is called electrostatics. The strength of particle’s electric interaction with objects around it depends on its electric charge, which can be either positive or negative. An object with equal amounts of two kinds of charge is electrically neutral, whereas one with an imbalance is electrically charged. In the table given below, if a body in the first column is rubbed against a body in the second column, the body in first column will acquire positive charge, while that in the second column will acquire negative charge. TABLE S.No. First Column Second Column 1. 2. 3. 4. 5.

Glass rod Flannes or cat skin Woollen cloth Woollen cloth Woollen cloth

Silk rod Ebonite rod Amber Rubber shoes Plastic objects

Electric Charge : Electric charge can be written as ne where n is a positive or negative integer and e is a constant of nature called the elementary charge (approximately 1.60 × 10–19 C). Electric charge is conserved, the (algebraic) net charge of any isolated system cannot be changed. Regarding charge following points are worth nothing : (a) Like charges repel each other and unlike charges attract each other. (b) Charge is a scalar and can be of two types; positive or negative. (c) Charge is quantized, i.e., the charge on any body will be some integral multiple of e, i.e., q = ± ne. where n = 1, 2, 3,........... 1  Charge on any body can never be  e , 1.5e etc. 3 (d) The electrostatic unit of charge is stat-coulomb and electromagnetic unit is ab-coulomb in CGS system. But in SI system the unit of charge is coulomb, 1 coulomb 

1 ab  coulomb  3  109 stat  coulomb. 10

NOTE : Recently, it has been discovered that elementary particles such as proton or  1  2 neutron are composed of quarks having charge  ±  e and  ±  e . However,, 3 3 as quarks do not exist in free state, the quantum of charge is still e. Example-1 : How many electrons are there in one coulomb of negative charge ? Sol. : The negative charge is due to presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is e = 1.6 × 10–19 C, the number of electrons q 1.0 n  e 1.6  1019 n = 6.25 × 1018

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1

ELECTROSTATICS COULOMB’S LAW The force acting between two point charges is directly proportional to the product of the charges and it is inversely proportional to the square of distance between them. Mathematically. F

q1q 2 r2

F

kq1q 2 r2

1  9.0  109 Nm 2 / C 2 4  0 where 0 = permittivity of free space 0 = 8.854 × 10–12 C2/N–m2 Regarding Coulomb’s law following points are worth noting : (a) When two charges exert forces simultaneously on a third charge, the total force acting on that charge is the vector sum of forces that the two charges would exert individually. This important property, called the principle of superposition of forces, holds for any number of charges. Thus,      Fnet  F1  F2  F3  .....  Fn (b) If some dielectric (K) is placed in the space between the charges, the net force acting on each charge is altered because charges are induced in the molecules of intervenning medium. k

q1

Fe

Thus,

or

Fe 

q2 r In vacuum

1 q1q 2 . 4 0 r 2

F'e 

1 qq . 12 2 4 0 k r

F'e 

1 q1q 2 . 4 r 2

Fe

(in vacuum) (In medium)

Here    0 K is called permittivity of the medium (c) The coulomb’s law expresses the force between two point charges at rest. In applying it to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its centre as that is true only for sphercal charged body, that is too for external point. (d) “A charged paritcle under the action of coulombian forces only can never be stable.” This statement is called Earnshow’s theorem. IMPORTANT FEATURE   1. Suppose the position vectors of two charges q1 and q2 are r1 and r 2 then electric force on charges q1 due to charge q2 is    1 qq F12  .  1 2 3 r1  r 2 4 0 r1  r 2





Similarly, electric force on q2 due to charges q1 is    1 qq F21  .  1 2 3 (r2  r1 ) 4 0 |r2  r1|

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2

ELECTROSTATICS

2.

Here q1 and q2 are to be subsitiuted with sign.   r 1 x1 i  y1 j  z1 k and r 2  x 2 i  y 2 j  z 2 k where (x1 , y1 , z1 ) and (x 2 , y 2 , z 2 ) are the co-ordinates of charges q1 and q2. Suppose two charges q1 and q2 are placed in vacuum at a distance r0 and the electric force between them is 1 q1q 2 F0  . 4 0 r02 Now, the same charges are placed in a dielecric medium of dielectric constant K at distance r ( R V

8.

Solid sphere of charge

q 4 0 R q 4 0 r

(a)

Inside 0  r  R

(b)

 r2  3   R2    Outside r  R V

R 2 6 0

V

q 4 0 r

Example-11 : In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are placed at (n – 1) corners. At the centre, the intensity is E and the potential is V. The ratio V/E has magnitude. (A) r n (B) r(n – 1) (C)(n – 1)/r (D) r(n – 1)/n q Sol. E  4  r 2 0

and



(n  1)q 4 0 r (n  1)q 4 0 r v   (n  1)r q E 4 0 r 2

v

–

– Example-12 : The figure shows a nonconducting ring which has positive and negative ++ – – + charge non uniformly distributed on it such that the total charge is zero. Which + – – of the following statements is true ? + – O – + (A) The potential at all the points on the axis will be zero. – axis + – + (B) The electric field at all the points on the axis will be zero. – + + – (C) The direction of electric field at all points on the axis will be along the axis. (D) If the ring is placed inside a uniform external electric field then net torque and force acting on the ring would be zero. –– q + V   0 + R 2 + x2 – Sol. (  q = 0) – + 4 0 R 2  x 2 – + – There for the potential at all the points on the axis will be zero. + – x O – + Hence (A) is correct. – + +

+

+

–

– –

b

Example-13 : Two concentric rings, one of radius ‘a’ and the other of radius ‘b’ have the charges +q and –(2/5)–3/2 q respectively as shown in the figure. Find the ratio b/a if a charge particle placed on the axis at z = a is in equilibrium.

a

–3/2 qB= –(2/5) q

z=a

qA=+q

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12

ELECTROSTATICS Sol. At equilibrium F=0  QE = 0  E=0 qA z 2

3 2 2



3 2 2

a  b   2a  2

3 2 2

 5    2

0

4 0 (z  b )



4 0 (a 2  a ) 3 2 2

3 2 2

2

4 0 (z  a ) qa

qBz

 2   5

3  2

qA z 

2

 a

a 2  b2



4 0 (a 2  b )

2

a

 

3 2

3 2 2

 2    5

q Bz



2

4 0 (z  a )

qa 3 2 2

3 2 2

4 0 (z  b ) 

3 2

b

3 2



2a2 + 2b2 = 10a2



b2 4 a2



a 2  b2 5  2a 2 2

3 2 2

–3/2 qB= –(2/5) q

a

z=a

qA=+q

2b2 = 8a2 b 2 a

Ans.

Example-14 : A circular ring of radius R with uniform positive charge density  per unit length is located in the yz plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point q P(R 3 , O, O) on the positive x-axis directly towards O, with an initial kinetic energy 4 . 0 (A) The particle crosses O and goes to infinity. (B) The particle returns to P (C) The particle will just reach O. (D) The particle crosses O and goes to –R 3 . Sol. According to machenical energy conser U i + Ti = Uf + Tf ...(i) q q1 l + + + R + 3R = x Ui  + R+ 4 0 (x) q + O + Where q1 is the charge on ringh + R 3 + + ++  q1 = (2R) q(2 R ) Ui  4 0 (2R) q Ui  4 0 q Ti  4 0 q q1 Uf  4 0 R q(2 R ) q Uf   4 0 R 2 0 1 Tf = mv2 2 where v is a velocity of charge putting the volues in eqn (1) 2

2

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13

ELECTROSTATICS q q q 1 2    mv 4 0 4 0 2 0 2 q q 1 2   mv  2 0 2 0 2

1 mv 2  0 2



v  0

The particle will just reach centre of ring. –q – – –– –– –

Example-15 : Find the electric field at centre of semicircular ring shown in figure. Sol.  But But 

E  E 20  E02  2 E0   E0  sin 2 0 R 4 l 2q K  , 4 0 R2 4Kq E directed negative x-axis. R2

Y ++ + q + + + R

Y E0

– – + + + +q – + – – + 45º 45º – + – 45º

–q

E

45º

O



X

E0

s+ Example-16 : A simple pendulum of length l and bob mass m is hanging in front of a large + + nonconducting sheet having surface charge density . If suddenly a charge +q is given + + to the bob & it is released from the position shown in figure. Find the maximum + + + angle through which the string is deflected from vertical. + Sol. The electric field intensity due to nonconducting sheet is  E ...(1) 2 0 According to work energy theorem Wnet = k = 0 Since initial and final velocity of bob is zero Therefore, WE + Wg = 0 ...(2) s+ WE = qE  sin  + + Wg = – mg (  –  cos) + qE + cosq qsinq Putting the values in eqn (2) + + v=0 + qE  sin – mg  (1–cos) = 0 mg + qE  sin = mg  (1–cos)  2 sin 2 qE 2 qE 1  cos       mg mg sin  2 sin cos 2 2 qE   qE    tan  tan 1   mg 2 2  mg 

 qE    2 tan 1   mg  Putting the value of E from eqn (1)  q    2 tan 1   2 0 mg 

X

 q    2 tan 1   2 0 mg 

l

Ans.

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ELECTROSTATICS Example-17 : A particle of mass m and negative charge q is thrown in a gravity free space with speed u from the point A on the large non conducting charged sheet with surface charge density , as shown in figure. Find the maximum distance from A on sheet where the particle can strike. Sol. Electric field intensity due to nonconducting sheet  E ...(1) 2 0 where ux = u sin Y + uy = u cos + ucos + + qE u + ax = + m + + 1 usin 2 A X x = uxT + a xT 2 where x = 0 1 1 qE 2 0 = u xT + a xT2 0   u sin   T  T  2 2 m 2umsin  1 qE 2 T T  u sin   T   qE 2 m y = u yT y = (u cos)T 2u msin  u 2 msin 2  u cos    y  y qE qE 2 2u m 0 sin 2 2u 2 0 m sin 2  y  y from eqn (1)  q q

y max 

+ + + + + + + +A

u

2u 2 0 m q

At ymax, sin2 = 1

+ + Example-18 : The figure shows three infinite non-conducting plates of charge perpendicular ++ + to the plane of the paper with charge per unit area +, +2 and –, Find the ratio of + + the net electric field at that point A to that at point B.

+ + A + + + + +

2.5m 2.5m

+s

Sol. The electric field intensity at point A due to plate x, y and z EA = Ex + Ey + Ez  2    2 0 2 0 2 0 EA = 0 ...(1) At point B  2  EB    2 0 2 0 2 0 4 EB  ...(2) 2 0 From eqn (1) and (2)

5m

A

+2s –s 5m

–s

+2s

+s

– – B – – – – –

B

2s s s 2Î 0 2Î 0 Î 0

x

y

z

EA =0 EB

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ELECTROSTATICS Example-18 : A metallic solid sphere is placed in a uniform electric field. 1 1 The lines of force follow the path(s) shown in figure as : 2 2 (A) 1 (B) 2 3 3 4 (C) 3 (D) 4 4 Sol. Electric field lines never enter a metallic conductor (E = 0, inside a conductor) and they fall normally on the surface of a metallic conductor (because whole surface is at same potential and lines are perpendicular to equipotential surface). Example-19 : A non-conducting solid sphere of radius R is uniformly charged. The magnitude of electric field due to the sphere at a distance r from its centre : (A) increases as r increases for r < R (B) decreases as r increases for 0 < r <  (C) decreases as r increases for R < r <  (D) is discontinuous at r = R (E) both (a) and (c) are correct r E Sol. r R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/

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ELECTROSTATICS 2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion. Sol. From machenical energy conservation U i + T i = Uf + Tf qV i + ½mv2 = qVf + 0 ...(1) 2 2 Q 1 Q(3R  r ) Vi  Vf  , R 4 0 r 2 4 0 R 3 v R +q O R where r = 2 r 2 2  R  Q  3R 2  Q 12R 2  R 2 4  1  Vf   Vf  3 32 0 R 3 2 4 0 R min



Vf 

Q(11) 32 0 R



Vf 



11Q 32 0 R

Putting the values of Vi and Vf in eqn. (1) qQ 1 11qQ  mv2  4 0 r 2 32 0 R  11qQ qQ mv 2    16 0 R 2 0 r qQ 11 R  v2    2m 0 R  8 r 

1 11qQ qQ mv2   2 32 0 R 4 0 r qQ  11 1  mv 2   2 0  8R r  2kqQ  3 R  v2  1  mR  8 r 

2kqQ  r  R 3  v2    mR  r 8

 2kqQ  r  R 3   2 v    8   mR  r

1



g

Example-22 : The diagram shows a small bead of mass m carrying charge q. The beam can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same +Q C plane a charge +Q has alos been fixed as shown. The potential at the point P due B a 4a to +Q is V. The velocity with which the bead should projected from the point P so that it can complete a circle should be greater than

6qV qV 3qV (B) (C) m m m Sol. According to conservation principal of mechanical energy. Ui + Ti = Uf + Tf 1 qQ qV  mv 20  0 ...(1) 2 4 0 a where v0 is velocity of at point ‘p’ the potential at the point p due to +Q is Q v 4 0 (4a) from eqn (1) (A)

x P

(D) none

g

x +Q C B a

v0 P

4a

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ELECTROSTATICS qv 

1 mv02  4qV 2

1 mv 20  3qV 2 v 20 

6qV m



v0 =

6qV m

Ans.

Example-23 : Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells at a distance d/2 from the center, on the line connecting the centers of the two shells, as shown in the figure. What is the net force on the charge q ? qQ qQ (A) 361 d 2 to the left (B) 361 d 2 to the right 0 0 362qQ 360qQ (C) 361 d 2 to the left (D) 361 d 2 to the right 0 0 Sol. Electric force on charge q due to sphre A is zero. Q But electric force due to sphere B on charge q is qQ d/2 A F towards left 2 F q  19  4 0  d  2  qQ  towards left 361 0d 2

Q

Q d d/2

10 d

Q B

19 d 2

Y

Example-24 : The diagram shows three infinitely long uniform line charges placed on the 3l X, Y and Z axis. The work done in moving a unit positive charge from (1, 1, 1) to (0, 1, 1) is equal to (A) ( ln 2) /20 (B) ( ln 2)/0 (C) (3 ln 2) /20 (D) None l Z Sol. Here r = 1+ x2 E1 = The magintude of electric field due to wire 3 along y-axis = directed paerpendicular to y-axis 2 0 1  x 2 E E2 = The magnitude of electric field due to wire 1+ x2 l along z-axis = directed perpendicular to z-axis. 1+ x2 The electric field due to wire along x-axis is directed perpendicular to x-axis. 3  cos  cos    The net component of electric field along x-axis is E x  2 0 1  x 2 2 0 1  x 2 where  = angle made by E1 with x-axis and  = angle made by E2 with x-axis x cos   cos   But 1  x2 4 x E  x  2 0 1  x 2



X 2l



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ELECTROSTATICS  –dV = Ex dx x 0

or   dV  v2

  dV  v1

4 x dx 2 2  1  x x 1 0



4 20



x0



x

 1 x

2

dx

x1

x 0

4 x dx  2 0 x 1 1  x 2 1 + x2 = z dz 2x = dx dz = 2 xdx

V1  V2 

Put or or

V1  V2 

2  0

x 0

 x 1 x 0

x dz  z 2x

 dz    0 x 1 z  x 0  ln z x1 0 2 0  ln(1 x)1 0  0  ln1 ln 21 0 v2  v1 

 ln 2 0

1 µC is projected towards 3 a non conducint fixed spherical shell having the same charge uniformly distributed on its surface. Find the minimum initial velocity of projection required if the particle just grazes the shall.

Example-25 : A particle of mass 1 kg & charge

2 2 2 m/s m/s (B) 2 (C) m / s 3 3 3 Sol. Apply conservation principle of angular momentum, m vd = mv0 r r   vd v v0      d   0.5 mm  2 r 2 Applying mechanical energy conservation principle. Ui + T i = Uf + T f 1 q2 1 2 0  mv   mv02 2 4 0 r 2 2 1 q 1 mv 2   mv02 or 2 4 0 r 2

(A)

V

from

1 mm

(D) none of these

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ELECTROSTATICS 2

2

or or or or or

1 q 1  v mv 2   m  2 4 0 r 2  2  1 mv 2 q2 mv 2   2 8 4 0 r 1 3 q2 2 mv   2 4 4 0 r 1 q2 mv 2  2 3 0 r 8q 2 v2  3  4 0 mr 1 8  9  109   1012 8 9   3 3  1  1  10 3 v=

8 3

=

2

2 m/s 3

ELECTRIC POTENTIAL ENERGY Two like charges repel each other while the two unlike charges attract each other. So, when the charges are moved away from each other or they are brought near each other, either some work is obtained or some work is done. This work is stored in the system of charges in form of electric potential energy. “The electric potential energy of a system of different point charges is equal to the work done in bringing those charges from infinity to form the system.” It is represented by U. The electric potential energy of a system of two point charges q1 and q2 in vacuum at a separation r is given by, 1 q1q 2 U J 4 0 r Electric Potential Energy of a System of Charges : The electric potential energy of a system of charges is given by qiq j 1 U  J 4 0 i  j rij This sum extends over all pairs of charges. We do not let i = j, because that would be an interaction of a charge with itself and we include only terms with i < j to make sure that we count each pair only once. For example, electric potential energy of four point charges q1, q2, q3 and q4 would be given by,

1  q 4q 3 q 4q 2 q 4q1 q 3q 2 q 3q1 q 2q1         4 0  r43 r42 r41 r32 r31 r21  Here, all the charges are to be substituted with sign. U

NOTE : Total number of pairs formed by n point charges are

n  n - 1 . 2 q4

Example-26 : Four charges q1 = 1µC, q2 = 2µC, q3 = –3µC and q4 = 4µC are kept on the vertices of a square of side 1 m. Find the electric potential energy of this system of charges. Sol. In this problem, q1

q3 1m 1m

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q2

20

ELECTROSTATICS r41 = r43 = r32 = r21 = 1m r42  r31  (1)2  (1)2  2 m Substituting the proper values with sign in the relation given above in the theory, we get  (4)(3) (4)(1) (4)(1) (3)(2) (3)(1) (2)(1)       U = (9.0 × 109) (10–6) (10–6)   1 1 1  2 2  1 5    9.0  103  12  2   = –7.62 × 10–2 J NOTE : Here negative sign of U implies that positive work has been done by electrostatic force in assembling these charges at respective distances from infinity.

and





+q

+q

+q

a

Example-27 : Consider the configuration of a system of four charges each of value +q. Find the work done by external agent in changing the configuration of the system from figure (i) to fig (ii).

a +q

a +q fig (i)

Sol. Ui = Electrical potential energy in square arrangement 4q2 2q2 Ui   4 0 a 4 0 2a Uf = Electric potential energy of circular arrangement of charge 4q2 2q2 Uf   4 0 2a 4 0 (2a)

+q

+q +q fig (ii)

q a

2a

q

q

a q

w ext  U  Uf  Ui  

q2 3 2 4 0 a





Example-28 : A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to take a test charge q from infinity to apex A of cone. The slant length is L. Sol. w ext  U  U f  U i  q(VA  V ) wext  q(VA  0)  qVA ...(1) Here VA = Electric potential at point A. The electric charge on the considered ring  dq   Electric potential due to consideredring is dq dV  4 0 x Q2 rdx  dV   RL4  x 0 r R  sin    x L

A L = AB B

Q2 rdx  RL

dx

x r

q

A

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ELECTROSTATICS 





Rx r L

R  2 Q  x  dx L  dV   RL4 0 x



VA   dv 

Q 2 0 L2

w ext  q VA 

L

Q

 dx = 2p Î 0

qQ 2 0 L

L 0

from eqn (1)

RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL In case of cartesian co-ordinates  E  E ˆi  E ˆj  E kˆ x

Here,

y

z

V   (partial derivative of V w.r.t. x) x V Ey     (partial derivative of V w.r.t. y) y Ex  

Ez  

V   (partial derivative of V w.r.t. z) z

  V ˆ V ˆ V ˆ  E   i j k y z   x This is also written as,   E   grad V  V



Example-29 : The electric potential in a region is represented as, V = 2x + 3y – z, Obtain expression for electric field strength.   V ˆ V ˆ V ˆ  E   i j k Sol. y z   x V    2x  3y  z  2 Here, x x V    2x  3y  z   3 y y V    2x  3y  z  1 z z  E  2 ˆi  3 ˆj  kˆ In polar co-ordinates V 1 V Er   and E    . r r  For example, electric potential due to a point charge q at distance r is 1 q V . 4 0 r  V   0  E = 0    1 q Er  . 2 and 4 0 r

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ELECTROSTATICS dV dr E = – (slope of v–r graph)

Thus, electric field E   or

 NOTE : • If electric field E is known, then electric potential can be determined from the relation as given below:   dV = -E . d r  b  Vb -Va = - E . d r or a  Here, dr  dx ˆi  dy ˆj  dz kˆ • In uniform electric field V = Ed

Example-30 : Find Vab in an electric field  N E  2 ˆi  3 ˆj  4 kˆ C   ˆ ˆ ˆ r a  i  2 j  k m and r b  2 ˆi  ˆj  2 kˆ m where













Sol. Here, the given field is uniform (constant). So using,   dV   E . dr  a  Vab  Va  Vb   E . dr or b

 

1,  2, 1

 2, 1,  2 1,  2,1

 

 2,1,  2

 2iˆ  3jˆ  4kˆ  .  dx ˆi  dy ˆj  dz kˆ   2dx  3dy  4dz 1,  2, 1

   2x  3y  4z  2,1,  2 =–1V

 Example-31 : A uniform electric field having strength E is existing in x–y plane as shown in figure. Find the p.d. between origin O & A (d, d, 0) (A) Ed(cos + sin) (B) –Ed(sin – cos) (C) 2 Ed (D) none of these  Sol. Here E  E cos  ˆi  E sin  ˆj     OA  d ˆi  d ˆj   v  E   v  Ed cos   Ed sin   Ed(cos   sin  )   (VA  V0 )  Ed(cos   sin  )  V0  VA  Ed(cos   sin  )

y

E 0) ,d, A(d

z

o

x

Example-32 : Uniform electric field of magnitude 100 V/m in space is directed along the line y = 3 + x. Find the potential difference between point A(3, 1) & B (1, 3) (A) 100 V (B) 200 2 V (C) 200 V (D) 0 Sol.  y=3+x  tan = 1

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23

ELECTROSTATICS   = 45º  E  100cos  ˆi  100sin  ˆj  100 ˆi  100 ˆj E 2 2   r  AB  ˆi  3jˆ  3iˆ  ˆj  2 ˆi  2 ˆj   V   E  r  100 ˆ 100 ˆ  V    i j  r  2 2   100 ˆ 100 ˆ V    i j  2 ˆi  2 ˆj  2 2  

V  100 2  100 2



VA - VB = 0



Example-33 : A, B, C, D, P and Q are points in a uniform electric field. The potentials a these points are V (A) = 2 volt. V(P) = V(B) = V(D) = 5 volt. V(C) = 8 volt. The electric field at P is (A) 10 Vm–1 along PQ (B) 15 2 Vm–1 along PA A –1 –1 (C) 5 Vm along PC (D) 5 Vm along PA V  VD 2  5 30 Ex  A    15 v / m Sol. 0.2 0.2 2 y VA  VB 2  5 30 E       15 v / m Also y AB 0.2 2 15 P  ˆ ˆ x’ 45º E = - 15 i - 15 j  A 15 | E |= ( - 15) 2 + ( - 15) 2 = 15 2 v / m

E PA  15 2

0.2m



C

B P A

Q D

0.2m

x

Hence (B) is correct.

Example-34 : Variation of electrostatic potential along x-direction is whosn in the graph. The correct statement about electric field is v (A) x component at point B is maximum (B) x component at point A is towards positive x-axis A B C (C) x component at point C is along negative x-axis x (D) x component at point C is along positive x-axis Sol. The negative slope v – x graph give x – component of electric field. In the given graph, slope at C is negative. Hence, x – component of electric field is positive. EQUIPOTENTIAL SURFACE Equipotential surface is an imaginary surface joining the points of same potential in an electric field. So, we can say that the potential difference +q between any two points on an equipotential surface is zero. The electric lines of force at each point of an equipotential surface are normal to the surface. Fig. shows the electric lines of force due to a point charge +q. This spherical surface will be the equipotential surface and the electrical lines of force emanating from the point charge will be radial and normal to the spherical surface.

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ELECTROSTATICS (a) (b) (c) (d) (e) (f) (g)

Regarding equipotential surface, following points are worth noting : Equipotential surface may be planar, solid etc. But equipotential surface never be a point. Equipotential surface is single valued. So, equipotential surfaces never cross each other. Electric field is always perpendicular to equipotential surface. electric lines of force cross equipotential surface perpendicularly. Work done to move a point charge q between two points on equipotential surface is zero. The surface of a conductor in equilibrium is equipotential surface. equipotential surface due to isolated point charge is spherical.

(h) Equipotential surface are planar in uniform electric field.

(i)

Equipotential surface due to line charge is cylindrical.

(j)

Equipotential surface due to an electric dipole is shown in the figure.

–q

+q



 

 



2 2 ˆ where a is Example-35 : The electric field in a region is given by : E  4axy z ˆi  2ax z ˆj  ax y / z k, a positive constant. The equation of an equipotential surface will be of the form (A) z = constant/[x3y2] (B) z = constant/[xy2] (C) z = constant/[x4y2] (D) none Sol. At equipotential surface potential is constant. Therefore - dv E= dr    dv   E . dr   E  .d r  constant    r  x ˆi  y ˆj  z kˆ dr  dxiˆ  dyjˆ  dzkˆ  ˆ = Constant E . (dx ˆi  dy ˆj  dz k)



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25

ELECTROSTATICS ˆ = constant z )iˆ  (2ax 2 z)ˆj  (ax 2 y / z)kˆ  . (dxiˆ  dyjˆ  dzk) ax 2 ydz 2  4axy z dx   2ax z dy   z = constant. 4ax 2 y z  2ax2 y z  2ax 2 y z = constant 2 2 6 ax y z = constant

 (4axy

z=

constant 6ax 2 y

constant x 4 y2

z 



Example-36 : The equation of an equipotential line an electric field is y = 2x, then the electric field vector at (1, 2) may be (A) 4 ˆi + 3 ˆj (B) 4 ˆi + 8 ˆj (C) 8 ˆi + 4 ˆj (D) –8 ˆi + 4 ˆj Sol. Electric field is perpendicular to equipotential line y= 2x  dy =2 or dx  m1 = 2  m1m2 = – 1 1 m2 =  2 1 tan     where  is angle with x-axis. 2 4 1 tan    In option (D), 8 2 ELECTRIC DIPOLE Electric dipole is a system in which two equal and opposite point +q –q P charges are placed at a small distance. The product of any of the 2l charges and distance between two charge is called electric dipole moment p. It is directed from negative charge to positive charge (fig.). The line joining the two charges is called axis of dipole. Let charges of an electric dipole are –q and +q and are separated by a small distance 2l. The dipole moment of such a dipole is given by p = q × 2l = 2ql Electric Potential and Field due to an Electric Dipole Let an electric dipole is placed along y-axis and its centre is at origin, then electric potential at point A(x, y, z) due to this dipole. 2

2

2

x + z + (y–l)

A (x, y, z)

y +q 2

l l z

2

2

x + z + (y+l) x

–q

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ELECTROSTATICS V

  q q   2  x 2  (y  l ) 2  z 2   x  (y  l )2  z 2 V V V Ex   , Ey   and E z   y x z

1 4 0

Special Cases : (i) On the axis of dipole (axial position) : x = 0, z = 0 1 p V (a) 2 4 0 (y  l 2 ) 1 p V If y = r 2 4 0 (r  l 2 ) 1 P Vaxis  or If r > > l 4 0 r 2 (b) Ex = 0 = Ez and 1 2py  E  Ey  (along ) 2 2 2 p 4 0 (y  l ) 1 2pr  If y = r 2 4 0 (r  l 2 )2 1 2p E axis  4 0 r3 (ii) On the perpendicular bisector of dipole (equatorical position) : Say along x-axis (it may be along z-axis also) y = 0, z = 0 (a) V bi sec tor  0 (b) Ex = 0, Ez = 0 1 p  (opposite to p ) 2 4 0 (x  l 2 )3/ 2  Here, negative sign implies that the electric field is along negative y-direction or antiparallel to p . Further at a distance r from the centre of dipole (x = r), then 1 p E . 2 2 3/ 2 4 0 (r  l ) 1 p E  bi sec tor  or if r >> l 4 0 r 3 (iii) In polar co-ordinates (r, ) : 1 p cos  V (a) 4 0 r 2 V  1 . 2p cos  E Er   (b) 4 0 r3 r r 1 V  1 p sin  P E   . 3 4 0 r r  O p 2 2 2 E  1  3cos    E  Er  E 4 0 r 3 tan  tan   (c) In Figure, 2

and

Ey  

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E Er A (r, )

27

ELECTROSTATICS y 3q

Example-37 : 4 charges are placed each at a distance ‘a’ from origin. The dipole moment of configuration is (A) 2qajˆ (B) 3qajˆ –2q ˆ ˆ (C) 2aq[i + j] (D) none Sol. Here

q

y

P1  2q  2 a P1 = 2 2 qa



–2q x

3q 2a P1 P2 45º –2q 45º 45º

P2  q  2a  2 qa

2a

P3  q  2a  2 qa

–2q

Px = P1 cos 45 - P2 cos 45 - P3 cos 45

Px = 2 2qa cos 45 -

2qa cos 45 -

P3

q

x

2a

2qa cos 45

Px = 0

Py = P1 sin 45 + P2 sin 45 - P3 sin 45

Py = 

2 2qa

+

2

2qa 2

-

2qa 2

= 2qa

 P  2 q ajˆ

Example-38 : Two point charges +3 µC and –3 µC are placed at a small distance 2 × 10–3 m from each other. Find: (a) electric field and potential at a distance 0.6 m from dipole in its equatorial position. (b) electric field and potential at the same distance from dipole as in (a) if the dipole is rotated through 90º. Sol. Dipole moment p = q × 2l = (3 × 10–6) (2 × 10–3) p = 6 × 10–9 C-m (a) Electric field in equatorial position, 1 p E . 3 4 0 r  6  109  E  9  109   (0.6)3  E = 250 N/C Electric potential, V = 0 (b) On rotating the dipole through 90º, the same point now will be in axial position. So, electric field 1 2p E .  500 N / C 4 0 r 3 and electric potential 1 p V 4 0 r 2 9 9 6  10 V  9  10 = 150 V (0.6)2









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28

ELECTROSTATICS Example-39 : A short electric dipole is situated at the origin of coordinate axis with its axis along x-axis and equator along y-axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distance r = 5 m from origin find the position vector of the point. Sol. Let P be such a point at distance r and angle  from equator. Now |EP| = |VP| KP K sin  1  3sin 2   P 2 or 3 r r 2 1  3sin  y  sin  or P 5 r q Squaring both nets x 1 + 3 sin2 = 5 sin2 1 sin   or 2 or = 45º  Positive vector of r point P is  5 ˆ ˆ r= i+j 2

( )

Electric Dipole in Uniform Electric Field : (i) Torque : When a dipole is placed in a uniform electric field as shown in Fig. the net force on it,   F  qE    q  E   0 while the torque, +q qE P

E

–q l qE

 = qE × 2l sin     pE sin     or pE From this expression it is clear that torque acting on a dipole is maximum (= pE) when dipole is perpendicular to the field and minimum (= 0) when dipole is parallel or antiparallel to the field. NOTE :If the electric field is not uniform the dipole will experience both a resultant force and a torque so its motion will be combined translatory and rotatory (if θ 0 or 180º)

≠

(ii) Work : Work done in rotating a dipole in a uniform field through a small angle d will be, dW = d= pE sin d So, work done in rotating a dipole from angular position 1 to 2 with respect to field, 2

W   pE sin d  pE [cos 1  cos 2 ] 1

So, if a dipole is rotated from field direction, i.e., 1 = 0 to position , i.e., 2 = . W = pE[1 – cos]

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29

ELECTROSTATICS –q

E =0 P Wmin = 0 (a)

+q

E = 180º

+q

–q

P Wmax = 2pE (b)

(iii) Potential energy : In case of a dipole (in a uniform field), potential energy of dipole is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction. i.e., U = (W – W90º) U = pE (1 – cos ) – pE (1 – cos 90º)   U   pE cos    p . E (iv) Angular SHM : A dipole when placed in a uniform electric field, align itself parallel to the field. If it is given a small angular displacement  about its equilibrium position, the restoring torque produced in it will be,  = –pE sin  = – pE ( sin   ) pE d 2 d2 2 2   I 2  pE or    or with I dt dt 2  2  This is standard equation of angular SHM with period T    . So, dipole will execute angular SHM with  time period

T  2

I PE

Interaction of Two Dipoles : If a dipole is placed in the field of other dipole, then depending on the positions of dipoles relative to each other, force, torque and potential energy are different. TABLE : Dipole-Dipole Interaction Relative position of Dipole Force F Torque 

S.No.

r

1.

2.

p1

F

F

p1

p2

F

p2

p1 r

1 6p1p 2 (along r) 4 0 r4

1 3p1p 2 4 0 r 4 (along r)

p2

r

F

3.



F



1 3p1p 2 4 0 r 4

Potential Energy U 1 2p1p 2 4 0 r 3

zero



zero

1 p1p 2 4 0 r 3

1 2p1p 2 on p1 , 4 r 3 0

zero

F

(perpendicular to r)

1 p1p 2 on p 2 , 4 r 3 0

Example-40 : Two point charges +3.2 ×10–19 C and –3.2 × 10–19 C are placed at a distance 2.4 × 10–10 m, from each other. This dipole is placed in a uniform electric field of 4.0 × 105 V/m. (a) Find the necessary torque required to rotate the dipole through 180º from its equilibrium position. (b) What is the work done in rotating the dipole through 180º ?

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30

ELECTROSTATICS (c) What is the potential energy of dipole in this position ? Sol. Electric dipole moment p = q × 2l Here, q = 3.2 × 10–19 C 2l = 2.4 × 10–10 m  p = (3.2 × 10–19) (2.4 × 10–10) p = 7.68 × 10–29 C-m (a) necessary torque  = pE sin  = 7.68 × 10–29 × 4 × 105 × sin 180º = 0 (  sin 180º = 0) (b) Work done in rotating the dipole through 180º is given by W = pE (cos1 – cos2) Here, 1 = 0, 2 = 180º, p = 7.68 × 10–29 C-m and E = 4.0 × 105 V/m  W = 7.68 × 10–29 × 4 × 105 × (cos 0º – cos 180º) W = 7.68 × 10–29 × 4 × 105 × (1 + 1) W = 6.14 × 10–23 J (c) The potential energy of dipole in this position is given by U = – pE cos  where is the angle between the axis of dipole and electric field. In equilibrium position, = 0  U0 = – pE or U0 = – (7.68 × 10–29) (4.0 × 105) U0 = –3.07 × 10–23 J Example-41 : A wheel having mass m has charges +q and –q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform –q vertical electric field E = mg mg mg tan q (A) (B) (C) (D) none q 2q 2q Sol. The torque of electric force about centre is balanced by torque due to friction about the centre. E r f = PE sin P 90  = PE sin +q But f = mg sin (for equilibrium) f -q mg  mgr sin = PE sin 90sin mgr mgr E  P q  2r P = q × 2r (dipole moment)  E

+q E

mg 2q

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31

ELECTROSTATICS Example-42 : A nonconducting ring of mass m and radius R is charged as shown. + The charged density i.e. charge per unit length is . It is then placed on a rough ++ + nonconducting horizontal surface plane. At time t = 0, a uniform electric  – field E = E 0i is swithced on and the ring start rolling without sliding. – – Determine the friction force (magnitude and direction) acting on the ring, – when it starts moving. Sol. Where f is friction surface. f = macm ...(1) Let d element on ring then dq   d y d + +  d   { } dq   Rd + R +    – p d  dP  E – – {  dp = (2Rdq)} d  2RdqE sin      – f d  2RE  R sin  d

y x

x

rough non-conducting surface

 /2 2

 E  2R  E  sin  d

dp

0

p- q

2

 E  2R  E 2R2 E – fR = mR2 2RE – f = mR 2RE – macm = mR for pure rolling, acm = R  2RE = macm + mR 2RE = 2 macm macm = RE

dq

E

dq q q

(from eqn (1))

–dq f

f  ma cm   RE 0

Example-43 : In the figure shown S is a large nonconducting sheet of uniform charge density . A rod R of length l and mass ‘m’ is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. Find the angular acceleration of the rod just after it is released.  Sol. Electric field due to non-conducting sheet is 2  . 0 d = x E dq × 2 = 2x Edq / 2

   d 

 0

  or

S s

R

xl -l

/2

 x2  2x E dx  2E    2 0

2E 2 E  2  8 4 E2 I  4 2 m E  2  12 4

S

+dq Edq x x

s Edq –dq

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ELECTROSTATICS E  12 3E   2 4 m m 2



 





3  2 0 m

ELECTRIC FLUX Electric flux through an elementary area dS is defined as the scalar product of area and field, i.e.,   n E dE  E . dS dE = EdS cos    dS  E   E . dS i.e., It represents the total lines of force passing through the given area. E Regarding electric flux following points are worth noting : n

(a) Electric flux will be maximum when cos = 1, i.e.,  = 0, i.e., electric field is normal to the area with (dE)max = EdS. (d E)max = EdS n

(b) Electric flux will be minimum when cos = 0, i.e.,  = 90º, i.e., field is parallel to the surface with (dE)min = 0.

E

(d E)min = 0

(c) For a closed body outward flux is taken to be positive [Fig. (a)] while inward negative [Fig. (b)] n

(a) Positive flux

(b) Negative flux

GAUSS’S LAW This law gives a relation between the net electric flux through a closed surface and the charge enclosed by the surface. According to this law, “the net electric flux through any closed surface is equal to the net charge inside the surface divided by 0.” In symbols it can be written as,  q  E   E . nˆ dS  in ...(i) 0 S but this form of Gauss’s law is applicable only under following two conditions : (i) The electric field at every point on the surface is either perpendicular or tangential. (ii) Magnitude of electric field at every point where it is perpendicular to the surface has a constant value (say E). Here, S is the area where electric field is perpendicular to the surface. Applications of Gauss’s Law : To calculate E we choose an imaginary closed surface (called Gaussian surface) in which Eqs. (i) or (ii) can be applied easily. In most of the cases we will use Eq. (ii). Let us discuss few simple cases.

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33

ELECTROSTATICS (i)

Electric field due to a point charge : q ES  in From Eq. (ii), 0 Here, S = are of sphere = 4r2 and qin = charge enclosing the Gaussian surface = q q E 4r 2   0 1 q E  4 0 r 2 It is nothing but Coulomb’s law. (ii) Electric field due to a linear charge distribution : Consider a long line charge with a linear charge density (charge per unit length) . We have to calculate the electric field at a point, a distance r from the line charge. We construct a Gaussian surface, a cylinder of any arbitrary length l of radius r and its axis coinciding with the axis of the line charge. Hence, we can apply the Gauss’s law as, q ES  in 0



q

E

r



+ + + +

r lE

E

+ + + +

E E

Curved Surface

Plane Surface

Here, and

S = area of curved surface = (2rl) qin = net charge enclosing this cylinder = l l E(2l )   0  E  2 0 r 1 E i.e., r (iii) Electric field due to a plane sheet of charge : From Gauss’s Law q ES  in + + + S0 + + + + + + 0 + + + + + + ( ) (S0 ) + + + E E E(2S0 )  + + + + +  + 0 + + + + + + + +  E  2 0 (iv) Electric field near a charged conducting surface : This is similar to the previous one the only difference is that this time charges are on both sides. Hence, applying q ES  in 0

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ELECTROSTATICS Here,  

S = 2S0 and qin = () (2S0) ( ) (2S0 ) E(2S0 )  0  E 0

E

++ ++ ++ ++ ++ ++ ++

+ + + + + + +

+ + + + + + +

+ ++++ ++++ +++ +

S0 E

+ ++ ++++ ++

Example-44 : Which of the following statements is/are correct ? (A) Electric field calculated by Gauss law is the field due to only those charges, which are enclosed inside the Gaussian surface. (B) Gauss law is applicable only when there is symmetrical distribution of charge. (C) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only. (D) None of these   E. Sol. Since, electric field at a point is equal to electric flux passing per unit area, therefore,   dS is the net flux emanating from a closed surface. Though net flux through the closed surface depends upon the charges enclosed in that surface only but electric field E at a point depends not only upon charges enclosed but it depends upon charges lying outside the surface also. Hence (A) is wrong. Gauss law is applicable to a closed surface. The surface may have any shape. It means, it is a general law. Hence (B) is wrong.   q Gauss law is  E. dS   . It means, net flux through a closed surface depends upon q. But it is equal 0 to net charge enclosed within the surface only. Hence (C) is correct. Obviously (D) is wrong. E

Example-45 : In a region of uniform electric field E, a hemispherical body is placed in such a way that field is parallel to its base (as shown in figure). The flux linked with the curved surface is :

R2 2 2 E (A) zero (B) –R E (C) R E (D) 2 Sol. The flux linked with the body is zero as it does not enclose any charge.  = cs + pb = 0 As field is parallel to base, the flux linked with base pb = E × R2 cos 90º = 0  paralled to base = 0 curved surface = 0

O C

E

n

Example-46 : If a point charge q is placed at the centre of a cube what is the flux linked (a) with the cube (b) with each face of the cube ? Sol. (a) According to Gauss’s Law, q q  total  in  0 0 (b) Now as cube is symmetrical body with 6-faces and the point charge is at its centre, so electric flux linked with each face will be  q ´ total  6 6 0

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ELECTROSTATICS NOTE : If charge is not at the centre of cube (but any where inside it), total flux will not change but the flux linked with different faces will be different. Example-47 : If a point charge q is placed at one corner of a cube, what is the flux linked with the cube ? Sol. By placing three cubes at three sides of given cube and four cubes above, the charge will be in the centre. So, the flux with each face will be one-eight of the flux (q/0). q  Flux through the cube  8 0 Example-48 : Calculate flux through the cube and flux through the each surface of cube when q is placed at one of its corner. Sol. To cover charged particle completly we have to use 7 extra cubes around the charged particle. q ( )  So, Flux Through cube 8 0  q Flux through ABCD  3  24  G F q 0 H  q E Flux through ABFE  3  24  C B 0 A D  q ADEH   Flux through 3 24 0 Flux through EFGH = 0 Flux through FGCB = 0 Flux through GCDH = 0

 E x  0 E i . Find the charge contained inside a cubical Example-49 : The electric field in a region is given by = l volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103N/C, l = 2cm and a = 1cm.  E x 0 ˆ i Sol. E   q Total flux   0 q EA  where A is area of surface 0 E 0a q E0a3 q  a2     0  0 E 0a 3 q 0  q 

5  103  8.85  104 2

 

q 



5  103  8.85  10 12  1  10 2



3

2  10 2

q  2.2 1012 C

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ELECTROSTATICS ELECTRIC FIELD AND POTENTIAL DUE TO CHARGED SPHERICAL SHELL OR SOLID CONDUCTING SPHERE Electric Field  The all points inside the charged spherical conductor or hollow spherical shell, electric field E  0, as there is no charge inside such a sphere. We can construct a Gaussian surface (a sphere) of radius r > R. From Gauss’s Law   q + q E + + C . ds  0 + + E + q 1 q + r R E(4r 2 )  E +  2 0 4 0 r + + + + Hence, the electric field at any external point is the same as if the total Gaussian surface charge is concentracted at centre. At the surface of sphere r = R, 1 q E  4 0 R 2 Thus, we can write, Einside = 0 1 q E surface  4 0 R 2 E 1 q 1 q = E outside  2 1 4 0 r 2 R E r2 The variation of electric field (E) with the distance from the centre (r) is as shown in fig. r O

R

NOTE : • At the surface graph is discontinuous. 1 q q/4πR 2 σ = = • Esurface = 4πε0 R 2 ε0 ε0 Potential : As we have seen, 1 q 4 0 r 2 1 q  dVoutside  dV   E     2    dr 4 0 r dr  V q r dr  (V = 0) 0 dVoutside   4 0  r 2 1 q 1 V or V   4 0 r r Thus, at external points the potential at any point is the same when the whole charge is assumed to the concentrated at the centre. At the surface of the sphere, r = R. 1 q V  4 0 R At some internal point electric field is zero everywhere, therefore, the potential is same at all points which is equal to the potential at surface. Thus, we can write, 1 q Vinside  Vsurface  4 0 R E outside 

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ELECTROSTATICS 1 q 4 0 r The potential (V) varies with the distance from the centre (r) as shown in Fig.

and

V

Voutside 

1

q = R R

O

V

R

1 r r

Example-50 : The diameter of hollow metallic sphere is 60 cm and charge on sphere is 500 µC. Find the electric field and potential at a distance 10 cm from centre of sphere. Sol. The point at 10 cm distance from centre of sphere will be inside the sphere. Inside hollow conducting sphere, electric field is zero every where and potential is uniform (same as at the surface). Hence, intensity of electric field E = 0 1 q and potential V  4 R 0 Here, q = 500 µC = 500 × 10–6 C and R = 30 cm = 0.30 m 500  106 9 V  9.0  10    0.30 7 V = 1.5 × 10 V





ELECTRIC FIELD AND POTENTIAL DUE TO A UNIFORMALY NON-CONDUCTING SPHERE Electric field : Let positive charge q is uniformly distributed throughout the volume of a solid sphere of radius R. We have to find the intensity of electric field due to this charged sphere at point P distance r from centre O. Applying Gauss’s law q ES  in ...(i) 0 4  q in  ()  r 3  Here, S = 4r2 and 3  q Here, ρ = charge per unit volume = 4/3R 3 Substituting these values in Eq. (i), we have 1 q E . 3 . r or Er 4 0 R At the centre r = 0, so, E=0 1 q E At surface r = R, so, 4 0 R 2 To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius r(>R). This surface encloses the entire charged sphere, so qin = q, and Gauss’s law gives. q 1 q 1 E(4r 2 )  E E 2 or 0 4 0 r 2 or r Thus, for a uniformly charged solid sphere, we have the following formulae for magnitude of electric field. 1 q E inside  . 3 .r E 4 0 R 1 q 1 q R2 E surface  . E r12 r 4 0 R 2 E 1 q r E outside  . R O 4 0 r 2

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ELECTROSTATICS The variation of electric field (E) with the distance from the centre of the sphere (r) is shown in fig. Potential : The field intensity outside the sphere is, 1 q E outside  . 2 4 0 r dVoutside   E outside dr V







V

or At r = R,



1 q . 2 dr  4 0 r

dVoutside   

1 4 0 1 V r 1 V 4 0

q r

r

as V  0

q R

1 q i.e., at the surface of the sphere potential is VS  4 R . 0 The electric intensity inside the sphere, 1 q E inside  . 3 .r 4 0 R dVinside   E inside dr V 1 q r  VS dVinside   40 R 3 R r dr r



1 q  r2  V  VS   4 0 R 3  2  R

Substituting VS 

1 q . , we get 4 0 R

1 q  3 1 r2  V .  4 0 R  2 2 R 2  3 1 q 3 At the centre r = 0 and VC  2  4 R   2 VS, i.e., potential at the centre is 1.5 times the potential at   0 surface. Thus, for a uniformly charged solid sphere we have the following formulae for potential. 1 q Voutside  4 0 r 1 q Vsurface  V 4 0 R q 3 1 1 q  3 1 r2   and 4 0 R  2 2 R 2  The variation of potential (V) with distance form the centre (r) is as shown in figure. Vinside 

2 1

R q R

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ELECTROSTATICS Example-51 : The radius of a solid metallic non-conducting sphere is 60 cm and charge on the sphere is 500 µC. Find the electric field and potential at a distance 10 cm from centre of sphere. Sol. The point 10 cm from centre of sphere will be inside the sphere. Hence, 1 qr E inside  . 3 4 0 R –6 Here, q = 500 × 10 C, r = 10 cm = 0.10 m R = 60 cm = 0.6 m 500  106  0.10 9 E  9.0  10   2  106 N / C  inside 3 (0.60) 2 1 q 3 1 r  Vinside   and 4 0 R  2 2 R 2 





6

)  3 1 (0.10) 2   2  2  (0.60) 2    3 1    7.5  106     2 72   107  7.5  106  72 7 = 1.1 × 10 V



 9  109

 10  (500(0.60)

Example-52 : A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge r density,   0 , where 0 is a constant and r is the distance from the centre of the sphere. Show that: R (a) the total charge on the sphere is Q = 0R3 and (b) the electric field inside the sphere has a magnitude given by, E  Sol. (a)

KQr 2 . R4

dv = 4r2dr

0r 4 r 2dr R 40 3  r dr R R 40 3 Q r dr R 0 dq  dV 

40 R 4   R 4 r 40 3 q in  r dr R 0 Q

(b)

R

r dr

Q  0 R 3

40 r 4 40 r 4   R 4 4R q in E 4 0 r 2 40 r 4 E 4R  4 0 r 2 Q = 0R3 

But

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ELECTROSTATICS  0  

Q R3

and

K

1 4 0

KQr 2 E R4

Example-53 : Three concentric metallic shells A, B and C of radii a, b and c (a < b < c) have surface charge densities , – and  respectively. (i) Find the potential of three shells A, B and C. (ii) If the shells A and C are at the same potential, obtain the relation between the radii a, b and c. Sol. The three shells are shown in figure. (i) Potential of A = (Potential of A due to + on A) + (Potential of A due to – on B) + (Potential of A due to + on C)  4 a 2 4 b2 4 c 2   K   b c   a    a  b  c c 0 Potential of B = (Potential due to + on A) + (Potential due to – on B) b + (Potential due to + on C) C B A a  4 a 2 4 b2 4 c 2  +s  K   b c   a -s

  a2    b  c +s 0  b  Potential of C = (Potential due to + on A) + (Potential due to – on B) + (Potential due to + on C)  4 a 2 4 b2 4 c 2   K   a b c  

(ii) Given that or or Solving we get,

   a 2 b2     c 0  b c  VA = VC     a 2 b2 a  b  c    c   0 0  b c  2 2 a b abc   c b c c = (a + b). 

Example-54 : An electric field converges at the origin whose magnitude is given by the expression E = 100 N/C, where r is the distance measured from the origin. (A) total charge contained in any spherical volume with its centre at origin is negative. (B) total charge contained at any spherical volume, irrespective of the location of its centre, is negative. (C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10–13 C. (D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10–9 C.

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ELECTROSTATICS Sol. Since, electric lines of forces are terminated into the sphere. Hence, total charge contained by should the sphere should be negative. So, options (A) and (B) are correct q E  4 0 r 2 Here r = 3 × 10–2C  q = 3 × 10–13 m  Hence, option (C) is correct. Example-55 : Figure shows a section through two long thin concentric cylinders of radii a & b with a < b. The cylinders have equal and opposite charges per unit length . Find the electric field at a distance r from the axis for (A) r < a (B) a < r < b (C) r > b

E E E

E

b a

b

Sol.

a

-l

l

b

(A) r < a





a E

0

r

 E  ds   C

0

E  2 r 

O 0

E=0

b

(B) a < r < b

a

  q C E  ds  0

E  2 r 

(C) r > b

E

 2 0 r





 0

   0 0 C E  2 r  0 E0

 E  ds 

1. 2. 3. 4. 5. 6. 7.

r

b a

r

THINKING TYPE PROBLEMS The electric charge of macroscopic bodies is actually a surplus or deficit of electrons. Why not protons? What are insulators and conductors? A charged rod attracts bits of dry cork dust which, after touching the rod, often jump away from it violently. Explain. A truck carrying explosives has a metal chaing touching the ground. Why ? How is the Colulomb force between two charges affected by the presence of a third charge ? A person standing on an insulating stool touches a charged insulated conductor. Is the conductor completely discharged ? Electric lines of fore never cross. Why ?

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ELECTROSTATICS 8. The electric field inside a hollow charged conductor is zero. Is this true or false ? 9. The electric field inside a hollow charged spherical conductor is the same at all points and is equal to the field at the surface. Is this true or false ? 10. The electric potential inside a hollow charged spherical conductor is the same at all points and is equal to the potential of the surface. Is this true or false ? 11. A charge A situated outside an uncharged hollow conductor experiences a force if another charge B is placed inside the conductor, but B does not experience any force. Why? Does it not violate the third law of motion ? 12. If only one charge is available, can it be used to obtain a charge many times greater than it in magnitude? 13. In the previous question, does it make a difference which face of B is touched in order to remove the free charge (include positive charge) from B? 14. Can a small spherical body of radius 1 cm have a static charge of 1 C? 15. A metal leaf is attached to the internal wall of an electrometer insulated from the earth. The rod and housing of the electrometer are connected by a wire, and then a certain charge is imparted to the housing. Will the leaves of the electrometer be deflected ? What will happen to the leaves if the wire is removed and the rod is then charged ? 16. A broad metal plate is connected to the earth through a galvanometer. A positively charged ball flies along a straight line above the plate at a distance much less than the linear dimensions of the plate. Draw an approximate diagram showing how the current flowing through the galvanometer depends on time. 17. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which it is taken. Is this true of false? Briefly explain your answer. 18. The figure shows lines of constant potential in a region in which on electric field is present. The values of the ptentials are written in brackets. Of the points A, B and C, the magnitude of the electric field is greatest at the point...... Give reasons for your answer in brief. 19. Can two balls with like charges be attracted to each other? 20. The housing of the electrometer of question no.15 is given a charge. Will the leaves of the electrometer be deflected in this case? Will the deflection change if the rod is earthed ? There is no connecion between the rod and the housing. 21. By touching different points of a metal bucket with a test ball B connected by a wire to an earthed electrometer it can be observed that deflection of the leaves of the electrometer is the same for any position of the ball. But if the wire is removed and the charge is transferred by the ball to the ball of the electrometer, the deflection depends on which surface of the bucket (external or internal) is touched before that. Why ? 22. The gravitational field strength is zero inside a spherical shell of matter. The electric field strength is zero not only inside an isolated charged spherical coonductor but inside an isolated conductor of any shape. Is the gravitational field inside, say, a cubical shell of matter zero ? If not, in what respect is the analogy not complete ? 23. Can two equipotential surfaces intersect ? 24. An isolated, conducting spherical shell of radius R carries a negative charge Q. What will happen if a small, positively charged metal object is placed inside and connected by a wire ? Assume that the positive charge is (a) less than, (b) equal to and (c) greater than Q. 25. An uncharged metal sphere suspended by a silk thread is placed in a uniform external electric field E. What is the magnitude of the electric field for points inside the sphere ? Will your answer change if the sphere carries a charge? 26. Is charge uniformly distributed over the surface of an insulated conductor of any shape ? If not, what is the rule for the distribution of charge over a conductor? 27. Are equipotential surfaces which arise due to a point charge and whose potentials differ by a constant amount (say 1 volt) evenly spaced in radius ? 28. Two point charges Q and 4Q are fixed at a distance of 12cm from each other. Sketch the lines of force and

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ELECTROSTATICS locate the neutral point, if any. 29. A charge +Q is fixed at a distance d in front of an infinite metal plate. Draw the lines of force indicating the direction clearly. 30. Indicate the surface distribution of charge on a square metal plate by using dots in such a way that the greater the surface density of charge, the farther away the dots are from the plate (shown in fig.). 31. A metal plate is moved with a constant acceleration a parallel to its plane. Does any charge develop on its surfaces which are perpendicular to its motion? 32. Is an electric field of the type shown by the electric lines in the figure physically possible? 33. Zero work is done when a charged particle is transferred from one equipotential surface to another. Is this true or false ? 34. A potential difference V is set up between a filament emitting electrons in a vaccum tube and a thin metallic ring. The electron beam passes past the ring through its central region without spreading. The kinetic energy of the electrons in the beam increases while the battery producing the potential difference V performs no work, since no current flows through the circuit. How can his fact be reconciled with the principle of conservation of energy? 35. There is an electric field near the surface of a conductor carrying direct current. Is this true or false? 36. Ordinary rubber is an insulator. But the special rubber tyres of aircrafts are made conducting. Why is this necessary? 37. A small sphere is charged to a potential of 50V and a big hollow sphere is charged to a potential of 100V. How can you make electricity flow from the smaller sphere to the bigger one? 38. An electric dipole is placed in a non-uniform electric field. Is there a net force on it? 39. A point charge is placed at the centre of a spherical gaussion surface. Does electric flux  e change (a) if the sphere is replaced by a cube of the same volume, (b) if a second charge is placed near, and outside, the original sphere, and (c) if a second charge is placed inside the gaussion surface? 40. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up, how does E vary for points (a) inside the balloon, (b) on the surface of the balloon, and (c) outside the balloon? THINKING PROBLEMS SOLUTION 1. Because protons are tightly bound in the nucleus. They cnnot be removed from there easily. 2. Conductors are materials in which there are a few free electrons per atom of the matter. Insulators are materials in which the electrons are not as free as in conductors. 3. A charged rod first attracts the dust by producing unlike charge at the near end and like charge at the far end. When the cork dust touches the rod, however, it acquires like charge and so is repelled strongly by the charge on the rod. 4. To conduct away the charge produced by friction. This charge, unless conducted away to the earth, may produce sparks causing the explosives in the truck to catch fire and explode. 5. The force on the charge due to another does not depend on the presence of other charges, that is, electrical forces are physically independent. 6. No, the man acts as a conducting body. He shares charge with the charged body. 7. Because if they cross, the electric field at the point of intersection will have two directions simultaneously, which is physically impossible. 8. It is true. The electric field inside a hollow charged conductor is zero. This follows from the condition that all points on conductor have the same potential.

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ELECTROSTATICS 9. It is false. The field inside is zero but on the surface, the field is finite. 10. It is true. 11. This is so because there is an electric field over A but there is no electric field over B as it lies inside a hollow conductor. No, this is not in violation of the third law of motion. In fact, force arises between A and the hollow sphere, the charge B is simply an internal part of the sphere. 12. Yes, by repeating the induction process. Place an insulated conductor B close to the charged body. A. y induction, an equal unlike charge is induced at the near end of B and equal like charge at the far end. Touch the body B with your finger tip in the presence of A. The like charge at the far end will flow to the earth. But the unlike charge at the near end will remain bound due to the force of attraction. Now remove the body B to a distant place. The unlike charge, equal in magnitude to the inducing charge, will spread over the surface of B as there is no force of attraction to keep it on one side. Next, deliver this charge to a big body by conduction, that is, by touching this body with B. Repeat the process after completely discharging B. To obtain a similar charge in large magnitude, first deliver unlike charge to a body C and then repeat the induction process between C and the body in which like charge is to be stored. 13. No. The free charge will go to the earth via the finger, irrespective of which part of B is touched, because the free positive charge in B exists at a higher potential due to the positive charge on A. It is a fact that positive charge always flows from higher to lower potential, irrespective of the path. 1 9  91013 N C1 . The field is every large. Air cannot sustain such a strong electric field, so a 14. E  910 0.01 small body cannot have a charge of 1 coulomb. 15. The housing and the rod connected together will have the same potential, so the leaves will not be deflected. After the wire is removed and the rod is charged, both the leaves will be deflected because of the potential difference between the rod and the housing. 16. As the charge approaches the plate, electrostatic induction causes the induced positive charge to pass into the earth, while the induced negative charge accumulates on the upper surface of the plate. A positive current pulse passes through the galvanometer. No current is produced when the charge moves above the plate. An opposite current is produced when the charge moves away from the plate. 17. It is false. the work done in carrying a point charge from one point to another does not depend on the path along which it is carried because electric field is conservative. 18. It is greatest at the point B. Since the electric field is the rate of fall of potential, the stronger the field, the closer the equipotential surfaces. In the figure, the equipotential surfaces are closest in the neighbourhood of B, so the field at B is the greatest. 19. They can, if the charge of one ball is much greater than that of the other. The forces of attraction caused by the induced charges may exceed the forces of repulsion. 20. When the housing of the electrometer is given some charge, the potential of the rod cannot be the same as that of the interior of the housing because the rod lies partly outside. On account of the potential difference between the rod and the housing, the leaves will be deflected. When the rod is earthed, the potential difference between the rod and the housing is increased further, so the deflection will be greater. 21. The electrometer measures the potential diference between the body and the housing of the electrometer. Since the surface of the bucket is equipotential, the leaves show the same deflection wherever the testing ball B may touch the surface of the bucket. But when the ball B is disconnected from A, the ball B collects charge by conduction. When it delivers this charge to A, the potential of the leaves (and also of the rod) is raised above the potential of the housing, so there is deflection of the leaves. When B touches the inner

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ELECTROSTATICS surface, it collects no charge because there is no potential difference between it and the inner surface of the bucket. Unless there is a potential difference, there can be no flow of charge from one body to another. So no deflection takes place when the ball B touches the inner surface and then touches A. But when it touches the outer surface, it collects a certain amount of charge, so there is deflection. 22. No, the field inside a cubical shell of matter is not zero. The analozy is not complete in respect of distribution of mass and charge. Mass is uniformly distributed but charge is not distributed uniformly. It is more concentrated at the edges and corners. 23. No, two equipotential surfaces can never intersect because if they intersect at a point, the electric field at that point can have two directions simultaneously, which is physically impossible. 24. For the sake of simplicity, let us consider the metal body to be a sphere placed at the centre of the shell. 1 Q  q VA (potential of the surface of the shell)  4 R 0 where R = radius of the shell and q is the charge of the metal body. 1  q Q  VB (potential of the metal body)  4  r  R  0

25. 26. 27.

28.

29.

q 1 1  V  V     B A  40  r R   VB  VA So charge will flow from the metal body to the shell. Since VB = VA at equilibrium, q must be reduced to zero, i.e., the entire charge on the metal body must flow to the outer shell. qQ V, the common potential  4 R . Thus, (a) when q < Q, V is negative, (b) when q = Q, V = 0, (c) when q 0 > Q, V is positive. E The field inside the metal sphere at all points is zero because  r is infinity for metals and  r  0 , So E = 0. E No, the answer does not change if the sphere carries a charge. Charge is not distributed uniformly over the surface of a conductor of any shape. Charge is distributed according to the curvature of the surface. The greater the curvature, the greater the surface density of charge. No. q  1 1  q r2  r1  V1  V2     40  r1 r2  4 0 r1r2 qr 1 or 4 0 r1 r1  r  where r  r2  r1 Obviously,  r is not constant. So the equipotential surfaces differing by 1 V are not equispaced. The lines of force are shown in the figure. Here, all the lines near the edge of the figure will appear to radiate uniformly from the point P, the centre of gravity of the charges. At the neutral point N, the total field is zero. Let it be at a distance x from 4Q. 1 4Q 1 Q  2 Then x = 8 cm 4 0 x 4 0 12  x 2 or The lines of force are shown in the figure. Explanation Since it is a metal plate, its surface is an equipotential surface. So lines of force must terminate normally on it. Note : One every important result follows from this map of electric flued due to a point charge and the induced charges on a metal plate. If a charge –Q is placed as far behind the plate as +Q

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46

ELECTROSTATICS is in front and the plate is removed, AB (position of the plate) is still an equipotential surface. So for interaction between +Q and the plate or for electric fields between them, we may use the formalism of replacing the plate by –Q. This is called the electrical image of +Q. 30. The distribution of charge is shown in the figure. Explanation Since the surface density of charge is proportional to curvature and corners have the greatest curvature and a plane surface has zero curvature, dots are equidistant from the straight portion and far away from the corners. 31. Since the plate is accelerated, the electrons inside are also accelerated with the same acceleration. A force may be applied on an electron, only when an electric field is created. So there must develop charges on the faces that are perpendicular to the direction of motion, so that there develops an electric field inside the metal plate. The front face will be charged positively, and the rear one negatively. The intensity of the electric field is given by eE = ma. Since E = /0, the surface density of charge developed is given by  = 0E = 0ma/e 32. No. The concentration of lines at the botom indicates a field which is stronger at the botoom than on top. Imagine a rectangular path with its two sides perpendicular to the electric lines. Now move a charge along this path. Some net work will be performed. But in an electric field, work done is essentially zero as it is a conservative field. Hence, an electric field of the type represented is physically impossible. 33. False, Work is always done by an external agent when a charged particle is transferred from one equipotential surface to another. The work done per unit charge = potential difference. 34. There is an electric field between the filament and the ring. As the electron is emitted from the filament, it experiences a force towards the ring and so it is accelerated. It electrical potential energy, which it acquires on emerging from the filament, is converted into its kinetic energy. Thus, though the battery does not supply any energy, the electron can gain velocity at the cost of its own potential energy. 35. True. There is an electric field inside a current-carrying conductor. This is equal to V/l, where V is the voltage across the conductor and l is the length of the conductor. Now, consider a closed path abcda as shown in the figure. Work done by an external agent along abcda is zero. Einside ab  0  Eoutside cd  0  0  E inside  E outside  Hence, there is a field near the surface of the conductor. 36. To conduct away electricity produced by friction. 37. Place the smaller one inside the bigger one. As potential inside a hollow conductor is the same as that of its surface, the potential inside the hollow conductor is 100V. The potential of the smaller conductor is now 100+50 = 150 V. Connecte the two conductors by a wire. Charge will flow from the smaller one to the bigger one as the smaller one is at 150 V and the bigger one at 100 V. E 38. Yes, there is a net force on the dipole given by F  p l   1 39. Total flux  E   E . s   q , by Gauss’s law.. 0 (a) No, the flux will not change as it depends only on the total charge inside the surface and not on the extent and shape of the surface. (b) No, it remains the same as the total flux is determined solely by the charge inside the surface and not the charge outside. (c) Yes, flux will change as the total charge inside the surface has changed. 40. (a) For points inside the balloon, E = 0, (b) E decreases as the surface density of charge decreases E   /  0  , (c) E remains constant because E  q / 4 0 r 2 , where r is the distance of the point from the centre of the balloon.

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47

ELECTROSTATICS

Reasioning Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as – (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : If there exists colomb attraction between two bodies, both of them may not be charged. Reason (R) : Due to induction effects a charged body can attract a neutral body. 2. Assertion (A) : A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric field. Reason (R) : X-rays emit photoelectrons and metal becomes negatively charged. 3. Assertion (A) : Electric current will not flow between two charged bodies when connected if their charges are same. Reason (R) : Current is rate of flow of charge. 4. Assertion (A) : The surface densities of two spherical conductors of radii r1 and r2 are equal. Then the electric field intensities near their surfaces are also equal. Reason (R) : Surface charged density = charge/area. 5. Assertion (A) : When charges are shared between two bodies, there occurs no loss sof charge, but there does occur a loss of energy. Reason (R) : In case of sharing of charges, the energy of conservation fails. 6. Assertion (A) : Two adjacent conductors, carrying the same positive charge have a potential difference between them. Reason (R) : The potential of a conductor depends upon the charge given to it. 7. Assertion (A) : Dielectric breakdown occurs under the influence of an intense light beam. Reason (R) : Electromagnetic radiations exert pressure. 8. Assertion (A) : The tyres of aircrafts are slightly conducting. Reason (R) : If a conductor is connected to ground, the extra charge induced on conductor will flow to ground. 9. Assertion (A) : Metallic shield in the form of a hollow shell, can be built to block an electric field. Reason (R) : In a hollow spherical shell, the electric field inside it is zero at every point. 10. Assertion (A) : A bird perches on a high power line and nothing happens to the bird. Reason (R) : The level of bird is very high from the ground. 11. A metal sphere is suspended from a nylon thread. Initially, the metal sphere is uncharged . When a positively charged glass rod is brought close to the metal sphere, the sphere is drawn toward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted, then repelled.

Level # 1. Objective Type Question 1.

Two copper spheres of same radii one hollow and other solid are charged to the same potential then (A) both will hold same charge (B) solid will hold more charge (C) hollow will hold more charge (D) hollow cannot be charged

2.

Through the exact centre of a hydrogen molecule, an  -particle passes rapidly, moving on a line perpendicular to the internuclear axis. The distance between the two hydrogen nuclei is b The maximum force experienced by the  -particle is

4e 2 (A) 3 3 0 b 2 3.

(B)

8e 2 3 0b 2

8e 2 (C) 3 3  0 b2

(D)

4e 2 3 0b 2

A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potential at the points A, B and C satisfy (A) VA > VB (B) VA < VC (C) VA < VB (D) VA > VC

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48

ELECTROSTATICS 4.

On an equilateral triangle of side 1 m there are three point charges placed at its corners of 1C, 2C and 3C. The work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m is (A) 19.8 x 1010 J (B) 39.6 x 1010 J (C) 9.9 x 1010 J (D) 4.45 x 1010 J

5.

The potential field depends on x and y coordinates as V = (x 2 – y2). Correspondence electric field lines in x.y plane as shown in figure.

(A)

6.

(B)

(C)

(D)

Two circular rings A and B, each of radius a = 130 cm are placed coaxially with their axes horizontal in a uniform electric field E = 105 NC–1 directed vertically upwards as shown in Figure. Distance between centers of these rings A and B is h = 40 cm, ring A has a positive charge q1  10C while ring B has a negative charge of magnitude q2  20C . A particle of mass m = 100 gm and carrying a positive charge q  10C is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance of 40 cm. (A)

7.

v  6 2ms 1

(C) v  7 ms 1

v  4 2ms 1

(D) v  32 ms 1

Three identical spheres each having a charge q and radius R, are kept in such a way that each touches the other two. the magnitude of the electric force on any sphere due to other two is

1 (A) 4 0 8.

(B)

 3  q 

2

1 (B) 4 0

4  R 

 2  q 

2

4  R 

1 (C) 4 0

 2  R  4  q 

2

1 (D) 2 0

 3  q 

2

4  R 

A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density

   r , where 

is a positive constant and r is the distance from the centre of the

sphere. Find the charge of the sphere for which the electric field intensity E outside is independent of r. (A)

 2 0

(B) 2

(C) 2 R 2 9.

 0

(D) None of the above

Three charges each of +q, are placed at the vertices of an equilateral triangle. The charge needed at the centre of the triangle for the charges to be in equilibrium is (A)

q 3

(B)  3q

(C)

(D)  3q

3q

10. Hollow spherical conductor with a charge of 500C is acted upon by a force 562.5 N. What is electric intensity at its surface? (A) zero (B) 1.125 x 106 N/C (C) 2.25 x 106 N/C (D) 4.5 x 106 N/C 11. A hemisphere of radius R is charged uniformly with surface density of charge centre? (A)

R 2 0

(B)

 4 0

(C)

 2 0

(D)



What will be the potential at

4 R 3 0

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49

ELECTROSTATICS B

12. A circular cavity is made in a conductor. A positive charge q is placed at the centre (A) The electric field at A and B are equal (B) The electric charge density at A = the electric charge density at B (C) Potential at A and B are equal (D) All the above.

A

13. An isolated metallic object is charged in vacuum to potential v 0, its electrostatic energy being W 0. It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant k. The electrostatic energy will be. (A) kW 0 14.

(B)

W0 k

(C)

W0 2k

(D) W 0.

A charge +q is fixed at each of the points x = x 0, x = 3x 0, x = 5x 0 .... ad inf. on the x-axis, and a charge – q is fixed at each of the points x = 2x 0, x = 4x 0, x = 6x 0 .... ad inf. Here x 0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/( 4    r ). Then, the potential at the origin due to the above system of charges is. (A) Zero

(B)

q 8   0 x 0 loge 2

(C) 

(D)

q loge 2 4  0 x 0

15.

A positively charged thin metal ring of radius R is fixed in the xy-plane with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z0) where z0 > 0. Then the motion of P is. (A) Periodic, for all values of z0 satisfying 0 < z0 <  (B) Simple harmonic, for all values of z0 satisfying 0 < z0  R (C) Such that P crosses O and continues to move along the negative z-axis towards z = –  (D) None of these

16.

For an infinite line of charge having charge density  lying along x-axis, the work done in moving charge from C to A arc CA is. q (A) 2   loge 2 0

q (B) 4   loge 2 0

q (C) 4   loge 2 0

q 1 (D) 2   loge 2 0

17.

A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speeds v A : v B will be (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

18.

Three point charges q1, q2 and q3 are taken such that when q1 and q2 are placed close together to form a single point charge, the force on q3 at distance L from this combination is a repulsion of 2 units in magnitude. when q2 and q3 are so combined the force on q1 at distance L is an attractive force of magnitude 4 units. Also q3 and q1 when combined exert an attractive force on q2 of magnitude 18 units at same distance L. The algebraic ratio of charges q1, q2 and q3 is. (A) 1 : 2 : 3 (B) 2 : – 3 : 4 (C) 4 : – 3 : 1 (D) 4 : – 3 : 2

19.

An electric potential is given by V = k(xy), where k is a constant. A particle of charge q0 is first taken from (0, 0) to (0, a) to (a, a), then directly from (0, 0) to (a, a) and lastly from (0, 0) to (a, 0) to (a, a). If W 1, W 2 and W 3 be the work done for the individual paths respectively then (A) W 1 = W 2 = W 3 = – q0ka2 (C) W 1 = W 3 > W 2

(B) W 1 = W 3 = – 2 q0ka2 (D) W 1 > W 2 > W 3

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50

ELECTROSTATICS 20.

Four charges 2C, – 3C, – 4C and 5C respectively are placed at all the corners of a square. W hich of the following statements is true for the point of intersection of the diagonals ? (A) Electric field is zero but electric potential is non-zero (B) Electric field is non-zero but electric potential is zero (C) Both electric field and electric potential are zero (D) Neither electric field nor electric potential is zero

21.

Two metallic identical spheres A and B carrying equal positive charge + q are a certain distance apart. The force of repulsion between them is F. A third uncharged sphere of the same size is brought in contact with sphere. A and removed. It is then brought in contact with sphere B and removed. What is the new force of repulsion between A and B ? (A) F

(B)

3F 8

(C)

F 2

(D)

F 4

22.

An electron of mass m e, initially at rest, moves through a certain distance in a uniform electric field in time t 1. A proton of mass m p , also initially at rest, takes time t 2 to move through an equal distance in this uniform electric field. Neglect the effect of gravity, the ratio t 2 /t 1 is nearly equal to (A) 1 (B) (mp /me)1/2 (C) (me /mp )1/2 (D) 1836

23.

Eight dipoles of charges of magnitude are placed inside a cube. The total electric flux coming out of the cube will be 8e (A)  0

24.

16 e (B)  0

e (C)  0

(D) Zero

A particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy attained by the particle after moving a distance x is. (A) q Ex 2 (B) q Ex 2 (C) q Ex (D) q2 Ex

25. Two point charges +q and –q are held fixed at (–d, 0) and (d, 0) respectively of a x-y coordinate system. Then (A) The electric field E at all points on the x-axis has the same direction. (B) Work has to be done in bringing a test charge from  to the origin [1995] (C) Electric field at all points on y-axis is along x-axis. (D) The dipole moment is 2qd along the x-axis. 26. Three charges Q, +q and +q are placed at the vertices of a ring-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to (A)

q 1 2

Q [2000]

2q (B)

(C) –2q

2 2

(D) +q

+q

+q a

27. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketches as in [2001]

(A)

(B)

(C)

(D)

28. Two equal point charges are fixed at x = -a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [2002] (A) x

(B) x 2

(D) x 3

(D) 1

x

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51

ELECTROSTATICS 29. A metallic shell has a point charge ‘q’ kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces? [2003]

(A)

(B)

(C)

(D)

Q

P 30. Six charges of equal magnitude, 3 positive and 3 negative are to U be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what is would have been if only one +ve charge is placed at R. [2004] (A) +, +, +, –, –, – (B) –, +, +, +, –, – (C) –, +, +, –, +, – (D) +, –, +, –, +, – 31. A Gaussian surface in the figure is shown by dotted lime. The electric field on the surface will be (A) due to q1 and q2 only (B) due to q2 only (C) Zero (D) due to all 32. Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is

2 ˆ k (A) 0 (C) 

(D) 

T

S

q1

q2

[2004] –q1

Z

[2005] 

4 ˆ k (B) 0

2 ˆ k 0

R

O

–2

4 ˆ k 0

Z=a P

–

Z=–a x Z=–2a

33. Two equal negative charges –q are fixed at points (0, –a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will [1984] (A) execute simple harmonic motion about the origin. (B) move to the origin remain at rest (C) move to infinity (D) execute oscillatory but not simple harmonic motion 34. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to: [1987] (A) 

Q 2

(B) 

Q 4

(C) 

Q 4

(D) 

Q 2

35. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is: [1989] (A) V (B) 2V (C) 4V (D) –2V

36. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in Figure as [1996] (A) a (B) 2 (C) 3 (D) 4

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52

ELECTROSTATICS 37. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its center [1998] (A) increases as r increases, for r < R (B) decreases as r increases, for 0 < r <  (C) is discontinuous at r = R. (D) None of these 38. An ellipsoidal cavity is carved within a perfect conductor Figure. a positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then [1999] (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is q

A B q

0 .

More Than One Choice Questions: 39. Three concentric spherical metallic shells A, B and C of radii a, b and c (a < b < c) have charge densities of  ,  , and  respectively. The potentials of A, B and C are:

1  a  b  c  (a) VA  0 (c) VC 

 1  a 2 b2    c  0  c c 

 1  a2   b  c  (b) VB  0  c  (d) VA  VB  VC 

1  a  b  c  0

40. When positively charged spheer is brought near a metallic sphere, it is observed that a force of attraction exists between two. It means: (a) metallic sphere is necessarily negatively charged. (b) metallic sphere may be electrically neutral. (c) metallic sphere may be negatively charged. (d) mothing can be said about charge of metallic sphere. 41. A conducting sphere of radius R has a charge. Then: (a) the charge is uniformly distributed over its surface, if there is no external electric field. (b) distribution of charge over its surface will be non-uniform, if an external electric field exists in the space. (c) electric field strength inside the sphere will be equal to zero only when no exxternal electric field exists. (d) potential at every point of the sphere must be same. 42. A small sphere of amss m and having charge q is suspended by a light thread, then: (a) tension in the thread may reduce to zero if anohter charged sphere is placed vertically below it. (b) tension in the thread may increase to twice of its original value if another charged sphere is placed vertically below it. (c) tension in the thread is greater than mg if another charged sphere is held in the same horizontal line in which first sphere stays in equilibrium. 43. Two point charge: Q and –Q/4 are separated by a distance x. Then: (a) potential is zero at a point on the axis which is x/3 on the right side of the charge –Q/4. (b) potential is zero at a point on the axis which is x/5 on the left side of the charge –Q/4. (c) electric field is zero at a point on the axis which is at a distance x on the right side of the charge –Q/4. (d) there exist two points on the axis, where electric field is zero. 44. Two equal and oppositely charged particles are kept some distance apart from each other. A spherical surface having radius equal to separation between the particles and concentric with their midpoint is considered. Then: (a) electric field is normal to the surface at two points only. (b) electric field is zero at no point. (c) electric potential is zero at every point of one circle only. (d) net electric flux through the surafce is zero.

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53

ELECTROSTATICS 45. Two identical charges +Q are kept fixed some distane apart. A small particle P with charge q is placed midway between them. If P is given a small displacement  , it will undergo simple harmonic motion if: (a) q is positive and  is along the line joining the charges. (b) q is positive and  is perpendicular to the line joining the charges. (c) q is negative and  is perpendicular to the line joining the charges. (d) q is positive and  is along the line joining the charges. 46. Select the correct statement(s): (a) Charge cannot exist without mass but mass can exist without charge. (b) Charge is conserved but mass is not. (c) Charge is independent of state of rest or motion. (d) None of these 47. Which of the following quantities do not depend on the choice of zero potential or zero potential energy? (a) Potential at a point. (b) Potential difference between two points. (c) Potential energy of a system of two charges. (d) Change in potential energy of system of two charges.

Fill in the blanks: 1.

Figure shows line of constant potential in a region in which an electric field is present. The value of the potential are written in brackets. Of the points A, B and C, the magnitude of the electric field is greatest at the point ............. [1984]

A B (50 V) (40 V) C (30 V) (20 V) (10 V)

2.

Two small balls having equal positive charges Q (coulomb0 on each are suspended by two insulating strings of equal length L (metre) from a hook fixed to a stand. The whole set up is taken in a satellite into space where there is no gravity (state of weightlessness). The angle between the two string is ............ and the tension in each string is ......... newtons. [1988] Y

3.

A point charge q moves from point P to point S along the path PQRS (figure) in a uniform electric field E pointing parallel to the positive direction of the X-axis. The coordinates of the points P, Q, R and S are (a, b, O), (2a, O, O) (a, –b, O) and (O, O, O) respectively. The work done by the field in the above process is given by the expression ..................... [1989]

P S

X

Q R

E

4.

The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x 2 volts. The electric field at the point (1m, 0.2 m) is ............... V/m. [1992]

5.

Five point charges, each of value +q coul, are placed on five vertices of a regular hexagon of side L metres. the magnitude of the force on the point charge of –q coul. Placed at he centre of the hexagon is .......... newton. [1992]

True / False : 6. 7. 8.

q

q 4

5 –q

3

6 1 q

q

2 L

q

The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which the point charge is carried. [1981] Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulombs and the other an equal negative charge. Their masses after charging are different. [1983] A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. [1983]

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54

ELECTROSTATICS 9.

A ring of radius R carries a uniformly distributed charge +Q. A point charge –q is placed in the axis of the ring at a distance 2R from the center of the ring and released from rest. The particle executes a simple harmonic motion along the axis of the ring. [1988]

Table Match 10.

Match List I and List II and select the correct answer using the codes given below the lists: The electric fields due to various charge distribution are (q is the total charge on the body,  is the surface charge density,



is the linear charge density)

List I I.

List II

At a distance x from the centre of a

  x 1  2  x  R2  

A.

E

 20

B.

E

 2 0 x

uniformly charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the ring II.

At a distance x from the centre of a uniformly charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the disc.

III.

At a distance x from an infinite sheet of

2r

C.

 E 2 0

uniform distribution of charge IV.

At a distance x from an infinite line of charge

(A) I-A, II-C, III-B, IV-D (C) I-C, II-B, III-D, IV-A

D.

E

r

d

q1 q2

1 qx 4 0  R 2  x 2  3 2 .  

(B) I-D, II-A, III-C, IV-B (D) I-B, II-D, III-A, IV-C

Passage Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE E is the electric field created by q1. V is the voltage at a given point in the field E. Assume that the electric field created by q2 is negligible compared to E. k is Coulomb’s Law constant. m 1 si the mass of q1, and m 2 is the mass of q2. 1.

2.

E can best be described as – (A) constant (B) decreasing as r increases (C) increasing as r increases (D) increasing as q2 increases What is the work done on q2 when it is moved at constant velocity along the distance d? (A) Zero

3.

(C) Vq2

(D) Edq2

Which of the following represents the work done on q2 when moved from its present position to a distance r from q1? (A)

4.

(B) Vq1

1 kq1q2 r 2

(B) kq1q2 r

(C) 2 kq1q2 r

(D) kq1q2 r 2

If q1 and q2 have opposite charges, then when q2 is moved from its present location to a distance r from q1, the force on q2 due to q1 – (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4

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55

ELECTROSTATICS 5.

6.

7.

If q1 is positive then when q2 is moved from its present location to a distance r from q1, the magnitude of the voltage experienced by q2 due to E– (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 The strength of the electric field E at r is – (A) half the field strength at 2r (B) the same as the field strength at 2r (C) twice the field strength at 2r (D) four times the field strength at 2r If q1 and q2 are both positively charged and q2 is released, what is the maximum velocity that can be achieved by q2? (A)

8.

(B)

kq1q2 m2 r 2

(C)

2kq1q2 m2r 2

(D)

2kq1q2 m2 r

Electrostatic precipitators use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coal-burning power plants. One form of precipitator consists of a vertical hollow metal cylinder with a thin wire, insulated from the cylinder, running along its axis. A large potential difference is established between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong radial electric field directed inward, producing a region of ionized air near the wire. Smoke enters the precipitator at the bottom, ash and dust in the smoke pick up electrons, and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius of the central wire is 80.0  m, the radius of the cylinder is 12.0 cm, and the potential difference between the wire and the cylinder is 60.0 kV. Assume that the wire and cylinder are both very long in comparison to the cylinder radius.

(a) (b)

9.

kq1q2 m2 r

What is the electric-field magnitude midway between the wire and the cylinder wall? What magnitude of charge must a 30.0-  g ash particle have if the electric field computed in part (a) is to exert a force ten times the particle’s weight?

A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. The device consists of a thin wire on the axis of a hollow metal cylinder and insulated from it. A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong radial electric field directed outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced ar accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audiable “click.” Suppose the radius of the central wire is 50.0  m and the radius of the hollow cylinder is 2.00 cm. What potential difference between the wire and the cylinder is required to produce an electric field of 6.00 x 104 N/C at a distance of 1.50 cm from the wire?

(Assume that the wire and cylinder are both very long in comparison to the cylinder radius.

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56

ELECTROSTATICS Level # 2 1.

2.

3.

4.

R As shown a solid spherical region having a spherical cavity whose diameter R is equal to the radius of spherical region C P that has a total charge Q. Find the potential at a point P, which Q is at a distance ‘x’ from C. A spherical water drop of radius a has a charge Q spread uniformly over its face. The drop is split into two identical spherical droplets (each of which has charge Q/2 spread uniformly over its face), which are kept very far from one another. (a) Compute the change in the electrostatic potential energy caused by the splitting. (b) Repeat the above calculation for the case in which the charge is uniformly distributed in the drop volume, before and after the splitting. An electric dipole is placed at distance x from centre O on the axis of a charged ring of radius R and charge Q uniformly distributed over it. (a) Find the net force acting on the dipole. (b) What is the work done in rotating the dipole through 180°. (c) If the dipole is slightly rotated about its equilibrium position, find the time period of oscillation. Assume that the dipole is linearly restrained. P Particle 1 located far from particle 2 and possessing the kinetic energy T 0 and mass m 1 strikes particle 2 of mass m 2 through 1 the aiming parameter  , the arm of the momentum vector relative r P0 to particle 2 as; in the figure. Each particle carries a charge +q.  Find the smallest distance between the particles when m 1 < < m 2. 2

5.

6.

7. 8. 9.

10.

An electric field line emerges from a positive point charge +q1 at an angle  to the straight line connecting it to a negative point charge q2. At what angle  will the field enter the charge – q2 ?

+ q1



 q2

Two point charges each carrying a positive charge of 5e P (e being the magnitude of the electronic charge) are sepaR A B rated by a distance 2d. An electron describes a circular O path due to the attraction of the charges in a plane bised d 5e 5e cting perpendicularly the line joining the two point charges. If the radius of the circular path described by the electron is R, determine the orbital speed of the electron. Five thousand lines of forces enter a certain volume of space, and three thousand lines emerge from it. What is the total charge in coulombs with in the volume ? Determine the strength E of the electric field at the centre of a hemisphere produced by charges uniformly distributed with a density  over the surface of this hemisphere. Two small identical balls lying on a horizontal plane are connected by a 2 1 weight-less-spring. One ball (ball 2) is fixed and the other (ball 1 is free. – + The balls are charged identically as a result of which the spring length O increases  = 2 times. Determine the change in frequency.. A semi-circular ring of mass M and radius R with linear charge density  hinged at its centre is placed in a uniform electric field as shown in the figure. (i) Find the net force acting on the ring. (ii) It the ring is slightly rotated about O and released, find its time period of oscillation (iii) What is the work done by an external agency to rotate it through an angle .

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57

ELECTROSTATICS 11.

12.

13.

14.

15.

Determine the force F of interaction between two hemispheres of radius R touching each other along the equator if one hemisphere is uniformly charged with a surface density,  1 and the other with a surface density  2. y Suppose in an insulating medium, having di-electric constant k = 1, volume density of positive charge varies with y-coordinate to law  = ay.. A A particle of mass m having positive charge q is placed in the medium v0 at point A(0, y0) and projected with velocity v = v 0 . i and assuming electric field strength to be zero at y = 0, calculate slope of trajectory of the particle as a function of y. x o A metal ball of radius r and density  is charged by direct contact from the Earth’s surface till it acquires its maximum value. What should be this charge Q max on the ball and the charge Q on the Earth (assuming it to be uniform sphere of mass m and radius R) such that it may be launched from the Earth’s surface with zero launch velocity. A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r1 and r2 respectively. Calculate angular momentum L of this particle. Three concentric, conducting spherical shells A, B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charge q2 = 4  C and q3 = 3  C are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. B 

16.

The system consists of a hemispherical dielectric with volume charge density ‘  ’. Find the potential difference between points A and B.

17.

 r A space is filled up with volume density of charge    0 e where 0 and  are positive constants, r is the distance from centre of the system. Find the magnitude of the electric field strength vector as a function of r. Two identical balls are suspended from the same point by two threads. The balls are given equal charges and immersed in kerosene. Determine the density of the material of the balls if the threads deflect equally in vacuum and kerosene. The density of kerosene 0 = 0.8 g/cm 3 and its relative permittivity 0 = 2. A charge ‘q’ is placed on the surface of an originally uncharged soap bubble of radius R0. Due to the mutual repulsion of the charged surface, the radius is increased to a somewhat large value R. Show that q =

3

18.

19.

A

1/ 2

 32 2   .0 PR 0R (R 2  R R 0  R 02 )   3  

in which P is the atmospheric pressure.

Level # 3 1.

2.

Two fixed charges –2Q and Q are located at the points with coordinates (–3a, 0) and (+3a, 0) respectively in the x-y plane. (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Given the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole xaxis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. [IIT 91] A blank to be filled appears in each of the following statements. Write in your answer book the subquestion number and write down against it your answer corresponding to each blank. In your answer, the sequence of the sub-questions should be the same as given in the question paper. (i)

If  0 and 0 are respectively, the electric permittivity and magnetic permeability of free space,  and  the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is .....

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58

ELECTROSTATICS (ii) The electric potential V at any point x, y, z (all in meters) in space is given by V = 4x 2 volts. The electric field at the point (1m, 0, 2m) is ...... V/m. q q (iii) Five point charges, each of value +q C, are placed on five vertices of a regular hexagon of side L metre. The magnitude of the force on the point charge of value – q C placed at the centre of the hexagon is ..... newton. [IIT 92] 3.

4.

5.

6.

7.

8.

9.

q q

A circular ring of radius R with uniform positive charge density  per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P(R 3 , 0, 0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. Consider the classical-model of an atom such that a nucleus of charge +e is uniformly distributed within a sphere of radius 2 Å. An electron of charge (– e) at a radial distance 1 Å moves inside the sphere. Find the force attracting the electron to the centre of the sphere. Calculate the frequency with which the electron would oscillate about the centre of the sphere. [REE 95 A charge 10–9 coulomb is located at origin in free space and another charge Q at (2, 0, 0). If the x-component of the electric field at (3, 1, 1) is zero calculate the value of Q. Is the y-component zero at (3, 1, 1) ? [REE 95] –2 A radioactive source in the form of a metal sphere of radius 10 m, emits beta particles at the rate of 5 × 1010 particles per s. The source is electrically insulated. How long will it take for its potential to be raised by 2 volt assuming that 40% of the emitted beta particles escape the source ? [REE 97] A non-conducting disc of radius a and uniform positive surface charge density  is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q/m = 4  0g /  . [IIT 99] (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. A particle of charge q and mass m moves rectilinearly under the action of an electric field E = A – Bx where B is a + ve constant and x is a distance from the point where the particle was initially at rest. Calculate (a) Distance travelled by the particle till it comes to rest and (b) Acceleration at that moment. Four point charges + 8  C, – 1 C, – 1  C and + 8  C are fixed at the points 27 / 2 m ,  3 / 2 m ,  3 / 2 m and  27 / 2 m respectively on the y-axis. A particle of mass 6 × 10–4 kg and of charge + 0.1  C moves along the – x direction. Its speed at x = +  is v 0. Find the least value of v 0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given 1/( 4   0 ) = 9 × 109 Nm 2/C2.

10.

q

[IIT 2000]

Eight point charges are placed at the corners of a cube of edge a as shown in figure. Find the work done in disassembling this system of charges. [JEE 2003]

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59

ELECTROSTATICS

Answer Key Assertion-Reason Que.

1

2

3

4

5

6

7

8

9

10

Ans.

A

C

D

B

C

B

B

A

A

C

Level # 1 Q. 1 Ans. A Q. 11 Ans. A Q. 21 Ans. B Q. 31 Ans. D Q. 41 Ans. ABD

2 C 12 A 22 B 32 C 42 ABC

3 4 A C 13 14 B D 23 24 D C 33 34 D B 43 44 ABC ABCD

5 A 15 A 25 C 35 A 45 AC

6 A 16 A 26 B 36 D 46 ABC

7 A 17 B 27 C 37 A 47 BD

8 C 18 D 28 B 38 C

9 A 19 A 29 C 39 BC

10 B 20 B 30 C 40 BC

Fill in the blanks, True-False & Match the column 1. B

1 Q2  2. 180°, 4 0 4 L2

3. –qEa

7. T

8. T

10. B

9. F

4. –8

5.

1 qq 4 0 L2

6. F

Passage Type Que. Ans.

1 B

2 A

8. (a) 1.37 x 105 volt/m

3 A

4 D

5 C

6 D

7 D

(b) 2.15 x 10–11 C

Level # 2 1. V 

Q 7  0

2 1    x 2 R 2  4 x 2

  

1   2. (a) The fractional change = –  1  2 / 3   2 

(b) The fractional change in this case is also the same,.

 aqQ R 2  2x 2  aqQx 3. (a)  2 2 5/ 2  (b)  0 (R 2  x 2 )3/ 2  2 0 (R  x ) 

4. rmin 

1  q2 (1  1  (2 T0 /  q2 )2 ,   4  0 2 T0

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60

ELECTROSTATICS  5. sin  2    

q1 sin  / 2 q2

 8. 4  0

R2 1  2 2 0

mr1 r2 Q q 2   0 (r1  r2 )

14.



0 1  er 17.

3 0 r

3

2  0 (R  d2 )3 / 2

qa

12.

7. 17.7 nc

2

3n  2 times n

9.

11. F 

5

6. V  Re

3 m  0 v 20

10. (i) RE (ii) 2

( y 3  y 30 )

13. Q max  4r

2

N (iii) 4ER2 2E

 0 Rg  0 gR 3 , Q  4 r 3 3 16. V 

15. q1 = – 3  C, Energy = 9.45 J



 R 2 2 2 R 3  0 3 0

18. 1.6 gm/cm 3

2

Level # 3 Q 1. (a) radius = 4a, centre dt (5a, 0) (b) V  4 0

particle eventually crosses the circle, V 

2. (i)

 00

(ii) – 8 V/m (iii)

5. Q = – 4.27 × 10

6. 700 F

–10

9  10 9 q2 (N)

C, E y  

2  10 9 4   0 11 11

(c) At x = 9a where V = 0, the charged

9q 8   0 ma

3. V 

L2

7. (a) H =

 1 2      | x  3a | | x  a | 

q 2 0m

0

a 4a and H = 0 (b) 0  3 3 –4

9. Vmin = 3 m/s, K.E at origin = 2.5 × 10 J

4. 9 × 1014 Hz

8. x = 0, x =

2A qA ,a=  B n

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61

 1 q2  10. W = 5.824  4   . a  0  

—X—X—X—X—

CURRENT ELECTRICITY

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

ELECTRIC CURRENT THEORY OF ELECTRIC CURRENT CURRENT ELECTRICITY The time rate of flow of charge through any cross-section is called current. Therefore, if through a crosssection, q charge passes in time t, the average electric current through that area q i av  t and instantaneous current q dq i  lim   t  0 t dt Regarding electric current following points are worth noting : (a) Current is assumed to be a fundamental quantity in physics with unit ampere and dimension [A]. The CGS unit of current is emu of current and is called biot (Bi), i.e., 1C (1/10)emu of charge 1A = = s s 1 1A = Bi 10 (b) The current is same for all cross-sections of a conductor of non-uniform cross-section. Similar to water flow, charge flows faster where the conductor is smaller in cross-section and slower where the conductor is larger in cross-section, so that charge rate remains unchanged. (c) Though conventionally a direction is associated with current (opposite to the i1 motion of electrons) it is not a vector as the direction merely represents the sense of charge flow and not a true direction. Further current does not obey i = i1 + i2 i2 the law of parallelogram of vectors, i.e., if two currents i1 and i2 reach a point we always have i = i1 + i2 whatever be the angle between i1 and i2. (d) By convention, the direction of current is taken to be that in which positive charge moves and opposite to the direction of flow of negative charge. (e) As charge is conserved and current is the rate of flow of charge, the charge entering at one end per second of a conductor is equal to the charge leaving the other end per second. (f) Current in different situations is due to motion of different charge carries. Current in conductors and vacuum tubes is due to motion of electrons, in electrolytes due to motion of both positive and negative ions, in discharge tube due to motion of positive ions and negative electrons and in semiconductors due to motion of electrons and holes. Example 1. The current in a wire varies with time according to the relation A  i  (3.0 A)   2.0  t  s

(a) How many coulombs of charge pass a cross-section of the wire in the time interval between t = 0 and t = 4.0 s? (b) What constant current would transport the same charge in the same time interval ? dq i Sol. (a) dt  



q

0

4

dq   idt 0

4

q   (3  2t)dt 0

4

q  3t  t 2  0  12  16  28 C

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1

ELECTRIC CURRENT (b)

i

q 28  7A t 4

Types of Electric Current According to its magnitude and direction current is usually divided into two types : (i) Direct current : If the magnitude and direction of current does not vary with time, it is said to be direct current (DC). Cell, battery or DC dynamo are its sources. (ii) Alternating current : If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC). AC dynamo is the source of it. NOTE : It is worthy to note here that rectifier converts AC into DC while inverter DC into AC. Rectifier AC

DC Inverter

Current Density  Current density J at at a point is defined as a vector having magnitude equal to current per unit area surrounding that point and normal to the direction of charge flow and direction in which current passes through that point.  So, if at point P current I passes normally through area S as shown in Fig., current density J at P will be given by dS  + i  J  lim n J P n  S 0  S – i  di  J n i.e., dS Regarding current density following points are worth noting : (a) If the cross-sectional area is not normal to the current, the cross-sectional area normal to the current in accordance with Fig., will be dS dS  dS cos   i.e., or i.e.,

di dScos  di = JdS cos    di  J . d S   i   J . dS J

J

i dS cos

(b) In case of conductors as V = iR and by definitions, V V and R   L S L  EL  i S i 1 J  E S  E

So, i.e., or

  JE

 1       

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2

ELECTRIC CURRENT (c) In case of uniform charge flow through a cross-section normal to it as i = nqvS  i  J  n  (nqv) n So, S    or J  nq v  v ()charge [()charge = nq] Drift Velocity The average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through it is called drift velocity. It is represented by ‘vd’. The current flowing through conductor i i  neAv d or vd  neA where n = number of moving electrons per unit volume A = area of cross-section NOTE : Some books have taken average drift velocity as half of initial and final drift velocities of an electron giving vd = (eE/2m). This is wrong as vd is the average of drift velocities of large number of electrons at same instant and as for each electron       v d = a f with a =  e E / m  constant , v d av =  e E / m .





  



Example 2. An n-type silicon sample of width 4 × 10–3m, thickness 25 × 10–5 m and length 6 × 10–2 m carries a current of 4.8 mA when the voltage is applied across the length of the sample. What is the current density ? If the free electron density is 1022 m–3, then find how much time does it take for the electrons to travel the full length of the sample ? Given that charge on an electron e = 1.6 × 10–19 C. Sol. By definition, i i J  [as s = (b × d)] S  b  d 4.8  103  4.8  103 A / m 2 So, 25  10 5  4  103 and as in case of electric conduction in metals J = nevd J vd  or ne 4.8  103 v d  22  3 m/s So, 10  1.6  1019 Hence, time taken by electron to travel the length L (= 6 × 10–2 m) of the conductor J



 



L 6  102 t   0.02 s. vd 3 Example 3. The area of cross-section, length and density of a piece of a metal of atomic weight 60 are 10–6 m2, 1 m and 5 × 103 kg/m3 respectively. Find the number of free electrons per unit volume if every atom contributes one free electron. Also find the drift velocity of electrons in the metal when a current of 16A passes through it. Given that Avogadro’s number NA = 6 × 1023/mol and charge on an electron e = 1.6 × 10–19 C.

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3

ELECTRIC CURRENT Sol. As according to Avogadro’s hypothesis,

N m N m N d  so n   N A  A NA M V VM M n





6  1023  5  103

 60  10  3

 5  10

28

m  as d  V 

/ m3

Now as each atom contributes one electron, the number of electrons per unit volume ne = 1 × n = 1 × 5 × 1028 = 5 × 1028/m3 Further as here, i 16  6  16  106 A / m 2 A 10 J = nevd J

and as

vd 

J 16  106   2  103 m / s 28 19 ne 5  10  1.6  10



 



NOTE : • From this example it is clear that : An electron will take 1/(2 × 10–3), i.e., 500 s = (8.3 min.) to travel 1 m length of wire if it can. • If resistivity of metal  is taken to 2 × 10–8 ohm–m, the electric field inside the metal E = J = (2 × 10–8) × (16 × 106) = 0.32 V/m and not zero as in electrostatics. RESISTANCE AND OHM’S LAW   For some materials, especially metals, at a given temperature, J is nearly directly proportional to E and the ratio of the magnitudes E and J is constant. This ratio is called the resistivity () and this relationship is called the Ohm’s law. E J SI units of resistivity are –m (ohm–metre). The reciprocal of resistivity is conductivity ().

Thus, resistivity  

1  A material that obeys Ohm’s law reasonably well is called an ohmic conductor or a linear conductor. Materials which show substantial departures from Ohm’s law are called non-ohmic or non-linear. Thus,



Resistance Suppose a conducting wire has a uniform cross-sectional area A and length l as shown in Fig. Let V be the potential difference between the ends of the   wire. If the magnitudes of the current density J and the electric field E are uniform throughout the conductor, the total current i is given by i = JA and the potential difference V between the ends is V = El.  Here, 

l E A

i J

–

+ V

V El l   i JA A l is constant for ohmic materials. This is called the resistance R. A

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4

ELECTRIC CURRENT V i The resistance R of a particular conductor is related to the resistivity  of its material by l R  A The equation, V = iR is often called Ohm’s law.

Thus,

R

V defines resistance R for any conductor, whether or not it obeys ohm’s law, but i V only when R is some constant we can correctly call this relationship Ohm’s law. Thus, for ohmic conductors V–i group is a straight line possing through origin. The slope of this line is equal to the resistance of the conductor.

NOTE : The equation R =

V R = = tan   i Reciprocal of resistance is called conductance (G),

i.e.

i

1 i G= = r V

SI unit of G is ohm–1 which is called mho. Example 4. Two copper wires of the same length have got different diameters, which wire has : (a) greater resistance and (b) greater specific resistance ? Sol. (a) For a given wire, R 

l A

1 A So, the thinner wire will have greater resistance. (b) Specific resistance () is a material property. It does not depend on l or A. So, both the wires will have same specific resistance.

i.e.

R

Example 5. A wire has a resistance R. What will be its resistance if it is stretched to double its length ? Sol. Let V be the volume of wire, then V = Al A

V l

Substituting this in R  

l , we have A

l2 R  V So, for given volume and material, (i.e., V and  are constant) R  l2 When l is doubled, resistance will become four times, or the new resistance will be 4R.

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5

ELECTRIC CURRENT VARIATION OF RESISTANCE WITH TEMPERATURE R The resistance of a conductor varies with temperature. The graph of variation of R Slope = R 0 0 resistance of pure metal with temperature is shown. Mathematically the dependence of resistance R on temperature is expressed as T R (T) = R0[1 + (T – T0)] O T0 In this equation R(T) is the resistance at temperature T and R0 is the resistance at temperature T0, often taken to be 0ºC or 20ºC. The factor  is temperature coefficient of resistivity. Example 6. The resistance of thin silver wire is 1.0 at 20ºC. The wire is placed in a liquid bath and its resistance rises to 1.2 . What is the temperature of the bath ? for silver is 3.8 × 10–3 perº C. Sol. R (T) = R0[1 + (T – T0)] Here, R (T) = 1.2 , R0 = 1.0 = 3.8 × 10–3 perº C and T0 = 20ºC Substituting the values, we have 1.2 = 1.0[1 + 3.8 × 10–3(T – 20)] or 3.8 × 10–3 (T – 20) = 0.2 Solving this, we get T = 72.6º C Example 7. A resistance R of thermal coefficient of resistivity =  is connected in parallel with a resistance = 3R, having thermal coefficient of resistivity = 2. Find the value of eff . Sol. The equivalent resistance at 0ºC is

R0 

R10 R 20 R10  R 20

...(i)

The equivalent resistance at tºC is R

R 1R 2 R1  R 2

...(ii)

R1 = R10 (1 + t) ...(iii) R2 = R20(1 + 2t) ...(iv) and R = R0(1 + efft) ...(v) Putting the value of (i), (iii), (iv), (v) in eqn (ii), But

5  4 Example 8. (a) The current density across a cylindrical conductor of radius R varies according to the equation  eff 

r  J  J 0 1   , where r is the distance from the axis. Thus the current density is a maximum J at the axis 0  R

r = 0 and decreases linearly to zero at the surace r = R. Calculate the current in terms of J0 and the conductor’s cross sectional area is A = R2. (b) Suppose that instead the current density is a maximum J0 at the surace and decreases linearly to zero at the axis so that J  J 0

r . Calculate the current. R

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6

ELECTRIC CURRENT Sol. (a) We concider a hollow cylinder of radius r and thickness dr. The cross-sectional area of considered element is dA = 2rdr The current in considerd element is r  dI  JdA  J 0 1   2rdr  R

or

r  dI  2J 0 1   rdr  R



R r I  2J 0   1   rdr 0  R

I  J0

R 2 A  J0 3 3

I  2J 0 

(b)

R

0

2J 0 I R

I

r2 dr R R

 r3  3  0

2J0 R 3 2A  J R 3 3 0

Example 9. A network of resistance is constructed with R1 & R2 as shown in the figure. The potential at the points 1, 2, 3,......, N are V1, V2, V3,........, VN respectively each having a potential k time smaller than previous one. Find:

V0

R1 I

V V1 = k0

R1

I1

I2

R2

V2 =

V0 k2

VN–1 I´

R2

VN =

I´1 R1 R2

V0 kN

VN I´2 R3

R1 R2 (i) R and R in terms of k 2 3

(ii) current that passes through the resistance R2 nearest to the V0 in terms V0, k & R3. V V1 = k0

R1

V0

I

Sol.

R1

V2 =

V0 k2

I´

I1 R2

VN–1

R2

VN =

I´1 R1 R2

V0 kN

VN I´2 R3

(i) According to kcL, I = I1 + I2 or

or

V0 V0 V0 V0 0  2 k  k  k k R1 R2 R1

V0 

 k  1 V0  kR 1

V0  k  1 V0  kR 2 k 2 R1

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7

ELECTRIC CURRENT 

R 1  k  1  R2 k

2

Also, I´ = I´1 + I´2 V0 V V0 V0  N01 0 0 N2 N 1 N 1 k k k k or   R1 R2 R1  R 3 After solving, R2 k  R3 k 1 V0 0 V1  0 (ii) Here, I1   k R2 R2 V V0  k  1 V0 I1  0   kR 2 k2R 3  k  k R3   k  1 Example 10. A rod of length L and cross-section area A lies along the x-axis between x = 0 and x = L. The material obeys Ohm’s law and its resistivity varies along the rod according to (x) = 0 e–x/L. The end of the rod at x = 0 is at a potential V0 and it is zero at x = L. (a) Find the total resistance of the rod and the current in the wire. (b) Find the electric potential in the rod as a function of x. Sol. (a) The resistance of considered element is

dx 0  x / L  e dx A A L  L  L R  0  e  x / Ldx  0  e x / L  0 A 0 A L  L  1 R   0  e 1  1  0 1   A A  e

dR   

R

 (b)  or

V0  0 V0 AeV0   R R 0 L  e  1 E = J I

E  0 e  x / L J

E



dx

0 L  e  1  A  e 

I 0  x / L E =  Ie A A



x

e x / L V0 L 1  e 1



V0  Vx x V0 – Vx = Ex

,

E

0 e x / L AeV0 A0 L  e  1



E

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8

ELECTRIC CURRENT or

Vx  V0  Ex 

  1  e 

V0 e  x / L  e 1 1

ELECTRIC CELL An electric cell is a device which maintains a continuous flow of charge (or electric current) in a circuit by a chemical reaction. In an electric cell, there are two rods of different metals called as electrodes. These electrodes are kept in a solution called electrolyte. On joining two electrodes by a wire the charge begins to flow in the wire, i.e., current flows in the wire. Inside the cell, a chemical reaction takes place in the electrolyte which maintains the charge on the electrodes and the flow of charge in the wire is continuously maintained. Thus, a cell converts the chemical energy into electrical energy. In the light of modern views in reference to a cell following terms need to be reviewed. Electromotive Force (EMF) The emf of a cell is defined as work done by the cell in moving unit positive charge in the whole circuit including the cell once. Therefore, if W is the work done by a cell in moving a charge q once around a circuit including the cell,

W q SI unit of emf is joule/coulomb and is called volt. The emf of a cell in a circuit is taken to be positive, if circuit current inside a cell, is from negative to positive terminal, (i.e., cell is discharging) otherwise negative as shown in Fig. emf E 

i – + – + i E1 E2 E = E1 + E2 (a)

i – + + – i

i – + + – i

E1 E2 E = E1 – E2 (b)

E1 E2 E = E2 – E1 (c)

NOTE : The term electromotive force is misleading introduced by Volta who thought it to be force that causes the current to flow. Actually emf is not a force but work required to carry unit charge from lower potential to higher potential inside the cell. Internal Resistance (r) and Terminal Potential Difference The potential difference across a real source in a circuit is not equal to the emf of the cell. The reason is that charge moving through the electrolyte of the cell encounters resistance. We cell this the internal resistance of the source, denoted by r. As the current moves through r, it experiences an associated drops in potential equal to ‘ir’. Thus, when a current is drawn through a source, the potential difference between the terminal of the source is, V = E – ir This can also be shown as below : E A

r i

B

VA – E + ir = VB or VA – VB = E – ir Following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a battery), then V = E + ir (ii) V = E, if the current through the cell is zero

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9

ELECTRIC CURRENT (iii) V = 0, if the cell is short circuited. This is because current in the circuit E r or E = ir  E – ir = 0 or V=0 Thus, we can summarise it as follows : i

E

Short Circuited

E

r

r V = E – ir or V < E

i E

r V = E + ir or V > E

i E

r

E i= r

V=E

if i = 0

V=0

if short circuited

r

E

COMBINATION OF RESISTANCES In Series Fig., represent a circuit consisting of a source of emf and two resistors connected in series. i

A R1

A V1

V

R

V R2

V2 B

B

Let equivalent resistor is R as shown. Then R = R1 + R2 This result can be readily extended to a network consisting of ‘n’ resistors in series.  R = R1 + R2 + ....... + Rn In Parallel Fig., represents a circuit consisting of a source of emf and two resistors connected in paralle. i

A i1

V

R1

If R be the equivalent resistance, then

i

A

i2 R2

B

R

V

B

1 1 1   R R1 R 2

This result can be extended to a network consisting of nwww.physicsashok.in resistors in parallel. The result is

10

ELECTRIC CURRENT 1 1 1 1    ..........  R R1 R 2 Rn

NOTE : In series combination, the same current flows through each resistance while in parallel combination, the voltage drop across each resistor is equal to the source voltage V. Example 11. Compute the equivalent resistance of the network shown in Fig., and find the current i drawn from the battery. 18V i

i 6 3

4

Sol. The 6  and 3  resistances are in parallel. Their equivalent resistance is, 18V

i

4

2

1 1 1   or R  2 R 6 3 Now this 2  and 4  resistances are in series and their equivalent resistance is 4 + 2 = 6 . Therefore, equivalent resistance of the network = 6 . 18V

i

6

Current drawn from the battery is,

i=

net emf 18 = =3A net resistance 6

Star-Delta () Conversion For Fig. (a)  (Delta), to be equivalent to Fig. (b) (Star) Y, A A R1

RA R1

R3

B

A

C

RB

RC

B R2

(a)

(b)

RA 

RB C

R2

RA

R3

RC C

B (c)

R1R 3 R1R 2 , RB  R1  R 2  R 3 R1  R 2  R 3

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11

ELECTRIC CURRENT

RC 

and

R 2R 3 R1  R 2  R 3

Example 12. Find the equivalent resistance between A and B in Fig. using Star-Delta Theorem. 10

A

10

C

10

10

10 B

10

D

10

Sol. Using Star-Delta theorem, the equivalent circuit can be drawn as shown in fig. 10 A

C

10 10

D

RC = 5

15

O RD = 2.5

9

5

RB = 5 B

O

A

B

A

5 B

22.5

RC 

R CB  R CD 20  10  5 R CB  R CD  R DB 20  10  10

RD 

R DC  R DB 10  10   2.5  R DC  R DB  R CB 10  10  20

RB 

R BC  R BD 20  10  5 R BC  R BD  R DC 20  10  10

15  22.5 15  22.5  9 15  22.5 37.5 RAB = R´ + 5 = 9 + 5 = 14  R´

Hence,

Example 13. What will be the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure. Sol.

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12

ELECTRIC CURRENT B

r r

r A

r

r

r A

r

r

r

B

r r

r r 2r

A

B

r

r

2r

r

A

r

r

B

R2 = Req = 3r But before added the conductors, the equivalent resistance is R1  R´eq  5r



R2 3  R1 5 .

KIRCHHOFF’S LAW Many electric circuits cannot be reduced to simple series-parallel combinations. For example, two circuits that cannot be so broken down are shown in fig. R1

B

A E1

R1

R2 D

A

C E2

E1 E

F R3 (a)

I

C R2

B R3

E2

E R5 (b)

D R4

F G H

E3

However, it is always possible to analyze such circuits by applying two rules, devised by Kirchhoff. First here are two terms that we will use often. Junction : A junction in a circuit is a point where three or more conductors meet. Junctions are also called nodes or branch points. For example, in figure (a) points D and C are junctions. Similarly, in figure (b) point B and F are junctions. Loop : A loop is any closed conducting path. For example, in figure (a) ABCDA, DCEFD and ABEFA are loops. Similarly, in figure (b), CBFEC, BDGFB are loops. Kirchhoff’s rules consists of the following two statements. (i) Junction rule : The algebraic sum of the currents at any junction is zero. i2 i2 That is,



i0

i3

junction

i

4 This law can also be written as, “the sum of all the curents directed towards a point in circuit is equal to the sum of all the currents directed www.physicsashok.in away from that point.”

13

ELECTRIC CURRENT Thus, in figure i1 + i2 = i3 + i4 The junction rule is based on conservation of electric charge. (ii) Loop rule : The angebraic sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero. That is,



V  0

closed loop

The loop rule is based on the fact that the electrostatic field is conservative in nature. In applying the loop rule, we need sign convention as discussed below :

E B

A

Path

(a) When we travel through a source in the direction from – to +, the emf is considered to be positive.

(a) E

B

A

(b) When we travel form + to –, the emf is considered to be negative.

Path (b)

(c) When we travel through a resistor in the same direction as the assumed current, the iR term is negative because the current goes in the direction of decreasing potential.

R

A

i

B

Path (c)

(d) When we travel through a resistor in the direction opposite to the assumed current, the iR term is positive because this represents a rise of potential.

R

A

i

B

Path (d)

Example 14. Find current in different branches of the electric circuit shown in figure. A

4

2V F

2

B 4V

2

E

6V 4

Sol. Applying Kirchhoff’s first law (junction law) at junction B, i1 = i2 + i3 ...(i) Applying Kirchhoff’s second law in loop 1 (ABEFA), –4i1 + 4 – 2i1 + 2 = 0 ...(ii) Applying Kirchhoff’s second law in loop 2 (BCDEB), –2i3 – 6 – 4i3 – 4 = 0 ...(iii) Solving Eqs. (i), (ii) and (iii), we get i1 = 1A 8 i2  A 3

C

D

4 A

i1 1

2V F

2

B

C

i3 i2 4V

2

i3

i1 2

E

6V D

4

5 i3   A 3 Here, negative sign of i3 implies that current i3 is in opposite direction of what we have assumed.

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ELECTRIC CURRENT IMPORTANT FEATURES 1. Distribution of current in parallel connections : When more than one resistances are connected in parallel, the potential difference across them is equal and the current is distributed among them in inverse ratio of their resistance, as

or

i

V R

i

1 for same value of V R

R i1 i

i3

2R i2

3R

e.g., in the figure, 1 1 1 : :  6:3:2 R 2R 3R 6  6  i1   i i   6  3  2 11 i1 : i 2 : i3 

2.



3  3  i2   i i   6  3  2 11

and

2  2  i3   i i   6  3  2 11

Distribution of potential in series connections : When more than one resistances are connected in series, the current through them is same and the potential distributed in the ratio of their resistance, as V = iR or V  R for same value of i. e.g., in the figure, i

R

2R

3R

V1

V2

V3

V1 : V2 : V3 = R : 2R : 3R = 1 : 2 : 3 

V  1  V1   V   1  2  3 6 2  V  V2   V  1  2  3 3

and

3  V  V3   V   1  2  3 2

COMBINATION OF CELLS Cells are usually grouped in following three ways : In Series Suppose ‘n’ cells each of emf E and internal resistancer are connected in series as shown in figure. Then

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15

ELECTRIC CURRENT E

r

E

r

E

r

i R

Net emf = nE Total resistance = nr + R 

Current in the circuit =

or

i=

Net emf Total resistance

nE nr + R

NOTE : If palarity of m cells is reversed, then equivalent emf = (n – 2m)E while total resistance is still nr + R 

i=

 n - 2m E nr + R E

In Parallel : Consider the following three cases : Ist case : Let ‘n’ cells each of emf E and internal resistance ‘r’ are connected in parallel. Net emf = E r Total resistance   R n net emf  Current in the circuit i = total resistance E i= or R + r/n IInd case : Let ‘n’ cells have different E and ‘r’. Net emf = Eeq =

E E

A i1 B

1 R  1/ r 

Hence,

or

i

r i R

  E/r   1/ r 

E eq

r

i

Total resistance = Req

i

r

i3

i2

E1

r1

E2

r2

E3

r3

F

i

i C

E

R

D

R eq

  E / r 1  R  1/ r 

IIIrd case : This is most general case of parallel grouping in which E and ‘r’ of different cells are different and the positive terminals of few cells are connected to the negative terminals of the others as shown. Net emf = Eeq

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16

CURRENT E1ELECTRIC r 1



Total resistance

i1

 1 1 1  r  r  r  1 2 3

i3

i

i

E2

r2

E3

r3 i

i R

1  1 1 1  r  r  r  1 2 3

E eq R eq

 E1 / r1    E 2 / r2    E3 / r3   1 1 1 1 R      r1 r2 r3 

In Mixed Grouping There are ‘n’ identical cells in a row and number of rows are ‘m’. Net emf = nE Total resistance = Req

Hence,

i2

= Req R

Hence,

 E1 / r1    E 2 / r2    E3 / r3 

nr R m nE i nr R m

E

r

i

i R

Example 15. Find the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in figure. 10V 2 6V 1 4V

2

Sol. The given combination consists of two batteries in parallel and resultant of these two in series with the third one. For parallel combination we can apply, E1 E 2 10 4   r1 r2 Eeq   2 2 3V 1 1 1 1   r2 r2 2 2 1 1 1 1 1     1 Further, req r1 r2 2 2  req = 1  Now this is in series with the third one, i.e.,

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17

ELECTRIC CURRENT 6V

3V 1

1

The equivalent emf of these two is (6 – 3) V or 3 V and the internal resistance will be, (1 + 1) or 2 . E=3V r=2

n

n

(E)

O

n

I/A

(D)

O

I/A

(C)

O

I/A

(B)

O

I/A

(A)

I/A

Example 16. A battery consists of a variable number n of identical cells having internal resistance connected in series. The terminals of the battery are short circuited and the current I measured. Which one of the graph below shows the relationship between I and n ?

O

n

n

nE nr

Sol. I=

nE E = nr r

I n

O

n

I/A

(D)

(E)

O

O

n

(C)

I/A

I/A

(B)

O

O

n

n

I/A

(A)

I/A

Example 17. In previous problem, if the cell had been connected in parallel (instead of in series) which of the above graphs would have shown the relationship between total current I and n ?

n

E r n

Sol. I=

I=

E nE = r r n nE r

I n

Example 18. Under what condition current passing through the resistance R can be increased by short circuiting the battery of emf E2. The internal resistances of the two batteries are r1 and r2 respectively. (A) E2r1 > E1(R + r2) (B) E1r2 > E2(R + r1) (C) E2r2 > E1(R + r2) (D) E1r1 > E2(R + r1) Sol. The current through R before short circuit

E1

E2 r2

r1

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R

18

ELECTRIC CURRENT I=

E1 + E 2 r1 + r2 + R

After short circuit :

E2 r2

E1

I' =

E1 r1 + R

r1 R

I’ > I E1 E + E2 > 1 r1 + R r1 + r2 + R

E1r2 > E 2 r1 + E 2 R



WHEATSTONE’S BRIDGE This is an arrangement of four resitances in which one resistance is unknown and rest known. The Wheatstone’s bridge is shown in fig. The bridge is said to be balanced when deflection in galvanometer is zero, i.e., ig = 0 and hence, the condition of deflection is

B

i1

ig = 0

E 1r 1 + E 1r 2 + E 1R > E 1r 1 + E 1R + E 2r 1 + E 2R

P A

i1

Q

C

G R

i2

S

i2

i

P R  Q S

D

E

NOTE : In Wheatstone’s bridge, cell and galvanometer arms are interchangeable. B

B Q

P A

C

G R

S

Q

P A

C R

S

D

D G

In both the cases, condition of balance bridge is,

P R = Q S P

Example 19. A hemisphere network of radius a is made by using a conducting wire of resistance per unit length r. Find the equavalent resistance across OP. Sol. Point (A and C) and (D and B) are symmetrically located with respect to points O and P. Hence, the circuit can be drawn as shown in figure. This is a balanced Wheatstone bridge between P and O Hence, r3 can be removed. And, R PO

Here and

r r  1 2 4

r1  R PB  R PD  r2 = ROB = (a)r

B O

A D

P r1

r1

2

 a  r 2

B

C

2

r2 2

O

r2

C

2

r3

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19

ELECTRIC CURRENT

R PO 

 2   ar 8

Ans. 2

Example 20. In the circuit shown, what is the potential difference VPQ? (A) +3V (B) +2V (C) –2V (D) none Sol. In ABCPA –2I + 4 – I + 2 = 0 6 I = = 2A A 3 I VP – VQ = –{algebriac sum of rise up 2V and drop up of voltage} P = – (2 – 2I) VP - VQ = 2 V = – (2 – 4) 

Q 4V

2V

1V

P 1

2

2

I1

B I

3

Q I1

(I–I 1)

4V

3

1V I1

1

D

I C 2

a Example 21. Two batteries one of the emf 3V, internal resistance 1 ohm and the other of emf 15 V, internal resistance 2 ohm are connected in series with a resistance R as shown. If the potential difference between a and b is zero the resistance of R in ohm is (A) 5 (B) 7 (C) 3 (D) 1 Sol. According to loop rule, 3 – I – IR + 15 – 2I = 0 3V 15V a b 18 I= 3+ R  Va – Vb = – (–3 + I) R or 0=3–I I  I =3A 18 3=  3+ R or 9 + 3R = 18 9 R  3  3

R1 Example 22. Consider an infinite ladder network shown . A in figure A voltage V is applied between the points R2 A and B. This applied velue of voltage is halved after each section. B (A) R1/R2 = 1 (B) R1/R2 = 1/2 (C) R1/R2 = 2 Sol. I = I1 + I2 v v v v v– –0 – 2  2 2 4  R1 R2 R1 v v v = + or 2R 1 2R 2 4R1

R1

R1

R2

b

R

R1

R2

R2

R1 R2

(D) R1/R2 = 3

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20

or

1 1 v = + R1 R 2 2R 1

or

1 1 = 2R 1 R 2

or

R1 1 = R2 2

R 2 I1 R 2 v=0

v=0

R2

R2

R2

v=0

v=0

v=0

v=0

6

Example 23. If the switches S1, S2 and S3 in the figure are arranged such that current through the battery is minimum, find the voltage across points A and B. Sol. For minimum current through battery, equivalent resistance across battery should be maximum. As we know, in series resistance increases, but in parallel, resistance decreases. From these points of view, all switches should be open. 6

ELECTRIC CURRENT

I v/2 I2 v/4

v

6

3

6

A S2

S1

1 9

B

4.5

I

A

I

1 9 24 V

1

1

S3

24V

7

A

3

0.5 1

1

24 V I B

According to loop rule, 24 – 7I – 4.5I – 0.5I = 0  I = 2A  VAB = VA – VB = 0.5 × I = 0.5 × 2 = 1 V Example 24. In the circuit shown in figure, calculate the following : (i) Potential difference between points ‘a’ and ‘b’ when switch ‘S’ is open. (ii) Current through ‘S’ in the circuit when ‘S’ is closed. Sol. (i) Here

I1 

6

3 S

a 3

b 6

36  0 4A 9

36  0 4A 9 Also, 36 – Va = I1 × 6 = 6 I1 or 36 – Va = 6 × 4 = 24 V  Va = 36 – 24 = 12 V 36 – Vb = 3 × 4 = 12 V and Vb = 36 – 12 = 24 V  Va – Vb = 12 – 24 = –12 V (ii) In loop ABCIHA. –6(I – I1) + 3I1 = 0 or –6I + 6I1 + 3I1 = 0 or 9I1 = 6I

Also

36V

I2 

I1

V0 = 36V I2

6

3 a

b

3

6 V=0

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21

A I

B

9 3 I  I1  I1  ...(i) 6 2 In loop CDEJIC, –3(I – I1 + I2) + 6(I1 – I2) = 0 or –3I + 3I1 – 3I2 + 6I1 – 6I2 = 0 or –3I + 9I1 – 9I2 = 0 ...(ii) In loop EFGHIJE, 36 – 3I1 – 6(I1 – I2) = 0 or 9I1 – 6I2 = 36 ...(iii) After solving eqn (i), (ii) and (iii) I2 = 3A from ‘b’ to ‘a’.

I1

(I – I1)

6 C

(I – I1 + I2)

3

G

ELECTRIC CURRENT H

I2

3 I 6 (I1 – I2)

D

E

J F

Example 25. An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to I1. When he joins a second identical cell is series with the first, the current becomes I2. When the cells are connected are in parallel, the current through the circuit is I3. Show that relation between the current is 3 I3 I2 = 2I1(I2 + I3) Sol. Let the equivalent resistance of circuit is R. The emf and internal resistance of cell is E and r respectively. 

I1 

E rR

and

I2 

2E 2r  R

E r/2R  3E  2E L.H.S. = 3I3 I2   r / 2  R  2r  R  

and

I3 

6E 2 3I3 I 2   r / 2  R  2r  R  R.H.S. = 2I1 (I2 + I3) E   2E   2E      r  R   2r  R r / 2  R 



6E 2  r / 2  R  2r  R 

Hence, L.H.S. = R.H.S. Example 26. Find the potential difference VA – VB for the circuit shown in the figure. 1V

1

1V B

1

1

1V

1

1V

1

1

1

1

1

1V

1V

1V

1V

A

Sol.

i1 + i2 = i 4i + i1 = 0

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x+4 i1



4 x 9



4 22  VA  VB    2     V  9 9

4

x+3 4i

3

B i xELECTRIC +2 x+1 CURRENT x 3i 2i i2 i1 2

A

1

0

Example 27. Power generated across a uniform wire connected across a supply is H. If the wire is cut into n equal parts and all the parts are connected in parallel across the same supply, the total power generated in the wire is : (A)

H n2

(B) n2H

(C) nH

Sol. Let the resistance of wire is R = r

(D)

H n

 A

r = the resistance of each piece r R = = nA n v2 H= t ...(i) R when all pieces are corrected in parallel, Then equivalent resistance is r R r0 = = 2 n n v2 n 2 v2 H1 = t = t   r0 R

n2H

from eqn. (i) R

Example 28. The ratio of powers dissipatted respectively in R and 3R, as shown is : (A) 9 (B) 27/4 (C) 4/9 (D) 4/27 Sol. P1 = Power dissiatted in resistance R

3R 2R

2

and



4  2I     R  I2 R  3 9 P2 = Power dissipatted in 3R = I2(3R) = 3I2R

P1 4 I2R  P2 9  3 I2 R



R 2I/3 I

P1 4 = P2 27

3R I

2R I/3

y

Example 29. In the figure shown the power generated in y is maximum when y = 5. 10V, (A) 2  (B) 6   (C) 5  (D) 3  Sol. According to KOL : 10 – 2 I – I y – I R = 0 y 10 I= I  2+y+R 10V R  P = I 2y I 100 y I P= or (2 + y + R) 2 For maximum power dissipatted,

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R 2R

23

ELECTRIC CURRENT

dP 0 dy  By solving, y=2+R or

R  3



5=2+R

Example 30. Find the current through 25V cell & power supplied by 20V cell in the figure shown. 5V

10V

20V

30V 11

5

10

5

25V

Sol. From figure, 25V 5V

15V

30V

V=0

5V 5

5

I4

25V 25V 20V 30V

I3 V=0

25V

55V

25V

I1

I=I1+I2+I3+I4

11

25V 10V

10

55  0  5A 11 50 I2   1A 5 30  0 I3   3A 10 15  0 I4   3A 5  Electric current through 25V cell = I1 + I2 + I3 + I4 = 5 + 1 + 3 + 3 = 12 A The power supplied by 20V cell is P = – 20I2= –20 × 1 = – 20 W I1 

I2 V=0

V=0

V=0

4

Example 31. The current I through a rod of a certain metallic oxide is given by I = 0.2 V5/2, where V is the potential difference across it. The rod is connected in series with a resistance to a 6V battery of negligible internal resistance. What value should the series resistance have so that : (i) the current in the circuit is 0.44 (the value of (2.2)2/5 = 1.37) (ii) the power dissipated in the rod is twice that dissipated in the resistance. Sol. (i) the potential difference across rod for current 0.44 is 2

 0.44  5 V´   0.2  2

V   2.2 5 The potential difference across connected resistor is V´´ = 0.44 R E = V´ + V´´ 2

or

6   2.2 5  0.44R

or

0.44 R  6   2.2 5

2

2

6  0.2 5  R  0.44 0.44 R = 13.64 – 3.12 R = 10.52  (ii) Total power supplied by battery is used by rod and resistor  E I = V I + I2 R

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24

ELECTRIC CURRENT But  or or  Also,

V I = power dissipated in rod V I = 2I2R EI = 2I2R + I2R = 3I2R E = 3IR 6 = 3IR IR = 2 ...(i) V´ + V´´ = E 2

 I 5    IR  6 0.2  2

or

 I 5   26 0.2 

or

 I 5   4 0.2 

2

2  I  5  4  25  32 or   0.2 or I = 32 × 0.2 = 6.4 A From eqn (i), IR = 2

2 20   0.3125  6.4 64 Remarks : In the case of rod, V – I graph is not straight line. So, ohm’s law is not applicable in the case of the given rod. R



Example 32. A person decides to use his bath tub water to generate electric power to run a 40 watt bulb. The bath tub is located at a height of 10 m from the ground & it holds 200 litres of water. If we install a water driven wheel generator on the ground, at what rate should the water drain from the bath tub to light bulb ? How long can we keep the bulb on, if the bath tub was full initially. The efficiency of generator is 90%. (g = 10 m/s–2) Sol. Power  P  gh

dm dt dm 90  dt 100

or

40  gh

or

40  0.9 gh



dm 40 40 4    kg / sec. dt 0.9 gh 9  10 9



 dm  m t  dt  0 t0 



dm dt

m 200   450 sec. 4  dm    9 dt 

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25

ELECTRIC CURRENT 3

4

Example 33. The circuit shown in figure is made of a homogeneous wire of uniform cross-section. 1234 is a square. Find the ratio Q12/Q34 of the amounts of heat 1 2 liberated per unit time in conductor 1–2 and 3–4. Sol. Let us represent the central junction of wires in the form of two junctions connected by the wire 5–6 as shown in figure. then in follows from symmetry that there is no current through this wire. Therefore, the central junction can be removed from the initial circuit. Further, R12 = R13 = R34 = R24 = r and

R15  R 25  R 36  R 46 

3

r 2





4 6

5 1

V2 Q12  r Current through 3 – 4,

3

6

Let V be the voltage between 1 and 2. Then the amount of heat liberated in conductor 1 – 2.

i 3 4 

4

5 2

1

2

...(i) V

r



Q3 4  i



2 3

2 3 4

V2

r r

Q1 2  Q 3 4







23



2

2

2  3  11  6 2

Ans.

Metre Bridge The metre bridge is the practical application of the Wheatstone’s network principle. Q

P B G J

R

S

A l

D

(100 – l)

C

In such a bridge, the ratio of two resistances say R and S, can be determined from the ratio of their balancing lengths. In Fig., AC is a 1 m long uniform wire, Let AD = l cm, then DC = (100 – l) cm Since, resistance  length

P AD l   Q DC 100  l If P is known, then Q can be determined. 

Example 34. The potentiometer wire AB is 100 cm long. When AC = 40 cm, no deflection occurs in the galvanometer, find R.

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26

ELECTRIC CURRENT 10

R G

A

B

10 AC  R CB

Sol.

 CB   100  40   60  R  10   (10)   10    15     AC   40   40 



Example 35. A battery of emf 0 = 10 V is connected across a 1 m long uniform wire having resistance 10/m. Two cells of emf 1 = 2V and 2 = 4V having A internal resistances 1 and 5 respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point P, find the distance of point P from the point A. Sol. The resistance in AP is R1 = x = 10 x where x is in metre. The resistance in PB is R2 = (1 – x)10 The equivalent circuit is 10V I According to loop rule 10 – IR1 – IR2 – 10I = 0 R1 

10 10 1 I   R 1  R 2  10 10  10 2

But

VA – VP = E = IR1



14 1   10x 6 2



x

A

10 0 =10V

P

B

1 1 =2V

G 5

2 =4V

I

P

R2 B

I E1 E2 5 E=r r + r = 1 2 6

2 4 + = 14 V 1 5 6

G r=

5 1×5 = 1+5 6

14 2   0.4667 m  46.67 cm 6 10

Example 36. In the figure shown for which values of R1 and R2 the balance point for Jockey is at 40 cm from A. When R2 is shunted by a resistance of 10 , balance shifts to 50 cm. Find R1 and R2. (AB = 1 m) : Sol. Let resistance perunit length of potentiometer is . Assume that P is contact point of potentiometer.  The resistance in AP is R3 = x and the resistance in PB is R4 = (1 – x) (  AB = 1 m) According to balance condition of wheatstone bridge, R1R4 = R2R3 or R1(1 – x) = R2 x or R1(1 – x) = R2x ...(i) when R2 is shunted.

R1

R2 G

A

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B

27

ELECTRIC CURRENT R 2  10 10R 2  R 2  10 R 2  10

Then

R2 



R1(1 – x) = R´2 x R1 1  x  

10R 2 x R 2  10

...(ii)

From eqn. (i) and (ii), R1 

10  3

and

R2 = 5.

IMPORTANT FEATURES 1. When battery and galvanometer arms of a Wheatstone’s bridge are interchanged, the balance position remains undisturbed while sensitivity of bridge changes. R

I – I1 A

I

(I – I 1 + IG )

G

IG

I1

(I1 – I G)

2.

I

S

P

B

Q

Two other common forms of balanced Wheatstone’s bridge are shown. G

P

Q A

R

B S

POTENTIOMETER Potentiometer is an ideal device to measure the potential difference between two points. It consists of a long resistance wire AB of uniform cross-section in which a steady direct current is set up by means of a battery. E1 i

i

L A

i

B C

i2 = 0 G

E2 , r

Potential gradient

k

Potential difference across AB Total length

VAB L iR AB k  i L k

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28

ELECTRIC CURRENT R AB = resistance per unit length of potentiometer wire L The emf of source balanced between points B and C R E 2  kl  i CB  l l E2 = iRCB

where  

Applications (i) To find emf of an unknown battery : E1

E1 i l1

A

i

l2

B C1

A

i

B C2

i2 = 0 G

i2 = 0 G

EK

EU

We calibrate the device by replacing E2 by a source of known emf EK and then by unknown emf EU. Let the null points are obtained at lengths l1 and l2. Then, EK = i( l1) and EU = i( l2) Here, = resistance of wire AB per unit length. E K l1   E U l2

l  EU  1  EK  l2  So, by measuring the lengths l1 and l2, we can find the emf of an unknown battery. (ii) To find the internal resistance of a call : Firstly the emf E of the cell is balanced against a length AD = l1. For this the switch S´ is left opened and S is closed. A known resistance R is then connected to the cell as shown. The terminal voltage V is now balanced against a smaller length AD´ = l2. Here, no switch S Is opened and S´ is closed. Then, or

l1

E l1  V l2

Since, or or

E Rr  V R R  r l1  R l2

l  r   1  1 R  l2 

l2

S

E

{ E = i(R + r) and V = iR}

D´

A

D

B

(E, r) G

S´ R

Example 37. A potentiometer wire of length 100 cm has a resistance of 10 . It is connected in series with a resistance and a cell of emf 2V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of external resistance ?

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29

ELECTRIC CURRENT Sol. From the theory of potentiometer, VCB = E, if no current is drawn from the battery or Here,

 E1  R  R

AB

  R CB  E

E1 = 2V, RAB = 10   40  R CB    10  4   100 

E1 i R

B

A C

E G

and E = 10 × 10–3 V Substituting in above, we get R = 790  MOVING COIL GALVANOMETER Moving coil galvanometer is a device used to detect small current flowing in an electric circuit. With suitable modifications, it can be used to measure current and potential difference. Conversion of galvanometer into an Ammeter An ammeter is an instrument which is used to measure current in a circuit in ampere (or milli-ampere or microampere). Hence, it is always connected in series in the circuit. Since, the galvanometer coil has some resistance of its own, S therefore, to convert a galvanometer into an ammeter, its is resistance is to be decreased so, to convert a galvanometer G a b i into ammeter a low resistance, called shunt (S) is connected in parallel to the galvanometer as shown in figure. Ammeter S i  i g Here, S  G   S G  RA  and  G  S where RA = resistance of ammeter S = resistance of shunt G = resistance of galvanometer Conversion of Galvanometer into Voltmeter A Voltmeter is an instrument which is used to measure the potential difference between two points of an electric circuit directly in volt G ig (or milli-volt or micro-volt). Hence, it is connected in parallel across R those two points of the circuit, between which the potential difference Voltmeter is to be measured. When it is connected. Since, the resistance of coil of galvanometer of its own is low, hence, to convert a galvanometer into a voltmeter, high resistance R is series is connected with the galvanometer. V RG where R + G = RV = resistance of voltmeter

Here,

ig 

Example 38. A moving coil galvanometer of resistance 10  produces full scale deflection, when a current of 25

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30

ELECTRIC CURRENT mA is passed through it. Describe showing full calculations, how will you convert the galvanometer into a voltmeter reading upto 120 V. Sol. Here, G = 10 , ig = 25 mA = 25 × 10–3 A To convert the galvanometer into voltmeter reading upto 120 V : To convert a galvanometer into voltmeter of range V, a large resistance R has to be connected in series to it. The value of R is given by

R

V G ig

Here,

V = 120 V



R

120  10  4790  25  103 , 10 mA

Example 39. A milliammeter of range 10 mA and resistance 9 is joined in a circuit as shown. The metre gives full-scale deflection for current I when A and B are used as its terminals, i.e., current enters at A and leaves at B (C is left isolated). The value of I is (A) 100 mA (B) 900 mA (C) 1 A (D) 1.1 A Sol. According to loop rule, –9 × 10 – 0.9 × 10 + 0.1 × (I – 10) = 0 10 mA 90 + 9 I - 10 = = 990 or I–10 0.1 I 1A  I = 1000 mA 

A

C

B

10 mA

I

Example 40. In the circuit shown in figure reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed. The reading of voltmeter is V3 when both S1 and S2 are closed then (A) V2 > V1 > V3 (B) V3 > V2 > V1 (C) V3 > V1 > V2 (D) V1 > V2 > V3 Sol. Step-I — When S1 is closed, R  I 4R 3R  v1  I  3R   3R  4R 3 v1  4 Step-II — When S2 is closed R  I 6R 7R  v2  I  6 R  6 v2  6R  7R 7 Step-III — When S1 and S2 both are closed. R  I 2R 3R v3 = I × 2R

3R R

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S1

6R V

S2

E

31

ELECTRIC CURRENT v3 

 2R  3R

2 3

1000

Example 41. In the circuit shown in figure the reading of ammeter is the same with both switches open as with both closed. then find the resistance R. (ammeter is ideal) Sol. Step-I : Discuss the circuit when both switches open : According to loop rule : 1.5 – 300 I – 100 I – 50 I = 0

R



100  R  I

R From eqn (1) and (2)

100 A I 50 I

R

...(2)

I

100

A



300

100  R 

R R = 600 

1.5 V

C

I

D

no current

I1 R

G

F (I1 –I)

50 I1

A 300

or

I

300

B

100  R  I  IR  1.5 300

1.5V

300

I

I1 

50

+ –

1.5 15 1   A 450 4500 300 Step-II : Discuss the circuit after closing the switch. In loop ABCDEA –IR + 1.5 – 300I1 = 0 or 300I1 + IR = 1.5 ...(1) In loop BCGFB –100I + (I1 – I) R = 0 or (I1 – I)R = 100 I I1R = (100 + R)I



A

I1

1.5V

E

1 R   1.5 300 300

Example 42. The battery in the diagram is to be charged by the generator G. The + generator has a terminal voltage of 120 volts when the charging current is G – 10 amperes. The battery has an emf of 100 volts and an internal resistance of 1 ohm. In order to charge the battery at 10 amperes charging current, the resistance R should be set at (A) 0.1  (B) 0.5  (C) 1.0  (D) 5.0  R Sol. VA – VB = – {algebraic sum of rise up and drop up of voltage} A 120 = – {–IR – 1 × I – 100} 120 = IR + I + 100 I B I or 20 = 10R + 10 100V R 1 

R –

+

Example 43. A galvanometer having 50 divisions provided with a variable shunt ‘s’ is used to measure the current when connected in series with a resistance of 90 and a battery of internal resistance 10 . It is observed that when the shunt resistance are 10, 50, respectively the deflection are respectively 9 & 30 divisions. What is the resistance of the galvanometer ?

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32

ELECTRIC CURRENT Sol. The electric current through galvanometer is proportional to its deflection. I× Ig 9 3    I´g 30 10 IR S I  g But R R



g

I

Also,

100 

S



E R gRS Rg  RS

ER S  R g RS  100  R  R  R g  R S  g S RS  R g RS  100  R g  RS  R g  R S  Ig   R´S I´g  R g R´S  100  R  R´  R g  R´S  g S 10  10R g  100    R g  10  R g  10  3  10     50   R  50  50R g  g 100   R g  50   Rg = 233.3 . Ig 



or















or













IMPORTANT FEATURES 1. (a) The reading of an ammeter is always lesser than actual current in the circuit. i´ i

R

i´

R

E (a)

A

E (b)

S

G i´ (c)

For example, in Fig. (a), actual current through R is, E i ...(i) R  GS  while the current after connecting an ammeter of resistance A   in series with R is,  G  S  E i´ ...(ii) RA From Eqs. (i) and (ii), we see that i´ < i and i´ = i ,When A = 0, i.e., resistance of an ideal ammeter should be zero.

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33

ELECTRIC CURRENT (b) Percentage error in measuring a current through an ammeter is 1  1   i  i´  R R A    100     100 1 i   R  A  % error    100 or  R  A  2. (a) The reading of a voltmeter is always lesser than true value. r i

r i

i

RV V G

R

For example, if a current ‘i’ is passing through a resistance r, the actual value is, V = ir ...(i) Now if a voltmeter of resistance RV(= G + R) is connected across the resistance ‘r’, the new value will be i   rR V  V´ r  RV ir V´ r or ...(ii) 1 RV From Eqs. (i) and (ii), we can see that V´ < V and V´ = V if RV =  Thus, resistance of an ideal voltmeter should be infinite. (b) Percentage error in measuring the potential difference by a voltmeter is,        1 1   V  V´  100   100 % error       100   or V  1 r   1 r  RV  RV    RC CIRCUITS In preceding sections we dealt only with circuits in which the current did not vary with time. Here, were discuss about time-varying currents. q = CV 0

C

Charging of a Capacitor First we consider the charging of a capacitor without resistance. Consider a capacitor connected to a battery of emf V through a switch S. On closing the switch, there is no time tag between connecting and charging. A charge q0 = CV comes in the capacitor as soon as switch is closed and q – t graph in this case is a straight line parallel to t=axis. If we employed a resistance in the circuit, charging takes some time Fig. The q – t equation in this case, q = q0(1 – e–t/)

S

+ –

+ –

+ –

V

V

q q0

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r

34

ELECTRIC CURRENT C

Here, q0 = CV and  = CR = time constant S q – t graph is an exponentially increasing graph. The charge q increases q q0 expponentially from 0 to q0. 0.632q 0 From the graph and equation we see that, At t = 0, q=0 and at t = , q = q0 At t = , q = q0(1 – e–1) = 0.632 q0 Here,  can be defined as the time in which 63.2% charging is over in a C – R circuit. Current flows in a C – R circuit during charging of capacitor. Once charging i as over or the steady state condition is reached the current becomes zero. i0 The current at any time t can be calculated bay differentiating q with respect to t. Hence, Charging current is, i = i0e–t/ i.e., current decreases exponentially with time. Time i – t graph is as shown in fig.

C

V

t

t

Discharging of a Capacitor Again we consider the discharging of capacitor with resistance. Suppose a capacitor has a charge q0. The positive plate has a charge +q0 and negative plate –q0. q0

q=0

+ –

q0

R

+ –

S

When the switch is closed, the extra electrons on negative plate immediately comes to the positive plate and net charge on both plates become zero. So, we can say that discharging takes place immediately. In case of C – R circuit discharging also takes time. The q – t equation in this case is,

C

S

q  q 0e  t / c

Thus, q decreases exponentially from q0 to zero, as shown in Fig. q At t = 0, q = q0 q0 At t = , q=0 In case of discharging definition of  is charged. 0.368q 0 At time t = , t q = q0 e–1 = 0.368q0 Hence, in this case  can be defined as the time when charge reduces to 36.8% of its maximum value q0. During discharging current flows in the circuit till ‘q’ becomes zero. This current can be found by differentiating ‘q’ with respect to ‘t’ but with negative sign because charge is decreasing with time. Hence, discharging current is, i i = i0e–t/ i0 This is an exponentially decreasing equation. Thus, i – t graph decreases exponentially with time from i0 to 0. The i – t graph is as shown in Fig. t

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35

ELECTRIC CURRENT C

Example 44. A capacitor C = 100 µF is connected to three resistor each of resistance 1 k and a battery of emf 9V. The switch S has been closed for long time so as to charge the capacitor. When switch S is opened, the capacitor discharges with time constant (A) 33 ms (B) 5 ms (C) 3.3 ms (D) 50 ms Sol.

1k S

1k 1k

9V

1k

1k

at t = 0

C

R = 0.5 k

 Time constant  = RC = 0.5 × 103 × 100 × 10–6 = 50 × 10–3 S = 50 m/s Example 45. In the circuit shown in figure C1 = 2C2. Switch S is closed at time t = 0. Let i1 and i2 be the currents flowing through C1 and C2 at any time ‘t’, then the ratio i1/i2 (A) is constant (B) increases with increase in time t (C) decreases with increase in time t (D) first increases then decreases

R C1 R C2 S V

t

Sol. Here

V I1  e RC1 R t

and

I2 

V RC 2 e R t RC1

t 1

1 



 I1 e R C C   t  e  1 2  I2 e RC 2



I1 e I2

 t  C1  C2  RC1C2

Hence option (B) is correct. Example 46. In the R–C circuit shown in the figure the total energy of 3.6 × 10–3 J is dissipated in the 10  resistor when the switch S is closed. The initial charge on the capacitor is

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S 2µF

10

36

ELECTRIC CURRENT (A) 60 µC

(C) 60 2 µC Sol. According to conservation principle of energy : Total energy stored on capacitor appears as heat in resistor.  or



(B) 120 µC

60 µC 2

(D)

q 20  H  3.6  103 2C q02 = 2C × 3.6 × 10–3 q02 = 2 × 2 × 10–6 × 3.6 × 10–3 q02 = 14.4 × 10–9 q0 = 12 × 10–5 q0 = 120 µC

+ – C

Example 47. The capacitors shown in figure has been charged to a potential difference of V volts, so that it carries a charge CV with both the switches S1 and S2 remaining open. Switch S1 is closed at t = 0. At t = R1C switch S1 is opened and S2 is closed. Find the charge on the capacitor at t = 2 R1C + R2C. Sol. When t < R1C q  CV e

At

S1 R2

t0 = R1C, q0

CV e

–q0 R1

when S2 is closed, t = t0 At instant t (t > t0)

R2 E

q  I  R1  R 2   0 C (R1 + R2)CI = EC – q dq  R1  R 2  C dt  EC  q q t dq  R1  R 2  CCV EC  q  t0 R1C dt e E

or or

S2

E

t R 1C

q 0  CV e 1 

or

R1

I +q

–q R1+R2

I

E

q

or

t  R1  R 2  C   ln  EC  q   CV   t R C e



or

 R 1  R 2  C ln  EC  

1

CV     ln  EC  q    t  R 1C e 

CV   EC   e   tR C R1  R 2  C  ln   1 or EC  q     But t = 2R1C + R2C Putting the value,

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37

ELECTRIC CURRENT  1  CV q  CE 1    2  e e

Example 48. In the figure shown initially switch is open for a long time. Now the switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time given that it was initially unchanged. Sol. Step-I : Discuss the circuit, when switch is open. I

Here

or or or or

or

C

C

V 2R I I

I

V R q I  1  0 2 2 C

or

R

V

CV q 0  CIR  Also, 2 Step-II : Discuss the circuit after closing the switch :

Also,

S

R

+q –q0

R R V

V R  q  q  q1  I  0 0 2 2 C

V R q q q I  0   1  0 2 2 C C C V R q q V IR I  0    0 2 2 C C 2 2 q q V  IR  0   0 C C q  q IR  V  0 C dq  q  q R  V 0  C  dt q t Rdq  0  q0  q  0 dt V  C 

q R 2

V 2

q 0+q –q 1

–(q0+q–q1)

q1 –q1

q

q

or

or

   q  q    RC ln V   0  t  C   0     q0  q    V   C   t RC  ln q0   V   C q0  q t C  e  RC q V 0 C

V or

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38

ELECTRIC CURRENT t q 0  RC  q0  q   V   V  e or     C   C But potential differ accross each capacitors are sume.

or 

...(i)

q 0  q  q1 q1  C C q0 + q = 2q1

q0  q 2 Putting the value of q, q1 

q1 

t  CV  1 RC 1  e   2  2 

THINKING PROBLEMS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

13. 14.

15. 16. 17.

Is a current-carrying conductor electrically charged? Is there an electric field near the surface of a conductor carrying direct current? Is current a scallar or vector? A potential difference V is applied to a copper wire of diameter d and length l. What is the effect on the electron drift velocity of (a) doubling l, (b) doubling V, and (c) doubling d? A current i enters the top of a copper sphere of radius R and leaves through the diametrically opposite point. Are all parts equally effective in dissipating Joule heat? Account for the increase in the resistance of metals with rise in temperature. Answer briefly — how can three resistances of values 2, 3 and 6 be connected to produce an effective resistance of 4 ? How can an electric heater designed for 220V be adopted for 110V without changing the length of the coil and also without a change in the consumed power ? The brilliance of lamps in a room noticeably drops as soon as a highpower electric iron is switched on and after a short interval, the bulbs regain their original brilliance. Explain. A current is passed through a steel wire which gets heated to a dull red. Then half the wire is immersed in cold water. The portion out of the water becomes brighter. Why? For manual control of the current of a circuit, two rheostats in parallel are preferable to a single rheostat. Why? In a hollow nonconducting pipe, there are two streams of ions in opposite directions. The ions of one stream are negatively charged and constitute a current of strength l and those of the other stream are positively charged and constitute a current of the same strength. What is the total current through the pipe? The drift velocity of electrons is quite small. How then does a bulb light up as soon as the switch is turned on, although the bulb may be quite far from the switch? Consider a voltaic cell in the open circuit with the copper plate at a higher potential with respect to the zinc plate. The electrolyte between the plates is a conducting medium. So the charges on the plates should be neutralized at once as it happens when two charged conductors are joined by a wire. Why are the charges not neutralized immediately? Does emf have electrostatic origin? Why is it easier to start a car engine on a warm day than on a chilly day? The resistance of the human body is about 10 k. If the resistance of our body is so large, why does one

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39

ELECTRIC CURRENT experience a strong shock from a live wire of 220 V supply? 18. Lay people have the notion that a person touching a high power line gets ‘stuck’ to the line. Is this true? If not, what is the fact? 19. When a direct current flows through a conductor, the amount of energy liberated is VQ, where Q is the charge passing through the conductor and V is the potential difference, while an energy of VQ/2 is liberated when a capacitor is discharged. Why? 20. An ordinary cell with a small emf can produce larger current than an electrostatic machine which generates thousands of volts? Explain. THINKING PROBLEMS SOLUTIONS

1. No, the amount of positive and negative charge in any elementary volume remains the same through the electrons are in motion. So a current-carrying conductor is electrically neutral. 2. There is an electric field inside the conductor. This is equal to the rate of fall of potential along the conductor. There is continuity of this field outside the surface of the conductor. 3. Current is a scalar but current density is a vector. 4. u = j/ne = E/ne where s is the conductivity and E is the electric field (by Ohm’s law j = E), or u = (/ne) (V/l) ( E = V/l). Obviously from this expression (a) u is halved if l is doubled, (b) u is doubled if V is doubled, and (c) u remains uncharged on doubling d. 5. No, the resistance of elements at right angles to the diameter varies from element to element but current is the same through all sections and so heating effect varies from section to section. 6. metals have mobile electrons. With increase of temperature, the lattice vibrations increase in amplitude. Thus, the probability of electrons striking ionic cores increases. This amounts to increase in resistance with temperature. 7. Connect the 3 and 6 resistors in parallel and the 2 resistor in series with the combination of 3 and 6. Then 3 and 6 in parallel will sum up to 8.

9.

10. 11.

12. 13.

14.

3 6  2 and this, with 2 in series, will work out to 36

4. Join the ends of the coil and apply the supply voltage (110V) between this common terminal and the midpoint of the coil as shown in the figure. Let P be the power of the entire coil when the coltage applied is V volts. T he P = V 2/R where R is the resistance of the coil. When it is connected as shown in the figure, the two parts are in parallel and the resistance of each part becomes R/2. Therefore, P1 = (V/2)2/(R/2)=P/2 Total power = P/2 + P/2 = P The cold resistance of the coil of the iron is much smaller than when it is hot. So when the iron is switched on, there is a large drop of voltage and consequently the bulbs do not receive the proper voltage. As the coil gets heated, its resistance increases and so the voltage drop is made up when it is fully heated. This is why after some time the bulbs region their brilliance. The resistance of the immersed portion decreases and so the current through the entire wire increases. This is why the portion outside the water becomes brighter. Suppose, to produce a certain current i in a circuit, the length of the single rheostat wire required is l. Then i = E/lr where E is the emf of the cell and r is the resistance per unit length of the wire. To produce the same current in the same circuit by using two rheostats in parallel, let l’ be the length of each rheostat wire. Then i = E/(l’r/2) = 2E/l’r. Obviously, l’ = 2l. Thus more length of each wire is required to produce the same current. This is definitely of advantage. The current constituted by the negative ions moving opposite to the positive ions has the same sense as the current constituted by the positive ions. Hence the total current is 2I and not zero. For the bulb to light up, it is not necessary for the same electron to travel from the switch to the bulb. When the switch is turned on, every electron in the circuit begins to move simultaneoulsy, including those www.physicsashok.in 40 in the filament of the bulb. The charges are prevented from being neutralized because the electrostatic field due to the charges on the plates is opposed by the charges of ions that migrate into the solution.

ELECTRIC CURRENT

REASONING TYPE QUESTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as – (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1.

Assertion (A) : Reason (R) :

When a wire is stretched so that its thickness is halved, its resistance would become 16 times. The data is insufficient to predict.

2.

Assertion (A) : Reason (R) :

A current flows in a conductor only when there is an electric field within the conductor. The drift velocity of electron in presence of electric field decreases.

3.

Assertion (A) : Reason (R) :

A current carrying wire should be changed. The current in a wire is due to flow of free electrons in a definite direction.

4.

Assertion (A) : Reason (R) :

The wires supplying current to an electric heater are not heated appreciably. Resistance of connected wires is very small and H  R .

5.

Assertion (A) : Reason (R) :

In meter bridge experiment, a high resistance is connected in series with the galvanometer. As resistance increases current through the circuit increases.

6.

Assertion (A) :

A 60 watt bulb has greater resistance than a 100 watt bulb.

Reason (R) :

V2 P R

7.

Assertion (A) : Reason (R) :

The conductivity of an electrolyte is very low as compared to a metal at room temperature. The number density of free ions in electrolyte is much smaller compared to number density of free electrons in metals. Further, ions drift much more slowly, being heavier.

8.

Assertion (A) : Reason (R) :

Terminal voltage of a cell is greater than emf of cell, during charging of the cell. The emf of a cell is always greater than its terminal voltage.

9.

Assertion (A) : Reason (R) :

Material used in the construction of a standard resistance is constantan or manganin. Temperature coefficient of these materials is very small.

10. Assertion (A) : Reason (R) :

If the current of a lamp decreases by 20%, the percentage decrease in the illumination of the lamp is 40%. Illumination of the lamp is directly proportional tot he current through lamp.

11. Assertion (A) : Reason (R) :

Heater wire must have high resistance and high melting point. If resistance is high, the electrical conductivity will be less.

12. Assertion (A) : Reason (R) :

The range of given voltmeter can be increased. By adjusting the value of resistance in series with galvanometer the range of voltameter can be adjusted.

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41

ELECTRIC CURRENT

LEVEL # 1 1.

The resistance of tungsten increases with increasing temperature. As a result, the relation between the current,  , flowing in the tungsten filament of an electric lamp and the potential difference, V, between itss ends is of the form.

I (A)

(B)

I

(C)

(D)

O

O O V V V V The current in a copper wire is increased by increasing the potential difference between its ends. Which one of the following statements regarding n, the number of charge carriers per unit volume in the wire, and Vd , the drift velocity of the charge carriers, is correct? (A) n is unaltered but v d is decreased. (B) n is unaltered but v d is increased. (C) n is increased but v d is decreased. (D) n is increased but v d is unaltered.

O

2.

I

I

3.

A potentiometer is to be calibrated with a standard cell using the circuit shown in the diagram. The balance point is found to be near L. To improve accuracy the balance point should be nearer M. This may be achieved by (A) replacing the galvanometer with one of lower resistance. (B) replacing the potentiometer wire one of higher resistance per unit length. (C) putting a shunt resistance in parallel with the galvanometer. (D) increasing the resistance R.

4.

When the switch 1 is closed, the current through the 8  resistance is 0.75 A. When the switch 2 is closed (only), the current through the 2  resistance is 1 A. The value of  is (A) 5 V

(B) 5 2 V

(C) 10 V

(D) 15 V R

5.

The time constant



(A) RC

(B)

for the shown RC circuit is

RC 2

(C) 2 RC

(D) not defined

R



6.

Two cells of equal emf of 10 V but different internal resistances 3  and 2  are connected in series to an external resistance R. The value of R that makes the potential difference zero across the terminals of one of any cells is (A) 5  (B) 6  (C) 1  (D) 1.5 

7.

The resistor in which the maximum heat is produced is given by (A) 2  (B) 3  (C) 4  (D) 12 

8.

n resistances each of resistance R are joined with capacitors of capacity C (each) and a battery of emf E as shown in the figure. In steady state condition ratio of charge stored in the first and last capacitor is E (A) n : 1 (B) (n–1) : R 2 2 (C) (n + 1) : (n – 1) (D) 1 : 1

R

C

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R

C

R

42

ELECTRIC CURRENT 9.

What is the equivalent capacitance between A and B in the circuit shown. (A) 6  F (B) 1.5  F (C) Zero (D) 2  F

10. A resistance R = 12  is connected across a source of emf as shown in the figure. Its emf changes with time as shown in the graph. What is the heat developed in the resistance in the first four second.



R = 12 

(Volt) 24

 Source

4 (A) 72 J

(B) 64 J

(C) 108 J

t(s)

(D) 100 J

11. A source of constant potential difference is connected across a A conductor having irregular cross-section as shown in the Figure. Then P Q (A) electric field intensity at P is greater than that at Q (B) rate of electric crossing per unit area of cross section at P is less than that at Q (C) the rate of generation of heat per unit length at P is greater than that at Q (D) mean kinetic energy if free electorn at P is greater than that at Q

B

12. Suppose a voltmeter reads the voltage of a very old cell to be 1.40 V while a potentiometer reads its voltage to be 1.55 V. What is the internal resistance of the cell (A) 20  (B) 30  (C) 10  (D) 40  13. In the above question, What is the current it would supply to a 5  resistor. Assume the voltmeter resistance be be 280  . (A) 44 A (B) 0.044 A (C) 4.4 A (D) None of the above. 14. If

 1 ,  2 and  3 are conductances of three conductors then their equivalent conductance when they are

joined in series will be (A)

1 +  2 +  3

(B)

1 1 1   1  2  3

(C)

 1 2 3 1   2   3

(D) None of these

15. A conductor is made of an isotropic material (resistivity  ) has rectangular cross-section. Horizontally dimension of the rectangle decreases linearly from 2x at one end to x at the other end and vertial dimension increases from y to 2y as shown in Figure. Length of the conductor along the axis is equal to  . A battery is connected across this conductor then (A) resistance of the conductor is equal to

4 9xy .

(B) rate of generation of heat per unit length is maximum at middle cross-section. (C) drift velocity of conduction electrons is minimum at middle section. (D) at the ends of the conductor, electric field intensity is same. 16. Find the current through 4  resistor just after making the circuit (A) 0 A (B) 6 A (C) 12 A (D) 2 A 17. 1 m long metallic wire is broken into two unequal parts P and Q. P part of the wire is uniformly extended into another wire R. Length of R is twice the length of P and the resistance R is equal to that of Q. Find the ratio of the resistance of P and R (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 1 : 1

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43

ELECTRIC CURRENT 18. Which of the following statements is/are correct for potentiometer circuit (A) Sensitivity varies inversely with length of the potentiometer wire (B) Sensitivity is directly proportional to potential difference applied across the potentiometer wire. (C) Accuracy of a potentiometer can be increased only by increasing length of the wire (D) Range depends upon the potential difference applied across the potentiometer wire. 19. In the given circuit the ammeter reading is zero. What is the value of resistance R ? (A) R = 100  (B) R = 10  (C) R = 0.1  (D) None of these

20. What is the equivalent resistance between A and B in the given circuit diagram. (A) 2  (B) 12  (C) 20  (D) 10  21. A cell of emf. 1.5 V and internal resistance 0.5  is connected to a (non-linear) conductor whose  -V graph is shown in Figure. Find the current drawn from the cell and its terminal voltage

(A) 1.5 A and 2 V

(B) 1 A and 1 V

(C) 1 A and 2 V

(D) 2 A and 1.5 V

22.

You are given several identical resistances each of value R = 10  and each capable of carrying a maximum current of 1A. It is required to make a suitable combination of these resistances to produce a resistance of 5  which can carry a current of 4 A. The minimum number of resistances required is (A) 4 (B) 10 (C) 8 (D) 20

23.

A conductor or area of cross section A having charge carriers, each having a charge q is subjected to a potential V. the number density of charge carriers in the conductor is n and the charge carriers (along with their random motion) are moving with a velocity v. If  is the conductivity of the conductor and  is the average relaxation time, then (A)  

24.

2

nq 

(B)  

m nq

(C)  

2

2 m nq

2

(D)  

m 2 nq 2

A vacuum diode consists of plane parallel electrodes separated by a distance d and each having an area A. On applying a potential V to the anode with respect to the cathode a current I flows through the diode. Assume that the electrons are emitted with zero velocity and they do not change the field between the electrodes. The electron velocity is v and the charge density is  at any point between the electrodes at a distance x from the cathode. If I is the equivalent current, m is the mass of each charge carrier, then (A) v 

25.

m

2 eVx md

(B)  

md 2 e VxA

2

(C) v 

eVx 2 md

2 md

(D)  

e VxA 2

A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the  radial distance r from the axis of the conductor as   2 ,  is a constant. The total resistance per unit r length of the conductor is R and the electric field strength in the conductor due to which a current I flow in it is  . (A) R 

2 A

2

(B) R 

4 A

2

(C)  

2  I A

2

(D)  

4I

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A2

44

ELECTRIC CURRENT

26.

A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. Now W 1, W 2 and W 3 are the output powers of the bulbs B1, B2 and B3 respectively. Then (A) W 1 > W 2 = W 3 (B) W 1 > W 2 > W 3 (C) W 1 < W 2 = W 3 (D) W 1 < W 2 < W 3 [JEE’ 2002 (Scr)]

27.

Variation of current passing through a conductor as the voltage applied across its ends is varied as shown in Fig. 9.55. If the resistance is determined at points A, B, C and D. We will find that : (A) Resistance at C and D are equal (B) Resistance at B is higher than at A (C) Resistance at C is higher than at B (D) Resistance at A is lower than at B

28.

The P. D. between the points A and B in the circuit shown here is 16 V. Which is/are the correct statement(s) out of the following ? 9V 3V (A) the current through the 2  resistor is 3.5 A 3 4 1 1 A B (B) the current through the 4  resistor is 2.5 A 2 (C) the current through the 3  resistor is 1.5 A (D) the P.D. between the terminals of the 9 V battery is 7 V.

29.

A and B are two points on a uniform ring of resistance R. The  ACB =  , where C is the centre of the ring. The equivalent resistance between A and B is R

(A) 30.

( 2   ) 

   (B) R 1  2    

 (C) R 2 

a3 R 6b

(B)

a3 R 3b

(C)

a3 R 2b

2   4

(D)

a3 R b

A resistance R carries a current I . The heat loss to the surroundings is  (T–T 0) where  is a constant, T is the temperature of the resistance, and T 0 is the temperature of the atmosphere. If the coefficient of linear expansion is , the strain in the resistance is (A) proportional to the length of the resistance wire (B) equal to (C) equal to

32.

(D) R

The charge flowing through a resistance R varies with time t as Q = at – bt 2. The total heat produced in R is (A)

31.

4 2

1 2 I R 2 

 2 I R 

(D) equal to   (IR)

The potential of the point O is (A) 3.5 V (B) 6.5 V (C) 3 V (D) 6 V

33. In the circuit shown in Figure the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor is (A) 1 calorie/sec (B) 2 calories/sec (C) 3 calories/sec (D) calories/sec [JEE’ 1981]

6

4

5

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45

ELECTRIC CURRENT 34. A battery of internal resistance 4  is connected to the network of resistance as shown. In order that the maximum power can be delivered to the network, the value of R in  should be [JEE’1995] (A)

4 9

(B) 2

(C)

8 3

E

R R

R 6R

R

4

4R

R

(D) 18

35. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is/are [JEE’1997] (A) current, electric field and drift speed. (B) drift speed only (C) current and drift speed (D) current only. Q P 36. In the circuit P  R, the reading of the galvanometer is S same with switch S open or closed. Then [JEE’1999] R (A)  R   G (B)  P   G G (C)  Q   G

(D)  Q   R

v 37. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount T in the same time t, the value of N is [JEE’2001] (A) 4 (B) 6 (C) 8 (D) 9

2R

38. The effective resistance between points P and Q of the electrical circuit shown in the figure is [JEE’2002]

2Rr (A) Rr

8R  R  r  (B) 3R  r

(C) 2 r + 4 R

5R  2r (D) 2

2R 2R

P

r 2R

2R

r

Q

2R

39. Express which of the following set ups can be used to verify Ohm’s law?

A V

(A)

(B)

A

V

A (C)

V

(D)

V

A

40. The three resistance of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation. [JEE’2003] (I) a (II) b (III) c (IV) c (A)

i

(B)

i

(C)

i

(D) i

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46

ELECTRIC CURRENT 41. In the given circuit, no current is passing through the galvanometer. If the crosssectional diameter of AB is doubled then for null point of galvanometer the value of AC would [JEE’2003 (Scr)] (A) x (B) x/2 (C) 2x (D) None

G A

x

B

C

P

42. Six equal resitances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum between [JEE’2004 (Scr)] (A) P and Q (B) Q and R (C) P and R (D) any two points

Q

R

B

43. For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between [JEE’2004 (Scr)] (A) B and C (B) C and D (C) A and D (D) B1 and C1

D

C

C1

B1

44. A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of n  with respect to time. If the resistance is changed to 2x, the new graph will be [JEE’2004] (A) P (B) Q (C) R (D) S

S R

ln I

Q P t

45. A moving coil galvanometer of resistance 100  is used as an ammeter using a resistance 0.1 . The maximum deflection current in the galvanometer is 100  A. Find the minimum current in the circuit so that the ammeter shows maximum deflection. [JEE’2005] (A) 100.1 mA (B) 1000.1 mA (C) 10.01 mA (D) 1.01 mA

46. In the figure shown the current through 2 resistor is (A) 2 A (B) 0 A (C) 4 A (D) 6 A [JEE’2005 (Scr)]

10

10V 5

20V

2

47. A galvanometer has resistance 100 and it requires current 100 µA for full scale deflection. A resistor 0.1  is connected to make it an ammeter. the smallest current required in the circuit to produce the full scale deflection is [JEE’2005 (Scr)] (A) 1000.1 mA (B) 1.1 mA (C) 10.1 mA (D) 100.1 mA A 48. Consider a cylindrical element as shown in the figure. Current flowing I through the element is I and resistivity of material of the cylinder is . 4r Choose the correct option out the following. (A) Power loss in second half is four times the power loss in first half. (B) Voltage drop in first half is twice of voltae drop in second half. (C) Current density in both halves are equal. (D) Electric field in both halves is equal.

B C 2r l/2

l/2

[JEE’2006]

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47

ELECTRIC CURRENT 49. A resistance of 2  is connected across one gap of a meter-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is [JEE’2007] (A) 3  (B) 4  (C) 5  (D) 6  MORE THAN ONE CHOICE MAY BE CORRECT 50. Capacitor C1 of capacitance 1 micro-farad and capacitor C2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. [JEE’1989] (A) The current in each of the two discharging circuits is zero at t = 0. (B) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C1, losses 50% of its initial charge sooner than C2 loses 50% of its initial charge. 3

51. In the circuit shown in Figure the current through (A) the 3  resistor is 0.50 A. (B) the 3  resistor is 0.25 A (C) the 4  resistor is 0.50 A (D) the 4  resistor is 0.25 A

9V

2

8

2

8

4

52. Which of the following is/are wrong? 2 2 2 (A) A current carrying conductor is electrically charged. (B) There is an electric field inside a current carrying conductor. (C) For manual control of the current of a circuit, two rheostats in series are preferable to a single rheostat. (D) None of these 53. A constant voltage is applied between the two ensd of a uniform metallic wire. Some heat is produced in it. The heat developed is doubled if: (A) both the length and radius of the wire are halved. (B) both the length and radius of the wire are doubled. (C) the radius of the wire is doubled. R1 R2 (D) the length of the wire is doubled and the radius of the wire is halved. 54. All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. The current through the galvanometer R3 R4 will still be zero if, (A) the emf of the battery is doubled. (B) all resistance are doubled. (C) resistance R1 and R2 are interchanged. (D) the battery and the galvanometer are interchanged. 55. Two cells of equal emf and having different internal resistance R1 and R2 (R2 > R1) are connected in series. If resistance of connecting wires is equal to R, which of the following statements is/are correct? (A) At a particular value of r, potential difference across second cell can be equal to zero. (B) If R = 0, negative terminal of second cell will be at higher potential than its positive terminal (C) Negative terminal of first cell can never be at higher potential than its positive terminal. (D) None of these 56. When electric bulbs of same power but with different marked voltages are connected in series across a power line, their brightness will be: (A) proportional to their marked voltages. (B) inversely proportional to their marked voltages. (C) proportional to the squares of their marked voltages. (D) inversely proportional to the square of their voltages. 57. Two conductors made of the same material have lengths L and 2 L, but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct? (A) The potential difference across the two conductors is the same. (B) The electric drift velocity is larger in the conductor of length 2L. (C) The electric field n the first conductor is twice that in the second. (D) The electric field in the second conductor is twice that in the first.

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48

ELECTRIC CURRENT 58. Mark correct statements: (A) Heat is always generated in a battery whether it charges or discharges. (B) When a battery supplies current in an external circuit, heat generated in it is always less than electrical power developed in it. (C) Potential difference across terminals of a battery is always less than its emf. (D) None of these. 59. A mocrometer has a resistance of 100  and a full scale range of 50  A . It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations. (A) 50 V range with 10 k  resistance in series.

(B) 10 V range with 200 k  resistance in series.

(C) 5 mA range with 1 resistance in parallel.

(D) 10 mA range with 1 resistance in parallel.

60. An electric current is passed through a circuit containing three wires arranged in parallel. If the length and radius of the wires are in ratio 2 : 3 : 4 and 3 : 4 : 5, then the ratio of current passing through wires would be: (A) 54 : 64 : 75 (B) 9 : 16 : 25 (C) 4 : 9 : 25 (D) 3 : 6 : 10. 61. When a galvanometer is shunted with a 4  resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2  wire, the further reduction in the deflection will be (the main current remains the same). (A) 8/15 of the deflection when shunted with 4 only. (B) 5/13 of the deflection when shunted with 4 only. (C) 3/4 of the deflection when shunted with 4 only. (D) None of these 62. During the charging of a storage battery, the current is 22 A and the voltage is 12 V. The rate of heat generated in the battery is 12 W. The rate of change of internal energy is: (A) 240 J/s (B) 252 J/s (C) 264 J/s (D) 126 J/s. 63. A cell of emf 5 V, internal resistance 1 will give maximum power output to: (A) a single resistor of 1 (B) two 1 resistors connected in series. (C) two 1 resistors connected in parallel. (D) two 2 resistors connected in parallel. 3

64. In the circuit shown in figure, the current through: (A) the 3 resistor is 0.50 A.

(B) the 3 resistor is 0.25 A.

(C) the 4  resistor is 0.50 A.

(D) the 4  resistor is 0.25 A.

65. For the circuit shown in the figure

9V

2

8

8

2

I

2k

R1

24V

6k

R2

RL

2

2 4 2

[JEE’2009]

1.5k

(A) the current I through the battery is 7.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9

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49

ELECTRIC CURRENT

FILL IN THE BLANKS 1. 2. 3.

An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 Watts is ........... ohms. (1997) The equivalent resistance between points A and B of the circuit (Figure) given below is ........... . In the circuit (Figure) shown above, each battery is 5 V and has an internal resistance of 0.2 ohm.

A

2R

2R

R

B

V

Fig - 1

Fig - 2

The reading in the ideal voltmeter V is ......... V.

[JEE’ 1997]

TRUE / FALSE 4.

Electrons in a conductor have no motion in the absence of a potential difference across it. [JEE’ 1982]

5.

The current-voltage graphs for a given metallic wire at two different temperatures T 1 and T 2 are shown in the figure. [JEE’ 1985] The temperature T 2 is greater than T 1.

T1 T2 I V

TABLE MATCH Column I

6. (A) (B) (C) (D) 7.

Current Current Density Electric field Resistance

Column II (P) (Q) (R) (S)

Two bulbs A and B consume same power P when operated at voltage VA and VB respectively. Bulbs are connected with a supply of d.c source then:

Column I (A) (B) (C) (D)

8.

Mircoscopic quantity Macroscopicquantity Parallel to the conductor boundaries Flux associated with current density

In series connection, the ratio of potential difference across A and B In series connection, the ratio of power consumed by A and B In parallel connection, the ratio of current in A and B In parallel connection, the ratio of power consumed in A and B

Column II (P)

RA/RB

(Q)

V2A/V2B

(R)

RB/RA

(S)

V2B/V2A

In a R-C circuit.

Column I

Column II

(A)

Charging current at tiem t = 0

(P)

1 CV 2 2

(B) (C) (D)

Discharging current at t = 0 While charging energy stored While charging energy dissipated as heat

(Q) (R) (S)

Maximum Capacitor becomes short circuit Exponential law

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50

ELECTRIC CURRENT 9.

A battery has an emf E and internal resistance r. A variable resistor R is connected across the terminals of the battery.

Column I

Column II

(A) (B)

Current in the circuit is maximum Potential difference across the terminals is maximum

(P) (Q)

R=0

(C)

Power delivered to the resistor is maximum

(R)

i

(D)

Power delivered to the load is zero

(S)

r=R

R

E r

10. Consider two identical cells each of emf E and internal resistance r connected to a load resistance R.

(A)

Column I

Column II

For maximum power transfer to load

(P)

E2 4r

(Q)

E2 2r Eeq  E , req 

if cells are connected in series (B)

For maximum power transfer to load if cells are connected in parallel

(C)

For series combination of cells

(R)

(D)

For parallel connection of cells

(S)

r 2 Eeq  2 E , req  2 r

PASSAGE TYPE QUESTIONS PASSAGE # I

1.

2.

3.

4.

5.

A physics instructor devises a simple electrical circuit setup in which one can easily insert various resistors and capacitors in series and parallel combinations. One can have only resistor combinations, only capacitor combinations, or capacitor-resistor combinations. The circuit is usually used for DC (direct current studies but can also be used for AC (alternating current) studies. The DC battery voltage is 6 volts. The AC rms voltage is 120 volts (at 60 Hz.) The student inserts the resistors and /or capacitors as instructed and has available suitable ammeters and voltmeters for both DC and AC experiments. (There are also three resistors, each of 2 ohms resistance. There are also three capacitors, each of 1 microfarad capacitance.) All three resistors are connected in series and the combination is connected to the 6-volt DC battery. What voltage drop occurs across each individual resistor as measured by the voltmeter? (A) 0.33 V (B) 1.0 V (C) 2.0 V (D) 6.0 V One capacitor and one resistor are connected in parallel. The ends of this combination are then connected t the 6-V DC battery, what are the final current and voltage, respectively, across the 1 microfarad capacitor? (A) 0 A, 6 V (B) 0.33 A, 3 V (C) 0.33 A, 6 V (D) 6 A, 6 V Two of the 2-ohm resistors are connected in parallel and the 120-V AC voltage is applied to the ends of this parallel combination. What current will the AC ammeter measure if connected so it measures only the current through one of the resistors? (A) 1 A (B) 2 A (C) 60 A (D) 120 A Two resistors are connected in parallel and the set of parallel resistors is then connected in series with the third resistor. If this series parallel combination is connected across the 6-volt battery. What total DC current is drawn from the battery? (A) 0.5 A (B) 2 A (C) 3 A (D) 6 A One capacitor and one resistor are connected is series. The 120 V, 60 Hz, A.C. voltage is then applied to this series “RC” circuit. What is the AC current through this series circuit? (A) 0.045 A (B) 0.50 A (C) 0.72 A (D) 40.0 A

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51

ELECTRIC CURRENT PASSAGE # II

7.

8.

9.

10.

PASSAGE # III In the laboratory, the voltage across a particular circuit element can be measured by a voltmeter. A voltmeter has a very high resistance and should be connected in parallel to the circuit element whose voltage is being measured. Connected improperly, the voltmeter will affect the circuit, interrupting it and preventing current from flowing through the circuit element that it is meant to measure. An experiment is conducted in which a voltmeter is used to investigate voltages in a circuit containing a capacitor and a light bulb. The bulb and the capacitor are connected in series with a battery and the voltmeter is placed in different position: in the first case across the capacitor in the second case across the light bulb, and in the third case across the battery (see figure 1). The voltmeter reading is recorded every 10 seconds. The voltage for Case 1 as a function of time is shown in figure 2. voltmeter V capacitor capacitor capacitor voltmeter V

bulb

bulb battery

bulb

battery

Case 2

Case 1

voltmeter

battery V

Case 3

Voltage

6.

A set of experiments in the physics lab is designed to develop understanding of simple electrical circuit principles for direct current circuits. The student is given a variety of batteries, resistors, and DC meters; and is directed to wire series and parallel combinations of resistors and batteries making measurements of the currents and voltage drops using the ammeters and voltmeters. The student calculates expected current and voltage values using Ohm’s law and Kirchoff’s circuit rules and then checks the results with the meters. A student connects a 6-volt battery and a 12-volt battery in series and then connects this combination across a 10-ohm resistor. What is the current in the resistor? (A) 0.8 A (B) 0.9 A (C) 1.8 A (D) 3.6 A Resistors of 4 ohms and 8 ohms are connected in series. A battery of 6 volts is connected across the series combination. How much power, in watts, is consumed in the 8-ohm resistor? (A) 0.67 W (B) 2 W (C) 12 W (D) 24 W Two 4-ohm resistors are connected in series and this pair is connected in parallel with an 8-ohm resistor. A 12 volt battery is connected across the ends of this parallel set. What power, in watts, is consumed in the 8-ohm resistor in this case? (A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules A 12-volt battery is connected across a 4-ohm resistor and the heat energy dissipated in the resistor in 10 minutes is used to heat 2 kg of water (which is thermally insulated so that no heat escapes). The initial temperature of the water is 20°C and the specific heat of the water is 4184 joule/kg-°C. What is the final temperature of the water at the end of the 10 minutes of heating? (A) 22.6 °C (B) 28.4°C (C) 34.2°C (D) 56.4°C

Time Fig. 2 A capacitor consists of two conducting plates separated by a nonconducting material. When a battery is connected to a circuit containing a capacitor and a light bulb in series, a current will flow, causing positive charge to accumulate on one capacitor plate and an equal amount of negative charge to accumulate on the other. After the current has flowed for a finite time, the capacitor will be fully charged. The ratio of the absolute amount of charge on one plate to the voltage across the plates is defined as the capacitance; this is constant for a given capacitor. A light bulb is a resistor that, when enough current flows through it, becomes hot enough to emit energy in the form of light. (Note : Assume that the battery has no internal resistance and that the resistance of the light bulb is constant).

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52

ELECTRIC CURRENT

(A)

(B) Time

Time

16.

Time

Time

(D) Time

Voltage

(C)

Voltage

(B) Time

15.

Time

Voltage

(A)

14.

(D)

After the experiment described in the passage is completed, the battery is taken out of the circuit and the wires are reconnected. Which of the following graphs, represents the voltage across the capacitor as a function of time –

Voltage

13.

(C)

Voltage

Which one of the following graphs could correctly represent the voltage across the battery as a function of time during the experiment described in the passage –

Voltage

12.

Voltage

In the circuit shows in figure 1, which of the following conditions would indicate that the capacitor was fully charged – I. A voltmeter connected across the capacitor reads a constant voltage. II. The light bulb in the circuit stops shining. III. The voltage across the bulb equals the voltage across the battery (A) I Only (B) III Only (C) I and II only (D) I, II, and III

Voltage

11.

Time

How will the voltage across the light bulb vary with time as the capacitor is charging – (A) It will decrease, because as the capacitor plates fill will charge, they will impede further charge, which will decrease the current and the voltage across the bulb. (B) It will remain the same, because as the capacitor plates fill with charge and impede the current, the voltage output of the battery will increase to keep the current constant. (C) It will increase, because as the capacitor plates fill with charge, they will induce further charge, which will create a greater voltage across the bulb. (D) It will increase, because as the capacitor plates fill with charge, the voltage across the capacitor will decrease, and therefore the voltage across the light bulb will increase. The light bulb shown in figure 1 is replaced first with two identical resistor in series, and then with the same two resistors in parallel. The total time taken for the capacitor to charge is measured in both cases, and found to be longer for the first case. It can be deduced that – (A) When the resistance of the circuit si increased, the capacitance of the capacitor increases. (B) the presence of resistors affects the final voltage across the capacitor plates. (C) the more charge is absorbed by the resistors as the resistance of the circuit increases. (D) the presence of resistors hinders the flow of charge, thus reducing the current in the circuit. In the diagram below, a voltmeter is connected in series to a circuit that includes a battery and two bulbs in series. The bulbs, which had been shining in the absence of the voltmeter immediately stop shining. How might the circuit be modified in order to make the bulbs shine steadily again with their former brilliance without removing the voltmeter – voltmeter V bulb

bulb battery V

V

(A)

(B)

V

(C)

V

(D)

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53

ELECTRIC CURRENT

LEVEL # 2 1.

In the figure shown,

3

2 1 F

3 F

3

2 6V, 0.5 

find the charge on 2 2.

 F each capacitor in steady state. [in  C ].

In the given circuit diaram, A

B

1

C

2

3

D

4

E

F

50 V find the current passing through wire CD [in Ampere] 3.

It is required to send a current of 8 A through a circuit whose resistance is 5  . What is the least number of cells which must be used for their purpose and how should they be connected? Emf of each cell is 2V and internal resistance is 0.5  . 100 V, 1

4.

In the circuit shown, the capacitor is charged by a battery of emf 100 V and 1 internal resistance by closing the switch. Calculate s

the heat generated across 99  resistance during the charging of capacitor. [in Joule].

99 

0.1 F

5.

In a Wheatstone’s bridge a battery of 2 volt and internal resistance 2 ohm is used. Find the value of the current through the galvanometer in that unbalanced condition of the bridge when P = 1 ohm, Q = 2 ohm, S = 30 ohm and resistance of galvanometer is 4 ohm.

6.

A 20 volt battery with an internal resistance of 6  is connected to a resistor of x ohms. If an additional 6  resistance is connected across the battery find the value of x so that external power supplied by battery remains the same. S

7.

Find how the voltage across the capacitor C varies with time t after capacitors C varies with time t after the shorting of the switch S at the moment t = 0.

8.

R

E

C

R

A homogeneous poorly conducting medium of resistivity  fills up the space between two thin coaxial ideally conducting cylinders. The radii of the cylinders are equal to a and b, (a < b) the length of each cylinder is  . Neglecting the edger effects. Find the resistance of the medium between cylinders.

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54

ELECTRIC CURRENT 9.

Find the current flowing through the resistance R in the circuit shown in figure. The internal resistance of the batteries are negligible.

R1

A 10.

11.

A circuit shown in figure has resistance R1 = 20  and R2 = 30  . At what value of the resistance Rx will the thermal power generated in it be practically independent of small variation of that resistance ? The voltage between the points A and B is supposed to be constant in this case.

R2

Rx

B

An ammeter and voltmeter are connected in series to a battery with an emf E = 6.0v. When a certain resistance is connected in parallel with the voltmeter, the reading of the latter decreases  = 2.0 times whereas the reading of ammeter increases the some number of time. Find the voltameter reading after the connecting of the resistance. E1 10

12.

In the circuit shown in the figure the current through 3  resistance is 2A. If E1 = 12V, E2 = 14V, what is the value of E ? Internal resistance of each battery is 1  .

1 E2

E3

1 3

13.

1.5

The circuit shows a capacitor C two batteries, two resistors and a switch S. Initially S has been open for a longtime. It is then closed for a long time by how much does the change on the capacitor change over this time period ? Assume C = 10  , E1 = 1.0 V, E3 = 3.0v, R1 = 0.20  , R2 = 0.40  .

S

10

 14.



In the given network switch S is closed at t = 0. Find the current through the 10 ohm resistor at t = 75  sec. 10V

20V

2F

15.

Two coils connected in series have resistance of 600  and 300  and temperature co-efficient of 0.001 and 0.004 (°C)–1 respectively at 20°C. Find resistance of the combination at a temperature of 50°C. What is the effective temperature co-efficient of combination. G

16.

The resistance of the galvanometer G in the circuit is 25  . The meter deflects full scale for a current of 10 mA. The meter behaves as an ammeter of three different ranges. The range is 0–10 A, if the terminals O and P are taken; range is 0–1 A between O and R. Calculate the resistance R1, R2 and R3.

R1

R2

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R3

55

ELECTRIC CURRENT 17. Calculate the steady current in the 2-ohm resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the conductor C is 0.2 microfarad.

OR Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. what will be the reading in the ammeter? similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will-be the reading in the voltmeter.

LEVEL # 3 1.

A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C (4  F) [JEE’ 1986]

2.

An infinite ladder network of resistances is constructed with a 1 ohm and 2 ohm resistances, as shown in figure. [JEE’ 1987]

The 6 volt battery between A and B has negligible internal resistance: (i) Show that the effective resistance between A and B is 2 ohms. (ii) What is the current that passes through the 2 ohm resistance nearest to the battery?

3.

An electrical circuit is shown in Figure. Calculate the potential difference across the resistor of 400 ohm, as will be measured by the voltmeter V os resistance 400 ohm, either by applying Kirchhoff’s rules or otherwise. [JEE’ 1996]

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56

ELECTRIC CURRENT 4.

Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistance r1 and r2 respectively, with polarities as shown in figure.

[JEE’ 1997]

5.

A leaky parallel plane capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity  = 7.4 x 10–12  –1 m –1. If the charge on the plane at instant t = 0 is q = 8.85 mC, then calculate the leakage current at the instant t = 12 s. [JEE’ 1997]

6.

In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0. [JEE’ 1998] (A) Find the charge Q on the capacitor at time t. (B) Find the current in AB at time t. What is its limiting value as t  :

7.

A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12  are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be X 12 made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [JEE’ 2002] A B C D (A) Are there positive and negative terminals on the galvanometer? (B) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at approxiate points. (C) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X. C

8.

9.

How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points out which output can be taken. [JEE’ 2003]

A

R

B

Draw the circuit diagram to verify Ohm’s Law with the help of a main resistance of 100  and two galvanometers of resistances 106  and 10–3  and a source of varying emf. Show the correct positions of voltmeter and ammeter. [JEE’ 2004]

10. An unknown resistance is to be determined using resistances R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [JEE’ 2005]

X

R R = R1 or R2 or R3

G A

B

C

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57

ELECTRIC CURRENT

ANSWER KEY Reasoning Type Que.

1

2

3

4

5

6

7

8

9

10

11

12

Ans.

C

C

D

A

C

A

A

C

A

D

B

A

Level # 1 Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans.

1 C 11 B 21 B 31 B 41 A 51 D 61 B

2 B 12 B 22 C 32 A 42 A 52 AC 62 B

3 D 13 B 23 B 33 B 43 C 53 B 63 AD

4 D 14 D 24 AB 34 B 44 B 54 ABD 64 D

5 A 15 CD 25 C 35 D 45 A 55 ABC 65 AD

6 C 16 A 26 D 36 A 46 B 56 C

7 C 17 A 27 A 37 B 47 D 57 AC

8 D 18 ABCD 28 ACD 38 A 48 A 58 AB

9 D 19 A 29 A 39 A 49 A 59 BC

10 B 20 C 30 A 40 A 50 BD 60 A

Fill in the Blanks / True–False / Match Table 3. 0 4. F 5. T R2 6. A  Q, R and S, B  P and R, C  P and R, D  Q 7. A  P and Q, B  P and Q, C  R and S, D  R and S 8. A  Q, R and S, B  Q and S, C  P and S, D  P and S 9. A  Q and R, B  P, C  S, D  Q, P and R 10. A  Q, B  Q, C  S, D  R 1. 20

2.

Passage Type Que. Ans. Que. Ans.

1 C 11 C

2 A 12 A

3 C 13 B

4 B 14 A

5 A 15 D

6 C 16 A

7 B

8 D

9 D

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10 A

58

ELECTRIC CURRENT

Level # 2 1.

72 µC 11

7. V =

2. 2 A

E 1 e2t RC 2



3. 160



4. 495 J

R1 R 2 10. Rx = R  R = 12  1 2

11.

13. decreases by 13.3  c

14. I = 20 mA

6. x = 7.5 

ER 2  R 3   E0R 3 9. I = RR  R   R R  2 3 2 3

 b n   2  a 

8.

1 Amp 9

5.

E = 2.0 V  1

12. E = 7V

15. 954  , 0.002 (°C)–1 17. 0.9A or 4.96 x 10–3 A, 1.95 V

16. R1 = 0.025  , R2 = 0.2275  , R3 = 2.5275 

Level # 3 1. 2.88 x 10–4 J

4.

2. (ii) 1.5 A

V1r2  V2 r1 r1 r2 , r1  r2 r1  r2

7. (a) No

3. 10/3 V

5. 0.199  A

6. (a)

CV V V 2 t 3 RC V 1  e 2 t 3 RC ; (b)  e ; 2 2R 6 R 2R





(c) 8 

(b) G

X A

8.

J

12

B

C

D

Battery should be connected across A and B. Out put can be taken across the terminals A and C or B and C. Voltmeter 6

G1

10

100 9.

Ammeter G2

103

E 10. This is true for r1 = r2; So R2 given most accurate value.

—X—X—X—X—

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59

MAGNETIC FIELD

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

MAGNETIC FIELD THEORY OF MAGNETIC EFFECT OF ELECTRIC CURRENT CONCEPT OF MAGNETIC FIELD The space around a current carrying conductor, in which its magnetic effect can be experienced, is called magnetic field. In magnetics, there are basically two methods of calculating magnetic field at some point. One is Biot Savert’s law which gives the magnetic field due to an infinitesimally small current carrying wire at some point and the other is Ampere’s law, which is useful in calculating the magnetic field of a highly symmetric configuration carrying a steady current. BIOT-SAVART’S LAW   According to this law, the magnetic field dB at the point P due to the small current element of length dl is given by

µ0 idl sin  Wb / m 2 or tesla 2 4 r where µ0 is a constant and is called, permeability of free space. µ0 = 4 × 10–7 Wb/A–m

P

dB 

i

Rules to Find the Direction of Magnetic Field (i) Right hand palm rule no. 1 : If we spread our right hand in such a way P that thumb is towards the direction of current and fingures are towards that point where we have to find the direction of field then the direction of field will be perpendicular to the palm

r

i Current carrying conductor

B

(ii) Maxwell’s right handed screw rule : If a right handed cork screw is rotated so that its tip moves in the direction of flow of current through the conductor, then the rotation of the head of the screw gives the direction of magnetic lines of force. i P2

Current carrying conductor × Magnetic line P1 of force

 NOTE : By convention the direction of magnetic field B perpendicular to paper going inwards is shown by  and the direction perpendicular to the paper coming out is shown by  .

Applications of Biot-Savart’s Law Let us consider few applications of Biot-Savart’s Law : (i) Magnetic field due to a straight thin conductor is

µi B  0  sin 1  sin  2  4d (a) For an infinitely long straight wire, 1 = 2 = 90º

i

2

d

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p

1

1

MAGNETIC FIELD

µ 0i 2d (b) When wire is semi-infinite, B



i

1  0 and  2 

 2

µ 0i    sin 0  sin  4d 2 µi B 0 4 d

d

B



1 , i.e., B–d graph for an infinitely long straight wire is a rectangular d hyperbola as shown in figure. (ii) Magnetic field on the axis of a circular coil having N turns is µ0 NiR 2 B 3/ 2 2 R2  x2 Here, R = radius of the coil x = the distance of point P from centre and i = current in the coil (a) At the centre of the loop, x = 0

(c) B 





B

d

P

O x R

µ0 Ni 2R 2 2 (b) For x > > R, x + R  x2



B



B

Here,

µ0 NiR 2  µ0   2NiR 2   µ0   2M      4   x 3   4   x 3  2x 3

M = magnetic moment of the loop M = NiA = Ni R2

NOTE : This result was expected since, the magnetic field on the axis of dipole is

µ0 2M 4π x 3

5 5 cm and 5 cm respectively carry current 5 Amp and Amp 2 2 respectively. The plane of B is perpendicular to plane of A and their centres coincide. Find the magnetic field at the centre. Sol. The magnetic field due to first coil is µI 4  107  5 B1  0 1  2r1 2  5  102 2 Example 1. Two circular coils A and B of radius

20  3.14  107 5  1.441  102 B1 = 8.88 × 10–5 web/m2 B1 

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2

MAGNETIC FIELD 7

and

B2 

µ0 I 2 4  10  5  2r2 2  5  102  2

B2 

2  3.14  105  4.44  105 1.414

B  B12  B22



B  (8.88) 2  (4.44) 2  10 5 weber / m 2

B  78.85  19.71  9.93 wb / m 2

Example 2. Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of co-ordinate system. If current ‘I’ is flowing through each ring then the magnitude of the magnetic field at the common centre is µ0 I (B) zero 2R Sol. The magnetic field due to ring in x–y plane is

(A)

3

(C)



 µ2RI

2 1

0

(D)



3 2

 µ2RI 0

 µ0 I B1  kˆ 2R the magnetic field due to ring in y–z plane is  µ0 I ˆi B2  2R and the magnetic field due to ring in x–z plane is  µ0 I ˆj B3  2R      B  B1  B2  B3  µ0i B kˆ  ˆi  ˆj 2R 3µ0 I B  2R Hence, option (A) is correct





i

(c) Magnetic field due to an arc of a circle at the centre is µ  i   µi B =   0 or B   0      2   2R  4   R  (iii) Field along the axis of a solenoid is

O

R Inwards

R 1

O

2

x

L

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3

MAGNETIC FIELD

µ Ni B  0  cos 2  cos 1  2 (a) For a long solenoid (L >>R), i.e., 1 = 180º and 2 = 0º B = µ0Ni (b) At the ends of solenoid, 2 = 0º, 1 = 90º 1 B  µ0 Ni (for L>>R) we get, 2 Example 3. In a high tension wire electric current runs from east to west. Find the direction of magnetic field at point above and below the wire. Sol. When the current flows from east to west, magnetic field lines are circular round it as shown in figure (a). And so, the magnetic field above the wire is towards north and below the wire towards south. B

N W

i

E S

(a)

(b)

Example 4. A 0.5 m long solenoid has 500 turns and has a flux density of 2.52 × 10–3 T at its centre. Find the current in the solenoid. Given, µ0 = 4 × 10–7 Hm–1. Sol. Here, B = 2.52 × 10–3 T; µ0 = 4 × 10–7 Hm–1 Length of the solenoid, l = 0.5 m; Total number of turns in the solenoid, N = 500 Therefore, number of turns per unit length of the solenoid, N 500 n   1000 m 1 l 0.5 If ‘i’ is the current through the solenoid, then B = µ0ni B 2.52  10 3 i   2.0 A or µ0 n 4  107  1000 Example 5. A rectangular polygon of ‘n’ sides is formed by bending a wire of total length 2R which carries a current ‘i’. Find the magnetic field at the centre of the polygon. Sol. One side of the polygon is, a

2R n

 2     n    2 n d  cot   a / 2 

d

i a

 a  R    d    cot    cot     2  n   n

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4

MAGNETIC FIELD All sides of the polygon produce the magnetic field at the centre in same direction (here  ). Hence, net magnetic field, B = (n) (magnetic field due to one side) µ i  B  n  0  sin   sin    4 d      µ   in   B  n 0   tan   2 sin     4    R   n  n

or

     sin     µ 0i    n  n  B  2R  cos   / n 

or

2

Example 6. Infinite number of straight wires each carrying current ‘I’ are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magntic field at point P is

µ0 I l n 2 ˆ k (A) 4 3a

a P

µ0 I l n 4 ˆ k (B) 4 3a

y

a 30º 30º

1

2

3

4

x

5 z

µ0 I l n 4 ˆ ( k) (C) 4 (D) Zero 3a  Sol. B   B1  B2  B3  B4 .........    kˆ Here

B1 

µ0 I  sin 30º  sin 30º  4 a

B1 

µ0 I  1 1    4a cos 30º  2 2 

B1  B2 

Similarly,

2µ0 I 4 3a 2µ0 I 8 3a

and so on

 2µ0 I  1 1 1  B   1    ........   2 3 4 4 3a  2µ0 I B ln 2 kˆ 4 3a  µ0 I B ln 4 kˆ 4 3a Hence, option (B) is correct.

Example 7. A long straight wire, carrying current I, is bent at its midpoint to form an angle of 45º. Magnetic field at point P, distance R from point of bending is equal to : (A)





2  1 µ0 I 4R

(B)





2  1 µ0 I 4R

(C)





2  1 µ0 I 4 2R

P I

45º

R

I

(D)





2  1 µ0I 2 2 R

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5

MAGNETIC FIELD Sol. µ0 I  sin 90º  sin135º  R 4 . 2 µI B 0 2 1 Hence option (A) is correct 4 R B





Example 8. Find the magnetic field at P due to the arrangement shown 2µ0i µ 0i  1   (A) (B) 1    2 d 2 d 2 (C)

µ 0i 2 d



(D)

  B  B1  B2

Sol.

B

45º

µ0 i  1   1   2 d 2

P

     sin  sin   2  4 2

90º

µ0 I  1  1  d  2 2 2

45º /4

d 2

2 µ0 I  2  1 B  2d  2 



B B

µ0 I 2 d



P

d 2



2 1 

µ 0i  1   1   2d 2

Hence option (A) is correct

Example 9. What is the magnitude of magnetic field at the centre ‘O’ of loop of radius 2 m made of uniform wire when a current of 1 amp enters in the loop and taken out of it by two long wires as shown in the figure.  Sol. B1 = magnetic field due to left wire is µI   B1  0  sin  sin  kˆ 4d  2 4  B2 = magnetic field due to right wire µI   ˆ  0  sin  sin  ( k) 4 d  2 4 In circular wire, I1  1  I  I2 2   R



d

º 45

B

µ0 I d 4 2

90º

   I1   I  2    2

O 45º

1 amp

90º

1 amp

O /4

d

I1 –

I2

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6

MAGNETIC FIELD

  B3 = magnetic field due to circular wire µ I  µ I (2  )  01  0 2 4r 4r µ I  µ (2  )  0 2  0 I2  0 4r 4r(2  )      B  B1  B2  B3  0 AMPERE’S CIRCUITAL LAW  It states that the line integral of B around any closed path or circuit is equal to µ0 times the total current crossing the area bounded by the closed path provided the electric field inside the loop remains constant. Thus,    B . dl  µ0  inet  C

Its simplified form is Bl = µ0 inet This equation can be used only under following conditions :   (a) at every point of the closed path B || dI (b) magnetic field has the same magnitude B at all places on the closed path. Applications of Ampere’s Circulatal Law (i) Magnetic field due to a long metal rod of radius R carrying a current ‘i’ :  µ 0i  (a) If r < R, B    r , i.e., B  r 2R 2 

(b) If r = R (i.e., at the surface) µi B 0 2R (ii) Magnetic field of a solenoid wounded in the form of a helix is B = µ0Ni NOTE : Ampere’s law is valid only for steady currents. Further more, it is useful only for calculating the magnetic fields of current configurations with high degrees of symmetry, just as gauss’s law is useful only for calculating the electric fields of highly symmetric charge distributions. d Example 10. Two long conductors are arranged as shown above to form overlapping cylinders, each of raidus ‘r’, whose centers are separated A by a distance ‘d’. Current of density J flows into the plane of the page along the shaded part of one conductor and an equal current flows out of the plane of the page along the shaded portion of the other, as shown. Vacuum What are the magnitude and direction of the magnetic field at point A ? (A) (µ0/2)dJ, in the +y-direction (B) (µ0/2)d2/r, in the +y-direction (C) (µ0/2)4d2J/r, in the –y-direction (D) (µ0/2)Jr2/d, in the –y-direction    Sol. B  B1  B2

 µ   d  µ B  0  j  ˆi   0 2 2  2

Conductor y x

  d ˆ   j    i    2 

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7

MAGNETIC FIELD  µ0  d  µ  d  B   jkˆ  ˆi   0   jkˆ  ˆi  2  2  2 2 

 µ0 µ B jd ˆj  0 jd ˆj 4 4  µ0 µ d j B jd ˆj  0 along y-axis 2 2

Hence, option (A) is correct.

MAGNETIC FIELD OF A MOVING POINT CHARGE    The magnetic field vector B at point P at position vector r , from the charge q moving with a velocity v is found to be  µ0 q   B  vr 4 r 3 Note down the following points regarding this equation. (a) Magnitude of B is,





µ qvsin  B 0 4 r 2 It is zero at  = 0º and 180º and maximum at  = 90º    (b) Direction of B is along v  r if q is   positive and opposite to v  r if q is negative.

p

v r q





(c) Suppose a charge q1 is moving with velocity and v1 another charge q2 is moving with velocity v 2 at

 position vector r relative to q1, then force on q2 will be,    F  q2 v2  B





   µ q     F   v 2   0 . 31  v1  r   4 r   

 





 µ qq    F  0 . 1 3 2  v 2  v1  r   4 r 

  



 This corresponds to Coulomb’s electrical force between the charges q1 and q2 moving with velocities v1 and  v 2 respectively relative to an observer at rest.  Example 11. For a charge ‘q’ moving with velocity v , find the relation between electric and magnetic fields.  q E Sol. ...(i) 4 0 r 3    µ0 qv  r B ...(ii) 4 r 3 1 c and µ0  0 or

c2 

1 µ0  0

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8

MAGNETIC FIELD From Eqs. (i) and (ii), we get    B  µ0  0 v  E    v  E B 2  c F i FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor is placed in a magnetic field, the conductor B experiences a force in a direction perpendicular to both the direction of magnetic field and the direction of current flowing in the conductor. This force is also called Lorentz force. i The direction of this force can be found out either by Fleming’s left hand rule or by right hand palm rule. The magnetic force is F = ilB sin     Fi l B In vector form, Where B = intesity of magnetic field i = current in the conductor l = length of the conductor and  = angle between the length of conductor and direction of magnetic field. Case : (i) If = 90º or sin = 1 then F = ilB (maximum) Therefore, force will be maximum when the conductor carrying current is perpendicular to magnetic field. (ii) If = 0º or sin = 0, Then F = ilB × 0 = 0 Thus, the force will be zero, when the current carrying conductor is parallel to the field.





Example 12. A straight current carrying conductor is placed in such a way that the P current in the conductor flows in the direction out of the plane of the paper. The R S S conductor is placed between two poles of two magnets, as shown. The Q conductor will experience a force in the direction towards. (A) P (B) Q (C) R (D) S Sol. The direction of magnetic field on the conductor is along SR.    But F  il  B  F  i l kˆ  (–B ˆi)  F  i l B ˆj  i l B(  ˆj) Hence, the direction of magnetic force on the wire is towards Q. Hence option (B) is correct

N

Example 13. In the figure shown a semicircular wire loop is placed in uniform magnetic field B = 1.0 T. The plane of the loop is perpendicular to the magnetic field. Current i = 2A flows in the loop in the direction shown. Find the magnitude of the magnetic force in both the cases (a) and (b). The radius of the loop is 1.0m. ×

×

×

×

×

× i = 2A × × 1m × ×

×

×

×

×

(a)

×

×

×

× × i = 2A × ×

×

×

×

×

×

×

×

×

×

×

(b)

×

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9

MAGNETIC FIELD Sol. Refer figure (a) : It forms a closed loop and the current completes the loop. Therefore, net force on the loop in uniform field should be zero. Refer figure (b) : In this case although it forms a closed loop, but current does not complete the loop. Hence, net force is not zero. × × C × ×B   FACD  FAD × × × ×      Floop  FACD  FAD  2 FAD × × × ×   D A  × × × × | Floop |  2| FAD |  (l = 2r = 2.0 m) | Floop |  2ilB sin  = (2) (2) (2) (1) sin 90º = 8 N × × × × Example 14. An arc of a circular loop of radius ‘R’ is kept in the horizontal plane × × I× × and a constant magnetic field ‘B’ is applied in the vertical direction as shown × × × in the figure. If the arc carries current ‘I’ then find the force in the arc. × × × 90º Sol. As we know, magnetic force on a closed loop placed in uniform magnetic field is zero. × × × ×    B1  B2  0 (1)   ˆ  B1   B2   IlB(  j)  R R B1  Il B ˆj /2  (2) B1  I B( 2R) ˆj   B1  2 IBR

Example 15. A conducting wire bent in the form of a parabola y2 = 2x carries a current  i = 2 A as shown in figure. this wire is placed in a uniform magnetic field B  4 kˆ Tesla. The magnetic force on the wire is (in newton) (A) 16 ˆi (B) 32 ˆi (C) 32 ˆi Sol. The net magnetic force on closed loop is zero.  Force on parabola + force on straight wire AB = 0  Force on the parabola = – force on straight wire AB    I ˆj  B



y(m)

x(m)

B

(D) 16 ˆi A

B

 

 2 4 ˆj  4 kˆ F  32 ˆi

A 2





× × B × × × × × × × ×

Hence option (C) is correct.

Example 16. A conductor of length ‘l ’ and mass ‘m’ is placed along the east-west line on a table. Suddenly a certain amount of charge is passed through it and it is found to jump to a height ‘h’. The earth’s magnetic induction is B. The charge passed through the conductor is : (B is horizontal) (A)

1 Bmgh

(B)

2gh Blh

(C)

gh Blh

(D)

m 2gh Bl

Sol. The magnetic force is F = I lB or

F

dq lB dt

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10

MAGNETIC FIELD or

dq 

Fdt lB

But

Fdt mv 0  lB lB 02 = v02 – 2gh



v 0  2gh



q 

or

q 

m 2gh lB

Hence option (D) is correct

Example 17. A U-shaped wire of mass m and length l is immersed with its two ends in mercury (see figure). The wire is in a homogeneous field of magnetic induction B. If a charge, that is, a current pulse q   idt , is sent through the wire, the wire will jump up. Calculate, from the height ‘h’ that the wire reaches, the size of the charge or current pulse, assuming that the time of the current pulse is

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× l

×

×

×

B

i

m

Hg

very small in comparision with the time of flight. Make use of the fact that impulse of force equals  Fdt , which equals mv. Evaluate ‘q’ for B = 0.1 Wb/m2, m = 10 gm, l = 20 cm & h = 3 meters. [g = 10 m/s2] Sol. IlB=F  Fdt = mv or I l Bdt = mv But

1 mv 2  mgh 2



v  2gh



Idt 



m 10  103 q 2gh  2  10  3  15 coulmb. lB 20  102  0.1

mv m  2gh lB lB

Example 18. A metal ring of radius r = 0.5 m with its plane normal to a uniform magnetic field B of induction 0.2 T carries a current I = 100 A. The tension in newtons developed in the ring is : (A) 100 (B) 50 (C) 25 (D) 10 F

Sol. We consider a small portion l of the loop. For equilibrium, sin    2T  T 2 2 or IlB = T or IrB = T  T = IrB = 100 × 0.5 × 0.2 = 10 N Hence option (D) is correct

Tcos /2

/2

/2

Tcos /2

F  2T

 T

T +

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B 11

MAGNETIC FIELD FORCE ON A MOVING CHARGE IN MAGNETIC FIELD : LORENTZ FORCE If a charged particle moves in a magnetic field, then a force acting on × × × × × × it is given by × × × × × × F = qvB sin  × × × × × ×    F × × × × × × Fq vB In vector form, × × × × × × v × × × × × × Cases : (i) If v = 0, then F = 0 i.e., no force is exerted on a stationary charge, in a magnetic field. (ii) If = 0, then F = 0 i.e., when the charge is moving parallel to the field then no force will be exerted by the field. (iii) If = 90º, then sin = sin 90º = 1 F = qvB × 1 = qvB i.e., when the charged particle is moving parpendicular to the field, the force exerted by the field will be maximum.





Rules of Find the Direction of Force (i) Right hand palm rule : If we stretch the right hand palm such that the fingers and the thumb are mutually perpendicular to each other and the fingers point the direction of magnetic field and the thumb points the direction of motion of positive charge, the direction of force will be along the outward normal on the palm. Force F

Field B

Current or motion of positive charge

(ii) Fleming’s left hand rule : If we spread the forefinger, central finger and thumb of our left hand in such a way that these three are perpendicular to each other then, if first forefinger is in the direction of magnetic field, second central finger is in the direction of current, then thumb will represent the direction of force. Force F

Field B Current or motion of positive charge v

NOTE : To learn this rule, remember the sequence of Father, Mother, child. Thumb  Father  Force Forefinger  Mother  Magnetic field Central finger  Child  Current or direction of positive charge   6 13 Example 19. When a proton has a velocity v  2 ˆi  3 ˆj  10 m / s it experiences a force F   1.28  10 kˆ N. When its velocity is along the z-axis, it experiences a force along the x-axis. What is the magnetic field ? Sol. Substituting proper values in,    Fq vB





We have,











 



 



 1.28  1013 kˆ  1.6  1019  2 ˆi  3 ˆj   B0 ˆj   106  

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12

MAGNETIC FIELD 

1.28 = 1.6 × 2 × B0

1.28  0.4 3.2  Therefore, the magnetic field is, B  0.4 ˆj T

or

B0 





MOTION OF CHARGED PARTICLE IN MAGNETIC FIELD   The path of charged particle in uniform magnetic field depends on angle between v and B . Therefore, following cases are possible : Case-I . When  is 0º or 180º : The magnetic force is F = Bqv sin 0º or sin 180º = 0. Hence, path of the charged particle is a straight line (undeviated) when it enters parallel or antiparallel to magnetic field. Case-II. When  = 90º : The magnetic force is F = Bqv sin 90º = Bqv. This magnetic force is perpendicular to the velocity at every instant. Hence, path is circle. The necessary centripetal force is provided by the magnetic force hence, if ‘r’ be the radius of the circle, then B

B

or q+

v

v

+ q

Fm = 0

mv mv r  qBv  qB r This expression of ‘r’ can be written in following different ways : 2

mv p 2Km    qB qB qB P = momentum of particle r

Here,

2qVm qB

p2 or p  2K m K = KE of particle  2m Further, time period of the circular path will be  mv  2   qB  2m 2r T   v v qB 2m T or qB or the angular speed () of the particle is  

2 qB  T m

qB m Frequency of rotation is, 1 qB f or f  T 2m





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13

MAGNETIC FIELD IMPORTANT FEATURES If angle  is other than 0º, 180º or 90º, then velocity of charged particle can be resolved in two components   one along B and another perpendicular to B . Let the two B

components be v| | and v  . Then v| |  v cos 

v cos

q, m +

v   v sin 

and

v sin

v

The component perpendicular to field ( v  ) gives a circular path and the component parallel to field ( v| | ) gives a straight line path. The resultant path is a helix as shown in figure. The radius of this helical path is,

mv  mvsin   qB qB Time period and frequency do not depend on velocity and so they are given by r

2m qB and f  qB 2m There is one more term associated with a helical path, that is pitch (p) of the helical path. Pitch is defined as the distance travelled along magnetic field in one complete cycle, i.e., p = v| |T T

or

p   v cos  



p

2m qB

2mv cos  qB

Example 20. What is the smallest value of B that can be set up at the equator to permit a proton of speed 107 m/s to circulate around the earth ? [R = 6.4 × 106 m, mp = 1.67 × 10–27 kg]. Sol. From the relation

R

mv qB

mv qR Substituting the values, we have We have

B

1.67  10 10   1.6  10 B 1.6  10 6.4  10  27

19

7

6

8

T q

Example 21. A block of mass m & charge ‘q’ is released on a long smooth inclined plane magnetic field B is constant, uniform, horizontal and parallel to surface as shown. Find the time from start when block loses contact with the surface. (A)

m cos  qB

(B)

m cosec θ qB

(C)

m cot θ qB

B

m

(D) none

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14

MAGNETIC FIELD Sol. For losing the contects N=0  qvB = mgcos

But or

mg cos  qB v = u + at v = 0 + gt sin 



gt sin  

v



mg cos  qB

m cot  Hence option (C) is correct. qB Example 22. An electron having kinetic energy T is moving in a circular orbit of radius R perpendicular to a uniform  magnetic induction B . If kinetic energy is doubled and magnetic induction tripled, the radius will become t



3R 2

3 R 2 mv P R  Sol. qB qB P2 But kinetic energy T   2m 2mT R  qB (A)



(B)

2m(2T) q(3B) 2 2 R´ R R 3 9

(C)

2 R 9

(D)

4 R 3

P  2mT

R´

Hence, option (C) is correct.

Example 23. A charged particle (charge q, mass m) has velocity v0 at origin in +x direction. In space there is a uniform magnetic field B in –z direction. Find the ‘y’ coordinate of particle when is crosses y–axis. Sol. The parth of charged particle is circular whose radius is + B mv0 r qB C 2mv0 y  2r   O qB Example 24. A mass spectrometer is a device which select particle of equal mass. An iron with electric charge q > 0 starts at rest from a source S and is accelerated through a potential difference V. It passes through a  hole into a region of constant magnetic field B perpendicular to the plane of the paper as shown in the figure. The particle is deflected by the magnetic field and emerges through the bottom hole at a distance ‘d’ from the top hole. The mass of the particle is qBd qB2 d 2 qB2 d 2 (A) (B) (C) V 4V 8V

S B V

(D)

qBd 2V

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15

MAGNETIC FIELD Sol. The speed of charged particle just before entering the magnetic field is V0. qV 

1 mv 20 2

2qV m The radius of circular path in magnetic field is v0 

r

mv0 qB

or

d mv0  2 qB

or

d 2 m 2 v02 m2 2qV   4 q 2 B2 q 2 B2 m

or

d 2 2mV  4 qB2



m

qB2d 2 8V

Hence, option (C) is correct

Example 25. A cyclotron is operating with a flux density of 3 Wb/m2. The ion which enters the field is a proton having mass 1.67 × 10–27 kg. If the maximum radius of the orbit of the particle is 0.5m, find : (a) the maximum velocity of the proton, (b) the kinetic energy of the particle, and (c) the period for a half cycle. Sol. (a) As in case of motion of a charged paricle in a magnetic field,

r

mv qBr i.e., v  qB m

So,

v max 

So,

v max 

(b) i.e.

qBrmax m

1.6  1019  3  0.5  1.43  108 m / s 27 1.67  10 2 1 1 K  mv 2   1.67  1027  1.43  108 2 2



K  1.71  1011 J 



1.71  1011  107 MeV 1.6  1019

(c) In case of circular motion, as T

2r 2  0.5   2.19  108 s v 1.43  108

So, time for helf cycle, T 

1 (T)  1.09  108 s 2

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MAGNETIC FIELD Example 26. The region between x = 0 and x = L is filled with uniform steady magnetic field B0 kˆ . A particle of mass ‘m’, positive charge ‘q’ and velocity v 0 ˆi travels along x-axis and enters the region of the magnetic field. Neglect the gravity throughout the question (a) Find the value of ‘L’ if the particle emerges from the region of magnetic field with its final velocity at an angle 30º to its initial velocity. (b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now expands upto 2.1 L. Sol. (A)  = 30º, L x=0 x=L sin   X + + + + R B = B0 k

mv 0 R qB0

Here

or  (b)

L mv0 qB0 1 qB0L  2 mv0 mv0 L 2qB0 In part (i) sin 30º 

or Now when

+

+

R

sin 30º 



+

C

L´ = 2.1 L

+

+

+

+

+L +

+

+

+

A + q× +

+ v0 + +

v0 P

+

Y

R × × v0 × B ×

L R

1 L  2 R

+

or

L = R/2

or

2.1 R 2

× ×

×

×

v0 ×

×

×

×

2.1 L´ = R > R 2

 L´ > R Therefore, deviation of the particle is  = 180º is as shown.    v f   v ˆi  v B 0

t AB

Example 27. A wire loop carrying a current ‘I’ is placed in the x–y plane as shown in fig. (a) If a particle with charge +Q and mass ‘m’ is placed at the centre  ‘P’ and given a velocity v along NP (see figure), find its instantaneous acceleration.  (b) If an external unfiorm magnetic induction field B  B ˆi is applied

M a I

120º

and

T m   2 qB0

V +Q P O

y x

N

find the force and the torque acting on the loop due to this field. Sol. The magnetic field at the centre P due to current in wire NM is µI B1  0 sin 60º  sin 60º  4r

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MAGNETIC FIELD y

µ I  3 3 B1  0    4 a / 2  2 2 

x

M Q

sin 30º 



120º

µ0 2I 3 4 a directed away from the reader perpendicular to the plane of paper. B1 

Q

V 60º

V cos60º

N

r a

M

a r 2

MS 

v

P

30º

30º a S

MS cos 30º  a



V sin60º

a 30º

P

r a

3a 2

N

 MN  3 a The magnetic field at the centre P due to current in arc MN is µ0 2I    µ0 2I  2 / 3  µ0 2I    2 a  2  4 a  2  4 3a directed towards the reader perpendicular to the plane of paper The net magnetic field B2 

B  B1  B2 

µ0 2 3I µ0 2I  4 a 4 3a

µ0 2I   µ 2I 3    0   (.68)  4a  3  4 a (directed away from the reader perpendicular to the plane of paper) The force acting on the charged particle Q when it has a velocity v and is instantaneously at the centre is F = QvB sin  = QvB sin 90º = QvB The acceleration produced B

A

F QvB Qv  µ0 2I      (0.68)   M m m  4 a 

A

0.11µ0 IQv ma y M Q Q

x

R

120º

v

P

B

N

  The direction of acceleration is given by the vecotr product v  B or by applying Fleming’s left hand rule

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MAGNETIC FIELD RPN = 90º and MPN = 120º  MPR = 120º – 90º = 30º Since, MPQ = 60º  RPQ = 30º i.e., wthe acceleration vector makes an angle of 30º with the negative x-axis. The torque acting on the loop in the magnetic field is giving.    MB where M = IA where A = (area of PMQNP) — (area of trangle PMN)



A

1 1 a 2   MN  PS 3 2

A

 a 2 1 a 3   3a   a 2    3 2 2 3 4 

 

  A  a2   3     Ia 2   3

3 ˆ k 4  3 ˆ ˆ kiB 4 

  3ˆ 2   BIa 2    j  0.614 BIa Jˆ 3 4  The force acting on the loop is zero. S

Example 28. An electron gun G emits electrons of energy 2 keV travelling in the positive X-direction. The electrons are requeired to hit the spot S where GS = 0.1 m, and the line GS make an angle of 60º with the x-axis as shown  in figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electrons hit S. Sol. Kinetic energy of electron, K

B G

60º

X

1 mv 2  2keV 2 S

 speed of electron,

B

v

2K m

v

2  2  1.6  1016 m/s 9.1  10 31

G

60º

v

v = 2.65 × 107 m/s   Since the velocity (v) of the electron makes an angle of  = 60º with the magnetic field B , the path will be a helix. So, the particle will hit S if GS = nP Here n = 1, 2, 3, .............

p  pitch of helix 

2 m v cos  qB

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19

MAGNETIC FIELD But for B to be minimum, n = 1 Hence,

GS  p 

2m v cos  qB

2mv cos  q(GS) Substituting the values, we have B  Bmin 



Bmin

 1 (2)(9.1  1031 )(2.65  107 )    2  tesla 19 (1.6  10 ) (0.1)

Bmin = 4.73 × 10–3 tesla

or

 Example 29. An electron moving with a velocity V1  2 ˆi m/s at a point in a magnetic field experiences a force   F1  2ˆjN . If the electron is moving with a velocity V 2  2ˆj m/s at the same point, it experiences a force   F2  2 ˆi N . The force the electron would experience if it were moving with a velocity V 3  2 kˆ m/s at the same point is   (A) zero (B) 2kN (C) 2kN (D) informationis insufficient    Sol. F1  qv1  B   2 ˆj   e(v1 )  B  2 ˆj   e(2 ˆi)  B  ˆj  2e( ˆi  B)



ˆj  eB( ˆi)  (  k) ˆ eB = 1  ˆ F2   e(2 ˆj)  (  B k) 2 ˆi  2eB ˆi eB = 1    F3   e(v3  B)   ˆ  (  B k) ˆ 0 F3   e(2 k)

Hence option (A) is correct

  Charged particle in uniform E & B

   When a charged particle moves with velocity v in an electric field E and magnetic field B , then. Net force experienced by it is given by following equation.     F  qE  q(v  B)  Combined force F is known as lorentz force    (i) E || B || v E B v In above situation particle passes underviated but its velocity will change due to electric field and magnetic force on it is zero.

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(ii)

MAGNETIC FIELD

  E || B and uniform, particle is released with velocity v0 at an angle . y v0

+q

v0

v0 sin +q

E, B

x E, B v0 cos

z

r

mv 0 sin  ; qB

T

2 m qB

x y

  1 qE 2  ˆ r   v 0 cos   t  t  i   R sin t  ˆj  R  R cos ˆt  kˆ 2 m  

 



R c

Cycloid motion   Suppose that B points in the x-direction, and E in the z-direction. z

v0 sin

z

E

0

a

c

b

y

B

x

      ˆ  Byz  ˆ   ma  m   F  q E  v  B  q  Ezˆ  Bzy yyˆ  zzˆ 





 qB  m E     z     y  y  z, B 

Their general solution is

y(t)  C1 cos t  C2 sin t  (E / B)t  C3   z(t)  C 2 cos t  C1 sin t  C4  y(t) 

E (t  sin t), B

z(t) 

E (1  cos t) B

E B (y – Rt)2 + (z – R)2 = R2 R

E B The particle moves as through it were a spot on the rim of a wheel, rolling down the y axis at speed, v. The  curve generated in this way is called a cycloid. Notice that the overall motion is not in the direction of E , but perpendicular to it. v  R 

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21

MAGNETIC FIELD Example 30. A particle of specific charge (charge/mass)  starts moving from the origin under the action of an   electric field E  E ˆi and magnetic field B  B kˆ . Its velocity at (x , y , 0) is (4 ˆi  3 ˆj) . The value of x is : 0

0

0

13 E0 (A) 2 B 0

16B0 E0

(B)

25 (C) 2E 0

0

0

5 (D) 2B 0

Sol. The workdone by magnetic force is WB = 0, because, magnetic force is always perpendicular to instantaneous displacement.   W  qE . S  qE ˆi . (x ˆi  y ˆj)  qE x E

0

0

0

0

0

The speed of particle at (x0, y0) is v  42  ( 3) 2  5m / s

According to work – energy theorm, W  T 

1 1 mv 2  mu 2 2 2

or

WE  WB 

or

qE 0 x 0 



x0 

1 m 52  0 2

25 m 2

25m 25  2qE 0 2E 0

 q       m 

Hence, option (C) is correct. Example 31. A particle of specific charge (q/m) is projected from the origin of coordinates with initial velocity [u ˆi  v ˆj] . Uniform electric magnetic fields exist in the region along the +y direction, of magnitude E and B. The particle will definitely return to the origin once if (A) [vB/2E] is an integer (C) [vB/E] in an integer ay 

Sol.

qE m

quB 

also,

(B) (u2 + v21/2 [B/E] is an integer (D) [uB/E] is an integer

mu 2 r

r

mu qB

T

2m qB

y   vt 

qE 2 t 2m

For origin, x = 0, and y = 0 0   vt 

qE 2 t 2m

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MAGNETIC FIELD

2mV qE For returning at the origin, 

t

nT 

2mV qE

2nm 2mV  qB qE



n

2mVqB qE(2m)

n

VB E

Hence, option (C) is correct

FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES Consider two long wires 1 and 2 kept parallel to each other at a distance ‘r’ and carrying currents i1 and i2 respectively in the same direction. The force per unit length of the wire 2 due to wire 1 is :

1

2

i1

i2 F

F µ0 i1i 2  l 2 r The same force acts on wire 1 due to wire 2.

l

r

NOTE : The wires attract each other if currents in the wires are flowing in the same direction and they repel each other if the currents are in opposite directions. Example 32. A conductor of length 2 m carrying current of 2 A is held parallel to an infinitely long conductor carrying current of 10 A at a distance of 100 mm. Find the force on small conductor. Sol. We know that force per unit length of short conductor due to long conductor is given by µ 2i i f 0 12 4 r  Total force on length l of the short conductor is µ 2i i F  fl  0 1 2 l 4 r 7 10  2  2  10  2 F  8  105 N 0.1 Force is attractive if the direction of current is same in both the parallel conductors and is repulsive if the direction of current is opposite in two parallel conductors. Example 33. A straight segment OC (of length L) of a circuit carrying a current ‘i’ is placed along the X-axis as shown in figure. Two infinitely long straight wires A and B, each extending from z = –  to + , are fixed at y = –a and y = + a respectively. as shown in the figure. If the wires A and B each carry a current ‘i’ into the plane of the paper, obtain the expression for the force acting on the segment OC. What will be force of OC if the current in the wire B is reversed ?

Y

×

B

O

i

C

X

×A Z

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MAGNETIC FIELD Sol. (a) Let us assume a segment of length dx at a point P, a distance x from the centre shown in figure. y I×B

× dx

a

i

O

x

C

a

×

x

BA

BB I

is the positive Z-direction and is the negative Z-direction and

Bnet L

A

y´

Magnetic field at P due to current in wires A and B will be in the directions perpendicular to AP and BP respectively as shown.    µ i | BA |  | BB |  | B |  0 2 AP Therefore, net magnetic field at P will be along negative y = axis as shown  µ  i  x  Bnet  2 | B | cos   2  0  and  2  AP  AP  µ ix  µ  i.x Bnet   0   0. 2 2    (AP)  (a  x 2 ) Therefore, force on the element will be (F = ilB) ix  µ dF  i  0 2 dx (in negative z-direction) 2    a x   Total force on the wire will be xL

L

µ i 2 xdx F   dF  0  2  0 x  a2 x0 µ0 i 2  L2  a 2  F ln 2  a 2 

(in negative z-axis)

 µ0i 2  L2  a 2  ˆ F   ln k Hence 2  a 2  (b) If current in wire B is reversed, then magnetic fields due to A and B will be in the directions shown in figure. i.e., net  magnetic field Bnet will be along positive x-axis and since current is also along positive x-axis, force on wire OC will be zero.  Note : BA is not necessarily parallel to BP..

CURRENT LOOP IN UNIFORM MAGNETIC FIELD    MB  |  | MBsin  ; where M = NIA Work done in rotating loop in uniform field from 1 to 2 W = MB (cos 1 – cos 2)

Y B

BB P

i

Bnet C

X

BA

A×

M

B

I

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MAGNETIC FIELD Example 34. The magnetic moment of a circular orbit of radius ‘r’ carrying a charge ‘q’ and rotating with velocity ‘v’ is given by qvr qvr (B) 2 2 Sol. The convection current is

(A)

(D) qvr2

(C) qvr

q qv  2 2r  The magnetic moment is I

M  Ir 2 



M

qv 2 r 2r

qvr 2

Hence option (B) is correct

Example 35. Q charge is uniformly distributed over the same surface of a right circular cone of semi-vertical angle  and height ‘h’. The cone is uniformly rotated about its axis at angular velocity . Calculated associated magnetic dipole moment. Sol. I = The moment of gnertia 



3 MR 2 10

Angular momentum  L  I 

But magnetic moment P 

QL 2m



P

Q 3 mR 2  2m 10



P

3Q 2 R  20



R h R = h tan



P

But

3 mR 2 10

tan  

3QR 2 3Qh 2 tan 2   20 20 y

Example 36. Figure shown a square current carrying loop ABCD of side 10 cm and  current i = 10 A. The magnetic moment M of the loop is 2 2 (A) (0.05) ˆi  3 kˆ A  m (B) (0.05) ˆj  kˆ A  m

 (C) (0.05) 

 3 ˆi  kˆ  A  m

2

Sol. The magnitude of magnetic moment is M = Il2 = 10 × (10 × 10–2)2 Am2 = 10 × 10–2 = 0.1 Am2

  (D)  ˆi  kˆ  A  m

B A

2

C i = 10 30º x D

z

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25

MAGNETIC FIELD The normal on the loop is in x – z plane. It makes 60º angle with x - axis.   M  M cos 60º ˆi  M sin 60º ˆj  

 M 3 ˆ M  ˆi  Mj 2 2  0.1 ˆi  3 ˆj M 2  M  (0.05) ˆi  3 ˆj Am 2









2n, 2a, 2I

Example 37. A rectangular coil PQ has 2n turns, an area 2a and carries a current 2I, (refer figure). The plane of the coil is at 60º to a horizontal uniform magnetic field coil of flux density B. The torque on the coil due to magnetic force is (A) BnaI sin 60º (B) 8 BnaI cos 60º (C) 4 BnaI sin 60º (D) none    Sol.   MB Here M = 2n(2I) (2a) M = 8 nIa   = MB sin(90º – 60º)  = MB cos 60º = 8 nIa cos 60º Hence option (B) is correct

60º B

Example 38. (a)A rigid circular loop of radius ‘r’ & mass ‘m’ lies in the ‘xy’ plane on a flat table and has a current    ‘I’ flowing in it. At this particular place, the earth’s magnetic field is B  B x ˆi  B y ˆj. How large must ‘I’ be before one edge of the loop will lift from table ?    ˆ (b) Repeat if, B  B x ˆi  B y k. Sol. (a) Torque due to magnetic force should be greater than torque due to weight. 

Ir 2 B  mgr

or

Ir 2 B2x  B2y  mgr

 (b) Since, Bz is parallel to dipolemoment 

I r 2 B x  mgr



I

mg rBx

Example 39. A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current i = 4 A. A horizontal magnetic field B B = 10T is switched on at time t = 0 as shown in figure. The initial angular acceleration of the ring will be (A) 40 rad/s2 (B) 20 rad/s2 (C) 5 rad/s2 (D) 15 rad/s2 Sol. As we know, if a coil or a closed of any shape is placed in uniform electric field, magnetic force on the coil or the loop is zero.

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26

MAGNETIC FIELD So, the centre of mass of the coil remains in rest. the torque on the coil is   = IAB = 4r2B or

mr 2   4r 2 B 2





8B 8  10   40rad / s 2 m 2

Hence option (A) is correct

Example 40. In the figure shown a coil of single turn is wound on a sphere of radius R and mass ‘m’. The plane of the coil is parallel to the plane and lies in the equatorial plane of the sphere. Current in the coil is ‘I’. The value of B if the sphere is in equilibrium is mg cos  IR Sol. For equilibrium,

(A)

(B)

mg IR

(C)

mg tan  IR

(D)

B

mg sin  IR P

mg sin  = f Also, net torque should be zero. The magnetic moment of the loop is perpendicualr to the plane.  The torque due to magnetic force is B = PB sin(180 – )  B = PB sin = IR2B sin This torque balances the torque due to friction about centre of mass.  FR = IR2Bsin or mg sin  = IRB sin  

B

mg IR

180º – B

Hence option (B) is correct

Example 41. A square current carrying loop made of thin wire and having a mass m = 10 g can rotate without friction with respect to the vertical axis OO1, passing through the centre of I the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneous magnetic field with an induction B = 10–1 T directed at right angles to the plane of the drawing. A current I = 2A is flowing in the loop. Find the period of small oscillations that the loop performs about its position of stable equilibrium. Sol.  = – PB sin or  = – IAB or I0 = – IAB

O+ B

O1

IAB I0

or





 



 2  

or



IAB I0

IAB I0

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27

MAGNETIC FIELD

or

2  T



T  2

Here

m l2 m l2 l l I0  m 0    m 0    0  0  2  2 12 12

IAB I0

I0 IAB 2

2

m 0 l 2 m 0l 2 I0   2 6 m  2.5  103 kg 4 l2 = A = area of loop, I = 2 amp After putting the value, T = 0.57 sec.

But

m0 

 Example 42. A uniform constant magnetic field B is directed at an angle of 45º to the X-axis in X – Y plane. PQRS is a rigid square wire frame carrying a steady current I0, with its centre at the origin O. At time t = 0, the frame is at rest in the position shown in the figure with its sides parallel to X and Y axes. Each side of the frame is of mass M and length L. Y  (a) What is the torque  about O acting on the frame due to the I0 S R magnetic field ? (b) Find the angle by which the frame rotates under the action of X this torque in a short interval of time t, and the axis about which this rotation occurs (t is so short that any variation in P Q the torque during this interval may be neglected). Given : the moment of inertia of the frame about an axis through its centre

4 ML2 . 3 Sol. Magnetic moment of the loop,  M  (iA)kˆ  (I0 L2 )kˆ  ˆ ˆ B (iˆ  ˆj) Magnetic Field, B  (B cos 45º )i  (Bsin 45º ) j  2 (a) Torque acting on the loop,    ˆ   B (iˆ  ˆj)    M  B  (I0 L2 k)  2     I0 L2 B   (ˆj  ˆi)  or |  | I0 L2B 2

perpendicular to its plane is

(b) Axis of rotation coincides with the torque and since torque is in ˆj  ˆi direction or parallel to QS. Therefore, the loop will rotate about an axis passing through Q and S as shown along-side :  ||  Angular acceleration, I

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28

MAGNETIC FIELD I = moment of inertia of loop about QS. IQS + IPR = IZZ (From theorem of perpendicular axis) But IQS = IPR 4 2 2IQS  IZZ  ML2 IQS  ML2   3 3  2 |  | I0 L B 3 I0 b    2 I 2 M 2  ML 3  Angle by which the frame rotates in time t is 1 3IB    . (t)2   0 . (t) 2 or 2 4 M

Y

Where

S

R X

P

Q

MOVING COIL OR SUSPENDED COIL OR D´ ARSONVAL TYPE GALVANOMETER Principle : When a current-carrying coil is placed in magnetic field, it experiences a torque. Construction : It consists of a narrow rectangular coil PQRS consisting of a large number of turns of fine insulated copper wire wound over a frame made of light, non-magnetic metal. A soft iron cylinder known as the core is placed symmetrically within the coil and detached from it. The coil is suspended between the two cylindrical pole-pieces (N and S of a strong permanent horse-shoe magnet) by a thin flat phosphor bronze strip, the upper end of which is connected to a movable torsion head T. The lower end of the coil is connected to a hair-spring ‘s’ of phosphor bronze having only a few turns. T

T2

T1 m S

P N

S

Core

R

Q s

Moving coil galvanometer

In order to eliminate air-disturbance, the whole arrangement is enclosed in a brass case having a glass window on the front. Levelling screws are provided at the base. The torsion head T is connected to a binding terminal T1. So, the phosphor-bronze strip acts as one ‘current lead’ to the coil. The lower end of the spring ‘s’ is connected to a binding terminal T2. A plane mirror or a concave mirror of larger radius of curvature is rigidly attached to the phosphor bronze strip. This helps to measure the deflection of the coil by lamp and scale arrangement. Radial magnetic field : The magnetic field in the small air gap between the cylinderical pole-pieces is radial. The magnetic lines of force within the S N air gap are along the radii. On account of this, the plane of the coil remains always parallel to the direction of the magnetic field i.e., the angle between Radial magnetic field the plane of the coil and the magnetic field is zero in all the orientations of the coil.

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MAGNETIC FIELD Theory : Let I = current flowing through the coil; B = magnetic field induction l = length of the coil; b = breadth of the coil N = number of turns in the coil; A(= l × b) = area of the coil Since the field is radial, therefore, the plane of the coil remains parallel to the magnetic field in all the orientations of the coil. So, the sides SP and QR remain parallel to the direction of the magnetic field. So, they do not experience any force. The sides PQ and RS remain perpendicualr to the direction of the magnetic field. These sides experience forces perpendicular to the plane of the coil. b S

P

F l F R

Q

Current-carrying loop in magnetic field

Force on PQ, F = NBIl Applying Fleming’s left hand rule, we find that this force is normal to the plane of the coil and directed outwards, i.e., towards the reader. Force on RS, F = NBIl Applying Fleming’s left hand rule, we find that this force is normal to the plane of the coil and directed inwards, i.e., away from the reader. The forces on the sides PQ and RS are : (i) equal in magnitude; (ii) opposite in direction; and (iii) act at different points. So, the two forces constitute a couple. This couple tends to deflect the coil and is known as deflecting couple. Moment of deflecting couple = NBIl × b = NBIA [The field is radial. The forces on the sides PQ and RS always remain perpendicular to the plane of the coil. So, the perpendicular distance between the forces is always equal to ‘b’ as in fig.] F P

S b

F

When the coil deflects, the suspension fibre gets twisted. On account of elasticity, a restoring couple is set up in the fibre. This couple is proportional to the twist. If  be the angular twist, then Moment of restoring couple = k where ‘k’ is the restoring couple per unit angular twist. It is also known as torsional constant. For equilibrium of the coil,  k  I  I  K NBIA = k or or  NBA   where K  

Now,

k   is the galvanometer constant. NBA  I 

or

 I

So, the deflection of the coil is proportional to the current flowing through the coil. This explains as to why we can use a linear scale in a galvanometer. The scale is calibrated to give direct values of current.

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MAGNETIC FIELD CURRENT SENSITIVITY OF A GALVANOMETER A galvanometer is said to be sensitive if it gives a large deflection for a small current. The current sensitivity of a meter is the deflection of the meter per unit current, i.e.,

 . It is given by I

 NBA  [ NBIA = k] I k The sensitiveness can be increased by increasing N, A and B and decreasing the value of ‘k’. But N and A cannot be increased much because this will increase the length and consequently the resistance of the coil. In that case, the galvanometer will not respond to weak electric currents. B can be increased by using a strong magnet. ‘k’ can be decreased by using phosphor-bronze for suspension. ‘k’ can be further reduced by using quartz suspension fibre. VOLTAGE SENSITIVITY OF A GALVANOMETER It is defined as the deflection of the meter per unit voltage, i.e., Now,

   V RI

 . V

 NBA  V kR

or

Advantages : (i) The galvanometer can be made extremely sensitive. (ii) Since the magnetic field B is very high, therefore, the external magnetic fields cannot appreciably after the deflection of the coil. So, the galvanometer can be used in any position. (iii) Since the deflection of the coil is proportional to current, therefore, linear scale can be used. (iv) Since the coil is wound over metallic frame, therefore, damping is produced by eddy currents. So, the galvanometer coil comes to rest quickly. This type of galvamometer is called aperiodic or dead beat galvanometer. The galvanometer can be made ballistic by winding the coil on a non-conducting frame of ivory or ebonite. (v) The lamp and scale arrangement used to measure the deflection of the coil makes the galvanometer very sensitive. POINTER TYPE OR WESTON OR PIVOTED MOVING COIL GALVANOMETER The suspended type moving coil galvanometers are very sensitive. They can measure currents of the order of 10–9 ampere. But these require very careful handling. So, for general use in the laboratory and for those experiments whose sensitivity is not required, pointer type galvanometers are used. 10

20 30

N T1

0

10

20

30

S T2

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31

MAGNETIC FIELD In this type of galvanometer, the coil is pivoted between two ball-bearings. A light aluminium pointer is attached to the moving coil. The controlling couple is provided with the help of a spring. NOTE : Numerical Examples based on Moving Coil Galvanometer Formulae used : 1.  = NBIA 2. k = NBIA 3. Current sensitivity,

α NBA = I k

α NBA = V kR 2 Units used : B in tesla, A in m , R in ohm, k in N m rad–1.

4. Voltage sensitivity,

EARTH’S MAGNETISM A freely suspended magnet always points in the north-south direction even in the absence of any other magnet. This suggests that the earth itself behaves as a magnet which causes a freely suspended magnet (or magnetic needle) to point always in a particular direction : north and south. The shape of earth’s magnetic field resembles that of a bar magnet of length one-fifth of earth’s diameter buried at its centre. Geographic axis Geographic N-pole

Magnetic S-pole

Eq ua to

S M eq agn ua et i to c r

Geographic S-pole

r

N

Magnetic N-pole Magnetic axis

The south pole of earth’s magnet is towards earth’s north pole (eographical north), while the north pole of earth’s magnet is towards earth’s south pole (geographical south). Thus, there is a magnetic S-pole near the geographical north, and a magnetic N-pole near the geographical south. The positions of the earth’s magnetic poles are not well defined on the globe, they are spread over an area. Magnetic equator : The great circle whose place is perpendicular to the earth’s magnetic axis is called earth’s magnetic equator. Geographical equator : The great circle whose plane is perpendicular to geographical axis is called geographical equator. Magnetic meridian : The line joining the earth’s magnetic poles is called the magnetic axis and a vertical plane passing through it is called the magnetic meridian. Geographical meridian : The line joining the geographical north and south poles is called the geographic axis and a vertical plane passing through it is called the geographical meridian. Magnetic Elements To have a complet knowledge of earth’s magnetism at a place, the following three elements must be known : (i) Angle of declination (ii) Angle of dip or inclination (iii) Horizontal component of earth’s field. (i) Angle of declination : The angle between the magnetic meridian and geographical meridian at a place is called the angle of declination (or simply the declination) at that place.

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MAGNETIC FIELD Geographic north

Geographical meridian

B´

Magnetic north

H

B

Magnetic meridian

C´ C

D

In fig. ABCD is the magnetic meridian and AB´C´D is the geographical meridian. The angle B´AB =  is the angle of declination. (ii) Angle of dip or inclination : The angle which the axis of needle makes with the horizontal, is called angle of dip (). In other words, the angle of dip at a place is the angle which the resultant magnetic field of earth at that place makes with the horizontal. H

B C i N

S

A V D

In fig. AC shows the direction of resultant magnetic field of earth and the angle BAC (=) between it and the horizontal AB is the angle of dip. (iii) Horizontal component of earth’s field : The direction of earth’s field at the magnetic poles is normal to the earth’s surface (i.e., in vertical direction) and at magnetic equator it is parallel to the earth’s surface, (i.e., in horizontal direction). Thus, the resultant earth’s field can be resolved in two components as shown in fig. (a) the horizontal component H along AB and (b) the vertical component V, along AD. From fig. Horizontal component H = Be cos  ...(i) and vertical component V = Be sin  ...(ii) 

V Be sin    tan  H Be cos 

or V = H tan  Again Eqs. (i) and (ii) give



H 2  V 2  B2e cos 2   sin 2 

or



Be  H 2  V 2

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MAGNETIC FIELD THINKING PROBLEMS



 

1. Of the three vectors in the equation F  qv  B, which pairs are always at right angles? Which may have any angle between them ? 2. If an electron is not deflected in passing through a certain space, can we be sure that there is no magnetic field in that region? 3. A bream of protons is deflected side ways. Could this deflection be caused (a) by an electric field? (b) by a magnetic field ? (c) If either is possible, how can you tell which one is present? 4. A rectangular current loop is in an arbitary orientation in an external magnetic field. Is any work required to rotate the loop about an axis perpendicular to its plane? 5. Will a tangent galvanometer work in the polar region ? 6. At that deflections is a tangent galvanometer most sensitive? 7. Why is the magnetic needle short in a tangent galvanometer ? 8. Why is the field in a moving coil galvanometer radial in nature? 9. What is the greatest disadvantage with a suspended-type moving coil galvanometer? 10. An ammeter is a Weston galvanometer whose resistance is made negligible by shunting the coil. Is this true or false? 11. A voltmeter is a Weston galvanometer of very high resistance. Is this true of False? 12. Which gives a more accurate value of a potential difference, a potentiometer or a voltmeter? 13. Why is an ammeter connected in series? 14. What is a faraday? 15. There is no charge in the energy of a charged particle moving in a magnetic field althrough a magnetic force is acting on it. Is this true or false? Give reasons in support of your answer. 16. A current-carrying circular conductor is placed in a uniform magnetic field with its plane perpendicular to the field. Does it experience any force? If it does, what is this force if its radius is a and the current passing through it equals? 17. How do you know that the current inside a conductor is constituted by electrons and not by protons? 18. A copper pipe is filled with an electrolyte. When a voltage is applied, the current in the electrolyte is constituted by the movement of positive and negative ions in opposite directions. Will such a pipe experience a force when placed in a magnetic field perpendicular to the current? 19. Two parallel wires currying current in the same direction attract each other while two beams of electrons travelling in the same direction repel each other. Explain why ? 20. Can a charged particle entering a uniform magnetic field normally from outside complete a circle? 21. A cylindrical electrolytic bath containing a solution of copper sulphate between two electrodes is mounted above the north pole of a strong electromagnet. One electrode is at the axis (–) and the other electrode is at the edge of the bath (+). What happens to the electrolyte in these circumstances? 22. A very strong current is made to flow for a short time through a solenoid. Will there be any change in its length and diameter? Explain. 23. Cosmic rays are charged particles that strike the atmosphere from some external source. It is found that more low-energy cosmic rays each the earth at the north and south magnetic poles than at the magnetic equator. Why is this so ?

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MAGNETIC FIELD

1. 2. 3.

4. 5. 6. 7. 8. 9. 10. 12. 13.

14. 15. 16. 17.

18.

19.

20. 21. 22.

23.

SOLUTION OF THINKING PROBLEMS       The pairs F , B and F , v are always at right angle. B and v may have any angle between them. No, we cannot be sure that there is no magnetic field because the force will be zero when the direction of motion is along the direction of the field. (a) Yes, it could be due to an electric field directed perpendicular to the motion. (b) Ys, it could be due to a magnetic field. (c) Stop the proton and keep it stationary. If it sill experiences a force in the same direction, it is due to an electric field. If the force vanishes of stopping the motion, there is a magnetic field. No, no work is done in rotating the coil because work done W = mB (1 –cos). Here there is no charge in  and so no work is done. No, because the earth’s magnetic field is vertical there. A tengent galvanometer is most sensitive when the deflections are near about 0º. The field due to the circular coil is uniform over a very small region about the centre of the coil. So the needle must be short so that it may be assumed to move in uniform megnetic fields. The field is made radial in order to have a linear relation between the current and the deflection. The greatest disadvntage is that this type of galvanometer is not portable. It is true. 11. It is true. A potentiometer gives a more accurate value of a potential difference than a voltmeter as it draws no current from the cell. Since the current following through the circuit has to pass through the ammeter, it must be connected in series because it is a characteristic property of series connection that the same current passes through all parts of the circuit. A faraday is the amount of charge required to liberate 1 gram equivalent of any sustance. True. A charged particle experiences a force at right angles to the velocity and so it moves in a circle with a constant speed. So there is no charge in the energy fo the particle though a magnetic force acts on it. It does not experience may force because the forces on the elements are directed radially away and sum up to zero. By the Hall effect. When a current-carrying flet conductor is placed in a magnetic field perpendicular to the flat face, a transverse voltage is developed. From the direction of this voltage called Hall voltage, it can be determined whether the current is due to negative or positive charge carriers. The current constituted by negtive ions in the positive direction is the same as the current formed by the positive ions and so both form current in the same direction; they do not cancel out each other. So the pipe will experience a force. A current in on wire produces only a magnetic field and on electric field (because a current-carrying conductor is electrically neutral) over the other current-carrying conductor and so only a magnetic force which is attractive is nature arises between them. But an electron beam is a source of both, an electric and a magnetic field. So there arise both magnetic force (attractive) and electric force (repulsive) between them. The repulsive force being in excess of the attractive force, they repel each other. No, the boundary line of the magneitc field has to be the diameter of the circle, whatever be the velocity of the charged particle. Hence, it will complete a semi-circle. It moves counterclockwise (viewed from above) The current in the adjacent turns flow in the same direction and so they attract each other. Thus there is contraction along the length of the solenoid. As the same currents in the diametrically opposite elements flow in opposite directions, they repel each other. On account of this repulsive force the diameter of the solenoid increases. Because at the poles, the magnetic field is parallel to the direction of motion of the cosmic ray particles, both being vertical. Hence the cosmic ray particles do not experience any force. The force experienced is given by   F = qvb sin( v , B ). At the equator these particles experience maximum deflecting force and hence low-energy gy particles cannot reach the earth.

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MAGNETIC FIELD

ASSERTION & REASON A statement of Statement-1 is given and a Corresponding statement of Statement-2 is given just below it of the statements, mark the correct answer as – (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 is NOT correct explanation of Statement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) If Statement-1 is false but Statement-2 is true. 1.

Statement-1 : Statement-2 :

Magnetic field interacts with a moving charge and not with a stationary charge. A moving charge produces a magnetic field.

2.

Statement-1 :

If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic field in this region. Force is directly proportional to magnetic field applied.

Statement-2 : 3.

Statement-1 : Statement-2 :

Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it. The average velocity of free electron is zero.

4.

Statement-1 : Statement-2 :

Electron cannot be accelerated by the cyclotron. Cyclotron is suitable only for accelerating heavy particles.

5.

Statement-1 : Statement-2 :

The coil is wound over the metallic frame in moving coil galvanometer. The metallic frame help in making steady deflection without any oscillation.

6.

Statement-1 : Statement-2 :

In electric circuits, wires carrying currents in opposite directions are often twisted together. If the wire are not twisted together, the combination of the wires forms a current loop. The magnetic field generated by the loop might affect adjacent circuits or components.

7.

Statement-1 :

If an electron and proton enter in an electric field with equal energy, then path of electron is more curved than that of proton. Electron has a tendency to form large curve due to small mass.

Statement-2 : 8.

Statement-1 : Statement-2 :

9.

Statement-1 : Statement-2 :

10. Statement-1 :

Statement-2 : 11. Statement-1 :

Statement-2 : 12. Statement-1 :

Statement-2 :

If a proton and an  -particle enter a uniform magnetic field perpendicularly, with the same speed, the time period of revolution of  -particle is double that of proton. In a magnetic field, the time period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to charge of particle. If an electron while coming vertically from outer space enter the earth’s magnetic field, it is deflected towards west. Electron has negative charge. An electron and proton enters a magnetic field with equal velocities, then, the force experienced by proton will be more than electron. The mass of proton is 1837 times more than the mass of electron. The magnetic field produced by a current carrying solenoid is independent of its length and cross sectional area. The magnetic field inside the solenoid is uniform. The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. [JEE 2008]

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MAGNETIC FIELD

Level # 1 1.

2.

An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of V volts between the cathode and the anode. The particle then passes through a uniform transverse magnetic field in which it experiences a force F. If the potential difference between the anode and the cathode is increased to 2 V, the electron will now experience a force (A) F 2 (B) F 2 (C) 2 F (D) 2 F In a hydrogen atom, an electron of mass m and charge e is in an orbit of radius r making n revolutions per second. If the mass of the hydrogen nucleus is M, the magnetic moment associated with the orbital motion of the electron is (A)

3.

n er 2 m M

(B)

n er 2 M m

(C)

n er 2 m (M  m)

(D)  n e r 2

An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus if B. The angular velocity  of the electron is 4 r B 0 e B 2  0 eB 2 r B (A)   (B)   (C)   (D)   r 0 e 0 e 4 r

D 4.

Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Figure. The force experienced by a 25 cm length of wire C is (A) 0.4 N (B) 0.04 N (C) 4 x 10–3 N (D) 4 x 10–4 N

C

30 A

G

10 A

3 cm 5.

10 cm 2

A charged particle of specific charge s passes through a region of space shown. (A) (B) (C)

 . B Work done to move the charged particle in REGION 1 is ZERO. The radius of the trajectory of the charged particle in

Velocity of the particle in REGION 1 is  

 REGION 2 is . s 0

1

v S

x x B

x

x

x Ex

x

x

7.

x x x xB x x x x 0 x x x x x x x x x x x x x x x x x x x x x x x x REGION 2

 REGION 1 The particle emerges from REGION 2 with a velocity  '   where  ' =  .  A wire of resistance R in the form of a semicircle lies on the top of a smooth table. A uniform magnetic field B is confined to the region as shown. The ends B=0 of the semicircle are attached to springs C and D whose other ends are fixed. below this line If r is the radius of the semicircle and k is the force constant for each spring, C then the extension x in each spring is 2EBR 2EBr EBr EBr E (A) x  (B) x  (C) x  (D) x  kr kR kR 2kR

(D) 6.

20 A

X B

D

An infinite collection of current carrying conductors each carrying a current  outwards perpendicular to paper are placed at x = x 0, x = 3x 0, x = 5x 0, ........ ad infinitum on the x axis. Another infinite collection of current carrying conductors each carrying a current  inwards perpendicular to paper are placed at x = 2x 0, x = 4x 0, x = 6x 0, ....... ad infinitum Here x 0 is a positive constant. The magnetic field at the origin due to the above collection of current carrying conductors is 0   0  loge 2 (A) ZERO (B) (C)  (D) 4x 0 loge 2 2x 0

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MAGNETIC FIELD Z D

8.

L A conductor ABCDEF, with each side of length L, is bent as shown. It is carrying a current  in a uniform magnetic induction (field) B, parallel to the positive y-direction. The force experienced by the wire is C (A) BIL in the positive y-direction. L (B) BIL in the negative z-direction. O (C) 3 BIL (D) zero X B

B

F

E

Y A

X

9.

The square loop ABCD, carrying a current  , is placed in a uniform magnetic field B, as shown. The loop can rotate about the axis XX’. The plate of the loop makes an angle  (  < 90°) with the direction of B. Through what Y angle will the loop rotate by itself before the torque on it becomes zero ? (A)  (B) 90° (C) 90° +  (D) 180° – 

Z

B 

y An electron is fired from the point A with a velocity V0 = 2 x 108 m/s. The magnitude and direction of magnetic field that will cause the V0 electron to follow a semicircular path from A to B is (A) 1.5 x 10–4 T out of the page. (B) 1.5 x 10–3 T into the page A (C) 1.5 x 10–2 T into the page (D) 1.5 x 10–2 T out of the page

B



A

10.

C

D

X’

B

x

15 cm A

11.

12.

Two long straight parallel wires 2 m apart, are perpendicular to the plane of the paper. Wire A carries a current of 10 A directed into the plane of the paper. Wire B carries a current such that the magnetic field at P at a distance of 0.8 m from this wire is zero.The magnitude and direction of the current in wire is zero. The magnitude and direction of the current in wire B is (A) 2.8 A into the page (B) 2.8 A out of the page (C) 2.8 x 10–6 A into the page. (D) 2.08 A out of the page.

1.6 m 2m S 1.2 m B P

Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2  . The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A)

q 2m

(B)

q m

(C)

q (D)  m

2q m

13. If a charged particle is describing a circle of radius r in a magnetic field with a time period T then, (A) T 2  r 3

(B) T 2  r

(C) T  r 2

(D) T  r 0



14. A non-conducting rod AB of length  has a linear charge density  . The rod is rotated about an axis passing through point A with constant angular velocity  as shown in the figure. The magnetic moment of the rod is (A)

  2

2

(B)

  3

2

(C)

3  2

3

(D)

A

  6

+++

B

3

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MAGNETIC FIELD 15. A charged particle of mass m and charge q is released from rest from (x 0, 0) along an electric field E0 jˆ . The angular momentum of the particle about origin (A) is zero (B) is constant (C) increases with time (D) decreases with time 16. Two particles Y and Z emitted by a radioactive source at P made tracks in a could chamber as illustrated in the figure. A magnetic field acted downward into the paper. Careful measurements showed that both tracks were circular, the radius of Y track being half that of the Z track. Which one of the following statements is certainly true? P (A) Both Y and Z particles carried a positive charge. (B) The mass of Z particle was one half that of the Y particle. (C) The mass of the Z particle was twice that of the Y particle. (D) The charge of the Z particle was twice that of the Y particle.

Y Z

17. The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre O is (A)

0 I 6a

(B)

0 I 3a

(C)

2 0 I 3 a

B

R

A

I

(D) Zero

2R a O 120° 120° C

R

18. A particle with a specific charge s is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for the particle not to hit the wall? (A) v/sd (B) 2v/sd (C) v/2sd (D) v/4sd 19. Current  flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at distance r from O is

0  r 20  (C) 4 r (A)

20  r 0  2 1 (D) 2 r

By Bx

r

(B)





2 1



x



45° O

20. Two circular coils X and Y having equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway be between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as By and that due to smaller coil X at O as Bx, then (A)

P

Y

1

By (B)

Bx

2

By (C)

Bx

X

I

Y d



1 2

X

By (D)

Bx



d

O

1 4

21. Two mutually perpendicular conductors carrying 1 and  2 lie in the x-y plane. Find locus of points at which magnetic induction is zero?

 22 x (A) y  2

(B) x 

1 y 2

(C) y 

1 x 2

2

2

(D) x  y 

1 2

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39

MAGNETIC FIELD 22. A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5 x 10–2 T. Find the couple acting on the coil when a current of 0.1 ampere is passed through it and the magnetic field is parallel to its plane. (A)

(B)

5 2N  m

(C) 5 3  107 N  m

N

S

5 2N  104 N  m

(D) 10 7 N  m



23. A particle of charge q and mass m starts moving from the origin under the action of an electric field E  E0iˆ







and magnetic field B  B0 kˆ . It s velocity at (x, 3, 0) is 4 iˆ  3 ˆj . The value of x is: (A)

36E0 B0 qm

(B)

25m 2q E0

(C)

10 m q E0

(D)

25 E0 B0 m

24. Two charged particles A and B enter a uniform magnetic field with velocities normal to the field. Their paths are shown in the Figure. The possible reasons are:

(A) The momentum of A is greater than that of B (C) The specific charge of A is greater than that of B

(B) the charge of A is greater than that of B. (D) the speed of A is less than that of B.

25. Two charged particle M and N enter a space of uniform magnetic field, with velocities, perpendicular to the magnetic field. the paths are as shown in the figure. The possible reason are:

(A) the charge of M is greater than that of N. (C) specific charge of M is greater than that of N.

(B) the momentum of M is greater than that of N. (D) the speed of M is less than that of N.

26. A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring, then: (A) there is no net force on the ring. (B) the ring may tend to expand. (C) the ring may tend to contact (D) none of these.



27. A charge q is moving with a velocity v1  1iˆ m s at a point in a magnetic field and experiences a force

 F1  q 1 ˆj  1 kˆ N. If the charge is moving with a velocity v2  1 ˆj ma/s at the same point, it experiences





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40

MAGNETIC FIELD 



a force F2  q 1iˆ  1kˆ  N . The magnetic induction B at that point is:





2 (A) iˆ  ˆj  kˆ Wb m







2 (B) iˆ  ˆj  kˆ Wb m



2 (C) iˆ  ˆj  kˆ Wb m





2 (D) iˆ  ˆj  kˆ Wb m

28. A parallel beam of electrons is shot into a uniform electric field, initially parallel to and against the field with a small initial speed. Then: (A) the beam will pass through the field accelerating down the field without changing its width. (B) the beam tends to spread out at the beginning and to narrow down later. (C) the beam tends to narrow down at the beginning and to spread out later. (D) the total energy of the beam is conserved. 29. The ratio of the energy required to set up in a cube of side 10 cm a uniform magnetic field of 4 wb/m 2 and a uniform electric field of 106 V/m is: (A) 1.4 x 107 (B) 1.4 x 105 (C) 1.4 x 106 (D) 1.4 x 103 30. A charged particle is fired at an angle  in a uniform magnetic field directed along the x-axis. During its motion along a helical path, the particle will: (A) never parallel to the x-axis. (B) move parallel to the x-axis once during every rotation for all values of (C) move parallel to the x-axis at least once during every rotation if (D) never move perpendicular to the x-direction.



.

– 45°.



31. A charged particle q enters a region of uniform B (out of the page) and is deflected a distance d after travelling a horizontal distance a. The magnitude of the momentum of the particle is:

 qB  a 2  d  (B) qBa (A)  2 d 2 

(C) Zero

(D) not possible to be determined as it keeps changing.

32. Two insulated rings, one of slightly smaller diameter then the other, are suspended along their common diameter as shown in the figure, initially the planes of the rings are mutually perpendicular. When a steady current is set up in each of them: (A) The two rings rotate towards a common plane. (B) The inner ring will oscillate about its initial position. (C) The inner ring stays stationary while the outer one moves into the plane of the inner ring. (D) The outer ring stays stationary while the inner one moves into the plane of the outer ring. 33. A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only on the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of B. Let T  2

m . The time spent QB

by the particle in the field will be: (A) T 

(B) 2T 

   2    2 

   2    2 

(C) T 

(D) T 

34. The current I flows through a square loop of a wire of side a. The magnetic induction at the centre of the loop is: (A)

2 0  a

(B)

2 2 0 a

(C)

2 0  a

(D)

2  0  a

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41

MAGNETIC FIELD 35. Suppose a uniform electric field and a uniform magnetic field exist along mutually perpendicular directions in a gravity free space. If a charged particle is released from rest, at a point in the space: (A) particle can not remain in static equilibrium. (B) first particle will move along a curved path but after some time its velocity will become constant. (C) particle will come to rest at regular interval of time. (D) acceleration of the particle will never become equal to zero. 36. A circular loop of radius R carries a charge q uniformly distributed on it. It is rotated at a frequency f about one of the diameters. A uniform magnetic field B exists along its diameter . The maximum and minimum torques acting on the loop due to the magnetic field are, respectively: (A)

 qfR 2 B, 0

(C) 2 qfR 2 B ,

(B) 0, 0

 qfR 2 B

(D) None of these

37. A semi-circular wire of radius R is connected to a wire bent in the form of a sine curve to form a closed loop as shown n the figure. If the loop carries a current  and is placed in a uniform magnetic field B, then the total force acting on the sine curve is: (A) 2BR (downward) (B) 2BR (upward) (C) B R (upward) (D) Zero





B



38. A charged particle of unit mass and unit charge moves with velocity of v  8 iˆ  6 ˆj m s in a magnetic field



of B  2 kˆ T . Choose the correct alternative(s): (A) The path of the particle be x 2 + y2 – 4x – 21 = 0. (C) The path of the particle may be y2 + z2 = 25.

(B) The path of the particle may be x 2 + y2 = 25. (D) The time period of the particle will be 3.14 s.

39. A proton, a deuteron and an  -particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp , rd , and r denote respectively the radii of the trajectories of these particles, then.

(JEE 1997)

(A) r  rp  rd

(B) r  rd  rp

(C) r  rd  rp

(D) rp  rd  r

40. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (JEE 1999) (A) straight line (B) circle (C) helix (D) cycloid 41. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX’ is given by (JEE 2000)

B (A)

B

X

X’ d

(B)

X

d

X’ d

B (C)

B

X

X’ d

d

(D)

X

d

X’ d

d

42. An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current  flows through PQR. The magnetic field due tot his current at the point M is H1. Now, another infinitely long straight conductor QS is connected at Q so that current is

 2 in QR as well as in QS, the current in PQ remaining unchanged.

The magnetic field at M is now H2. The ratio H1 H 2 is given by (A) ½

(B) 1

(C)

23

(JEE 2000) (D) 2

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42

MAGNETIC FIELD 43. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then (JEE 2000) (A) positive ions deflect towards +y direction and negative ions towards –y direction (B) all ions deflect towards +y direction. (C) all ions deflect towards –y direction (D) positive ions deflect towards –y direction and negative ions towards +y direction. 44. A non-planar loop of conducting wire carrying a current  is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P(a, 0, a) points in the direction (JEE 2001) (A)

1  ˆj  kˆ 2





(B)

1 ˆ ˆ ˆ i  jk (C) 3



1  ˆj  kˆ  iˆ 3



1 ˆ ˆ i k (D) 2





z

y

I



x 2a



45. Two particles A and B of masses m A and m B respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are VA and VB respectively and the trajectories are as shown in the figure. Then (JEE 2001) (A) m A v A < mB v B (B) m A v A > mB v B (C) m A < mB and v A < v B (D) m A = mB and v A = v B

A B

46. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current  passes through the coil, the magnetic field at the center is (JEE 2001)

0 N  (A) b

0  N b ln (D) 2 b  a  a

0 N  b (C) 2 b  a ln a  

2 0 N  (B) b

47. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (JEE 2002) (A)

qbB m

(B)

q b  a  B m

(C)

qaB m

(D)

q b  a  B 2m

48. A long straight wire along the z-axis carries a current  in the negative z direction. The magnetic vector field

 B at a point having coordinates (x, y) in the z = 0 plane is 0  y iˆ  x ˆj 0  x iˆ  y ˆj

(A)

 2  x

2

 y2

 

(B)

 2  x

2

 y2

 

(C)

(JEE 2002)

0   x ˆj  y iˆ  2  x 2  y 2 

49. For a positive charged particle moving in a x-y plane initially along the x-axis, there is a sudden change in its path due tot he presence of electric and/or magnetic fields beyond P. The curved path is shown in the x-y plane and is found to be non-circular. Which one of the following combinations is possible? (JEE 2003)

















(A) E  0; B  b iˆ  c kˆ

(B) E  a iˆ; B  c kˆ  a iˆ

(C) E  0; B  c ˆj  b kˆ

(D) E  a iˆ; B  c kˆ  b ˆj

0   x iˆ  y ˆj  (D)

2  x 2  y 2 

y

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P

x

43

MAGNETIC FIELD Y B 50. A conducting loop carrying a current  is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to (JEE 2003) (A) contract (B) expand (C) move towards +ve x-axis (D) move towards –ve x-axis.

x I

51. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV, arrange them in the decreasing order of Potential Energy (JEE 2003)

nˆ

nˆ

B B

B

nˆ

I

(A) I > III > II > IV

II

III

(B) I > II > III > IV

B

nˆ

(C) I > IV > II > III

IV

(D) III > IV > I > II

52. An electron traveling with a speed u along the positive x-axis enters

of the region with speed v then (A) v = u at y > 0 (C) v > u at y > 0

B

y

into a region of magnetic field where B   B0 kˆ  x  0  . It comes out

e

(JEE 2004) (B) v = u at y < 0 (D) v > u at y < 0

–

u x

53. A magnetic needle is kept in a nonuniform magnetic field. It experiences (A) a force and a torque (B) a force but not a torque (C) a torque but not a force (D) neither a force nor a torque

(JEE 1982)

54. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively. This region of space may have:(JEE 1985) (A) E = 0, B = 0 (B) E = 0, B  0 (C) E  0, B = 0 (D) E  0, B  0 55. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If steady current  is established in the wire as shown in the figure, the loop will : (JEE 1985) (A) rotate about an axis parallel to the wire (B) move away from the wire (C) move towards the wire (D) remain stationary

i I

56. Two particle X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is (JEE 1988) 12

(A)  R1 R2 

(B) R2 R1

(C)  R1 R2 

2

(D) R1 R2

57. A particle of charge +q and mass m moving under the influence of a uniform electric field E iˆ and uniform magnetic field B kˆ follows a trajectory from P to Q as shown in Figure. The velocities at P and Q are v iˆ and 2 ˆj which of the following statement(s) is/are correct? (JEE 1991)

3  mv 2  E  (A) 4  qa 

P

V

E B

a

2a

Q x 2v

3  mv 3  (B) Rate of work done by the electric field at P is  4  a  (C) Rate of work done by the electric field at P is zero (D) Rate of work done by the electric field at Q is zero

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44

MAGNETIC FIELD 58. A current 1 flows along the length of an infinitely long, straight, thin-walled pipe. Then: (A) The magnetic field at all points inside the pipe is the same, but not zero (B) The magnetic field at any point inside the pipe is zero. (C) The magnetic field is zero only on the axis of the pipe (D) The magnetic field is different at different points inside the pipe.

(JEE 1993)

59. H+ , He+ and O ++ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O 2+ are 1 amu, 4 and 16 amu respectively. The: (JEE 1994) (A) H+ will be deflected most. (B) O 2+ will be deflected most (C) He+ and O 2+ will be deflected equally (D) All will be deflected equally. 60. Two very long, straight, parallel wires carry steady currents  and –  respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires, Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (JEE 1998) (A)

0 qv 2 d

(B)

0 qv d

(C)

20 qv d

(D) 0 C

61. An infinite current carrying wire passes through point O and in perpendicular to the plane containing a current carrying loop ABCD as shown in the figure. Choose the correct option (s) (JEE 2006) (A) Net force on the loop is zero (B) Net torque on the loop is zero. (C) As seen from O, the loop rotates clockwise. (D) As seen from O, the loop rotates anticlockwise.



B O D



B0

62. A magnetic field B  B0 ˆj exists in the region a < x < 2a and B   B0 ˆj , in the region 2a < x < 3a, where B0 is a positive constant. A positive point charge  moving with a velocity v  v 0 ˆj , where v 0 is a positive constant, enters the magnetic field at x = 0. The trajectory of the charge in this region can be like, (JEE 2007)

z

z

z a

(A)

a

2a

3a x

(B)

2a

O´ A

0

2a 3a

x

–B0

z

3a

a x (C)

a

a

2a

(A) The particle enters Region III only if its velocity v 

ql B m

(B) The particle enters Region III only if its velocity v 

ql B m

3a

3a x (D)

Region I 63. A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice(s)

2a

0

v

(JEE 2008)

(C) Path length of the particle in Region II is maximum when velocity v 

x

Region II × × × × × × × × × × × × × × × × × × × × × × × × l

Region III

ql B m

(D) Time spent in Region Ii is same for any velocity V as long as the particle returns to Region I

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45

MAGNETIC FIELD

FILL IN THE BLANKS X

1.

2. 3.

A neutron a proton, and an electron and alpha particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper. The tracks of the particles are labelled in Figure. The electron follows track ________ and the alpha particle follows track ________. (JEE 1984)

X

X

X X

X

X

X X

X

X X

X

X

X X

X

X

X X

X

X

X X

X

X

X

X X

X

X

X X

X

X

X

X X

X

X

X X

X

X

X

X X

X

X

X X

X

BX

X

C

D

A wire of length L meters carrying a current i amperes is bent in the form of a circle. The magnitude of its magnetic moment is ________ in MKS units. (JEE 1987) In a hydrogen atom, the electron moves in an orbit of radius 0.5Å making 1016 revolutions per second. The magnetic moment associated with the orbital motion of the electron is ________ . (JEE 1988)

I 4.

I

The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current 1 as shown. The magnitude of the magnetic induction at the centre C is ________ . (JEE 1988)

R2 R 1 Q R C D Z

S

I

C 5.

A wire ABCDEF (with each side of length L) bent as shown in figure and carrying a current  is placed in a uniform magnetic induction B parallel to the positively-direction. The force experienced by the wire is ________ in the ________ direction. (JEE 1990)

F E

Y

B X

6.

P

A



A metallic block carrying current  is subjected to a uniform magnetic induction as B as shown in Figure.



The moving charges experience a force F given by _____ which results in the lowering of the potential of the face _____ Assume the sped of the carriers to be v. (JEE 1996)

7.

B

A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles. With a field B Tesla, it is found that the filter transmits  -particles each of energy 5.3 MeV. The magnetic field is increased to 2.3 B Tesla and deuterons are passed into the filter. the energy of each deuteron transmitted by the filter is _____ MeV. (JEE 1997)

Source

Detector

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46

MAGNETIC FIELD

TRUE / FALSE 8.

No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. (JEE 1981)

9.

There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting on it. (JEE 1983)

10. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. The path of the particle is a circle. (JEE 1983) 11. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular paths of the same radius. (JEE 1985)

TABLE MATCHING 12. (A) (B) (C) (D)

Column I Electric field Magnetic field Electric force Magnetic force

Column II (P) Stationary charge (Q) Moving charge (R) Changes the kinetic energy (S) Does not change kinetic energy

13. Regarding the trajectory of a charged particle, match the following: Column I Column II (A) In electric field (P) Straight line path (B) In magnetic field (Q) Circular path (C) In crossed field (R) Helical path (D) In mutually perpendicular electric and (S) Parabolic path magnetic field, charge being at rest (T) Parabolic path 14. A current flows along length of a long thin cylindrical shell: Column I (A) Magnetic field at all points lying inside the shell (P) (B) (C) (D) 15.

Magnetic field at any point outside the shell Magnetic field is maximum Magnetic field on the axis of the shell Column I

(Q) (R)

Column II Inversely proportional with distance from axis of shell Zero Just outside the shell

Column II

(A)

Unit of magnetic field

(P) Am 2

(B)

Unit of magnetic permeability (µ0)

(Q)

(C)

Unit of magnetic flux ()

(R) N A2

(D)

Unit of magnetic dipole moment

(S)

N Am

Nm A

16. Two long parallel wires carrying equal currents in opposite directions are placed at x =  a parallel to Y-axis with z = 0. Then: Column I Column II (A)

Magnetic field B1 at origin O

(B)

Magnetic field B2 at P (2a, 0, 0)

(C)

Magnetic field at M (a, 0, 0)

(D)

Magnetic field at N (–a, 0, 0)

0 i 3 a  0i (Q)  4 a 0i (R)  a (P)

(S) Zero

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47

MAGNETIC FIELD 17. Two wires each carrying a steady current I are shown in four configurations in Column-I. Some of the resulting effects are deseribed in Column-II. Match the statements is Column-I with the statements in Column-II. Column I Column II [JEE 2007] (A) (B)

(C)

(D)

P

Point P is situated midway between the wires. Point P is situated at the mid-point of the joining the centers of the circular wires, which have same radii. Point P is situated at the mid-point of the line joining the centres of the circular wires, which have same radii. Point P is situated at the common center of the wires.

P

(P)

The magnetic fields (B) at P due to the currents

(Q)

in the wires are in the same direction. The magnetic fields (B) at P due to the currents in the wires are in opposite directions.

(R)

There is no magnetic field at P

(S)

The wires repel each other.

P

P

18. Six point charges, each of the same magnitude q, are arranged in different manners as shon in Column-II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. [JEE 2009] Column I Column II (A) E = 0 (P) – Charges are at the corners of a regular hexagon. + Q M is at the centre of the hexagon. PQ is – M + perpendicular to the plane of the hexagon. – P + P (B) V  0 (Q) – + – + – + Charges are on a line perpendicular to PQ at equal intervals. M is the mid-point between the M two innermost charges. Q (C)

B=0

(R)

–

+ –

+

Q

Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is rependicular to the plane of the rings.

–

M

P + (D)

µ  0

(S)

–

+

–

+

P

Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the longer sides. M is at the centre of the rectangle. PQ is parallel to the longer sides.

– Q

M

–

P –

+

(T)

M

+

+

– Q

–

Charges are placed on two coplanar, insulating rings at equal intervals. M is the mid-point between the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings.

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48

MAGNETIC FIELD

PASSAGE TYPE PASSAGE = 1. A cyclotron is a device used by physicists to study the properties of subatomic particles. Small charged particles are deposited at high speed into a circular pipe. electric and magnetic field then accelerate the particle to an even higher speed. Finally, when the particle reaches the desired speed, a magnetic field keeps it moving in a circle at constant speed. In figure 1, a magnetic field pointing “into the page” keeps a charged particle travelling in a counterclockwise circle at constant speed inside the cyclotron. The magnetic force on the particle points towards the center of the circle, and has strength F mag = qvB where q is the particle’s charge, v is its speed, and B is the magnetic field strength. As a result of this force, the particle moves in circles in the cyclotron at frequency

f 

qB 2 m

North West

particle

East South

where m denotes the particle’s mass. The frequency (in hertz) is the number of revolutions completed by the particle per second. In the following questions, neglect gravity. 1. 2. 3.

In figure 1, what is the direction (if any) of the particle’s acceleration? (A) North (B) East (C) West (D) It has no acceleration If the magnetic field in figure were turned off, in which direction would the particle travel (until crashing)? (A) North (B) East (C) West (D) Now here; it would stop moving An alpha particle has charge 2e and mass 4 amu. A proton has charge e and mass 1 amu. Let f alpha and f proton denote the frequencies with which those particles circle a cyclotron. If both particles experience he same

f alpha magnetic field in the same cyclotron, what is (A) 2 4.

5.

(B) 1

(C) 1

2

f proton ? (D) 1

4

In order for a cyclotron to work properly, the magnetic field must make the particle move in a circle. Which of the following particles would not work in a cyclotron? (A) lithium atom (Li) (B) positive lithium ion (Li +) + (C) negative lithium ion (Li ) (D) all of the above particle would “work”. Which of the following cannot create a magnetic field? (A) Electrical current flowing through a well-insulated straight metal wire. (B) A beam of electrons moving across a cathode ray tube. (C) Electric current flowing around a superconducting ring. (D) Static electricity (i.e., extra electrons) built up on a stationary door knob.

PASSAGE = 2. Magnetically levitated (MAGLEV) trains are considered to be important future travelling machines. The idea of MAGLEV transportation has been in existence since the early 1900s. The benefits of eliminating friction between the wheel and the rail to obtain higher speeds and lower maintenance costs has great appeal. The basic idea of a MAGLEV train is to levitate it with magnetic fields so that there is no physical contact between the train and the rails. For comparison, ‘bullet’ trains in Japan have a maximum speed of about 250-300 km/h while a MAGLEV train under development has reached a speed of 411 km/hr. The MAGLEV train uses powerful on board superconducting electromagnets with zero electrical resistance to support the train above the rails.

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49

MAGNETIC FIELD

1.

2.

3. 4. 5.

1. 2. 3.

The walls along the track contain a continuous series of vertical coils of ordinary wire. As the train passes each coil, the superconducting magnet on the train induces a current in these coils and makes them electromagnets. The electromagnets on the train and outside produce forces that levitate the train and keep it centered above the rails. In addition, electric current flowing through coils outside the train propels the train forward, as shown in Figure. What is the major advantage of using super conductors in the electromagnets (A) very low resistance for electric current (B) small size (C) low hysteresis loss (D) zero electrical resistance. Choose the correct statement (A) MAGLEV trains have high maintenance cost (B) They have low maintenance cost but high running cost (C) they have low maintenance cost but high energy efficiency (D) they have low maintenance and running cost as well as low installation cost. MAGLE trains are operating in which country (A) Japan (B) USA (C) China (D) None of these Super conducting electromagnets are installed in (A) inside the train (B) on the railroads (C) on the walls (D) on the station Coils installed on the wall will carry currents (A) only if train is moving very fast (B) all times whether train is standing or running (C) only if train is moving irrespective of whether fast or slow. (D) None of these. PASSAGE = 3. A charged particle +q of mass m is placed at a distance d from another charged particle –2q of mass 2 m in a uniform magnetic field B as shown in figure. If the particles are projected towards each other with equal speeds v. Find the maximum value of projection speed v m so that the two particles do not collide. Find the time after which collision occurs between the particles if projection speed equals 2v m. Assuming the collision to be perfectly inelastic find the radius of particle in subsequent motion. (Neglect the electric force between the charges)

x q, m x V x x

x -2q, 2m x V x

x

x

x

x

x

x

x

x

x

d

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50

MAGNETIC FIELD

Level # 2 1.

 A particle of charge q and mass m is projected from the origin with velocity v =  0 ˆi in a nonuniform  magnetic field B  B0 xkˆ . Here  0 and B0 are positive constants of proper dimensions. Find the maximum positive x coordinate of the particle during its motion.

2.

A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is  . The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity  . Find the torque on the disc if it is placed in a uniform magnetic field B directed perpendicular to the rotation axis.

3.

A wire loop ABCDE carrying a current  is placed in the x-y plane as shown in figure. A particle of mass m and charge q is projected V0 ˆ ˆ i  j m/s. Find the 2 (a) instantaneous acceleration.

from origin with velocity V 

E

 

90°

B

(B) If an external magnetic field B  B 0 ˆi is applied, find the force and torque acting on the loop due to this field. 4.

y

A

x

O D

r/2

Z

C A long wire of radius a is placed along z-axis and carries current i as indicated in the figure. y-axis is taken perpendicular to the plane of paper directed into the paper. An electron escapes from the surface of the wire with velocity  0 directed along x-axis. Determine the maximum distance from the wire along x-axis up to which electron can move.

i

e O

X V0

5.

A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30 A, as shown in figure. Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations.

A C

B D

6.

A current  flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. Find the magnetic induction at the centre of the polygon.Analyse the obtained expression at n   .

7.

Find the magnetic induction at the centre of rectangular wire frame whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal to  = 30°; the current flowing in the frame equals  = 5.0 A.

8.

A thin conducting strip of width h = 2.0 cm is tightly wound in the shape of a very long coil with crosssection radius R = 2.5 cm to make a single-layer straight solenoid. a direct current  = 5.0 A flows through the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance r from its axis.

9.

Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to R = 100 mm and the magnetic induction at its centre is equal to B = 6.0  T..

10.

Calculate the magnetic moment of a thin wire with a current  = 0.8 A, wound tightly on half a tore (Figure). The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500.

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51

MAGNETIC FIELD 11.

A non-conducting thin disc of radius R charged uniformly over one side with surface density  rotates about its axis with an angular velocity  . Find : (a) the magnetic induction at the centre of the disc ; (B) the magnetic moment of the disc.

12.

A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in Figure. The cross-section area of the coil is S = 1.0 cm 2, the length of the arm OA of the balance beam is  = 30 cm. W hen there is no current in the coil the balance is in equilibrium. On passing a current  = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass m = 60 mg on the balance pan. Find the magnetic induction at the spot where the coil is located.

13.

A square frame carrying a current  = 0.90 A is located in the same plane as a long straight wire carrying a current  0 = 5.0 A. The frame side has a length a = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is  = 1.5 times greater than the side of the frame. Find : (a) Ampere force acting on the frame ; (B) the mechanical work to be performed in order to turn the frame through 180° about its axis, with the currents maintained constant.

14.

15.

In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B (Figure). A current  is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. Find the gauge pressure produced by the pump if B = 0.10 T,  = 100 A, and a = 2.0 cm. A proton accelerated by a potential difference V = 500 kV flies through a uniform transverse magnetic field with induction B = 0.51 T. The field occupies a region of space d = 10 cm in thickness (Figure). Find the angle  through which the proton deviates from the initial direction of its motion.

B

I a

+B  d

16.

A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference V propagates from a point A along the axis of a straigth solenoid. The beam is brought into focus at a distance  from the point A at two successive values of magnetic induction B1 and B2. Find the specific charge q/m of teh particles.

17.

A non-relativistic proton beam passes without deviation through the region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E = 120 kV/m and B = 50 mT. Then the beam strikes a grounded target. Find the force with which the beam acts onthe target if the beam current is equal to  = 0.80 mA.

18.

A beam of non-relativistic charged particles moves without deviation through the region of space A (Figure), where there are transverse mutually perpendicular electric and magnetic fields with strength E and induction B. When the magnetic field is switched off, the trace of the beam on the screen S shifts by x . Knowing the distances a and b, find the specific charge q/m

S

A



a

b

of the particles.

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52

MAGNETIC FIELD

Level # 3 1.

B

nˆ An electron in the ground state of hydrogen atom is revolving in anticlock-wise 30° direction in a circular orbit of radius R. (JEE 1996) (i) Obtain an expression for the orbital magnetic dipole moment of the electron. (ii)



The atom is placed in a uniform magnetic induction B such that the plane-normal of the electron-orbit makes an angle of 30° with the magnetic induction. find the torque experienced by the orbiting electron.

2.

D A current of 10 A flows around a closed path in a circuit which is in the horizontal r 2 C plane as shown in the figure. The circuit consists of eight alternating arcs of radii r1 = 0.08 m and r1 = 0.12 m. Each arc subtends the same angle at the center. A (A) Find the magnetic field produced by this circuit at the center. (JEE 2001) r1 (B) An infinitely long straight wire carrying a current of 10 A is passing the center of the above circuit vertically with the direction of the current being into the plane of the circuit. What is three force acting on the wire at the center due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the center? d

3.

A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T 0. If the breaking tension of the strings are 3T 0/2, find the maximum angular velocity

0 with which the wheel can be rotated.

T0

T0 0 B

(JEE 2003) y

4.

A pair of stationary infinitely long bent wires are placed in the x-y plane as shown in Figure. The wires carry currents of  = 10 A each as shown. The segments LR and MS are along the x- axis. the segments RP and SQ are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of the magnetic induction at the origin O. (JEE 1989)

Q L

 M

O

R

S

 P

x





5.

A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. (JEE 1990)

6.

Three infinitely long thin wires, each carrying current in the same direction are in the x-y plane of a gravity free space. The central wire is along the y-axis while the other two are along x =  d. (JEE 1997) (a) Find the locus of points for which the magnetic field B is zero. (B) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wire is  , find the frequency of oscillations.

7.

A proton and an alpha particle, after being accelerated through same potential difference, enter a uniform magnetic field the direction of which is perpendicular to their velocities. Find the ratio of radii of the circular paths of the two particles. (JEE 2004)

8.

In a moving coil galvanometer, torque on the coil can be expressed as b = ki, whre ‘i’ is current through the wire and ‘k’ is constant. The rectangular coil of the galvanometer having numbers of turns N, area A and moment of inertia I is placed in magnetic field B. Find ‘k’ in terms of given parameters N, I, A and B. the torsional constant of the spring, if a current i0 produces a deflection of /2 in the coil in reaching equilibrium position. the maximum angle through which coil is deflected, id charge Q is passed through the coil almost instantaneously. (Ignore the damping in mechanical oscillations) (JEE 2005)

(a) (b) (c)

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53

MAGNETIC FIELD

Answer Key ASSERTION & REASON Q. Ans. Q. Ans.

1 A 10 E

2 E 11 B

3 A 12 C

4 B

5 A

6 A

7 D

8 A

9 B

Level # 1 Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans.

1 C 11 B 21 C 31 A 41 B 51 A 61 AC

2 D 12 A 22 C 32 A 42 C 52 B 62 A

3 C 13 D 23 B 33 C 43 B 53 A 63 ACD

4 D 14 D 24 BCD 34 B 44 D 54 ABD

5 ABCD 15 C 25 ACD 35 ACD 45 B 55 C

6 C 16 A 26 ABC 36 B 46 C 56 C

7 D 17 D 27 A 37 B 47 B 57 ABD

8 B 18 A 28 BD 38 ABD 48 A 58 B

9 C 19 D 29 C 39 A 49 B 59 AC

10 C 20 C 30 AD 40 A 50 B 60 D

FILL IN THE BLANKS / TRUE-FALSE / MATCH TABLE iL2 4

3. 1.25 x 10–23 Am 2

1. D, B

2.

5.

6. evB; ABCD

lB ; +z direction

7. 14.0185 eV 8. True

12. [(A—PQ), (B—Q), (C—PQR), (D—QS)] 14. [(A—Q), (B—P), (C—R), (D—Q)] 16. [(A—R), (B—P), (C—Q), (D—Q)] 18. [(A—PRS), (B—RS), (C—PQT), (D—RS)]

4. 9. True

0 I  1 1     4  R1 R2  10. False

11. False

13. [(A—PT), (B—PQR), (C—P), (D—S)] 15. [(A—Q), (B—R), (C—S), (D—P)] 17. [(A—QR), (B—P), (C—QR), (D—QS)]

PASSAGE TYPES Passage = 1. 1. C Passage = 2. 1. D Passage = 3. 1. Vm 

Bqd 2m

2. A

3. C

4. A

5. D

2. C

3. D

4. A

5. C

2. t 

m 6Bq

3. R  3d.

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54

MAGNETIC FIELD

Level # 2 1.

2mv 0 B0q

3. (a) a 

2.

 BR 4 4

 0 q 8

   4 2  B V  ˆj  ˆi  2rm 0

5. 0.2 Sec.

6.

(B) Zero,

0

nµ0 I  tan 2R n



4.  m  ae

(2mv 0 )  0 ei

7. B  4 0  / d sin   0.10 mT.

  / h (1  (h / 2R )2  0.3 mT, r  R, B   0 8.   0 / 42 / r, r  R.

10. pm  1/ 2 Nd2  0.5 A.m2

r 2  1     B0 4  2

9. pm  2R 3B /  0  30 mA.m2

(b) pm  1/ 4 R 4 .

11. (a) B  1/ 2  0 R

12. B  mg / NS  0.4 T.



13. (a) F  2 0  0 /  42  1  0.40 N (b) A  ( 0 a 0 / ) ln [2  1) /(2  1)]  0.10  J.  q  15.   arcsin dB 2mV 

14. p  B / a  0.5 kPa.

17. F = mE  /qB = 20  N.

18. q / m 

   30.  

16. q / m 

8 2 V  (B 2  B1 )2 2

a(a  2b)B 2 2Ex

Level # 3 1. (i) M 

eh 4m

(ii)

ehB   directed perpendicular to the both M and B . 8 m

2. (a) 6.54 x 10–5 Tesla (b) 0, Force on arc AC = 0, Force on segment CD = 8.1 × 10–6 N (inwards) 3. Wman 

DT0 BQr 2

6. (a) x  

d 3

4. 1 x 10–4 T or Wb m –2.

rp

12

, z  0. (b)

8. (a) k = NAB, (b)

C

  0    2  d    

7.

r



5.  = 5.9 x 10–6 Nm.

m p q 1 .  m q p 2

NAB 2i 0 NAB ; (c) Q  21i0 

—X—X—X—X—

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55

ATOMIC PHYSICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

ATOMIC PHYSICS MASS OF A PHOTON Rest mass of photon is zero. Effective mass : If we assume that m is the effective mass of photons in moving conditions, then according to Einstein’s theory. Energy of photon, E = mc2

but

E c2 E = h = pc,



m=

m=

p = photon – momentum

E h p 2 = 2 = c c c

C1: Draw the graph of (a) momentum versus effective mass of photon (b) Energy versus effective mass, (c) wavelength versus momentum, (d) Effective mass versus wavelength for a photon. Sol:

p

E 

O

 m

m

O

(a) p = mc  straight line tan = c

(b)

E = mc2  straight line tan = c2 m



p

(c)de-broglie wavelength of photon  = p = h  Rectangular hyperbola



h p

(d)

Effective mass of photon m=

E h hc 2 = 2 = c c c 2

m =

h  Rectangular hyperbola c

Power of a light source Suppose P = Power of the light source,  = frequency of emitted photons, Energy of single photon = h Let, n = numbers of photons emitted per sec by the source, then Thus, Energy emitted per sec. by the source = W = h Intensity of light source at a point The amount of energy incident on a point per unit area at that point in a unit time is called the intensity of light at that point.

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1

ATOMIC PHYSICS Thus, intensity at a point is energy incident per unit area unit time. Suppose, I = Intensity of light at a point  = Frequency of photon n0 = no. of photon incident / sec / Area  I = n0h SI unit of intensity is J/second m2. Intensity of light at a distance r : Let, W = Power of a point source Amount of energy received by spherical surface per second = Amount of energy emitted by light source per second = W Hence, Intensity at distance

Thus,

r=

W W = Area 4r 2

I=

W 4r 2

P r

Example 1: The intensity of sunlight on the surface of earth is 1400 W/m2. Assuming the mean wavelength of sunlight to be 6000 Å, calculate the number of photons emitted from the sun per second assuming the average radius of earth’s orbit around sun is 1.49 × 1011 m. Sol: Average radius, r = 1.49 × 1011 m. intensity of sunlight received by earth = I = 1400 W/m2,  = mean wavelength = 6000 Å. Energy emitted per second by the sum = Power of the sun = W but, Power, W = nh, where n is number of photons emitted by sun per sec. .....(i) Again, intensity at a distance ‘r’ from a point source of power W. I=

W 4r 2

I=

nh 4r 2

I  4r 2 I  4r 2 n= = h hc 

[From equation (i)] hc [ h = ] 

Sun

r

Earth

Power = P

4r 2I 4 (1.49)2 1022  6 107 1400 n= = hc 6.63 10 34  3 108 n = 1.18 × 1045 photons per sec.

Photon Flux (i.e. photon/sec) Suppose, W = Power of a source. If A is the area of the metal surface on which radiations are incident, then the power received by the plate is  W  W = IA =  A 4r 2  If  is the frequency of radiation, then the energy is photon is given by h.

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2

ATOMIC PHYSICS If n is the number of photons incident on the plate per second, then, we have W = n h  W    A = N h 4r 2 



Area(A)

 W   4r 2  A  n =    h   

P Plate r

n´ is called photon flux. Photon flow density (n0) : The number of photons incident on unit area of the plate in one second is called the photon flow density. n' W i.e., Photon flow density, n0, = = A 4  r 2 h Photon-concentration : The number of photons per unit volume of the space. If n is the number of photons incident per unit area per second, then n Photon - concentration = , where c is the speed of light. c Example 2: The power of light emitted by the sun is 3.9 × 1026 W. Assuming the mean wavelength of sunlight to be 6000 Å, calculate the photon flux arriving at 1 m2 area on earth perpendicular to light radiations. The average radius of earth’s orbit around sun is 1.49 × 1011 m. Sol: Power of light emitted by the sun, W = 3.9 × 1026 W r = earth sun mean distance = 1.49 × 1011 m  = mean wavelength of sun light = 6000 Å = 6 × 10–7 m W 4r 2 Power received by area A on earth, W = I × A W  W = ×A 4r 2 but, W = n h, where n is the number of photon incident per sec or photon-flux.

Intensity of light an earth, I =





P Sun

r

Earth

 W A  2   W'  4r  n = = h  h    n =

WA 4r 2 hc

[h = hc/]

(3.9 10 26 )  (6 107 ) 1 n´ = photon /sec 4(1.49)2 1022  6.63 1034  3 108 = 4.22 × 1021 photon/sec

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3

ATOMIC PHYSICS PRESSURE EXERTED BY PHOTONS OR RADIATIONS PRESSURE (a) Each photon carries energy and momentum. Hence when photons of light is incident on a surface, the light is either absorbed or reflected or both is done by the surface. (b) change in momentum of light takes place, which causes impulse or force on the surface resulting in a pressure called radiation pressure. Let light of intensity I is incident on a surface. Each photon carries energy h and momentum =

h . c

Energy incident on unit area in unit time = I(by the definition of intensity). Number of photons incident on unit area in unit time. N I = , At h Momentum delivered to unit area in unit time,

i.e.

.....(i)

 N  P =  × (change in momentum of each photon) At  At  Using equation (i) & (ii), we get

i.e.,

P At

 I  =   × (change in momentum of each photon)  h 

but,

P At

=



Pr = Radiation pressure =

Force exerted , A = Radiation pressure

[ force exerted =

.....(ii)

P ] t

I × (change in momentum of each photon) .....(iii) h

Radiation pressure for perfectly absorbing surface : In this case, change of momentum suffered by each photon =

h . c

hv c

Using the equation (iii), we get Radiation pressure =

Perfectly absorbing surface

I h I × = h c c

 Radiation pressure =

I for a perfectly absorbing surface. c

Radiation pressure for a perfectly reflecting surface :

hv c

h  h  2h In this case, change in momentum of each photon = –   = c c  c 

Hence, using the equation (iii), we get

Perfectly reflecting surface

hv c

 I   2h  2I Radiation pressure =   ×  = c  h   c 

 Radiation pressure =

2I for a perfectly reflecting surface. c

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ATOMIC PHYSICS Radiation pressure for a surface of reflection coefficient () : In this case, Momentum of incident photon = Momentum of reflected photon =  h – c Thus, using equation (iii), we get,

 Change in momentum =

 Radiation pressure = (1 + )

h c

hv c

h ,  = reflection coefficient c

surface

 hv c

 h  h    = (1 + ) c  c 

I for photons falling normally on a surface. c

NOTE : (i) For a perfectly absorbing surface  = 0 I  Radiation pressure = c (ii) For a perfectly reflecting surface  = 1



Radiation pressure = (1 + 1)

I 2I = c c

C2: A laser emits a light pulse of duration  = 0.13 ms and energy E = 10J. Find the mean pressure exerted by such a light pulse when it is focussed into a spot of diameter d = 10 m on a surface perpendicular to the beam and possessing a reflection -coefficient  = 0.50. Sol: Laser energy, E = 10J  2 d 4 Pulse duration,  = 0.13 s Pressure exerted by light pulse i.e.,

Plate area =

Laser

I Radiation pressure = (1 + ) , by the definition of radiation pressure by photons of light c Here,  = reflection – coefficient = 0.50 I = Intensity of light c = speed of light = 3 × 108 m/s

d

d = 10 m

.....(i)

Laser energy By the definition of intensity of light, I = (pulse duration)  (Area of plate) E .....(ii)  2  d  4  4E From (i) & (ii) Radiation pressure = (1 + ) c d 2

i.e. 

I=

4 10 3 10  (0.13  103 )100 1012 = 4.9 × 107 Nm–2 = 490 atm = (1 + 0.50)

8

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ATOMIC PHYSICS Example 3: 1 A current flows in an x-ray tube operated at 10,000 V. The target area is 10–4 m2. Find the pressure on the target, assuming that the electrons strike the target normally and the photons leave the target normally. Consider the ideal situation where each incident electron gives rise to a photon of the same energy. Sol: Energy of each electron = energy of each photon = E = 104 eV Momentum delivered by each electron = p1 = 2mE E p1 – e c p2 Change of momentum due to each electron-photon pair = p1 – (–p2) = p1 + p2 (photon) Current incident on target = i = 10–6A  Number of electrons incident per second = i/e = no. of photons emitted per second. Plate

Momentum taken away by each photon = p2 =

i Total momentum change per second = force =   (p1 + p2) e



Pr, Pressure = force / area = =



 i    (p + p2)  Ae  1 E  i     2mE   c  Ae  

 104 1.6 1019  106 31 4 19 1/ 2 (2  9.1  10  10  1.6  10 )  Pr =   3 108 104 1.6 1019   –6 –2 = 3.7 × 10 Nm

FORCE EXERTED BY PHOTONS ON A SURFACE We know that, when photons of light is incident on a surface, the change in momentum of photons takes place resulting in a force exertion by the photons on the surface. Let light of power W is incident on a surface. h c Energy incident in unit time = W(by the definition of power) But, power is given by

Each photon carries energy h and, momentum =

W =

N (h), t

.....(i) Power = W

Surface

where N = total no. of photons incident in time t.

W N = h t Momentum delivered in unit time.

.....(ii)

P  N  =   × (change in momentum of each photon) t  t 

= but,

P (change in momentum of each photon) h

P = rate of change of momentum t

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ATOMIC PHYSICS P = F = Force exerted by the photons. t

Hence, Thus,

 P  force =   × (change in momentum of each photon)  h 

...(iii)

Force exerted by photons on a perfectly absorbing surface : In this case, change of momentum suffered by each photon =

h c

using equation (iii), we get hv c

 W   h  W force =   ×   = h c  c 

thus,

F=

Perfectly absorbing surface

W for a perfectly absorbing surface. c

Force exerted by photons on a perfectly reflecting surface: In this case, change of momentum suffered by each photon =

hv c

h  h  2h –   = c c  c 

using equation ( iii), we get

 hv c

 W   2h  2W Force =   ×  = h c  c 

thus, F =

surface

2P , for a perfectly reflecting surface. c

Force exerted by photons on a surface of reflection coefficient : In this case momentum of incident photon = Momentum of reflected photon = 

h , c

hv c

h c

surface

 hv c

h   h  h –    = (1 + ) c c c   Thus, using equation (iii), we get

 Change in momentum =

   W h Force =   × (1 + ) = (1 + ) c c  h 

i.e.

F = (1 + )

W for photons falling normally on a surface. c

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ATOMIC PHYSICS C3: If a point source of light of power W is placed at the centre of curvature of a hemispherical surface, whose inner surface is completely reflecting, then

W (Source)

W . 2c C4: If a perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture and having an intensity I, then the

the force on the hemisphere due to the light falling on it is given by F =

πr 2 I force exerted by the beam on the sphere is given F = . c Note that force is equal to the product of (I/c) and the projected area of the sphere. C5: If photon of light of power W falls at an angle  to a perfectly reflecting

I

2W surface, then net force exerted on the plate is given by Fnet = cos. c

r

 

C6: A laser emits a light pluse of duration T = 0.10 ms and energy E = 10 J. Find the mean pressure exerted by such a pulse when it is focused on a spot of diameter d = 10 µm on a surface perpendicular to the beam and with a reflection coefficient  = 0.5. Sol. p, momentum of a photon = hv/c = E/c Momentum of reflected photon = (E/c)  change of momentum = (E/c) – (–E/c) = (1 + )E/c Force exerted = [(1 + ) E/c]/T Pressure = 4 [(1 + ) E/c]/Td2 4(1  )E 4(1  0.5)  10  2 cTd   3  108  104  1010 = 6.37 × 106 Nm–2

=

C7: A plane light wave of intensity I = 0.50 W cm–2 falls on a plane mirror of reflection coefficient  = 0.8. The angle of incidence  = 45º. Find the normal presure exerted by light on that surface. Sol. If S is the area of the mirror on which light falls, the transverse section of the incident beam is S cos . Momentum of the incident photons = (I/c)(S cos ). Normal component of momentum flux = (IS cos /c) cos = IS cos 2/c. Momentum of reflected photons = (IS cos2/c) in the opposite direction.  rate of change of momentum = force = (1 + ) IS cos2/c  normal pressure = force/area = (1 + ) I cos2/c  required pressure = (1 + 0.5) (0.5 × 104 cos2 45º)/3 × 108 = 1.25 × 10–5 Nm–2 Example 4: An isotropic point source of radiation power P is placed on the axis of an ideal mirror. The distance between the source and the mirror is n times the radius of the mirro. Find the force that light exerts on the mirror.

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ATOMIC PHYSICS Sol. Energy flux through the annular space between two cones of half-angles  and  + d = (P/4) (2 sin  d)  momentum of flux = P sin  d/2c  rate of change of momentum along the normal. 2(P sin  d/2c) cos  = P sin  cos  d/c

A R

d

S

nR

p  sin  cos  d c 0 where  = half-angle of the cone subtened by the mirror = tan–1(1/n)  F = P sin2  / 2c = P/2(n2 + 1)c

 force on mirror =

B

B

Example 5: Figure shows a small, plane strip of mass m suspended from a fixed support through a string. A continuous beam of monochromatic light is incident horizontally on the strip and is completely absorbed. The energy falling on the strip per unit time is W. find the deflection of the string from the vertical, if the strip stays in equilibrium. Sol: Force exerted by the photons of light = (number of incident photons per sec) × (change in momentum of each photon) Let, Number of incident photons per sec = N h c hence force exerted by the photons of light,

change in momentum of each photon =

Light

h (Nh) = c c W = Nhv, by the definition of power

F = N but,

W c If the string makes an angle  with the vertical when the strip is at equilibrium, then for the equilibrium of the strip, Tsin = F, in horizontal direction, and Tcos = mg, in vertical direction.  T = tension in the string Dividing the above equations, we get



F =

W W/c F tan = = = , mg c mg mg  tan =

 F

 W  W  = tan–1  c mg  c mg 

T

m

mg

Example 6. A plane light wave of intensity I = 0.80 W cm–2 illuminates a sphere of radius R = 5.0 cm. Find the force that the light exerts on the sphere. Sol. p, momentum of the incident pulse = E/c.  p  (E / c) (sin  ˆi  cos  ˆj) with respect to a frame of reference with the outward normal as the y-axis i

and a line perpendicular to it and lying in the plane of the mirror as the x-axis.

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ATOMIC PHYSICS



 pf = E/c and pf  (E / c) (sin  ˆi  cos  ˆj)     p  p f  pi  (  1)E / c sin  ˆi  (  1) cos E / c ˆj  |  p |  E / c (  1)2 sin 2   (  1)2 cos 2   E / c 1  2  2 cos 2

 |  p |  10 / 3  108





1  0.82  2  0.8 cos 60º

= 5.2 × 10–8 Nm–2

Impulse applied by photon on a surface Let hn be the energy of photons of a light incident normally on a surface, h . c Change in momentum of the photon takes place due to impact of the photons with the surface. This change in momentum of the photons causes impulse. Using impulse – momentum theorem, we get Impulse = Total change in momentum of photons = (Total number of photons) × (change in momentum of each photon)  Impulse = N × (change in momentum of each photon) ......(i)

Momentum of an incident photon =

Impulse on a totally absorbing surface: In this case, change in momentum of each photon =

h c

 using equation (i), we get Impulse =  Impulse =

N ' h E = , where E is the total energy of the light c c E for a perfectly absorbing surface. c

Impulse on a totally reflecting surface: Impulse = N × (change in momentum of each photon)  h  h   2(N ' h) 2E = N ×       = = c c   c c  

Impulse =

[ E = total energy = Nh]

2E for a perfectly reflecting surface. c

Impulse on a surface of reflection-coefficient : In this case, change in momentum of each photon =

h   h  h –    = (1 + ) . c  c c 

Using equation (i), we get

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ATOMIC PHYSICS Impulse = N(1 + )

 N ' h  h E = (1 + )   = (1 + ) , c c  c 

 Impulse = (1 + )

E for photons falling normally on the surface. c

[E = total energy]

Example 7: A small perfectly reflecting mirror of mass m = 10 mg is suspended by a weightless thread of length  = 10 cm as shown in the figure. Find the angle through which the thread will be deflected when a short laser with energy E = 13J is shot in horizontal direction at right angles to the mirror. (g = 10 m/s2).



m

2E Sol: Impulse applies by laser on the mirror, Impulse = , as mirror is perfectly c reflecting. Initial momentum of the mirror = 0 Let, final momentum of the mirror = mv change in momentum of the mirror due to impact = 0 – (–mv) = mv, where v is the speed of mirror just after impact. But, Impulse momentum theorem gives, Impulse = change in momentum

Laser m





v=0 2E 2E = mv or v= .....(i) c mc Total mechanical energy of the mirror will be conserved after impact. Thus, Loss in kinetic energy = Gain in potential energy

(1 – cos) = H v

1 2 mv = mgh  v2 = 2gH 2

4E 2 = 2gH, m 2c2 2E 2 = gH m 2c2

[as v = 

2E from equation (i)] mc

2E 2 = g(1 – cos), m 2c2

[ H =  (1 – cos)]

2E 2 1 – cos = 2 2 m c g 2E 2  2sin   = 2 2  2  m c g

 [ 1 – cos = 2sin2   ] 2

2

E2  sin = 2 mc g



sin

Here E = 13J, m = 10 × 10–6 kg,  = 0.1 m

13  = 5 = 0.0043 2 10  3 108 10  0.1

or

 = 0.5º

PHOTONS UNDER GRAVITY

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ATOMIC PHYSICS Photons can be considered as a particle of mass m. If m = mass of photon v = frequency of photon E = energy of photon, then mc2 = E = hv  m =

h E 2 = c2 c

Thus, a photon of frequency v acts gravitationally like a particle of mass

h . c2

C8: If a photon frequency v falls on the surface of earth from a height h, then what will be its frequency on the surface of earth. v, E = hv + mgH Sol: Change in frequency of the photon takes place. Let v be the frequency of photon on the surface of earth. H

h v, E = hv Mass of photon = 2 c Ground Mass of the photon depends on its frequency, but we will consider the mass to be constant as difference in v´ and v is very small. Energy conservation gives Initial energy = final energy  hv + mgH = hv h gH = hv c2



hv +



 gH  v = v  1  2  c  

Example 8: A planet of mass M and radius R emits a photon of frequency v. What will be the frequency of photon when it covers an infinite distance. Sol: Let vbe the frequency photon at a infinite distance. v G = Gravitational constant v R

h Mass of the photon, m = 2 c

at 

M

Energy of photon on the surface = hv – G

GM  h  Mm = hv –   R  c2  R

Energy of photon at infinity = hv Energy must be conserved, 

hv –

GM  h    = hv´ R  c2 

NOTE:



 GM  v = v 1  2   Rc 

 GM = is called frequency shift.  Rc 2

PHOTOELECTRIC EFFECT

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ATOMIC PHYSICS Ejection of electrons from a metal plate when illuminated by light or any other radiation of suitable wavelength (or frequency) is called photoelectric effect. This phenomenon was first discovered by Heinrich Hertz in 1887. One year later, Hallwachs made the important observation that when a negatively-charged zinc plate is irradiated with ultra-violet light, it loses its negative charge. Afterwards, it was discovered that alkali metals like lithium, sodium, potassium, rubidium and caesium eject electrons when visible light falls on them. Ten years later, J.J. Thomson and P.Lenard showed that the action of light was to cause the emission of free electrons from metal surface. Although these electrons are no different from all other electrons, it is customary to refer to them as photoelectrons. Experimental Study of Photoelectric Effect Quartz Photoelectric phenomenon can be studied with the help of a simple apparatus shown in figure. It consists of two photosensitive surfaces – C + E E and C enclosed in an evacuated quartz bulb. In the absence of light, there is no flow of current in the circuit and the ammeter A A reads zero. When plate E is exposed to some source of V monochromatic light, current starts flowing. However, no current is found to flow when light falls on plate C. The explanation of the above behaviour lies in the fact that when E is irradiated with light, the incident photons eject electrons by collision with its atoms. These photoelectrons are immediately attracted by the collector plate C thereby starting current flow as indicated by the ammeter. When C is irradiated, even then photoelectrons are produced but they are not allowed to leave plate C (i) firstly, because of the pulling effect of positive potential of C and (ii) secondly, due to repulsion from the negative plate E. Hence, no current is found to flow in the circuit. Einstein’s Photoelectric Equation: Following Planck’s idea that light consists of photons, Einstein proposed an explanation of photoelectric effect as early as 1905. According to this explanation when a single photon is incident on a metal surface, it is completely absorbed an imparts its energy hf to a single electron. The photon energy is utilised for two purposes: (i) Partly for getting the electron free from the atom and away from the metal surface. This energy is known as the photoelectric work function of the metal and is represented by W0. (ii) the balance of the photon energy is used up in giving the electron a kinetic energy of 1/2mv2.  hf = W0 + 1/2mv2 .....(i) It is known as Einstein’s photoelectric equation. In case, the photon energy is just sufficient to liberate the electron only then no energy would be available for imparting kinetic energy to the electron. Hence, the above equation would reduce to hf0 = W0 .....(ii) where f0 is called the threshold frequency. It is defined as the minimum frequency which can cause photoelectric emission. For frequencies lower than f0, there would be no emission of electrons whereas for frequencies greater than f0, electrons would be ejected with a certain definite velocity (and hence kinetic energy). Substituting this value of W0 in equation (i) above, the Einstein’s photoelectric equation becomes 1 mv2 2 hf = hf0 + K.E.

hf = hf0 + or

Long Wavelength Limit ( 0)

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ATOMIC PHYSICS It is the wavelength corresponding to the threshold frequency f0. Its physical significance is that radiations having wavelength longer than 0 would not be able to eject electrons from a given material whereas those having  < 0 will. In other words, it represents the upper limit of wavelength for photoelectric phenomenon. By analogy, it is also referred to as threshold wavelength. Now,

c = f0 0

Also, W0 = hf0 

0 =

c f0

1 h = f0 W0

ch 0 = W 0



(i) When W0 is in joules 3  108  6.625  1034 0 = W0 19.875  1026 = metre W0 (ii) When W0 is electron volts (eV) 3  108  6.625  1034 0 = 1.602  1019 W0 12, 400 12.4  107 = metre = W Å W0 0

Kinetic energy of Photoelectrons Einstein’s photoelectric equation can be used to find the velocity and hence the kinetic energy of an ejected photo-electron. 1 mv2 2 = hf0 + K.E. K.E. = hf – hf0 = h(f – f0)

hf = W0 +



Now, f 

=

c c and f0 =   0

1 1  K.E. = ch      0   1 1  = 3 × 108 × 6.625 × 10–34      0   1 1  = 19.875 × 10–26      joules 0  

 and 0 in metres

1 1  = 19.875 × 10–26      joules  and 0 in Å 0   www.physicsashok.in

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ATOMIC PHYSICS

=

19.875 1016  1  1    eV 1.602 1019    0 

 and 0 in Å

1 1  K.E. = 12,400      eV  and 0 in Å 0   Incidentally, it may be noted that this also represents the maximum value of the kinetic energy a photoelectron can have.  Emass = h(f – f0) = h.f joules 

1 1  = 12,400      eV 0  

 and 0 in Å

Photoelectric Work Function As explained above it is defined as the energy which is just sufficient to liberate electrons from a body with zero velocity. Its value is given by W0



ch 3  108  6.625  1034 = hf0 =  = 0 0

=

19.875  1026 joules 0

0 in metres

=

19.875  1016 joules 0

0 in Å

=

19.875  1016 eV 1.602  1019  0

0 in Å

12, 400  0 eV

0 in Å

W0 =

Laws of Photoelectric Emission Quartz A C V

B R

The apparatus shown in figure was used by Millikan to study the photoelectric effect and the various laws governing it. S is a source of radiations of variable but known frequency f and intensity I. E is the emitting electrode made of the photosensitive material and C is the collecting electrode. Both these electrodes are enclosed in an evacuated glass envelope with a quartz window that permits the passage of ultraviolet and

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ATOMIC PHYSICS visible light. As shown, any potential difference can be established between the two electrodes. A reversing switch RS enables the polarities of the two electrodes to be reversed. If the tube is in the dark, then no photoelectrons are emitted and the microammeter A read zero. However, if ultraviolet or visible light is allowed to fall on the emitting electrode, electrons are liberated and circuit current is set up. From the experimental data collected by Richardson and Compton in 1912, photoelectric emission was found to be governed by the following laws: (i) Photoelectric current (i.e., number of electrons emitted per second) is directly proportional to the intensity of the incident light (or radiation). Photoelectric current

Frequency constant

Light Intensity I

Photoelectric current

This can be verified by increasing the intensity of light and measuring the corresponding photoelectric current while holding the frequency of the incident light frequency of the incident light and the accelerating potential V of the collecting electrode C constant. The graph is similar to one shown in figure. Increase in intensity means more photons and hence greater ejection of electrons. (ii) For each photosensitive surface, there is a minimum frequency of radiation (called threshold frequency) at which emission begins.

As seen from Einstein’s photoelectric equation of 1 mv 2max = h(f – f0) 2

or Emax  f

Emax

Intensity Light of lower frequency (or longer wavelength), however strong, constant will not be able to produce any electron emission. This fact can be verified by keeping the light intensity constant while varying the frequency. The graph so obtained is shown in figure. The current is found to increase (though non-linearly) with the frequency of the incident light. f0 Moreover, it is seen that there is a limiting or critical frequency below Frequency which no photoelectons are emitted. It is called threshold frequency and its value depends on the nature of the material irradiated because for each material there is a certain minimum energy nessecary to liberate an electron. This energy is known as photoelectric work function or threshold energy W0. As seen from f0 = W0/h. The wavelength corresponding to be threshold frequency f0 is called threshold wavelength or long wavelength limit. No photo-electrons are emitted for wavelength greater than 0, no matter how long the light falls on the surface or how greater is its intensity. The photoelectric or quantum yield (which is defined as the photoelectric current in amperes per watt of incident light) depends on the frequency (and not intensity) of the incident light. Intensity (iii) The maximum velocity of electron emission (and hence kinetic energy) varies constant linearly with the frequency o the incident light but is independent of its intensity.

 f0 Frequency

Hence, increase in the frequency of the incident light increase the velocity with which photoelectrons are ejected. The same fact is illustrated by figure. Incidentally, it may be noted that the slope of the curve gives the value of Planck’s constant h. If the intensity of the incident light is increased (keeping frequency constant), more photons will be incident

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ATOMIC PHYSICS on the metal surface, each photon having the same energy. Hence, more photo electrons will be ejected. Since an electron can absorb only one photon, each photoelectron will have the same energy and will be ejected with the same velocity. Obviously, increase in intensity only increases the number of photo-electrons ejected but not their kinetic energy. (iv) The velocities of emitted electrons have values between zero and a definite maximum. The proportion of the electrons having a particular velocity is independent of the intensity of radiation. (v) Electrons are emitted almost instantaneously even when the intensity of incident radiations is very low. The time lag between the incidence of radiation and emission of first electrons is less than 10–8 second. (vi) For a given metal surface, stopping potential V0 is directly proportional to frequency but is independent of the intensity of the incident light. i Suppose in figure, the frequency and intensity of incident light are H held constant but the potential difference V between the two electrodes E and C is increased. Up to some stage as this p.d. is b a I increased photoelectric current is also increased. However, soon some value of V is reached when the current reaches a limiting or saturation value when al the photoelectrons emitted by E are v0 v O immediately collected by C. Further increase in V hardly produces any appreciable increase in current as shown by the flat portion of curve I in figure. If V is reversed with the help of the reversing switch RS i.e. E is made positive and C negative, the current i does not immediately drop to zero proving that electrons are emitted from E with some definite velocity. This velocity is such that it gives enough kinetic energy to the electrons so as to surmount the retarding electric field between the two electrodes. Hence, some electrons do succeed in reaching C despite the fact that the electric field opposes their motion. When reversed V is made large enough, a value V0 (called stopping or inhibiting potential) is reached when current is reduced to zero. Stopping potential is that value of the retarding potential difference between the two electrodes which is just sufficient to half the most energetic photoelectrons emitted. As seen from curve II of figure, doubling the intensity of the incident light merely doubles the current but does not affect the value of V0. Now, if vmax is the maximum velocity of emission of a photoelectron and V0 the stopping potential, then

1 mv 2max 2

or

= eV0 or vmax =

vmax =

2.eV0 m

2 1.76 1011  V0

= 5.93 × 105

V0 m/s

Obviously Emax = eV0 joules = V0 electron-volt If however, the experiment is repeated by varying the frequency of the light, it is found that the stopping potential varies linearly with frequency as shown in figure. Below threshold frequency, no electrons are emitted, hence stopping potential is zero for that reason. But as frequency is increased above f0, the stopping potential varies linearly with the frequency of incident light.

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ATOMIC PHYSICS

hf = W0 +

1 mv 2max 2 1 mv 2max = eV0 2

Now,

W 0 = hf0 and



hf = hf0 + eV0 or V0 =

Now,

f

=

V0 =

h(f  f 0 ) e

c c and f0 =   0

f0 Frequency

ch  1  1    e   0 

1 1  = 12.4 × 10–7      volt 0   

Stopping potential, V0

Einstein’s photoelectric equation may be expressed in terms of stopping potential as given below:

1 1  V0 = 12,400      volt 0  

 and 0 in metres

 and 0 in Å

C9: Photoelectrons with a maximum speed of 7 × 105 m/sec are emitted from a metal surface when light of frequency 8 × 1014 Hz falls on it. Calyculate the threshold frequency of the surface. Sol: Emax = h(f – f0)  or 

1 mv 2max = h(f – f0) 2 1 × 9.1 × 10–31 × (7 × 105)2 = 6.625 × 10–34 (8 × 1014 – f0) 2 f0 = 4.635 × 1014 Hz

C10: A tungsten cathode whose threshold wavelength 2300 Å is irradiated by ultraviolet light of wavelength 1800 Å. Calculate (i) the maximum energy of the photoelectrons emitted and (ii) the work function for tungsten, both in electron-volts. Sol: (i)

W0 =

12, 400 12, 400 = = 5.4 eV 0 2300

1 1  (ii) Emax = 12,400      eV 0   1   1  = 12,400   = 1.5 eV  1800 2300  C11: If light of  = 6000 Å falls on a metal surface and emits photoelectrons with a velocity of 4 × 105 m/s, what is photoelectric threshold wavelength ?

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ATOMIC PHYSICS 1 (4 105 )2 –31 Sol: K.E. of photoelectrons = × 9.1 × 10 × = 0.445 eV 2 1.6 1019

Energy content of photon of  = 6000 Å = 

12, 000 = 2.07 eV 6000

W0 = 2.07 – 0.445 = 1.625 eV 0 =

12, 000 = 7631 Å 1.625

C12: Calculate the threshold frequency for gold having photoelectric work function equal to 4.8 eV. If light of wavelength 2220 Å falls on gold, what will be maximum kinetic energy of the photoelectrons coming out? 12, 000 = 5.58 eV. Out of this, 4.8 eV would be used for dislodging the 2220 electron and the balance would represent its kinetic energy. Emax = 5.58 – 4.8 = 0.78 eV

Sol: Energy of the light photon =

Alternatively, 0 = 

12, 400 = 2583 Å. Hence, we may use 4.8

1 1  Emax = 12,400      eV.. 0  

C13: When violet light of  = 4000 Au strikes the cathode of a photocell a retarding potential of 0.4 V is required to stop emission of electrons. Calculate (i) light frequency (ii) photon energy (iii) work function (iv) threshold frequency and (v) net energy after the electron leaves the surface. Sol: (i)

c 3 108 =  4000 1010 = 7.5 × 1014 Hz

f =

(iii)

E = hf = 6.625 × 10–34 × 7.5 × 1014 = 4.95 × 10–19 J = 3.1 eV W0 = hf – K.E. = hf – V0 = 3.1 – 0.4 = 2.7 eV

(iv)

f0 =

(ii)

W0 2.7 1.6 1019 = h 6.625 1034 = 6.5 × 1014 Hz

(v) Net energy hf – W0 = 3.1 – 2.7 = 0.4 eV = 6.4 × 10–20 J Example 9: A photon of wavelength 3310 Å falling on a photo cathode ejects an electron of energy 3 × 10–19 J and one of wavelength 5000 Å ejects an electron of energy 0.972 × 10–19 J. Calculate the value of Planck’s constant and the threshold wavelength for the photo cathode. Sol:

hf = W0 + K.E. or

hc = W0 + K.E. 

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ATOMIC PHYSICS

h  3 108 In the first case, = W0 + 3 × 10–19 3310 1010 h  3 108 In the second case, = W0 + 0.972 × 10–19 5000 1010 Subtracting one from the other, h = 6.62 × 10–34 J-s Substituting this value of h, W0 = 3 × 10–19 J ch 3 108  6.62 10 34 0 = W = 3 1019 0



vmax

= 6620 × 10–10 m. = 7.12 × 105 m/s

Example 10: A certain metal has a threshold wavelength of 6525 Å. Find the stopping potential when the metal is irradiated with (a) monochromatic light having a wavelength of 4000 Å. (b) light having twice the frequency and three times the intensity of that in (a) above. (c) If a material having double the work function were used, what would be the answer to (a) and (b) above? 1   1  Sol: (a) V0 = 12,400   = 1.2 volt  4000 6525  (b) Stopping potential is independent of the intensity of the incident light but varies directly as frequency f provided it is more than f0. Since frequency is twice, the wavelength of the light is half i.e. 2000 Å 1   1  V0 = 12,400   = 4.3 V  2000 6525  6525 (c) If work function is double, then 0 is reduced to half i.e. 0 = = 3262.5 Å. Since the incident light 2 has  = 4000 Å, it would not be able to produce photoemission. In the second case,  = 2000 Å. 1   1   V0 = 12,400   = 2.4 volt  2000 3262.5 

C14: A certain metallic surface is illuminated by monochromatic light of variable wavelength. No photoelectrons are emitted above a wavelength of 5000 Å. With an unknown wavelength, a stopping potential of 3.1 V is necessary to stop photoelectric current. Find the unknown wavelength. Sol: Here, 0 = 5000 Å, V0 = 3.1 V,  = ? 1 1  Now, V0 = 12,400      0    

1  1 3.1 = 12,400      5000   = 2,222 Å

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ATOMIC PHYSICS C15: Light of wavelength 2000 Å falls on an aluminium surface. In aluminium, 4.2 eV are required to remove an electron. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photoelectron (iii) stopping potential and (iv) cut-off wavelength for aluminium. Sol: Photon energy of the incident light is 12400 = 6.2 eV 2000 (i) Emax = (6.2 – 4.2) eV = 2 eV = 2 × 1.6 × 10–19 = 3.2 × 10–19 J (ii) Emin = 0 (iii) Ve = Emax = 2 V

=

(iv)  0 =

12400 12400 W0 = 4.2 = 2952.4 Å

C16: The stopping potential is 4.6 V for light of frequency 2 × 1015 Hz. When light of frequency 4 × 1015 Hz is used, the stopping potential is 12.9 V. Calculate the value of Planck’s constant. Sol: eV0 = h(f – f0) Substituting the two given values, we get 4.6e = h (2 × 1015 – f0) 12.9e = h(4 × 1015 – f0) Subtracting one from the other, we have 8.3e = 2h × 1015 8.3 × 1.6 × 10–19 = 2h × 1015 

h =

8.3 1.6 10 19 = 6.44 × 10–34 Js 2 1015

Example 11. 10–3 W of 5000 Å light is directed on a photoelectric cell. If the current in the cell is 0.16 µA, the percentage of incident photons which produce photoelectrons, is (A) 0.4% (B) .04% (C) 20% (D) 10% Sol. The percentage of incident photons which produce photoelectrons is = 

n / t  100 N / t

I

...(1)

ne t

n I 0.16  1016   t e 1.6  1019 n  1012 t N hc W and  t  1012 Persentage = 16  4  100 10



Persentage  0.04%

...(2)

N W 1016   t hc 4

...(3)

[from eq. (1), (2) and (3)] Hence (B) is correct.

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ATOMIC PHYSICS Example 12. In a photo-emissive cell, with exciting wavelength , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to (A) 3K/4 Sol.

(B) 4K/3

3 the the kinetic energy of the fastest emitted electron will be : 4 (C) less than 4K/3 (D) greater than 4K/3

hc K 

...(1)

hc    K´ 3 / 4 Substracte eqn. (1) from eq. (2)

...(2)

4hc hc   K´K 3  4hc  3hc  K´ K 3 hc  K´ K 3 K  K´ K 3

K´

K K 3

K´

4K 3

Hence (D) is correct.

Example 13. Let K1 be the maximum kinetic energy of photoelectrons emitted by a light of wavelength 1 and K2 corresponding to 2. If 1= 22, then : (A) 2K1 = K2 Sol.

(B) K1 = 2K2

(C) K1
2K2

P2 K.E. = 2m

P  2mK P K

h  K 



1 K

1 K2  2 K1

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ATOMIC PHYSICS 

1 = 22



2 2  2 4



K2 K1

K2 K1

K1 

K2 4

K1 

K2 2

Hence (C) is correct.

Example 14. Radiation of two photon energies twice and five times the work function of metal are incident sucessively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted is the two cases will be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol. E1 = 2  E2 = 5  E1 =  + K1 2  =  + K1 K1 =  and E2 =  + K2 5  =  + K2 K2 = 4  

K1 1  K2 4

1 mv 2max 1 1 2  1 2 mv max 2 4 2 v 2max1 v

2 max 2



v max 1 v max 2

1 4 

1 2

Hence (A) is correct.

Example 15. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy Ta eV and de-Broglie wavelength a. the maximum kinetic energy of photoelectrons liberated from another metal B by photones of energy 4.7 eV is Tb = (Ta – 1.5) eV. If the De-Broglie wavelength of these photoelectrons is b = 2a, then find (a) The work function of a (b) The work function of b is (c) Ta and Tb

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ATOMIC PHYSICS 1 h  A  mv 2Amax 2

Sol. and

a 

h PA



PA 

h a



1 P2 mv 2A max  A 2 2m



h  A 

or

4.25 = A 

where m is mass of electron.

PA2 2m h2 2m 2a

...(1)

For B, 4.7 = B + Tb h2 4.7   B  2m 2b But b = 2a and Tb = Ta – 1.5 After solving (a) 2.25 eV (b) 4.2 eV

...(2)

(c) 2 eV and 0.5 eV

C17: An isolated metal body is illuminated with monochromatic light and is observed to become charged to a steady positive potential 1.0 V with respect to the surrounding. The work function of the metal is 3.0 eV. the frequency of the incident light is __________. Sol. h  =  + eV or h  = 3 + 1 = 4 eV 

4 eV 4  1.6  10 19   h 6.63  1034  = 0.96 × 1015 Hz

Example 16. 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5 × 109 incident photons is absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is _______. N = no. of photon incident per second. t

Sol. 

663  103 

N hc t 

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ATOMIC PHYSICS



N 663  103 663  103   hc 1242 nmeV t  540



n / t 1  N / t 5  109



n 1 N 1 663  103  540     t 5  109 t 5  109 1242 nmeV



I

ne  5.76  1011 A t

Example 17. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage V0 to prevent a electron from reading collector. In the same setup, light of wavelength 220 nm, ejects electrons which require twich the voltage V0 stop them in reaching a collector. Find the numerical value of voltage V 0. (take plank’s constant, h = 6.6 × 10–34 Js and 1 eV = 1.6 × 10–19 J) hc    eV0 1

Sol. and

hc    2eV0 2

Here 1 = 330 nm and 2 = 220 nm After solving, V0 

15 volt 8

Example 18. A small 10W source of ultraviolet light of wavelength 99 nm is held at a distance 0.1 m from a metal surface. The radius of an atom of the metal is approximately 0.05 nm. Find (i) the average number of photons striking an atom per second. (ii) the number of photoelectrons emitted per unit area per second if the efficiency of liberation of photoelectrons is 1%.

10 4(0.1) 2 w = power incident on atom

Sol. (i) I = intensity = 

2 = I r 

10  0.05  10 9 2 4  10





2

n hc t 



w 



n w 5   t hc /  16

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ATOMIC PHYSICS (ii) no. of photons incident per unit area per second I hc /   no. of ejected electrons per unit area per second 

I 1 1020   hc /  100 80 

Example 19. The surface of cesium is illuminated with monochromatic light of various wavelengths and the stopping potentials for the wavelengths are measured. The results of this experiment is plotted as shown in the figure. estimate the value of work function of the cesium and Planck’s constant.

supporting potential (volt)



2 1 0 –1 –2

0.49 0.5 1.0 1.5

v × 10 15 Hz

h  =  + eVs

Sol. or

Vs 

h   e e

From graph h 2  e 0.49  1015



But 

2  1.6  1019 15 0.49  10 h = 6.53 × 10–34 Js h

  2 e  = 2 eV 

Example 20. In a photoelectric effect set-up a point source of light of power 3.2 × 10–3 W emits monoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0 × 10–3 m. The efficiency of photoelectron emission is one for every 106 incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission. (a) Calculate the number of photoelectrons emitted per second. (b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons emitted. (c) It is observed that the photoelectron emission stops at a certain time t after the light source is switched on why? (d) Evaluate the time t. Sol. (a) Energy of emitted photons E1 = 5.0 eV = 5.0 × 1.6 × 10–19 J E1 = 8.0 × 10–19 J Power of the point source is 3.2 × 10–3 watt or 3.2 × 10–3 J/s s 0.8 m r = 8.0×10–3 m Therefore, energy emitted per second, E2 = 3.2 × 10–3 J. Hence number of photons emitted per second

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ATOMIC PHYSICS n1 

E2 E1

or

n1 

3.2  103 8.0  1019

n1 = 4.0 × 1015 photons/sec. Number of photons incident on unit area at a distance of 8.0 m from the source S will be

n1 4.0  1015 n2 = = 5.0 × 1014 photon/sec – m2. 4 (0.8)2 4(0.64)  The area of metallic sphere over which photons will fall is : A = r2 = (8 × 10–3)2 m2  2.01 × 10–4 m2 Therefore, number of photons incident on the sphere per second are n3 = n2 A = (5.0 × 1014 × 2.01 × 10–4)  10111 per second But since one photoelectron is emitted for every 106 photons hence number of photoelectrons emitted per second,

n3 1011 n = 6 = 6 = 105 per second 10 10 5 or n = 10 per second (b) Maximum kinetic energy of photoelectrons Kmax = Energy of incident photones – work function Kmax = (5.0 – 3.0) eV = 2.0 eV = 2.0 × 1.6 × 10–19 J Kmax = 3.2 × 10–19 J The de-Broglie wavelength of these photoelectrons will be 1 

h  p

h 2 K max m

(Here h = Planck’s constant and m = mass of electron) 6.63  10 34    1 2  3.2  10 19  9.1  10 31 1 = 8.68×10–10 m = 8.68 Å 12375 Wavelength of incident light 2 (in Å) = E (in eV) 1 12375 Å = 2476 Å 5 Therefore, the desired ratio is

or

2 =

 2 2475   285.1 1 8.68

(c) As soon as electrons are emitted from the metal sphere, it gets positively charged and acquires positive potential. The positive potential gradually increases as more and more photoelectrons are emitted from its surface. Emission of photoelectrons is stopped when its potential is equal to the stopping potential required for fastest moving electrons. (b) As discussed in part (c), emission of photoelectrons is stopped when potential on the metal sphere is equal to the stoppeing potential of fastest moving electrons. Since Kmax = 2.0 eV

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ATOMIC PHYSICS Therefore, stopping potential V0 = 2 volt. Let q be the charge required for the potential on the sphere to be equal to stopping potential or 2 volt. Then 2

1 q 9 .  9.0  109 4 0 r 8.0  103





 q = 1.78 × 10–12 C Photoelectrons emitted per second = 105 [part a] –19 5 or charge emitted per second = (1.6 × 10 ) × 10 C = (1.6 × 10–14) C Therefore, time required to acquire to charge q will be

q 1.728  102 sec  sec 1.6  1014 1.6 t  111 second t

or

Example 21. Monochromatic radiation of wavelength 1 = 3000 Å falls on a photocell operating in saturation mode. The corresponding spectral sensitivity of photocell is J = 4.8 mA/W. When another monochromatic radiation of wavelength 2 = 1650 Å and power P = 5 mW is incident. It is found that maximum velocity of photoelectrons increases to n = 2 times. Assuming efficiency of photo-electron generation per incident photon to be same for both the cases, calculate (i) threshold wavelength for the cell and (ii) saturation current in second case. [Given, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1 and e = 1.6 × 10–19 coul.] hc – W0, where  is wavelength of incident  radiation and W0 is work function of the surface on which radiation is incident.  Maximum kinetic energy of photoelectrons emitted by radiation of wavelength 1 is

Sol. Maximum kinetic energy of photoelectrons is given by Ek =

given by

1 hc mv12 = – W0 2 1

 hc  mv12  2   W0  ...(1)  1  where m is mass of an electron and v1 is maximum velocity of photoelectrons. or

Similarly, for radiation of wavelength 1, But

1 hc mv 22   W0 2 2

...(2)

v2 = 2v1, therefore from equation (2), 2mv12 

hc  W0 2

...(3)

From equations (1) and (3),

 hc  hc 4   W0    W0  1  2 or

W0 = 3 eV

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ATOMIC PHYSICS hc But work-function W0 =  where 0 is threshold wavelength. 0 hc  0 = W = 4125 Å Ans. (i) 0 In saturation mode, spectal sensitivity with wavelength 1 = 3000 Å is J = 4.8 mA/W or 4.8 mC/J. It means when 1 joule radiation of wavelength 1 = 3000 Å is incident, a charge of 4.8 mC flows in saturation mode 4.8 mC or electrons are ejected. e hc Energy of each photon of wavelength 1 is E1 =  1  Number of photons in 1 joule radiation of wavelength 1 1 1   E1 hc J 4.8  103 = = 3 × 1016 e 1.6  10 19  Efficiency of photo-electron generation per incident photon, 3  1016   0.0198 . (1 / hc) Energy of each photon of wavelength 2, hc E2 =  2  Rate of incidence of photons of wavelength 2 in a radiation of power P

No. of electrons ejected by these photons =



P P 2  E2 hc per second

Since, efficiency  of photo-electron generation is same for both the case, therefore, rate of ejection of electrons in later case P   . 2 per second hc  P 2  Rate of flow of charge in saturation mode = e Cs–1 = 13.2 µ Cs–1 hc But rate of flow of charge is current. Hence, saturation current is second case = 13.2 µA. Ans. (ii) A Example 22. A monochromatic point source S radiating wavelength  = 6000 Å with power P = 2 watt, an aperture A of radius R = 1 cm and a large screen are placed as shown in fig. A photoemissive detector S D L D of surface area S = 0.5 cm2 is placed at centre of the screen. Efficiency of detector for photoelectric emission per incident photon 60cm is  = 0.9. 6m (i) calculate photon flux at centre of screen and photo current in the detector. (ii) If a convex lens L of focal length f = 30 cm is inserted in the aperture as shown, calculate new value of photon flux and photo current assuming a uniform average transmission of 80% from the lens. (iii) If work function of photo-emissive surface is W0 = 1 eV, calculate value of stopping potential in two cases (without and with the lens in aperture). Given, h = 6.625 × 10–34 J-S, c = 3 × 108 ms–1.

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ATOMIC PHYSICS Sol. Photon flux is rate of incidence of photons per unit are of detector. Therefore, to calculate photon flux, first rate of emission of electrons from the source should be calculated. hc  Rate of emission of energy from the source is E

Energy of each photon is

P = 2 watt = 2 Js–1

 Rate of emission of photons from the source is n 

(i)

P P  E hc

or n = 6.04 × 1018 photons per second Distance of detector from source is r1 = 6 m  Photon flux at detector, 1 

n  1.33  1016 photons/m2s 4r12

Rate of incidence of photons on detector = 1 . S Rate of emission of electrons from detector = 1S per second Since, current is charge flowing per second, therefore photo current = (1S)e = 9.6 × 10–8 amp (ii) When a concave lens is inserted in the aperture, it refracts incident rays. Therefore, photon flux and hence photo-current changes. Distance of lens from source is r2 = 0.60 m n  Photon flux at lens is ´ = 4r 2 = 1.33 × 1018 photons/m2s 2

Considering a very small area A of the lens, Rate of incidence of photons on this area of lens = A´ Now considering refraction through the lens, u = – 60 cm, f = + 30 cm v = ? 1 1 1   , v = + 60 cm v u f Since, average transmission from lens is 80%, therefore, rate of transmission of photons from area A of lens = 0.8 A´ But these photons are transmitted in a solid angle subtended by the area A at P as shown in fig.

Using lens formule,

A A  2 v (0.6)2  Rate of photons transmitted per unit solid angle is 

This solid angle,

T

(0.8 A´) = (0.8 ´) (0.6)2 = 0.288 ´ 

D

P

S v

4.80m

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30

ATOMIC PHYSICS Solid angle subtended by unit area of detector at P, ´ =

1 (4.80)2

 Photon flux at D, 2 = T´ = 1.67 × 1016 photons/m2s Rate of incidence of photons on detector = 2S Rate of emission of electrons from detector = 2S  Photo current = (2S)e = 1.20 × 10–7 amp  hc  (iii) Since, stopping potential V0 is given by e.V0 =   W0  and  and W0 both remain unchanged, therefore,  stopping potential is same for both the cases. 1  hc  V0    W0   1.07 volt   e 

PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT According to wave theory, when light falls on a metal surface, energy is continuously distributed over the surface, energy is continuously distributed over the surface. All the free electrons at the surface receive light energy. An electron may be ejected only when it acquired energy more than the work function. If we use a low-intensity source, it may take hours before an electron acquires this much energy from the light. In this period, there will be many collisions and any extra energy accumulated so far will be shared with the remaining metal. This will result in no photoelectron. This is contrary to experimental observations. No matter how small is the intensity, photoelectrons are ejected and that too without any appreciable time delay. In the photon theory, low intensity means less number of photons and hence less number of electrons get a change to absorb energy. But any fortunate electron on which a photon falls, gets the full energy of the photon and may come out immediately. In figure, we illustrate an analogy to the wave the particle behaviour of light. In part (a), water is sprayed from a distance on an area containing several plants. Each plant receives water at nearly the same rate. It takes time for a particular plant to receive a certain amount of water. In part (b) of the figure, water is filled in identical, loosely-tied water bags and a particle physicist throws the bags randomly at the plants. When a bag collides with a plant, it sprays all its water on that plant in a very short time. In the same way, whole of the energy associated with a photon is absorbed by a free electron when the photon hits it.

(a)

(b)

The maximum kinetic energy of a photoelectron does not depend on the intensity of the incident light. This fact is also not understood by the wave theory. According to this theory, more intensity means more energy and the maximum kinetic energy must increase with the increase in intensity which is not true. The dependence of maximum kinetic energy on wavelength is also against the wave theory. There should not be any threshold wavelength according to the wave theory. According to this theory, by using sufficiently intense light of any wavelength, an electron may be given the required amount of energy to come out. Experiments, however, show the existence of threshold wavelength.

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ATOMIC PHYSICS DUAL NATURE OF LIGHT (a) Wave nature: Wave nature of light can be explained on the basis of reflection, refraction, interference, diffraction and polarization. (b) Particle Nature: Energy is transported by energy particles, photons. It could be explained by photoelectric effect, Zeeman effect, Compton effect etc. MATTER WAVE THEORY OR DE-BROGLIE’S THEORY (a) This theory was given on the basis of duel nature of light. (b) According to de-Broglie theory each and every moving particle has some wave nature associated with itself which is called matter waves. (c) Thus, moving particles like e–, proton, neutron, -particle etc, also behave like waves. (d) These waves are waves of probability. (e) The wavelength associated with a moving particles is given by  = h/p, where p is the momentum of the particle. (f) This wavelength is known as the de-Broglie wavelength of the particle. DE BROGLIE WAVELENGTHS For Photon

For moving particle

Rest mass

Zero

m

Effective mass

m=

Energy (kinetic)

E = hv =

Momentum

p=

h E h = = c c 

p = mv =

2mE

Wavelength

=

h hc = p E

=

h = p

h 2mE

h E 2 = c2 c hc 

E=

m0 1  v2 / c2 1 p2 mv2, E = 2 2m

h 2mE K.E. = qV =

Wavelength for charged particle accelerated by V volts Speed

m=

c = 3 × 108 m/s

De-Broglie’s explanation for stable Bohr’s orbits : (a) De-Broglie suggested that non-radiation of energy by the electrons circling in a Bohr’s orbit can be explained on the basis on the basis of the formation of stationary waves by the electrons in circular motion in Bohr’s orbits. (b) Comparing this to the vibrations of a wire loop such stationary waves would be formed if each wave joins smoothly with the next.

v=

2E / m =

2qV / m





r O

Fourth Bohr-orbit

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32

ATOMIC PHYSICS (c) In other words the number of wavelengths must be an integer. (d) Condition for stable orbits : An election can circle around an atomic nucleus without radiation energy if the circumference of its orbit is an integral multiple of the electrons wavelength. i.e. 2r = n condition for stable orbits. 2r = n

h nh nh  pr =  mvr = p 2 2

[ p = mv]

nh 2 As mvr is the angular momentum of the circling electron. Bohr’s postulate is justified.

mvr =

C18: Calculate the de-Broglie wavelength associated with the motion of earth (mass = 6 × 1024kg) orbiting around the sun at a speed of 3 × 106 ms–1. Sol: 

6.63  1034 (Js) h = = (6  1024 )  (3  106 )(kg  ms1 ) mv  = 3.68 × 10–65 m

NOTE: The wavelengths associated with the motion of macroscopic objects like earth, train etc, are negligible compared to their sizes. This is why the wave-like character of these objects is not observable in our daily life. C19: Calculate the de-Broglie wavelength of an -particle of mass 6.576 × 10–27 kg and charge 3.2 × 10–19 coulomb, accelerated though 2000 V. Sol: E = kinetic – energy of a-particle = qV  E = 3.2 × 10–19 × 2000 J = 6.4 × 10–16 J de Broglie wavelength, h = , [ 2mE = momentum of photon] 2mE =

6.63  10 34 (J  s) 2  6.576  10

27

 6.4  10

16

(J  kg)

= 2.28 × 10–13 m

NOTE: The wavelength associated with -particles is of the order of size of the -particle. That is why the wave like character of a-particle is observable. C20: A particle of mass m and charge q is accelerated through a potential difference V. Find (a) its kinetic energy (b) momentum, and (c) de-Broglie wavelength associated with its motion. Sol: When the particle is accelerated through a potential difference V, gain in kinetic energy is given by K = qV. (a) Thus, kinetic energy, K = qV. p2 (b) Momentum of the particle (p) is given by K = 2m p = 2mK = 2m qv 

Momentum = 2m qv

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33

ATOMIC PHYSICS (c)

De- Broglie wavelength is given by, =

h = p

h 2m qV

Thus, wavelength =

h 2m qV

Example 23. Assume that the de-Broglie was associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the type described above can form. Sol. From the figure it is clear that 2Å p . (/2) = 2 Å (p + 1) . /2 = 2.5 Å N N  /2 = (2.5 – 2.0) Å = 0.5 Å p-loops /2 or = 1 Å = 10–10 m (i) de Broglie wavelength is given by (p + 1) loops 2.5 Å

h h   p 2 km K = kinetic energy of electron

h2 (6.63  1034 )2 K   2.415  1017 J 2 31 10 2 2m 2(9.1  10 )(10 )  2.415  1017  K eV  1.6  1019   (ii) N 

K = 150.8 eV N The least value of d will be when only one loop is formed dmin = /2

or

dmin = 0.5 Å

C21: Find the de-broglie wavelength associated with an electron accelerated through a potential difference of 30 kV. Sol: Kinetic energy of the electron K = qV = e(30 kV) = 30 keV = 30 × 103 × 1.6 × 10–19 J = 4.8 × 10–15 J Now, momentum of the electron =  De-broglie wavelength,  =

2mK = 2  9.1 1034  4.8  1015 kg.J = 2.96 × 10–24 kg.m/s

h Momentum

6.63  1034 J  s = 2.96  1024 Kg m / s = 2.24 × 10–10 m = 2.24 A

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34

ATOMIC PHYSICS

ATOMIC STRUCTURE ATOMIC MODEL By now, it is well-known that matter, electricity and radiation etc. are all atomic in character. Although, no one has so far seen individual atoms, there is no doubt that they really exist. In 1895, it was discovered by J.Perrin in Paris that the cathode rays consist of negatively-charged practices called electrons. In 1897, J.J. Thomson measured the e/m ratio for an electron whereas its charge was measured by Millikan in 1906 by his famous oil-drop experiment. Mass of the electron was found by dividing charge e by the ratio e/m. Discovery of positive rays during the latter part of 19th century indicated that a normal atom consisted of both negative and positive charges. But how these charges are distributed in an atom was not known at that time. Consinuous efforts have been made since then to study the physical structure of an atom such as its extra-nuclear electronic structure chiefly with the help of spectral properties of atoms. To account for the spectroscopic data obtained experimentally over the years, several theories regarding atomic structure have been proposed from time to time which are called the atomic models. Various atomic models proposed by scientists over the last few decades are: (i) Thomson’s Plum pudding model, (ii) Rutherford’s Nuclear model, (iii) Bohr’s model (iv) Sommerfeld’s Relativistic model (v) Vector model and finally (vi) Wave-mechanical model. These different models have been suggested one after the other in an effort to get a satisfactory interpretation of the experimental data which, it is hoped, will ultimately lead to a perfect and complete understanding of the physical structure of an atom. + + + + +

+ + +

+

+ + + +

+ +

++ +

+ + + + + + + +

+ + + +

+

+

+ + + + + +

+

+ +

++ + ++

+

+ +

+

+ + + +

+

+ +

+ +

+ + +

+

+

+ + + + +

+ + + + + Thomson’s Plum Pudding Model + + +– + + + + + According to this model, the atom is regarded as a heavy sphere of positive + + + + – + – charge seasoned with enough electron ‘plums’ to make it electrically neutral. + + + + Thomson visualized the positive charge of an atom being spread out uniformly – – throughout a sphere of about 10–10 metre radius with electrons as smaller – particles distributed in circular shells as shown in figure. Whereas the net force exerted by the positively-charged sphere on each electron is towards the centre of the sphere, the different electrons experience mutual repulsion and get arranged in the form of circular shells. This atomic model was given up after some time because it could not provide any satisfactory mechanism for explaining the large deflection suffered by -particles in Rutherford’s experiment.

+ + +

Rutherford’s Experiment on -particle Scattering As shown in figure high-speed a-particle (i.e. helium nuclei each with a charge of +2e) from some radioactive material like radium or radon, confined to a narrow beam by a hole in a lead block, were made to strike a very thin gold foil G. While most of the -particles went straight through the foil as if there were nothing there (and produced scintillations on a fluorescent screen), some of them ‘collided’ with the atoms of the foil and were scattered around at various angles -a few being turned back towards the source itself.  

Lead block

 15 0º

45º 60º

 



Beam of  -particles

Radium

+

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ATOMIC PHYSICS Although some small angle scattering could be expected from Thomson’s model, large-angle scattering was absolutely not expected at all. Further detailed experiments in Rutherford’s laboratory by Geiger and Marsden showed conslusively that large-angle scattering can be expected only if one assumes that a massive positive point charge exists at the centre of each gold atom as shown in figure. According to this model proposed by Rutherford in 1911, the positive massive part of an atom is assumed to be concentrated in a very small volume at its centre. This central core, now called nucleus, is surrounded by a cloud of electrons which makes the entire atom electrically neutral. The large-angle scattering of positively-charged -particles Trajectory P could be easily explained on this atomic model as shown in of -particle  Nucleus   figure. This scattering is due to the mutual repulsion (as per Coulomb’s law) between the -particles and the concentrated +Ze positive charge on the nucleus. The -particle approaches the positively-charged nucleus along AO. If there were no Asymptote of trajectory repulsion from the nucleus, it would have passed at a distance of p from it. However, due to coulombic force of repulsion, the -particle follows a hyperbola with nucleus as its focus. The lines AO and AO are the asymptotes of the hyperbola and present approximately the initial and final directions of the particle when it has passed out of the effective range of the nuclear electric field. As seen, the -particle is deflected through an angle . The perpendicular distance from the nucleus to the line AO is called the impact parameter and is denoted by p. The Rutherford’s scattering formula is  Q1Q2 = 2pE 2 Charge of the incoming -particle Charge of the scattering nucleus Kinetic energy of the incident -particle Impact parameter

tan

where

Q1 = Q2 = E = p =

Distance of Closest Approach Suppose that an -particle approaches a positively-charged nucleus for a head on collision with a kinetic energy of K. As shown in figure at point A, the repulsive force of the nucleus is so strong as to stop the particle momentarily. At this point, all the kinetic energy of the -particle is converted into potential energy. Let D be the distance of closest approach of the -particle. The potential at point A due to nuclear charge Ze is Ze = 4  D 0



Potential energy of the -particle when at point A is

A

2

Ze.2e 2Ze = 4  D = 4 0 D 0



+

D

2Ze 2 2Ze 2 K= or D = 4 0 D 4 0 K

Since -particles are generally obtained from natural radioactive substances, their kinetic energy K is known. Hence, value of D can be found easily from the above equation.

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36

ATOMIC PHYSICS C22: In Geiger-Marsden experiment on -particle scattering from gold foil, the kinetic energy of -particles used was 7.68 MeV. Calculate the distance of closest approach of -particle if atomic number of gold in 79. 2Ze 2 D= 4 0 K

Sol: Here,

Z = 79, e = 1.6 × 10–19 C; 0 = 8.854 × 10–12 F/m –13 K = 7.68 MeV = 7.68 × 1.6 × 10 J

2  79  (1.6 10 –19 )2 4 8.854 1012  7.68 1.6 1013 = 2.96 × 10–14 m

D=

Major Deficiencies in Rutherford’s Nuclear Model It was found later on that Rutherford’s model had two serious drawbacks concerning (i) distribution of electrons outside the nucleus and (ii) the stability of the atom as a whole. It can be shown that electrostatic forces between the positive nucleus and the static negative extra-nuclear electrons are not enough to produce equilibrium in such a nuclear atom. For example, consider the case of an atom* having two electrons and a nucleus with a charge of +2e. Suppose the electrons are symmetrically placed at a distance of r from the nucleus and are stationary. The force of attraction between the nucleus and each of the electrons is F = e.

2e 2e 2 = while the force of repulsion between the two electrons is 4   r 2 4  r 2

e2 e2 = . Since the force of attraction is eight times the force of repulsion, the electrons will 4 4r 2 16  r 2 fall into the nucleus thereby destroying the stable structure of the atom. – + 2e +

+ +

+ 2e +

– v

m v r e– + +

2e +

–

Photon

(hv)

– (b)

(a)

(c)

To overcome this difficulty, Rutherford suggested that stability can be achieved (as in a solar system) by assuming that electrons, instead of being static, revolve round the nucleus with such a speed that the centrifugal force balances the attractive force exerted by the nucleus on the electrons. As seen from figure (b), condition for stability is achieved when 2e.e 2e 2 mv 2 2 = 4  r 2 or mv r = 4 0 r 0 In general, if Z is the atomic number, then nuclear charge is Ze, so that the above relation becomes Ze.e mv 2 = 4  r 2 r 0

Ze 2 or mv r = 4 0 2

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ATOMIC PHYSICS Incidentally, it may be noted that according to the above relation, it is possible to have an infinite number of orbits in which electrons can rotate. But this assumption of revolving electrons led to serious difficulty from the point of view o the electromagnetic theory according to which an accelerated charge must continuously emit electromagnetic radiation or energy. Since an electron revolving in a circular orbit has centripetal acceleration (= v2/r), it must radiate energy as per the laws of classical electrodynamics. Due to this continuous loss of energy, the electrons will gradually approach the nucleus by a spiral path and finally fall into it as shown in figure (c). Hence, it is seen that the orbital motion of the electron destroys the very purpose for which it was postulated i.e. the stability of the atom. Obviously, either Rutherford’s nuclear atomic model with revolving electrons is defective or the classical electromagnetic theory fails in this particular case. This dilemma was solved in 1913 by Neils Bohr who proposed an improved version of Rutherford’s atomic model. Bohr’s Atomic model This model (first proposed for hydrogen atom but later applied to other atoms as well) retains the two essential features of Rutherford’s planetary model i.e. (i) the atom has a massive positively-charged nucleus and (ii) the electrons revolve round their nucleus in circular orbits the centrifugal force being balanced, as before, by the electrostatic pull between the nucleus and electrons. However, he extended this model further by utilizing Planck’s Quantum Theory. He made the following three assumptions: (iii) an electron cannot revolve round the nucleus in any arbitrary orbit but in just certain definite and discrete orbits. Only those orbits are possible (or permitted) for which the orbital angular momentum (i.e. moment of h nh i.e. orbital angular momentum = where 2 2 n is an integer and h is Planck’s constant. Such orbits are also known as stationary orbits. (iv) while revolving in these permitted stationary (or stable) orbits, the electron does not radiate out any electromagnetic energy. In other words, the permissible orbits are non-radiating paths of the electron. (v) the atom radiates out energy only when an electron jumps from one orbit to another. I f E2 and E1 are the energies corresponding to two orbits before and after the jump, the frequency of the emitted photon is given by the relation E2 – E1 = hv or E = hv where v is the frequency of the emitted radiations.

momentum) of the electron is equal to an integral multiple of

C 23: If I is the moment of inertia of an electron and  its angular velocity, then as per assumption (iii) given above I =

nh 2

or

(mr2) =

nh 2

n.h n.h (mr 2 )v = or mvr = 2 2 r Alternatively, since the momentum of the revolving electron is mv, its moment about the nucleus is = mvr

or

Hence

n.h mvr = 2

.....(i)

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v e– +ze r

m

38

ATOMIC PHYSICS where, n = 1, 2, 3 for the first second and third orbits respectively. It is called the principal quantum number and because it can take whole number values only, it fixes the sizes of the allowed orbits (also called Bohr’s circular orbits). E3 E2

r1 r2

E1

E3 E2

E

E1

r3

+Ze

+Ze

n=1 n=2 n=3

Permitted orbits (a)

K L M

Electron Jump (b)

Let the different permitted orbit have energies of E1, E2, E3 etc. as shown in figure (a). The electron can be raised from n = 1 orbit to any other higher orbit if it is given proper amount of energy. When it drops back to n = 1 orbit after a short interval of time, it gives out the energy difference E in the form of a radiation as shown in figure (b). The relation between the energy released and frequency of the emitted radiation is E2 – E1 = hv or E = hv Expressions for velocity, radius, energy of electron and orbital frequency in Bohr’s orbit Here it should be kept in mind that Bohr’s model is valid only for hydrogen atom and hydrogen-like ions. In other words, we can say that is applicable to hydrogen atom and ions having just one electron. Examples of such ions are He+, Li++, Be+++ etc. According to Bohr’s 2

First postulate,

Ze mv 2 = 4 0 r r

1 where 4  = 9 × 109 Nm2 C–2 0

and



 0 = 8.854 × 10–12 C2/Nm2 = Absolute permittivity of vacuum or free space

Ze2 r= 4 0 mv 2

ze + r

2

–e, m

v

v = velocity of electron m Z r e n

= = = = =

mass of an electron atomic number radius of the orbit magnitude of charge on an electron principal quantum number i.e. orbit number

.....(i)

From fourth postulate of Bohr, we have nh .....(ii) 2 where, n = 1, 2, 3, 4, ........ i.e., n is a positive integer. From equations (i) and (ii), we get

mvr =

Ze2 ze 2 nh mv × =  v = 4 0 mv 2 2 0 nh 2 Now, let us substitute the value of v in equation (ii),

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ATOMIC PHYSICS ze 2  r nh m× = 2 0 nh 2 Note that for fixed n,

0 n 2 h 2 r mZe 2

v  Z,

r 

1 Z

1 , r  n2 n If V1 is the speed of the electron in the 1st orbit

v

For fixed z,

Z c , where V1 = , where c is speed of light. n 137 If a0 = first Bohr radius = 0.53 Å, then

Then, Vn = V1

rn = a0

n2 Z

mZ2 e 4 1 kZe 2 2 K.E. of the electron = mv = =K= , 8 20 h 2 n 2 2 2r Potential energy of the atom = –

Ze 2 kZe 2 =U=– 4 0 r r

[k = 1/40]

mZ2 e 4 U=– 4 02 h 2 n 2  Total energy of the atom, E = K + U E=–

mZ2 e 4 8 20 h 2 n 2

In general, En = –13.6

Z2 in eV.. n2

Orbital frequency for the electron, Ze 2  mZe 2 mZ2 e 4 V v= =  v= 2 2 0 hn0 h 2 n 2 4 02 n 3 h 3 2r Time period of revolution (T) is given by T = T1

n3 Z2

where T1 = time period of revolution in the 1st orbit = 1.52 × 10–16 s =

2a 0 v1

NOTE : It is assumed that the acceleration of the nucleus is negligible on account of its large mass. Some important results for H-atoms when n = 1 1. Bohr radius, a0 = 0.53 Å rn = a0n2/Z 2. v1 = 2.18 × 106 ms–1  c/137 3.

E1 = –13.6 eV =

U1 2

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40

ATOMIC PHYSICS U1 2

4.

K 1 = 13.6 eV = –

5. 6. 7. 8.

U1 = –27.2 eV v1 = 6.6 × 1015 Hz 1 rydberg = –13.6 eV B1 = 12.5 tesla, where B1 is the magnetic field at the centre of Bohr atom due to the current generated by the motion of electron in 1st orbit.

Ground state and excited states The state of an atom with the lowest energy is called its ground state or normal state. For ground state, n = 1. The states with higher energies are called excited states. For the first excited state, n = 2; for the 2nd excited state, n = 3 and so on. For mth excited state, n = m + 1 Ionization energy and ionization potential The minimum energy needed to ionize an atom is called ionization energy. The potential difference through which an electron should be accelerated to acquire the value of ionization energy is called ionization potential. The value of ionization energy of H-atom in ground state is 13.6 eV that of ionization potential is 13.6 eV. Binding energy Binding energy of a system is the energy needed to separate its constituents to large distances or it may be defined as the energy released when its constituents are brought from infinity to form the system. The value of binding energy of H-atom is 13.6 eV, identical to its ionization energy. Excitation energy and excitation potential The energy needed to take the atom from its ground state to an excited state is called the excitation energy of that excited state. The potential difference through which an electron should be accelerated to acquire the value of excitation energy is called excitation potential. NOTE : (A) I.E. = E – E1 = –E1 = Binding energy of the H-atom. (B) Movement of electron in circular orbits in a Bohr atom causes electric current in the orbit. This current will lead to self-generated magnetic field in the atom and also magnetic current (m) (a) Magnetic field(B) If B is the magnetic field generated at the centre of atom, then –e  0i B = 2a , where i is the current due to motion of the electron 0

and a0 is the 1st Bohr-radius. Hence, i = ev

a0 +e H-atom

i

 0 e hence B = 2a 0

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ATOMIC PHYSICS Putting the values of v and a0, we get B = 12.5 Tesla  Magnetic movement vector (  )

(b)





 = iA



 = iA = ev a 20 

(c)



Relation between  & L (angular momentum vector) µ = iA = ev a 20 ,

[i = ev, A =  a 20 ]

L = mV1a0 = m2v a 20

[V1 = 2va0]

ea 20  e  e  =  = 2 = m2a 0 L 2m L 2m Vectorially, e  L 2m



 =

Example 24: The quantum number of Bohr orbit in H-atom whose radius is 0.01 mm is (a) 223 (b) 435 (c) 891 (d) none of these Sol: (b) We know that rn = a0n2 rn n2 = a 0

n =

0.01 10 3 0.529  1010



n = 435

Example 25: The quantum number n in the Bohr’s model of H-atom specifies: (a) radius of the electron (b) energy of the electron (c) angular momentum of the electron (d) all of these Sol: (d)

 0 h 2  rn =  me 2  × n2   En = –

me 4 , 8 20 h 2 n 2

 h  Ln = n    2 

Example 26: The radius Bohr’s orbit in ground state for H-atom is (Take 0 = 8.86 × 10–12 C2/Nm2, h = 6.6 × 10–34 J-s) (a) 0.528 Å (b) 0.0528 Å (c) 5.28 × 10–10 m Sol: (a)



a0 =

0 h 2 me 2

(d) 5.28 × 10–10 cm

(in this case, Z = 1, n = 1)

(6.6 1034 )2  (8.86 10 12 ) a0 = 3.14  (9.11031 )  (1.6 1019 )2

= 0.528 Å

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ATOMIC PHYSICS Example 27: The speed of the electron in the first Bohr orbit of H-atom is (Take c = speed of light in vacuum) c c (a) c (b) (c) (d) 137 c 13.6 137 Sol: (c) For H-atom, we know that e2 v = [since n this case, Z = 1 and n = 1] 2 0 h e2 c = 2 0 hc e2 1 = = fine structure constant 2 0 hc 137

But 

v =

c 137

Example 28: In a H-atom, binding energy of the electron in the ground state is E1. Then the frequency of revolution of the electron in the nth orbit is 2E1 2mE1 2E1n 3 (b) (c) 3 nh h n 3h Sol: (b) The frequency of revolution of the electron in the nth orbit is given by

(a)

(d) none of these

mZ2 e 4 mZ2 e 4 2E1 v = = where E1 = 4 02 h 3 n 3 8 20 h 2 hn 3 C24: Calculate the energy of a He+ ion in its first excited state. Sol:

(13.6eV)Z2  En = – n2 Here, Z = 2, n = 2 13.6  2 2  En = eV = –13.6 eV n2

C25. An electron in a hydrogen like atom is in an excited state. It has a total energy of –3.4 eV. Calculate : (i) the kinetic energy (ii) the de-Broglie wavelength of the electron Sol. (i) Kinetic energy of electron in the orbits of hydrogen and hydrogen like atoms = | Total energy |  Kinetic energy = 3.4 eV (ii) The de Broglie wavelength is given by h h   (K = kinetic energy of electron) P 2 Km Substituting the values, we have (6.6  1034 J  s)  2(3.4  1.6  1019 J)(9.1  1031 kg)  = 6.63 × 10–10m or  = 6.63 Å

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ATOMIC PHYSICS Example 30. The electron in a hydrogen atom makes transition from M shell to L. The ratio of magnitudes of initial to final centripetal acceleration of the electron is (A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81 2 Sol.  rn × n and

vn ×

1 n

an 

v2n rn

1 n  n2 1 an  4 n an 

an 

2

K n4

a M  a3 

K 81

{ M = 3}

aL  a2 

K 16

{ L = 2}

a M 16  a L 81

Hence (D) is correct.

Example 31. The angular momentum opf an electron in the hydrogen atom is The kinetic energy of this electron is : (A) 4.53 eV (B) 1.51 eV Sol.

(C) 3.4 eV

3h . Here h is Planck’s constant. 2

(D) 6.8 eV

nh 3h  2 2   n=3 In the electronic third orbit, the energy of electron L

E  13.6

Z2 n2

E  13.6  E

1 9

13.6  1.51 eV 9

Hence (B) is correct.

Example 32. A particle of mass m moves along a circular orbit in a centrosymmetrical potential field kr 2 U(r) = . Using the Bohr’s quantization condition, find the permissible orbital radii and energy levels of 2 that particle.

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ATOMIC PHYSICS Sol.

dU = – kr dr Negative sign implies that force is acting towards centre. The necessary centripetal force to the particle is being provided by this force F. Hence

F=–

mv 2 = kr r

...(1)

h    where h   2  solving equations (1) and (2) we get

and

mvr = n h

r = rn =

nh m

where  =

...(2)

k m

1 kr 2 and total energy E = U + K = + mv2 2 2 Substituting the values, we get E = n h  = En

Example 33. In a hypothetical system a particle of mass m and charge –3q is moving around a very heavy particle having charge q. Assuming Bohr’s model to be true to this system, the orbital velocity of mass m when it is nearest to heavy particle is 3q 2 (A) 2 0 h

3q 2 (B) 4 0 h

Fe 

Sol.

3q (C) 2 h 0

3q (D) 4 h 0

mv 2 r

q  3q mv2  4 0 r 2 r 3 q2   mvr  v 4 0 h 2



mvr 



3 q2 h   v 4 0 2 3 q2 v 2 0 h

Hence (A) is correct.

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ATOMIC PHYSICS SPECTRUM Dispersed light arranging itself in a pattern of different wavelength is referred to as a spectrum. Light coming from a source may be dispersed by a prism or by any other dispersing medium. When white light falls on a prism and the transmitted light is collected on a white wall or white paper then a spectrum is obtained which consists of different colours from red to violet. Kinds of spectra : (A) Emission spectra: When a light beam emitted by certain source is dispersed to get the spectrum, it is called an emission spectrum.

(b) Line spectrum: The atoms and molecules can have certain fixed energies. An atom or molecule, in an excited state, can emit light to lower its energy. Light emitted in such a process has certain fixed wavelengths. When such a light is dispersed, certain sharp bright lines on a dark background is obtained. Such a spectrum is called line emission spectrum.

Energy

An emission spectrum can be three types : (a) Continuous spectrum: That emission spectrum which is obtained by continuously varying wavelength, is called continuous emission spectrum. In this case, when light is dispersed, a bright spectrum continuously distributed on a dark background is obtained. Light emitted from an electric bulb, a candle or a red hot iron piece comes under this category.

 Line spectrum Atomic energy levels

For example, when electric discharge is passed through sodium vapour, they vapour emits light of the wavelength 589.0 nm and 589.6 nm. When this light is dispersed by a high resolution grating , one obtains two bright yellow lines on a dark background.

Energy

(c) Band spectrum : The wavelengths emitted by the molecular energy levels which are generally grouped into several bunches, are also grouped; each group being well separated from the other. The spectrum looks like separate bands of varying colours. Such a spectrum is called band emission spectrum.

Molecular energy levels

Band spectrum



(B) Absorption spectrum : When white light is passed through an absorbing material, the material may absorb certain wavelengths selectively. When the transmitted light is dispersed, dark lines or bands at the positions of the missing (absorbed) wavelengths are obtained. Such type of spectrum is called absorption spectrum.

White light Absorbing material Absorption spectrum

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ATOMIC PHYSICS An absorption spectrum may be of two types : (a) Line Absorption spectrum : Light may be absorbed by atoms to take them from lower energy states to higher energy states. In the similar way when white light is passed through a gas, the gas is found to absorb light of certain wavelength. The absorption spectrum consists of dark lines on bright background. Such a spectrum is called a line absorption spectrum. When light coming from the sun is dispersed, it shows certain sharply defined dark lines. This shows that certain wavelengths are absent. There missing lines are called Fraunhofer lines. (b) Band Absorption spectra : If absorbing media is polyatomic such as H2, CO2 or KMnO4 solution, instead of dark lines we get few characteristic dark bands (against coloured background) called band absorption spectra. Hydrogen Spectra : If hydrogen gas enclosed in a sealed tube is heated to high temperature, it emits radiation. This radiation consists of components of different wavelengths which deviate by different amounts. The radiation with different amounts of deviation forms H-spectrum. Explanation of hydrogen spectra by Bohr (a) The electron in a H-atom if not disturbed remains in the ground state (i.e. n = 1 state). When the electron receives energy from outside, it is elevated to any one of the higher permitted states (excited state). (b) The electron remains only for a short interval of time (generally in the order of 10–8 s) in the excited state and comes back to the ground state finally. (c) The electron can reach the ground state from any one of the excited states in many ways. As a result, many electron transitions take place. (d) According to Bohr, all electron transitions terminating at a particular state give rise to a particular spectral series. n= n=5 n=4 n=3

E=0

n=2

Energy Series limit

n =1 Lyman series Balmer Paschen Brackett series series series

1 1   2  2  n = 2, 3, 4,.......... 1 n 

(e) Lyman:

nf = 1:

1 E =– 1  ch

Balmer:

nf = 2:

1 E  1 1  = – 1  2  2  n = 3, 4, 5, ..........  ch  2 n 

Paschen: nf = 3:

1 E 1 1  = – 1  2  2  n = 4, 5, 6, ..........  ch  3 n 

Brackett: nf = 4:

1 E  1 1  = – 1  2  2  n = 5, 6, 7, .........  ch  4 n 

Pfund:

nf = 5:

1 E1  1 1  =–  2  2  n = 6, 7 , 8,......  ch  5 n 

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ATOMIC PHYSICS (f)

If an electron makes a jump from the nith to nfth orbit (ni > nf), the extra energy Ei – Ef is emitted as a photon of electromagnetic radiation. The corresponding wavelength is given by 1 Ei  Ef = where c = speed of light in vacuum.  hc According to Bohr, we can write

mZ2e4  1 1  1 = 2 3  2  2  8 0 ch  n f n i  1 = RZ2 

1 1  2  2  nf ni 

me 4 where R = is called the Rydberg constant. 8 02 ch 3 (g) The value of R is 1.0973 × 107 m–1

NOTE: – En =

RhcZ2 , n2

1 rydberg = –13.6 eV,

Rhc = 13.6 eV

(i)

1 2  is called wave number (  ) of the line and is called angular wave number of the line.   Photon energy = Ep = hv

(j)

Photon Energy Momentum of a photon = p = speed of light

(h)



p=

Ep c

Important points regarding H-spectra (a) The sharply defined, discrete wavelengths exist in the emitted radiation in the H-spectrum. (b) A hydrogen sample emits radiation with wavelengths less than those in the visible range (i.e. uv light) and also with wavelengths more than those in the visible range (i.e. infrared). (c) the lines may be grouped in separate series. (d) In each series, the separation between the consecutive wavelengths decreases as we move from higher wavelength to lower wavelength. (e) A particular minimum wavelength in each series approach a limiting value known as series limit. (f) The series corresponding to uv region, visible region and infrared region are known as Lyman, Balmer and Paschen series respectively. (g) In the Balmer series of hydrogen, the H line (3  2) is red, the H line (4  2) is blue, the H ( 5  2) and H(6  2) lines are violet, and the other lines are in the near ultraviolet (uv). n(n  1) . 2 Theoretically possible no. of absorption spectral lines = (n – 1)

(h) Theoretically possible no. of emission spectral lines = (i)

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ATOMIC PHYSICS (j)

Approximate range of wavelength for different colours of visible light Colour Wavelength Range Violet + Indigo 3800 Å to 4500 Å Blue 4500 Å to 5000 Å Green 5000 Å to 5500 Å Yellow 5500 Å to 6000 Å Orange 6000 Å to 6500 Å Red 6500 Å to 7200 Å Infrared rays: 720 nm to 50 mm Ultraviolet light: 10 Å to 3800 Å

Limitations of Bohr’s Model (a) It is valid only for one electron atom and hydrogen-like ions e.g. : H, He+, Li+2, Na+10 etc. (b) Orbits were taken as circular but according to SOMMER field these are elliptical. (c) (d) (e) (f)

Intensity of spectral lines could not be explained. Nucleus was taken as stationary but it also rotates on its own axis. It could not be explained the minute structure in spectrum line. This does not explain the ZEEMAN effect (splitting up of spectral lines in magnetic field) & Stark effect (splitting up in electric field). (g) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890 Å to 5896 Å). C26.

Total number of emission lines from some excited state n1 to another energy state n2(< n1) is given by

(n1  n 2 )(n1  n 2  1) n(n  1) . For example total number of lines from n1 = n to n2 = 1 are . 2 2 C27. As the principal quantum number n is increased in hydrogen and hydrogen like atoms, some quantities are decreased and some are increased. The table given below shows which quantities are increased and which are decreased. Table Increased Decreased Radius Speed Potential energy Kinetic energy Total energy Angular speed Time period Angular momentum

1  Whenever the force obeys inverse square law  F 2  , and potential energy is inversely proportional r to r, kinetic energy (K), potential energy (U) and total energy (E) have the following relationships.

C28.

K=

|U| U and E = –K = . 2 2

If force is not proportional to

1 1 , the above relations do not 2 or potential energy is not proportional to r r

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ATOMIC PHYSICS hold good. In JEE problems, this situation arises at two places, in an atom (between nucleus and electron) and in solar system (between sun and planet). C29. Total energy of a closed system is always negative and the modulus of this is the binding energy of the system. For instance, suppose a system has a total energy of –100 J. It means that this system will separate if 100 J of energy is supplied to this. Hence, binding energy of this system is 100 J. Thus, total energy of an open system is either zero or greater than zero. C30. Kinetic energy of a particle can’t be negative, while the potential energy can be zero, positive or negative. It basically depends on the reference point where we have taken it zero. It is customary to take zero potential energy when the electron is at infinite distance from the nucleus. In some problem suppose we take zero potential energy in first orbit (U1 = 0), then the modulus of actual potential energy in first orbit (when reference point was at infinity) is added in U and E in all energy states, while K remains unchanged. Example 33: The value of series limit in Lyman series is: (a) 121.6 nm (b) 91.2 nm Sol:

1  min

 1 = R 1    

1  min

= R[1 – 0]

min

=

(c) 656.3 nm

(d) 365.0 nm

1 = 91.2 nm 1.097  107

C31: Find the longest wavelength present in the Balmer series of hydrogen: Sol: In the Balmer series, nf = 2. The longest wavelength in this series corresponds to the smallest energy difference between energy levels. Hence the initial state must be ni = 3. 

 1 1   1 1 1 1 1 = R  n2  n2  = R  2  2  = R     2 3  4 9  f i  1 5R =  36 36 = 6.56 × 10–7 m 5  1.097  107  = 656 nm (near the red end of the visible spectrum)

 =

C32: A hydrogen atom emits uv radiation of 102.5 nm. Calculate the quantum numbers of the states involved in the transition. Sol: The uv radiation 102.5 nm lies in the lyman region of spectrum. Thus nf = 1  1  1  = R 1  n 2    i  

1 1 2 = 1 – ni R



1 1 1 = 1 – 2 = 1 –  9 7 ni 102.5  10  1.097  10 1.124

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ATOMIC PHYSICS 1 n i2 = 0.1 n i2 = 10 ni  3 Hence transition is from 3  1.

C33: What is the unit of reciprocal of Rydberg constant in S.I. units ?

 1 1  1 = R  n 2  n 2  the term in the bracket is unitless.   f i  Now, we can write

Sol: We know that

1 =R 

Hence, the unit of



1 = R

1 will be metre i.e. m. R

C34: How many different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n ? Sol: The total number of possible transitions is n(n  1) (n – 1) + (n – 2) + (n – 3) + ............. + 2 + 1 = 2 C35: Consider the following two statements: (A) Line spectra contain information about atoms only (B) Band spectra contain information about molecules (a) Both A and B are wrong (b) A is correct but B is wrong (c) B is correct but A is wrong (d) Both A and B are correct Sol: (c) Line spectra contain information about atoms and molecules both. C36. Ultraviolet light of wavelength 800 Å and 700 Å when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and 4.0 eV respectively. Find the value of the Planck constant. Sol. hv = E0 + T where E0 = ground level energy and T = kinetic energy of electron hc  E0  T  hc  = E0 + 1.8 × 1.6 × 10–19 800  1010 hc and = E0 + 4.0 × 1.6 × 10–19 700  1010 hc  1 1  Subtracting    = 2.2 × 1.6 × 10–19 108  7 8  2.2  1.6  10 27  56 or h= = 6.57 × 10–34 Js 3  108

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ATOMIC PHYSICS C37: The excitation energy of a hydrogen like ion in its first excited state is 40.8 eV. Find the energy needed to remove the electron from the ion. Sol: The excitation energy in the first excited state is 1 1  E = 13.6Z2  2  2  1 2  3 40.8 = (13.6 eV) × Z2 × 4  Z=2

13.6Z2 Now, ionization energy = = –4 × (13.6 eV) 12 Eion = –54.4 eV C38: The first ionization potential of some hydrogen like Bohr atom is x V. Then the value of the first excitation potential for this atom will be: x 3 (a) xV (b) V (c) xV (d) 20 xV 2 4 Sol: The value of first excitation potential is given by  1 x = x  1   V  4 3x x = V 4 Example 34: A doubly ionised lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third Bohr orbit. (Take ionization energy of H-atom equal 13.6 eV). E3 n=3 Sol: For E1, Z = 3, n = 1 E n=2 2

E

2

13.6  9 13.6Z =– = 122.4 eV 2 1 n For E3, Z = 3, n = 3 13.6  32  E3 = – = –13.6eV 32 We have E + E1 = E3  E = E3 – E1 = 108.8 eV

 E1 = –



=

E1

n=1

hc 12400 = Å = 114 Å E 108.8

Example 35: In Bohr’s model of hydrogen atom when the electron is moving in one of the stationary orbits then: –e (a) velocity of the electron is fixed and no emission of energy takes place v2 v1 (b) velocity changes continuously but no emission of energy takes place (c) energy is emitted but the velocity does not change (d) energy is emitted and the velocity also changes. –e –e  Sol: (b) According to the second postulate of Bohr.     | v1| = | v 2 | = | v3| = | v 4 | = v v3 –e

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ATOMIC PHYSICS Example 36. An electron in an unexcited hydrogen atom acquired an energy of 12.1 eV. To what energy level did it jump? How many spectral lines may be emitted in the course of transition to lower energy levels? Calculate the shortest wavelength.

Z2 Sol. E = – E0 2 where E0 = 2.18 × 10–18 J = 1 rydberg n E (energy gap between unexcited state (n = 1) and an excited state (n = m)) 1  1  E 0 Z2  2  2  1 m 



n=3

12.1 × 1.6 × 10–19 = 2.18 × 10–18 × 12(1 – 1/m2)

n=2

1 n=1 or m=3 2 = 0.888 m Three lines are emitted. The shortest wavelength corresponds to the greatest energy gap.

or

1–



hc 1 1 = E – E = E  2  2  3 1 0  min 1 3

or

min =

6.6  1034  3  108  9 = 1.02 × 10–7 m = 1020 Å 2.18  1018  8

Example 37. A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 and 17.0 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z. (lonization energy of H atom = 13.6 eV) Sol. From the given conditions En – E2 = (10.2 + 17) eV = 27.2 eV ...(1) and En – E3 = (4.25 + 5.95) eV = 10.2 eV ...(1) Equation (1) and (2) gives E3 – E2 = 17.0 eV 2 or Z (13.6)(1/4 – 1/9) = 17.0  Z2 (13.6) (5/36) = 17.0  Z2 = 9  Z=3 From equation (1) Z2 (13.6) (1.4 – 1/n2) = 27.2 or (3)2(13.6) (1/4 – 1/n2) = 27.2 or 1/4 – 1/n2 = 0.222 or 1/n2 = 0.0278 or n2 = 36  n=6

ATOMIC EXCITATION WITH THE HELP OF COLLISION (a) An atom can be excited to an energy above its ground state by a collision with another particle in which part of their joint kinetic energy is absorbed by the atom. (b) An excited atom returns to its ground state in an average of 10–8 s by emitting one or more photons.

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ATOMIC PHYSICS (c) Energy transfer is a maximum when the colliding particles have the same mass. (d) The energy used in this process will be of discrete nature. (E = E2 – E1 = hv =

hc ). 

NOTE: If the joint kinetic energy of colliding particles is less than 20.4 eV (considering particles as Hatoms or one neutron and one H-(atom) then the nature of collision will be necessarily elastic. Example 38: A neutron moving with speed v makes a head-on collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10–27 kg. Sol: Let us suppose that neutron and H-atom move at speeds v1 and v2 after the collision. Suppose an energy E is used in this way. On the basis of conservation of linear momentum and energy, we can write mv = mv1 + mv2 .....(i) 1 1 1 2 2 mv2 = mv1 + mv 2 + E 2 2 2 From equation (i) we have

v2 = v12 + v 22 + 2v1 v2 Now, from equation (ii) 2E m  From equation (iii) and (iv)

v2 = v12 + v 22 +

.....(ii)

.....(iii)

.....(iv)

2E m 2 (v1 – v2) = (v1 + v2)2 – 4v1v2

2v1v2 = Hence,

= v2 –

4E m

Since v1 – v2 is always real, v2 –

4E 0 m

mv2  4E 1 mv2  2E 2 The minimum kinetic energy of the neutron needed for an inelastic collision corresponds to transition from n = 1 to n = 2



 

1 mv2min = 2 × 10.2 eV 2 Kmin = 20.4 eV

Kmin =

Example 39. Consider an excited hydrogen atom in state n moving with a velocity v (v