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edit Vol. XXIII

No. 4

April 2015

t was a very highly inspiring advice given to the teachers and students by the Hon’ble prime Minister, Shri Narendra Modi ji himself in a recent radio-talk.

plot 99, Sector 44 institutional area, Gurgaon -122 003 (Hr). Tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in

Many books have been written on education by very learned authors. These are addressed mainly to higher level students and professors. But the difficulties

Regd. Office 406, Taj apartment, Near Safdarjung Hospital, ring road, New Delhi - 110029.

: :

how to Excel in any competitive Exams, at any Level

i

Corporate Office :

Managing Editor Editor

of the students are not spelt out. The prime Minister’s advice reaches directly the students of all levels.

Mahabir Singh anil ahlawat (BE, MBa)

The first is the fear of exams which all of us have felt. To combat this, the advice is to hold a week-long examination festival, two times a year, with

contents Physics Musing (Problem Set-21)

8

satirical poems on exams, cartoon contents and lectures on the psychological effects of exams with debates to pepper the lectures. Comparisons with others is wrong. Competition should be with yourself.

Thought Provoking Problems

11

Compete with yourself,

BITSAT Full Length

14

Compete for speed,

Practice Paper 2015 JEE Advanced

Compete to do more,

31

Brain Map

46

AIPMT

48

Practice Paper 2015 57

Practice Paper 2015 CBSE Board Class XII

Compete to achieve newer heights, Focus on doing better every-time.

Practice Paper 2015

AIIMS

rial

66

The former Ukrainian pole vault champion Sergey Bubka had broken his own record 35 times! perhaps you might have seen the film ‘Bhaag Milkha Bhaag’. Milkha Singh, the famous champion runner was breaking his own record every time. Even after winning, he runs round the track at the same speed waving to the crowds! The final advice given by the prime Minister is valid not only to the students but also to everyone working in any field. “live in the present, struggle with the present. Victory will walk alongside”, Go ahead, students! Victory is yours.

Solved Paper 2015

Anil Ahlawat

Core Concept

80

Physics Musing (Solution Set-20)

83

You Ask We Answer

84

Crossword

85

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physics for you | april ‘15

7

PHYSICS

P

MUSING

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their solutions. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams. By : Akhil Tewari

21 1. A uniform rod of mass m and length l starts rotating with constant angular acceleration a in a horizontal plane about a fixed vertical axis passing through one end. The horizontal component of the net force exerted on the rod by the axis when it has rotated by an angle p/2, is l l (a) ma (b) ma 1 + p2 2 2 m l pa (c) (d) none of these 2 2. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring substends an angle q at the centre. The value of the magnetic induction at the centre due to the current in the ring is (a) proportional to 2(180° – q) (b) inversely proportional to r (c) zero, only if q = 180° (d) zero, for all values of q. 3. A particle is projected with a speed u in air at angle q with the horizontal. The particle explodes at the highest point of its path into two equal fragments, one of the fragments moving up straight with a speed u. The difference in time in which the two particles fall on the ground is (Assume it is at a height H at the time of explosion.) u 2 (a) 2u (b) u − 2 gH g g 8

(c)

1 2 u + 2 gH g

(d)

2 2 u + 2 gH g

4. Consider the cube shaped carriage ABCDEFGH of side l and a mass M and it can slide over two frictionless rails PQ and RS. A shot of mass m is thrown from corner A such that it lands at corner F. The angle of projection as seen from the carriage is 45°. While the shot is in the air, the velocity of carriage as seen from the ground is H D

A

G

l

C

E R

F B S Q

P

(a)

m gl 2 2(M + m)

(b)

m 2 gl 2(M + m)

(c)

m gl (M + m)

(d)

m 2 gl (M + m)

5. A particle of mass m kept at the origin is subjected  ^ to a force F = ( pt − qx) i where t is the time elapsed and x is the x co-ordinate of the position of the particle. Particle starts its motion at t = 0 with zero initial velocity. If p and q are positive constants, then

physics for you | april ‘15

Page 8

(a) the acceleration of the particle will continuously keep on increasing with time (b) particle will execute simple harmonic motion (c) the force on the particle will continuously keep on decreasing with time (d) the acceleration of particle will vary sinusoidally with time. 6. Two rods of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 2:3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of (a) 3 : 2 (b) 3 : 2 (c) 3 3 : 2 2 (d) 9 : 4

µ0 I (3p + 4)  8pR µ I (c) 0 (3p − 4)  8pR

µ0 I (3p + 4) ⊗ 8pR µ I (d) 0 (3p − 4) ⊗ 8pR

(a)

(b)

10. Two blocks of masses 2 kg and 4 kg are connected through a massless inextensible string. The co-efficient of friction between 2 kg block and ground is 0.4 and the coefficient of friction between 4 kg block and ground is 0.6. Two forces F1 = 10 N and F2 = 20 N are applied on the blocks as shown in the figure. Calculate the frictional force between 4 kg block and ground. (Assume initially the tension in the string was just zero before forces F1 and F2 were applied.)

7. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is

(a) 24 N (c) 18 N

(b) 8 N (d) 10 N 

solution of march 2015 crossword 1

(a)

2GM (4 2 − 5) 7R

GM (c) 4R

(b) − (d)

2GM (4 2 − 5) 7R

2GM ( 2 − 1) 5R

8. An ideal gas is expanded so that amount of heat given is equal to the decrease in internal energy. Find the adiabatic exponent if the gas undergoes the process TV1/5 = constant. (a) 7/5 (b) 6/5 (c) 8/5 (d) None of these 9. The magnetic field at the point P is given by

F L U I D I T Y 4 G E E 7 R S 8 Y M A 9 10 U L T I C E L I N E C T 11 12 E HA L L E F F E N X I 15 U C Y C L O N E 16 S C A I E T T 19 H Y P O T H E S I O N O 24 23 M O M E N B 26 R I P P L E 27 A E 29 C P D 31 I K O H 32 VO L C A P B L C O H 34 B L I N D S P O T 35 U R Y G I N R C F 36 L M E A N F R F P E E 38 R H E Y

2 T O D E S I C T A R A S O N I C

C T

L A R S T

N

C E F

14

F O L I A T 17

18

F A H R

L 20 L U A 22 T W O F I N E R T I T U G 30 R E D S H I F E T L O 33 E L E C T S A N D E S C E N 37 E E T I M E S F L U X S P E

3 C B I O M 6A S S H M E A S L I G 13 O A L M I O N N E N H E I T A 21 R I C A N T W O 25 M A A V P E 28 R E A P S O L S D A Y S O M A G N E T I A M C C E I I A S T I C T S Y D 5

B I

T

R N L E

Winner (March 2015) neha Gupta Solution Senders Sandeep Kumar rana Atriz roy

10

physics for you | april ‘15

Page 10

By : Prof. Rajinder Singh Randhawa*

I

II

III –Q

2. Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball (ball 2) is fixed at O and the other (ball 1) is free. The balls are charged identically as a resultant of which the spring length increases h = 2 times. Determine the change in frequency? O

2 +

+ + +

+

4. A non-conducting ring y of mass m and radius R, the charge per unit length l is shown in – figure. It is then placed – – on a rough non– – x conducting horizontal O  ^ plane. At time t = 0, a uniform electric field E = E0 i is switched on and the ring starts rolling without sliding. Find the frictional force acting on the ring. 5. A non-conducting solid cylindrical rod of length L and radius R has uniformly distributed charge Q. Find the electric field at point P, a distance L from the centre of the rod.

R

O

P

L

1 +

Q2 3. A point charge Q1 = –125 mC is fixed at the centre of an insulated h disc of mass 1 kg. The disc rests Q1 on a rough horizontal plane. Another charge Q2 = 125 m C is fixed vertically above the centre of the disc at a height h = 1 m. After the disc is displaced slightly in the horizontal direction, find the time period of oscillation of disc.

++

1. An infinitely long conducting wire of charge density +l and a point charge –Q are at a distance from each other. In which of the three regions (I, II or III) are there points that (a) lie on the line passing through point charge perpendicular to the conductor and (b) at which the field is zero?

Solution

1. In the region II, the electric field of wire and point charge point in the same direction, +ve x-axis. So no point can exist where the field is zero. Now, we take a point to the right of the point charge at a distance x from it. Resultant field at this point is  ER =

Q λ ^ ^ i+ (− i ) 2 2 πε0 (x + a) 4 πε0 x

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699 physics for you | april ‘15

11

Page 11

Resultant field is zero if λ Q = ( x + a) 2 x 2

Q Qa =0 or 2lx2 = Qx + Qa or x 2 − x− 2λ 2λ On solving the quadratic equation in x, we get Q Q 2 Qa ± + 4λ 16 λ2 2 λ Here, there is only one value of x (with +ve sign) because –ve sign would mean that the point is to the left of point charge. Now we take a point to the left of wire at a distance x from it. The resultant field is  Q λ ^ ^ ER = i (− i ) + 2 2 πε0 x 4 πε0 (a + x ) The two fields point in the opposite directions, so resultant field can be zero if, Q λ = 2 πε0 x 4 πε0 (a + x )2 x=

Q  or x 2 −  − 2a  x + a2 = 0  2λ  On solving the quadratic equation in x, we get 2

1 Q 1Q   x =  − 2a  ± − 2a  − a2   4  2 λ 2  2λ If the discriminant of the quadratic equation is real, we have two points where the field is zero. Discriminant is +ve for Q ≥ 8al. 2. When the balls are uncharged, the frequency of oscillation is 1 k 2π m where k is the force constant of the spring and m = mass of the oscillating ball (ball 1). When the balls are identically charged, q2 1 ...(i) = k(ηl − l ) = kl(η − 1) 4 πε0 (ηl )2 where l is the natural length of spring and hl is the new length of spring after its extension. q2 ... (ii) or l 3 = 4 πε0 η2 (η − 1)k υ0 =

When the ball 1 is displaced by a small distance x from the equilibrium position to the right, the unbalanced force to the right is given by

12

q2 1 − k ( ηl + x − l ) 4 πε0 (ηl + x )2 Using Newton’s law, Fr =

−2

x 1 q2  m 1 +  − kl(η − 1) − kx =  2 2 2 4 πε0 η l  ηl  dt Expanding binomially and using eqn. (i), we get   1 2q2 d2 x m = − ⋅ + k x dt 2   4 πε0 η3l 3 Using eqn. (ii), we get   1 2q2 ⋅ + k  d2 x = −  4 πε0 3 m q2 x 2 η dt   2 4 πε0 η (η − 1)k    3η − 2   2(η − 1)  k + k x = −  kx = −  η   η   3η − 2  k d2 x ...(iii) ∴ = − x 2  η  m dt d2 x

By definition of simple harmonic motion, d2 x 2

= −ω2 x

...(iv)

dt From eqns. (iii) and (iv), we get  3η − 2  k 1  3η − 2  k ω2 =  ⇒ υ=   η m 2 π  η  m ∴

υ 3η − 2 = υ0 η

3η − 2 times. η Here, h = 2 so frequency increases 2 times. Thus frequency is increased

3. Let the radius of the Q2 disc be R. If the disc F is displaced x, then h  q = x/R. The restoring Fsin torque t about the point x of contact of the disc with ground, tP = (Fsinq)R = Ia   MR2 = + MR2  α   2 Q2 where, F = 4πε0 (h2 + x 2 ) and sin θ =

Fcos Q1

P

x 2

h + x2

physics for you | april ‘15

Page 12

Hence,

or α =

R

2

Q x 2

6 πε0 MR(h + x ) Q x

For x < < h, α ≈ −

2

Q θR

=−

6 πε0 MRh 6 πε0 MRh3 Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other. Thus, α = −

3

Q2θ

6 πε0 Mh3

Hence, ω =

or T = 2 π

Q2 6 πε0 Mh3 6 πε0 Mh3 Q

h 6 πε0 Mh Q

N +

dF

–

– –

+ L /2   + L /2 (L − r )dr dr − ∫  ∫ 2 2 2 1/2  2 πR Lε0  − L /2 − L /2 [(L − r ) + R ]  The second integral can be evaluated by substituting (L – r)2 + R2 = t. Differentiating both sides, we get, –2(L – r)dr = dt  + L /2 1 dt  Q ∴ E= [r ]− L /2 + 2 ∫ 1/2  2 t  2 πR Lε0 

E=

+

+ + +

A force of same magnitude but in opposite direction acts on a corresponding element in the region of negative charge. \ Equation of motion for pure rolling is π /2

2 ∫ 2 λRdθE0 R sin θ − fR = (mR )α

and f = ma and a = Ra Solving eqns. (i), (ii) and (iii), we get f = lRE0 along +ve x-axis.

Q

=

dF = lRdqE0

0

 (Q / L)dr  (L − r ) 1−   πR2 (2ε0 )  [(L − r )2 + R2 ]1/2 

E=

f

mg

or 2 λR2 E0 − fR = mR2 α

L

 (Q / L)dr  (L − r ) 1−  2 2 2 1/2  − L /2 πR (2ε0 )  [(L − r ) + R ] 

4. Consider a differential element subtending an angle dq at the centre and at angle q as shown in figure.

– –

+L/2

+ L /2

T = 0.6 s



r dr

E = ∫ dE = ∫

On substituting the given values, we get

d

P

Consider a disc of radius R of thickness dr at a distance r from the centre O of the cylinder. Charge on the disc, Q Q dq = × πR2dr = dr 2 L πR L ∵ Electric field due to disc along its axis  x σ  Ex = 1 − 2  2ε0  (x + R2 )1/2  Hence, dE =

= 2π

2

O

–L/2

2 3/2

2

L

5.

  MR2 = + MR2  α 4 πε0 (h2 + x 2 )3/2  2  Q 2 xR

Q 2 πR2 Lε0

[L + t ]

+ L /2   L + [(L − r )2 + R2 ]1/2  − L /2      2 πR2 Lε0 

Q

1/2 1/2   2   9L2   L 2 2  E= L+ +R  − +R   2  4 4 2 πR Lε0         

Q

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

physics for you | april ‘15

13

Page 13

EXAM from th

th

14 to 29 MAY

BITSAT FULL LENGTH PRACTICE PAPER physics 1.

2.

3.

4.

5.

14

Suppose speed of light (c), force (F) and kinetic energy (K) are taken as the fundamental units, then the dimensional formula for mass will be (a) [Kc–2] (b) [KF–2] –2 (c) [cK ] (d) [Fc–2] A sand bag of mass M is suspended by a rope. A bullet of mass m is fired at it with speed v and gets embeded in it. The loss of kinetic energy of the system is M v2 M mv 2 (a) (b) 2(M + m) 2(M + m) 2 2 m v 1 (c) (d) (M + m) v 2 2 (M + m) 2 A steel wire with cross-section 3 cm2 has elastic limit 2.4 × 108 N m–2. The maximum upward acceleration that can be given to a 1200 kg elevator supported by this cable wire if the stress is not to exceed one-third of the elastic limit is (Take g = 10 m s–2) (a) 12 m s–2 (b) 10 m s–2 –2 (c) 8 m s (d) 7 m s–2 A body of density r at rest is dropped from a height h into a lake of density s where s > r. Neglecting all dissipative forces, find the maximum depth to which the body sinks before returning to float on the surface. hρ h (a) (b) σ σ−ρ hρ hσ (c) (d) σ−ρ σ−ρ Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original volume adiabatically. The ratio of final pressure of gas in B to that of gas in A is

(a) 2

g–1

 1  (c)   1 − γ 

2

1 (b)   2

γ −1

 1  (d)   γ − 1 

2

6.

Soap water drips from a capillary. When the drop breaks away, the diameter of its neck is 1 mm. The mass of the drop is 0.0129 g. Find the surface tension of soapy water. (Take g = 9.8 m s–2) (a) 12.9 × 10–3 N m–1 (b) 31.2 × 10–3 N m–1 (c) 40.3 × 10–3 N m–1 (d) 58.6 × 10–3 N m–1

7.

An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth. The height of the satellite above the Earth’s surface is (Take radius of Earth = 6400 km) (a) 6000 km (b) 5800 km (c) 7500 km (d) 6400 km A needle placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of the lens. Its focal length and the size of image if the size of the needle is 5 cm are respectively (a) – 30 cm, 10 cm (b) + 30 cm, – 10 cm (c) – 20 cm, 15 cm (d) + 20 cm, – 15 cm

8.

9.

In Young’s double slit experiment distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 Å. Then, angular position of the first dark fringe is (a) 0.08° (b) 0.16° (c) 0.20° (d) 0.32°

10. When light of wavelength 400 nm is incident on

the cathode of a photocell, the stopping potential recorded is 6 V. If the wave of the incident light is increased to 600 nm, then the new stopping potential is (a) 1.03 V (b) 2.42 V (c) 4.97 V (d) 3.58 V

physics for you | april ‘15

Page 14

11. Two particles A and B describe S.H.M. of same

amplitude a and frequency u along the same straight line. The maximum distance between two particles is 3 a. The initial phase difference between the particles is (a) 2p/3 (b) p/6 (c) p/2 (d) p/2

12. A racing car moving towards a cliff sounds its

horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If v be the velocity of sound, the velocity of the car is (a) v/ 2 (b) v/2 (c) v/3 (d) v/4

kinetic energy equals the original kinetic energy of the lighter particle. What is the original speed of the heavier particle ? (a) (2 − 2 ) m s–1 (b) 2(1 + 2 ) m s–1 (c) (2 + 3 2 ) m s–1

(d) 4(1 − 2 ) m s–1

18. An equilateral triangle of side length l is formed from

a piece of wire of uniform resistance. The current I is fed as shown in the figure. The magnitude of the magnetic field at its centre O is Q

13. The rate of cooling at 600 K, if surrounding

temperature is 300 K is R. Assume that the Stefan’s law holds. The rate of cooling at 900 K is 16 2 (a) (c) 3R (d) R R (b) 2R 3 3 14. The ratio of specific heat of gas at constant pressure to that at constant volume is g. The change in internal energy of one mole of gas when volume changes from V to 2V at constant pressure P is R (a) (b) PV ( γ −1) γ PV PV (c) (d) . γ −1 ( γ −1)

O P

(a) (c)

16. A thin uniform rod AB of mass M and length L

is hinged at one end A to the level floor. Initially it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity of the rod, when its end B strikes the floor is (g is acceleration due to gravity)  Mg  (a)   L 

 Mg  (b)   3L 

g (c)   L

(d)  3 g   L 

1/2

1/2

17. A particle of mass m has half the kinetic energy of

another particle of mass m/2. If the speed of the heavier particle is increased by 2 m s–1, its new

I

3 µ0 I

(b)

2 πl µ0 I

3 3 µ0 I 2 πl

(d) zero

2 πl

19. Two inductors L1 and L2 are connected in parallel

and a time varying current flows as shown in figure. The ratio of current I1/I2 at any time t is

15. A boy throws a ball upwards with velocity

u = 15 m s–1. The wind imparts a horizontal acceleration of 3 m s–2 to the left. The angle q with vertical at which the ball must be thrown so that the ball returns to the boy’s hand is (Take g = 10 m s–2) (a) tan–1 (0.4) (b) tan–1 (0.2) (c) tan–1 (0.3) (d) tan–1 (0.15)

R

I

I1

L1

I

I I2

(a) (c)

L2 L1

L2

(b) L22

(d)

L1 L2 L21

(L1 + L2 )2 (L1 + L2 )2 20. An a.c. source is connected across an LCR series circuit with L = 100 mH, C = 0.1 mF and R = 50 W. The frequency of a.c. to make the power factor of the circuit, unity is 10 4 (a) 2π Hz

103 (b) 2π Hz

10 −4 (c) 2π Hz

10 −3 (d) 2π Hz

21. The electric field (in N C–1) in an electromagnetic

wave is given by E = 50 sin w(t – x/c). The energy stored in a cylinder of cross-section 10 cm2 and physics for you | april ‘15

15

Page 15

length 100 cm along the x-axis will be (a) 5.5 × 10–12 J (b) 1.1 × 10–11 J –11 (c) 2.2 × 10 J (d) 3.3 × 10–11 J 22. In an interference experiment using waves of same

amplitude, path difference between the waves at a point on the screen is l/4. The ratio of intensity at this point with that at the central bright fringe is (a) 1 (b) 0.5 (c) 1.5 (d) 2.0

23. A plane mirror is placed along the x-axis facing

negative y-axis. The mirror is fixed. A point object ^

^

is moving with 3 i + 4 j in front of the plane mirror. The relative velocity of image with respect to its object is y

(c) Ex = 33 cos p × 1011 (t –

x ) c

x Bx = 11 × 10–7 cos p × 1011 (t – ) c x 11 (d) Ey = 66 cos 2p × 10 (t – ) c x –7 Bz = 2.2 × 10 cos 2p × 1011 (t – ) c 27. The three stable isotopes of neon

10Ne

20

,

21 10Ne

and 10Ne22 have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u respectively. The average atomic mass of neon is (a) 11.18 u (b) 15.18 u (c) 20.18 u (d) 10.18 u 28. Light rays of wavelength 6000 Å and of photon

x

^

^

(a) − 8 j ^

(b) 8 j ^

(c) 3 i − 4 j

^

(d) − 6 i

24. A surface irradiated with light of wavelength 480 nm

gives out electrons with maximum velocity v m s–1, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2v m s–1 if it is irradiated by light of wavelength (a) 325 nm (b) 360 nm (c) 384 nm (d) 300 nm

25. The ratio of the de Broglie wavelengths of proton

and alpha particle which have been accelerated through same potential difference is (a) 2 3 (b) 3 2 (c) 2 2 (d) 3 3

26. A plane electromagnetic wave travelling along the

x-direction has a wavelength of 3 mm. The variation in the electric field occurs in the y-direction with an amplitude 66 V m–1. The equations for the electric and magnetic fields as a function of x and t are x (a) Ey = 33 cos p × 1011 (t – ) c x Bz = 1.1 × 10–7 cos p × 1011 (t – ) c x (b) Ey = 11 cos 2p × 1011 (t – ) c x By = 11 × 10–7 cos 2p × 1011 (t – ) c

16

intensity 39.6 W m–2 is incident on a metal surface. If only 1% of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be (Take h = 6.64 × 10–34 J s, c = 3 × 108 m s–1) (a) 12 × 1018 (b) 10 × 1018 (c) 12 × 1017 (d) 12 × 1016

29. A sample contains 10–2 kg each of the two

substances A and B with half-lives 4 s and 8 s respectively. Their atomic weights are in the ratio of 1 : 2. Find the ratio of the amounts of A and B after an interval of 16 seconds. (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1

30. A diode having potential difference 0.5 V across

its junction which does not depend on current, is connected in series with resistance of 20 W across source. If 0.1 A current passes through resistance, then what is the voltage of the source? (a) 1.5 V (b) 2.0 V (c) 2.5 V (d) 5 V

31. If ground state ionisation energy of H-atom is

13.6 eV, the energy required to ionize a H-atom from second excited state is (a) 1.51 eV (b) 3.4 eV (c) 13.6 eV (d) 12.1 eV

32. Two satellites S1 and S2 revolve around a planet in

coplanar circular orbits in the same sense. Their periods of revolution are 1 h and 8 h respectively. The radius of orbit of S1 is 104 km. When S2 is closest to S1, the speed of S2 relative to S1 is (a) p × 104 km h–1 (b) 2p × 104 km h–1 4 –1 (c) 3p × 10 km h (d) 4p × 104 km h–1

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fragments A, B and C. The momentum of A is P i and that of B is 3 P j where P is positive number. The momentum of C is (a) (1 + 3 ) P in a direction making 120° with A (b) 2 P in a direction making 150° with A (c) 2 P in a direction making 150° with B (d) (1 + 3 ) P in a direction making 150° with B.

34. One end of a uniform

A

rod of length l and B mass m is hinged at A. It is released from rest from horizontal C position AB as shown in figure. The force exerted by the rod B′ on the hinge when it becomes vertical is 3 5 (a) mg (b) 3mg (c) 5mg (d) mg 2 2 35. Two bulbs 40 W and 60 W and rated voltage 240 V are connected in series across a potential difference of 420 V. Which bulb will work at above its rated voltage? (a) 40 W (b) 60 W (c) Both 40 W and 60 W (d) None of the bulbs 36. The three resistances A, B and C have values 3R, 6R

and R respectively. When some potential difference is applied across the network, the thermal powers dissipated by A, B and C are in the ratio 3R A 6R B

R C

(a) 2 : 3 : 4 (b) 2 : 4 : 3 (c) 4 : 2 : 3 (d) 3 : 2 : 4 37. The masses of three wires of copper are in the ratio 1 : 3 : 5 and lengths are in the ratio 5 : 3 : 1. Then the ratio of their electrical resistances are (a) 1 : 3 : 5 (b) 5 : 3 : 1 (c) 1 : 15 : 25 (d) 125 : 15 : 1 38. A pendulum bob of mass m carrying a charge q is at rest with its string making an angle q with the vertical in a uniform horizontal electric field E. The tension in the string is mg qE mg qE (a) sin θ and cos θ (b) cos θ and sin θ qE mg (c) mg (d) qE

39. A modulating signal is a square wave as shown in

figure. The carrier wave is given by c(t) = 2sin(8pt) volt. What is the modulation index? 1 O

m(t) V

33. An object initially at rest explodes into three

0.5

(a) 0.2 (c) 0.4

1

1.5

2.0

t(s)

(b) 0.3 (d) 0.5

40. A Zener diode when used as a voltage regulator is

connected (i) in forward bias (ii) in reverse bias (iii) in parallel with load (iv) in series with load (a) (i) and (ii) are correct (b) (ii) and (iii) are correct (c) (i) only correct (d) (iv) only correct chemistry

41. The correct order of first ionisation potential

among the following elements Be, B, C, N, O is (a) B < Be < C < O < N (b) B < Be < C < N < O (c) Be < B < C < N < O (d) Be < B < C < O < N

42. Amongst TiF62–, CoF63–, Cu2Cl2 and NiCl42– (At.

nos. of Ti = 22, Co = 27, Cu = 29, Ni = 28). The colourless species are (a) CoF63– and NiCl42– (b) TiF62– and CoF63– (c) Cu2Cl2 and NiCl42– (d) TiF62– and Cu2Cl2

43. Identify ‘Z’ in the sequence :

(a) C6H5CN (c) C6H5COOH

(b) C6H5CONH2 (d) C6H5CH2NH2

44. The oxidation number of S in Caro’s acid (H2SO5) is

(a) + 5 (c) + 2

(b) + 3 (d) + 6

45. The solubility of sulphates in water down the IIA

group follows the order Be > Mg > Ca > Sr > Ba. This is due to (a) increase in melting point (b) increasing molecular mass (c) decreasing lattice energy (d) high heat of solvation of smaller ions. physics for you | april ‘15

17

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46. A hydrocarbon P of the formula C7H12 on

ozonolysis gives a compound Q which undergoes aldol condensation giving 1-acetylcyclopentene. The compound P is (a)

(b)

(c)

(d)

47. Identify the pair of species in which the central

atom has the same type of hybridisation. (a) BF3 and NCl3 (b) H2S and BeCl2 (c) NCl3 and H2S (d) SF4 and BeCl2 compound is

CH3

(b) CH3 CH2 CH CH2 (c) CH3 C CH2

(c) 0 , 1, 1

(d) 0 , 1 , 1

n

n

55. The equilibrium constant of the reaction,

1 O 2 2(g) is 0.15 at 900 K. The equilibrium constant for 2SO2(g) + O2(g) 2SO3(g) is SO2(g) +

(b) 49.72 mol–1 L (d) 44.44 mol–1 L

56. In the Hoope’s process for refining of aluminium,

CH3

(d) CH3 CH CH2 CH2 CH3

49. The cubic unit cell of aluminium (molar mass

27.0 g mol–1) has an edge length of 405 pm and density 2.70 g cm–3. What type of unit cell is it? (a) Face-centred (b) Body-centred (c) Simple cubic (d) None of these.

50. Al2O3 on heating with carbon in an atmosphere of

Cl2 at high temperature produces (a) Al + CO2 (b) Al + CO2 + NO (c) Al4C3 + CO2 (d) AlCl3 + CO following

does

not

show

(b) CH3CHO (d) C6H5COC(CH3)3

52. Which of the following has least covalent P H

(b) P2H6 + (d) PH6

53. What products are expected from the dispropor-

tionation reaction of hypochlorous acid? (a) HClO3 and Cl2O (b) HClO2 and HClO4 (c) HCl and Cl2O (d) HCl and HClO3

18

(b) 1 , 1, 0

n

(a) 52.52 mol–1 L (c) 63.34 mol–1 L

CH3

bond? (a) PH3 (c) P2H5

(a) 1, 1 , 0

SO3(g)

(a) CH3 CH2 CH2 CH2 CH2

of the tautomerism? (a) C6H5COCH3 (c) CH3COCH3

x, y and z are respectively

n

48. The structure of neo pentyl group in an organic

51. Which

54. In the given Freundlich adsorption isotherm plot,

the fused materials form three different layers and they remain separated during electrolysis also. This is because (a) there is a special arrangement in the cell to keep the layers separate (b) the three layers have different densities (c) the three layers are maintained at different temperature (d) the upper layer is kept attracted by the cathode and the lower layer is kept attracted by the anode. 57. Which of the following complexes will give white precipitate with BaCl2(aq)? (a) [Co(NH3)4SO4]NO2 (b) [Cr(NH3)4SO4]Cl (c) [Cr(NH3)5Cl]SO4 (d) Both (b) and (c). 58. At 25°C, the molar conductivity of 0.001 M hydrofluoric acid is 184.5 W–1 cm2 mol–1. If its L°m is 502.4 W–1 cm2 mol–1, then equilibrium constant at the given concentration is (a) 3.607 × 10–4 M (b) 5.404 × 10–4 M (c) 2.127 × 10–4 M (d) 6.032 × 10–4 M 59. Zinc on reacting with cold, dil. HNO3, gives

(a) ZnNO3 (c) NO2

(b) NH4NO3 (d) NO

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60. Which of the following is aromatic?

(a)

+

(c)

(b)

–

(d)

61. Which of the following statements is not correct?

(a) Some antiseptics can be added to soaps. (b) Dilute solutions of some disinfectants can be used as an antiseptic. (c) Disinfectants are antimicrobial drugs. (d) Antiseptic medicines can be ingested.

62. Which molecule/ion out of the following does not

contain unpaired electrons? (a) N2+ (b) O2 (c) O22–

(d) B2

63. Energy of an electron in hydrogen atom is given by E=

13.6 n2

eV. Which one of the following statements

is true if n is changed from 1 to 3? Energy will (a) decrease three times (b) increase three times (c) increase nine times (d) decrease nine times. 64. Structure of the compound whose IUPAC name is

3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is (a)

(b)

(b) (c) (d)

+ Zn/Hg and conc. HCl + CrO2Cl2 in CS2 followed by H3O+ + H2 in presence of Pd-BaSO4

68. A reaction having equal energies of activation for

forward and reverse reactions has (a) DH = 0 (b) DH = DG = DS = 0 (c) DS = 0 (d) DG = 0 69. Amongst the acids, (i) CH CCOOH (ii) CH2 CHCOOH and (iii)CH3CH2COOH, the acid strength follows the sequence (a) (i) < (ii) > (iii) (b) (i) > (ii) > (iii) (c) (i) = (ii) = (iii) (d) (i) = (ii) > (iii)

70. Ionization potential of hydrogen atom is 13.6 eV.

Hydrogen atom in ground state is excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr’s theory will be (a) one (b) two (c) three (d) four.

71. Which of the following reactions will yield

2-propanol? H+ → I. CH2 CH CH3 + H2O  (i) CH MgI

3 → II. CH3 CHO  (ii) H O

(i) C H MgI

2

2 5 → III. CH2O  (ii) H O 2

Neutral KMnO

(c)

(d)

65. The value of Planck’s constant is 6.63 × 10–34 J s.

The speed of light is 3 × 1017 nm s–1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s–1? (a) 50 (b) 75 (c) 10 (d) 25

66. Identify the correct order of solubility in aqueous

medium. (a) Na2S > CuS > ZnS (b) Na2S > ZnS > CuS (c) CuS > ZnS > Na2S (d) ZnS > Na2S > CuS

67. Reaction by which benzaldehyde cannot be

prepared (a)

+ CO + HCl in presence of anhydrous AlCl3

4 → IV. CH2 CH CH3  (a) I and II (b) II and III (c) III and I (d) II and IV 72. The heat liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25°C and it increases the temperature of 18.94 kg of water by 0.632°C. If the specific heat of water at 25°C is 0.998 cal/g-deg, the value of the heat of combustion of benzoic acid is (a) 881.1 kcal (b) 771.12 kcal (c) 981.1 kcal (d) 871.2 kcal 73. Which of the following combinations illustrates the law of reciprocal proportions? (a) N2O3, N2O4, N2O5(b) NaCl, NaBr, NaI (c) CS2, CO2, SO2 (d) PH3, P2O3, P2O5 74. Which of the following is formed, when benzaldehyde reacts with alcoholic KCN? (a) Benzoin (b) Benzyl alcohol (c) Benzoic acid (d) Ethyl benzoate

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75. The standard electrode potentials for Pb2+| Pb and

Zn2+ | Zn are – 0.126 V and – 0.763 V respectively. The e.m.f. of the cell Zn | Zn2+(0.1 M) || Pb2+(0.1 M)|Pb is (a) 0.637 V (b) < 0.637 V (c) > 0.637 V (d) 0.889 V

76. Which of the following will be most acidic?

(a)

(b)

(c)

(d)

KCN forms yellow, white and reddish-brown precipitate. (X) gives insoluble complex with excess of KCN and no ppt. upon passing H2S gas. (Y) also gives insoluble complex with excess of KCN but gives yellow ppt. on passing H2S gas. (Z) gives yellow solution with excess of KCN. Then X, Y and Z respectively are (a) Cu2+, Cd2+, Fe3+ (b) Cu2+, Fe2+, Cd2+ (c) Pb2+, Cd2+, Cu2+ (d) Fe3+, Pb2+, Fe2+

78. Which of the following represents the correct order

of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl? (a) O < S < F < Cl (b) F < S < O < Cl (c) S < O < Cl < F (d) Cl < F < O < S

79. Potassium permanganate has intense purple colour

due to (a) weak d-d transitions (b) metal to ligand charge transfer (c) ligand to metal charge transfer (d) both metal and ligand transitions.

TsCl

H

(b) N

CH NO

HO

O

HO HO

N

(d)

H

20

Q

Sn/HCl, 

H

(c)

82. Let R be a relation in N defined by

R = {(x, y) : x + 2y = 8}, then range of R is (a) {2, 4, 6} (b) {1, 2, 3, 4, 6} (c) {1, 2, 3} (d) none of these.

N H

set B = {1, 3, 5} i.e. (a, b) ∈ R ⇔ a < b, then RoR–1 equals (a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3, 3), (5, 3), (3, 5), (5, 5)} (c) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} (d) {(4, 5), (3, 4), (3, 3)}.

84. The straight lines joining the origin to the points of

intersection of the straight line hx + ky = 2hk and the curve (x – k)2 + (y – h)2 = c2 are at right angles, then (a) h2 + k2 + c2 = 0 (b) h2 – k2 – c2 = 0 2 2 2 (c) h + k – c = 0 (d) none of these.

85. The exhaustive range of values of a such that the

angle between the pair of tangents drawn from (a, a) to the circle x2 + y2 – 2x – 2y – 6 = 0 lies in the range (p/3, p) is (a) (1, ∞) (b) (–5, –3) ∪ (3, 5) (c) (−∞, − 2 2 ) ∪ (2 2 , ∞) (d) (–3, –1) ∪ (3, 5)

86. If a = tan27q – tanq and

sin θ sin 3θ sin 9θ + + , then cos 3θ cos 9θ cos 27θ (a) a = b (b) a = 2b (c) b = 2a (d) none of these.

2 P NaOH,3 MeOH

R

(a)

(a) 2 sin x sin y sin z (b) 2 cos x cos y cos z (c) 2 sin x cos y cos z (d) 2 sin x sin y cos z

β=

80. Identify ‘R’ in the following series of reactions. OH Pyridine

81. If x + y = π + z, then sin2x + sin2y – sin2z is equal to

83. If R be a relation ‘ 0 (b) a = 0, b > 0 (c) b > 0, a < 0 (d) b < 0, a > 0 2 − 1 1 2 , then det(Adj(Adj A)) =   1  2 −1  1

100. If A =  −1

(a) 144 (b) 143 2 (c) 14 (d) 14 101. The existence of the unique solution of the system of equations x + y + z = l, 5x – y + mz = 10, 2x + 3y – z = 6 depends on (a) m only (b) l only (c) both l and m (d) neither l nor m 102. If A and B are square matrices such that

B = –A–1BA , then (a) AB + BA = 0 (b) (A + B)0 = A2 + B2 + AB (c) (A + B)2 = A2 + 2AB + B2 (d) (A + B)2 = A + B 1

1 103. If A = 2

2 1 2

3 a  (a) a = 2, b = 1 (c) a = 2, b = –1

2 − 2 is an orthogonal matrix, then  b (b) a = –2, b = –1 (d) a = –2, b = 1 physics for you | april ‘15

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104. If A and B are square matrices of order 3 × 3 such

that A is an orthogonal matrix and B is a skew symmetric matrix, then which of the following statements is true? (a) |AB| = 1 (b) |AB| = 0 (c) |AB| = –1 (d) none of these

105. The least value of the expression

2log10 x – logx(0.01) for x > 1 is (a) 10 (b) 2 (c) – 0.01 (d) 4

y = 2x – 2x2 is (a) 7/12 (b) 5/2

(c) 1/12 (d) 4

49 113. sin 51x (sin x) dx equals

∫

(a)

sin 50x (sin x)50 50 cos 50x (sin x)50

+C

x x  2 107. If f (x) = (a − 3a + 2)  cos2 − sin2  + (a − 1)x + sin 1  4 4

+C 50 cos 50x (cos x)50 (c) +C 50 sin 50x (sin x)51 (d) +C 51 114. If f be a polynomial function satisfying f (x2 + x + 3) + 2 f (x2 – 3x + 5) = 6x2 – 10x + 17 ∀ x ∈R then (a) f is a decreasing function (b) f(x) = 0 has a root in (0, 2) (c) f(x) is an odd function (d) no such polynomial exist

108. If x ∈ (– ∞, –2) and y3 – 3y + x = 0, then

115. If |f ′′(x)| ≤ 1 ∀ x ∈ R and f(0) = 0 = f ′(0) then

106. The expression {x + (x 3 − 1)1/2}5 + {x − (x 3 − 1)1/2}5is

a polynomial of degree (a) 5 (b) 6 (c) 7

(d) 8

possesses critical points, then the set of values of ‘a’ are (a) (– ∞, 0] ∪ [4, ∞) (b) (– ∞, 0] ∪ [4, ∞) ∪ {1} (c) (– ∞, 0) ∪ (4, ∞) ∪ (4, ∞) ∪ {1} (d) none of these (a) y is not a function of x (b) y is not a monotonic function (c) y is an increasing function of x (d) y is a decreasing function of x

1  1  ln x, x ∈  , 3  , 2  3  then greatest value of f (x) is π 1 π 1 (a) − ln 3 (b) + ln 3 6 4 6 2 1 π π (c) ln 3 − (d) + ln 3 4 6 6 4

−1 109. If f (x) = cot x +

1 −1 [x] + [−x] + 3− | x | + 2 110. The range of f (x) = tan x

is (where [.] denotes G.I.F) 1  1 (a)  , ∞  (b)   ∪ 2, ∞ ) 4  4 1 5   5  (c)  , , 1 + 2  (d)  , 1 + 2  9 4  4   111. The solution of differential equation x2dy – y2dx + xy2(x – y)dy = 0 is (a) ln

x + y y2 xy y2 + =C + = C (b) ln x−y 2 x−y 2

2 2 (c) ln x − y + y = C (d) ln x − y + y = C 2 xy x+y 2

22

112. The area of the region bounded by y = x ln x,

(b)

which of the following cannot be true? 1 1  1 1 (a) f   = (b) f  −  = 3 5  3  12 1 (c) f (3) = 4 (d) f (−3) = 3  1  f ( x ) = 116. Let  sin{x}  , where [.] and {.} denote   the greatest integer and fractional part function respectively. The range of f is (a) the set of integers = I (b) the set of natural numbers = N (c) the set of whole numbers = W (d) {2, 3, 4, ....} 117. The chord of contact of tangents from any point

of circle x2 + y2 = a2 with respect to the circle x2 + y2 = b2 touches the circle x2 + y2 = c2 where (a, b, c > 0) then a+c (a) b < 2 1 1 1 (b) , , are in A.P. 1 + log a 1 + log b 1 + log c (c) a, b, c are in A.P. (d) b > ac

118. The locus of mid-point of the chord of the circle with

diameter as minor axis of the ellipse

x2

y2 + =1 a 2 b2

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(a > b) which subtend right angle at centre of ellipse is 2

a +b 2 2 b 2 2 (c) 2(x2 + y2) = b2 (d) x + y = 4 119. Let a, b, g are the roots of the equation (b) x 2 + y 2 =

(a) x2 + y2 = 2b2

2

 x2 + x + 1

x →∞

 29 23  (c) circumcentre of the triangle is  , −   8 16  (d) centroid of the triangle is  1 , − 1   4 sec2 x

∫ (sec x + tan x)9/2 dx

equals (for

some arbitrary constant K) 1 1 1 (a) − − (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x) (b)

1

(sec x + tan x)

(c) − (d)

11/2

1

} } } }

1 1 − (sec x + tan x)2 + K 11 7

11/2

(sec x + tan x)

1

{

{

{

{

colours can be distributed among 3 persons so that each person gets at least one ball is (a) 75 (b) 150 (c) 210 (d) 243

124. If lim  

1   1  1  x3 + 3x2 – 6x – 8 = 0. If  , α  ,  , β  and  , γ  α  β  γ  are the vertices of the triangle, then  3  (a) centroid of the triangle is  − , − 1  4   1  (b) orthocentre of the triangle is  − , − 8   8 

120. The integral

123. The total number of ways in which 5 balls of different

1 1 + (sec x + tan x)2 + K 11 7

1 1 + (sec x + tan x)2 + K 11/2 11 7 (sec x + tan x)

121. Let z be a complex number such that the imaginary

part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value 1 3 1 (a) –1 (b) 3 (c) (d) 4 2

 π  x 2 cos , x ≠ 0 , x ∈ R, then f is 122. Let f (x) =  x  x =0 0, (a) differentiable both at x = 0 and at x = 2 (b) differentiable at x = 0 but not differentiable at x=2 (c) not differentiable at x = 0 but differentiable at x=2 (d) differentiable at neither at x = 0 nor at x = 2

x +1

(a) a = 1, b = 4 (c) a = 2, b = –3

 − ax − b  = 4, then  (b) a = 1, b = – 4 (d) a = 2, b = 3

125. The function f : [0, 3] → [1, 29], defined by

f(x) = 2x3 –15x2 + 36x + 1, is (a) one-one and onto (b) onto but not one-one (c) one-one but not onto (d) neither one-one nor onto

english and logical reasoning

Directions (Questions 126 to 128) : Read the passage and answer the following questions. Books are, by far, the most lasting product of human effort. Temples crumble into ruin, pictures and statues decay, but books survive. Time does not destroy the great thoughts which are as fresh today as when they first passed through their author’s mind. These thoughts speak to us through the printed page. The only effect of time has been to throw out of currency the bad products. Nothing in literature which is not good can live for long. Good books have always helped man in various spheres of life. No wonder that the world keeps its books with great care. 126. Of the product of human effort, books are the

most (a) Permanent (c) Enjoyable

(b) Important (d) Useful.

127. Time does not destroy books because they contain

(a) Useful material (b) Subject matter for education (c) High ideals (d) Great ideas.

128. “To throw out of currency” means

(a) Destroy (c) Extinguish

(b) Put out of use (d) Forget.

Directions (Questions 129 to 130) : Pick out the correct synonyms for each of the following words. 129. Eradicate

(a) Dedicate (c) Complicate

(b) Eliminate (d) Indicate physics for you | april ‘15

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Page 23

130. Myopic

(a) Astigmatic (c) Blind

(b) Cross-eyed (d) Short-sighted

Directions (Questions 131 to 132) : In each of the following questions, an idiomatic expression/a proverb has been given, followed by some alternatives. Choose the one which best expresses the meaning of the given idiom or proverb. 131. To take the wind out of another’s sails

(a) To defeat the motives of another (b) To anticipate another and to gain advantage over him (c) To manouevre to mislead another on the high seas (d) To cause harm to another

132. To keep the ball rolling

(a) To keep the conversation going (b) To make the best use of (c) To earn more and more (d) To work constantly

Directions (Questions 133 to 137) : Rearrange the given five sentences A, B, C, D and E in the proper sequence so as to form a meaningful paragraph and then answer the questions given below them. A. The alternative was a blitz by the health workers to popularise preventive measures. B. This information was considered inadequate. C. People have been reading about AIDS in the mass media. D. Nobody is sure as to how effective this would be. E. People were also not being influenced enough to take preventive measures. 133. Which sentence should come first in the

paragraph? (a) C (b) D

(c) B

(d) A

134. Which sentence should come second in the

paragraph? (a) C (b) D

(c) B

137. Which sentence should come last in the

paragraph? (a) A (b) D

(c) B

(d) E

Directions (Questions 138 to 140) : In each of the following questions, a word has been written in four different ways out of which only one is correctly spelt. Find the correctly spelt word. 138. (a) Garuntee

(b) Guaruntee (d) Guarantee

139. (a) Benefited

(b) Benifitted (d) Benifited

140. (a) Efflorescance

(b) Eflorescence (d) Efflorescence

(c) Gaurantee (c) Benefeted (c) Efflorascence

141. There is a certain relationship between two

given words on one side of : : establish a similar relationship on another side of : : by selecting a word from the given options. Doctor : Patient : : Politician : ? (a) Masses (b) Voter (c) Power (d) Chair

142. In the following question, four words have been

given, out of which three are alike in some manner, while the fourth one is different. Choose out the odd one. (a) Seminar (b) Semicolon (c) Semifinal (d) Semicircle 143. Find the missing character in the following : ?

235

4 7

117 59

(a) 327

(b) 386

29

(c) 438

15

(d) 469

144. Select a figure from amongst the Answer Figures

which will continue the same series as established by the five Problem Figures. Problem Figures

(d) E

135. Which sentence should come third in the

paragraph? (a) C (b) D

(c) B

(d) E

136. Which sentence should come fourth in the

paragraph? (a) C (b) D

24

(c) B

(d) A

(a)

(b)

(c)

(d)

physics for you | april ‘15

Page 24

145. In the following question, find out which of the

figures (a), (b), (c) and (d) can be formed from the pieces given in fig. (X).

150. In the following number series, one term is wrong.

Find out the wrong term. 1, 3, 10, 21, 64, 129, 356, 777 (a) 21 (b) 129 (c) 10

(d) 356

solution x

(a)

(b)

(c)

(d)

146. Select the missing term.

A, D, H, M, ?, Z (a) T (b) G

(c) N

(d) S

147. In the following question, find out which of the

answer figures (a), (b), (c) and (d) completes the figure matrix ?

2. (a) : According to law of conservation of linear momentum, mv = (M + m) V or V = m v / (M + m) 1 Initial KE of the system = mv 2 2 1 Final KE of the system = (M + m) V2 2

?

(a)

(b)

(c)

(d)

148. In the following question, a set of figures carrying

certain characters, is given. Assuming that the characters in each set follow a similar pattern, find the missing character. 15 5

(a) 35

80

2

9

6

4

(b) 48

65

7

13

6

11

(c) 72

?

16 8

(d) 120

149. In the following question, choose the set of figures

which follows the given rule. Rule : Closed figures gradually become open and open figures gradually become closed. (a) (b) (c) (d)

1. (a) : Let M = kc F y K z where k is a dimensionless constant. \ [M1 L0 T0] = [LT–1]x [MLT–2]y [ML2 T–2]z = M y+z Lx+y+2z T –x–2y–2z Applying principle of homogeneity of dimensions, y+z=1 ...(i) x + y + 2z = 0 ...(ii) – x – 2y – 2z = 0 ...(iii) Adding (ii) and (iii), we get, y = 0 Now, from (i) z = 1 – y = 1 from (ii) x = – y – 2z = 0 – 2 \ [M] = [c–2 F 0 K1] = [Kc–2]

2 2 1 1 (m v )  mv  = (M + m)  =  M + m  2 2 (M + m) 2 1 1 (m v ) Loss of KE = m v 2 − 2 2 (M + m)

mv 2  M + m − m  m M v 2 = = 2  M + m  2 (M + m) 3. (b) : Maximum tension an elevator can tolerate is 1 T = stress × area of cross-section 3 1 = × (2.4 × 108) × (3 × 10–4) = 2.4 × 104 N 3 If a is the maximum upward acceleration of elevator then T = m (g + a) or 2.4 × 104 = 1200 (10 + a) On solving, a = 10 m s–2. 4. (c) : The speed of the body just before entering the liquid is u = 2 gh . The buoyant force FB of the lake (i.e., upward thrust of liquid on the body) is greater than the weight of the body W, since s > r. If V is the volume of the body and a is the acceleration of the body inside the liquid, then FB – W = ma physics for you | april ‘15

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Page 25

sVg–rVg=rVa

(σ − ρ) g or (s – r) g = r a or a = ρ Using the relation, v2 = u2 + 2as, we have hρ 2 (σ − ρ) 0 = 2 gh − 2 g s or s = ρ σ−ρ

(

)

5. (a) : For isothermal compression of gas A, PV P f = i i = 2Pi Vf

1 (∵ Vf = Vi ) 2

For adiabatic compression of gas B,

9. (b) : Angular position of first dark fringe, λ λ θ1 = (2 × 1 − 1) = 2d 2d 5460 × 10 −10 = = 2730 × 10 −6 rad −3 2 × 0.1 × 10 180° = 2730 × 10 −6 × = 0.16°. π 10. (c) : As Kmax = hn – f0 hc hc φ or eV0 = − φ 0 or V0 = − 0 λ eλ e \ DV0 = (V0)1 – (V0)2  hc φ 0   hc φ 0  hc  1 1  − =  −  = − −  eλ1 e   eλ 2 e  e  λ1 λ 2 

γ

V  P f′ = Pi  i  = 2 γ Pi  Vf  P f′

2 γ Pi Thus, = = 2 γ −1 2Pi Pf 6. (c) : When the drop breaks away from the capillary, weight of the drop = force of surface tension acting on it due to capillary, i.e., mg mg = (πD) × T or T = ...(i) πD Here, m = 0.0129 g = 1.29 × 10–5 kg, g = 9.8 m s–2, D = 1 mm = 10–3 m From eqn. (i), T=

(1.29 × 10 −5 kg )(9.8 m s −2 ) 3.14 × 10 −3 m

= 40.3 × 10 −3 N m −1

v As v 0 = e , 2

⇒

2 gR gR 2 gR 2 gR = or = , 2 R+h R+h 2 h = R = 6400 km

8. (b) : Here, u = – 45 cm, v = + 90 cm Using thin lens formula, 1 1 1 1 1 1+ 2 = − = + = f v u 90 45 90 \ f = + 30 cm h2 v h 90 or 2 = = Magnification, m = 5 − 45 h1 u \ 26

Size of image, h2 = – 10 cm.

\

11. (a) : y1 = a sin wt and y2 = a sin (wt + q) y2 – y1 = a 3 = a sin (wt + q) – a sin wt (ωt + θ) + ωt (ωt + θ) − ωt sin 2 2 = 2a cos (wt + q / 2) sin q / 2 For maximum value, cos (wt + q / 2) = 1 \ 3 a = 2 a sin q / 2 or

or

3 a = 2 a cos

sin θ / 2 =

3 π θ π 2π = sin or = or θ = . 2 3 2 3 3

12. (c) 13. (a) : Rate of cooling is proportional to (T 4 – T 40), as per Stefan’s law. R′ 900 4 − 300 4 \ = R 600 4 − 300 4

gR 2 7. (d) : v 0 = and v e = 2 gR R+h

\

6.6 × 10 −34 × 3 × 108  1 1  −   −19 −7  4 × 10 1.6 × 10 6 × 10 −7  = 1.03 V (V0)2 = (V0)1 – 1.03 = 6 – 1.03 = 4.97 V =

[∵ h1 = 5 cm]

=

9 4 − 34 4

4

=

34 (34 − 1) 4

4

=

80 16 = 15 3

6 −3 3 (2 − 1) 16 or R′ = R. 3 CP C − CV 14. (c) : As \ P =γ = γ −1 CV CV C − CV R or CV = P = γ −1 γ −1 RdT P dV ∆U = nCV dT = n = ( γ − 1) γ − 1 P (2V − V ) PV = = γ −1 γ −1

physics for you | april ‘15

Page 26

15. (c) : Here, uy = u cos q = 15 cos q ux = u sin q = 15 sin q y

u = 15 m s–1

θ

x

Time of flight of the ball is T=

2u y

=

2 × 15 cos θ

= 3 cos θ

...(i) g 10 The boy will catch the ball if in time T, displacement of the ball in horizontal direction should also be 1 zero. So 0 = uxT − a xT 2 2 2ux 2(15 sin θ) or T = ...(ii) = = 10 sin θ ax 3

From (i) and (ii), 3 cos q = 10 sin q 3 or tan θ = = 0.3 or θ = tan −1 (0.3). 10 16. (d) : As the rod is hinged at one end, its moment of ML2 inertia about this end is I = . 3 Total energy in upright position = total energy on striking the floor 0+

MgL 1 2 1 ML2 2 = Iω + 0 = ω 2 2 2 3

g=

3g Lω 2 or ω = 3 L

17. (b) 18. (d) :

P I

O r 2I/3

I/3

e1 = e2  dI   dI  L1  1  = L2  2   dt   dt  Integrating both sides, we get I L L1I1 = L2 I 2 ∴ 1 = 2 I 2 L1 20. (a) 21. (b) : Energy contained in a cylinder U = average energy density × volume 1 = ∈0 E02 × Al 2 1 = × (8.85 × 10 −12 ) × (50)2 × (10 × 10 −4 ) × 1 2 = 1.1 × 10–11 J 22. (b) : As IR = I1 + I2 + 2 I1I 2 cos φ 2π λ π Here, φ = × = , I =I =I λ 4 2 1 2 π \ I R = I + I + 2I cos = 2 I 2 At the central bright fringe, I′ = 4I \ I R = 2 I = 0. 5 I ′ 4I ^ ^  23. (a) : Velocity of object, v ob = 3 i + 4 j

^ ^ ^ ^ ^    v rel = v image − v ob = (3 i − 4 j) − (3 i + 4 j) = − 8 j

R I

The magnetic field induction at O due to current through PR is B1 =

19. (a) : As the inductors are in parallel, therefore, induced e.m.f. across the two inductors is the same i.e.

^ ^  Velocity of image v image = 3 i − 4 j Relative velocity of image with respect to its object

Q I/3

µ 0 (I / 3) µ 2I [sin 30° + sin 30°] = 0 . 4π r 4 π 3r It is directed inside the paper. \ Resultant magnetic field induction at O is B1 – B2 = 0. B2 = 2 ×

µ0 2I / 3 µ 2I [sin 30° + sin 30°] = 0 . 4π r 4π 3 r

It is directed outside the paper. The magnetic field induction at O due to current through PQR is

24. (d) 25. (c)

26. (d) : E0 = 66 V m–1 E 66 = 2.2 × 10 −7 T B0 = 0 = c 3 × 108 Since electromagnetic wave is of transverse nature, hence if electric field is along y-axis the magnetic field must be along z-axis, since the propagation of wave is along x-axis. Thus the equations given in option (d) are correct. physics for you | april ‘15

27

Page 27

27. (c) : The masses of three isotopes are 19.99 u, 20.99 u, 21.99 u. Their relative abundances are 90.51%, 0.27% and 9.22%. \ Average atomic mass of neon is 90.51 × 19.99 + 0.27 × 20.99 + 9.22 × 21.99 m= (90.51 + 0.27 + 9.22) 1809.29 + 5.67 + 202.75 2017.7 = = = 20.18 u 100 100 28. (c) : Useful intensity for the emission of electron is 1 × 39.6 = 0.396 W m −2 I ′ = 1% of I = 100 hc Energy of each photon = λ =

(6.64 × 10

−34

8

) × (3 × 10 )

−19

= 3.32 × 10 J 6000 × 10 −10 Number of photoelectrons emitted per second per unit area 0.396 = ≈ 12 × 1017 3.32 × 10 −19 29. (a) : Let N0 be the initial amount of a radioactive substance. Then the amount left after n half-lives n 1 will be N = N 0   2 16 s t = =4 For A : nA = T1/2 4 s 4 1 N A = 10 −2 kg   = 6.25 × 10 −4 kg 2 16 s For B : nB = =2 8s 2 1 \ N B = 10 −2 kg   = 2.5 × 10 −3 kg 2 NA 1 = NB 4 30. (c) : The circuit diagram is as shown below :

GM mv 2 GMm = ⇒ v2 = 2 R R R 2πR 4 π2R 2 GM Also, v= ⇒ v2 = = 2 T R T 4π 2R 3 2 \ T = GM If T1 and T2 are the time periods for satellite S1 and S2 respectively

32. (a) :

2

 T1   R1   T  =  R  2 2

T  ⇒ R2 =  2   T1 

2 /3

R1

Here, T1 = 1 h, T2 = 8 h, R1 = 104 km \

8 R2 =   1

2 /3

× 10 4 km = 4 × 10 4 km

2πR1 2π × 10 4 = = 2π × 10 4 km h −1 T1 1 2πR2 2π × 4 × 10 4 v2 = = = π × 10 4 km h −1 T2 8 v1 =

Relative velocity of S2 with respect to S1 is v = v2 – v1 = (p × 104 – 2p × 104) km h–1 |v| = p × 104 km h–1 33. (c) : Here Px = P and Py =

\

3P

Resultant momentum of A and B P = Px2 + Py2 = P 2 + ( 3 P)2 = 2 P

It is along OC′. As is clear from figure

Y

C

B 

X

20  V

3

O C

0.1 A

A

X

Y

BC ′ OA P 1 = = = or q = 30° OB OB 3P 3 As the object was initially at rest, the vector sum of linear momenta of A, B and C must be zero. Therefore, momentum of C = 2 P along OC opposite to OC′. It makes an angle with B = ∠YOC = ∠YOX′ + ∠X′OC = 90° + 60° = 150° tan θ =

V = V′ + IR = 0.5 + 0.1 × 20 = 0.5 + 2.0 = 2.5 V 31. (a) : Here E∞ – E1 = 13.6 or E1 = – 13.6 eV For second excited state, E 13.6 E3 = 21 = − = −1.51 eV 9 3 Energy required to ionise H-atom from second excited state = E∞ – E3 = 0 + 1.51 = 1.51 eV. 28

34. (d) : As the rod rotates about A, therefore, from

conservation of mechanical energy, decrease in potential energy = increase in rotational kinetic energy about A

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Page 28

1  ml  2 3g l 1 mg   = IA ω 2 =   ω or ω 2 = 2 2 2 3 l Centripetal force of centre of mass of the rod in this l 3g 3 mg position is = mrw2 = m = . 2 l 2 If F is the force exerted by the hinge on the rod 3 mg (upwards), then F − mg = 2 3 mg 5 F= + mg = mg . 2 2 (240)2 Ω 35. (a) : Resistance of 40 W bulb, R1 = 40 (240)2 Resistance of 60 W bulb, R2 = Ω 60 When bulbs are in series, the effective resistance 2

1  (240)2 1 R = R1 + R2 = (240)2  +  = Ω  40 60  24 420 × 24 21 Current I = = A 120 (240)2

Potential difference across 40 W bulb 21 (240)2 = × = 252 V 120 40 Potential difference across 60 W bulb 21 (240)2 = × = 168 V 120 60 Since potential difference of 40 W is greater than 240 V, so it will work at above its rated voltage.

mg qE ; Tsinq = qE or T = cos θ sin θ 39. (d) : Here, Am = 1 V, Ac = 2 V, Am 1 Modulation index, µ = A = 2 = 0.5 c 40. (b) : The zener diode when used as a voltage regulator is connected in reverse bias and a load is connected in parallel to zener diode for output voltage. Tcosq = mg or T =

41. (a) : Due to the extra stability of half-filled p-orbitals of N, its first ionisation potential is higher than those of O and C. Further because of higher nuclear charge, first ionisation potential of C is higher than that of Be and B. Amongst Be and B, the first ionisation potential of Be is higher than that of B because in case of Be, an electron is to be removed from 2s2 orbital while in case of B, an electron is to removed from 2p1 orbital. Thus, the overall order is B < Be < C < O < N. 42. (d) : Colour of salts is a property of partially filled d-orbitals. Since TiF62– has completely empty and Cu2Cl2 has completely filled d-subshells, hence these are colourless salts. 43. (c) :

36. (c) :

44. (d) : 2

4  2I  Thermal power in A, PA =   3R = I 2R 3 3 2 2 I Thermal power in B, PB =   6R = I 2R 3 3 2 Thermal power in C, PC = I R. 4 2 PA : PB : PC = I 2R : I 2R : I 2R = 4 : 2 : 3 3 3

37. (d) 38. (b) :

θ

T Tsinθ

Tcosθ θ

+q

mg

From figure at equilibrium,

CH3

46. (a) :

O Ozonolysis

1-Methylcyclohexene (P)

qE E

Two oxygen atoms are in peroxo linkage i.e., oxidation number = –1 3 (–2) + 2(–1) + x + 2(+1) = 0 ⇒ – 8 + x + 2 = 0 ⇒ x = + 6 45. (d) OH–

H2C—C—CH3 Aldol O condenH2C C sation H H2C—CH2 (Q)

O— —C—CH3 C H2C

CH

H2C —CH2 1-Acetylcyclopentene physics for you | april ‘15

29

Page 29

47. (c) : N atom in NCl3 and S atom in H2S are sp3 hybridised. 48. (c)

K=

Cα 2 0.001 M × (0.367 )2 = = 2.127 × 10–4 M 1− α 0.633

59. (b) : 4Zn + 10HNO3

4Zn(NO3)2 + NH4NO3 + 3H2O 60. (b) : Aromaticity can be predicted by the use of Huckle’s rule which says that (4n + 2) p-electrons are required in delocalisation system to give it aromaticity. 61. (d) 62. (c) (dilute)

49. (a) : ρ = Z=

Z× M N0 × a

3

or Z = ρ × N0 × a

3

M

23

2.7 × 6.023 × 10 × ( 405 × 10 −10 )3 =4 27.0

i.e., number of atoms per unit cell is 4. Hence, unit cell is face-centred type. 50. (d) : Al2O3 + 3C + 3Cl2

1000°C

2AlCl3 + 3CO

51. (d) : C6H5COC(CH3)3 does not contain an a-hydrogen and hence does not show tautomerism. 52. (d) : Due to the +ve charge on P, it attracts the electrons of the P—H bond towards itself. As a result, it has some ionic character. In other words, the P—H bond in PH6+ is least covalent. 53. (d) : 3HClO(aq) HClO3(aq) + 2HCl(aq) It is a disproportionation reaction of hypochlorous acid where the oxidation number of Cl changes from +1 (in ClO–) to +5 (in ClO3–) and –1 (in Cl–). 54. (a)

Kc = 1

Kc = 2

On making square,

… (i)

…(ii)

[SO3 ] 1 = K c [SO2 ][O2 ]1/2 1

2

 1  [SO3 ]2 = Kc   = 2 2  K c1  [SO2 ] [O2 ] \ 56. (b)

[By Eq. (ii)]

2

 1  Kc =  = 44.44 mol −1 L 2  0.15 

57. (c) : Sulphate ion is present outside the coordination sphere so it can form white ppt. of BaSO4 with BaCl2(aq). 58. (c) : α = 30

Λm Λ°m

=

184.5 = 0.367 502.4

65. (a)

66. (b) : Sodium sulphide is soluble in water. The solubility product (and hence solubility) of ZnS is larger than that of CuS. 67. (b) : Reduction in presence of Zn-Hg and conc. HCl is useful for aldehyde and ketone but carboxylic acid group remains unaffected. 68. (a) : DH = (Ea)f – (Ea)b = 0 69. (b) : Since sp-hybridized carbon is more electronegative than a sp2-hybridized carbon which in turn is more electronegative than sp3-hybridized sp2

[SO3 ]2

By reversing Eq. (i),

i.e., when n = 3; E decreases nine times.

acid than CH2 CH COOH which in turn, is a

2SO3(g)

[SO2 ]2[O2 ]

n2

sp

[SO2 ][O2 ]1/2 = 0.15 [SO3 ]

For 2SO2(g) + O2(g)

64. (d)

1

carbon, therefore, CH C COOH is a stronger

1 SO2(g) + O2(g) 2

55. (d) : For SO3(g)

63. (d) : E ∝

sp3

stronger acid than CH3 CH2 COOH . Thus, the overall order of acid strength is (i) > (ii) > (iii). 70. (c) : Since ionization potential of hydrogen atom is 13.6 eV. \ E1 = – 13.6 eV Now, En − E1 = −13.6 n2

−13.6 n2

− ( −13.6) = 12.1

+ 13.6 = 12.1 \ n = 3

After absorbing 12.1 eV the electron in H atom is excited to 3rd shell. Thus, possible transitions are 3 i.e., 3 → 2, 2 → 1 and 3 → 1. 71. (a) 72. (b) : Given : Weight of benzoic acid = 1.89 g; Temperature of bomb calorimeter = 25°C = 298 K; Mass of water (m) = 18.94 kg = 18940 g; Increase in temperature (t) = 0.632°C and specific heat of water (s) = 0.998 cal/g-deg. We know that heat gained by water or heat liberated by benzoic acid (Q) = msDt = 18940 × 0.998 × 0.632 = 11946.14 cal …Contd. on Page no. 76

physics for you | april ‘15

Page 30

paper-1 Section-1

one or More Than one options correct Type This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE THAN ONE are correct.

1. A small charged bead can slide on a circular frictionless, insulating wire frame. A point like dipole is fixed at the centre of circle, dipole moment  is p. Initially the bead is on the plane of symmetry of the dipole. Bead is released from rest. Ignore the effect of gravity. Select the correct options. 

p

r

Charged bead mass = m charge = Q

Dipole

(a) Magnitude of velocity of bead as function of its Qp cos q . angular position is 2 pe mr 2 0

(b) Normal force exerted by the string on bead is zero. (c) If the wire frame were not present bead executes circular motion and returns to initial point after tracing a complete circle. (d) Bead would move along a circular path until it reaches the opposite its starting position and then executes periodic motion. 2. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power of the same rate. The wavelength lB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.0 mm. If the temperature of A is 5802 K,

(a) (b) (c) (d)

the temperature of B is 1934 K lB = 1.5 mm the temperature of B is 11604 K lB = 2.0 mm

3. The two balls A and B as shown in figure are of masses m and 2m, respectively.

u m A

2m B

The ball A moves with velocity u toward right while B is at rest. The wall at extreme right is fixed. Coefficient of restitution for collision between two 1 balls is and between either ball and wall is 1. Then 2 speeds of A and B after all possible collisions are u u (a) vA = (b) vA = 2 4 u u (c) vB = (d) vB = 8 4 4. Consider two satellites A and B B r of equal mass m, moving same A circular orbit about earth, but in Earth opposite sense as shown in figure. The orbital radius is r. The satellites undergoes 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 GMm system just before collision is − . r (b) The total energy of the two satellites plus earth 2GMm system just after collision is − . r (c) The total energy of the two satellites plus earth GMm system just after collision is − . 2r (d) The combined mass (two satellites) will fall towards the earth just after collision. physics for you | april ‘15

31

 5. A charged particle with velocity v = xi + y j moves  in a magnetic field B = yi + x j. The force acting on the particle has magnitude F. Which one of the following statements is/are correct? (a) No force will act on charged particle if x = y. (b) If x > y, F ∝(x2 – y2). (c) If x > y, the force will act along z-axis. (d) If y > x, the force will act along y-axis. 6. One mole of an ideal gas is carried through a thermodynamic cycle as shown in the figure. The cycle consists of an isochoric, an isothermal and an adiabatic processes. The adiabatic exponent of the gas is g. Choose the correct option(s). Pressure (P) 3P0

C B

P0 P0 /2

A VC V0

Volume (V)

ln 6 ln 5 (b) g = (a) g = ln 3 ln 3 (c) BC is adiabatic (d) AC is adiabatic  7. The torque τ 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   dL (a) is perpendicular to L at all instants 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. 8. A metallic sphere of radius r remote from all other bodies is irradiated with a radiation wavelength l which is capable of causing photoelectric effect. Mark out the correct statement(s). (a) The maximum potential gained by the sphere will be independent of its radius. (b) The net positive charge appearing on the sphere after a long time will depend on the radius of the sphere. (c) The kinetic energy of the most energetic electrons emanating from the sphere will keep declining with time. (d) The kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the sphere. 32

physics for you | april ‘15

9. When a charge of amount Q is given to an isolated metal plate X of surface area A, its surface charge density becomes s1. When an isolated identical plate Y is brought close to X the surface charge density on X becomes s2. When Y is earthed the surface charge density on X becomes s3. Then Q Q (a) s1 = (b) s1 = 2A A Q (c) s1 = s2 (d) s 3 = A 10. An inductor of inductance 2 mH is connected across a charged capacitor of capacitance 5 mF and the resulting LC circuit is set oscillating at its natural frequency. Let Q denote the instantaneous charge on the capacitor and I the current in the circuit. It is found that the maximum value of Q is 200 mC. (a) When Q = 100 mC, then the value of dI = 104 A s–1. dt (b) When Q = 200 mC, then the value of I = 2 A. (c) Maximum value of I = 3 A. (d) When I is equal to one-half the maximum value then the value of |Q| is 173 mC. Section-2

one integer Value correct Type This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

11. Two identical steel cubes each of mass 50 g and side 1 cm collide head-on face to face with a speed of 10 cm s–1 each. The maximum compression of each cube is n × 10–7 m, then find n. [Young’s modulus for steel = 2 × 1011 N m–2] 12. The internal energy of monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. The temperature rises through 2°C. The distance moved by the piston is given as a × 10b m in scientific notation. Find the value of a + b. 25 [Take R = in SI unit, atmospheric pressure 3 = 100 kPa] 13. A coil of inductance L = 2 mH and resistance R = 1 W is connected to a source of constant emf e = 3 V as shown in the figure.

L, R

R0 S

(r = 0)

14.

15.

16.

17.

A resistance R0 = 2 W is connected in parallel with the coil. Find the amount of heat generated (in mJ) after the switch S is disconnected. A cylindrical wooden float whose base area is 4 m2 and height 1 m drifts on the water surface in vertical position. Density of wood is 500 kg m–3 and that of water is 1000 kg m–3. What minimum work (in kJ) must be performed to take the float out the water ? Light of wavelength 627 nm illuminates two slits. The minimum path difference between the waves from the slits for the resultant intensity to fall to 25% of the central maximum is (200 + n)nm. Find the value of n. Consider two point masses m and lm located at points, x = a and x = ma respectively. Assuming that the sum of the two masses is constant, what is the value of l for which the magnitude of the gravitational force is maximum? A X-ray tube is working at a potential difference of 20 kV. The potential difference is decreased to 10 kV. It is found that difference of the wavelength of Ka X-ray and the most energetic continuous X-ray becomes 4 times of the difference prior to the change of voltage. The atomic number of the target element is 11x. Find the value of x.

18. Find the ratio of the fundamental tone frequencies of two identical strings after one of them was stretched by n1 = 2% and the other by n2 = 9%. Also the tension is assumed to be proportional to the elongation. 19. Electromagnetic radiation whose electric component varies with time as E = C1(C2 + C3 coswt) cosw0t, here C1, C2 and C3 are constants, is incident on lithium and liberates photoelectrons. If the kinetic energy of most energetic electrons be 2.6 eV, the work function of lithium is (in eV). [Take : w0 = 2.4p × 1015 rad s–1 and w = 0.8p × 1014 rad s–1, Plank’s constant h = 6.6 × 10–34 MKS] 20. A thin rod of mass m and length 2l is placed horizontal and perpendicular to a horizontal rough nail, as shown in figure and set free. The point of contact of the rod with the nail is l/3 distance away from the centre of rod. If the rod starts slipping when it forms an angle q with the horizontal and the coefficient of friction of rod with nail is m, then m . find tan q

paper-2 Section-1

only one option correct Type This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONlY ONE is correct.

1. What is the minimum attainable pressure of the gas in the process T = T0 + aV2, where T0 and a are the positive constants, and V is the volume of one mole? (a) 2Ra T0 (b) 2R aT0 (c) R 2aT0

(d) 0

2. The differential equation of charging of a capacitor is as given below : 1 dq + K 2q = K 3 K1 dt The time constant and steady state charge are respectively K 1 1 and K 3 and 3 (a) (b) K1 K1 K 2 K2 (c)

K2 and K 2 K 3 K1

(d)

K 1 and 3 K1 K 2 K1

3. A chain of mass M and length l is suspended vertically with its lower end touching a weighing scale. The chain is released and falls freely onto the scale. Neglecting the size of the individual links, what is the reading of the scale when a length x of the chain has fallen? (a)

Mgx l

(b)

2Mgx l

3Mgx 4Mgx (d) l l 4. A point mass m is suspended at the end of massless wire of length L and cross-sectional area A. If Y is the Young’s modulus of elasticity of the material of wire. The frequency of simple harmonic motion along the vertical line is (c)

(a)

1 A 2p mLY

(b)

1 YA2 2 p mL3

(c)

1 YA 2p mL

(d)

1 YL 2p m

physics for you | april ‘15

33

smooth

(a) 26 J (c) 52 J

4 kg a = 5.2 m s–2 12 kg

(b) 12 J (d) 48 J

7. What should be the minimum value of refractive index of a prism of refractive angle A, so that there is no emergent ray irrespective of angle of incidence ?

S a

(a) I

(b)

9I (d) 2 I 10 10. Inside a long straight uniform wire of round crosssection there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from latter by a distance d. If a direct  current of density J flows along the wire, then magnetic field inside the cavity will be   1 (a) 0 (b) m0 J × d 2     3 (c) m0 J × d (d) m0 J × d 2 Section-2

comprehension Type (only one option correct) This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Paragraph for questions 11 and 12 A thin ring of radius R metres is placed in x-y plane such that its centre lies on origin. The half ring in region x < 0 carries uniform linear charge density +lC m–1 and the remaining half ring in region x > 0 carries uniform linear charge density –lC m–1. y

++++ – ––––– ++ –– + +

34

physics for you | april ‘15

++ + ++

9. A point source of light S is placed in front of a perfect reflecting mirror as shown in the figure. X is a screen. The intensity at the centre of screen is found to be I. If the mirror is removed, then the intensity at the centre of screen would be

+

x

– +–

–––

–– –

bg + g a2 + b2

– – – – – ––

ag + g a2 + b2 (d)

x

–––

(c)

(b) ag

10 I 9

––

(a) bg

X

(c)

A (a) sin 2

A (b) cos 2 A A (c) cosec (d) sec 2 2 8. A particle is projected from the origin in such a way that it passes through a given point P(a, b). The minimum required speed to do so is

a

++ ++++ +++++ ++ + +

5. A curved rectangular bar forms a resistor. The curved sides are concentric circular arcs. If r is the resistivity of the material of bar, l0 is the length of inner arc of radius r0, (r0 + b) is the radius of the outer arc, and a is the width of the bar. The electric resistance of the bar across its rectangular ends is rl0 rl0 (a) (b) ar0  b ar0 ln 1 +   r0  2rl0 (c) (d) None of these  b ar0 ln 1 +   r0  6. A 4 kg block is placed on top of a long 12 kg block, which is accelerating along a smooth horizontal table at a = 5.2 m s–2 under application of an external constant force. Let minimum coefficient of friction between the two blocks which will prevent the 4 kg block from sliding is m0 and coefficient of friction between blocks is only half of this minimum value (i.e., m0/2). Find the amount of heat generated due to sliding between the two blocks during the time in which 12 kg block moves 10 m starting from rest.

––

y

11. The electric potential (in volts) at point P whose R coordinates are (0 m, + m) is 2 1 l (a) (b) 0 4 pe0 2 1 l (c) (d) cannot be determined 4 pe0 4

physics for you | april ‘15

35

12. The dipole moment of the ring in C m is (a) −(2 pR2 l)i (b) (2 pR2 l)i

2 (c) −(4 R l)i (d) (4 R2 l)i Paragraph for questions 13 and 14

A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M at a distance L from the top surface is as shown in the figure. The atmospheric pressure is P0.

(c)

(a) (c)

14. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is  P pR2 − Mg   2P pR2  0 (a)  (2L) (b)  0 2  (2L)  2  pR P0   pR P0 + Mg 

 P pR2   P pR2 + Mg  0 0 ( 2 ) L (d)  (2L)  2  2  pR P0   pR P0 − Mg 

(c) 

Paragraph for questions 15 and 16 A ray of light goes from point A in a medium where the speed of light is v1 to a point B in a medium where the speed of light is v2 as shown in the figure. The path of the rays in the two is shown in figure.

v1 a X

l i

P v2

36

1 i

O

Q

r

physics for you | april ‘15

v2a sin r v1b sin r

(d)

a sec i b sec r + v1 v2

16. The slope of i – r curve is

13. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be P0 (a) P0 (b) 2 P0 Mg P0 Mg + − (c) (d) 2 2 pR 2 pR2

A

15. The time taken for the light to go from the point A to the point B in the figure is a sin i b sin r (a) (b) v1 v2

r

b B

X 2

−b cos2 i 2

a cos r a sin2 r 2

b sin i

(b)

a cos r b cos i

(d)

−a tan i b tan r

Section-3

Matching List Type (only one option correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (a), (b), (c) and (d), out of which one is correct.

17. In case of motion of a particle of mass m in a vertical circle of radius r with the help of a string, velocity at lowest point is v0. Assume the vertical distance of any said point from lowest point to be h. Match the Column I and Column II. Column I P. Q.

Tension in the 1. string at any point Minimum value of 2. v0 for looping the loop

R.

Difference of 3. tensions at the lowest and the highest points

S.

Difference of 4. tensions at the lowest and the highest points in case the motion is uniform

Code : (a) P - 4, Q - 3, R - 2, S - 1 (b) P - 1, Q - 3, R - 4, S - 2 (c) P - 2, Q - 3, R - 1, S - 4 (d) P - 3, Q - 2, R - 4, S - 1

Column II 2 mg 6 mg

5gr

m 2 (v − 3 gh + gr ) r 0

18. Match the quantities given in Column-I with their values given in Column-II. P.

Column I orbital radius in 1. Bohr model

Column II

nth

Q.

nth orbital speed in 2. Bohr model

−me 4

R.

nth orbital total 3. energy in Bohr model

S.

nth orbital potential 4. energy in Bohr model

e2 2 ∈0 nh 8 ∈20 n2h2 −me 4

S.

Speed of image of bird 4. relative to the fish looking downwards in the mirror

Code : (a) P - 4, Q - 1, R - 2, S - 3 (b) P - 1, Q - 2, R - 4, S - 3 (c) P - 3, Q - 1, R - 2, S - 4 (d) P - 3, Q - 4, R - 1, S - 2 20. A network consisting of three resistors, three batteries, and a capacitor is shown in figure. A

4 ∈20 n2h2 ∈0 n2h2 pme

19. A bird in air is diving vertically over a tank with speed 6 cm s–1. Base of tank is silvered. A fish in the tank is rising upward along the same line with speed 4 cm s–1. Water level is falling at rate of 2 cm s–1. (Take : m (water) = 4/3)

B

C

3 5 F

10 V

5

2

Code : (a) P - 4, Q - 1, R - 2, S - 3 (b) P - 1, Q - 3, R - 4, S - 2 (c) P - 2, Q - 3, R - 1, S - 4 (d) P - 3, Q - 2, R - 4, S - 1

3 cm s–1

8V

E

F

12 V

4 D

Column I Current in branch EB is Current in branch CB is Current in branch AF is Current in branch EC is

P. Q. R. S. Code : (a) P - 4, Q - 1, R - 2, S - 3 (b) P - 2, Q - 1, R - 4, S - 3 (c) P - 2, Q - 3, R - 1, S - 4 (d) P - 3, Q - 2, R - 1, S - 4

Column II 1. 2 A 2. 1.5 A 3. 0.5 A 4. 0 A

SolutionS

PaPer-1 1. (a,b,d) : Applying energy conservation principle, increase in KE of charged bead + decrease in electrostatic potential energy = 0

P.

Q.

R.

Column I Column II Speed of the image of 1. 12 cm s–1 fish as seen by the bird directly Speed of the image of fish 2. 4 cm s–1 formed after reflection from the mirror as seen by the bird Speed of image of bird 3. 9 cm s–1 relative to the fish looking upwards

 p cos q  1 2 mv + Q  =0 2 2  4 pe0r  or

v=

−2Qp cos q 4 pe0mr

p   ≤ q ≤ p 2 

2

Circular motion of bead requires a centripetal force. ∂V 2 p cos q Er = − = ∂r 4 pe0r 3 mv 2 r 4 p ∈0 r 3 thus wire frame does not exert any force on the bead \ QEr =

2 pQ cos q

=

physics for you | april ‘15

37

to sustain circular motion. Bead will reach the point opposite its starting position and then repeatedly retrace its path executing a periodic motion. 4

2. (a,b) : Radiant power of body A = eAsTAA 4

Radiant power of body B = eBsTB A The two powers are equal. \

eB sTB4 A = e A sTA4 A

or

e  4  0.01  4 TB4 =  A  TA4 or TB =   × (5802)  0 81 .  eB  4

4 5802 1 TB4 =   × (5802)4 or TB = = 1934 K 3 3 \ Option (a) is correct.

or

According to Wien’s displacement law, lT = constant \ lATA = lBTB l T l 1934 1 = or l B = 3lA or A = B or A = lB TA lB 5802 3 Given : lB – lA = 1.0 × 10–6 m or 3lA – lA = 10–6 or 2lA = 10–6 or lA = 0.5 × 10–6 m \ lB = 3 × 0.5 × 10–6 = 1.5 × 10–6 m Hence option (b) is correct. 3. (a,d) : After 1st collision between the 2 balls v1 m

2m

v2

mv1 +2mv2 = mu ...(i) v2 − v1 u 1 = − ⇒ v2 − v1 = and ...(ii) 0−u 2 2 u From (i) and (ii), v2 = , v1 = 0 2 After collision of ball B with wall, direction of velocity is interchanged only. Finally after collision of balls v2 v1 u mv1′ + 2mv2′ = 2m × 2m m 2 ⇒ v1′ + 2v2′ = u ...(iii) v′ − v′ 1 Also 1 2 = − u 2 0− 2 u ⇒ v1′ − v2′ = ...(iv) 4 From (iii) and (iv), u u v2′ = , v1′ = 4 2 \ Speeds of A and B after all possible collisions, u u u vA = 0, ; vB = , 2 2 4 38

physics for you | april ‘15

4. (a,b,d) : Just before collision, the total energy of two satellites is, GMm GMm GMm E=− − =− 2r 2r r Let orbital velocity be v, then from momentum conservation, mv – mv = 2m × v1 ⇒ v1 = 0 As velocity of combined mass just after collision is zero, the combined mass will fall towards earth. At this instant, the total energy of the system only consists of the gravitational potential energy given GM × 2m by U = − r    5. (a, b, c) : If x = y, then v || B, i.e., F = 0.  F = q(x 2 − y 2 )k If x > y, the force is along z-axis. 6. (a, d) : For isothermal process PV = constant and for adiabatic process PV g = constant where g > 1. First we assume that BC is isothermal and CA is adiabatic. V ⇒ P0V0 = 3P0VC ⇒ VC = 0 [for process BC] 3 g P0 g  V0  ln 6 and 3P0   = V0 ⇒ g = [for process CA]  3  2 ln 3 Now, we assume that process BC is adiabatic and CA is isothermal. PV V 3P0VC = 0 0 ⇒ VC = 0 [for process CA] 2 6 g

V  and 3P0  0  = P0V0g ⇒ 6 g = 3 [for process BC]  6  ln 3 ⇒ g= < 1[not possible] ln 6 ln 6 Hence, process AC is adiabatic and g = ln 3     dL   7. (a,b,c) : τ = A × L i.e., = A×L dt  dL This relation implies that is perpendicular to dt   A and L. Therefore, option (a) is correct.   L ⋅ L = L2 Differentiating with respect to time, we get     dL dL   dL dL dL L ⋅ + ⋅ L = 2L ⇒ 2 L. = 2L dt dt dt dt dt    dL  dL Since L ⊥ so, L. =0 dt dt

dL = 0 and hence L does not change with dt time. So option (c) is correct. Since L is not changing with time, therefore it is  the case when direction of L is changing but its   magnitude is constant and τ is perpendicular to L at all points.

=

Therefore

8. (a,b,c,d) : Maximum potential will be equal to the stopping potential which depends on l and nature of material. kQ rV As V = ⇒Q = r k Since V and k are constant, maximum positive charge appearing depends on r. As the sphere gets charged (which goes on increasing), it applies a force on the emanating electrons thus reduces the velocity of emanating electrons. Initially the sphere is uncharged, thus KEmax of emanating electron is independent of radius of sphere. Q 9. (b, c, d) : Initially charge Q will be + + + + distributed uniformly on plate X of + + + + surface area A, a plate has 2 surfaces. + + + + Q + + = s \ 1 2A X As plate Y is brought closer, the charge is induced on it but there is no effect on the plate X. Q Charge on outer surface of X = 2 ⇒ distribution of charges on X is

s2 =

Q = s1 2A

+ + + Q/2 ++ + +

Q/2

X

+ + + + Q/2 + + +

– – – – – – –

Y

+Q/2 + + + + + +

When Y is earthed, the new charge distribution + – will be as shown in the figure + – + – Q X + – Y \ s3 = + – A + – + –

10. (a, d) : The charge stored

L = 2 mH

in the capacitor oscillates simple harmonically. Here Q0 = maximum value of charge = 200 mC = 2 × 10–4 C 1 and w = LC

C = 5 F

1 (2 × 10

−3

H)(5 × 10

−6

F)

= 104 s −1

Let at t = 0, Q = Q0 then, Q(t) = Q0 cos(wt) ⇒

I (t ) =

dQ = −Q0 w sin(wt ) and dt

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

dI (t ) ...(iii) = −Q0 w2 cos(wt ) dt Q When Q = 100 mC = 0 2 1 (from eqn. (i)) cos(wt ) = , 2 dI 1 = (2 × 10−4 C)(104 s −1 )2   \ 2 dt ⇒

dI = 104 A s −1 dt

When Q = 200 mC = Q0, cos(wt) = 1, i.e., wt = 0, 2p... At this time, I(t) = –Q0wsin(wt) ⇒ I(t) = 0 (sin0° = sin2p = 0) I(t) = –Q0wsin(wt) The maximum value of I is Q0w ⇒ Imax = Q0w = (2 × 10–4 C)(104 s–1) = 2 A (d) From energy conservation, 1 2 1 1 Q2 LI max = LI 2 + 2 2 2 C

⇒

Q = LC(I 2max − I 2 )

I For I = max = 1 A, we get 2

Q = (2 × 10−3 )(5 × 10−6 )(22 − 12 )

⇒

Q = 3 × 10−4 C =1.732 × 10–4 C ≈173 mC

F DL where A is the =Y A L surface area and L is length of the side of the cube. If k is spring or compression constant, then F = kDL A \ k = Y = YL L 1 2 −4 Initial KE = 2 × mv = 5 × 10 J 2 1 2 Final PE = 2 × k(DL) 2

11. (5) : From Hooke’s law :

\

DL =

KE KE 5 × 10−4 = = = 5 × 10−7 m 11 k YL 2 × 10 × 0.01 physics for you | april ‘15

39

12. (1) : The change in internal energy of the gas is 25 DU = 1.5nR(DT) = 1.5 × 1 × × 2 = 25 J 3 The heat given to the gas = 42 J The work done by the gas is DW = DQ – DU = 42 J – 25 J =17 J If the distance moved by the piston is x, the work done is DW = (100 kPa)(8.5 cm2)x Thus, (105 N m–2 )(8.5 × 10–4 m2)x = 17 J or x = 0.2 m = 2 × 10–1m 13. (3) : Initially, after a steady current is set up, the current, flowing is as shown. e e In steady condition I20 = , I10 = R R0 When the switch is R L disconnected, the current through R0 I2 changes from I10 to R0 I I1 the right to I20 to the left.  (The current in the inductance cannot change suddenly). We then have the equation, dI L 2 + (R + R0 )I2 = 0 dt This equation has the solution I2 = I20e −t (R + R0 )/ L The heat dissipated in the coil is, ∞

∞

0

0

2 Q = ∫ I22 Rdt = I20 R ∫ e −2t (R + R0 )/ L dt 2 = RI20 ×

L Le2 = = 3 mJ 2(R + R0 ) 2R(R + R0 )

W = mgh − =

mg = FB

a=0 h

⇒

dSH × g = rSh × g ⇒ h =

dH r

Now when force F is applied, for minimum work a=0 ( for a = 0, F is minimum) F x

f 15. (9) : As I = I0 cos2   , 2 I Here I = 0 4 f p 2p f 1 \ cos   = ⇒ = ⇒ f = 2 2 2 3 3 or

a=0

W = ∫ Fdx = ∫ (mg − rSxg )dx = mg ∫ dx − rSg ∫ xdx physics for you | april ‘15

Dx 2p 627 × 2p = ⇒ Dx = nm l 3 3 Dx = 209 nm = (200 + 9)nm

16. (1) : Gravitational force, F =

Gm1m2 r2

Here, m1 = m, m2 = lm, r = ma – a = a(m – 1) Also, m1 + m2 = constant (C) C m + lm = C; m = . (1 + l) Gm2 l GC 2 l So, F = = ⋅ 2 2 2 2 a (m − 1) a (m − 1) (1 + l)2 dF GC 2 (1 + l)2 × 1 − l × 2(1 + l) = ⋅ d l a2 (m − 1)2 (1 + l)4 For F to be maximum,

dF =0 dl

\ (1 + l)2 – 2l(1 + l) = 0 or (1 + l)(1 + l – 2l) = 0 or (1 + l)(1 – l) = 0 \ l=1 4

17. (5) : l K = a

2

and l min =

( l ≠ −1)

hc eV

3R(Z − 1) As per question, l min(10 kV) − l K = 4 (l min(20 kV) − l K ) a

F – mg + rSxg = 0; F = mg – rSxg

40

2

rSgh2 rSg  dH  Sgd 2 H 2 = = 2 2  r  2r

Here S = 4 m2, H = 1 m, d = 500 kg m–3, and r = 1000 kg m–3 Putting these values, we get W = 5 kJ

14. (5) : Applying Newton’s law in vertical direction initially

rSgh2 rSgh2 = (rSh) gh − 2 2

⇒

hc

4 4 1  20 − 10  = 3 ×  e × 10  3R(Z − 1)2 3

On solving, we get Z = 55 \ x=5

a

18. (2) : Frequency of stretched wire is given by u=

1 T 1 Tl = 2l m 2l m

Angular acceleration of rod at an angle q N

m   m = l 

Here, For the first wire, new length l1 = l + n1l new tension, T1 = an1 l New frequency of first wire (an1l )(l + n1l ) 1 u1 = 2(l + n1l ) m

Fr mg

a=

l mg   cos q 3 (4 / 9)ml 2

Similarly, u2 = \

(an2l )(l + n2l ) 1 2(l + n2l ) m

⇒

n2 (1 + n1 ) 0.09 (1 + 0.02) = = 2.05 n1 (1 + n2 ) 0.02 (1 + 0.09)

≈2 19. (4) : The given wave is superposition of 3 waves with frequency, w0, w + w0, w0 – w wmax = (w0 + w) \

Emax = humax = h

(w0 + w) 2p

f= =

6.6 × 10 2p

N=

T = T0 + a 15

(2.4 p + 0.8 p) × 10 1.6 × 10

3mg cos q 4 F m Now, m = r = 2 tan q; = 2. N tan q ⇒

1. (b) : T = T0 +

h (w + w) − KEmax 2p 0 −34

3g cos q 4 l

−19

eV − 2.6 eV

or, a

R2T 2

R2T 2 P2

a V2

P2

(as, V = RT/ P for one mole of gas)

= T − T0 or, P 2 =

aR2T 2 T − T0

or, P = a RT (T − T0 )−1/2

= 4 eV 20. (2) : Till the rod does not slip, the centre of mass of the rod is in circular motion around the nail. Thus from conservation of energy l 1 mg sin q = I w2 3 2

Condition for minimum pressure,

4ml 2 ml 2 4ml 2 + = 12 9 9

\

thus w =

3g sin q 2l

1 or, (T − T0 ) − T = 0. 2 ...(i)

...(i)

After differentiating, we get, dP 1   = a R (T − T0 )−1/2 − T (T − T0 )−3/2  dT 2  

where I =

l 2 Now, m w = Fr − mg sin q 3 3 or Fr = mg sin q 2

...(ii)

PaPer-2

Now, humax = KEmax + f ⇒

a=

and linear acceleration of centre of mass l From pure rolling, a =   a 3 g thus a = cos q 4 Also mg cosq – N = ma

u2 (1 + n1 ) n2 (1 + n2 ) = u1 (1 + n2 ) n1 (1 + n1 ) =



dP =0 dT

1   aR (T − T0 )−1/2 − T (T − T0 )−3/2  = 0 2  

or, T = 2T0 From (i) and (ii), we get,

...(ii)

Pmin = a R.2T0 (2T0 − T0 )−1/2 = 2R aT0 physics for you | april ‘15

41

2. (b) : Consider the circuit q e − − IR = 0 C and I = ⇒ ⇒ ⇒

dq dt

+q –q R

C 

I

dq ⋅ RC + q dt dq RC + q = eC dt

eC =

...(i)

−t / RC ) q = eC(1 − e −t /RC ) = q0 (1 − e

Comparing eqn. (i) with we get τ =

1 dq ⋅ + K 2 q = K 3, K1 dt

K 1 , q0 = 3 K1 K 2 K2

\

}

M . l The descending part of the chain

3. (c) : Mass per unit length, l =

x

is in free fall, also its every point has descended by a distance x. So, speed of each point, v = 2 gx Assume a very small distance dx falls in a short internal of time dt. Normal exerted on the falling part, dp −(0 − (ldx )v ) N =− x = dt dt

F

mg

42

physics for you | april ‘15

x

YA = 2 pu or mL

w=

u=

1 YA 2 p mL

5. (b) : Let us consider an elemental portion of the resistor. The element considered is a circular arc of radius r and thickness dr. The resistance of this element would be r × rq adr here l0 = r0q dR =

dx

v



l0

rr × l0 ar0 × dr

So, dR =

= lv2 = l(2gx) = 2lgx Normal due to x part of the chain on the weighing machine, N′ = lgx Reading of the scale W = N + N′ = 3lgx 3Mgx = l 4. (c) : Let l be the increase in the length of the wire due to the force F = mg. Then F l Stress = , Strain = A L Stress mgL YlA Y= = \ F = mg = Strain lA L

m

This force is acting upward in the equilibrium state. If the mass is pulled down a little through a distance x, such that the total extension in the string is (l + x), then force in wire acting upwards is YA(l + x ) = L and downward force is F = mg. The restoring force is the net downward force. Restoring force YA YAl YA YAx = mg − (l + x ) = − (l + x ) = − L L L L Force YAx \ Acceleration of the mass = =− mass mL

r0

a

b

 O

If we divide the entire resistor in these elemental portions, then these elemental resistors are joined in parallel, equivalent resistance of which is given by r +b 1 1 r0 +b ar0dr ar0 =∫ = ∫ = × ln 0 R dR r rl0r rl0 r0 0

⇒

R=

rl0

 b ar0 ln 1 +   r0  6. (c) : First assume that blocks have common acceleration, for both blocks to move together acceleration of 4 kg block must be 5.2 m s–2. f

4 kg 12 kg

f

a = 5.2 m s–2

a = 5.2 m s–2 \ f = 4 × 5.2 kg m s–2 or m0mg = m(5.2 m s–2) m0 = 0.52 1 If m = (0.52) = 0.26, the acceleration of 4 kg block 2 due to friction is a1 = mg = 2.6 m s–2

As there is relative motion between blocks 1 Srel = urel t + arel t 2 2

arel = a1 – a = 2.6 – 5.2 = –2.6 m s–2, 1 2 \ Srel = (−2.6)t 2 Time of motion can be determined from motion of lower block. 1 For 12 kg block, S = (5.2)t 2 = 10 m (given) 2 20 ⇒ t= s 5. 2 \ Srel = –5 m Work done by friction is given by Wf = mmgSrel = 0.26 × 4 × 10 × (–5) = –52 J \ Heat generated = 52 J 7. (c) : If the ray just emerges from face AC, e = 90° and r2 = C ...(i) From Snell’s law at face AB, we have 1sini = nsinr1 ...(ii) and A = r1 + r2 = r1 + C ...(iii) From eqn. (ii) n is minimum when r1 is maximum, i.e., r1 = C. In this case i = 90°. A

i

r1 r2 = C

90°

\

umin = bg + g a2 + b2

9. (c) : Let the power of light source be P, then intensity at any point on the screen is due to light rays directly received from source and that due to after reflection from the mirror P P ⇒ I= + 2 4 pa 4 p × (3a)2 =

P

P  10   1 1 + 9  =   4 pa 4 pa2  9  2

When mirror is taken away, P 9I I1 = = 2 10 4 pa     10. (b) : From Ampere’s law, ∫ B ⋅ dl = m0 ∫ J ⋅ ds Inside the conductor at a distance r from its axis B(2pr) = m0 J(pr2)  1   1 B = m0 Jr or B = m0 J × r 2 2 Now, conductor has cavity so we can assume that current in it is the superposition of positive current and negative current (in place of cavity). M

From eqn. (iii), A = 2C or C = A/2 1 A 1 From eqn. (ii), sin C = ; sin = n 2 n A \ n = cosec 2

O d

8. (d) : Equation of trajectory of a particle is y = x tan a −

gx 2

2u2 cos2 a

If projectile passes through the point(a, b),

ga2tan2a

or u ≥ bg + g a2 + b2

C

B

b = a tan a −

Thus Discriminant ≥ 0 i.e., 4a2u4 – 4ga2(ga2 + 2bu2) ≥ 0 or u4 – 2gbu2 – g2a2 ≥ 0 or u4 – 2gbu2 + b2g2 ≥ b2g2 + a2g2 or (u2 – bg)2 ≥ (b2 + a2)g2

ga2

2

2

2u cos a

= a tan a −

2au2tana

(ga2

ga2

2u

2

(1 + tan2 a)

or – + + 2bu2) = 0 This quadratic equation in tan a must give real roots for a particle to pass through (a, b).

P

Required magnetic field at point M  BN = Magnetic field at M due to whole conductor – Magnetic field at M due to cavity shaped conductor    1   1 BN = m0 ( J × OM ) − m0 ( J × PM ) 2 2 1    1   = m0 J × (OM − PM ) = m0 J × d . 2 2 11. (b) : Consider two small elements of ring having charges +dq and –dq symmetrically located about y-axis. physics for you | april ‘15

43

Mg + P(pR2) = P0pR2 Also, P0(2LpR2) = P(xpR2)

y

d

– – – – – ––

x



–––

d

\

––

++ ++++ +++++ ++ + +

+dq ++++ ––––– –dq –– ++ – ++

+

x

–

–––

–– –

+–

––

 P pR 2  0 x=  ( 2 L) 2  P0 pR − Mg 

15. (d) : Total time taken = time taken (t1) from A to O in medium 1 + time taken (t2) from O to B in medium 2. A a

++ + ++

y

The potential due to this pair at any point on y-axis is zero. The sum of potential due to all such possible pairs is zero at all points on y-axis. Hence  R potential at P  0,  is zero.  2 12. (c) : Consider two small elements of ring having charge + dq and –dq as shown in figure.

P

++ ++++ +++++ ++ + +

– – – – – ––

+–

–––

––

++ + ++

The pair constitutes a dipole moment. dp = dq 2R The net dipole moment of system is vector sum of dipole moments of all such pairs of elementary charges. By symmetry the resultant dipole moment is along negative x-direction. \ Net dipole moment + p /2  p = − ∫ (dp cos q)i − p /2 + p /2

= − ∫ (2 lR2 cos qdq)i = −4 R2 li − p /2

14. (d) Let the pressure of the trapped air at the equilibrium be P. Draw the forces on the piston. For the equilibrium of the piston, 44

AO a sec i = v1 v1

t2 =

BO b sec r = v2 v2 a sec i b sec r + v1 v2

–

y

13. (a)

t1 =

t = t1 + t2 =

x

–– –

+dq

–––

+

––

x

physics for you | april ‘15

Q

B

++++ – ––––– –dq ++ –– + + d

O r

b

y



i

16. (a) : To optimize time,

dt =0 dr

⇒

d  a sec i b sec r  + =0 dr  v1 v2 

⇒

a sec i tan i di b sec r tan r + =0 v1 dr v2

⇒

v1 di b sec r tan r =− × dr v2 a sec i tan i b v sin r cos2 i =− × 1 × × a v2 sin i cos2 r m b m b cos2 i cos2 i =− ⋅ 2 × 1 × =− a m1 m2 cos2 r a cos2 r

17. (a) : P → 4, Q → 3, R → 2, S → 1 18. (a) : P → 4, Q → 1, R → 2, S → 3 19. (d) : P → 3, Q → 4, R → 1, S → 2 20. (c) : P → 2, Q → 3, R → 1, S → 4



physics for you | april ‘15

45

BRAIN

WAVES

Wave Motion 2n æt xö y = A sin2p ç + ÷ = A sin (vt + x) l èT l ø

s1 =

v =u 2L

xö Ït f = 2p ç + ÷ + f 0 èT l ø nv 2L Df =

2n Dt. T

Df =

2n Dx. l

u=

s1 =

Stationary Waves

dy 2nA æt xö = cos 2p ç + ÷ dt T èT l ø

C h 3 r

i + v=

2n A = wA T

u0 =

du 4n 2 æt xö a= = - 2 A sin2p ç + ÷ dt èT l ø T

a0 =

4n 2

A = w2A

T2

v;

T , m

I=

§;

Ğ r

§=

i r

v=

i x⁄ ~ = r

v=

k ps xp = r

v ù u = nê ú ë 2(l + 1.2 r) û v ù u = nê ú ë 4(l + 0.6 r) û

eP , r

s=

nv n T = 2L 2L m

s0 =

v 1 T = , 2L 2L m

1 u 2 A2 k 2p2 k 2 2 P0Av P02 = A v = = 2 v v 2k 2r v

1 u 2 A2T = 2p2mvA2 u2 2 v

Stationary Waves

Stationary Waves

P . r

1 u 2 A2T = 4 v

Pav =

v =u 4L

v×

1 r

v2 p1 = v1 r2

v×

T

Ï 1ö n+ ÷ 2ø T 1 ö v çè æ u = çn + ÷ = 2 2 L 2L m è ø

vt T 273 ) t = = v0 T0 273 s0 =

Stationary Waves

y = 2 A sin

x = 0,

x=

2n x 2pt cos l T

j 3l , l, .... 2 2

j 3l 5l , , ... 4 4 4

u¢ =

v Ûv m ± v o ´u v ± v m }v s

u¢ =

v Ûv o ´u v }v s

v 1 T = 4L 4L m

Target

AIPMT 1. A car starts from station and moves along the horizontal road by a machine delivering constant power. The distance covered by the car in time t is proportional to (a) t2 (b) t3/2 (c) t2/3 (d) t3

5. If the position vector of a particle is given by :  r = (4 cos 2 t ) i + (4 sin 2 t )j + (6t ) k m, its acceleration at t = p/4 s in m s–2 is

2. The dimensions of a rectangular block measured with callipers having least count of 0.01 cm are 5 mm × 10 mm × 5 mm. The maximum percentage error in the measurement of the volume of the block is (a) 5% (b) 10% (c) 15% (d) 20%

6. Three identical blocks, each having a mass m are pushed by a force F on a frictionless table as shown in figure. (i) What is the acceleration of the blocks?

3. The figure shows the velocity (v) of a particle plotted against time (t). B

10 0

O D

C T 2T

t (s)

10 A

Mark out the incorrect statement. (a) The displacement of the particle in time 2T is zero. (b) The initial and final speeds of the particle are the same. (c) The acceleration of the particle remains constant throughout the motion. (d) The particle does not change its direction of motion. 4. Two stones having different masses m1 and m2 are projected at angles q and (90° – q) with same velocity from the same point. The ratio of their maximum heights is (a) 1 : 1 (b) 1 : tan q (c) tan q : 1 (d) tan2q : 1 48

Practice Paper

Physics For you | april ‘15

(a) 16 i

F

(b) −16 k (c) −16 j (d) 16 (i + j)

A

B

C

(ii) What is the net force on the block A? (iii) What force does A apply on B? (iv) What force does B apply on C? (i) (ii) (iii) (iv) (a) F/3m F/3 2F/3 F/3 (b) F/4m F/3 2F/3 F/3 (c) F/3m F/4 2F/3 F/3 (d) F/3m F/3 2F/4 F/3 7. A large plane sheet of charge having surface charge density 5.0 × 10–16 C m–2 lies in the X-Y plane. The electric flux through a circular area of radius 0.1 m, if the normal to the circular area makes an angle of 60° with the Z-axis is [Given : e0 = 8.85 × 10–12 C2 N–1 m–2.] (a) 4.44 × 10–7 N m2 C–1 (b) 3.33 × 10–7 N m2 C–1 (c) 2.22 × 10–7 N m2 C–1 (d) 1.11 × 10–7 N m2 C–1 8. A 4 mF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 mF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation? (a) 2.67 × 10–2 J (b) 3.67 × 10–2 J –4 (c) 2.67 × 10 J (d) 5.67 × 10–4 J

9. The equivalent resistance of the arrangement of resistances shown in the given figure between points P and Q is 8 16 

20 

16  P

9 18 

(a) 6 W

(b) 8 W

Q 6

(c) 24 W (d) 16 W

10. A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, then new value of the magnetic field is (a) B (b) 2B (c) 4B (d) B/2

15. Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulum will be one behind the other is (a) T/4 (b) T/3 (c) 4T/3 (d) 4T 16. A denotes the cross-sectional area of a cubical tank, h the depth of an orifice of area of cross-section a, below the liquid surface. The velocity of the liquid flowing through the orifice is (a)

2gh

(b)

 A2  2 gh  2 2   A −a 

(c)

 A  2gh  A−a

(d)

 A2 − a 2  2 gh   A2 

11. A Carnot engine operates with a source at 500 K and sink at 375 K. Engine consumes 600 kcal of heat per cycle. The heat rejected to sink per cycle is (a) 250 kcal (b) 350 kcal (c) 450 kcal (d) 550 kcal

17. The electric potential V at any point (x, y, z) in space is given by V = 4x2 V. The electric field at the point (1 m, 0, 2 m) in V m–1 is

12. A wire of cross section 4 mm2 is stretched by 0.1 mm by a certain weight. How far will the wire of same material and length but of area 8 mm2 stretch under the action of same force? (a) 0.05 mm (b) 0.10 mm (c) 0.15 mm (d) 0.20 mm

18. A convex lens of focal length f is placed somewhere in between an object and a screen. The distance between object and screen is x. If numerical value of magnification produced by lens is m, focal length of lens is mx mx (a) (b) 2 (m −1)2 (m +1)

13. A light particle moving horizontally with a speed of 12 m s–1 strikes a very heavy block moving in the same direction at 10 m s–1. The collision is one dimensional and elastic. After the collision, the particle will 12 m s–1

10 m s–1

(a) move at 2 m s–1 in its original direction (b) move at 8 m s–1 in its original direction (c) move at 8 m s–1 opposite to its original direction (d) move at 12 m s–1 opposite to its original direction. 14. A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is (a) 32000 km (b) 36000 km (c) 3400 km (d) 4800 km

(a) − 8i

(b) + 8i

(c) −16i (d) 16 k

(m −1)2 (m +1)2 x (c) (d) x m m 19. A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents are respectively (a) 40 A, 16 A (b) 16 A, 40 A (c) 20 A, 40 A (d) 40 A, 20 A 20. The coefficient of friction between the ground and the wheels of a car moving on a horizontal road is 0.5. If the car starts from rest, what is the minimum distance in which it can acquire a speed of 72 km h–1 without slipping? (Take g = 10 m s–2). (a) 10 m (b) 20 m (c) 30 m (d) 40 m 21. A block of mass 2 kg initially at rest is dropped from a height of 1 m into a vertical spring having force constant 490 N m–1. The maximum distance through which the spring will be compressed, is (a) 0.11 m (b) 0.33 m (c) 0.22 m (d) 0.44 m Physics For you | april ‘15

49

22. A body of mass m is raised to a height h from the surface of the earth where the acceleration due to gravity is g. If R is the radius of the earth and h n1), draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the lens maker’s formula.  25. (a) An electric dipole of dipole moment p consists of point charges +q and –q separated by a distance 2a apart.  Deduce the expression for the electric field E due to the dipole at a distance x from the centre of the dipole on its axial line  in terms of the dipole moment p . Hence show  that in the limit x >> a, E 3 p /(4pe0x3).

y (b) Given the electric field in  the region E = 2 xi , find the electric flux through the cube and the charge x a enclosed by it. z OR (a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility. (b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point Q charge is placed at its centre C and an other 2

68

charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (i) the force on the charge at the centre of shell and the point A, (ii) the electric flux through the shell. 26. (a) State ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius ‘r’, having ‘n’ turns per unit length and carrying steady current I. (b) An observer to the left of a solenoid of N turns each of cross section area ‘A’ observes that a study current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA. A

N

OR (a) Define mutual inductance and write its S.I. units. (b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other. (c) In an experiment, two coils C1 and C2 are placed close to each other. Find out the expression for emf induced in the coil C1 due to a change in the current through the coil C2. solution 1. According to Gauss’s law, net flux through a closed

surface,   q φE =  ∫ E ⋅ ds = εen0 Total charge enclosed, qen = 0 as net charge on dipole is zero. 0 \ fE = =0 ε0

2. Focal length of a concave lens is negative.

Using lens maker’s formula,  1 1  µl 1  = − 1  −  f  µm   R1 R2 

Here, μl = 1.5, μm = 1.65 µ µ  Also, l < 1, so  l − 1 is negative and focal µm µ  m  length of the given lens becomes positive. Hence, it behaves as a convex lens.

physics for you | april ‘15

Page 68

where m is the mass of the electron, and v, its speed in an orbit of radius r. Bohr’s quantisation condition for angular momentum is nh nh L = mvr = or r = ...(iii) 2π 2π mv From equation (ii) and (iii), we get

3. Side bands are produced during the modulation.

When a message signal is super imposed on a carrier wave then there exists the sum and difference of the f – f f f + f c m c c m two frequencies of different waves. These are called side bands. In amplitude modulation, side bands are : Lower side band frequency = fc – fm Upper side band frequency = fc + fm 4. (i) Region DE has negative resistance property because current decreases with increase in voltage or slope of DE is negative. (ii) Region BC obeys Ohm’s law because current varies linearly with the voltage. 5. Capacitor reactance is meaningful for an a.c. electrical circuit. It is the resistance offered by a capacitor when it is connected to an electrical circuit. Mathematically, XC = 1 2πυC Where u = frequency of a.c. source C = capacitance of the capacitor. Ohm (W) is the SI unit of capacitor reactance. 6. According to Bohr’s theory, a hydrogen atom consists of a nucleus with a positive charge Ze, and a single electron of charge –e, which revolves around it in a circular orbit of radius r. Here Z is the atomic number and for hydrogen Z =1. The +Ze r e–, m electrostatic force of Nucleus attraction between the hydrogen nucleus and the electron is F=

k e.e

k e2

= r2 r2 To keep the electron in its orbit, the centripetal force on the electron must be equal to the electrostatic attraction. Therefore, mv 2 ke 2 = r r2

ke 2 or mv = r 2

or r =

ke 2 mv 2

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

ke 2

nh 2 π ke 2 or v = ...(iv) mv 2 2 π mv nh Substituting this value of v in equation (iii), we get r=

=

nh nh n2h2 ; ⇒ r= . 2 πm 2 π ke 2 4π2mke 2

∴ r ∝ n2

7. Intrinsic Semiconductors

Extrinsic Semiconductors

1.

These are pure semiconducting tetravalent crystals.

2.

Their electrical conductivity is low. There is no permitted energy state between valence and conduction bands.

These are semiconducting tetravalent crystals doped with impurity atoms of group III or V. Their electrical conductivity is high. There is permitted energy state of the impurity atom between valence and conduction bands.

3.

8. From mirror formula,

1 1 1 = − v f u

Now for a concave mirror, f < 0 and for an object on the left of the mirror, u < 0 1 1 1 > > 2f u f

\

2f < u < f or

or

−

or

1 1 1 1 1 1 1 1 or < >p). On illumination, let the excess electrons and holes generated be Dn and Dp, respectively: n′ = n + Dn p′ = p + Dp. Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carrier concentrations when there is no illumination. Remember Dn = Dp and n >>p. Hence, the fractional change in the majority carriers (i.e., Dn/n) would be much less than that in the minority carriers (i.e., Dp/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity.

12.5 10 7.5 5 2.5 0

Base current (IB)

60 A 50 A 40 A 30 A 20 A 10 A

1 1.5 2 2.5 3 3.5 4 0.5 Collector to emitter voltage (VCE) in volts

From output characteristic, find the ratio of change in collector current for the corresponding change in the base current. It gives the current amplification factor. λ λ 17. (a) : Angular width, q = or d = θ d –7 Here, l = 600 nm = 6 × 10 m 0. 1 × π π q = 0.1° = rad = rad 180 1800 d=? 6 × 10−7 × 1800 \ d= = 3.44 × 10–4 m π (b) : Frequency of a light depends on its source only. So, the frequencies of reflected and refracted light will be same as that of incident light. Reflected light is in the same medium (air) so its wavelength remains same as 500 Å.

physics for you | april ‘15

Page 72

λ Wavelength of refracted light, lr = µw μw = refractive index of water. So, wavelength of refracted wave will be decreased. 18. Inductive reactance, XL = wL

fo = 15 m 



h Eye lens

Objective lens

Height of object

L

E

~

Impedance of the circuit, X L2 + R2 = ω2 L2 + R2 (i) When the number of turns in a inductor coil decreases then its inductance L decreases. So, the net impedance of the circuit decreases and current through the bulb (circuit) increases. Hence brightness (I2R) of bulb increases. (ii) When an iron rod is inserted in the inductor, then its inductance L increases. So, Z will increase and current through the bulb will decrease. Hence, brightness of the bulb will decrease. (iii) A capacitor is connected in the series in the circuit, so its impedance, Z=

2 2 Z = ( X L − XC ) + R Z=R

3.48 × 106 m 2 Rm = 1.74 × 106 m Distance of object = Radius of lunar orbit R0 = 3.8 × 108 cm Distance of image for objective lens is the focal length of objective lens, fo = 15 m Radius of image of moon by objective lens can be calculated. R h tan θ = m = R0 fo i.e., Radius of moon Rm =

R

(Q XL = XC)

This is the case of resonance so maximum current will flow through the circuit. Hence brightness of the bulb will increase. 19. (a) : Microwave are suitable for radar systems used in aircraft navigation. These wave are produced by special vacuum tubes, namely klystrons, magnetrons and Gunn diodes. (b) Infrared waves are used to treat muscular pain. These waves are produced by hot bodies and molecules. (c) X-ray are used as a diagnostic tool in medicine. These are produced when high energy electrons are stopped suddenly on a metal of high atomic number. 20. (i) : Here, fo = 15 m = 1500 cm and fe = 1.0 cm angular magnification by the telescope in normal adjustment 1500 cm f m= o = = 1500 fe 1.0 cm (ii) The image of the moon by the objective lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length.

h=

Rm × fo R0

=

1.74 × 106 × 15 3.8 × 108

h = 6.87 × 10–2 m Diameter of the image of moon, DI = 2h = 13.74 × 10–2 m = 13.74 cm 21. Einstein’s photoelectric equation 1 Kmax = mv2 = hu – f0 = hu – hu0 …(i) 2 Here, Kmax = Maximum Kinetic energy of photoelectron with speed v u = frequency of incident light f0 (= hu0) = work function of the metal Important features of photoelectric equation. (i) Kmax depends linearly on u and is independent of intensity of incident light. This happen due quantum nature of light. (ii) Since Kmax must be positive, equation implies that photoelectric emission is possible only if hu > f0 = hu0 or u > u0 Thus there exists a threshold frequency u0 (= f0/h) for the metal surface, below which no photoelectric emissions possible. From eqn. (i) hc Kmax = – f0 λ According to question, Kmax = 2Kmax =

hc – f0 λ1

hc – f0 λ2 physics for you | april ‘15

…(ii) …(iii) 73

From eqn. (ii) and (iii),  hc  hc − φ0 2  − φ0  =  λ2  λ1 f0 = Also, or

2hc hc 1  2 − = hc  −  λ1 λ2  λ1 λ2 

hc 1  2 = hc  −  f0 = hc \ λ0  λ1 λ2  λ0 λ1λ2 1 2 λ2 − λ1 = ; λ0 = 2 λ2 − λ1 λ0 λ1λ2

22. Trajectory of a particles in coulomb field of target

nucleus. Nucleus Only a small fraction of  > 90° the number Incident + of incident -particles a-particles (1 in  < 90° 8000) rebound back. This shows that the number of a-particles undergoing head-on collision is small. This implies that the entire positive charge of the atom is concentrated in a small volume. So, this experiment is an important way to determine an upper limit on the size of nucleus. mass of nucleus Density of nucleus = volume A × 1 amu ρ= 4 3 πR 3 where R = R0 A1/3 A × 1 amu 1 amu Density ρ = = 4 3 4 3 πR0 A πR 3 3 0 r = 2.97 × 1017 kg m–3 so, nuclear density is constant irrespective of mass number or size. OR Nuclear fission : It is the process in which a heavy nucleus (A > 230) when excited gets split up into two smaller nuclei of nearly comparable masses. For example, 235 1 141 92 1 U + 0n Ba + 36Kr + 30n + Q 92 56 Nuclear fusion : It is the process of fusion of two

smaller nuclei into a heavier nucleus with the liberation of large amount of energy. For example, 2 2 H + 1H 1 2 3 1H + 1H

74

4 He + 24 2 4 2He + n

MeV

Mass deffect, Dm = (2.014102 + 3.016049) – (4.002603 + 1.008665) = 0.018883 Energy released, Q = 0.018883 × 931.5 MeV = 17.589 MeV 23. (i) : Transformer is a device which is used to bring the high voltage down to low voltage of a.c. current. It works on the principle of mutual induction of two coils in a transformer. (ii) Transformer does not work for dc voltage. A dc current gives constant magnetic field and constant magnetic flux through the coil of fixed area of cross-section. As there is no change in magnetic flux so there is no induced emf in the coil. (iii) The value displayed by the students are gaining knowledge and curiosity to learn new things. The values displayed by the teacher are providing good education and helpful. 24. (a) Waves diffract when they encounter obstacles. Applying Huygens principle it becomes clear. A wavefront impinging on a barrier with a slit in it, only the points on the wavefront that move into the slit can continue emitting forward moving waves but because a lot of the wavefront has been blocked by the barrier, the points on the edges of the hole emit waves that bend round the edges.

Before the wavefront strikes the barrier the wavefront generates another forward moving wavefront. Once the barrier blocks most of the wavefront the forward moving wavefront bends around the slit because the secondary waves they would need to interfere with to create a straight wavefront have been blocked by the barrier. Each point on the wavefront moving through the slit acts like a point source. We can think about some of the effect of this if we analyse what happens when two point sources are close together and emit wavefronts with the same wavelength and frequency. These two point sources represent the point sources on the two edges of the slit and we can call the source A and source B as shown in the figure.

physics for you | april ‘15

Page 74

Each point source emits wavefronts from the edge of the slit. In the diagram we show a series of wavefronts emitted from each point source. The continuous lines show peaks in the waves emitted by the point sources and the dotted lines represent troughs. We label the places where constructive interference (peak meets a peak or trough meets a trough) takes place with a solid diamond and places where destructive interference (trough meets a peak) takes place with a hollow diamond. When the wavefronts hit a barrier there will be places on the barrier where constructive interference takes place and places where destructive interference happens.

A

B

The measurable effect of the constructive or destructive interference at a barrier depends on what type of waves we are dealing with. (b) Refer point 6.14 (4 (ii)), (MTG Excel in Physics) (c) On increasing the value of n, the part of slit contributing to the maximum decreases. Hence, the maximum becomes weaker. OR (a) Refer point 6.5 (5 (ii) and (iii)), (MTG Excel in Physics) (b) Refer point 6.6 (1), (MTG Excel in Physics) 25. (a) Refer point 1.4 (3), (MTG Excel in Physics)  (b) E = 2 xi So flux passes through y faces of cube which are perpendicular to x-axis. The magnitude of electric field at the x a left face (x = 0), z EL = 0 The magnitude of electric field at the right face, (x = a), ER = 2a   So, net flux, f = E ⋅ ∆s = EL Ds cos 180° + ERDs cos0° = 0 + 2a × a2 = 2a3 Assume enclosed charge is q. q Use Gauss’s law, f = ; q = e0f ε0

q = 2a3e0 OR (a) When a conductor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such \

a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static situation is achieved i.e., when the two fields cancel each other and the net electrostatic field in the conductor becomes zero. Dielectrics are non-conducting substances i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by reorienting molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produce a field that opposes the external field, unlike a conductor in an external electric field. However, the opposing field so induced does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric. The effect of electric field on a conductor and a dielectric is shown below : E0

� + E0 � + � � free E free++ E0 � in + � � E0 + Ein = 0++

E0

� �

�

�

E0

+

+ + E0 Ein + � E0 + Ein  0+

Conductor

Dielectric

The dipole moment per unit volume is called polarisation and is denoted by P. For linear isotropic dielectrics, P = cE where c is electric susceptibility of the dielectric medium. (b) The electric field inside a spherical conducting shell is zero. Q So force experienced by the charge at the 2 point C. Q F= E=Q×0=0 2 Q Q 2

C

x

A 2Q

Force experienced by charge 2Q at point A,  Q  FA = 2Q E A = 2Q  4πε x 2  0

26. (a) Refer point 3.2 ((1), (4)), (MTG Excel in Physics) OR (a) Refer point 4.2 (2(i)), (MTG Excel in Physics) (b) Refer point 4.2 (2(vi)), (MTG Excel in Physics) (c) Refer point 4.2 (2(ii)), (MTG Excel in Physics)  physics for you | april ‘15

75

…Contd. from Page no. 30

Since 1.89 g of acid liberates 11946.14 cal of heat, therefore heat liberated by 122 g (mol. wt. of benzoic acid) of acid = 73. (c)

11946.14 × 122 = 771126.5 cal = 771.12 kcal 1.89

74. (a) : 2C6H5CHO Benzaldehyde

alc. KCN 

C6H5CH(OH)COC6H5 Benzoin

The reaction is known as benzoin condensation. 75. (a) : The cell is given as Zn|Zn2+(0.1 M)||Pb2+(0.1 M)|Pb \ E°cell = E°right – E°left = – 0.126 – (– 0.763) V = 0.637 V Now, using Nernst equation 2+ Ecell = E°cell – 0.059 V log [Pb 2 + ]

2

Ecell = 0.637 V – \

[Zn ]

0.059 V [0.1] log 2 [0.1]

Ecell = 0.637 V

76. (b) : Compound which can stabilise the charge after removal of proton, will be more acidic.

77. (a) :

2+

X = Cu

KCN

quickly decomposes

Cu(CN)2 Yellow ppt.

[Cu(CN)4]3–

in excess KCN

CuCN + (CN)2

Colourless soluble complex H2S passed

Y = Cd

Cyanogen

No ppt.

2+ KCN

Cd(CN)2

in excess KCN

White ppt. H2S passed Yellow ppt.

CdS

3+

Z = Fe

KCN

Fe(CN)3

2–

[Cd(CN)4]

Colourless soluble complex

in excess KCN

Reddish-brown ppt.

K3[Fe(CN)6]

Yellow solution

78. (a) : 3rd period elements have more DHeg than 2nd period elements and in a period it increases. 79. (c)

80. (a)

81. (d) : sin2 x + sin2 y – sin2 z = sin2 x + sin (y + z) sin (y – z) = sin2 x + sin (y + z) sin (π – x) = sin x [sin x + sin (y + z)] = sin x [sin (π + z – y) + sin (y + z)] = sin x [sin (y + z) – sin (z – y)] = 2 sin x cos z sin y 76

82. (c) : Let R = {(x, y) : x + 2y = 8, x, y ∈ N}

\

x + 2y = 8 (which must be a natural number) 8−x \ x = {2, 4, 6} \ y = {3, 2, 1} ⇒ y = 2 \ R ={(x, y) : x + 2y = 8} ⇒ R = {(2, 3), (4, 2), (6, 1)}\Range of R = {1, 2, 3}

83. (b) : A = {1, 2, 3, 4}, B = {1, 3, 5}

R = {(a, b) : a < b, a ∈A, b ∈A} = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} \ R–1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} Now (3, 1) ∈ R–1, (1, 3) ∈ R \ (3, 3) ∈ RoR–1.

(3, 1) ∈ R–1, (1, 5) ∈ R ⇒ (3, 5) ∈ RoR–1 (5, 1) ∈ R–1, (1, 3) ∈ R ⇒ (5, 3) ∈ RoR–1 (3, 2) ∈ R–1, (2, 3) ∈ R ⇒ (3, 3) ∈ RoR–1 (3, 2) ∈ R–1, (2, 5) ∈ R ⇒ (3, 5) ∈ RoR–1 (5, 2) ∈ R–1, (2, 3) ∈ R ⇒ (5, 3) ∈ RoR–1 (5, 2) ∈ R–1, (2, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 (5, 3) ∈ R–1, (3, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 (5, 4) ∈ R–1, (4, 5) ∈ R ⇒ (5, 5) ∈ RoR–1 \ RoR–1 = {(3, 3), (3, 5), (5, 3), (5, 5)}. 84. (c) : Making the equation of the curve homogeneous with the help of the line, we get  hx + ky  x 2 + y 2 − 2(kx + hy)   2hk  2  hx + ky  + (h2 + k 2 − c 2 )  = 0  2hk  or 4h2k2x2 + 4h2k2y2 – 4hk2x(hx + ky) – 4h2ky(hx + ky) + (h2 + k2 – c2)(h2x2 + k2y2 + 2hkxy) = 0 This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if coefficient of x2 + coefficient of y2 = 0. (h2 + k2)(h2 + k2 – c2) = 0 since [h2 + k2 ≠ 0] ⇒ h2 + k2 = c2. 85. (d) : The centre of the circle is (1, 1) and radius = 2 2. From (a, a) must lie outside the circle, so 2a2 – 4a – 6 > 0 ⇒ a < – 1 or a > 3 θ 2 2 . Now, tan = 2 2a2 − 4a − 6

physics for you | april ‘15

Page 76

π π θ π As

⇒ a2 − 2a − 3 < 2 3 2 3 2a − 4a − 6 2 ∴ a − 2a − 15 < 0 ⇒ −3 < a < 5 ∴ a ∈ (−3, −1) ∪ (3, 5). sin 9θ cos 9θ sin θ cos θ sin 3θ cos 3θ 86. (b) : β = + + cos 3θ cos θ cos 9θ cos 3θ cos 27θ ⋅ cos 9θ sin 6θ sin 18θ sin 2θ + + ∴ 2β = cos 3θ cos θ cos 9θ cos 3θ cos 27θ ⋅ cos 9θ sin(3θ − θ) sin(9θ − 3θ) sin(27θ − 9θ) = + + cos 3θ cos θ cos 9θ cos 3θ cos 27θ cos 9θ = tan 3q – tanq) + (tan 9q – tan3q) + tan27q – tan9q) = tan27q – tanq = a \ a = 2B. 2 87. (a) : f (x ) = cos x cos(x + 2) − cos (x + 1) = cos(x + 1 – 1)cos(x + 1 + 1) – cos2(x + 1) = cos2(x + 1) – sin21 – cos2(x + 1) by using cos2A – sin2B = cos(A + B) ⋅ cos(A – B) f ( x) = − sin 2 1

(constant)

π ⇒ x = ⇒ f (x) = − sin2 1 2 ⇒ f(x) represents a straight line through π 2   , − sin 1 which is parallel to x-axis as 2 f(x) = – sin21 is a constant. 88. (d) : Each coupon can be selected in 15 ways. The total number of ways of choosing 7 coupons is 157. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8. 97 − 87  3  7  8  7 =  =  Desired probability is  5   15  157 89. (c) : The total numbers of arrangements is 11! 11! = 2 !2 !2 ! 8 The number of arrangements in which C, E, H, I, S 11! 11 6 ! = appear in that order =  5    2 !2 !2 ! 8 ⋅ 5! 11! 11! 1 1 ∴ Required Probability = ÷ = = 8 ⋅ 5 ! 8 5 ! 120 90. (c) :

30

∑ n(Ai ) = 5 × 30 = 150 i =1

Suppose S has m elements ⇒ 150 = 10m ⇒ m = 15 n

∑ n(Bi ) = 3n = 9m ⇒ n = 3m = 45 i =1

91. (d) : 50C6 – 5C1 40C6 + 5C2 30C6 – 5C3 20C6 + 5C4 10C6 = coefficient of x6 in [5C0 (1 + x)50 – 5C1(1 + x)40 + 5C2(1 + x)30 – 5C3(1 + x)20 + 5C4(1 + x)10 – 5C5(1 + x)0] = coefficient x6 in [(1 + x)10 – 1]5 = 5C1 · (10C1)4 (10C2) = 2250000 3 92. (b) : y = 2 x +1 ⇒ x2 = 3 − y ...(i) y When x = 0, y = 3; x = ∞, y = 0 x2 = 2 ⇒ x = 2

1 1 2(x2 + 1) = 3 ⇒ x 2 = ⇒ x = 2 2 1    1  − 0 ⋅2 Area =  2 −  ⋅1 +    2 2 1 2 3 ⇒ Definite integral = + = 2 2 2 93. (c) 94. (a) : f (x + 4) = f (x + 2) – f (x) ⇒ f (x + 6) = f (x + 4) – f (x + 2) = f (x + 2) – f (x) – f (x + 2) = – f (x) f (x + 12) = f (x) λ +12

∫

1

95. (b) :

f (x)dx =

λ

12

∫ f (x)dx 0

sin xdt

∫ sin2 x + (t − cos x)2

−1

1

sin x  t − cos x  = tan −1   sin x  −1 sin x x x   = tan −1  tan  + tan −1  cot    2 2 x π 0 0 ⇒ y > 2 dy 1 =