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Physics

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Problems in

Physics DC Pandey

ARIHANT PRAKASHAN (Series), MEERUT

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Arihant Prakashan (Series), Meerut All Rights Reserved

© Author No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon. All disputes subject to Meerut (UP) jurisdiction only.

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For further information about the books published by Arihant log on to www.arihantbooks.com or email to [email protected]

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PREFACE The primary objective of preparing this book, Problems in Physics, is to help students develop a command over the basic concepts and principles of Physics through problem solving. With more than a decade of distinguished career in teaching Physics to JEE aspirants, I can firmly say that the best way to grapple with the concepts of the Physics is to learn with the problems themselves. The problems included in the book with their solutions, aim to give you the mastery required for solving the intricate problems asked in the exams such as JEE. Critical analysis of the situation is required whatever be format of the questions. This book just gives emphasis on this, and it is hoped that it will give a boost to all the meritorious and hard-working students. I am thankful to my student, Vineet Jain for the help in editing of this book. I would also like to thank Ankit Kapoor and Bhaskar Sharma for their valuable suggestions. Above all, I shall like to thank Mr Yogesh Chand Jain, Chairman, Arihant Group, for bringing the book in this nice form.

Suggestions for further improvement of the book are welcome.

DC Pandey This book is dedicated to my honourable grandfather

(LATE) SH. PITAMBER PANDEY a Kumaoni poet, and a resident of village Dhaura (Almora), Uttarakhand

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CONTENTS 1. Kinematics

3

2. Laws of Motion

11

3. Work, Power and Energy

18

4. Centre of Mass, Conservation of Momentum, Collision & Impulse

21

5. Rotation

28

6. Gravitation

37

7. Simple Harmonic Motion

39

8. Solids and Fluids

45

9. Waves

52

10. Heat and Thermodynamics

57

11. Optics

67

12. Current Electricity

80

13. Electrostatics

85

14. Magnetics 92 15. Electromagnetic Induction 16. Modern Physics

98 108

Answers

117

Solutions

131

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CHAPTER

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1

Problem 1.

KINEMATICS The coordinates of a particle moving in a plane are given by

x = 4 cos 6t and y = 6 sin 6t (a) find the equation of the path of the particle →

→

(b) find the angle between position vector r and velocity vector v at time t = π/ 12 (c) prove that the acceleration of the particle is always directed toward a fixed point. Problem 2. A particle is projected vertically upwards with speed u in a medium in which the resistance to motion is proportional to the square of the speed. Initially net force on the particle is two times its weight downwards. Find the time of ascent and distance ascended by the particle.

A car starts from rest with an acceleration of 6 m/s 2 which decreases to zero linearly with time, in 10 second, after which the car continues at a constant speed. Find the time required for the car to travel 400 m from the start. Problem 3.

Problem 4. A charged particle of mass m and charge q is moved from rest at x = 0 by an electric field → E = (E 0 – αx ) i$ where α is a positive constant and x is the displacement of the particle in time t along x-axis. Find the distance moved by the particle when it again is brought to rest and the acceleration of the particle at that instant. Describe the motion of the particle. → Hint : Force on a charge ‘q’ placed in an electric field E is given by → → Fe = qE

Problem 5. A particle of mass m moving in a straight line is acted upon by a force F which varies with its velocity as F = – kv n . Here k is a constant. For what values of n the average value of velocity of the 1 ortime, 2) till it stops, is one third the initial velocity. ( particle averaged over n ≠ the Problem 6. A lift of total mass M kg is raised by cables from rest to rest through a height h. The greatest tension which the cables can safely bear is nMg newtons. Find the shortest interval of time in which the ascent can be made (n > 1). Problem 7. A particle moves in a straight line with constant acceleration ‘a’. The displacements of particle from origin in times t1 , t 2 and t 3 are s 1 , s 2 and s 3 respectively. If times are in A.P. with common ( s1 – s 3 )2 . difference d and displacements are in G.P. Then prove that a = d2 Problem 8. A ball of mass 2 kg is dropped from a height of 80 m on a floor. At each collision with the floor the ball loses half of its speed. Plot the velocity-time, speed-time, and kinetic energy-time graphs of its motion till first two collisions with the floor (Take g = 10 m/s 2 ).

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4

Problems in Physics

Problem 9. The acceleration-displacement graph of a particle moving in a straight line is as shown alongside. Initial velocity of particle is zero. Find the velocity of the particle when displacement of the particle is, s = 12 m.

a(m/s2) 4 2 8

2

10

s(m)

12

Problem 10. An aircraft flies with constant airspeed (speed of aircraft in still air) 200 km/hr from position A to position B, which is 100 km north-east of A and then flies back to A.Throughout the whole flight the wind velocity is 60 km/hr from the west. Find total time of flight from A to B and back. Problem 11. A motor boat going downstream overcame a raft at point A. After one hour it turned back and after some time it met the raft again at a distance 6 km from point A. Find the river velocity. Problem 12. A particle A is projected with an initial velocity of 60 m/s at an angle 30° to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of 60 m/s 50 m/s 50 m/s from a point at a distance of 100 m from A. If the particles collide in air, find the angle of projection α of B, time α 30° B when collision occurs and the distance of point P from A, where A collision takes place. 100 m (g = 10 m/s 2 )

A particle is projected with speed v at an angle θ to the horizontal from the point x = 0, y = 0. If x and y-axes are horizontal and vertically upwards respectively and point of projection is the lowest point on the inner surface of a bowl formed by rotating the curve x 2 = 4 ay, where a is a Problem 13.

positive constant. Show that the particle strikes the bowl at a horizontal distance x =

4 av 2 tan θ v 2 + 2ag + 2ag tan 2 θ

A particle is released from a certain height H = 400 m. Due to the wind the particle gathers the horizontal velocity component v x = ay where a = 5 s –1 and y is the vertical displacement of Problem 14.

the particle from point of release, then find : (a) the horizontal drift of the particle when it strikes the ground (b) the speed with which the particle strikes the ground (Take g = 10 m/s 2 ) A particle of mass m is attached by a light inextensible string of length 2R to a fixed point O. When vertically below O at point A, the particle is given a horizontal velocity u. When the string becomes horizontal, it hits a small smooth nail C, at a distance R from O and the particle continues to rotate about C. Find the minimum value of u so that the particle just describes complete circle abou Problem 15.

O

R

C

R

B

2R

A

u

Problem 16. A particle moves in a vertical circle. Its velocity at topmost point is half of its velocity at bottommost point. Find the magnitude of acceleration of the particle at the moment when its velocity is directed vertically upwards. (g = 10 m/s 2 )

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5

Problems

A uniform electric field of strength 10 6 N/C is directed vertically downwards. A particle of mass 0.01 kg and charge 10 –6 C is suspended by an inextensible thread of length 1 m. The particle is displaced slightly from its mean position and released. Calculate the time period of its oscillation. What minimum velocity should be given to the particle at bottom so that it completes a full circle. Calculate the maximum and minimum tensions in the thread. (g = 9.8 m/s 2 ) Problem 17.

Hint :

→ → → → Electrostatic force Fe on a charge q in electric field E is Fe = qE

A particle moves along a straight line and its velocity depends on time as v = 3t − t 2 . Here v is in m/s and t in second. Find: (a) average velocity and (b) average speed for first five seconds. Problem 18.

Problem 19. At the initial moment three points A, B and C are on a horizontal straight line at equal distances from one another. Point A begins to move vertically upward with a constant velocity v and point c vertically downward without any initial velocity but with a constant acceleration a. How should point B move vertically for all the three points to be constantly on one straight line. The points begin to move simultaneously.

Two particles A and B start from positions shown in figure and move with constant speeds v and u (> v). A moves along x-axis and B moves such that its velocity is always aimed at A. Let r be the distance between them and θ be the angle made by the trajectory of B with x-axis at sometime t. Prove that,

A

Problem 20.

u

x

d

+1

u

(sin θ)v r = d (1 − cos θ)u / v

B

A boat is moving in a river with a speed v w.r.t. water. The water is flowing with a speed u. At time t = 0 the boat is at the origin of a co-ordinate system with x-y axes in the horizontal plane and positive x-axis in the opposite direction of the flow of water. The boat has to reach the point P (x , y) as shown in the figure. Show that the boat has to start in a direction inclined at an angle u y sin −1 to the line joining O to P. 2 2 v x +y

Y

Problem 21.

P(x, y)

u m/s X

O Boat

Also find the time taken by the boat to reach the point P. Problem 22. A fighter plane enters inside the enemy territory, at time t = 0, with velocity v o = 250 m / s and moves horizontally with constant acceleration a = 20 m / s 2 (see figure). An enemy

tank at the border, spot the plane and fire shots at an angle θ = 60° with the horizontal and with velocity u = 600 m / s. At what altitude H of the plane it can be hit by the shot ?

v

600 m/s H θ = 60°

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6

Problems in Physics

A 3 m long arm OA rotates in a plane such that θ = 0.5 t where θ is the angle with x-axis in radian and t is in second. Problem 23.

A

2

B

A slider collar B slides along the arm in such a way that its distance from the hinge O is given by r = 3 – 0.4 t 2 where r is in meters. Find the velocity of the collar at an instant the arm has turned to θ = 30°.

r

O

θ

X

B

A

cam

°

acceleration of point B on the rod at the instant of interest as shown in figure.

Rod follower v = 5 cm/s a = 10 cm/s2

30

Problem 24. A rod follower AB is subjected to a vertical up and down movement while resting on the circular contour of radius 30 cm of a cam. The cam moves to the right with a velocity of 5 cm/s and an acceleration of 10 cm/s 2 . Find the velocity and

30 cm

Two rods of equal length are lying one along x-axis and the other along line x = y. → They intersect at origin at their mid point. The first rod moves with velocity v1 = v $j and the second with → v $ v $ velocity v 2 = i – j . Find the velocity of point of intersection of two rods. 2 2 Problem 25.

Problem 26. A particle of mass 1 kg which moves along the x-axis is subjected to an accelerating force which increases linearly with time and a retarding force which increases directly with displacement (constant of proportionality being one with proper dimensions in both the cases). At time t = 0, displacement and velocity both are zero. Find the displacement as a function of time t. Problem 27. A particle moves along the x-axis according to the equation x = A cos ωt. Find the distance travelled by the particle during the time interval t = 0 to t = t.

(i) The points A and B are moving with the same speed u in the positive direction of the x-axis and y-axis respectively. Find the magnitude of velocity relative to A of a point C, which is mid point of AB, and show that it is reverse of the velocity of C relative to B. (ii) A particle P moves on the circle x 2 + y 2 = 1 with constant speed v. Show that each instant when the acceleration of P is parallel to the line x + y = 0, the velocities of P relative to points A and B of part (i) are equal in magnitude. Find v in terms of u if the maximum value of the velocity of P relative to C is u.

Problem 28.

Problem 29. A river of width ' a ' with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes Ox and Oy are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and crosses the v river. If the boat is steered due east and u varies with x as : u = x (a − x ) 2 . a Find : (a) equation of trajectory of the boat (b) time taken to cross the river (c) absolute velocity of boatman when he reaches the opposite bank (d) the displacement of boatman when he reaches the opposite bank from the initial position.

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7

Problems Problem 30.

A river of width w is flowing such that the stream velocity varies with y as 3−1 v R = v o 1 + y w

where y is the perpendicular distance from one bank. A boat starts rowing from the bank with constant velocity v = 2 v o in such a way that it always moves along a straight line perpendicular to the banks. (a) at what time will he reach the other bank? (b) what will be the velocity of the boat along the straight line when he reaches the other bank ? v

A α

Problem 31. Two points A and B move with speeds v and 2v in two concentric circles with centre O and radii 2 rand r respectively. If the points move in the same sense and if ∠OAB = α , when the relative motion is along AB, find the value of α.

O

2v

B

r

2r

Problem 32. Two parallel straight lines are inclined to the horizon at an angle α. A particle is projected from a point midway between them so as to graze one of the lines and strikes the other at right angles. Show that if θ is the angle between the direction of projection and either of the lines, then

tan θ = ( 2 − 1) cot α A regular hexagon stands with one side on the ground and a particle projected so as to graze its four upper vertices. Show that the ratio of its maximum velocity to that of its minimum velocity is 31 . 3 Problem 33.

Problem 34. Two stones are projected simultaneously with equal speeds from a point on an inclined plane along the line of its greatest slope upwards and downwards respectively. The maximum distance between their points of striking the plane is double that of when they are projected on a horizontal ground with same speed. If one strikes the plane after two second of the other, find : (a) the angle of inclination of plane. (b) the speeds of their projection (Take g = 9.8 m/s 2 ). R

A river of width ω is flowing with a uniform velocity v. A boat starts moving from point P also with velocity v relative to the river. The direction of resultant velocity is always perpendicular to the line joining boat and the fixed point R. Point Q is on the opposite side of the river and P, Q and R are in a straight line. If PQ = QR = ω, find : (a) the trajectory of the boat, (b) the drifting of the boat and (c) the time taken by the boat to cross the river. Problem 35.

Q

P

y

AB is an inclined roof and a body is projected from origin towards the roof as shown in figure. Find ‘ h’ for which body will just touch the roof. Given : θ = α = 45° and u = 10 m / s, g = 10 m / s 2 .

A

Problem 36.

h u a q O

B

x

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8

Problems in Physics

Problem 37. A particle is projected from point G, such that it touches the points B, C, D and F of a regular hexagon of side ‘ a’. Find its horizontal range GH.

C

D

B

E

G

A

F

H

Problem 38. The benches of a gallery in a cricket stadium are 1 m high and 1 m wide. A batsman strikes the ball at a level 1 m about the ground and hits a ball. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit. y

A particle is dropped from point P at time t = 0. At the same time another particle is thrown from point O as shown in the figure and it collides with the particle P. Acceleration due to gravity is along the negative y-axes. If the two particles collide 2 sec after they start, find the initial velocity of particle which was projected from O.

P

2m

Problem 39.

10m v0 O

Problem 40. In the vertical plane shown two particles ‘ P’ and ‘ Q’ are located at points ‘ A’ and ‘ B’. At t = 0, the particle ‘ P’ is projected perpendicular to the inclined plane ‘ OA’ with velocity 2v o and simultaneously the particle Q is projected horizontally in ‘ BO’ direction. What is the necessary value of v o (in terms of ‘ d’ and θ) so that both the particles meet each other between the points ‘ O’ and ‘ B’.

P A

x

2v0

θ C

v0

O

d

Q B

3d

Y C

A projectile is launched from point ‘ A’ with the initial conditions shown in the figure. Determine the ‘ x ’ and ‘ y’ co-ordinates of the point of impact. Problem 41.

v0 = 68 m/s

A

150 m

B 30° 123 300 m

X

IIT JEE PROBLEMS Problem 42. In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data. (JEE 2004)

.5 m

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9

Problems

Problem 43. The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm 2 ) of the wire in appropriate number of significant figures. (JEE 2004) Problem 44. N-divisions on the main scale of a vernier callipers coincide with N + 1 divisions on the vernier scale. If each division on the main scale is of a units, determine the least count of the instrument. (JEE 2003) Problem 45. On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant velocity of ( 3 − 1)m/s. At a particular instant, when the line OA makes an angle of 45° with the x-axis, a ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. (a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame. 4θ (b) Find the speed of the ball with respect to the surface, if φ = . 3

An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the + X direction with an acceleration 1.5 m/s 2 . At the same instant a stone is projected from the origin with a Problem 46.

→

A

45° O x

(JEE 2002)

Y A P

1.25m

velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle →

of 45° to the horizontal. All the motions are in X-Y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s 2 .

y

u O

3.0m

X

(JEE 2000) –2

A particle of mass 10 kg is moving along the positive X-axis under the influence of k where k = 10 –2 Nm 2 . At time t = 0, it is at x = 1.0 m and its velocity is v = 0. a force F( x ) = – 2x 2 (a) find its velocity when it reaches (b) find the time at which it reaches x = 0.25 m (JEE 1998) Problem 47.

Problem 48. A large heavy box is sliding without friction down a smooth plane of inclindation θ. From a point P on the bottom of the box, a particle is projected inside the box. The inital speed of the particle with respect to the box is u and the direction of projection makes an angle α with the bottom as shown in the figure. (a) find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands (assume that the particle does not hit any other α Q surface of the box. Neglect air resistance.) P (b) if the horizontal displacement of the partcle as seen by an observeer on the ground is zero, find the speed of the box θ with respect to the ground at the instant when the particle was projected. (JEE 1998)

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10

Problems in Physics

Problem 49. A cart is moving along X-direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in Y-Z plane making an angle of 30° with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine : (i) the speed of the combined mass immediately after the collision with respect to an observer on the ground. (ii) the length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass. (g = 9.8 m/s 2 ) (JEE 1997)

Problem 50. Two guns situated on the top of a hill of height 10 m fire one shot each with the same speed 5 3 m/s at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60° with the horizontal. The shots collide in air at point P. Find (a) the time interval between the firings and (b) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane. (g = 10 m/s 2 ) (JEE 1996) Problem 51. Two towers AB and CD are situated a distance d apart as shown in figure. AB is 20 m high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of 10 m/s towards CD. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle of 60° to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other. (i) calculate the distance d between the towers, (ii) find the position where the objects hit the ground. (g = 9.8 m/s 2 ) (JEE 1994)

2m

C

60o

A m

d B

D

Problem 52. A bullet of mass M is fired with a velocity 50 m/s at an angle θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob of mass 3 M suspended by a massless 10 string of length m and gets embedded in the bob. After the collision the string moves through an 3 angle 120°. Find : (a) the angle θ (b) the vertical and horizontal coordinates of the initial position of the bob with respect to the point of firing of the bullet (g = 10 m / s 2 ). (JEE 1988)

CHAPTER

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2

LAWS OF MOTION

Problem 53. Two men of masses M and M + m start simultaneously from the ground and climb with uniform accelerations up from the free ends of a massless inextensible rope which passes over a smooth pulley at a height h from the ground. (a) Which man reaches the pulley first. (b) If the man who reaches first takes time t to reach the pulley. Find the distance of the second man from the pulley at that instant. Problem 54.

In the arrangement shown in figure, all pulleys are smooth and massless.

When the system is released from rest, accelerations of blocks 2 and 3 relative to 1 are 1 m/s 2 downwards and 5 m/s 2 downwards. Acceleration of block 3 relative to 4 is zero.

1

Find the absolute accelerations of blocks 1, 2, 3 and 4. 4

3

2

F

Problem 55.

The system shown in figure is in equilibrium. Find the force F and

C

60°

magnitude of total force on the bearing of pulley C. Each pulley is free to rotate about its bearing, and the weights of all parts are small compared with the load. (g = 10 m/s 2 ) B

A 1000 kg

Problem 56.

Determine the constraint equation which relates the

accelerations of bodies A, B and C. A

C

B

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12 Problem 57. Under the action of a force F the constant acceleration of block B is 3 m/s 2 to the right. At

Problems in Physics A

B

C

F

the instant when the velocity of B is 2 m/s to the right, determine the velocity of B relative to A; the acceleration of B relative to A and the absolute velocity of point C of the cable. A small bead of mass m is free to move inside a smooth vertical semicircular ring of 2mg and natural radius R. The bead is attached to one end of a massless spring of force constant k = R 3R lenght l o = . The other end of the spring is fixed at the centre of the ring. How R 60° 4 m does the normal reaction on the bead and tangential acceleration of it varies with angle θ (show graphically), where θ is the angle which the spring makes with a vertical line passing through the centre of the ring. The bead is released from the position shown in figure. Problem 58.

O is the centre of a circular disc of mass 50 kg which rests in a vertical plane on two rough pegs A and B, the coefficient of friction with each 1 being . AO makes 60 °and BO makes 30 °with the vertical. Find the maximum 2 Problem 59.

F O 60°

30°

A

force F, which can be applied tangentially at the highest point of the disc without causing rotation of the disc. (g = 10 m/s 2 )

B

F = 20 t

Problem 60. Three blocks shown in figure have the masses m A = 1 kg, m B = 2 kg and m C = 1 kg. A time varying force (in newtons) F = 20 t is applied on the pulley as shown in figure (here t is in seconds). Find the relative velocity between blocks B and A, when block C has acquired a velocity of 2.5 m/s. (g = 10 m/s 2 ) Both the pulleys are massless and friction is absent everywhere.

A

B

C

A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.3 s with a constant acceleration of 2 m/s 2 . The belt then moves with a constant Problem 61.

deceleration and comes to stop after a total displacement of 2.2 m. The coefficient of friction between the package and the belt are µ s = 0.35 and µ k = 0.25. Determine the displacement of the package relative to the belt as the belt comes to stop. Take g = 10 m/s 2 A car travelling at 28 m/s has no tendency to slip on a track of radius 200 m banked at an angle θ. When the speed is increased to 35 m/s, the car is just on the point of slipping up the track. Calculate the coefficient of friction between the car and the track. (g = 9.8 m/s 2 ) Problem 62.

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13

Problems Problem 63.

Two blocks of mass 10 kg and 2 kg respectively are

connected by an ideal string passing over a fixed smooth pulley as shown in figure. A monkey of 8 kg started climbing the string with a constant acceleration 2 m / s with respect to the string at t = 0. Initially the system is in

2m

2

8 kg 10 kg

equilibrium and the monkey is at a distance of 2 m from the pulley. Find the time taken by the monkey to reach the pulley.

2 kg

Problem 64. If masses of the blocks A and B shown in figure (a) and (b) are 10 kg and 5 kg respectively, find the acceleration of the two masses. Assume all pulleys and strings are ideal.

A

A

B

B (a)

(b)

Problem 65.

Two identical blocks each having a mass of 20 kg are connected to each other by a light F = 40 t inextensible string as shown and are placed over a rough 20 kg 20 kg surface. Pulleys are connected to the blocks. Find acceleration of the blocks after one second after the application of the time varying force of 40t N, where t is in m= 0.4 m= 0.4 second. Problem 66. An insect lying on the bottom of the hemi-spherical bowl tries to come out from it. The coefficient of static friction between insect and bowl is 0.5. How high up does the insect go without slipping? Now if the bowl starts rotating about axis as shown in figure. At what angular speed ω will the insect just be able to come out of the bowl? (Radius of the bowl 5 cm)

O

w

O'

There is a parabolic-shaped bridge across a river of width 100 m. The highest point of the bridge is 5 m above the level of the banks. A car of mass 1000 kg is crossing the bridge at a constant speed of 20 ms −1 . Problem 67.

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14

Problems in Physics m

v h

d

Using the notation indicated in the figure, find the force exerted on the bridge by the car when it is at the highest point of the bridge. (Ignore air resistance and take g as 10 ms −2 ) Problem 68. A sphere rests between a smooth wall and a smooth wedge of mass M whose inclination to the horizontal is α = 60 °. Mass of sphere is m and its radius is R. The wedge initially touches the right R wall. The vertical side of the wedge is connected to the mg k side wall with the help of light spring of force constant mg , where η is a positive constant. Find the k =η M R 60° minimum value of ηfor which the sphere does not collide with the horizontal surface; if the spring is let go in the position shown and spring is initially compressed. Neglect friction. Also find the normal reaction between the sphere and the right side vertical wall in critical case. B

Block B of mass 10 kg rests as shown on the upper surface of a 22 kg block A. Find acceleration of block A and magnitude of acceleration of block B relative to A. Neglect friction. Problem 69.

Wedge is fixed (g = 10 m/s 2 ).

A

30°

30°

A 25 kg block A rests on an inclined surface and a 15 kg counter weight B is attached to a cable as shown. Neglecting friction, determine the acceleration of A, acceleration of B and tension in the cable after the system is released from rest. Cable is parallel to the plane. Problem 70.

Take g = 10 m/s 2 .

Problem 71. A 6 kg block B rests as shown on the upper surface of a 15 kg block A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of A (b) the acceleration of B relative to A. Take g = 10 m/s 2 . (wedge is fixed)

B A

30°

B A

30°

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Problems Problem 72. The 10 kg solid cylinder is resting in the inclined V-block. If the coefficient of static friction between the cylinder and the block is 0.5, determine (a) the frictional force F acting on the cylinder at each side before force P is applied (b) the value of P required to start sliding the cylinder up the incline (g = 9.8 m/s 2 )

In the given figure, assume that there is no friction between block B and the surface on which it moves and the coefficient of friction between blocks A and B is µ. (a) find the minimum value of M so that block A starts sliding over B. (b) if M is two times that obtained in part (a), find the time when the block A topples from B.

45

° 45

°

En

dv

iew P

30°

l

Problem 73.

A B

m 4m 4l M m

A wedge of mass M = 4 kg with a smooth quarter circular plane is kept on a rough horizontal plane. A particle of mass Problem 74.

m = 2 kg is released from rest from the top of the wedge as shown in figure. Find the minimum value of coefficient of friction between the wedge and the horizontal plane so that the wedge does not move during complete journey of the particle. Problem 75. A block A of mass m and length l is placed on a horizontal floor. A rectangular box B is used to cover A. The distance between interior of the walls of B is L (> l ) and the mass of B is also m. The coefficient of friction between A and floor is µ 1 and that between B and floor is µ 2 (µ 2 > µ 1 ). Initially the left end of A touches the left wall of B as shown in figure and both A and B move with velocity v 0 towards the right. All collisions between A ad B are elastic and contact time during each collision is very short. Find an expression for the period between two consecutive collisions.

C

M

B L l A

Problem 76. A point mass of 0.5 kg moving with a constant speed of 5 m/s on an elliptical track experiences an inward force of 10 N when at either end point of the major axis and a similar force of 1.25 N at each end of the minor axis. How long are the axes of the ellipse. Problem 77. A disc of mass m and radius 1 m is hinged at its centre on a frictionless horizontal surface. It has a massless wall of m short height around the circumference. A small particle of mass 2 is projected with velocity 10 m/s keeping it in contact with the wall and base of the disc. If coefficient of friction between the small particle and the base of the disc in 0.5 and the wall is smooth. Find the angular displacement of the mass after 2 sec.

O

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Problems in Physics

Problem 78. In the above problem, the coefficient of friction between the particle and the wall is 0.5 and the base of disc is smooth. Find the time after which relative motion between the two is stopped. Problem 79. Two blocks A and B connected to each other by a string and spring. The string passes over a frictionless pulley as shown in the figure. When the block C is moving on an inclined plane with g acceleration upward, block B of mass 2 kg slides over the top surface 2 of block C and block A slides over the side wall of C, both with the same uniform speed. Coefficient of friction between all the blocks is 0.2. The force constant of the spring is 1800 N/m. Find: (a) the mass of the block B. (b) energy stored in the spring.

B

A

C g/2 m/s2

30°

IIT JEE PROBLEMS 2m

Two blocks A and B of equal masses are released from an inclined plane of inclination 45° at t = 0. Both the blocks are initially at rest. The coefficient of kinetic friction between the block Problem 80.

A B

A and the inclined plane is 0.2 while it is 0.3 for block B. Initially the block A is 2 m behind the block B. When and where their front faces will come in a line. (Take g = 10 m/s 2 )

Problem 81.

(JEE 2004)

45°

A

B

In the figure, masses m 1 , m 2 and M are 20 kg, 5 kg and 50 kg respectively. The

coefficient of friction between M and ground is zero. The coefficient of friction between m1 and M and that between m 2 and ground is 0.3. The pulleys and the strings are massless. The string is perfectly horizontal between P1 and m1 and also between P2 and m 2 .The string is perfectly vertical between P1 and P2 . An external horizontal force F is applied to the mass M. Take g = 10 m / s 2 .

P1

P2

m1

M

F

m2

(a) Draw a free body diagram of mass M, clearly showing all the forces.

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Problems

(b) Let the magnitude of the force of friction between m1 and M be f1 and that between m 2 and ground be f 2 . For a particular F it is found that f1 = 2 f 2 . Find f1 and f 2 . Write equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. (JEE 2000)

Two blocks of mass m1 = 10 kg and m 2 = 5 kg connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of turn table. The coefficient of friction between the table and m1 is 0.5 while there is no friction between m 2 and the table. The table is rotating with an angular velocity of 10 rad/s about a vertical axis passing through its centre O. The masses are placed along the diameter of the table on either side of the centre O such that the mass m1 is at a distance of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turn table. ( g = 9.8 m / s 2 ) Problem 82.

(i) Calculate the frictional force on m1 . (ii) What should be the minimum angular speed of the turn table so that the masses will slip from this position. (iii) How should the masses be placed with the string remaining taut so that there is no frictional (JEE 1997) force acting on the mass m1 . A smooth semicircular wire track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3R/4 is attached to the lowest point O of the wire track. A small ring of mass m which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle 60° with the vertical. The spring constant k = mg/R. Consider the instant when the ring is released. (i) Draw the free body diagram of the ring. (ii) Determine the tangential acceleration of the ring and the normal reaction. (JEE 1996) Problem 83.

C

P 60°

O

Problem 84. A hemispherical bowl of radius R = 0. 1 m is rotating about its own axis (which is vertical) with an angular velocity ω. A particle of mass 10 –2 kg on the frictionless inner surface of the bowl is also rotating with the same ω. The particle is at a height h from the bottom of the bowl. value ofrelation in order to (a) Obtain the between ω. What is the minimum ω needed, have a non-zero value of h? (b) It is desired to measure g (acceleration due to gravity) using this set-up, by measuring h accurately. Assuming that R and ω are known precisely, and that the least count in the measurement of h is 10 −4 m, what is the minimum possible error ∆ g in the measured value

of g?

(JEE 1993)

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WORK, POWER AND ENERGY D

O

A θ

3

0. m

Problem 85. A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 40 N/m and undeformed length equal to arc of the circle AB. A 0.2 kg collar C, not attached to the spring, can slide without friction along the rod. The collar is released from rest at an angle θ with the vertical. (a) Make the equation for minimum value of θ for which the collar will pass through D and reach point A. (b) Determine the velocity of collar as it reaches point A for minimum value of θ. (Take g = 10 m/s 2 )

C B

Problem 86. A body of mass m slides down a plane inclined at an angle α. The coefficient of friction is µ. Find the rate at which kinetic plus gravitational potential energy is dissipated at any time t.

The two particles of mass m and 2m respectively are connected by a rigid rod of negligible mass and slide with negligible friction in a circular path of radius r on the inside of the vertical circular ring. If the system is released from rest at θ = 0° determine (a) the velocity v of the particles when the rod passes the horizontal position, (b) the maximum velocity vmax of the particles and (c) the maximum value of θ. Problem 87.

m

θ

θ 2m

C

A person rolls a small ball with speed u along the floor R from point A. If x = 3R, determine the required speed u so that the ball u returns to A after rolling on the circular surface in the vertical plane B A x from B to C and becoming a projectile at C. What is the minimum value of x for which the game could be played if contact must be maintained at point C. Neglect friction. Problem 88.

Problem 89. If the system is released from rest, determine the speeds of both masses after B has moved 1 m. Neglect friction and masses of the pulleys. (g = 10 m/s 2 )

10

kg A

30°

B

8 kg

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Problems Problem 90. The free end of a flexible rope of length L and mass λ per unit length is released from rest in the position shown in the (a)-part of the figure. Determine the velocity v of the moving portion of the rope in terms of y.

L — 2

y

v (a)

A disc slides up without friction a hill of height h whose profile depends only on the x-coordinate (figure a). At the bottom the disc has the velocity v1 whose direction forms the angle α 1 with the X-axis (see figure b, top view). Find the direction of motion of the disc after it reaches the top, i.e., find the angle α 2 . Also describe the condition that the disc cannot overcome the hill. Problem 91.

(b)

Z

h x1

X

x2

Y

(a) v2

α2

v1 α1 x1

Problem 92. A small object loops a vertical loop of radius R in which a symmetrical section of 2α has been removed. Find the maximum and minimum heights from which the object after loosing contact with the loop at point A and flying through the air, will reach point B. Find the corresponding angles of the section removed for which this is possible.

(b)

x2

X

B H

αα

A

Problem 93. A car of mass 4m holds a block of mass m which is v0 attached to the former by means of a spring of spring constant k, as shown k m in the diagram. All surfaces are frictionless and the wheels are massless. The system is at rest. A bullet of mass m is fired at the first block with a horizontal velocity v 0 and sticks to it. Find: (a) the velocity of the car at the moment when the spring undergoes maximum compression. (b) the maximum compression of the spring.

A block of mass 2 kg approaches a spring of stiffness k = 80 N/m on a smooth horizontal plane with a speed of 10 m/s. Find the time in which kinetic energy of the block reduces to 50% of its initial kinetic energy. Assume t = 0 when the block just touches the spring, for the first time. Problem 94.

4m

10 m/s 2 kg

Problem 95. A pendulum bob of mass m is suspended at rest. A constant horizontal force F = mg starts acting on it. Find: (a) the maximum angular deflection of the string. (b) the maximum possible tension in the string.

k = 80 N/m

mg

F = mg

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Problems in Physics

Problem 96. A small bob ‘ P’ of mass ‘ m’ is attached with a thread of length ‘ l ’. If the bob ‘ P’ is given a velocity in such a way, that it just completes a vertical circle, then find the possible value of velocity, with which we can project an another bob ‘ Q’ of same mass ‘ m’, which can hit the bob ‘ P’ at the topmost point of their trajectory and falls Q vertically down after collision (The collision between ‘ P’ and ‘ Q’ is perfectly inelastic). Also find the angle with the A horizontal and the distance from the bottommost point of d bob ‘ P’ from where the bob ‘ Q’ should be projected. (The bob ‘ Q’ was not necessarily projected, when ‘ P’ was at the bottom of its trajectory).

O l

P

IIT JEE PROBLEMS Problem 97. A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure). The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less then d. All θ Sphere B surfaces are frictionless. The ball is given a gentle push d (towards the right in the figure). The angle made by the O radius vector of the ball with the upward vertical is denoted R by θ (shown in the figure). (a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle θ. Sphere A (b) Let N A and N B denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of N A and N B as function of cos θ in the range 0 ≤ θ ≤ π by drawing two separate graphs in your answer book, taking cos θ on the horizontal axes. (JEE 2002) A

O

Problem 98. A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance of L/8 from O as shown. The particle is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. (JEE 1999)

L/8 L

u

B

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CENTRY OF MASS, CONSERVATION OF MOMENTUM, COLLISION, IMPULSE

Problem 99. A 60 kg man and a 50 kg woman are standing on opposite ends of a platform of mass 20 kg. The platform is placed on a smooth horizontal ground. The man and woman begin to approach each other. Find the displacement of the platform when the two meet in terms of the displacement x 0 of the man relative to the platform. The length of the platform is 6 m. Problem 100. A drinking straw of length 3a / 2 and mass 2m is placed on a square table of side ' a'

parallel to one of its sides such that one third of its length extends beyond the table. An insect of mass m / 2 lands on the inner end of the straw (i. e., the end which lies on the table) and walks along the straw until it reaches the outer end. It does not topple even when another insect lands on top of the first one. Find the largest mass of the second insect that can have without toppling the straw. Neglect friction. Problem 101. A cannon of mass m slides down a smooth inclined plane forming the angle α with the

→ horizontal. At the moment when the velocity of cannon reaches v it fires a shell in a horizontal direction → with the result that the cannon stops and the shell carries away the momentum p . Suppose that the firing duration is ∆ t. What is the reaction force of the inclined plane averaged over the time ∆ t.

Problem 102. A car of mass 1000 kg and running at 25 m/s holds three men each of mass 75 kg.

Each man runs with a speed of 5 m/s relative to the car and jumps off from the back end. Find the speed of the car if the three men jump off (i) in succession (ii) all together. Neglect friction between the car and the ground. Problem 103. A ring of mass m can slide on a smooth horizontal wire. The ring is attached to a particle of mass 3m by a string of length l. A horizontal velocity v o is given to the ring. Find the maximum angle the string will make with the vertical in subsequent motion. Ring 0

Problem 104. Two bodies of mass m1 = 1 kg and m 2 = 2 kg move towards each other in mutually perpendicular directions at velocities v1 = 3 m / s and v 2 = 2 m / s. As a result of collision, the bodies stick together. Determine the amount of heat liberated as a result of collision.

v

Particle

Problem 105. A particle of mass m moving with a velocity (3i$ + 2$j ) m/s collides with a stationary

m 1 body of mass M and finally moves with a velocity (−2i$ + $j ) m / s. If = , find: M 13

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Problems in Physics (a) the impulse received by each. (b) the velocity of the M. (c) the coefficient of restitution.

Problem 106. A hemisphere of mass M and radius R is at rest. One solid sphere of mass 2M and radius R, moving with a velocity v o , collides with the hemisphere. If e is the coefficient of restitution, find the speed of the hemisphere and the solid sphere after collision.

2M, R v0

R µ=0

Problem 107. You are given three billiards-tables of different lengths and the same width. Balls are struck simultaneously from the edge of one of the long sides of each table (fig.) with velocities which are equal in direction and magnitude. Is it possible that these balls should not return to the side from which they started at exactly the same moment? Problem 108. A rope thrown over a pulley has a ladder with a man A on one of its ends and a

→ counterbalancing mass M on its other end. The man whose mass is m, climbs upwards by ∆ r relative to the ladder and then stops. Ignoring the masses of the pulley and the rope, as well as the friction in the pulley axis, find the displacement of the centre of mass of this system. Problem 109. Two identical particles are projected at the same instant from points A and B at the same level, the first from A towards B with velocity u at 45 ° above AB and the second from B towards A with velocity v at 60 ° above BA. If the particles collide directly when each reaches its greatest height, v2 find the ratio 2 and prove that u 2 = ag (3 − 3) where a is the distance AB. After the collision the first u particle falls vertically. Show that the coefficient of restitution between the particles is ( 3 − 1)/ ( 3 + 1). Problem 110. A boy throws a ball with initial speed 2 ag at an angle θ to the horizontal. It strikes a smooth vertical wall and returns to his hand. Show that if the boy is standing at a distance ‘a’ from the 1 . Also show that θ wall, the coefficient of restitution between the ball and the wall equals (4 sin 2 θ − 1) cannot be less than 15°. Problem 111. A ball falls freely from a height onto an smooth inclined plane forming an angle α with the horizontal. Find the ratio of the distance between the points at which the jumping ball strikes the inclined plane. Assume the impacts to be elastic. Problem 112. A ball is projected from a point A on a smooth inclined plane which makes an angle α to the horizontal. The velocity of projection makes an angle θ with the plane upwards. If on the second bounce the ball is moving perpendicular to the plane, find e in terms of α and θ. Here e is the coefficient of restitution between the ball and the plane. Problem 113. Two identical smooth balls are projected towards each other from points A and B on the horizontal ground with same speed of projection. The angle of projection in each case is 30 °. The distance between A and B is 100 m. The balls collide in air and return to their respective points of projection. If coefficient of restitution is e = 07 . , find

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Problems (a) the speed of projection of either ball. (b) coordinates of point with respect to A where the balls collide. (Take g = 10 m/s 2 ) 30 m/s

Problem 114. A ball strikes the wall at point A (e1 = 0.5) and then hits the ground at B (e 2 = 0.3). Find the distance x from the point O, where should a fielder be to catch the ball at a height of 1.2 m from the ground. Here e is coefficient of restitution. Take g = 10 m/s 2 .

45°

A 2.4 m

1.2 m B

O

x

Problem 115. Two smooth spheres A and B of equal radius but of masses m and M respectively are free to move on a horizontal table. A is projected with speed u towards B which is at rest. On impact, the line joining their centres is inclined at an angle θ to the velocity of A before impact. If e is the coefficient of restitution between the spheres, find the speed with which B begins to move. If A deviates by 90° from its initial path, find the angle θ. Problem 116. A shell of mass m = 1 kg is fired from a gun of mass M = 10 kg from x = 0 with an initial speed u relative to the gun as shown in figure. The gun is situated on a horizontal floor. The gun strikes a wall at x = L moving towards negative x-direction with velocity v. Find the coefficient of restitution between the shell and the wall so that the shell returns to the point of the gun from where it was started. Take u = 10 m/s, L = 4 m and v = 2 m/s. Neglect friction. (g = 10 m/s 2 ) v m 45° Y

M x=0

x=L

X

Problem 117. Two identical cubes A and B of same mass 2m and side 2d are placed one over the other as shown in figure. B is attached to one end of a spring of force constant k. The other end of the spring is attached to a wall. The system is resting on a smooth horizontal surface with the spring in the relaxed state. A small object of mass m moving horizontally with speed v at a height d above the horizontal surface hits elastically the block B along the line of their centre of mass. There is no A k m v friction between A and B : B (a) find the minimum value of v (say v 0 ) such that block A will topple d B A over block B (b) if v = v 0 / 2, find the amplitude of oscillation of block spring system. Problem 118. Three identical particles A, B and C lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ∠ ABC is 135°. An impulse J is applied to the particle C in the direction BC. Find the initial speed of each particle. The mass of each particle is m.

J C

135° A

B

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Problems in Physics

Problem 119. A mass 2m rests on a horizontal table. It is attached to a light inextensible string which passes over a smooth pulley and carries a mass m at the other end. If the mass m is raised vertically through a distance h and is then dropped, find the speed with which the mass 2m begins to rise.

m h

2m

m

Problem 120. A small ball of mass m is connected by an inextensible massless

string of length l with an another ball of mass M = 4m. They are released with zero tension in the string from a height h as shown in figure. Find the time when the string becomes taut for the first time after the mass M collides with the ground. Take all collisions to be elastic.

Problem 121. A 2 kg sphere A is connected to a fixed point O by an inextensible cord of length 1.2 m. The sphere is resting on a frictionless horizontal surface at a distance of 0.5 m from O when it is given a velocity v 0 in a direction perpendicular to the line OA. It moves freely until it reaches position A′ when the cord becomes taut. Determine (a) the maximum allowable velocity v 0 if the impulse of the force exerted on the cord is not to exceed 3 N-s (b) the loss of energy as the cord becomes taut, if the sphere is given the maximum allowable velocityv 0 .

M h

A′ 2

1.

m

v0 O

A 0.5 m

C

Problem 122. Ball B is hanging from an inextensible cord BC. An identical ball A is

released from rest when it is just touching the cord and acquires a velocity v 0 before striking ball B. Assuming perfectly elastic impact (e = 1) and no friction, determine the velocity of each ball immediately after impact.

A

v0 B

Problem 123. Figure shows a small block of mass m = 1 kg which is given a horizontal velocity v 0 = 10 m/s on the horizontal part of the bigger block of mass M = 9 kg placed on a horizontal floor. The curved part of the surface shown is semicircular of radius R = 1 m. Find the distance from point B where the block m lands finally after looping the semicircular part BCD. Neglect friction everywhere. Assume that the horizontal portion AB is long enough. (g = 10 m/s 2 )

D C v0

R M

B

m A

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Problems Problem 124. A railroad car of length L and mass m 0 when empty is moving freely on a smooth horizontal track while being loaded with sand dm from a stationary chute at a rate = q. Knowing that the car was dt approaching the chute at a speed v 0 . Determine (a) the speed of the car v f at the instant when the car has cleared the chute. (b) the mass of the car and its load at that instant.

v0

L

Problem 125. A particle whose initial mass is m 0 is projected vertically upwards at time t = 0 with speed gT, where T is a constant. At time t the mass of the particle has increased to m 0 e t / T . If the added

mass is at rest relative to particle when it is acquired, find the time when it is at highest point and mass at that instant. Problem 126. A rain drop falls from rest in an atmosphere saturated with water vapour. As it falls, water vapour condenses on the drop at the rate of mass µ per second. If initial mass of drop is m 0 , how much distance the drop falls in time t. urel = u

Problem 127. A cannon with bullets of total mass M 0 is

kept on a rough horizontal surface. The coefficient of friction between the cannon and the surface is µ. If the cannon fires bullets with constant frequency with a relative velocity u, find the velocity of the cannon when its mass with remaining bullets become M after time t. Initially cannon was at rest. Assume that the thrust force is greater than the limiting friction right from the beginning.

M0 cannon

v = 1 m/s

Problem 128. A long thin carpet is laid on the floor. One end of the

carpet is bent back and then pulled backwards with constant velocity O x=1m v = 1 m/s, just above the part of the carpet which is still at rest on the floor. (a) Find the speed of centre of mass of the moving part. (b) What is the minimum force needed to pull the moving part, if the carpet has unit length and unit mass. Problem 129. A light inextensible thread passes over a small frictionless pulley. Two blocks of masses m = 1 kg and M = 3 kg respectively are attached with the thread as shown in the figure. The heavier block rests on a horizontal surface. A shell of mass 1 kg moving upward with a velocity 10 ms −1 collides and

sticks with the block of mass m as shown in the figure at t = 0. If the inclined plane is smooth. Calculate: (a) maximum height ascended by M. (b) time ‘ t’ at that instant.

m

q = 30°

M

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Problems in Physics Au

x

Problem 130. A particle is projected horizontally with a speed of

u = 10 m/s from point A. A steel frame is rigidly fixed as shown in the figure. The frame may be considered as an arc of circle with centre at A and radius R = 21 m. At which point will the particle strike the frame? If the particle rebounds elastically from the frame will it again strike it? Take g = 10 m/s 2

21 Frame

y

IIT JEE PROBLEMS Problem 131. A particle of mass m, moving in a circular path

of radius R with a constant speed v 2 is located a point ( 2 R, 0) at time t = 0 and a man starts moving with a velocity v1 along the positive y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. the man as a function of time. (JEE 2003)

y v1

v2

R (0, 0)

x m

Problem 132. Two point masses m1 and m 2 are connected by a spring of natural length l0 . The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity v 0 along positive x-axis. When the system reaches

the origin the string breaks (t = 0). The position of the point mass m1 is given by where A and ω are constants. x1 = v 0t − A(1 − cos ωt ) Find the position of the second block as a function of time. Also find the relation between A and l0 . (JEE 2003)

Problem 133. A car P is moving with a uniform speed of 5 3 m/s towards a carriage of mass 9 kg at rest kept on C the rails at a point B as shown in figure. The height AC is P 120 m. Cannon balls of 1 kg are fired from the car with an initial velocity 100 m/s at an angle 30° with the horizontal. The first cannon ball hits the stationary carriage after a time t 0 and sticks to it. Determine t 0 . At A B t 0 , the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second cannon ball also hits and sticks to the carriage. What will be the horizontal velocity of the carriage just after the second impact ? Take g = 10 m/s 2 (JEE 2001)

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Problems Problem 134. Two blocks of mass 2 kg and M are at rest on an inclined plane and are separated by a distance of 6.0 m as shown. The coefficient of friction between each block and the inclined plane is 0.25. The 2 kg block is given a velocity of 10.0 m/s up the inclined plane. It collides with M, comes back and has a velocity of 1.0 m/s when it reaches its initial position. The other block M after the collison moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M. [take sin θ ≈ tan θ = 0.05 and g = 10 m/s 2 ]

M kg

2

m 6.0

θ

(JEE 1999)

Problem 135. A large open top container of negligible mass and uniform cross-sectional area A

has a small hole of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m 0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate (i) the acceleration of the container, and (ii) its velocity when 75% of the liquid has drained out. (JEE 1997) Problem 136. A small sphere of radius R is held against the inner surface of larger sphere of radius 6R (as shown in figure). The masses of large and small spheres are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the large sphere, when the smaller sphere reaches the other extreme position. (JEE 1996) Y

6R M (L,0)

X

R

4M

Problem 137. A cylindrical solid of mass 10 –2 kg and cross-secional area 10 –4 m 2 is moving

parallel to its axis (the X-axis) with a uniform speed of 10 3 m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with stationary dust particles of uniform density 10 –3 kg/m 3 . When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimensions of the cylinder remain practically unchanged, and that the dust sticks only to the front face of the cylinder, find the x-coordinate of the front of the cylinder at t = 150 s. (JEE 1993)

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5

ROTATION Y

Problem 138. A ring of radius R rolls without slipping on a

horizontal plane with constant velocity v. Find the position, velocity and acceleration of any point P on the circumference of the ring at any time t. Assuming that point P was at origin at time t = 0.

v

X

O

Problem 139. A 4m long rod AB slides down the plane with v A = 4 m/s to the left and a A = 5 m/s 2

to the right. Angle θ at this moment is 30°. Determine the angular velocity and angular acceleration of the rod at this instant of time. B

4m

A

θ

135° O

P

Problem 140. A cylinder of diameter 0.3 m and mass 25 kg rests on a rough surface as shown in figure. The coefficients of static and kinetic frictions are µ s = 0.4 and µ k = 0.35. Determine the minimum value of P to be applied to roll the cylinder without slip = 10over g / the step ( )

30°

0.3m

0.025m

Two heavy and light cylindrical rollers of diameters D and d respectively rest on a horizontal plane as shown in figure. The larger roller has a string wound round it to which a horizontal force P can be applied as shown. Assuming that the coefficient of friction µ has the same value for all surfaces of contact, determine the limits of µ so that the larger roller can be pulled over the smaller one.

P

Problem 141.

D d

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Problems Problem 142. A wheel of radius r rolls to the left without slipping and at the instant considered the centre O has a velocity v 0 and acceleration a 0 to the left. Determine the magnitude of acceleration of points A and C on the wheel for the instant considered.

ω α

r0 a0

v0

θ

A

O r

C

Problem 143. A sphere of radius r and mass m has a linear velocity v 0 directed to the left and no angular velocity as it is placed on a belt moving to the right with a constant velocity v1 . If after sliding on the belt the sphere is to have no linear velocity relative to the ground v0 as it starts rolling on the belt without sliding. In terms of v1 and the v1 coefficient of kinetic friction µ k between the sphere and the belt, determine (a) the required value of v 0 (b) the time t1 at which the sphere will start rolling on the belt (c) the distance the sphere will have moved relative to the ground at time t1 . O

Problem 144.

Two identical rods of same mass and length ‘ l ’

are hinged at P as shown. Initially rod 2 was perfectly horizontal. Find the angular acceleration of rod 1 at the moment the string connecting the rod 2 in cut.

1 37° P

2 y A x

Problem 145. A ring of radius R is rolling (without slipping) over a rough horizontal surface with a velocity v 0 . Two points are located at A and B on the rim of the ring. Find the velocity of A w.r.t. B.

A disc of radius R is given a linear velocity v o and an vo angular velocity and placed on a rough surface as shown in the figure. R Assume coefficient of kinetic friction = µ. (a) Will the disc return? (b) Plot a curve for linear velocity of disc with respect to time, indicating appropriate values of velocities and time. Problem 146.

A uniform bar of length L and weight W stands vertically touching a vertical wall (y-axis). When slightly displaced, its lower end begins to slide along the floor (x-axis). Obtain an expression for the angular velocity (ω) of the bar as a function of θ. Determine the distance moved by the lower end at which the bar no longer touches the vertical wall. Neglect friction everywhere. Problem 147.

v0 R

B

v0 R

ω

v0

Y

A

L/2 C L/2 W

O

θ B

X

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30 A rectangular lamina of length l breath b, and → mass M rotates with angular velocity ω about an axis ACB in the plane of lamina as shown in figure. Axis of rotation passes through centre of mass C of the lamina making an angle θ with one of the sides. Find the angular momentum of the lamina about C and its component about axis of rotation.

Problems in Physics Y

Problem 148.

A ω

l

θ

b

X

C

B

Problem 149. A thin uniform bar of mass m and length 2L is held at an angle 30° with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing determine the tension in the string at the left end and the angular acceleration of the bar immediately after string breaks. 30° B

A uniform slender bar AB of mass m and length L supported by a frictionless pivot at A is released from rest at its vertical position as shown in figure. Calculate the reaction at the pivot when the bar just acquires the horizontal position shown dotted. If at this instant, the bar is released from its support gently and allowed to move for t second further, estimate its angular speed and the velocity of the centre of mass at that instant. Problem 150.

A uniform thin rod AB of mass M = 0.6 kg and length l = 60 cm stands at the edge of a frictionless table as shown in figure. A particle of mass m = 0.3 kg flying horizontally with velocity v 0 = 24 m/s strikes the rod at point P at a height 45 cm from base and sticks to it. The rod is immediately driven off the table. Determine the co-ordinates of centre of mass (COM) of the combined system with edge of the table as origin when rod becomes horizontal for the first time (g = 10 m/s 2 ).

A

A

Problem 151.

A block of mass m height 2h and width 2b rests on a flat car which moves horizontally with constant acceleration ' a' as shown in figure. Determine : (a) the value of the acceleration at which slipping of the block on the car starts, if the coefficient of friction is µ (b) the value of the acceleration at which block topples about A, assuming sufficient friction to prevent slipping and

m P

v0

Y X B

2b

Problem 152.

2h A a

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Problems

(c) the shortest distance in which it can be stopped from a speed of 72 km/hr with constant deceleration so that the block is not disturbed. The following data are given b = 0.6 m, h = 0.9 m µ = 0.5 and g = 9.8 m/s 2 b Problem 153. A block of mass M = 4 kg of height ' h' and breadth ‘b’ is placed on a rough plank of same mass M. A light inextensible string is connected to the upper end of the block and passed h M through a light smooth pulley as shown in figure. A mass m = 1kg is hung to the other end of the string. M (a) What should be the minimum value of coefficient of friction between the block m and the plank so that, there is no slipping between the block and the wedge ? (b) Find the minimum value of b/h so that the block does not topple over the plank, friction is absent between the plank and the ground.

Problem 154. A beam of mass m1 = 100 kg supported on two solid cylindrical rollers each of mass m 2 = m 3 = 20 kg and radius R = 01 . m are moved up the inclined plane by a force F applied at an angle 30° with the incline as shown in figure. The inclined plane makes an angle of 30° with horizontal. Find the magnitude of F if the beam is moving up with an acceleration of a = 1 m/s 2 . There is no slipping at points of contact.

m

1

F 30° m3

m2

30°

(g = 10 m/s 2 ).

sphere

A homogeneous cylinder and a homogeneous sphere of equal mass m = 20 kg and equal radii R are connected together by a light frame and are free to roll without slipping down the plane inclined at 30° with the horizontal. Determine the force in the frame. Assume that the bearings are frictionless. Take g = 10 m/s 2 . Problem 155.

Problem 156. A cylindrical pipe of diameter 1 m is kept on the truck as shown in figure. If the truck now starts moving with constant acceleration of 1 m/s 2 the pipe rolls backward without

slipping on the floor of the truck and finally falls on the road. If the pipe moves a total length of 4m on the floor of the truck. Find how much distance the pipe moves on the road before it finally stops. The coefficient of friction between the pipe and the road is 0.4. (g = 10 m/s 2 )

cylinder

30° 30°

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Problems in Physics

Problem 157. A rod AB of mass 3 m and length 4 a is falling freely in a horizontal position and C is a point distant a from A. When the speed of the rod is u, the point C collides with a particle of mass m which is moving vertically upwards with speed u. If the impact between the particle and the rod is perfectly elastic find (a) the velocity of particle immediately after impact (b) the angular velocity of the rod immediately after impact. (c) the speed of B immediately after impact.

G

C

A

u m

B u

Problem 158. A uniform ring of mass m, radius a and centre C lies at rest on a smooth horizontal table. The plane of the ring is horizontal. A point P on the circumference is struck horizontally and it begins to move in a direction at 60° to PC. If the magnitude of impulse is mv 7, find the initial speed of point P. Problem 159. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through D and perpendicular to face ABCD.A bullet of mass m and 4a as shown in figure. The bullet becomes speed v is shot at a height of 3 embedded in the cube. Find the minimum value of v required to topple the cube. Assume m µ ′ ). Problem 368.

The faces of prism ABCD made of glass of refractive index µ form dihedral angles ∠ A = 90° , ∠ B = 75° , ∠ C = 135° and ∠ D = 60° . A beam of light falls on face AB and after complete internal reflection from face BC, escapes through face AD. Find the range of µ and angle of incidence α of the beam, if a beam that has passed through the prism in this manner, is perpendicular to the incident beam.

µ′

µ

µ

α

B

Problem 369.

C

α

A

A ray of light is incident on the surface of a sphere of refractive index 7 / 2. Other half of the sphere is silvered. After refraction it is reflected and then refracted out of the sphere again such that the total deviation is minimum. Find : (a) the angle of incidence of the ray

D

Problem 370.

(b) total deviation of the ray. sin 41° =

i

3 7

Problem 371. A point object is kept at a distance of 100 cm from a parabolic reflecting surface x = 8y 2 . An equiconvex lens of focal length 20 cm is kept at a distance of 80 cm from origin as shown in

figure. Find the position of the image after reflection from the surface.

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Problems

71 Y

X

P

O

80 cm 100 cm

A transparent cylinder of radius R = 2.00 m has a mirrored surface on its right half as shown in figure. A light ray travelling in air is incident on the left side of the cylinder. The incident ray and the exiting ray are parallel and at a distance d = 2.00 m. Determine the refractive index of the material.

Incident ray

Problem 372.

2m

Exiting ray

Problem 373. When the object is placed at 4 cm from the objective of a compound microscope, the final image formed coincides with the object and is at the least distance of distinct vision (25 cm). If the magnifying power of the microscope is 14, calculate the focal length of the objective and the eye piece. Problem 374. Three right angled prisms of refractive indices µ 1 , µ 2 and µ 3 are joined together so that the faces of the middle prism are in contact each with one of the outside prisms. If the ray passes through the composite block undeviated, show that

µ 12 + µ 23 – µ 22 = 1 Problem 375. A lake is lit by an underwater isotropic lamp. If the surface of the lake is covered by a layer of oil of refractive index 1.2. Calculate the percentage of light (a) escaping from the lake surface (b) totally internally reflected in the water-oil layer. Assume that the lake is absolutely calm and that both the oil and water are 100% transparent. The refractive index of water is 1.33. Problem 376. A point source of light is placed on optical axis of a concave lens having focal length 30 cm at a distance of 90 cm from the lens. A thin prism having prism angle 5° and made of two materials having refractive indices 3/2 and 4/3 respectively, is placed at a distance of 44.5 cm from the lens. Find the co-ordinates of images formed assuming position of source as origin and optical axis of lens as x-axis, also state whether the images are real or virtual.

y

S

θ = 5°

n1

x 45.5 cm

44.5 cm n2 f0 = 30 cm

In the arrangement shown, the angle of thin prism is 1°. The image of the object kept π at origin O is formed at a point 10 cm, − cm . Find 24 Problem 377.

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Problems in Physics

72 y \ \\\\\

\\\\\\\\\\\\\\\\\\\\\

O

x 30 cm 15 cm

(a) radius of curvature of the mirror, (b) refractive index of the prism. Consider only paraxial rays. f1 = d

A point object is kept at O(0, 0). A1 and A 2 are the optical centers of the two lenses shown. Find the co-ordinates of final image. Given : d = 10 cm. The shaded portion of first lens is opaque.

f2 = 3d 2

Problem 378.

∆ = 0.005 cm

A1

Y X

A2

O 2d d

Problem 379. An equilateral prism has an angle of deviation 30° when the angle of incidence is 60°. Find the angle of deviation if a ray is incident normally on a surface.

A glass prism in the shape of a quarter-cylinder lies on a horizontal table. A uniform, horizontal light beam falls on its vertical plane surface, as shown in the figure. If the radius of the cylinder is R = 5 cm and the refractive index of the glass is n = 1.5, where, on the table beyond the cylinder, will a patch of light be found? Problem 380.

Light

R n

Problem 381. The refractive index of the medium within a certain region, x > 0, y > 0, changes with y. A thin light ray travelling in the x-direction strikes the medium at right angles and moves through the medium along a circular arc. How does the refractive index depend on y? What is the maximum possible angular size of the arc?

Two slits s 1 and s 2 are on a plane inclined at an angle of 45° with horizontal. The distance between the slits is 2 mm. A monochromatic point source S of wavelength, λ = 5000 Å is placed at a distance 1/ 2 mm from the midpoint of slits as shown in figure. The screen is placed at a distance of 2 m. Find the fringe width of interference pattern on the screen.

y

Problem 382.

45o s1 S s2

1 2

2m

Screen mm

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Problems

73

Problem 383. Parallel beam of a mono- chromatic light of wavelength λ is incident on a converging lens containing two identical parts L1 and L 2 having refractive indices µ and µ – ∆µ (∆µ > d), the radii of the ith and kth dark fringes are ri and rk . Find the wavelength of the light used in the experiment. Given that ri and rk b. O A

a

m

b

B

When the particle is held at B, the portion AO of the string is stretched while portion OB is slack and so when the particle is released it moves towards O. Let the particle be at any point P (at a distance x from O between O and B) at any time t. x A O

T

P m

B

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239

Solutions Tension in the string AP is x T = Y , action towards O. a Tension in PB is zero as it is slack. Equation of motion is d 2x Y d 2x Y m 2 =− x or =− x 2 a am dt dt

…(1)

This represents a simple harmonic motion with centre at O and amplitude OB. Let the particle take t1 time in reaching O from 1 B. Then t1 = × time period of SHM 4 represented by (1) 1 2π π am = ⋅ = 4 Y / am 2 Y The particle will have certain kinetic energy at O, due to which it will move towards A, stretching the portion OB. The segment AO becomes slack. Let the particle be at Q (at a distance x 1 from O) at any time t in the segment AO. x1

C

Q

219. This is an interesting problem because the maximum displacement of the particle on either side of the mean position is not same through the time taken to cover these distances is same. Consider Eq. (1) of solution 218 d 2x Y =− x am dt 2 dx Multiplying both sides by 2 and then dt integrating, we have 2 Y 2 dx x + k, =− dt am where k is a constant. dx At point B, x = b and = 0. dt Y 2 Yb2 ∴ 0= − b + k or k = am am 2 Y dx Hence, = (b2 − x 2) dt am At the mean position, i. e., at O, x = 0 x

B

A T

Tension in the string OB =

A

O

Yx1 towards O. b

Tension in AO = 0. d 2x 1 Yx 1 Hence, =− m 2 b dt 2 d x1 Yx1 or …(3) =− bm dt 2 The Eq. (3) represents SHM with centre at O and time period 2 1 bm 2π = 2π Y Y / bm Let the amplitude of this simple harmonic motion be OC. Time taken to reach C from O is bm π bm 1 ⋅ 2π = 4 Y 2 Y The required periodic time for complete oscillation between B and C is π am π bm m 2× + = π ( a + b) 2 Y Y 2 Y

O

T

P

B

m

dx Y …(1) =v= b dt am This is the velocity of the particle at O. Now from Eq. (3) of solution 218, we have d 2x 1 Yx1 =− 2 bm dt dx Multiplying both sides by 2 1 and integrating dt 2 Y dx x 1 + D, =− dt bm where D is a constant. But at O, x 1 = 0 2 Y 2 dx 1 2 [from (1)] and b =v = dt am Y 2 Y 2 ∴ b b = 0 + D or D = am am 2 Y 2 Y 2 dx Hence, 1 = − b x1 + dt am bm Y b3 2 = − x1 bm a or

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Problems in Physics If the extreme postion of the particle is point C such that OC = c, dx then at x 1 = c, 1 = 0 dt b Y b3 or 0 = − c 2 or c = b a bm a The particle thus oscillates through a distance OC + OB = b

b( b + a ) b + b= a a

Alternatively This problem can be sovled easily by energy considerations. When the string segment of length a is stretched by a distance x, force Yx . developed in it will be F = a Work done in stretching it to a distance b is Y b Y b2 x dx = a ∫a a 2 This is the potential energy stored in the string when the particle is held at B. This energy remains conserved. When the particle is at C(OC = c) the potential energy stored in the segment of length b(OB)is

F = − (extra upthrust) = − (extra volume immersed) (ρL ) (g ) (a = acceleration) or ma = − (πR 2) × ρg ∴

4 πR 3 3

ρ 2 a = − (πR ρg ) x 2 3g a=− x 2R

∴

As a ∝ − x , motion is simple harmonic. Frequency of oscillation, f =

1 2π

a 1 = x 2π

221. Let S be the area of cross-section of the rod. In the displaced position, as shown in figure, weight (W ) and upthrust (FB ) both pass through its centre of gravity G.

W=

Y c2 2 b Since, the particle’s velocity is zero at both B and C, we have from energy conservation Y c 2 Y b2 = 2 b a 2

or c = b

b a

The particle thus oscillates through a distance b+ b

b b( b + a ) = a a

220. Half of the volume of sphere is submerged. For equilibrium of sphere, weight = upthrust V Vρs g = (ρ) (g ) ∴ 2 ρ ρs = ∴ 2 When slightly pushed downwards by x, weight will remain as it is while upthrust will increase. The increased upthrust will become the net restoring force (upwards).

3g 2R

FB Q

r1 G

θ W P

Here W = (volume) × (density of rod) × g W = (SL)(d1)g FB = (volume) × (density of liquid) × g = (SL)(d 2)g Given that d1 < d 2. Therefore, W < FB Therefore, net force acting at G will be : F = FB – W = (SLg )(d 2 – d1) upwards. Restoring torque of this force about point P is : τ = F × r1 = (SLg )(d 2 – d1) (QG) L or τ = – (SLg )(d 2 – d1) sin θ 2 Here negative sign shows the restoring nature of torque SL2g (d 2 – d1) …(1) or τ=– θ 2 As sin θ ≈ θ for small values of θ From equation (1), we see that τ∝–θ Hence motion of the rod will be simple harmonic.

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241

Solutions Rewriting equation (1) as SL2g (d 2 – d1) d 2θ …(2) I 2 =– θ 2 dt Here I = moment of inertia of rod about an axis passing through P. ML2 (SLd1)L2 I= = 3 3 Substituting this value of I in equation (1), we have 3 g (d 2 – d1) d 2θ =– θ 2 d1L dt 2 Comparing this equation with standard differential equation of SHM i.e., d 2θ = – ω 2θ dt 2 The angular frequency of oscillation is 3 g( d 2 – d1 ) ω = 2d1 L 222. (i) Given : Mass of each block A and B, m = 01 . kg Radius of circle, R = 0.06 m A

B

x=Rθ

x=Rθ θ

θ

O θ=π/6

Restoring torque of this force about origin, τ = – F ⋅ R = – (4kx )R = – (4kRθ)R or …(1) τ = – 4kR 2 ⋅ θ Since τ ∝ – θ, each ball executes angular SHM about origin O. Equation (1) can be rewritten as Iα = – 4kR 2θ 4k or (mR 2)α = – 4kR 2θ ⇒ α = – θ m ∴ Frequency of oscillation, f =

1 2π

acceleration displacement

=

1 2π

α 1 = θ 2π

4k m

Substituting the values, we have f =

1 2π

4 × 01 . 1 = Hz 01 . π

(ii) In stretched position, potential energy of the system is 1 P. E. = 2 k { 2x } 2 = 4kx 2 2 and in mean position, both the blocks have kinetic energy only. Hence 1 K. E. = 2 mv 2 = mv 2 2 From energy conservation: P. E= K. E. ∴ ∴

v = 2x

m

= 2Rθ

m

Substituting the values

4k

x

Natural lengthkof spring, k (Half circle) l 0 = 0.06 π = πR and spring constant, k = 01 . N/m In the stretched A position elongation in each spring (x = Rθ) θ Let us draw FBD of A O Spring in lower side is stretched by 2x and on upper side compressed by 2x. Therefore, each spring will exert a force 2kx on each block. Hence a restoring force, F = 4kx will act on A in the direction shown in figure.

4kx 2 = mv 2

F=

v = 2(0.06)(π/ 6) or

. 01 . 01

v = 0.0628 m/s

(iii) Total energy of the system, E = P. E. in stretched position or

= K. E. in mean position E = mv 2 = (01 . )(0.0628)2 J

or

E = 3.9 × 10 –4 J

CHAPTER

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8

SOLIDS AND FLUIDS

223. Consider an element of length dl and mass dm of the ring. Let S be the cross-sectional area of wire and T the tension. dl = R . dθ dθ F = 2T sin ≈ T (dθ) 2

A

dl

B

dθ

dθ dθ ≈ sin 2 2

P dV is the work done per unit volume. The V negative sign implies that a decrese in pressure gives rise to increase in volume and vice-versa. Hence volume density of elastic potential energy

T . dθ = (dm)Rω 2 T = ρω 2R 2 S

or

u= ⇒

= (R. dθ. S. ρ)Rω 2 …(i)

P . dP 1 (hρg )2 = B 2 B

Substituting the values, we obtain

This F provides the necessary centripetal force. Hence,

hρg

∫0

u=

1 (103)2 (103)2 (9.8)2 2 2 × 109

u = 2.4 × 10 4 J/m 3

225. Tension in the plank varies linearly from F1 to F2. Therefore, tension in the plank at a distance x from the front edge is :

Let ∆R be the increase in radius. Then

A

dx

dθ T cos ( — ) 2

dθ T cos ( — ) 2 F

T

l T

∆L ∆(2πR) ∆R longitudinal strain = = = L 2π R R stress Now T. dx Y= = strain ∆R/R Y= Hence

∆R =

ρω 2R 3 ∆R ρω 2 R 3 Y

T = F1 – (F1 – F2)

or

x l

change in length in element dx is : dl =

or

Fl ∆l = AY

SY

x F1 – (F1 – F2) l dl = . dx SY

∴ total change in length will be

224. Bulk modulus is given by dV dP dP or =− B=– V B (dV/V) P . dV P dP =− V B

T x=x

B dθ

T

x=0 F1

F2

or

x =l

∆l =

∫ x = 0 dl

∆l =

x F1 – (F1 – F2) l ∫0 SY l

dx

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243

Solutions or

∆l =

( F1 + F2 ) l 2SY

F – mg – T = ma T = F – m (g + a) ρ = m w (g + a) – m (g + a) ρs

or

226. (a) Let ∆l be the displacement of the joint towards right. l α t – ∆l Strain on first rod = 1 1 l1 or force exerted by the first rod on the joint is l α t – ∆l F1 = Y1 A 1 1 l1

…(1)

l 2 α 2t + ∆l l2 or force exerted by second rod on the joint is

Strain on second rod =

l α t + ∆l F2 = Y2 A 2 2 l2

…(2)

In equilibrium F1 = F2 Solving this, we get l l t(Y α – Y2 α 2) ∆l = 1 2 1 1 (Y1l 2 + Y2 l 1)

…(3)

Solving (1), (2) and (3), we get At ( l1α1 + l 2α 2 ) F1 = F2 = l1 l + 2 Y2 Y1

ρ = (g + a) m w – 1 ρs Substituting the values, we get 1000 T = (10 + 2) ( 2) – 1 = 24 N 500 (b) When thread snaps, T =0 ∴ absolute acceleration of sphere upwards, F – mg as = m ρ m (g + a) w − mg ρs or as = m ρ or a s = (g + a) w – g ρs 1000 2 = (10 + 2) – 10 = 14 m/s 500 ∴ Upward acceleration of sphere relative to tank

(b) l ′1 = l1 + ∆l = l1 +

a r = a s – a = 14 – 2 a r = 12 m / s 2

l1 l 2t ( Y1α1 – Y 2α 2 ) ( Y1 l 2 + Y 2 l1 )

228. (a) Level will be maximum when

and l ′ 2 = l 2 – ∆l = l2 –

Rate = α

l1 l 2t( Y1α1 – Y 2α 2 ) Y1 l 2 + Y 2 l1

227. (a) The forces acting on the sphere are shown in the figure.

A

F = upthrust force h

a

F = Vρw g eff = Vρw (g + a) m = . ρw (g + a) ρs

m3 s

v = 2gh a

mg

T

Here ρw = density of water = 1000 kg/m 2 and ρs = density of sphere = 500 kg/m 3 Equation of motion of sphere is

rate of inflow of water = rate of outflow of water i.e. ⇒

α = av or α = a 2g hmax hmax =

α2 2 ga 2

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Problems in Physics

dh 2gh

=

t

∫0

dt A

= 0.9 i.e. 90% volume of ice is ρw immersed in water. 2cm

A ag

α α – a 2 gh ln – 2 gh α a

A

B

Level of water remains the same when ice melts. Hence Earlier the centre of mass of ice was at a height of 10 cm from the line AB and when it melts, it will be at 9 cm. Hence

229. Let h f be the final level in the two tanks, then A

G B

A

Solving this, we get t =

G

18cm

9cm

h

∫ 0α – a

or

ρi

230.

10cm

(b) Let at time t, the level of water be h. Then dh A = α – a 2gh dt

2A

∆h = – 1 cm = − 10−2 m or ∆U = mg ∆h

h

= – (l 3)(ρi) g (10–2)

h′

a

= – (0.2) (900)(10)(10–2)

v

or ( A + 2 A) h f = H ⋅ A H hf = 3

or

(m = mass of ice)

3

∆U = – 0.72 J

231. Pressure at the bottom of the vertical cylinder is …(1)

Let at time t, level of left tank is h and that of the right tank is h′. Ah + 2 Ah ′ = AH H –h h′ = ∴ 2 H – h 3h – H ∴ ∆h = h – h ′ = h – = 2 2

dh r

a

Then

r dh

h=0 P = P0 + ρωgh

v = 2g (∆h)

∴

dh A – = av w dt 0

P = P + ρ gh

3h – H = a 2g 2 – dh

or or

a

g (3 h – H ) –∫

dh

H/ 3 H

g (3h – H )

=

a dt A

=

a t dt A ∫0

Solving this, we get t =

2A 3a

2H g

…(1)

The same is the pressure on the lower part of the horizontal cylinder. Now let us consider the cross-section of the horizontal cylinder. Let us consider two symmetrically located strips of width dh at a distance ‘a’ from the centre as shown in figure. Force exerted by water on the upper strip is F1 = { P – ρw g (r + a)} ∆S and the force on the lower strip is F2 = { P – ρw g (r – a)} ∆S Here ∆S is the area of these two strips.

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245

Solutions

is inside the water or 3 cm of its length is immersed in water as shown.

Sum of these two forces, F = F1 + F2 = 2∆S (P – ρw g r ) is independent of a. Hence the total force on piston will be ( P – ρw g r ) ⋅ πr 2 (Σ 2 ∆s = πr 2) In equilibrium, this should be equal to P0 ⋅ πr 2 ( P – ρw gr ) πr = P0πr 2

i.e.

or { P0 + ρw g (h – r )} πr = P0πr 2

2

2

∴ h= r ρ1 1 i.e., in equilibrium, block is half 232. = ρ2 2 immersed in water. Let h′ be the increase in level when the block is pressed by an amount h. Then

When it is depressed by x, net upward force will be the extra upthrust force i.e., F = π (R)2(x ) ρw g = π (10–2)2 x (103)(10) = πx newton The same force will have to be applied downwards. HenceW = ∫ =

x = 10–2 m x =0

∫x = 0

πx ⋅ dx

π × 10– 4 = 1.57 × 10 –4 J 2

(b) Reduction in the force = weight of cylinder = (πR 2)(h)ρc = 9.42 × 10 –2 N

h

234. Net force on the block at a height h from the bottom is

( A – l 2) h ′ = l 2 ⋅ h Substituting the values, we get h ′ = 0.5 h We have to immerse further l/ 2. Hence l = h + h ′ = 1.5 h or h = l/ 3 2 When the block is depressed by h, extra upthrust F = (h + h ′ )ρ2 l 2 ⋅ g = 1.5 l 2ρ2 g h Work has to be done against this upthrust. Hence l/ 3

∫0

x = 10–2

= (π)(10–2)2(4 × 10–2)(075 . × 103)(10)

h′

W=

F dx =

Fdh =

l/ 3

∫0

(1.5 l 2ρ gh) dh

Substituting the values, we get W = 6.75 J

x h=

h0 2

Fnet = upthrust – weight

(upwards)

m 3h = ρ0 4 – g – mg h0 5/2 ρ0 Fnet = 0 at h = So, h =

h0 2

h0 is the equilibrium position of the 2

block.

233. (a) The specific gravity of the cylinder is 0.75, i.e. 75% of its volume or 75% of its length 1 cm 3 cm

For

h>

h0 , weight > upthrust 2

i.e., net force is downwards and for h

∆x 1 and below point P, ∆x 1 > ∆x 2 Now let P1 and P2 be the two minimas on either side of central maxima. Then for P2

O

x2 =∆

1

∆x 2 – ∆x 1 = λ/ 2

In this case, net path difference, ∆x = ∆x1 ~ ∆x2

s =d

∆x 1 = ∆x 2

or

y1 θ1

So or

P1

α

OP1 = y1 and OP2 = y2

(D = 1 m)

and as minima can be on either side of centre O. Therefore, there will be four minimas at positions ± 0.26 m and ± 1.13 m on the screen. α s 1

O

s2

So, the position of minima will be : 1 y1 = tan θ1 = m = 0.26 m 15 3 y 2 = tan θ 2 = m = 1.13 m 7

sin

y = 0.58 m

30°

y = D tan θ = tan θ

=d 1

P1

θ2 θ θ1

central maximum ∆x = 0

θ2

or or or

=

∆x 2

θ2

∴

O

∆ x 2 = ∆ x 1 + λ / 2 = λ + λ / 2 = 3λ / 2 d sin θ 2 = 3λ / 2 3λ (3) (0.5) 3 = = sin θ 2 = 2d (2) (1.0) 4 tan θ 2 =

y2

s1 P2 In this case, ∆x = ∆x1 + ∆x2

(b) When α = 30° path difference between the rays before reaching s1 and s 2 is ∆x 1 = d sin α = (1.0) sin 30° = 0.5 mm = λ

or

y2 =

3 7

3 7

=

y2 D

m = 1.13 m

Similarly for P1 ∆x 1 – ∆x 2 = λ/ 2 or ∆x 2 = ∆x 1 – λ/ 2 = λ – λ/ 2 = λ/ 2 or

d sin θ1 = λ / 2

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Solutions or

sin θ1 =

∴ or

λ 0.5 1 = = 2d (2) (1.0) 4

tan θ1 = y1 =

1 15

1 15

=

y1 D

∆ = λ /6 i. e.

m = 0.26 m

∆x = (µ 2 – µ 1) t = (17 . – 1.4) t = 0.3 t

2

(∆x = 0.3 t)

or

t = 9.3 × 10– 6 m = 9.3 µm

409. (i) Path difference due to the glass slab, ∆x = (µ – 1) t = (1.5 – 1) t = 0.5 t t µ

Due to this slab, 5 red fringes have been shifted upwards. Therefore, ∆x = 5λ red

O s2

= (31/ 6) λ = 0.3 t

31 λ (31) (5400 × 10– 10) ∴ t= = m 6 (0.3) 1.8

6th Minima

5th

∆x = 5λ + λ/ 6

…(1)

Now since 5th maxima (earlier) lies below O and 6th minima lies above O. 1

…(4)

From equations (3) and (4), we find that

Therefore, y-coordinates of the first minima on either side of the central maximum are and y1 = 0.26 m y 2 = 1.13 m In this problem sin θ ≈ tan θ ≈ θ is not valid as θ is large. 408. µ 1 = 1.4, µ 2 = 17 . and let t be the thickness of each glass plate. Path difference at O, due to insertion of glass plate, will be

s1

3 φ = cos2 2 4

or

Maxima

or

0.5 t = (5) (7 × 10– 7 m)

∴

t = thickness of glass slab = 7 × 10 – 6 m

This path difference should lie between 5λ and 5λ + λ/ 2 So, let where

∆x = 5λ + ∆

…(2)

∆ < λ /2

Due to the path difference ∆x, the phase difference at O will be 2π 2π ∆x = (5λ + ∆) φ= λ λ 2π …(3) = 10π + . ∆ λ 3 Imax and since 4 φ I (φ) = Imax cos2 2 3 2 φ Imax = Imax cos 2 4

Intensity at O is given,

∴

(ii) Let µ ′ be the refractive index for green light then ∆ x ′ = (µ ′ – 1) t Now the shifting is of 6 fringes of red light. Therefore, ∆x ′ = 6 λ red ∴ ∴

(µ ′ – 1) =

⇒ (µ ′ – 1) t = 6λ red (6) (7 × 10– 7 ) 7 × 10– 6

= 0.6

µ ′ = 1.6

(iii) In part (i), shifting of 5 bright fringes was equal to 10– 3 m. Which implies that 5ω red = 10– 3 m ∴

ω red =

Now since

[Here ω = Fringe width]

10– 3 m = 0.2 × 10– 3 m 5

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Problems in Physics ω=

ω red ω green = ω red

411. (i) For the lens, u = – 015 . m;

or ω ∝ λ

ω green

∴ ∴

λD d

=

λ green S

2

λ red

ω green = 0143 . × 10– 3 m

S2

1 1 1 – = we have v u f 1 1 1 1 1 = + = + v u f (– 015 . ) (010 . )

Therefore, using

∆ ω = – 5.71 × 10 – 5 m

v = 0.3 m

or

410. (i) Given λ = 6000 Å

Linear magnification, m = First minima

θ θ D

Let b be the width of slit and D the distance between screen and slit. First minima is obtained at b sin θ = λ or (sin θ ≈ θ) bθ = λ λ or θ= b 2λ Angular width of first maxima = 2θ = ∝λ b Angular width will decrease by 30% when λ is also decreased by 30%. Therefore, new wavelength, 30 λ ′ = (6000) – 6000 Å 100 λ ′ = 4200 Å (ii) When the apparatus is immersed in a liquid of refractive index µ, the wavelength is decreased µ times. Therefore, 6000 Å 4200 Å = µ µ=

6000 4200

or

0.25 mm 0.50 mm

1.30 m

= (0143 . – 0.2) × 10– 3 m

∴

A

D = 1.0 m

0.15 m 0.30 m

∆ ω = ω green – ω red

b

0.50 mm 0.25 mm

O

λ green

5 × 10– 7 = (0.2 × 10– 3) 7 × 10– 7 ∴

S1

1

λ red

f = + 010 . m

µ = 1.429

v 0.3 = =–2 . u – 015

Hence two images S1 and S2 of S will be formed at 0.3 m from the lens as shown in figure. Image S1 due to part 1 will be formed at 0.5 mm above its optic axis (m = – 2). Similarly, S2 due to part 2 is formed 0.5 mm below the optic axis of this part as shown. Hence d = distance between S1 and S2 = 1.5 mm D = 1.30 – 0.30 = 1.0 m = 103 mm

λ = 500 nm = 5 × 10– 4 mm Therefore, fringe width, λD (5 × 10– 4) (103) = mm d (1.5) 1 = mm 3 Now as the point A is at the third maxima OA = 3ω = 3 (1/ 3) mm or OA = 1 mm ω=

Note: The language of the question is slightly confusing. The third intensity maximum may be understood as second order maximum (zero order, first order and the second order). In that case OA = 2ω = 2 (1/3) = 0.67 mm.

(ii) If the gap between L1 and L2 is reduced, d will decrease. Hence the fringe width ω will increase or the distance OA will increase.

CHAPTER

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CURRENT ELECTRICITY

412. Since wires MACN and MBDN are connected in parallel between points M and N, the fall of potential is one and the same for the whole length of these wires. The fall of potential is distributed uniformly along each of these wires. Therefore if length MA equals length MB, the potentials at A and B will be the same and current will not pass along wire AB. The same thing can be said of sections MC and MD. If they are of the same length as one another, current will not pass along CD. But wires AB and CD will be at different potentials and if points E and F upon these wires are connected, current will pass along EF and consequently also along AE, BE, FC and FD. This will happen wherever points E and F are chosen along wires AB and CD, since the potential at E will always be higher than that at F. 413. If we connect a voltmeter between points A and B, we find which of the two points has the higher potential. Suppose that we find that the potential at A is higher than that at B. Then we bring the magnetic needle, mounted on a vertical pivot, up under the corresponding wire, e. g ., the upper one. The deflection of the magnetic needle's North pole tells us the direction of flow of current 1 2 through the wire. For example, if the needle's North pole is deflected towards us from the plane of the paper, the current in this wire is flowing through A from right to left. Hence it follows that the source of current in our example is to the right of A. 414. The galvanometer should be connected to the arm to which is normally connected the unknown resistance Rx , and a switch should replace the galvanometer on the bridge’s diagonal. Resistances r1 and r2 should be selected to that the galvanometer shows the same deflection whether the switch is open or

closed. This will mean that there is no current across the bridge’s diagonal and consequently the ratio obtains : RG r1 = , R r2 therefore

RG = R

r1 r2

415. As you know, a voltmeter connected directly to the source of current shows not the e.m.f. but the potential at the source's terminals V = E − Ir , Where r is the internal resistance of the source. Since r is unknown, it is not possible to determine the e.m.f. from the readings of the voltmeter and ammeter with the sliding contact of the rheostat in only one position. But if the sliding contact is moved and the current and potential difference are measured for this new position then we shall obtain the self-evident equation E = V1 + I1r = V2 + I 2r . V − V2 Hence r= 1 I 2 − I1 and the unknown e.m.f. E = V1 + I1r = V1 + I1

V −V I 2 − I1

E2 r2 < . If one cell of E1 (r1 + R) e.m.f. E1 and internal resistance r1 be connected to an external resistance R, then the current passing through the circuit E1 I1 = r1 + R

416. It is possible if

If a second cell of e.m.f. E 2 and internal resistancer2 is then added in series, the current

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Problems in Physics I2 =

E1 + E 2 r1 + r2 + R

P,Q

Clearly E + E2 E1 , < I 2 < I1 if 1 r1 + r2 + R r1 + R

In other words for this inequality to obtain, the internal resistance of the second cell must be sufficiently great. 417. There are two circuits in parallel connecting points A and B, each consisting of half one side of the hexagon, a rhombus and then half another side of the hexagon, connected in series. The resistance of each rhombus is R, therefore the resistance of one circuit is 2R. Therefore the resistance of the whole frame is R. The fact that the vertices of the rhombi are linked at O plays no part in this, since the fact that the two circuits under consideration are identical means that the vertices of both rhombi are at the same potential. Therefore we can calculate the resistances of the two circuits without taking into consideration their link at O. 418. When a positive charge is introduced into a capacitor, a negative charge is induced on the inside faces of the capacitor plates and it remains there, while a positive charge appears on the outside faces. The positive charge on the earthed plate escapes into the earth. The charge on the other plate also escapes into the earth, passing through the galvanometer. So 390 5 will a150 the galvanometer show deflection. But when the positive charge begins to escape from the capacitor, the negative charge will begin to flow into the earth from the plates. The charge flowing from one of the plates will pass through the galvanometer, which will then show a deflection in the opposite direction. 419. The simplified circuit can be redrawn as follows i=

10 E = 63 63 r 144 144

P,Q r r 2

r 2

r

3r 5

r

E

3r 5

r

E r

r

E2 r2 < E1 r1 + R

hence

∴

1440 A 63

= 22.85 A or

r

r 3r 5 E

r

r

63 r 144

21r 13

E

(a)

r = 1Ω

as

Now current distribution in different branches would be as shown in figure (b). 3/ 5 r 39 (i 2 + i 3) Here = = 21/ 13 r 105 ii or

39 1440 (i 2 + i 3) = 105 + 39 63 =

390 A 63

105 i1 = 105 + 39

and

1440 63

1050 A 63 i 2 (3/ 5 + 1) r 8 = = 5 i3 r =

also ∴

8 390 240 i2 = A = 8 + 5 63 63

and

A i3 = = 8 + 5 63 63 i2+i3

P,Q

i 3r 5 i1

r

3r 5

i2

i3

E i2 + i3 r (b)

r

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Solutions Finally currents in all the branches are follows as shown in figure : 810 A 63 10 A 3

E 10A

20 A 3

10 A 3

10 A 21

20 A 21

30 A 21

240 63 A 1440 63 A

810 63 A

390 63 A (c)

Solving this equation we get r = 30 Ω (b) i =

420. (a) Heat produced per second in the resistance is H = i 2r = (0.3)2(500) = 45 J As temperature remains constant, whole heat supplied to the gas is used in doing work against gravitation and atmospheric pressure. If v is the speed of piston, then work done per unit time is W = (P0. S + mg ) v or

45 = (P0. S + mg ) v 45 v= P0S + mg 45 = 5 –3 (10 )(10 ) + (10)(10) = 0.225 m/s

(b) In this case, whole heat is used to raise the temperature of the gas. Hence

422. Let RBP = x then RPC = (a – x ) B E

P A

R AB = c, R AC = b

and

R AP = d

(i rt) 3 n R 2

or

dθ =

(45)(30) = 54.15 ° C 3 (2) (8.31) 2

1 1 1 1 = + + =y R c + x b + (a – x ) d

(say)

current in the circuit will be minimum when or y is minimum or or

−1 (c + x )2

+

dy =0 dx

1 (b + a − x )2 2

or

c+x =1 b + a – x

or

c+x =1 b+ a– x

i 2r = 45 J/sec

but

(given)

Hence the total resistance R of the circuit will be given by 1 1 1 1 = + + R R AB + RBP R AC + RPC R AP

i rt = nC V dθ 2

C

Also

2

dθ =

or

E 1.55 = A = 0.044 A R + r 5 + 30

150 63 A

Hence the current in branch PQ is 10 10 240 i PQ = + + 3 21 63 480 A = 7.62 A = 63

Hence

1.55 1.40 = 1.55 – ⋅r 280 + r

So

10 A 21

Q

V = E – ir

Now

40 A 21

p

421. (a) EMF of the cell = 1.55 V(reading of potentiometer)

or or

2x = (a + b) – c (a + b) – c x= 2

=0

1 R

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Problems in Physics So, minimum value of 1 = R min

1 is R

This circuit can be simplified as : 200V

1 a + b – c c+ 2 3000Ω

1 1 + a + b – c d b+ a – 2

+

A

i

Hence minimum current in the circuit would be E (a + b + c + 4d) 1 imin = E = R min (a + b + c) d 423. (a) (i) When switch S is open, the equivalent circuit is as follows :

OR C,D

A

B

∴

VAC = VCB V1 = V2 = 100 V

or

i=

and 2000Ω B

C

3000Ω

2000Ω

Potential difference across AB is 200 V VAC V1 3000 3 and = = = VCB V2 2000 2

1 A 12 1 A will be distributed as (b) Current i = 12 shown in figure i

2000Ω

2000Ω B 3000Ω

D

3000Ω i2

3 1 3 i1 = A = 3 + 2 12 60 and

C

B i1

D

200V

A

C 2000Ω

i2

2000Ω

(ii) When switch S is closed, equivalent circuit is as follows :

3000Ω

200V

3000Ω A i1

2 V2 = (200) = 80 V 3 + 2

and

200 1200 + 1200

=

3 V1 = (200) = 120 V 3 + 2

Hence

1200Ω

1200Ω

200V

3000Ω

3000Ω 200V

4d + a + b + c d (a + b + c)

A

B

2000Ω

2 2 1 1 or = + + R min a + b + c a + b + c d =

2000Ω C,D

2 1 2 i2 = A = 3 + 2 12 60

Hence current in CD will be : 1 A from D to C. i1 – i 2 = 60

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Solutions 424. (a) Current time graph in the first case will be a straight line as shown in figures. i i0

(dq) = i (dt)

hence area under i – t graph gives the charge flown through the resistance. 1 Hence q = (i 0) t0 2 2q or i0 = t0 i 0 2q = 2 t0 t0

or

So

i = i 0 e – λt

Now

q=

∫ 0 i dt

or

q=

∫ 0 i0 e

or

i q = 0 [e − λt ]∞0 –λ

t

t0

Now since

this with Comparing disintegration. Half life is t1/ 2 = t0 (given) Hence, disintegration constant ln(2) ln(2) λ= = t1/ 2 t0

Therefore, H =

…(1) =

Now from the graph, i – t equation can be written as

H =

i i = i0 – 0 t t0

=

2q 2q i= – 2 t t0 t0

or

dH = i Rdt H =

Hence

t0

∫0

dH

2q 2q – 2 ⋅ t R dt t t dT 0 V Solving this we get H =

t0

∫0

H =

∞

∫0

dt

q ln (2) t0

∞ 2

∫0 i

– λt

…(3)

R ⋅ dt 2

q ln(2) –2λt e R ⋅ dt t0

q2{ln (2)} 2 t02 q2{ln (2)} 2 t02

1 ⋅R 2λ 1 t ⋅R ⋅ 0 2 ln (2)

425. Let at time t, temperature of conductor be T . 2

or

∞

2 1 q R = ln ( 2) 2 t0

2

Since

…(2)

∞

i 0 = qλ =

or

radioactive

4 q2 R 3 t0

Then

2 dT V C⋅ –q = dt R

or

C⋅ – α (T – T0) = dt R

2

dT

or

(b) In the second case current decreases exponentially from a peak value i 0 to zero. i

or

T

∫T

V2 – α (T – T0) R dT 2

=

dt C

=

∫0 C

V – α (T – T0) R solving this equation we get, 0

i0

t

t

dt

– αt V2 T = T0 + 1 – e C αR

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Problems in Physics

426. Let ρ be the resistance per unit length of the wire. If the fault exists at a distance x from the end A and r is the resistance of the leak. A

C

B l–x

x

A

C

4αd 23 (Tmelt – T0) ρ

i 2 = 8 i1 or i 2 = 80 A

l–x

428. Let the current in different branches be i1, i 2 and i1 – i 2 as shown in figure.

r

2Ω A

(a) When B is earthed : The resistance between A and the earth through the cable is 1 (given) xρ + = l 1ρ 1/r + 1/ (l – x )ρ 1 1 1 = + ρ (l 1 – x ) r (l – x ) ρ

or

…(1)

(b) When A is earthed : The resistance between B and the earth through the cable is 1 (l – x ) ρ + = l2ρ 1/r + 1/ρx 1 1 1 = + ρ (l 2 – l + x ) r xρ

or

…(2)

Eliminating r from equations (1) and (2) we get 1 1 1 1 – = – ρ (l 1 – x ) (l – x ) ρ ρ (l 2 – l + x ) ρx or

x = l– x

l1 ( l – l 2 ) l 2 ( l – l1 )

Hence proved.

427. The condition required for melting of the wire should be i 2R = αS (Tmelt – T0) Now or

i12R1 = αS1 (Tmelt – T0) i12 (ρl ) d12

= α (4 d1)l (Tmelt – T0)

(ρ = specific resistance of the wire) or

i12 =

and similarly

4αd13 (Tmelt – T0) ρ

…(2)

From equations (1) and (2) we get,

B

x

r

i 22 =

…(1)

2Ω

B

C

i1 3Ω

7V

1V

(i1–i2)

G

F

i2

D

Applying Kirchhoff’s second law in loop ABFGA we get 7 = 2i1 + 3 (i1 – i 2) or

7 = 5i1 – 3i 2

…(1)

Similarly applying Kirchhoff’s second law in loop ACDGA, we get 7 – 1 = 2i1 + 2i 2 3 = i1 + i 2

or

…(2)

solving these two equations we get i1 = 2 A

and

i2 = 1 A

now power supplied/consumed by a battery, is given by P = ± Ei Plus sign is used when power in supplied (i.e. when current flows from negative terminal to positive terminal inside the battery) and negative sign is used when power is consumed (i.e. when current flows from positive terminal to negative terminal inside the battery) Here power will be supplied by the battery of emf E1 = 7 V and which is equal to + 7i1, or + 14 watt and will be consumed by the battery of emf E 2 = 1 V which is equal to –1i 2 or –1 watt. Note: Here we can see that power is supplied by battery E1 while it is consumed by the battery E 2 and all the three resistances in the circuit and power supplied = power consumed

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Solutions 429. We know that net emf of a battery is equal to the potential difference across the terminals of the battery when no current is drawn from the battery i.e.

VQ = VQ ′ ∴ On superimposing P and P ′ and Q and Q′ etc. we get the simplified arrangement of resistors as shown in figure,

2V

Q

1Ω

4V 0.5Ω

A

B 1Ω

i=0

Q'

P'

R

P

i=0

R'

6V

E=V

when i = 0

In this case also let the current drawn from the battery is zero i.e., 2+ 6 But internal current i = = 4 A will flow in 1+1 the circuit. E = VA – VB

Now

VA – 4 – 2 + (1)(4) = VB

where

A (+)

Req =

Hence

B

47 r0 105

432. (a) Let i1 and i 2 be the currents in two loops respectively ∴ 10 − 10i1 − R(i1 − i 2) + 5 = 0

(for loop 1)

(10 + R) i 2 − Ri1 = − 5

(for loop 2)

VA – VB = 2V

or

O (–)

K 5V

7Ω

Hence equivalent emf is 2V. 430. (a) Current flowing through resistance 5 Ω is 11 ampere

10V

power dissipated = i 2 R

i1

R

3Ω

5V

i2

4Ω

= (121)5 = 605 watt (b) VB + 8 V + 3 V + 12 V − 12 V − 5 V = VC VB + 11 V − 5 V = VC 6 V = VC − VB (c) Both batteries are being charged. 431. After connecting A and O to the terminals of a battery we find, there is symmetry about OA. Therefore the current distribution will the 2 in the 25 same as shown figure.

solving

i2 P

B O

A

R' i1

15 + 2R , 10 + 2R

i2 =

5 + 2R 10 + 2R

P = (i1 − i 2) R =

i2

(b)

5 = 15 −

Q'

It is obvious that potentials, VP = VP ′ , VR = VR ′

×R

⇒

dP =0 dR

R= 5Ω dR R = R0 − ∆θ dθ

a P'

(5 + R)2

⇒ For maximum power dissipation ⇒

i1

i1 =

Power dissipated in R,

R

Q

6Ω

1 ∆θ 2

∆θ = 20° C

⇒ temperature at that instant = 30° C

n–

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Problems in Physics (c) According to Newton’s law: dθ = − k (θ − 20° ) dt 30 dθ ∫ 50 θ − 20 = − kt

436. (i) Equivalent emf (V ) of the battery : Potential difference across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit :

− ln 3 t = − ln 3 100 t = 100 sec 10–3Ω

433.

Ammeter

G2

G1

106Ω

X =8Ω

or

=

∴

X 40 2 = = 12 60 3

or

Voltmeter

i

i=0

i=0 r2

100 Ω

V2

A

B i r1

Variable D.C. voltage

434. The rheostate is as shown in figure. Battery should be connected between A and B and the load between C and B. C A

Load

A

G X

Positive i.e. if A VrVrVV >=

J

Therefore, potential difference between A and B would be VA – VB = V1 – ir1

B

435. (a) There are no positive and negative terminals on the galvanometer because only zero deflection is needed. (b)

A

Current in the internal circuit Net emf V + V2 = 1 i= Total resistance r1 + r2

C

B

B

12Ω C

D

V + V2 VA – VB = V1 – 1 r1 r1 + r2

∴

=

(c)

AJ = 60 cm BJ = 40 cm

If no deflection is taking place. Then the Wheatstone bridge is be said to be balanced. Hence X R = BJ 12 R AJ

V1r2 – V2 r1 r1 + r2

So, the equivalent emf of the battery is V r – V2 r1 V = 1 2 r1 + r2 V1r2 − V2 r1 then V = 0

Note that if 12

∴

V1

2 1

A

B

side of the equivalent battery will become the positive terminal and vice-versa. (ii) Internal resistance ( r) of the battery: r1 and r2 are in parallel. Therefore, the internal resistance r will be given by

or

1/r = 1/r1 + 1/r2 rr r= 12 r1 + r2

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Solutions

P

0Ω

=

20

10

=

0Ω

Q

i2 i1

Rv = 400 Ω V R

0Ω

437. The given circuit actually forms-a balanced Wheatstone bridge (including the voltmeter) as shown in figure

A

20 S

0Ω

B

=

10

100 Ω

=

400 Ω 200 Ω

100 Ω

10 V

100 Ω

(c)

(Bridge is balanced) 10 V (a) Q=200 Ω P = 100 Ω

A

100 Ω

B S = 200 Ω

R = 100 Ω

Therefore, resistance between A and B can be ignored and equivalent simple circuit can be drawn as shown in figure (c). The voltmeter will read the potential difference across resistance Q. 10 Currents i1 = i 2 = 100 + 200 =

10 V (b)

Here we see that

P R = Q S

1 A 30

∴ Potential difference across voltmeter 20 1 = Qi1 = (200) V = V 30 3 Therefore, reading of voltmeter will be

20 V 3

CHAPTER

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ELECTROSTATICS

438. The work goes on increasing the accumulator’s energy. Since the plates of the capacitor are connected all the time to the accumulator terminals, the potential difference applied to them remains constant and consequently their charge must decrease. For the capacitor’s charge ε SV Q = CV = 0 d where C is the capacitor’s capacity, V is the potential difference, S is the area of the capacitor plates and d is the distance between them. If V remains constant and d increases, Q − − must decrease. A + + partial discharge of the capacitor takes place and current flows through the circuit in the direction shown by the arrow in figure, as a result of which the accumulator is charged. When the capacitor plates are moved apart, the energy W of the electric field in the 1 capacitor decreases. In fact W = CV 2 and if 2 V is constant and C decreases, then W also decreases. The energy which is released in the capacitor also goes on charging the accumulator. Thus all the work expended on moving the plates apart goes on increasing the accumulator’s energy and so too does a part of the energy stored in the capacitor. 439. For an electron that is inside the disc at a distance r from the axis to move along a circle, there should be a force putting it to the axis. According to Newton’s second law F = mω 2r

This force is generated by a radial electric field caused by the redistribution of the electrons in the disc and is such that the force acting on the electron is— F = eE = mrω 2 Substituting we have or

V2

∫V

1

Hence

E=–

dV dr

dV = –

mω 2 rdr e

dV = –

mω 2 e

V1 – V2 =

R

∫ 0 r dr

mω 2 R 2 2e

i.e. V1 > V2 or potential at centre is more than the potential at edge. → → 440. (i) Let v1 and v 2 be the velocities of the first and second balls after the given time → interval say ∆t. The angle between v1 and → the initial velocity v is 60°. Hence → → …(1) | ∆ p 1 | =| q1 E ∆ t| = 2m1v sin 60° Here we use the condition that v1 = 2v which → implies that the change in momentum ∆ p1 of the first ball occurs in a direction → perpendicular to the initial velocity v . Because after time ∆t component of momentum in the direction of initial velocity → v is 2m1v cos 60° or m1v which is equal to the initial momentum. So, we can conclude that the electric field is in a direction perpendicular to the initial velocity of both. For the second ball also, momentum in initial direction should not change. Hence

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Solutions m2v 2 cos 30° = m2v 2 v 2 = v sec 30° = v 3

or

oscillations are simple harmonic in nature. 3 2 Now τ = Iα I = MR 2

(ii) For the second ball → → 2 | ∆ p2| =| q2 E ∆ t| = m2v sin 30° 3

or …(2)

Dividing equation (1) by equation (2) we get 3m1 sin 60° q1 = q2 m2 sin 30° q sin 30° 1 q2 = 1 ⋅ ⋅ m2 m1 sin 60° 3

or

=

441. In the figure shown θ h

∴

T = 2π

or

T = 2π h

F cosθ F r

F sinθ

=

x

F=

(r 2 = x 2 + h 2)

1 Qq ⋅ 4 πε 0 (x 2 + h 2)

1 Qq x R or τ = – ⋅ 2 2 2 4 πε 0 x + h x + h 2 x T2) For translational equilibrium of ring in vertical direction, …(2)

For rotational equilibrium, (T1 − T2)

D ωBQR 2 =τ= 2 2

T1 − T2 =

ωBQR 2 D

→

= − 3IabB0 $j

ωBQR 2 τ = MB sin 90° = 2

T1 + T2 = mg

→

=M×B = (− Iabk$) × {(3i$ + 4k$) B0}

ω M = iA = Q (πR 2) 2π

→

→

=r ×F

mga $ a = i$ × (−mgk$) = j 2 2 We see that when the current in the wire PQ is →

→

from P to Q, τ1 and τ 2 are in opposite direction. So they can cancel each other and the loop may remain in equilibrium. So the direction of current I in wire PQ is from P to Q. Further for equilibrium of the loop : →

mga 2 mg I = 6 bB0

3IabB0 =

or

mg ωBQR 2 + 2 2D

As T1 > T2 and maximum values of T1 can be 3T0 , we have 2

→

| τ1| = | τ 2|

…(3)

Solving Eqs. (2) and (3) we have T1 =

y

R

x

2 1 = 1 2

Magnetic moment,

or

Q I

S

496. In equilibrium,

or

DT0 BQR 2

497. (a) and (c) Let the direction of current in wire PQ is from P to Q and its magnitude be I.

towards the top of the page. 495.

ωmax =

mg = T 0 2

(b) Magnetic force on wire RS is : →

→

→

F = I ( l × B)

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Solutions = I [(− bj$) × {(3i$ + 4k$) B0}] or

→

F = IbB0 (3 k$ − 4i$)

and i = 10 A, r1 = 0.08 m 498. (a) Given . m. Straight portions i. e., CD etc. r2 = 012 will produce zero magnetic field at the centre. Rest eight arcs will produce the magnetic field at the centre in the same direction i. e. , perpendicular to the paper outwards or vertically upwards and its magnitude is

Force on CD : Current in central wire is also i = 10 A Magnetic field at P due to central wire, µ i B= 0. 2π x ∴

Magnetic force on element dx due to this magnetic field i µ dF = (i) 0 . . dx (F = ilB sin 90° ) 2π x µ dx = 0 i2 2π x

D C

r1 r2

Therefore, net force on CD is A

F=

x = r2

∫ dF =

µ 0i2

x = r1

2π

0.12

dx µ 0 2 3 = i ln 2 2π x 0. 08

∫

D

i

B i P

B = Binner arcs + Bouter arcs 1 µ i 1 µ 0 i = 0 + 2 2r1 2 2r2

C

x

r + r µ = 0 (πi) 1 2 4π r1r2 Substituting the values, we have

Substituting the values,

(10–7 ) (314 . ) (10) (0.08 + 012 . ) tesla B= (0.08 × 012 . )

or

B = 6.54 × 10

–5

tesla

(Vertically upward or outward normal to the paper) centre is also direction. (b)in vertical Force on AC : Therefore, portions of the circuit i. e. , AC etc. due to the wire at the centre will be zero because magnetic field due to the central wire at these arcs will be tangential (θ = 180° ) as shown.

dx

F = (2 × 10– 7 ) (10)2 ln (1.5) F = 8.1 × 10 – 6 N

(inwards)

Force on wire at the centre : Net magnetic field at the centre due to the circuit is in vertical direction and current in the wire in net force on the wire at the centre will be zero. (θ = 180° ). Hence (i) Force acting on the wire at the centre is zero. (ii) Force on arc AC = 0. (iii) Force on segment CD is 81 . × 10– 6 N (inwards).

→ 499. (a) Magnetic field ( B ) at the origin = Magnetic field due to semicircle KLM + Magnetic field due to other semicircle KNM

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Problems in Physics

or

1 B0qL = 2 mv 0

∴

L=

x=0 +

+

+

+

+

+

+ +

+

+ +

C

+

i

+

+

+

+

D

+

B

+

A

X

X

Z

+

A q

+

Y

+

R

+

+

+

+

+

+

v0 +

+

+

+

+

+

+

+

+

+

L

P

v0

Y

or

A

21 . Now when L′ = 21 . L or R 2 ⇒

+

B

+

v0

L R 1 L = 2 R R L= 2

sin 30° = or

+

R

(b) In part (i)

→ | F ADC| = I ( AC)B

From this we can conclude that net force on a current carrying loop in uniform magnetic field is zero. In the question, segments KLM and KNM also form a loop and they are also placed in a uniform magnetic field but in this case net force on the loop will not be zero. It would had been zero if, the current in any of the segments was in opposite direction.

+

v0

+

If a current carrying wire ADC (of any shape) → is placed in a uniform magnetic field B . perpendicular to the plane ADC. → → Then F ADC = F AC or

x=L +

θ

or Total force on the Loop, → → → → F = F1 + F2 or F = 4BIRi$

mv 0 2B 0 q

B=B0k C

→ → F1 = F2 = 2BIRi$

∴

L mv 0 B 0q

+

and

→ F KM = BI (2R) i$ = 2BIR i$

sin 30° =

+

→ µ qv I F = – 0 0 k$ 4R → → → (b) F KLM = F KNM = F KM

∴

+

µ I = q {(– v 0 i$) × (– i$ + $j )} 0 4R

Here

+

∴ Magnetic force acting on the particle → → → F = q( v × B)

L R mv 0 R= B 0q

sin θ =

+

µ0 I $ $ (– i + j ) 4R

θ = 30°,

500. (a)

+

($j )

+

→ µ I µ I B = 0 (– i$) + 0 4R 4R → µ I µ I B = 0 i$ + 0 $j = 4R 4R

+

∴

2.1 L′ = 2 R > R

L′ > R

Therefore, deviation of the particle is θ = 180° is as shown. → → ∴ v f = – v 0 i$ = v B and

t AB =

πm T = 2 B0 q

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Solutions → → → → → E v × B B $ v0 $ $ or k = 0 501. j = or :i = B v0 E v0 B Force due to electric field will be along Y-axis. Magnetic force will not affect the motion of charged particle in the direction of electric field (or Y-axis). So,

→ qB ∴ v = v 0 cos t m

qB + v 0 sin m

Y E and B

Z

ay =

…(1)

The charged particle under the action of magnetic field describes a circle in x-z plane → (perpendicular to B ) with Z

2πm Bq

Magnetic Field, → B = (B cos 45° ) i$ + (B sin 45° ) $j =

vx X

v0

T =

The path of the particle will be a helix of increasing pitch. The axis of the helix will be along Y-axis. 502. Magnetic moment of the loop, → M = (iA) k$ = (I 0L2) k$

v0 θ

or ω =

2π qB = T m

Initially (t = 0) velocity was along X-axis. → → Therefore, magnetic force (Fm ) will be along → → → positive Z-axis [Fm = q (v 0 × B )]. Let it makes an angle θ with X-axis at time t, then θ = ωt qB …(2) ∴ v x = v 0 cos ωt = v 0 cos t m qB t v z = v 0 sin ωt = v 0 sin m From (1), (2) and (3) → ∴ v = v x i$ + v y $j + v z k$

→ → v × B t 0 v 0B

→ → qB v 0 × B + sin t m B

X

vz

→ E t E

→ qB → q → or v = cos t (v 0 ) + t ( E ) m m

v0

Fe qE = = constant. m m qE Therefore, v y = a y t = .t m

→ v 0 qE + m v0

B $ $ (i + j ) 2

(a) Torque acting on the loop, → → → B $ τ = M × B = (I 0L2k$) × (i + 2 → I L2 B ( $j – i$) τ = 0 ∴ 2 or

$j )

| τ | = I 0L2B

(b) Axis of rotation coincides with the torque and since torque is in $j – i$ direction or Y S

R

…(3)

X P

Q

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Problems in Physics parallel to QS. Therefore, the loop will rotate about an axis passing through Q and S as shown alongside : → | τ| Angular acceleration, α = I

i

i

i x =x x

O

d–x d+x

Where I = moment of inertia of loop about QS.

1

I QS + I PR = I ZZ

3

2 V

B=0

B=0

(From theorem of perpendicular axis) But ∴ ∴

∴

I QS = I PR 4 2I QS = I ZZ = ML2 3 2 I QS = ML2 3 → | τ | I 0L2B 3 I 0B α= = = 2 I ML2 2 M 3

x=– d 3 z=0

∴ Angle by which the frame rotates in time ∆t is 1 θ = α . (∆t)2 2 3 I 0B or . ( ∆t) 2 θ= 4 M 503. (i) Magnetic field will be zero on the y-axis i. e., x = 0 = z I

Y

II i

x = –d

i O

III

d x=– 3

B=0

z=0

Let magnetic field is zero on line z = 0 and x = x . Then magnetic field on this line due to wires 1 and 2 will be along negative z-axis and due to wire 3 along positive z-axis. Thus B1 + B2 = B3 µ i µ i µ i or 0 + 0 = 0 2π (d + x ) 2π x 2π (d − x ) 1 1 1 + = d+x x d–x

or

This equation gives x=±

IV i X x = +d

d 3

Hence there will be two lines d d and x = – x = 3 3

(z = 0)

where magnetic field is zero. (ii) In this part we change our coordinate axes system, just for better understanding. Magentic field can not be zero in region I and region IV because in region I magnetic field will be along positive z direction due to all the three wires, while in region IV magnetic field will be along negative z-axis ⊗ due to all the three wires. It can be zero only in region II and III.

Z

x y-axis

1

2

3

x

x

x

x = –d

x=0

x=d

X

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Solutions There are three wires 1, 2 and 3 as shown in figure. If we displace the wire 2 towards the z-axis, then force of attraction per unit length between wires (1 and 2) and (2 and 3) will be given by F=

µ 0 i2 2π r

=

1 2π i 2π d

f =

or

µ0 πλ

a 1 i = z 2π d µ0 πλ

504. Given that x >> a. Z

k

Y

The components of F along x-axis will be cancelled out. Net resultant force will be towards negative z-axis (or mean position) and will be given by

j

v

i a

X

x

2 x F

θ

θ

F

r

r z

1 x

3 x d

d

µ i2 z Fnet = 2F cos θ = 2 0 2π r r Fnet =

µ0 i2 .z π (z 2 + d 2)

If z Z A,

…(ii)

Though a photon has no rest mass, it acts as if hν it has a mass of 2 . If the photon falls through c a height H , its energy E shall increase by mgH . Since, the photon cannot have speed greater than c, the increase in energy is a consequence of increase in ν. Let the new frequency be ν′. Then or

mv = mv1 + mv 2 …(iii)

Z A = 1, Z B = 2 ∴ A is

and B is

(Deuterium)

4 2 He (Helium)

[Remember, it is given that number of neutrons is same as number of protons.] 554. The photon manifests any increase in its energy by increase in its frequency because its speed is already c which cannot increase any further. This is a fact which students must know. At times this assumption may not be written clearly in the problem.

H E = hν + mgH

hν c hν

Momentum of photon, p =

p = c c2

v = v1 + v 2

…(1)

Squaring (1) we have v 2 = v12 + v 22 + 2v1v 2 Comparing this with (2) 2∆ E 2v1v 2 = m Hence, (v1 − v 2)2 = (v1 + v 2)2 − 4v1v 2 = v2 −

E = hν

Mass of photon, m =

or

From energy conservation, we have 1 1 1 mv 2 = mv12 + mv 22 + ∆ E 2 2 2 2∆ E 2 2 2 …(2) or v = v1 + v 2 + m

Solving (ii) and (iii) 2 1H

[from (1)]

555. The collision will be inelastic if a part of the kinetic energy is used to excite the atom. Let us assume that an energy ∆ E is used in this way. Also, let the neutron and the hydrogen atom move at speeds v1 and v 2 after collision. (a) Using conservation of linear momentum

equation (i) can be revised as Z B2 − Z 2A = 3

hν ′ = hν + mgH gH ν ′ = ν 1 + 2 c

…(1)

4∆ E m

For (v1 − v 2) to be real 4∆ E 4∆ E ≥ 0 or v 2 ≥ v2 − m m The minimum energy that can be absorbed by the hydrogen atom in ground state to go to an excited state is 10.2 eV. Thus, the minimum velocity of neutron for the collision to be inelastic is

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Solutions 2 = vmin

4 × 10.2 × 1.6 × 10−19 1.67 × 10

= 39.08 × 108 or

vmin = 6.25 × 10 m/s

(b) From the law of momentum conservation

φ = hν −

mv − mv = mv1 + mv 2 v1 = − v 2

…(3)

From conservation of energy, we have 1 1 1 1 mv 2 + mv 2 = mv12 + mv 22 + ∆ E 2 2 2 2 2∆ E [using (3)] or 2v 2 = 2v12 + m ∆E or v12 = v 2 − m ∆E 2 Since v1 is real v ≥ m Minimum value of v is obtained when

1 1 Bey mv 2 = m 2 2 πm

1 Bey m 2 πm

1 B 2y 2 e = 4.9 − electron-volt 2 4 π 2 m = (4.9 − 0.4) eV = 4.5 eV Note: The maximum pitch will be corresponding to the electrons having maximum speed or maximum kinetic energy.

h

λ de =

557.

10.2 × 1.6 × 10−19 J

2 vmin =

or

vmin = 3.13 × 104 m/s

2

2 1 m Bey = 4.9 − electron-volts 2 e πm

2me (E ph − W )

∆ E = 10.2 eV ∴

2

Using Einstein’s equation 1 mv 2 = hν − φ 2

4

or

KE =

∴

−27

λ Kα =

1.67 × 10−27 kg

4 3R(Z − 1)2

Given that λ de = (1.09)2 10λ Kα

556. Pitch of the helix is given by y = (v cos θ) . T = v(cos 60° ) T =

vT 2

Where time period of circular motion 2πm T = eB

v cos 60° v

Substituting the values we have, Z − 1 = 23 ∴

Z = 24

558. We have for B dN B = P − λ 2 NB dt t NB dN ⇒ ∫0 P − λ 2BNB = ∫0 dt ⇒

P − λ 2 NB ln = − λ 2t P

60°

⇒ v sin 60°

2π m πmv Pitch y = v cos 60° = Be Be

NB =

P(1 − e − λ 2 t ) λ2

The number of nuclei of A after time t is N A = N 0e − λ1 t

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Problems in Physics Thus ⇒

dN c = λ 1 N A + λ 2 NB dt dN c = λ 1 N 0e − λ1 t + P(1 − e − λ 2 t ) dt

e − λ 2 t − 1 ⇒ N c = N 0 (1 − e − λ1 t ) + P t + λ2

dU = – kr dr Negative sign implies that force is acting towards centre. The necessary centripetal force to the particle is being provided by this force F. Hence

560. F = –

mv 2 = kr r

559. (i) Frequency of electron in nth orbit is given by K 2mz 2e 4

νn = Here

h=

h 2π

…(1)

2π n 3 h 3 K=

and

Similarly ν n + 1 =

2

K mZ e

2π (n + 1)3 h 3

r = rn = …(2)

and

E n+1 = –

mK 2Z 2e 4

2 (n + 1)2 h 2

or

(2n + 1) 2n 2 (n + 1)2

1 2 + K mZ e n ν= 2 3 3 2π n h 1 2 1 + n 2

2 4

…(3)

From equations (1), (2) and (3) we can see that

Centripetal force to electron is provided by the electrostatic force. So, mr2ω 2 =

(ii) For large values of n and

1 → 0, n

So, from equations (1) (2) and (3) we see that ν = νn = νn + 1

kr 2 1 + mv 2 2 2

561. Given that M is the mass of the nucleus (proton) and m the mass of electron. Proton and electron both revolve about their centre of mass (COM) with same angular velocity (ω) but different linear speeds. r1 com r2 m M Let r1 and r2 be the distances of COM from proton and electron. Let r be the distance between the proton and the electron. Then Mr1 = mr2 r1 + r2 = r mr Mr and r2 = ∴ r1 = M +m M +m

νn + 1 < ν < νn n +1≈n

k m

E = nhω = E n

mK 2Z 2e 4

2πh 3

where ω =

Substituting the values, we get

2n 2h 2

mK 2Z 2e 4

nh mω

E =U + K =

Therefore, frequency of photon ν when it jumps from (n + 1)th energy state to nth state will be given by E n + 1 – E n mK 2Z 2e 4 1 1 ν= = 2 – 3 2π h 4 πh (n + 1)2 n =

h …(2) where h = 2π

and total energy

Energy of electron in nth orbit is En = –

mvr = n h

solving equations (1) and (2) we get

1 4 πε 0

2 4

and

…(1)

or or

1 e2 4 πε 0 r 2

Mr 2 1 e2 m ω = 4 πε 0 r 2 M + m Mm 3 2 e2 r ω = M + m 4 πε 0

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Solutions Substituting

If motion of nucleus is not taken into account, in that case

Mm =µ M +m

(reduced mass of proton and electron) e2 µr 3ω 2 = 4 πε 0

E b′ = …(1)

Moment of inertia of the atom about COM is : I = Mr12 + mr22 Mm 2 2 I= r =µr M + m

or

…(2)

According to Bohr’s theory

or

Iω = n h

h where h = 2π

µr 2ω = nh

…(3)

solving equations (1) and (3) for r we get r=

4 πε 0 n 2h 2 µe 2

…(4)

Electrical potential energy of the system is U=–

e2 4πε 0 r

and kinetic energy is 1 1 K = Iω 2 = µr 2ω 2 2 2 But from equation (1) 2

ω =

e2 4 πε 0 µr 3

e2 ∴ K= 8πε 0 r

∴ Total energy of the atom is E = K +U = –

e2 8πε 0 r

Substituting value ofr from equation (4) we get E=–

µe 4 32π 2ε 20 n 2h 2

For ground state n = 1 E1 = –

µe 4 32π 2ε 20 h 2

or binding energy E b = | E1| =

µe4 32π 2 ε 02 h 2

me 4 32π 2ε 20 h 2

,

m > µ ∴ E b′ > E b ∴ percentage increase =

E b′ – E b × 100 Eb

mM m – m – µ M + m = × 100 × 100 = mM µ M +m m = × 100 = 0.00055 × 100 = 0.055% M 562. The product nucleus 198 Hg is in excited state and possesses extra 1.088 MeV energy. If 198 Hg would had been in ground state, the kinetic energy available to electron and antineutrino must have Q = (mAu – mHg ) 931 MeV = (197.968233 – 197.966760) 931 MeV = 1.3714 MeV Since 198 Hg is in excited state, actual kinetic energy available to electron and antineutrino is K = (1.3714 – 1.088) MeV = 0.2834 MeV As β-ray and antineutrino has continuous spectrum starting form zero value, therefore, this is also the maximum kinetic energy of the electron emitted. 563. Let M be the total mass of Uranium mixture. Then the masses of the isotopes 92 U 234 and 92 U

238

in the mixture are

M 1 = 01 . M

and

M 2 = 0.9 M

The mass number of isotopes are A1 = 234

and

A2 = 238

Number of molecules of these isotopes in the mixture are M M N1 = 1 N A and N 2 = 2 N A A1 A2

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Problems in Physics Where N A = Avogadro number. Activity of a radioactive sample ln 2 (t1/ 2 = half life) R = λN = ⋅N t1/ 2 R1 N1 (t1/ 2)2 = R2 N 2 (t1/ 2)1

∴

or

4.5 × 105 = 0.2 2.5 × 105

92 U

t

∫ dt 0

N2 ≈ N0 e – λ 2t

0 .2 = × 100 = 16.67% 0 .2 + 1

and percentage activity of 238

d N2 = λ 1 N0 – λ 2 N2

Note: To find the actual dependence of N 2 on t, we integrate (1). The solution has the form λ1N0 (e – λ1t – e – λ 2t ) N2 = λ 2 – λ1

∴ Percentage activity of 234

∫

which after integration gives λ N 2 = 1 N 0 (1 – e– λ 2t ) λ2

Substituting the values, we have

92 U

N2 0

M A (t ) = 1 2 1/ 2 2 M 2 A1 (t1/ 2)1 R1 01 . M 238 = R2 0.9 M 234

This transforms equation (1) into dN 2 = – (λ 2 N 2 – λ 1 N 0) dt

= (100 – 16.67)% = 83.33%

564. In time interval dt, number of increase of daughter nuclei are

and N2 ≈

for λ1 > > λ 2

λ1 N0 (1 – e – λ 2t ) λ2

565. (a) Electric field is parallel to magnetic field and initial velocity is perpendicular to both the fields. Hence the resultant path will be a helix of increasing pitch. Velocity of α- particle at any ztime t, would be

dN 2 = λ 1 N1dt – λ 2 N 2dt or

x-axis

dN 2 = λ 1 N 0e – λ1 t dt – λ 2 N 2dt ( N1 = N 0 e – λ1 t )

or

dN 2 + λ 2 N 2 = λ 1 N 0e – λ1 t dt (t1/ 2)1 < < (t1/ 2)2

(t1/ 2 = half life)

We can assume that N 20 ≈ N 0 so that N 2 = N 0e – λ 2 t (N 20 = number of daughter atoms at time t = 0) Physically this means that parent nuclei practically instantly transform into daughter nuclei, which then decay according to the law of radioactive decay with decay constant λ 2. Case 2 : When λ 1 < < λ 2 i.e.

θ v0

→ q E v = α 0 t $i + v 0 cos θ $j – v 0 sin θ k$ mα q B where θ = ωt and ω = α mα (b) Speed of α-particle at time t is → | v |=

2

qα E 0 t + v 02 mα

→ Given| v | = 2v 0 at, m t = 3 × 10 7 α sec qα E 0

(t1/ 2)1 >> (t1/ 2)2 In this case number of parent nuclei can be assumed to remain constant over a sizable time interval and is equal to N 0.

v0

y

…(1)

Case 1 : When λ 1 >> λ 2 i.e.

for λ1 < < λ 2

So, 4v 02 = ( 3 × 107 )2 + v 02 or

v 0 = 107 m/s

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Solutions (c) When α-particle is emitted with velocity v 0 from a stationary nucleus X, daughter nucleus Y recoils with speed v which is given by mY v = mα v 0 m v v= α 0 mY

or

=

Similarly activity of B after t = 20 days (four half lives of B) is 4

1 RB = RB 0 = 0.0625 RB 0 2 Now it is given that

4.003 × 107 m/s 221.03

R A + RB = 20% of 1010

= 1.81 × 105 m/s ∴ Total energy released during α-decay of a nucleus X 1 1 E = mYv 2 + mα v 02 2 2 4.003 221.03 = × (1.81 × 105)2 + × (107 )2 2 2 1.67 × 10–27 MeV –13 1.6 × 10 = 2126 . MeV ∴ Mass lost during α-decay 2126 . ∆m = u = 0.00228 u 931

or 0.25 R A0 + 0.0625 RB 0

= 0.2 × 1010 dps

…(2)

Solving equations (1) and (2) we get RA 0 = 0.73 × 1010 dps and (b) ∴

RB0 = 0.27 × 1010 dps R A0 RB 0 N A0 N B0

=

λ A N A0 λ B NB 0

=

(t1/ 2)B N A0 ⋅ (t1/ 2) A N B 0

RA 0 (t1 / 2 ) A = RB0 (t1 / 2 ) B 0.73 10 = = 5 .4 0.27 5

567. Let E n and E m be the energies of electron in

∴ Mass of nucleus X

nth and mth states. Then

mX = mY + mα + ∆m

E n – E m = hν 0

= 221.03 + 4.003 + 0.00228 mX = 225.035 u Mass defect in nucleus X. or ∆m ′ = [92m p + (225 – 92) mn – mX ]

mm

2

1 R A = R A0 = 0.25 R A0 2

In the second case when the atom is moving with a velocity v. Let v′ be the velocity of atom after emitting the photon. Applying conservation of linear momentum,

= 1.9348 u

∴ Binding energy per nucleon of nucleus 1.9348 × 931 X= MeV 225 = 8.00 MeV 566. (a) Let R A0 and RB 0 be the initial activities of A and B. Then 10

R A0 + RB 0 = 10

dps

…(1)

Activity of A after time t = 20 days (two half lives of A) is

…(1)

v′

v

mv = mv ′ +

ν

hν c

(m = mass of hydrogen atom) hν or v ′ = v – mc Applying conservation of energy 1 1 E n + mv 2 = E m + mv ′ 2 + hν 2 2

…(2)

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Problems in Physics or hν = (E n – E m ) +

1 m (v 2 – v ′ 2 ) 2

= hν 0 +

2 1 2 hν m v – v – 2 mc

= hν 0 +

h 2ν 2 2hνv 1 2 m v – v 2 – 2 2 + 2 mc m c

= hν 0 +

hνv h 2ν 2 – c 2mc 2

Here the term neglected

2 2

h ν

2mc 2

1 N = N0 2

So energy released per day ln (2) –13 = N (8.4 × 10 ) J t1/ 2 But only 10% of this energy is used as electric power. So,

hνv c

hν = hν 0 +

or

v ν 1 – = ν 0 c

10 ln (2) –13 7 N { 8.4 × 10 } = 1.2 × 10 J 100 t1/ 2

–1

or

v ν = ν 0 1 – c

or

v ν ≈ ν 0 1 + c

as

v t1. Some more negative charge reaches the upper plate which repels further movement of electrons from lower plate to upper plate. That is why current is now decreasing. At time t = t2, the potential difference between the plates is equal to the stopping potential. Hence current becomes zero. Therefore, potential difference between the plates for t > t2 is equal to the stopping potential or 1.5 V (corresponding to maximum kinetic energy of electrons) 573. Energy of a photon is given by E = pc Applying conservation of linear momentum, we get P2 P1

Initial position

θ φ electron (Rest)

p2 = p12 + p22 – 2 p1 p2 + 2 p1m0c – 2 p2m0c

Ic

C

VCE

B

800Ω Ib RB

RL 0.8v

E Ie

VBB

VCC = 8V

collector current voltage drop across RL 0.8 Ic = = RL 800 or

I c = 10–3 A

(i) Collector emitter voltage is VCE = 8 – 0.8 = 7.2 V (ii) Current gain β = or

P Final position

…(5)

From equations (3) and (5) we can show that 1 1 1 – = (1 – cos θ) p2 p1 m0c 574. The circuit is shown in figure

t1 = 10 9 s

photon

…(4)

or

Ib =

Ic Ib

Ic 10–3 = A β 25/ 26

I b = 1.04 × 10 –3 A

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Solutions (iii) Voltage gain Routput 25 800 =β⋅ = × = 3.846 26 200 Rinput

N

M R1 = 500kΩ

B

= (current gain) (voltage gain) 25 = (3.846) = 3.698 26

E ie

575. Potential difference across RL is

A

I c RL = 8 – VCE = (8 – 4) V = 4 V 4 4 = = 10 3 Ω ∴ RL = I c 4 × 10–3

RB

Similarly applying Kirchhoff’s law between AEBCNP we get VCE + i c R2 = Vc or VCE = Vc – i c R2 = 5.5 – (5.2 × 10–3)(103)

RL IC

Ib

E

VBE

Further β =

Ic Ib

∴

Ic β

Ib =

8V

C

B

VCE

4 × 10–3 = = 4 × 10–5 A 100 Now potential difference across RB is I bRB = 8 – VBE = 8 – 0.6 = 7.4 V 7.4 7.4 ∴ RB = = Ω Ib 4 × 10–5

…(2) = 0.3 V From equation (1) VBE = VB – VE = 0.5 V and from equation (2) VCE = VC – VE = 0.3 V ∴ VBC = VB – VC = (0.5 – 0.3) V = 0.2 V This is a NPN transistor, and VBE , VBC both are positive i.e., VB > VE and VB > VC or emitter-base and collector-base both are forward biased. Hence the circuit can not be used as an amplifier. The transistor is in saturation mode. 577. λ AB = λ A + λ B ∴

1 1 1 (Here T = half life) = + T AB T A TB

or

1 = 2+ 4 = 6 T AB

N0

5

or RB = 1.85 × 10 Ω 576. Applying Kirchhoff’s law between ABMNP we get VBE + i bR1 = Vc

A

1 hr = T A 2

N0 2

B

1 hr = 4TB

…(1)

5

5

A+B 1 hr = 3T AB 2

or VBE = Vc – i bR1 = 5.5 – (10 × 10–6) (500 × 103) = 0.5 V

Vc = +5.5V

R2 = 1kΩ ic C

ib

(iv) Power gain

P

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Problems in Physics ∴

1 hr 6

T AB =

E4 = − 8

1 Therefore, after 2 hours N 0 nuclei will be 2 left. 578. Given, R1 = 2.5% of R2 Here R1 is the activity of old sample and R2 the activity of 10 years old bottle. 2.5 − λt Now, R0e − λt1 = R e 2 100 0 100 = 40 2.5 λ (t1 − t2) = ln (40) = 3.7 0.693 or (t1 − t2) = 3.7 20 ∴ t1 − t2 = 106.8 Given t2 = 10 years ∴ t1 ≈ 116.8 years 579. Let N 0 be the initial number of nuclei of 238 U. After time t ∴

e λ( t1

− t2 )

=

1 NU = N0 2 Here

∴

E2 = −

(4)2 13.6 (2)2

= − 0.85 eV = − 3.4 eV

∆ E = E 4 − E 2 = 2.55 eV Kmax = Energy of photon − work function = 2.55 − 2.0 = 0.55 eV

1 1 581. ∆ E = hν = Rhc (Z − b)2 2 − 2 n1 n 2 For K-series b = 1

L

n2 = 2 ∆E n1 = 1

K

1 1 ν = Rc (Z − 1)2 2 − 2 n1 n 2 Substituting the values 4.2 × 1018 = (1.1 × 107 ) (3 × 108) (Z − 1)2

n

1 1 × − 1 4

n = number of half lives =

t t1/ 2

=

1.5 × 109 4.5 × 109

1 NU = N0 2

=

1 3

1/ 3

NU = N Pb

1 2

1 2

1/ 3

1/ 3

1 1− 2

1/ 3

∴

(Z − 1)2 = 1697

or

Z − 1 ≈ 41 or Z = 42

582. Let n 0 the number of radioactive nuclei at time

and N Pb = N 0 − N U = N 0 1 − ∴

and

13.6

t = 0. Number of nuclei decayed in time t are given by n 0 (1 − e − λt ), which is also equal to the number of beta particles emitted during the same interval of time. For the given condition,

= 3.846

580. Wavelengths corresponding to minimum wavelength (λ min ) or maximum energy will emit photoelectrons having maximum kinetic energy. λ min belonging to Balmer series and lying in the given range (450 nm to 750 nm) corresponds to transition from (n = 4 to n = 2). Here,

n = n 0 (1 − e −2λ )

…(1)

(n + 0.75n) = n 0 (1 − e −4 λ )

…(2)

Dividing (2) by (1) we get 1.75 =

1 − e −4 λ 1 − e −2 λ

or

1.75 − 1.75e −2λ = 1 − e −4 λ

∴

1.75e −2λ − e −4 λ =

3 4

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Solutions Let us take e −2λ = x

584. (a) Total 6 lines are emitted. Therefore, n (n − 1) =6 2

Then the above equation is, x 2 − 1.75x + 0.75 = 0 or or

So, transition is taking place between m th energy state and (m + 3)th energy state. ∴

∴ From Eq. (3) either e −2λ = 1 or e −2λ =

n=4

or

1.75 ± (1.75)2 − 4 (0.75) x= 2 3 x = 1 and 4

E m = − 0.85 eV Z2 −13.6 2 = − 0.85 m

or

3 4

but e −2λ = 1 is not accepted because which means λ = 0. Hence 3 e −2 λ = 4

Z = 0.25 m

or

E m + 3 = − 0.544 eV

Similarly or

−13.6

−2λ ln (e) = ln (3) − ln (4)

or ∴

= 0.14395 s−1 1 = 6.947 sec λ

583. Maximum kinetic energy photoelectrons would be

of

the

i

8µA

(a) (b)

= − 0.544 …(2)

(b) Smallest wavelength corresponds to maximum difference of energies which is obviously E m + 3 − E m ∴ ∆Emax = − 0.544 − (−0.85) = 0.306 eV hc ∴ λ min = ∆Emax =

1240 = 4052.3 nm 0.306

585. Area of plates A = 5 × 10−4 m 2 distance between the plates d = 1 cm = 10−2 m

4 µA –2V

(m + 3)2

Solving Eqs. (1) and (2) for Z and m we get, m = 12 and Z = 3

Substituting the given values, 1 λ = 0.6931 − × (1.0986) 2

∴ Mean life tmean =

Z2

Z = 0.2 m+3

or

= ln (3) − 2 ln (2) 1 λ = ln (2) − ln (3) 2

…(1)

Anode potential V

Kmax = E − W = (5 − 3) eV = 2 eV Therefore, the stopping potential is 2 volt. Saturation current depends on the intensity of light incident. When the intensity is doubled the saturation current will also become two fold. The corresponding graphs are shown in figure.

(a) Number of photoelectrons emitted upto t = 10 s are (number of photons falling in unit time) × (Area × time) n= 106 1 = 6 [(10)16 × (5 × 10−4 ) × (10)] 10 = 5.0 × 107

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Problems in Physics Rates

(b) At time t = 10 s charge on plate A,

X

Y

q A = + ne

λXNX

= (5.0 × 107 ) (1.6 × 10−19) = 8.0 × 10−12 C

(ii) Given N Y (t) =

and charge on plate B, = 25.7 × 10−12 C ∴ Electric field between the plates (q − q A) E= B 2 Aε 0 E=

(25.7 − 8.0) × 10

or λ X ( N 0e – λ X t ) = λ Y

−12

N 0λ X λX – λY [e – λ Y t − e − λ X t ]

2 × (5 × 10−4 ) (8.85 × 10−12)

λ X – λ Y e– λ Y t = –λ t – 1 λY e X

or

Increase in energy of photoelectrons or

= (eEd) joule = (Ed) eV = (2 × 103) 10−2 eV = 20 eV = (20 + 3) V = 23 eV 586. (i) Let at time t = t, number of nuclei of Y and Z are N Y and N Z . Then

20

N0 = 10 NX = N0e–λxt

Zero NY

Z

t=

dN Y = λXNX – λY NY dt

…(2)

dN Z = λY NY dt

…(3)

t = 16.48 s

or

(iii) The population of X at this moment N X = N 0e – λ X t = (1020) e –( 0.1)(16. 48) N X = 1.92 × 1019

Rate equation of the populations of X , Y and Z are …(1)

λ 1 ln X (λ X – λ Y ) λ Y

Substituting the values of λ X and λ Y , we have 1 . 01 t= ln = 15 ln (3) (01 . – 1/ 30) 1/ 30

Zero NZ

dN X = – λXNX dt

– λY) t

λ (λ X – λ Y ) t ln (e) = ln X λY

or

Energy of photoelectrons at plate B

Y

λX = e( λ X λY

or

= E − W = (5 − 2) eV = 3 eV

X

…(4)

[From equation 2]

(c) Energy of photoelectrons at plate A

and

N 0λ X [e – λ Y t − e − λ X t ] λX – λY

λ X NX = λ Y NY

i.e.

= 2 × 10 3 N/C

t=0 t=t

λYNY

For N Y to be maximum dN Y (t) =0 dt

qB = (33.7 × 10−12 − 8.0 × 10−12)

or

Z

NY =

NX λ X λY

= (1.92 × 1019)

(From equation 4) (01 .) (1/ 30)

= 5.76 × 1019 ∴ NZ = N0 – NX – NY = 1020 – 1.92 × 1019 – 576 . × 1019 or

N Z = 2.32 × 1019

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Solutions 587. The reactor produces 1000 MW power or 109 W power or 109 J/s of power. The reactor is to function for 10 years. Therefore, total energy which the reactor will supply in 10 years is

K=

P m

But since the efficiency of the reactor is only 10%, therefore, actual energy needed is 10 times of it or 31536 . × 1018 J. One Uranium atom liberates 200 MeV of energy or 200 × 1.6 × 10–13 J or 3.2 × 10–11 J of energy. = 0.9855 × 1029

Substituting the values, we get K=

(1151 . × 10–19)2 2 (4.002 + 223.610) (1.67 × 10–27 ) (4.002 × 1.67 × 10–27 ) (223.61 × 1.67 × 10–27 )

K = 10–12 J

or number of kg-moles of uranium needed are n=

6.02 × 1026

= 1637 .

m = (n)M = (1637 . )(235) kg m ≈ 38470 kg

or

m = 3.847 × 10 4 kg

588. (i) Given mass of α-particle, m = 4.002 a.m.u. and mass of daughter nucleus a.m.u. de-Broglie M = 223.610 wavelength of α-particle, λ = 576 . × 10

–15

or

h 6.63 × 10–34 kg-m/s = λ 576 . × 10–15

. × 10 p = 1151

–19

kg-m/s

10–12 1.6 × 10–13

MeV = 6.25 MeV

K = 6.25 MeV

(ii) Mass defect, 6.25 amu = 0.0067 amu ∆m = 931.470 Therefore, mass of parent nucleus = mass of α-particle + mass of daughter nucleus + mass defect (∆m) = (4.002 + 223.610 + 0.0067) amu = 227.62 amu Hence mass of parent nucleus is 227.62 amu. 589. (a) Let ground state energy (in eV) be E1 Then from the given condition

m

So, momentum of α-particle would be p=

K= or

Hence total mass of uranium required is or

M

1 amu = 1.67 × 10–27 kg

So, number of uranium atoms needed are

0.9855 × 10

K2

p

= 31536 . × 1017 J

29

2

p2 M + m 2 Mm K1

= (109 J/s) (10 × 365 × 24 × 3600 s)

3.2 × 10–11

2

p 1 1 p p + = + 2m 2M 2 m M

K = K1 + K 2 =

E = (Power) (time)

. × 1018 31536

2

…(1)

From law of conservation of linear momentum, this should also be equal to the linear momentum of the daughter nucleus (in opposite direction). Let K1 and K 2 be the kinetic energies of α-particle and daughter nucleus. Then total kinetic energy in the final state is :

or or and or

E 2 n – E1 = 204 eV E1 – E1 = 204 eV 4n 2 1 E1 2 – 1 = 204 eV 4n E 2n – E n = 40.8 eV E1 E – 21 = 40.8 eV 2 4n n

…(1)

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Problems in Physics – 3 E1 2 = 40.8 eV 4n

or

…(2)

From equation number (1) and (2) 1 1– 4n 2 = 5 or 1 = 1 + 15 3 4n 2 4n 2 4n 2

4

or

n2

or = 1 or

n=2

and Kmax = E i – work function = (10.6 – 5.6) eV = 5.0 eV ∴

E1 = – 217.6 eV E1 = – (13.6)Z 2 E1 –217.6 = = 16 –13.6 –13.6

∴

Z2 =

∴

Z =4

Emin = E 2n – E 2n – 1 E E1 = 12 – 4n (2n – 1)2

n=2

E1 E1 7 E1 =– – 144 16 9 7 =– (–217.6) eV 144

=

∴

Emin = 10.58 eV

(b) Energy of incident photon E i = 10.6 eV = 10.6 × 1.6 × 10–19 J = 16.96 × 10–19 J Energy incident per unit area per unit time (intensity) = 2 J ∴ Number of photons incident on unit area in 2 unit time = = 118 . × 1018 16.96 × 10–19 Therefore, number of photons incident per unit time on given area (1.0 × 10–4 m 2) = (118 . × 1018) (1.0 × 10–4 ) = 118 . × 1014

n = 6.25 × 1011 Kmin = 0

From equation (2) 4 E1 = – n 2 (40.8) eV 3 4 = – (2)2 (40.8) eV 3 or

But only 0.53% of incident photons emit photoelectrons ∴ Number of photoelectrons emitted per second (n) 0.53 . × 10 14) n= (118 100

Kmax = 5.0 eV

(i) From the past experience it has been observed that one question in modern physics is usually asked from Bohr’s theory + photoelectric effect. Sometimes only one question is asked mixing both the theories and otherwise they are asked separately as two parts of the same question. (ii) In Bohr’s theory questions may be asked in near future based on dependence of rn, v n and E n on mass of the electron. For this remember that 1 rn ∝ m En ∝ m and v n is independent of mass of electron. Sometimes electron is replaced by some other particle say µ-meson, which is 210 times heavier than electron. Now again two cases are possible. Case 1 : When mass of nucleus >> mass of µ-meson or nucleus is assumed to be stationary. In that case rn decrease 210 times, whereas | E n| increases 210 times and v n remains the same. Case 2 : When mass of nucleus is comparable to mass of µ-meson or any other particle. In that case m is replaced by reduced mass of nucleus and µ-meson; which will definitely be less than 210. Let us say it comes out to be 205. Then rn will decrease by 205 times while | E n| will increase 205 times.

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Solutions 590. Given : work function W = 1.9 eV Wavelength of incident light

Hence the energy of emitted photons in the range of 2 eV and 4 eV are 3.376 eV during combination and 3.94 eV and 2.64 after combination. 591. (a) Let at time ‘t’, number of radioactive nuclei are N.

λ = 400 nm ∴ Energy of incident light, hc E= = 31 . eV λ

A

Rate of producton = α

(Substituting the values of h, c and λ) Therefore, maximum kinetic energy photoelectron

Rate of decay = λN

of

Kmax = E – W = 31 . – 1.9 = 1.2 eV

= 1.2 eV

+

He in fourth excited state or n = 5 (Z = 2) E5 = – 2.176 eV

α-particles +2 (He ions)

He

t=t N=N n = 5, E5 = – 2.176 eV

Now the situation is as shown below : .. .. .. .. .. .. .. .. .. .. – e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . Kmax

or

+

+

Energy of electron in 4th excited state of He (n = 5) will be E 5 = – 13.6

Z2 n2

E 5 = – (13.6)

eV 2

(2)

(5)2

. eV = – 2176

Therefore, energy released during the combination = 1.2 – (–2176 . ) = 3.376 eV Similarly energies in other energy states of He+ will be E 4 = – 13.6 E 3 = – 13.6 E 2 = – 13.6

(2)2 (4)2

(2)2 (3)2 2

(2)

22

= – 3.4 eV = – 6.04 eV

Net rate of formation of nuclei of A. dN dN = α – λN or = dt dt α – λN N

∫N

0

t dN = ∫ dt 0 α– N

Solving this equation, we get 1 N = [α – (α – λN 0 ) e– λt ] λ

…(1)

(b) (i) Substituting α = 2λN 0 ln (2) and t = t1/ 2 = λ in equation (1), we get 3 N = N0 2 (ii) Substituting α = 2λN 0 and t → ∞ in equation (1), we get α N = = 2 N 0 or N = 2N 0 λ 592. From the figure it is clear that 2A

= – 13.6 eV

N

N p-loops

The possible transitions are ∆ E 5 → 4 = E 5 – E 4 = 1.3 eV < 2 eV

(p + 1) loops

∆ E 5 → 3 = E 5 – E 3 = 3.94 eV

2.5 A

∆ E 5 → 2 = E 5 – E 2 = 11.5 eV > 4 eV

p ⋅ (λ/ 2) = 2 Å

∆ E 4 → 3 = E 4 – E 3 = 2.64 eV ∆ E 4 → 2 = E 4 – E 2 = 10.2 eV > 4 eV

( p + 1) ⋅ λ/ 2 = 2.5 Å ∴

λ/ 2 = (2.5 – 2.0) Å = 0.5 Å

λ/2

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Problems in Physics λ = 1 Å = 10–10 m

or

Therefore, energy released per second

(i) de Broglie wavelength is given by h h λ= = p 2 Km

= (0.08 × 107 × 200 + 0.92 × 107

K = kinetic energy of electron

∴ Power output (in watt) = Energy released per second (J/s)

∴ K=

h

. ) MeV × 5136 8

= 2.072 × 10 MeV

2

= (2.072 × 108) (1.6 × 10–13) (J/s)

2mλ 2

= 2.415 × 10–17 J =

∴ Power output = 3.32 × 10 – 5 watt 594. t1 / 2 = 4.5 × 10 9 years

(6.63 × 10–34 )2 2 (91 . × 10–31)(10–10)2

∴ Disintegration constant λ =

2.415 × 10 = eV 1.6 × 10–19 –17

∴ (ii) ∴

or

K = 150.8 eV N The

N

least value of d will be

when only one loop is formed λ or dmin = 0.5 Å dmin = 2

593. The reaction involved in α-decay is 248 96 Cm

→

244 94 Pu

+ 42He

Mass defect ∆m = Mass of

248 96Cm

– Mass of

244 94 Pu

– Mass of 42He = (248.072220 – 244.064100 – 4.002603) u

E α = (0.005517 × 931) MeV = 5136 . MeV Similarly,

Efission = 200 MeV

Here N 0 = Number of atoms of U 238 at time t=0 and N = Number of atoms of U 238 at time t = t But from N = N 0 e – λt we have N0 = e λt N

tmean = 10

1 s= λ

∴ Disintegration constant λ = 10–13 s–1 Rate of decay at the moment when number of nuclei and N (= 1020) = λ N = (10–13) (1020) = 107 disintegration per second. Of these disintegrations, 8% are in fission and 92% are in α-decay.

…(2)

Equating (1) and (2), we have e λt = 4 / 3 or

13

years–1

Ratio of U 238 to Pb 206 is 3 : 1 i.e. 3 parts of U 238 and 1 part of Pb 206. We may assume that at t = 0, all 4 parts were of U 238. Therefore, we can say that N0 4 (of U 238) …(1) = 3 N

(given)

Mean life is given as

0.693 4.5 × 109

= 1.54 × 10–10 years–1

= 0.005517 u Therefore, energy released in α-decay will be

λ=

0.693 t1/ 2

or ∴

λt ln (e) = ln (4 / 3) λt = 0.29 0.29 t= λ 0.29 = 1.54 × 10–10 = 1.88 × 10 9 years

Therefore, age of ore is approximately 1.88 × 109 years. 595. (a) (i) Kinetic energy of electron in the orbits

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Solutions of hydrogen and hydrogen like atoms = |Total energy| ∴

Kinetic energy = 3.4 eV

(ii) The de Broglie wavelength is given by h h λ= = P 2Km

Power of the point source is 3.2 × 10–3 watt or 3.2 × 10– 3 J/s Therefore, energy emitted per second, E 2 = 3.2 × 10–3 J. Hence number of photons emitted per second

(K = kinetic energy of electron)

n1 =

Substituting the values, we have λ=

2 (3.4 × 1.6 × 1019 J) (91 . × 10–31 kg) λ = 6.63 × 10

m

n2 =

(b) (i) In 10 second, number of nuclei has been reduced to half (25% to 12.5%). Therefore, its half life is t1/ 2 = 10 s Relation between half life and mean life is t 10 tmean = 1/ 2 = s ln (2) 0.693 tmean = 14.43 s

2

4 π(0.8)

=

4.0 × 1015 4 π(0.64)

A = πr 2 = π (8 × 10–3)2 m 2 ≈ 2.01 × 10–4 m 2 Therefore, number of photons incident on the sphere per second are n3 = n2 A = (5.0 × 1014 × 2.01 × 10–4 ) ≈ 1011 per second

t1 / 2

→ 12.5% → 6.25% ∴

n1

The area of metallic sphere over which photons will fall is :

100% → 50% → 25% t1 / 2

8.0 × 10–19

≈ 5.0 × 1014 photon/sec - m 2

(ii) From initial 100% to reduction till 6.25%, it takes four half lives. t1 / 2

3.2 × 10–3

Number of photons incident on unit area at a distance of 0.8 m from the source S will be

λ = 6.63 Å

or

or n1 =

n1 = 4.0 × 1015 photons/sec.

(6.6 × 10–34 J - s) –10

E2 E1

t1 / 2

t = 4 t1/ 2 = 4(10) s = 40 s

But since one photoelectron is emitted for every 106 photons hence number of photoelectrons emitted per second,

t = 40 s

n=

596. (a) Energy of emitted photons E1 = 5.0 eV = 5.0 × 1.6 × 10–19 J = 8.0 × 10–19 J

or

n3 10

6

=

1011 106

= 105 per second

n = 10 5 per second

(b) Maximum kinetic energy of photoelectrons Kmax = Energy of incident photons – work function

S

0.8 m

= (5.0 – 3.0) eV = 2.0 eV r = 8.0 × 10–3 m

= 2.0 × 1.6 × 10–19 J Kmax = 3.2 × 10–19 J The de-Broglie wavelength photoelectrons will be

of

these

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Problems in Physics λ1 =

h = p

= (1.6 × 10–19) × 105 C

h 2Kmax m

= (1.6 × 10–14 ) C

(Here h = Planck’s constant and m = mass of electron) ∴ λ1 =

6.63 × 10–34 2 × 3.2 × 10

= 8.68 × 10

–10

–19

× 91 . × 10

–31

m = 8.68 Å

Wavelength of incident light λ 2 (in Å) 12375 = E1 (in eV ) or

12375 Å = 2475 Å λ2 = 5

Therefore, the desired ratio is 2475 λ2 = = 285.1 8.68 λ1 (c) As soon as electrons are emitted from the metal sphere, it gets positively charged and acquires positive potential. The positive potential gradually increases as more and more photoelectrons are emitted from its surface. Emission of photoelectrons is stopped when its potential is equal to the stopping potential required for fastest moving electrons. (d) As discussed in part (c), emission of photoelectrons is stopped when potential on the metal sphere is equal to the stopping potential of fastest moving electrons. Since

Kmax = 2.0 eV

Therefore, stopping potential V0 = 2 volt. Let q be the charge required for the potential on the sphere to be equal to stopping potential or 2 volt. Then q q 1 ⋅ = (9.0 × 109) 2= 4 πε 0 r 8.0 × 10–3 ∴

q = 178 . × 10–12 C

Photoelectrons emitted per second = 105 [part a] or charge emitted per second

Therefore, time required to acquire the charge q will be t=

q 1.6 × 10

–14

sec =

1728 . × 102 sec 1.6

t ≈ 111 second

or

597. From the given conditions E n – E 2 = (10.2 + 17) eV = 27.2 eV and

…(1)

E n – E 3 = (4.25 + 5.95) eV = 10.2 eV

…(2)

Equation (1) and (2) gives E 3 – E 2 = 17.0 eV Z 2 (13.6)(1/ 4 – 1/ 9) = 17.0

or ⇒

Z 2 (13.6)(5/ 36) = 17.0

⇒

Z2 = 9 ⇒ Z = 3

From equation (1) Z 2 (13.6)(1/ 4 – 1/n 2) = 27.2 or

(3)2 (13.6)(1/ 4 – 1/n 2) = 27.2

or

1/ 4 – 1/n 2 = 0.222 1/n 2 = 0.0278

or

n 2 = 36 ∴ n = 6

or

598. λ = Disintegration constant =

0.693 0.693 = hrs–1 = 0.0462 hr –1 t1/ 2 15

Let R0 = Initial activity = 1 microcurie = 37 . × 104 disintegrations per second. r = Activity in 1 cm 3 of blood at t = 5 hrs =

296 disintegration per second 60

= 4.93 disintegrations per second, and R = Activity of whole blood at time t = 5 hrs Then, total volume of blood should be R R e – λt V= = 0 r r

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Solutions Substituting the values, we have

∆ E can have the following values

37 . × 104 –( 0. 0462)( 5) V= cm 3 e 4.93 3

V = 5.95 × 10 cm

∆E1 = – 13.6 – (–54.4) eV = 40.8 eV Substituting in (5), we get K1 + K 2 = 24.2 eV

3

Solving (4) and (6), we get

or V = 5.95 litre 599. (i) Let K1 and K 2 be the kinetic energies of neutron and helium atom after collision and ∆E be the excitation energy. K1

X

pi = p f 2Km = 2 (4m)K 2 cos θ

…(1)

( p = 2Km ) Similarly, applying conservation of linear momentum in Y-direction, we have 2K1m = 2 (4m) K 2 sin θ

…(2)

Squaring and adding (1) and (2), we get K + K1 = 4 K 2

…(3)

4 K 2 – K1 = K = 65 eV

…(4)

– 3.4 eV

n=4

– 6.04 eV

n=3

– 13.6 eV

n=2

∆E1

∆E = ∆E 2 = {–6.04 – (–54.4)} eV

∆E2

…(7)

K1 = 0.312 eV and K 2 = 16.328 eV Similarly, when we put

From conservation of linear momentum along X-direction

or

Similarly, when we put

Solving (4) and (7), we get

θ K2

⇒

and K 2 = 17.84 eV

K1 + K 2 = 16.64 eV

Y

4m

K1 = 6.36 eV

= 48.36 eV in equation (5), we get

K = 65 eV m

…(6)

∆E3

∆ E = ∆ E 3 = {– 3.4 – (–54.4)} = 511 . eV in equation (5), we get K1 + K 2 = 14 eV

…(8)

Now solving (4) and (8), we get K1 = – 1.8 eV

and

K 2 = 15.8 eV

But since the kinetic energies can’t have the negative values, the electron will not jump to third excited state or n = 4. Therefore, the allowed values of K1 (K.E. of neutron) are 6.36 eV and 0.312 eV and of K 2 (K.E. of the atom) are 17.84 eV and 16.328 eV and the electron can jump upto second excited state only ( n = 3). (ii) Possible emission lines are only three as shown in figure. The corresponding frequencies are 1

n=3

2

n=2 n=1

– 54.4 eV He+ (z = 2)

3

Now during collision, electron can be excited to any higher energy state. Applying conservation of energy, we get K = K1 + K 2 + ∆ E or

65 = K1 + K 2 + ∆ E

…(5)

n=1

ν1 =

(E 3 – E 2) h

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Problems in Physics =

{– 6.04 – (–13.6)} × 1.6 × 10–19 6.63 × 10–34

E 2 = – 3.4 eV

Hz

E 3 = – 1.5 eV

ν2 = =

Since E 4 – E 2 = 2.55 eV

(E 3 – E1) h {– 6.04 – (–54.4)} × 1.6 × 10

–19

6.63 × 10–34

Hz

= 11.67 × 1015 Hz (E – E1) and ν 3 = 2 h =

{– 13.6 – (–54.4)} × 1.6 × 10–19 6.63 × 10–34

Hz

= 9.84 × 1015 Hz Hence the frequencies of emitted radiations are 1.82 × 1015 Hz, 11.67 × 1015 Hz and 9.84 × 1015 Hz 600. (a) From Einstein’s equation of photoelectric effect, Energy of photons causing the photoelectric emision = Maximum kinetic energy of emitted photons + work function or E = Kmax + W = (073 . + 1.82) eV or

E 4 = – 0.85 eV

and

= 1.82 × 1015 Hz

E = 2.55 eV

(b) In case of a hydrogen atom,

Therefore, quantum numbers of the two levels involved in the emission of these photons are 4 and 2 ( 4 → 2) (c) Change in angular momentum in transition from 4 to 2 will be h h ∆L = L2 – L4 = 2 – 4 2π 2π ∆ L= –

or

h π

(d) From conservation of linear momentum |Momentum of hydrogen atom| P = mv

..... ............ ..... ......

E = 2.55 eV

Hydrogen atom

=|Momentum of emitted photon| E or mv = (m = mass of hydrogen atom) c E or v = mc =

2.55 × 1.6 × 10–19 J (1.67 × 10–27 kg) (3.0 × 108 m/s)

v = 0.814 m/s

E1 = – 13.6 eV

❒❒❒