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Solutions to '. Irodov Problems in General Physics Volume I Mechanics t Heat t Electrodynamics SECOND EDITION ABHAY KUMAR SINGH Director Abhay's I.I.T. Physics Teaching Centre Patna-6 CBS PUBLISHERS & DISTRIBUTORS 4596/1 A, 11 DARYAGANJ, NEW DELHI -110 002 (INDIA)

Dedicated to my Teacher Prof. (Dr.) J. Thaknr (Department of Physics, Patna University, Patna-4) ISBN :81-239-0399-5 First Edition: 1995 Reprint: 1997 Second Edition: 1998 Reprint: 2000 Reprint: 2001 Reprint: 2002 Reprint: 2003 Reprint: 2004 Reprint: 2005 Copyright © Author & Publisher All rights reserved. No part of this book may- be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage arid retrieval system without permission, in writing, from the publisher. Published by S.K. Jain for CBS Publishers & Distributors, 4596/1 A, 11 Darya Ganj, New Delhi - 110 002 (India) Printed at : India Binding House, Delhi - 110 032

FOREWORD Science, in general, and physics, in particular, have evolved out of man's quest to know beyond unknowns. Matter, radiation and their mutual interactions are basically studied in physics. Essentially, this is an experimental science. By observing appropriate phenomena in nature one arrives at a set of rules which goes to establish some basic fundamental concepts. Entire physics rests on them. Mere knowledge of them is however not enough. Ability to apply them to real day-to-day problems is required. Prof. Irodov's book contains one such set of numerical exercises spread over a wide spectrum of physical disciplines. Some of the problems of the book long appeared to be notorious to pose serious challenges to students as well as to their teachers. This book by Prof. Singh on the solutions of problems of Irodov's book, at the outset, seems to remove the sense of awe which at one time prevailed. Traditionally a difficult exercise to solve continues to draw the attention of concerned persons over a sufficiently long time. Once a logical solution for it becomes available, the difficulties associated with its solutions are forgotten very soon. This statement is not only valid for the solutions of simple physical problems but also to various physical phenomena. Nevertheless, Prof. Singh's attempt to write a book of this magnitude deserves an all out praise. His ways of solving problems are elegant, straight forward, simple and direct. By writing this book he has definitely contributed to the cause of physics education. A word of advice to its users is however necessary. The solution to a particular problem as given in this book is never to be consulted unless an all out effort in solving it independently has been already made. Only by such judicious uses of this book one would be able to reap better benefits out of it. As a teacher who has taught physics and who has been in touch with physics curricula at LIT., Delhi for over thirty years, I earnestly feel that this book will certainly be of benefit to younger students in their formative years. Dr. Dilip Kumar Roy Professor of Physics Indian Institute of Technology, Delhi New Delhi-110016.

FOREWORD A. proper understanding of the physical laws and principles that govern nature require solutions of related problems which exemplify the principle in question and leads to a better grasp of the principles involved. It is only through experiments or through solutions of multifarious problem-oriented questions can a student master the intricacies and fall outs of a physical law. According to Ira M. Freeman, professor of physics of the state university of new Jersy at Rutgers and author of "physic—principles and Insights" — "In certain situations mathematical formulation actually promotes intuitive understand- understanding Sometimes a mathematical formulation is not feasible, so that ordinary language must take the place of mathematics in both roles. However, Mathematics is far more rigorous and its concepts more precise than those of language. Any science that is able to make extensive use of mathematical symbolism and procedures is justly called an exact science". I.E. Irodov's problems in General Physics fulfills such a need. This book originally published in Russia contains about 1900 problems on mechanics, thermody- thermodynamics, molecular physics, electrodynamics, waves and oscillations, optics, atomic and nuclear physics. The bookjias survived the test of class room for many years as is evident from its number of reprint editions, which have appeared since the first English edition of 1981, including an Indian Edition at affordable price for Indian students. Abhay Kumar Singh's present book containing solutions to Dr. I.E. Irodov's Problems in General Physics is a welcome attempt to develop a student's problem solving skills. Tlie book should be very useful for the students studying a general course in physics and also in developing their skills to answer questions normally encountered in national level entrance examinations conducted each year by various bodies for admissions to profes- professional colleges in science and technology. B.P. PAL Professor of Physics LIT., Delhi

PREFACE TO THE SECOND EDITION Nothing succeeds like success, they say, Now, consequent upon the warm welcome on the part of students and the teaching fraternity this revised and enlarged edition of this volume is before you. In order to make it more up-to-date and viable, a large number of problems have been streamlined with special focus on the complicated and ticklish ones, to cater to the needs of the aspiring students. I extend my deep sense of gratitude to all those who have directly or indirectly engineered the cause of its existing status in the book world. Patna June 1997 Abhay Kumar Singh

PREFACE TO THE FIRST EDITION When you invisage to write a book of solutions to problems, one pertinent question crops up in the mind that—why solution! Is this to prove one's erudition? My only defence against this is that the solution is a challenge to save the scientific man hours by channelizing thoughts in a right direction. The book entitled "Problems in General Physics" authored by I.E. Irodov (a noted Russian physicist and mathematician) contains 1877 intriguing problems divided into six chapters. After the acceptance of my first book "Problems in Physics", published by Wiley Eastern Limited, I have got the courage to acknowledge the fact that good and honest ultimately win in the market place. This stimulation provided me insight to come up with mv second attempt—"Solutions to I.E. Irodov's Problems in General Physics." This first volume encompasses solutions of first three chapters containing 1052 problems. Although a large number of problems can be solved by different methods, I have adopted standard methods and in many of the problems with helping hints for other methods. In the solutions of chapter three, the emf of a cell is represented by § (xi) in contrast to the notation used in figures and in the problem book, due to some printing difficulty. I am thankful to my students Mr. Omprakash, Miss Neera and Miss Punam for their valuable co-operation even In my hard days while authoring the present book. I am also thankful to my younger sister Prof. Ranju Singh, my younger brother Mr. Ratan Kumar Singh, my junior friend Miss Anupama Bharti, other well wishers and friends for their emotional support. At last and above all I am grateful to my Ma and Pappaji for their blessings and encouragement. ABHAY KUMAR SINGH

CONTENTS Foreword iii Preface to the second edition v Preface to the first edition vi PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS 1.1 Kinematics 1-34 1.2 The Fundamental Equation of Dynamics 35-65 1.3 Laws of Conservation of Energy, Momemtum, and Angular Momentum 66-101 1.4 Universal Gravitation 102-117 1.5 Dynamics of a Solid Body 118-143 1.6 Elastic Deformations of a Solid Body 144-155 1.7 Hydrodynamics 156-167 1.8 Relativistic Mechanics 168-183 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS 2.1 Equation of the Gas State. Processes 184-195 2.2 The first Law of Thermodynamics. Heat Capacity 196-212 2.3 Kinetic theory of Gases. Boltzmann's Law and Maxwell's Distribution 213-226 2.4 The Second Law of Thermodynamics. Entropy 227-241 2.5 Liquids. Capillary Effects 242-247 2.6 Phase Transformations 248-256 2.7 Transport Phenomena 257-266 PART THREE ELECTRODYNAMICS 3.1 Constant Electric Field in Vacuum 267-288 3.2 Conductors and Dielectrics in an Electric Reid 289-305 3.3 Electric Capacitance. Energy of an Electric Field 306-324 3.4 Electric Current 325-353 3.5 Constant Magnetic Field. Magnetics 354-379 3.6 Electromagnetic Induction. Maxwell's Equations 380-407 3.7 Motion of Chaiged Particles in Electric and Magnetic Fields 408-424

PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS 1.1 KINEMATICS 1.1 Let v be the stream velocity and v' the velocity of motorboat with respect to water. The motorboat reached point B while going downstream with velocity (v + v') and then returned with velocity (v' - v) and passed the raft at point C Let t be the time for the raft (which flows with stream with velocity v^) to move from point A to C, during which the motorboat moves from A to B and then from B to C. Therefore On solving we get v ■ -— 1/ & v 1.2 Let 5 be the total distance traversed by the point and tx the time taken to cover half the distance. Further let 2f be the time to cover the rest half of the distance. Therefore f - vori or h' TT C1) Z L Vq and x- (vi + v^r or 2f« —-— B) Hence the sought average velocity s s 13 As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances as well as the times taken are same in these phases of motion. Let At be the time for which the car moves uniformly. Then the acceleration / deceleration tune is —-— each. So,

or <v>x I (x - Af)' 2W 4 (x-AQ —-—L Hence Af* x V 1- wx 15 s. 1.4 (a) Sought average velocity s <v> - - t 200 cm — zus 10 cm/s 20 (b) For the maximum velocity, — should be ds maximum. From the figure — is maximum for all points on the line ac9 thus the sought maximum velocity becomes average velocity for the line ac and is equal to : 6c 100 cm ab 4s mmm —rf ■■Mi -*• > ■** Maw 11 VI ,'YJ /I / / 4/ b M ■■■ Ml 10 20t,S 25 cm/s (c) Time r0 should be such that corresponding to it the slope — should pass through the ds s point O (origin), to satisfy the relationship -^ * —. From figure the tangent at point d at t0 passes through the origin and thus corresponding time t« /0 - 16 s. 1.5 Xet the particles collide at the point A (Fig.), whose position vector is r^(say). If t be the time taken by each particle to reach at point A, from triangle law of vector addition : -* A V SO, Vt therefore, f= , _* -,. lv2-vil From Eqs. A) and B) B) 1 2 2 2 V I1* -* V2-Vl o *7 n or, rl~r2 r v I rl " r2 I I V2 ~ Vl I 77, which is the sought relationship. X 1.6 We^have v ■v-v0 A) From the vector diagram [of Eq. A)] and using properties of triangle

and or V Vq + v2 + 2 v0 v cos <p - 39.7 km/hr . A v sin cp or, sin 8* j-*- B) sin (ji - cp) sin 6 . _ i (v sin Using B) and putting the values of v and d 8 - 19.r 1.7 Let one of the swimmer (say 1) cross the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1. , (where AB • d is the width of the river) Vv'2-V20 A) For the other swimmer (say 2), which follows the quickest path, the time taken to cross the river. B B) B ■*■ X i. i s V In the time t2, drifting of the swimmer 2, becomes Vn Xm V2- 77* («sing Eq. 2) C) If r3 be the time for swimmer 2 to walk the distance x to come from C to B (Fig.), then (using Eq. 3) fe-i vii According to the problem fx « t2 + or, On solving we get Vv'2 - u 3 km/hr. (l-vt) 0 -1

1.8 Let / be the distance covered by the boat A along the river as well as by the boat B acre the river. Let v0 be the stream velocity and v' the velocity of each boat with respect water. Therefore time taken by the boat A in its journey / / v' + v0 v' - v0 and for the boat B Hence, On substitution 2/ 'o Vv'2-vg Vti2-1 1-8 I where r\ = — 1.9 Let v0 be the stream velocity and v' the velocity of boat with respect to water. A — ■ T] ■ 2 > 0, some drifting of boat is inevitable. Let v* make an angle 0 with flow direction. (Fig.), then the time taken to cross the rive t * . . - (where d is the width of the river) v sin 0 v In this time interval, the drifting of the boat jc* (v'cos0 + v >' cos 0 + vj , . . - (cot 0 + ti cosec 0) d ^ v sin 0 For jc • (minimum drifting) dQ (cot 0 + K] cosec 0) * 0, which yields cos 0 - - — « - — n 2 I Hence, 0 - 120° 1.10 The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration : r12" r0A2) + v0A2) or, But (because wu~ 0 and r^U)- 0) ' A) - v So, from properties of triangle 0A2) - V v2 + v2-2vovocos(ji/2-0o) Hence, the sought distance £|- voV2(l-sin0) r= 22m.

-11 Let the velocities of the paricles (say vl and v2 ) becomes mutually perpendicular after time t. Then their velocitis become As v1 lv2 so, v1 • or, 0 0 or 0 ■Vttr** Hence, V vi g C) Now form the Eq. r12 - 1 * t + — >v12 ^21" I ^12) I *> (because here m^ - 0 and Hence the sought distance 0) 8 (as I v '0A2) V2) 1.12 From the symmetry of the problem all the three points are always located at the vertices of equilateral triangles of varying side length and finally meet at the centriod of the initial equilateral triangle whose side length is a, in the sought time interval (say t). 6^720° Let us consider an arbitrary equilateral triangle of edge length / (say). Then the rate by which 1 approaches 2, 2 approches 3, and 3 approches 1, becomes : -dl dt V - VCOS On integrating : 3v 2 a a 3 vt so t 2a —

6 1.13 Let us locate the points A and B at an arbitrary instant of time (Fig.). If A and B are separated by the distance s at this moment, then the points converge or -ds point A approaches B with velocity —r- - v - u cos a where angle a varies with time. On intergating, 0 r - J ds « J (v - u cos a) dt9 1 o (where T is the sought time.) T t or /« J (v - u cos a) df As both A and 2? cover the same distance in jt-direction during the sought time interval, so the other condition which is required, can be obtained by the equation So, Solving A) and B), we get T *x-fvxdt T uT» I vcos adit o ul B) -5 One can see that ifu~v, or u < v, point A cannot catch B. 1.14 In the reference frame Gxed to the train, the distance between the two events is obviously equal to /. Suppose the train starts moving at time t« 0 in the positive x direction and take the origin (x - 0) at the head-light of the train at t» 0. Then the coordinate of first event in the earth's frame is and similarly the coordinate of the second event is The distance between the two events is obviously. -0-242 km in the reference frame fixed on the earth.. For the two events to occur at the same point in the reference frame Ky moving with constant velocity V relative to the earth, the distance travelled by the frame in the time interval T must be equal to the above distance. Thus Vx - / - wx( t + x/2) / x So, 4-03 m/s The frame K must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point xl on the earth at time t, it coincides with the point x2 at time t + x.

1.15 (a) One good way to solve the problem is to work in the elevator's frame having the observer at its bottom (Fig.). Let us denote the separation between floor and celing by h - 2-7 m. and the acceleration of the elevator by *v« 1-2 m/s From the kinematical formula ±v2 A) Here and O,yo 'Qy 0 So, 0 or, 0-7 s. (b) At the moment the bolt loses contact with the elevator, it has already aquired the velocity equal to elevator, given by : v0- A-2) B)- 2-4 m/s In the reference frame attached with the elevator shaft (ground) and pointing the y-axis upward, we have for ^ the displacement of the bolt^ v0T* 1 i t V 1 f \ or, Ay « B-4) @-7) + \ (- 9-8) @-7J * - 0-7 m. Hence the bolt comes down or displaces downward relative to the point, when it loses contact with the elevator by the amount 0-7 m (Fig.). Obviously the total distance covered by the bolt during its free fall time 1.16 Let the particle 1 and 2 be at points B and A at t - 0 at the distances lx and l2 from intersection point O. Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity i£2- vj*- v£ and its trajectory is the straight line BP. Obviously, the minimum distance between the particles is equal to the length of the perpendicular AP dropped from point A on to the straight line BP (Fig.).

B From Fig. (b), v12 - V v\ + v2 , and tan 0 - — The shortest distance AP- AM sinG- (OA - GM) sin 0 - (/2 -/x cot 0) sin 0 or AP Vvf (using 1) The sought time can be obtained directly from the condition that (lx -vlt) + (^- v20 is minimum. This gives t - —* *— . 1.17 Let the car turn off the highway at a distance x from the point D. So, CD = jc, and if the speed of the car in the field is v, then the time taken by the car to cover the distance AC - AD - x on the highway and the time taken to travel the distance CB in the field B) So, the total time elapsed to move the car from point A to B AD-x yfWi x\ v For t to be minimum , - 0 or - ax v 0 or t|2jc2» /2+jc2 or jc

9 1.18 To plot x (t\ s (t) and wx (t) let us partion the given plot vx (t) into five segments (for detailed analysis) as shown in the figure. For the part oa : w =* 1 and v * t - v Thus, Putting f= 1, we get, For the part ab : o y- sx(t) 1 .t — unit 7 mmmtmmt 1 2 MM 3 4 6 s t A7' = O and v = v= constants 1 Thus Putting f= 3, Ax2= 52» 2 unit For the part b4 : w^ ■ 1 and vx - 1 - (t - 3) ■ 4 -1) ■ v .2 Thus Putting For the part 4d : So, Thus Putting Similarly 15 1 f = 4, A jc3 - x3 - — unit x m - 1 and vx = - A - 4) - 4-1 « t-4 for ^>4 Ax4(t)-f(l-t)dt r- 6, A*4= -1 Putting For the part d 7 : Now, Putting t- 6, *4» wx" 2 and v-|v,|- 6 Ax@-/2 (r t= 4, Ax5 : 2 unit Vxm ~2 2G-0 -7)A- s - 1 + 2 for r2- (r-6)« 2(r-7) r<-7 14r + 48 Similarly Putting s5[t)- f 2G-t)dtm \4t-t2- 48 t- 7, s. On the basis of these obtained expressions w^ (f), jc (r) and 5 (t) plots can be easily plotted as shown in the figure of answersheet

10 1.19 (a) Mean velocity Total distance covered 50 cm/s A) Time elapsed m £ xR f" X (b) Modulus of mean velocity vector 2R \<v>\ 32 cm/s B) At x (c) Let the point moves from i to f along the half circle (Fig.) and v0 and v be the spe at the points respectively. dt " W* We have or, v * v0 + wt t (as wt is constant, according to the problem) t So, 0 So, from A) and C) xR x Now the modulus of the mean vector of total acceleration w> o Ar (see Fig.) Using D) in E), we get : w> x2 1.20 (a) we have So, and (b) From the equation r - a t A - a t) v«-«a{l-2a0 w a -at), 0, at t * 0 and also at r = Ar a So, the sought time Ar = a As So, v- \v] 2at) a A - 2 a t) forf > 1 2a 1 2a

11 Hence, the sought distance 1/2 a I/a s = f v dt= J a(l-2at)dt+ J a B at 0 1/2 a Ddt Simplifying, we get, s - a 2a 1.21 (a) As the particle leaves the origin at t = 0 So, Ajc - x « J vx dt As V= Vr x where v0 is directed towards the +ve x-axis So, From A) and B), T 0 1- 2x Hence x coordinate of the particle at t = 6 s. 10x6 1- 2x5 24cm= 0-24 m Similarly at jc= 10x10 10 s 1- 10 2x5 = 0 and at 20 s 10x20 1- 20 2x5 -200 cm = -2m A) B) C) (b) At the moments the particle is at a distance of 10 cm from the origin, x = ± 10 cm. Putting x m + 10 in Eq. C) 10= 10t(l-j^\ or, f2-10f+10=0, So, Now putting 10 ± V 100 - 40 ,.,« x = - lOin Eqn C) s On solving, t - 5 ± V35T As t cannot be negative, so, )s

12 Hence the particle is at a distance of 10 cm from the origin at three moments of time t m 5 ± VU s, 5 + V3F s (c) We have So, So v-vo[l-- f or t s x for t > x t* x m v0 r A - and vox [1 + A - ftJ] /2 for t >x 4 (A) And f or t - 8 s o 8 0 x ' 5 On integrating and simplifying, we get s m 34 cm. On the basis of Eqs. C) and D), x (t) and s (t) plots can be drawn as shown in the answer sheet 1.22 As particle is in unidirectional motion it is directed along the Jt-axis all the time. As at t = 0, x = 0 So, Therefore, or, Ax= jc= s9 and v= aVx - aVs dv a ds a As, w dt 2Vs dt 2 Vs a v aaVs ot^ 2Vs 2Vs ~ dv c^ dt" 2 A) On integrating, ^-dt or, v* ^- B) o 0

13 (b) Let s be the time to cover first s m of the path. From the Eq. s = Jv dt s = 2 2 (using 2) o or The mean velocity of particle 2\S/a /* iVs/a 1.23 According to the problem or, a vv (as v decreases with time) -I Vv dv- aj ds o On integrating we get s - — Vq Again according to the problem /2 dv dt avv or - or, J dv a\dt Thus a 1.24 (a) As So, and therefore r*= att^-bt /* x= atyy= -bt -bx2 a C)

14 which is Eq. of a parabola, whose graph is shown in the Fig. (b) As r^ ati^-bt2j~* = ai-2bt) A) So, v= Diff. Eq. A) w.r.t. time, we get So, | v? (c) cos a v w or, cos a = so, tan a > 2bt or, a= tan * 2bt (d) The mean velocity vector < v > - — ai-btj fdt t Hence, I < v*> I - Va^+ (-btJ - VaJTbTtJ 1.25 (a) We have x- at and y - a t A - a t) A) Hence, y (x) becomes, y = I 1 I « x - — x (parabola) a ^ a j a (b) Diiferentiating Eq. A) we get vx = a and vy = a A - 2 a f) B)

15 So, ; Diff. Eq. B) with respect to time wx= 0 and wy = -2aa So, w« V wx2 + m^2 - 2aa (c) From Eqs. B) and C) We have v^« a j"+ a ( 1 - 2a r ) 7"* and i?« 2 a a F So, cos - = 4 4 V2 vm> aVl + (l-2aroJ2aa On simplifying. 1 - 2 a r0 « ±1 As, r0 * 0 , r0 m — 1.26 Differentiating motion law : jc * a sin co r, y * a(l-coscot), with respect to time, v = a co cos cor, v » a co sin cor •* • So, v^ a co cos cor J+ acosincorj* A) and v - a co * Const B) Differentiating Eq. A) with respect to time w ■ -r-- -a co smcori + aco cos cor; C) at (a) The distance s traversed by the point during the time x is given by s - J v dr«Jacocfr«acox (using 2) 0 0 (b) Taking inner product of v and w We get, v • w * (a co cos cor i + a co sin cor; )• (a co sin cor ( - i) + a co cos cor -; ) So, v^ hT« - a2 co2 sin cor cos cor + a2 co3 sin cor cos cor - 0 Thus, vtL v?, i.e., the angle between velocity vector and acceleration vector equals —. 1.27 Accordiing to the problem dVx dVv So, w,- -^r- 0 and wr- -^- -w A) Differentiating Eq. of trajectory, y* ajc-fejc2, with respect to time ^2bx dt dt dt

16 So, dt a x-0 dx dt *-0 Again differentiating with respect to time 2" 7^£\2_ dt1 dt * J 2bx dt or, -w- a(Q)-2b or, dt -2bx(Q) (using 1) (using Using C) in B) dy_ dt x- 0 •VI Hence, the velocity of the particle at the origin v = Hence, c- 1.28 As the body is under gravity of constant accelration g, it's velocity vector and displacemen vectors are: ^ T' (i; and V + \gt2 (F*« OaU = 0) So, <v> over the first t seconds C) Hence from Eq. C), <v> over the first t seconds D) For evaluating tt take or, v 2 V 0 Lt=o) But we have v = v0 at ^ - 0 and Also at t - x (Fig.) (also from energy conservation) B; Vo

17 Hence using this propety in Eq. E) As x * 0, so, x « - Putting this value of x in Eq. D), the average velocity over the time of flight " V0~g 2 g 1,29 The body thrown in air with velocity v0 at an angle a from the horizontal lands at point P on the Earth's surface at same horizontal level (Fig.). The point of projection is taken as origin, so, Ax = x and Ay - y (a) From the Eq. Ay - v0 J + T w t2 0 1 vosinax--gx As x * 0, so, time of motion x = 2 v0 sin a 8 (b) At the maximum height of ascent, vy = 0 so, from the Eq. + 2 w Ay 0= (v0sinaJ-2g/f q sin2 a t\ Hence maximum height During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation or, when 1 i = v0cosax- — @)x = v0cosax = sin 2 a sin2 2 sin2 sin2 a v2 sin2 a g 2g or tan a = 4, so, a = tan" * 4 (c) For the body, x (t) and y (f) are x = v0 cos a

18 1 9 and y* vosinar--gr B) Hence putting the value of t from A) into B) we get, i ■ vosina a 2 I v0 cos a x tana- 2 2 2vftcos a 0 Which is the sought equation of trajectory i.e. y (x) (d) As the body thrown in air follows a curve, it has some normal acceleration at all the moments of time during it's motion in air. At the initial point (x ■ 0, y ■ 0), from the equation : w* = ~R y (w^ere ^ is the radius of curvature) vo v g cos a - — (see Fig.) or Ro Ro v °7 u g cos a At the peak point v = 0, v = vx ■ v0 cos a and the angential acceleration is zero. Now from the Eq. wn R cos2 a Vq cos2 a Note : We may use the formula of curvature radius of a trajectory y (x\ to solve part (d)y ? (dy/dx) 1.30 We have, vx = v0 cos a, vy = v0 sin a - gt As vj t wr all the moments of time. Thus v2= v,2-2gfvosina + g2f2 Now, wf= ^ Hence | wt\ Now V^/ V224 v * or wM - g — (where vv n vr V x

19 As t v* > during time of motion On the basis of obtained expressions or facts the sought plots can be drawn as shown in the figure of answer sheet 1.31 The ball strikes the inclined plane (Ox) at point O (origin) with velocity v0 = V 2gh A) As the ball elastically rebounds, it recalls with same velocity v0, at the same angle a from the normal or y axis (Fig.). Let the ball strikes the incline second time at P, which is at a distance / (say) from the point Oy along the incline. From the equation 1 ... .2 1 2 0 * v0 cos a x - — g cos a r where x is the time of motion of ball in air while moving from O to P. As 0, so, x = 2v, 8 B) Now from the equation. X — Vq^ t + M 1 . 2 v0 sin a x + y g sin a x SO, vosina f2 \ 'o 4 v0 sin a g (using 2) Hence the sought distance, / - 1.32 Total time of motion 2 v0 sin a 8 sin a o Sh sui a (Using Eq. 1) or 9-8 and horizontal range R = v0cos ax or cos a From Eqs. A) and B) 2 v0 2 x 240 /? 5100 _ 85_ v0 x * 240 x * 4 x A) B) On simplifying D80J +Dx2J x4 - 2400 x2 + 1083750 = 0

20 Solving for x we get : 2 2400 ± V 1425000 2400 ± 1194 Thus x - 42-39 s - 0-71 min and x * 24-55 s « 0-41 min depending on the angle a. 1.33 Let the shells collide at the point P (jc, y). If the first shell takes t s to collide With second and Ar be the time interval between the firings, then x - v0 cos Q11 - v0 cos 02 (t - A t) A) 1 2 B) Vr and From Eq. A) t v 7 Arcos 0, cos 02 - cos C) From Eqs. B) and C) 2 v0 sin @X - 02) g (cos 02 + cos 0X 1.34 According to the problem as 0 00 v0 or dy = vodt Integrating y s 0 or And also we have — - ay or dx=*aydt~avotdt (using 1) x t 2 /r i 2 10 v d^=avojr^r, or, *«-avor ---^-(using 1) 0 0 U (b) According to the problem vy- v0 and v^- ay So, Therefore 2 2 ay Jv a y dfy / ss s — ^ —ll- r dt Vu? + flv2 t/r l + /ay/vo\ Diff. Eq. B) with respect to time. d v dv jc t/y 0 and -^— - vvv - a -f- - a vn Jr * ^ ° (i) B) So, av 0

21 Hence 1.35 (a) The velocity vector of the particle v * a i + fox; SO, From A) J dx - a j dt or, jc - at B) o o And dy~ bxdt= batdt y t Integrating J dy - ab \tdt or, y= —abt2 C) o o From Eqs. B) and C), we get, y^—x2 D) (b) The curvature radius of trajectory y (jc) is : 3 (dyldx) ' 2 12 b 2 Let us differentiate the path Eq. y - — jc with respect to *, dy b Sy b -f-m —x and —*rm - F) dx a dx2 ^ From Eqs. E) and F), the sought curvature radius : 3 2 12 L36 In accordance with the problem But wt - —r~ or vav wt = a • x So, or, vdv** ai-dr= adx (because a is directed towards the jc-axis) V JC So, \vdv= a\ dx Hence v ■ 2ox or, v - Vlax

22 1.37 The velocity of the partide v « at So, —- wtm a A) 2 2 2 And wn- j- ±1- (usingv«ar) B) From * « f v dr t a\vdt~ -at2 o *2 From Eqs. B) and C) wH t . .,.n Hence w * V v?l + y?i I2 - 0-8 m/s2 1.38 According to the problem 2 ForvW, ^--5- Integrating this equation from v0^ vi v and 0 £ t£ f o 2 N jw for v (s), - ~~ * — , Integrating this equation from v0 s v ^ v and 0 s s s s SO, | —- -~| ^y or, In ~ vo Hence v- voe"^ B) (b) The normal acceleration of the point R 2) And as accordance with the problem l- \wn\ and A n so, V2

23 1.39 From the equation v * a Vs dv a ds a /-a2 T * "zn= T * ^= avs » —, and A 2VJdr 2Vs 2 v2 a2* As ^ is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction. Hence the angle between v and w is equal to between wti*t an w, and a can be found by means of the formula : tan a = jr « = 1^1 a2/2 1.40 From the equation / - a sin go t dl v« a o) cos cor So, w,- —« -a co sin cor, and A) v2 a2 oJ cos2 o) f (a) At the point / * 0, sin co t - 0 and cos co t« ± 1 so, cor - 0, n etc. co2 a2-2 Hence w~ wH Similarly at / - ± a9 sin cor * ± 1 and cos cor = 0, so, wn * 0 i i 2 Hence w « | ^ | * a o> 1.41 As w, = a and at t ■ 0, the point is at rest 1 2 So, v(r) and s(t) are, v* ar and .s* —at A) Let i? be the curvature radius, then v2 a2r2 2 But according to the problem wn-bt* So, 6r4= 4 Therefore w - Vm£ + m£ - V a2 + Bas/RJ - Va2 + ( 4 6j2 / a2J (using 2) Hence w = a V 1 + D 6s2/a3J

24 1.42 (a) Let us differentiate twice the path equation y (x) with respect to time. dy -f- dt lax dx (fy — ; —*r dt dt2 2a dt dt dlx 2 Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point x * 0 it coincides with the direction of derivative d2y/dt2. Keeping in mind that at the point x * 0, — We get dx dt dt v, 2 n x- 0 So, 2 V 1 Wn.2av-j, or R-- Note that we can also calculate it from the formula of problem A.35 b) (b) Differentiating the equation of the trajectory with respect to time we see that , 2 dx 2 dV rx /1\ bx**ay*~° A) which implies that the vector (b xi + a yj) is normal to the velocity vector V the normal and the normal component of acceleration is clearly = — f*+ -*£) which, of course, is along the tangent. Thus the former vactor is along ,2 b x dt2 dt1 on using Wn " ( b V ♦ a V I/2 /i / | /t|. At jc « 0,y* ± t and so at x = 0 x- 0 Differentiating A) '(* A fit) = 0 Also from A) 0 at x - 0 So I — I « ± v (since tangential velocity is constant « v ) Thus dt2 and This gives R * a2/b

25 1.43 Let us fix the co-ordinate system at the point O as shown in the figure, such that the radius vector F*bf point A makes an angle 0 with x axis at the moment shown. Note that the radius vector of the particle A rotates clockwise and we here take line ox as reference line, so in this case obviously the taking angular velocity co * ""~7~ anticlockwise sense of angular displacement as positive. Also from the geometry of the triangle OAC R r sin 0 sin (n - 2 0) Let us write, or, r « 2 R cos 0. 0 Differentiating with respect to time. at or or, v- 2R dt at [sin20i-cos20;] ~dFJ or, v*« 2R a) (sin 2 0 T- cos2qJ So, |v| or v« 2o)jR«0-4m/.s. dv As a) is constant, v is also constant and wt * — * 0 So, w* w. Alternate : From the Fig. the angular velocity of the point A, with respect to centre of the circle C becomes dBQ) -dQ) 2 co = constant dt y dt Thus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius R with constant angular velocity 2 co. Hence v* 2coi? and Bco/Q2 2 n R R 1.44 Differentiating (p (t) with respect to time dtp dt For fixed axis rotation, the speed of the point A: toz- 2at A) v«toi?« 2atR or R 2at B)

26 Differentiating with respect to time dv wtm —* 2aR= ~, (using 1) But So, R v/2at 2 a tv (using 2) 1.45 The shell acquires a constant angular acceleration at the same time as it accelerates linearly. The two are related by (assuming both are constant) / 2nn Where w = linear acceleration and 0 =_angular acceleration Then, co « V2-02n;n« V2-yB3inJ But 2 CO 2w/, hence finally 231/1 V 1,46 Let us take the rotation axis as z-axis whose positive direction is associated with the positive direction of the cordinate <p, the rotation angle, in accordance with the right-hand screw rule (Fig.) (a) Defferentiating <p (t) with respect to time. -~* a - 3 bt2 = co, dt z A) and d2<p dt2 dt B) From A) the solid comes to stop at A t * t The angular velocity co* a-3bt29 for 0st So, < co > f(a-3bt)dt Similarly 0 - | 0 | - 6fr f for all values of

27 So, <p (b) From Eq. B) Pz* -66f So, Hence P« (PJ / I » 2V3a& 1.47 Angle a is related with | wt | and wH by means of the fomula : tana* :—r, where w* co R and wJ« QR A) where R is the radius of the circle which an arbitrary point of the body circumscribes. From the given equation p « -r— ■ at (here p * -r— > as P is positive for all values of Integrating within the limit f Jco = a \ tdt or, cosra^ /i2f4 bo, 2 and |wr|« PA- Putting the values of | wt \ and wH in Eq. A), we get, a2tAR/4 at3 tan a * =— ■ —r- or, t atR 4 1.48 In accordance with the problem, Pz < 0 \D\* 1 — tan a 1/3 Thus - -t— ■ A: VaT, where A: is proportionality constant or, - I -,==m k I dt or, vco - vcoo - •*?- A) V to i 2 When co ■ 0, total time of rotation t ■ r

28 Average angular velocity < a) > I J7 Hence < co > @ 0 12 o 1.49 We have to * co0 - a cp ~ Integratin this Eq. within its limit for (cp) t d cp (o0-/ /:(p J dt or, In (xr 0 0), .2.2 dt co0/3 Hence (b) From the Eq., (o * co0 -* it(p and Eq. A) or by differentiating Eq. A) 1.50 Let us choose the positive direction of z-axis (stationary rotation axis) along the vector . In accordance with the equation dm. or dt rz " ™z dy nz Z^K or, toz d (oz - Pz J cp * P cos cp d cp, Integrating this Eq. within its limit for to or, I rfaJ* poj coscpdcp o o 0) or, Y* Posin(P Hence coz * ± V 2 po sin cp The plot (oz (cp) is shown in the Fig. It can be seen that as the angle cp grows, the vector of first increases, coinciding with the direction of the vector j^ (coz > 0), reaches the maximum at cp * cp/2, then starts decreasing and finally turns into zero at cp » n. After that the body starts rotating in the opposite direction in a similar fashion (coz < 0). As a result, the body will oscillate about the position cp « cp/2 with an amplitude equal to jc/2.

29 1.51 Rotating disc moves along the x-axis, in plane motion in x - y plane. Plane motion of a solid can be imagined to be in pure rotation about a point (say 7) at a certain instant known as instantaneous centre of rotation. The instantaneous axis whose positive sense is directed along co of the solid and which passes through the point/, is known as instantaneous axis of rotation. Therefore the velocity vector of an arbitrary point (P) of the solid can be represented as : coxr p/ A) On the basis of Eq. A) for the C. M. (C) of the disc c a V / According to the problem vctt* QJTtt^*i'c« (&±-x-y plane, so to satisy the Eqn. B) r£7 is directed along (- j ). Hence point / is at a distance rCI = yy above the centre of the disc along y - axis. Using all these facts in Eq. B), we get 0 or y« — c @ C) (a) From the angular kinematical equation co- On the other hand x * v f, (where x is the x coordinate of the CM.) or. From Eqs. D) and E), co Using this value of co in Eq. C) we get y • — co ( hyperbola ) 1 2 —wt and vc« wt (b) As centre C moves with constant acceleration w, with zero initial velocity So, Therefore, Hence vc- w co co V2jc (parabola) E)

30 1.52 The plane motion of a solid can be imagined as the combination of translation of the CM. and rotation about CM. So, we may write vA (l)and ' x "* " 0] WAm WC+WAC B) is the position of vector of A with respect to C* In the problem vc = v = constant, and the rolling is without slipping i.e., vc = v = coi?, So,' wc ■ 0 and p * 0. Using these conditions in Eq. B) A —* Here, wA c is the unit vector directed along rA c. Hence h>a * — and wA is directed along (- uA c ) or directed toward the centre of the wheel. (b) Let the centre of the wheel move toward right (positive Jt-axis) then for pure tolling on the rigid horizontal surface, wheel will have to rotate in clockwise sense. If co be the vc v angular velocity of the wheel then a) * — * —. Xv Xv Let the point A touches the horizontal surface at t * 0, further let us locate the point A at t * f, When it makes 0 = co f at the centre of the wheel. From Eqn. A) v^ * v£ + of x ^ or, vA « v i + co R \ cos co t (- i) + sin * (v - cos cor) i + v sin cor / (asv* oaR) So, v. * V (v - v cos cor) + (v sin corJ - vV 2(l-coscof) « 2vsin(cor/2) Hence distance covered by the point A during T« 2 tc/co 2 v sin(cof/2) dr - — co 0 1.53 Let us fix the co-ordinate axis xyz as shown in the fig. As the ball rolls without slipping along the rigid surface so, on the basis of the solution of problem 1.52 : vn * v. + co x r ■ * 0 Thus D . -♦* a t J\ -• v * (oR and co j t (- *) as v

31 and and At the position corresponding to that of Fig., in accordance with the problem, M> « W9 SO V « Wt and co----and p (using 1) /77J777 (a) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the Fig. As point 0 is the instantaneous centre of rotation of the ball at the moment shown in Fig. so, Vo«O, Now, So, Similalry 2vci » 9 9 vc i + co (- k) x R (j ) * (vc + i (using 1) + of x r^, ■ vc i + co (- A:) x R (i) So, vB * V2 vc * V2 >vt and v£ is at an angle 45° from both i and ; (Fig.) oc - u oc where uoc is the unit vector along roc 2 2 2 so, w>0«~«^— (using 2) and is directed towards the centre of the ball Now w wc + co2 (- — (-;) (using 1) R So, Similarly V 4>V + r- 2>v 2\2 H-co2i? (-i) + P (-T) xR (T) V- -;) (using 1)

32 w- R —w w i + w(-j) (using 2) So, yB V 2\2 1.54 Let us draw the kinematical diagram of the rolling cylinder on the basis of the solutioi of problem 1.53. 0 0 As, an arbitrary point of the cylinder follows a curve, its normal acceleration and radius of curvature are related by the well known equation v2 so, for point A, or, Similarly for point B, W, ;2 c (because vc« cor, for pure rolling) w ■<•> 2 co r cos 45 co r 1.55 The angular velocity is a vector as infinitesimal rotation commute. Then the relative angular velocity of the body 1 with respect to the body 2 is clearly. - co as for relative linear velocity. The relative acceleration of 1 w.r.t 2 is dT

33 where S' is a frame corotating with the second body and 5 is a space fixed frame with origin coinciding with the point of intersection of the two axes, but r dt aJ x Since 5 ' rotates with angular velocity w2 . However with constant angular velocity in space, thus dt 0 as the first body rotates Is co1 x co2. Note that for any vector 5f the relation in space forced frame (k) and a frame (k) rotating with angular velocity oTis dt dt ICC > ""* 1 " .5ft We have a) « ati + bt j So, (o A) (o- V (atJ + (fr2J , thus, *,,,_ 1Os Differentiating Eq. A) with respect to time 7.81 rad/s dm i-* —« ai + 2btj B) So, and (b) V a2 + B rad/s cos a @ Putting the values of (a) and (&);and taking t * 10s, we get a- 17° 1.57 (a) Let the axis of the cone (OC) rotates in anticlockwise sense with constant angular velocity oT and the cone itself about it's own axis (OC) in clockwise sense with angular velocity co^ (Fig.). Then the resultant angular velocity of the cone. As the rolling is pure the magnitudes of the vectors <o andco0 can be easily found from *<J \rJ (&'! Fig. ^O a)'- V , (o0- v/fl B) 0 H cot a u v y u As a) X (Oq, from Eq. A) and B) A

34 co U /? cot a (b) Vector of angular acceleration p Rcosa 2-3 rad/s 7 (as co * constant.) The vector co*j which rotates about the OO' axis with the angular velocity co*, retains i magnitude. This increment in the time interval dt is equal to I d co01» co0* co' dt or in vector form d co0 » (co x co0) dt. Thus P** co* x co*) The magnitude of the vector jTis equal to p » co' co0 (as co X co0 So, P V V v — « —r tan a ~ 2-3 rad/s RcotaR R2 1.58 The axis AB acquired the angular velocity A) Using the facts of the solution of 1.57, the angular velocity of the body co» Vco? + co/2 0-6 rad/s And the angular acceleration. 0 =^ t But 'o to x co0, and dt So, As, P^lco*3 so, p« V (co0 po 02 + Po 0-2 rad/s2

35 THE FUNDAMENTAL EQUATION OF DYNAMICS Let R be the constant upward thurst on the aerostat of mass my coming down with a constant acceleration w. Applying Newton's second law of motion for the aerostat in projection form Fy - mwy mg-R* mw Now, if Am be the mass, to be dumped, then using the Eq. Fy = mwy R-(m- Am) g « (m - Am) w> From Eqs. A) and B), we get, Am * B) 1.60 Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of x and y axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value of acceleration (say w) mow B) and T2-km2g* m2w C) The simultaneous solution of Eqs. A), B) and C) yields, g i T, and ut f ft \ / / fxnfff'rtrf' fr W'$ 77?2 * fn. As the block m0 moves down with acceleration w, so in vector form 1.61 Let us indicate the positive direction of jr-axis along the incline (Fig.). Figures show the force diagram for the blocks. Let, R be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration w.

36 Newton's second law of motion in projection form along jc-axis for the blocks gives : m1gsina-k1m1g cos a + R « m1 w A) m2 g sin a - R - k^ m2 g cos a « m2 w B) Solving Eqs. A) and B) simultaneously, we get 1 ^1 * ^ ^2 w * g sin a - g cos a and cos a (b) when the blocks just slide down the plane, w « 0, so from Eqn. C) 1 1 2 2 g sin a - g cos a * 0 Ifl + /W * + or, (m! + /W2) sin a * (kx mx + k>± m2) cos a Hence tan a = 1.62 Case 1. When the body is launched up : Let fc be the coefficeint of friction, u the velocity of projection and / the distance traversed along the incline. Retarding force on the block - mg sin a + kmgcosa and hence the retardation - g sin a + kg cos a. Using the equation of particle kinematics along the incline, 0* w2-2(gsina + JfcgcosaO ' 2(gsina + A:gcosa) and 0 « u - (g sin a + kg cos a) t or, u « (g sin a + kg cos a) t B) Using B) in A) / - \r (g sin a + kg cos a) t2 C) Case B). When the block comes downward, the net force on the body « mg sin a - km g cos a and hence its acceleration « g sin a - k g cos a Let, t be the time required then, /= |(gsina-*gcosa)r'2 D) From Eqs. C) and D) since -A: cos a sin a + k cos a But — - — (according to the question), Hence on solving we get a» 016 1)

37 1.63 At the initial moment, obviously the tension in the thread connecting ml and m2 equals the weight of m2. . (a) For the block m2 to come down or the block m2 to go up, the conditions is and J- where T is tension and fr is friction which in the limiting case equals long coscl Then or m2g-m1 sin a > km1gcosa or > (k cos a + sin a) (b) Similarly in the case or, m1 g sin a - m^ g > km1 g cos a or, < (sin a - k cos a) (c) For this case, neither kind of motion is possible, and fr need not be limiting. Hence, 2 (k cos a + sin a) > — > (sin a - k cos a) mi 1.64 From the conditions, obtained in the previous problem, first we will check whethet the mass m2 goes up or down. Here, m2/m1 * r\ > sin a + k cos a, (substituting the values). Hence the mass m2 will come down with an acceleration (say w). From the free body diagram of previous problem, and T - ml g sin a - km1 g cos a * m1 w Adding A) and B), we get, m2g-m1gsh\a-kmlg cos a - (ml + m2) w {m2/mx - sin a - k cos a) g fa _ sm a _ k cos a) B) A + m^/m^ 1 + T| Substituting all the values, w- 0-048g-005g As m2 moves down with acceleration of magnitude w = 0.05 g > 0, thus in vector form acceleration of m2: 1 +T| 1.65 Let us write the Newton's second law in projection form along positive Jt-axis for the plank and the bar /r- mxwl9 fr~ m2w2 A)

38 At the initial moment, fr represents the static friction, and as the force F grows so does the friction force fr, but up to it's limiting value i.e. fr =* Unless this value is reached, both bodies moves as a single body with equal acceleration. But as soon as the force fr reaches the limit, the 777//////////////////////// bar starts sliding over the plank i.e. w2 2 vvr Substituting here the values of wx and h>2 taken from Eq. A) and taking into account that km2 fr= km^ g, we obtain, (at -km2g)/m2z g, were the sign "=" corresponds to the moment m t0 (say) Hence, If t £ tQy then km2g (constant), and On this basis m>2» (at-km2g)/m2 (t) and w2 (r), plots are as shown in the figure of answersheet. 1.66 Let us designate the x-axis (Fig.) and apply Fx* mwx for body A : mgsina-ifcmgcosa* mw or, w = g sin a - kg cos a Now, from kinematical equation : /seca- 0 + (l/2)wr2 or, (using Eq. A)). i.e. t* V 2 /sec a/(sin a - Jtcos a)g « V 2 // (sin 2 a/2 - kcos1 a) g j ( sin 2 a , 2 ) a\ —-—-kcos a da 0 2 cos 2 a + 2k cos a sin a - 0 N or, tan2a- -T=>a« 49° k and putting the values of a, k and / in Eq. B) we get tain * Is. 1.67 Let us fix the x - y co-ordinate system to the wedge, taking the x - axis up, along the incline and the y - axis perpendicular to it (Fig.).

39 Now, we draw the free body diagram for the bar. Let us apply Newton's second law in projection form along x and y axis for the bar : Tcosp-mgsina-/r» 0 A) Jsinp +N-mgcosa« 0 or, N=* mgcosa-JsinP B) But fr** kN and using B) in A), we get J« m g sin a + kmg cos a/(cos P For T the value of (cos P + it sin P) should be maximum C) So, d (cos ft + fcsin ft) 0 or tan P « k Putting this value of P in Eq. C) we get, _, m g (sin a + &cos a) mm i/Vi+*:2+jfc2/Vi+jfc2 m g (sin a + k cos a) VTTP" 1.68 First of all let us draw the free body diagram for the small body of mass m and indicate x - axis along the horizontal plane and y - axis, perpendicular to it, as shown in the figure. Let the block breaks off the plane at t * t0 i.e. N ■ 0 So, or. /wg-afosina« 0 mg A) a sin a From Fx ■ mwxy for the body under investigation : m d yj/dt = at cos a ; Integrating within the limits for v (t) V ml dvx» a cos a I tdt (using Eq. 1) So, ds a cos a 2 F* 2m B) Integrating, Eqn. B) for s (t) a cos a t * 2m 3 Using the value of t« r0 from Eq. A), into Eqs. B) and C) C) m g~ cos a , m2 g3 cos a 2 a sin2 a and 5- 6 a2 sin3 a

40 1.69 Newton's second law of motion in projection form, along horizontal or x - axis i.e. Fx = mwx gives. dv F cos (as) « m v — (as a * as) as or, F cos (as) ds* mvdv Integrating, over the limits for v (s) 00 —J cos (as) ds « y or v sin a ma - V2gsina/3fl (using ^ which is the sought relationship. 1.70 From the Newton's second law in projection from : For the bar, T-2kmg~ Bm)w For the motor, T-kmg= mW Now, from the equation of kinematics in the frame of bar or motor : A) B) C) From A), B) and C) we get on eliminating T and W t~ V71/{kg + 3w) fr ' w-»..fr 1.71 Let us write Newton's second law in vector from F ** mwt for both the blocks (in the frame ofcround). 2T+ /wxg*~ r"i *i A) T*+m2g*-m2w2 B) These two equations contain three unknown quantities vv^, \v2 and T. The third equation is provided by the kinematic relationship between the accelerations : M^« Wq + W y M^« Wq-W C) p i where w is th acceleration of the mass mv with respect to the pulley or elevator car. Summing up termwise the left hand and the right-hand sides of these kinematical equations, we get 7772 T Vjo N

41 = 2w 0 D) The simultaneous solution of Eqs*(l), B) and D) yields x - m2) g*+2m2w0 mum2 Using this result in Eq. C), we get, 0*1 "*" and 2 m^ m 0*1 "*" Using the results in Eq. C) we get v? GT- (b) obviously the force exerted by the pulley on the celing of the car 4m1m2 F=-27= Note : one could also solve this problem in the frame of elevator car. 1.72 Let us write Newton's second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of xx and x2 as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline Tl-mlg sin a = m1w1 m2w A) B) For the pulley, moving in vertical direction from the equation Fx» ntwx (as mass of the pulley mp - 0) or 7\« 2 T2 C) As the length of the threads are constant, the kinematical relationship of accelerations becomes w- 2wx D) Simultaneous solutions of all these equations yields : N w 2g f 2 ( 4 tn "*2 mi —— 0*1 \ -sin a + 1 2 g B T] - sin a) As T| > 1, iv is directed vertically downward, and hence in vector form 2 g*B T] - sin a) w

42 1.73 Let us write Newton's second law for masses ml and m2 and moving pully in vertical direction along positive x - axis (Fig.) : 8 - m or m2g-T = m2w2x TX-2T= 0(asm = 0) T, « 2J A) B) ////////V77/77 C) Again using Newton's second law in projection form for mass m0 along positive x1 direction (Fig.), we get Tx - m0 w0 D) The kinematical relationship between the accelerations of masses gives in terms of projection on the x - axis Wlx + W2x " ^ W>0 E) Simultaneous solution of the obtained five equations yields [4 m1 m2 + m0 {m1 - m^) ] g In vector form / / / / / x a 1.74 As the thread is not tied with my so if there were no friction between the thread and the ball m, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously in the given problem it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem wM>wm and as both are directed downward so, relative acceleration of M - wM-wm and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives : ///////, •2 A) 7v\ Newton's second law in projection form along vertically down direction for both, rod and ball gives, Mg-fr-MwM B) mg-fr- mwm C) Multiplying Eq. B) by m and Eq. C) by M and then subtracting Eq. C) from B) and after using Eq. A) we get 2lMm -TP a I T fr (M-m)t

43 1.75 Suppose, the ball goes up with accleration wx and the rod comes down with the acceleration w2. As the length of the thread is constant, 2wp w2 A) From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives, T - m g * mwl and Mg - T'= Mw2 but T = 2 T (because pulley is massless) From Eqs. A), B), C) and D) B-TQg B) C) D) m + AM r| + 4 (in upward direction) and (downwards) From kinematical equation in projection form, we get T as, wx and w2 are in the opposite direction. Putting the values of xvx and w2> the sought time becomes T TM fM /=V2/(T] + 4)/3B-T])g= 1 -4 s 1.76 Using Newton's second law in projection form along x - axis for the body 1 and along negative x - axis for the body 2 respectively, we get m1w1 A) For the pulley lowering in downward direction from Newton's law along x axis, < 0 (as pulley is mass less) 7,-21. or, 7\ = 2 T2 C) As the length of the thread is constant so, h>2= D) The simultaneous solution of above equations yields, 2 (m2 - 2m2) g 2 (ti - 2) , "*i ////////////// 4 m + 4 (as m E) v2 ~r i**] ij tt #f»2 Obviously during the time interval in which the body 1 comes to the horizontal floor covering the distance hy the body 2 moves upward the distance 2h. At the moment when the body 2 is at the height 2/i from the floor its velocity is given by the expression : 2 w2 B/*) - 2 T] + 4 T] +4 After the body m1 touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it's subsequent motion.

44 Owing to the velocity v2 at that moment or at the height 2h from the floor, the body 2 further goes up under gravity by the distance, 2g r| + 4 Thus the sought maximum height attained by the body 2 : 4hQn - 2) 6r\h + 4 H 2h (Tl + 4) Let us draw free body diagram of each body, i.e. of rod A and of wedge B and also draw the kinemetical diagram for accelerations, after analysing the directions of motion of A and B. Kinematical relationship of accelerations is : tana w B Let us write Newton's second law for both bodies in terms of projections having taken positive directions of y and x axes as shown in the figure. mAg-Ncos a * n*AwA (?) and Nswa = ?nBwB Simultaneous solution of A), B) and C) yields : mA g sin a « mA sin a + mB cot a cos a A + t| cot2 a) and C) tan a (tan a + r\ cot a) Note : We may also solve this problem using conservation of mechanical energy instead of Newton's second law. 1.78 Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of M and m on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.). As the length of threads are constant so, dsmM « dsM and as vmM and vM do not change their directions that why w mM w*M I * w (say) and fmM TT vm an<* WM tt Y

45 U/ro so, from the triangle law of vector addition wm=* yflw From the Eq. Fx « mwx > for the wedge and block : T-N= Mwy and N *= mw Now, from the Eq. Fy « mw , for the block mg-T-kN = mw Simultaneous solution of Eqs. B), C) and D) yields : mg A) B) C) D) w (bn + 2m+M) (k + 2 +M/m) Hence using Eq. A) w M 1.79 Bodies 1 and 2 will remain at rest with repect to bar A for wa^n zwz where the sought minimum acceleration of the bar. Beyond these limits there will be a relative motion between bar and the bodies. For Osn'i w^, the tendency of body 1 in relation to the bar A is to move towards right and is in the opposite sense for w fc m^. On the basis of above argument the static friction on 2 by A is directed upward and on 1 by A is directed towards left for the purpose of calculating w^. Let us write Newton's second law for bodies 1 and 2 in terms of projection along positive x - axis (Fig.). T-frx = mw or, frx= T-mw A) N2 « mw B) As body 2 has no acceleration in vertical direction, so fc-mg-T C) From A) and C) But frl +fr2 * k (Nx or fr1+fr2*k (mg + mw) E) 7/////////////////////////

46 From D) and E) m(g-w) or w A+*) Hence mm N 1.80 On the basis of the initial argument of the solution of 1.79, the tendency of bar 2 with respect to 1 will be to move up along the plane. Let us fix (jc - y) coordinate system in the frame of ground as shown in the figure. From second law of motion in projection form along y and x axes : m g cos a-N= m wsin a or, N = m(g cos a - w sin a ) A) m g sin a + fr * m w cos a or, fr-m(w cos a - g sin a ) B) but fr x kN, so from A) and B) ( w cos a- gsina)* k(g cos a + w sin a ) or, w ( cos a-fcsina)sg(fc cos a + sin a) ( cos a + sin a) ' cos a - k sin a ' So, the sought maximum acceleration of the wedge : (k cos a + sin a) g (k cot a + 1) g x max where cot a > k cos a - k sin a cot a - k 1.81 Let us draw the force diagram of each body, and on this basis we observe that the prism moves towards right say with an acceleration wl and the bar 2 of mass w, moves down the plane with respect to 1, say with acceleration w21, then, w2 « w21 + wx (Fig.) Let us write Newton's second law for both bodies in projection form along positive y2 and xx axes as shown in the Fig. m2g cos a -AT- m2w2{yj= m2 I w2i{y2) + wHy2)\ " m2 \ ° + wi sina 1 or, m2 g cos a-N= m2w1 sin a A) and N sin a « 0*1*^ B) Solving A) and B), we get m2 g sin a cos a g sin a cos a (m1/m2) + sin a mx + m2 sin a

47 1.82 To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in the figure. On the basis of force diagram, it is obvious that the wedge M will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the floor. Hence dsmM « ds^. As v^ and v*M do not change their directions and acceleration that's why w*mM TT v^a/ an(* ™m TT V (say) and accordingly the diagram of kinematical dependence is shown in figure. anc^ wmMssWMssW v////////////////////////// m VtM As w m + wu, so from triangle law of vector addition. -2wmMwMco&a *vV2(l-cosa) A) From Fx = m wx, (for the wedge), J* Jcosa+^sina* Mw B) For the bar m let us fix (x - y) coordinate system in the frame of ground Newton's law in projection form along x and y axes (Fig.) gives mg sinct - T - m wm m ^ w] m w A - cosa) mgcos a-N» mwn Solving the above Eqs. simultaneously, we get _ fflgsina C) D) 2m(l-cosa) Note : We can study the motion of the block m in the frame of wedge also, alternately we may solve this problem using conservation of mechanical energy. 1.83 Let us sketch the diagram for the motion of the particle of mass m along the circle of radius R and indicate x and y axis, as shown in the figure. (a) For the particle, change in momentum A/? * mv (- i) - m v (/) so, | A/T| * Vz mv and time taken in describing quarter of the circle,

48 Hence, < F > = (b) In this case Pi = 0 and xR 2v A? At 2>/2mv2 so mwtt Hence, « mw>, 1.84 While moving in a loop, normal reaction exerted by the flyer on the loop at different points and uncompensated weight if any contribute to the weight of flyer at those points. (a) When the aircraft is at the lowermost point, Newton's second law of motion in projection form Fn = m wn gives 2 N-mg = mv ~R or , N = mg + m v R = 2-09 kN (b) When it is at the upper most point, again from Fn - mwn we get N" + mg R N - -mg*= 0-7kN (c) When the aircraft is at the middle point of the loop, again fromFn- mwK m v = ^-» 1-4 kN g The uncompensated weight is mg. Thus effective weight = VW +m g =l-56kN acts obliquely. .85 Let us depict the forces acting on the small sphere m, (at an arbitrary position when the thread makes an angle 0 from the vertical) and write equation F = m w via projection on A A the unit vectors ut and un. From Ft = m wn we have . Q dv mg sin 0 * m — vdv v dv ds (as vertical is refrence line of angular position)

49 or vdv* -glsin Integrating both the sides : v 6 fvdv* -g/f sin 9dQ 0 a/2 or, — - g I cos 9 Hence — » 2 g cos 9 - wn (Eq. A) can be easily obtained by the conservation of mechanical energy). From Fn - mwn T -mg cos 9 = m v I Using A) we have T = 3 mg cos 9 Again from the Eq. Ft - m tv,: wg sin 9 = m wt or vvf = g sin 9 B) C) HI 11IIIII II* Hence w - V wr2 + n>n2 * V (g sin 9 J + Bg cos 9 J (using 1 and 3 ) * gVl+3cos29 (b) Vertical component of velocity, v ■ v sin 9 So, .2-2 For maximum Vy = vzsinz9- 2g / cos 9 sin z 9 (using 1) 2 d (cos 9 sin 9) which yields cos 9 V3 Therefore from B) T = 3wg -j= - VT mg wt ut + wn un thus n un (c) We have hT- wt ut ^ But in accordance with the problem w = 0 + w n(y) or, g sin 9 sin 9 + 2g cos 2 9 (- cos 9) - 9 or, cos 9 or, 9-54T

50 1.86 The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let v be the speed of the ball at its lowest position and / be the length of the thread, then according to the problem y-gsina A) where a is the maximum deflection angle From Newton's law in projection form : Ft * mwt dv 7////////////A or, - g I sin 0 d 0 - v dv On integrating both the sides within their limits. a 0 -g/J sin0d0« J o v dv or, v2 = 2gl A - cos a) B) Note : Eq. B) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity. From Eqs. A) and B) with 0 * a 2g/ A - cos a) * lg cos a or, cos a * — so, a * 53* 1.87 Let us depict the forces acting on the body A^(which are the force of gravity mg and the normal reaction N ) and write equation F= mw via projection on the unit vectors u, and un (Fig.) From Ft» mwt or, gRsinQdB = mg sin I vdv Integrating both side for obtaining v or, e J gRsinBdQ 0 From Fn = mwn Uv2- 2g* mgcos 0 9« m * m 0) V -/' 0 dt vdv ds dv A - cos 0) v2 mR vdv mRdQ / / f f 1/ 1 <9 B) At the moment the body loses contact with the surface, N becomes v2 * gR cos 0 0 and therefore the Eq. B) C)

51 where v and 0 correspond to the moment when the body loses contact with the surface. -l Solving Eqs. A) and C) we obtain cos 0 « - or, 0 - cos B/3) and v - V 2gR/3 . 1.88 At first draw the free body diagram of the device as, shown. The forces, acting on the sleeve are it's weight, acting vertically downward, spring force, along the length of the spring and normal reaction by the rod, perpendicular to its length. Let F be the spring force, and A/ be the elongation. From, Fn ■ mwn : #sm0+,Fcos0« mco2r where r cos 0 * (/0 + A/). Similarly from Ft « mwt NcosQ-FsinQ* 0 or, N - F sin 0/cos 0 From A) and B) F (sin 0/cos 0) • sin 0 + F cos 0 - m co r - m to2 (/0 + A/)/cos 8 On putting F « k A /, KA/sin20 + KA/cos20* mco2(/0 + A on solving, we get, A/= ma/ o I 0 <r A) B) k - m co (k//w co - 1) and it is independent of the direction of rotation. 1.89 According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from the equation Fn « m vvn mv or mv or or v = A) For vmax, we should have or, / r2s 0 -^-» 0, so r- /?/2 K 1.90 As initial velocity is zero thus 2wts A) As wt > 0 the speed of the car increases with time or distance. Till the moment, sliding starts, the static friction provides the required centripetal acceleration to the car. Thus /r* /mv, but frz kmg

52 So, w2 & k2g2 or, or, Hence vmax max 1 / /fcp \ so, from Eqn. A), the sought distance 5 « -— «~'yj-fi- -1 * 60 m. , 2 I iv, I 1.91 Since the car follows a curve, so the maximum velocity at which it can ride without sliding at the point of minimum radius of curvature is the sought velocity and obviously in this case the static friction between the car and the road is limiting. Hence from the equation F - mw mv2 kmg * -—r— or vi VkRg so v^ - Vfc/^g. A) We know that, radius of curvature for a curve at any point (x, y) is given as, f 1 + (dy/dxf ]3/2 ((fy)/dx2 For the given curve, B) a ■ — cos I — ax a la 2 and d y -a . x dx2 a 2 Substituting this value in B) we get, [1 + (a2/a2) cos2 (jc/a) ]3/2 (a/a2) sin (*/a) *^^ *V1P For the minimum /?, — ■ — ' a 2 and therefore, corresponding radius of curvature Hence from A) and B) 1.92 The sought tensile stress acts on each element of the chain. Hence divide the chain into small, similar elements so that each element may be assumed as a particle. We consider one such element of mass dmy which subtends angle d a at the centre. The chain moves along a circle of known radius R with a known angular speed co and certain forces act on it We have to find one of these forces. From Newton's second law in projection form, Fx * mwx We get 2 Jsin (da/2) - dN cos 0 - dma?R and from Fz « mwz we get dN sinQ = gdm Then putting dm » mda/2 n and sin (da/2) * da/2 and solving, we get, m(@2£+gcote) 2jc

53 1.93 Let, us consider a small element of the thread and draw free body diagram for this element, (a) Applying Newton's second law of motion in projection form, Fn - mwn for this element, (T+ dT)sm(dQ/2) + Tsm(dQ/2) -dN- dm«>2R- 0 27 sin (d 9/2)- dN9 [negelecting the term(dT sin d Q/2) ] . dB djl 2 or, or, dN, as A) Also, dfr- kdN From Eqs. A) and B), dT)-T- dT kTdO- dT or dT kdQ In this case Q**n so, T 2 or, or, In — = k n C) So, 1 . T2 1 as m2 g m B) T+dT m. (b) When — * y\, Svhich is greater than r]^ the blocks will move with same value of m acceleration, (say w) and clearly m2 moves downward. From Newton's second law in projection form (downward for m2 and upward for mx) we get : and m2g-J2- m2w T1-m1gm mxw D) E)

54 Also = T»o F) Simultaneous solution of Eqs. D), E) and F) yields (m2 - g g as = T] 1.94 The force with which the cylinder wall acts on the particle will provide centripetal force necessary for the motion of the particle, and since there is no acceleration acting in the horizontal direction, horizontal component of the velocity will remain constant througout the motion. So v0 cos a * ^—^ ru ^ h^ ^ \oc Using, Fn = mwnt for the particle of mass m, 2 2 2 mvx mv0 cos a = ~R~~ = R ' which is the required normal force. 1.95 Obviously the radius vector describing the position of the particle relative to the origin of coordinate is r = xT% yj = a sin co t i + b cos co tj Differentiating twice with respect the time : ,2- —=■ dt2 = - co2 (a sin co t i + b cos co t j) = - co2 F A) Thus 1.96 (a) We have F = mw= -mco r f F^dt jmg*dt mgt A) 0 (b) Using the solution of problem 1.28 (b), the total time of motion, x = 2(vo-g) Hence using t = x in A) =mgx -2m(vo-g)/g 1.97 From the equation of the given time dependence force F vanishes, - a t(x-t) at f = x, the force (a) Thus />« I Fdt 4 0

55 or, but -t)dt ax' o p « m v so v ax 6m (b) Again from the equation F dt or, aXtx-t2)dt'= mdv* Integrating within the limits for v\t), t V* \ a\tx-t2)dt- ml or, Thus 0 —I a_ m 2 3 ml dv o a*i2 Ix _ t m 2 [x t . . v- --- I for t* x m \ 2 3 Hence distance covered during the time interval t - x, o f J 2 3 m 12 1.98 We have F - F{ or On integrating, When jsin m — i7 CO cor dT dt * r -cos Fosin cor + C, r- 0, cor or mdv* (where C is v*0, so C» « ,Fosincordr integration a mco -* o Hence, v = cos cor + mco m co ii As I cos cor £ 1 so, v « A - cos cor) 1 mco

56 Thus s o fvd, Fot F0sin(ot Fo . (cor-sincor) mco mco (Figure in the answer sheet). 1.99 According to the problem, the force acting on the particle of mass m is, F * Fo cos (of dv*~~* —* M) So, m —7- * Fn cos (of or d v « — cos (at m Integrating, within the limits. V t /F C - dv*= — I cos (of dt or v = —— sin (at m J m(a o o It is clear from equation A), that after starting at f « 0, the particle comes to rest fro the first time at f * —. O) From Eq. A), v * I vl - —^- sin (at for f £ — B) n v ' ■ ■ m(a (o v ' Thus during the time interval f = jc/(o, the sought distance 2F I sin (of dt ntw J maJ From Eq. A) max m(a — as | sin (of | d v "~* 1.100 (a) From the problem F » -rv so m — * - rv Thus m~TXi -rv or, — » -—df On integrating lnv = - — t + C But at f * 0, v * v0, so, C « In v( v r -£. or, In — = - — f or, v = v0 e m Thus for f -* oo, v * 0 (b) We have m — **-rv so dv* — ds K ' dt m

57 Integrating within the given limits to obtain v (s): V S /j r C. rs ,^. av--- Ids or v-v0-— A) ° Thus for v - 0, /wv0 total /\t* ^ mdv dv - r (c) Let we have ■ -rv or — ■ — v ' v v m i m — I dt y or, In ■ - — J v m J T| v0 m or, •/ v m « 0 -m ln(l/Ti) bo f * a "• ■ r r Now, average velocity over this time interval, „ * 1 fvdt o <V>" /* " 1.101 According to the problem dv , : nt~7"~* -kv Integrating, withing the limits, p t 0 To finu the valu6 of /:, rewrite dv , j ds On integrating rf m i —lnri r or, m r« V dt dv V2' m k -- VO(T1-1) -** (vo-v) vov k A — ds m v So, Putting the value of k from vo B) in ^ __ * ■■ m A), *( II u u ln~ V we vln > am get v) ■MM V B)

1.102 From Newton's second law for the bar in projection from, Fx= mwx along x direction we get mg sin a - kmg cos a - mw or, or, or, dv / t \ v — - gsma- axg cos a, (as k - ax), v dv m (g sin a - axg cos a) dx V X \ v dv ■ g J (sin a - x cos a ) dx o 0 So, —= g (sin a jc - — a cos a ) ^> ^> From A) v = 0 at either 2 jc = 0, or jc = — tan a a As the motion of the bar is unidirectional it stops after going through a distance of 2 A) tan a. a From A), for vmax, — (sin ax - — a cos a ) * 0, which yields jc - — tan a dx 2 J a Hence, the maximum velocity will be at the distance, jc - tan a/a Putting this value of jc in A) the maximum velocity, max A/gsinatana a 1.103 Since, the applied force is proportional to the time and the frictional force also exists, the motion does not start just after applying the force. The body starts its motion when F equals the limiting friction. Let the motion start after time r0 , then or, a So, for t = £ r0, the body remains at rest and for t > t0 obviously t = a (t- t0) or, mdv** a(t- tG)dt Integrating, and noting v = 0 at t » t0, we have for t > tQ I mdv= a I (t -10 )dt or v = a 0 Thus c* «•

1.104 While going upward, from Newton's second law in vertical direction : vdv , .9. vdv m ds kv ) or -ds m At the maximum height /*, the speed v = 0, so 0 h f Integrating and solving, we get, m 2> m8 j 59 A) When the body falls downward, the net force acting on the body in downward direction equals (mg-kv2), Hence net acceleration, in downward direction, according to second law of motion vdv kv2 vdv . g-— or, -- ds ds m m Thus /vdv p-ifcv2, g-kv /m o d o Jds Integrating and putting the value of h from A), we get, / VTTI vo/mg. 1.105 Let us fix x - y co-ordinate system to the given plane, taking Jt-axis in the direction along which the force vector was oriented at the moment t * 0, then the fundamental equation of dynamics expressed via the projection on x and y-axes gives, F cos co t« m and F sin cor* m dt dv.. dt (a) Using the condition v@) « 0, we obtain v. F . x met) sin cor and B) C) V m ( 1 - COS @ t ) y mco D) Hence, V + V • vx y 2F ffllO sin

60 (b) It is seen from this that the velocity v turns into zero after the time interval Af, which can be found from the relation, co — = Jt. Consequentely, the sought distance, is fvdt SF o moo Average velocity, < v > 0 2n/a> So, <v> mco sin o ~\dtj Bjio))- - 4F 1.106 The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane Fx - mg sin a and the friction force fr « kmg cos a. In our case k = tan a and therefore fr * Fx * mg sin a Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the x-axis : mwt= Fx cos qp - fr « mg sin a ( cos qp - 1) m wx « Fx- fr cos cp « mg sin a A- cos cp ) It is seen fromthis that wt ■ - wx, which means that the velocity v and its projection vx differ only by a constant value C which does not change with time, i.e. where vx * v cos cp. The constant C is found from the initial condition v = v0, whence C « v0 since (p « — initially. Finally we obtain v = v0 / A + cos qp). In the cource of time cp -* 0 and v -> Vq/2. (Motion then is unaccelerated.) 1.107 Let us consider an element of length ds at an angle qp from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence, dN- Xds cos qp= 0, where X is the linear mass density of the chain Let T and T+dT be the tension at the upper and the lower ends of ds. we have from, = mwt or, = Xdswt

61 If we sum the above equation for all elements, the term J dT = 0 because there is no tension at the free ends, so l/R "KgR I sin cp d op - Xwt f o Hence i As wn = a at initial moment So, w - I wt 1.108 In the problem, we require the velocity of the body, realtive to the sphere, which itself moves with an acceleration w0 in horizontal direction (say towards left). Hence it is ad visible to solve the problem in the frame of sphere (non-inertial frame). At an arbitary moment, when the body is at an angle 0 with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form n n mv or, mg cos 8 - N - mw0 sin 8 - —jr- A) At the break off point, N - 0, 8 = 80 and let v* v0so the Eq. A) becomes, From, Ft - mwt . vdv vdv mg sin 8 - mw0 cos 8 « m —r— - m or, v dv = R (g sin 8 + w0 cos 8 ) d 8 e 0 Integrating, I v dv = I R (g sinO + m>0 cos8) J 8 0 0 - cos90) + wQ sin 8 0 C) Note that the Eq. C) can also be obtained by the work-energy theorem A= AJ (in the frame of sphere) 1 i therefore, mgR A - cos 80) + mwQR sin 80 * [here mwQk sin 80 is the work done by the pseudoforce (- mwo)] or, 2R = g A - cos 80) + w0 sin 80

62 Solving Eqs. B) and C) we get, v0 * y/2gR/3 and Hence o , where k * — g 1.109 This is not central force problem unless the path is a circle about the said point. Rather here Ft (tangential force) vanishes. Thus equation of motion becomes, vtm vo" constant and, mv o — for We can consider the latter equation as the equilibrium under two forces. When the motion is perturbed, we write r - r0 + x and the net force acting on the particle is, 2 0 — A r* nx 1 — m voI _ x 0 m v, o 0 mv, This is opposite to the displacement xy if n < 1- — A force while is the1 inward directed external force). is an outward directed centrifugul 1.110 There are two forces on the sleeve, the weight Fx and the centrifugal force F2* We resolve both forces into tangential and normal component then the net downward tangential force on the sleeve is, mgsinG 1- (D2R 8 cos 8 This vanishes for 0= Oand for 8 « 80 - cos -l g_ , which is real if > g. If co2jR < g, then 1 - co2/? g cos 8 7nu)zRS/7?O=Fz r/: m$CosQ+mu)tRSi7)°' is always positive for small values of 8 and , hence the net tangential force near 0=0 Q; opposes any displacement away from it. ' 0 = 0 is then stable. 2 If co R> gy 1 cos8 is negative for small 8 near 8*0 and 0 = 0 is then unstable. However 8 « 80 is stable because the force tends to bring the sleeve near the equilibrium position 8 = 9q. If co R = gy the two positions coincide and becomes a stable equilibrium point.

63 1.111 Define the axes as shown with z along the local vertical, x due east and y due north. (We assume we are in the northern hemisphere). Then the Coriolis force has the components. Fcor* -2m(o)xv) >z sinG) i - vx cosQj + vx cosG k 1 rhen the direction in which tn< 2mo> (v^ cosG - vz sinQ) i 2mco I v%, cosG - v, le gun is fired is due north. Thus the equation of motion (neglecting centrifugal forces) are x = 2/nco (vy sincp - vz coscp), y * 0 and z = -g CO Integrating we get y * v (constant), z « -gt * 2 and x - 2co v sincp t + cog f coscp Finally, 2 r si sincp+ r- coscp Now v » gr in the present case, so, x** covsincp| —| « cosinqp *• 7 cm (to the east). 1.112 The disc exerts three forces which are mutually perpendicular. They are the reaction of the weight, mg, vertically upward, the Coriolis force 2mv' co perpendicular to the plane of the vertical and along the diameter, and mco r outward along the diameter. The resultant force is, F= m V i + co4 r2 + Bv' co): 1.113 The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it. The equation is, 2 u dr mv ■ mco r where v * — dt so, or, Butv* v —* —| -v dr dry 2 — v * --co r + constant v0 being the initial velocity when r = 0. The Coriolis force is then, 2mco Vvq + co2 r2 = 2mco2 r Vl + v2/co2 r2 2-83 N on putting the values.

64 1.114 The disc OBAC is rotating with angular velocity co about the axis OO' passing through the edge point O. The equation of motion in rotating frame is, mw M o1 B where F^ is the resultant inertial fore ^pseudo force) which is the vector sum of centrifugal and Coriolis forces. (a) At A, Ffo vanishes. Thus 0 * - 2mco2 Rn + 2mv' co n A where n is the inward drawn unit vector to the centre from the point in question, here A. Thus, v'- co/? so, (b) AtB V'2 V'2 2 _ w= —* — ■ co /?. Fk - mcoz OC + its magnitude is mco \4R - r , where r * OB. 1.115 The equation of motion in the rotating coordinate system is, mw * F + mco2/?+ 2m(v xco) Now, v** R 6 £T+/? sin0 cp ? and w cos -w' sin 0 e e 9 0 7? 6 /? sin 6 <p co cos 0 - co sin 0 0 2m « e£((oR sin 0 9) + wR sul 8 cos 6 9 ^e"" ^ ® cos ® Now on the sphere, + (R 6' - R sin 0 cos 0 cp2) + (i? sin 0<p* + 2/? cos 0 6 (j>) i* Thus the equation of motion are, m(-RQ2-R sin2 0 cp2) * N- mg cos 0 + mco2/? sin2 0 + 2m(oR sin2 0 cp m (/? 0 - iR sin 0 cos 0 <p ) * mg sin 0 + mco /? sin 0 cos 0 + 2mco R sin 0 cos 0 cp mORsinOcp'+ 2J? cos 0 6<p)« -2mcoi?6cos0 From the third equation, we get, <p » - co A result that is easy to understant by considering the motion in non-rotating frame. The eliminating cp we get, m/?02* mgcosQ-N m R 6' * mg sin 0 Integrating the last equation, ^2« mg(l-cos0)

65 Hence N - C - 2 cos 9) mg -12 So the body must fly off for 0 * 0O « cos " -, exactly as if the sphere were nonrotating. 2 /s"" 2 Now, at this point Fcf * centrifugal force * mco R sin 80 * y- mco R Fcor= Vco2i?292cos29 + (co2R2Jsin20 x2m x2m« \ 3 co2/* 1.116 (a) When the train is moving along a meridian only the Coriolis force has a lateral component and its magnitude (see the previous problem) is, 2m co v cos 0 « 2m co sin X So, (Here we have put R 6 -* v) 54000 VT 86400" 3600 X 2 2 x 2000 x 103XT-TTX-rrrr-x * 3-77 kN, (we write Xfor the latitude) (b) The resultant of the inertial forces acting on the train is, _^ Fw * - 2minR 6 cos 0 i£ + (mco2 /? sin 0 cos 0 + 2m co R sin 0 cos 0 cp) i + (mw2 R sin2 0 + 2m co i? sin2 0 cp ) i* This vanishes if 6=0, cp « - — co Thus v *= v e , v * - (We write X. for the latitude here) Thus the train must move from the east to west along the 60* parallel with a speed, * 115-8 m/s « 417 km/hr (o/?cosX tx^ZTx 2 4 8-64 1.117 We go to the equation given in 1.111. Here vy « 0 so we can take y * 0, thus we get for the motion in the x z plane. and Integrating, x * - 2co vz cos 0 z - -g 2* 2 JC - CO g COS <pt So '2*' Tcogcoscp/3* ~cogcoscp| — 3/2 COS There is thus a displacement to the east of |x~^64x500xlx ** 26 cm.

66 1.3 Laws of Conservation of Energy, Momentum and Angular Momentum. 1.118 As F is constant so the sought work done or, A- 1.119 Differentating v (s) with respect to time dv a ds a dt 2Vs dt 2Vs C r ( i^s 17 J a2 2 (As locomotive is in unidrectional motion) Hence force acting on the locomotive F * m w ma Hence the sought work, A= Fs mg2 (a2/2) 2 4 Let, ar v - 0 at t - 0 then the distance covered during the first t seconds maY 8 las2 m 2 as m 1.120 We have or, v A) Differentating Eq. A) with respect to time 2vm\ 4 as m v or, >vr B) Hence net acceleration of the particle h> = '2<w\2 m 2 or r 2 as mR m V 1 Hence the sought force, F = mw = 2as Vl + (s/R)z 1.121 Let i7 makes an angle 6 with the horizontal at any instant of time (Fig.). Newton's second law in projection form along the direction of the force, gives : F = kmg cos 0 + mg sin 8 (because there is no acceleration of the body.) As F | T d r*the differential work done by the force F, dA= F-dr*= Fds, (where ds * | dr*\ ) = kmg ds (cos Q) + mgds sin 6 F N ■ kmgdx + mgdy. I h kmg I dx + mg I dy Hence, A - A7wg | or + mg o o = kmgl + mgh= mg(kl

67 1.122 Let s be the distance covered by the disc along the incline, from the Eq. of increment of M.E. of the disc in the field of gravity : AT + tJJ * Afr 0 + (- mgs sin a) * - kmg cos a s - kmg I kl or, s« -7- . sin a - k cos a Hence the sought work Ajr * - kmg [s cos a + / ] klmg A) [Using the Eqn. A)] u - -0.05 J *r 1 - k cot a On puting the values 1.123 Let x be the compression in the spring when the bar m2 is about to shift Therefore at this moment spring force on m2 is equal to the limiting friction between the bar m2 and horizontal floor. Hence kx * km2g [where k is the spring constant (say)] A) For the block m1 from work-eneigy theorem : A * A7 * 0 for minimum force. (A here indudes the work done in stretching the spring.) 1 2 x so, Fx- — kxr-kmgjc* 0 or k — * F-kmlg B) From A) and B), F= kg 1.124 From the initial condition of the problem the limiting fricition between the chain lying on the horizontal table equals the weight of the over hanging part of the chain, i.e. X r\ Ig * k X A - rj) Ig (where X is the linear j\J mass density of the chain) So, a) Let (at an arbitrary moment of time) the length 'of the chain on the table is x. So the net friction force between the chain and the table, at this moment : /r« kN= kkxg B) The differential work done by the friction forces dA -k\xg(-dx) C) (Note that here we have written ds * - dx., because ds is essentially a positive term and as the length of the chain decreases with time, dx is negative) Hence, the sought work done o J x8T^{xdx= -A-^ ^

68 1.125 The velocity of the body, t seconds after the begining of the motion becomes v** v*+ g*i. The power developed by the gravity (m g*) at that moment, is mjrv*= mOT' Vo + g2'K8 mg (gf - v0 sin a) A) As nig* is a constant force, so the average power x x where Ar is the net displacement of the body during time of flight As, nig^l Ar* so <P> * 0 1.126 We have wn - ^* at2, or, v - VaT ty t is defined to start from the begining of motion from rest So, m>,« — * vaR —* —^ A A A Instantaneous power, P~F-v= m(wtut + wnut ) • (va/? rMr ), A A (where w, and ut are unit vectors along the direction of tangent (velocity) and normal respectively) So, P« mwtVaRt= maRt Hence the sought average power t t ma J o / o Hence <P> « maRt2 maRt It 2 1.127 Let the body m acquire the horizontal velocity v0 along positive x - axis at the point O. (a) Velocity of the body t seconds after the begining of the motion, r-^+sr*-(v0-*gor* (i) Instantaneous power P - F • v - (- kmg i )'(vo-kgt)i=* - long (v0 - kgt) From Eq. A), the time of motion x * v^/kg Hence sought average power during the time of motion T I - kmg (v0 -kgt)dt <P> m = - —|-^ = - 2 W (On substitution) X £* From Fx * mwx -kmg= mwx= mvx-£ or, vxdvx= -kgdx * - agxdx

69 To find v (x), let us integrate the above equation J vxdvx~ - a8 j xdx or, v2- v^-a A) Now, P= jp'v** -majcgVvQ-agjc2 B) For maximum power, — (vv 0x - Xgx ) « 0 which yields x - , Putting this value of jc, in Eq. B) we get, 1.128 Centrifugal force of inertia is directed outward along radial line, thus the sought work r2 A= \ mco2rdr= -mco2(r\-r^\ = 0-20T (On substitution) 1.129 Since the springs are connected in series, the combination may be treated as a single spring of spring constant Kl From the equation of increment of M.E., A T + A U * A ext 1 . .i 1 A, or, A _ .». —• -«, -~.», .«. _ t2/ A/ 2 1.130 First, let us find the total height of ascent At the beginning and the end of the path of velocity of the body is equal to zero, and therefore the increment of the kinetic energy of the body is also equal to zero. On the other hand, in according with work-energy theorem AT is equal to the algebraic sum of the works A performed by all the forces, i.e. by the force JF and gravity, over this path. However, since AT=» 0 then A = 0. Taking into account that the upward direction is assumed to coincide with the positive direction of the y - axis, we can write (i7 + /ngj' d r** I (F - o o h 88 w# I A - 2 ay) dy * mgh A - a/i) * 0. o whence h « I/a. The work performed by the force F over the first half of the ascent is h/2 h/2 c c Ap** I Fvdy** 2mg I A - ay) dy « 3 mg/4a. J y J o o The corresponding increment of the potential energy is A G= mgh/2- mg/2a.

70 dU 1.131 From the equation Fr » - — we get Fr 2a bj\ r3%2 2a (a) we have at r » r0, the particle is in equilibrium position. i.e. Fr - 0 so, r0 - — To check, whether the position is steady (the position of stable equilibrium), we have to satisfy d2U dr We have d2U dr2 la >0 r6a 261 r4"V Putting the value of r « r0 « ~,we get b d2U b4 dr So, 8a: , (as a and b are positive constant) d2U dr2 8a: which indicates that the potential eneigy of the system is minimum, hence this position is steady. (b) We have dr dF. 2a b 1 r 3 """r2 For F, to be maximum, -r— - 0 ar 3a -fc3 So, r - y and then Fr(max) - As Fr is negative, the force is attractive. 1.132 (a) We have ATT • - 2 a x and 7 ay So, F» 2axi-2$yi and, F For a central force, r x F « 0 Here, rxF~ (xi+yj )x(-2xxxi-2$yj ) Hence the force is not a central force, (b) As So, A) According to the problem F - 2 V a2 x2 + p2 y2 * C (constant)

71 or, x2 v2 C2 or, —+ ^r- — « A; (say) p2 a2 2a2p2 y Therefore the surfaces for which F is constant is an ellipse. For an equipotential surface U is constant So, a jc + p y » Co (constant) x2 y2 or, /— + -y=m —~* Ko(constant) VP2 Va2 a P Hence the equipotential surface is also an ellipse. 1.133 Let us calculate the work performed by the forces of each field over the path from a certain point 1 (xx, yx) to another certain point 2 (x2, y2) *2 (i) dA « F• dr*** ayi'dr^ aydx or, A * a I y<£t (ii) dA « F -dr~ (axi + byi ) • dr « axdx + 2 2 Hence A * I a xdx + I bydy In the first case, the integral depends on the function of type y (jc), i.e. on the shape of the path. Consequently, the first field of force is not potential. In the second case, both the integrals do not depend on the shape of the path. They are defined only by the coordinate of the initial and final points of the path, therefore the second field of force is potential. 1.134 Let s be the sought distance, then from the equation of increment of M.E. AJ+AC/- A/r 1 i\ 0 - -r mvn + (+ mgs sina)« - kmg cos a s , 2 / or, s** — I (sin a + k cos a) -kmvn Hence Afr ■ - kmg cos a s 2 (k + tan a) 1.135 Velocity of the body at height /*, vh - \2gjH- h), horizontally (from the figure given in the problem). Time taken in falling through the distance h. t« Y -— (as initial vertical component of the velocity is zero.) Now s= vht

72 For W (Hh ~ h2)« 0, which yields h « y Putting this value of /* in the expression obtained for s, we get, 1.136 To complete a smooth vertical track of radius Ry the minimum height at which a particle starts, must be equal to — R (one can proved it from energy conservation). Thus in our problem body could not reach the upper most point of the vertical track of radius R/2. Let the particle A leave the track at some point O with speed v (Fig.). Now from energy conservation for the body A in the field of gravity : mg + sin 8) 1 2 or, v2 - gh A - sin 8) A) From Newton's second law for the particle at the point O\ F * mw , N + mg sin 0 mv (A/2) A But, at the point O the normal reaction N - 0 So, 2 sin 0 B) From C) and D), sin 0 ■ r- and v - After leaving the track at 0, the particle A comes in air and further goes up and at maximum height of it's trajectory in air, it's velocity (say v') becomes horizontal (Fig.). Hence, the sought velocity of A at this point vcos(9O-0)« vsin6« j 1.137 Let, the point of suspension be shifted with velocity vA in the horizontal direction towards left then in the rest frame of point of suspension the ball starts with same velocity horizontally towards right Let us work in this, frame. From Newton's second law in projection form towards the point of suspension at the upper most point (say B) : mv mv iivy n rr»v mg+T- — or, T- — B -mg Condition required, to complete the vertical circle is that 7"i0. But A) B) mg So, C)

73 From A), B) and C) j^ -mgfcO or, vA* V5g/ From the equation Fn * rnn^ at point C \T vc mv D) \ Again from energy conservation From D) and E) 1 2 1 2 , - -mvc + mgl T- 3mg E) 1.138 Since the tension is always perpendicular to the velocity vector, the work done by the tension force will be zero. Hence, according to the work energy theorem, the kinetic energy or velocity of the disc will remain constant during it's motion. Hence, the sought time f = —, where s is the total distance traversed by the small disc during it's motion. Now, at an arbitary position (Fig.) ds* (lo-RQ)dQ, l/R SO, S 0 or, r s * — - R 2R2 2R Hence, the required time, t I 1A 2Rv o It should be clearly understood that the only uncompensated force acting on the disc A in this case is the tension T, of the thread. It is easy to see that there is no point here, relative to which the moment of force T is invarible in the process of motion. Hence conservation of angular momentum is not applicable here. 1.139 Suppose that A/ is the elongation of the rubbler cord. Then from energy conservation, - 0(as AJ« 0) or, or, 1^ 2 0 ~k A/2-/nghl-mgl 4* 0

74 or. A/ V 4 x y V 1+V 1± 2 k/ mg Since the value So, 2k/ 1 +:r:r:- is certainly greater than 1, hence negative sign is avoided. mg A/ = 1 + v \ 2 k/ 1.140 When the thread PA is burnt, obviously the speed of the bars will be equal at any instant of time until it breaks off. Let v be the speed of each block and 0 be the angle, which the elongated spring makes with the vertical at the moment, when the bar A breaks off the plane. At this stage the elongation in the spring. A/* /Osec0-/O = /O(sec0-1) A) Since the problem is concerned with position and there are no forces other than conservative forces, the mechanical energy of the system (both bars + spring) in the field of gravity is conserved, i.e. AT + AU « 0 So, 2 From Newton's second law in projection form along vertical direction : mg = N+kIo(sec0-1)cos 0 But, at the moment of break off, N « 0. Hence, k /o (sec 0-1) cos 0 - mg k /0 - mg 0 B) or, cos 0 i0 C) Taking k * , simultaneous solutionNof B) and C) yields : WO V 32 1-7 m/s. 1.141 Obviously the elongation in the cord, A/ * /0 (sec 0 - 1), at the moment the sliding first starts and at the moment horizontal projection of spring force equals the limiting friction. So, kx A/ sin 0 - kN A) (where kx is the elastic constant). From Newton's law in projection foim along vertical direction : or, From A) and B), mg. N- mg - Kj A/ cos 0 A/ sin 8 * k (mg - kx A/ cos

75 bng 1 A/ sm 0 + k A/ cos 0 From the equation of the increment of mechanical energy : AG+AT* Afr or. 7r or, hngU 2A/(sin6+A:cos0) bng L (sec 0 - Thus Afr ■ u— ~- 0 09 J (on substitution) fr 2 (sin 0-it cos 0) v ; 1.142 Let the deformation in the spring be A/, when the rod AB has attained the angular velocity co. From the second law of motion in projection form Fn * mwn . ~ ma) L kA/= mo>2(/0 + A0 or, A/*- ^ k -mo From the energy equation, A^ » ^r 1 2 On solving mco /0 k -ma) mo2 /02 K-mco ~, where ma) £» A — Tl) K- 1.143 We know that acceleration of centre of mass of the system is given by the expression. m w, Since Now from Newton's second law F - mw, for the bodies ml and m2 respectively. and T+m2g- m2\v2= -m2wx C) Solving B) and C) 1 2'2- D) y//////

76 Thus from A), B) and D), (m1 + m2) 1.144 As the closed system consisting two particles ml and of m2 is initially at rest the CM. of the system will remain at rest. Further as m2 « mj/2, the CM. of the system divides the line joining m1 and m2 at all the moments of time in the ratio 1 : 2. In addition to it the total linear momentum of the system at all the times is zero. So, pT * - 5t and therefore the velocities of m1 and m2 are also directed in opposite sense. Bearing in mind all these thing, the sought trajectory is as shown in the figure. 1.145 First of all, it is clear that the chain does not move in the vertical direction during the uniform rotation. This means that the vertical component of the tension T balances gravity. As for the horizontal component of the tension 7, it is constant in magnitude and permanently directed toward the rotation axis. It follows from this that the CM. of the chain, the point C, travels along horizontal circle of radius p (say). Therefore we have, T cos 0 - mg and T sin 0 - mto p Thus and gtan0 no p » a—-— ~ 0-8 cm (O mg cos 0 - 5N 7T?2 1.146 (a) Let us draw free body diagram and write Newton's second law in terms of projection along vertical and horizontal direction respectively. N cos a-mg + fr sin a ■ 0 fr cos a - N sin a - mo / A) B) From A) and B) sin a fr cos a - cos a (- fr sin a + mg) = moo / r

77 So, /r« mg CD2/ sin a + cos a = 6N C) g ' (b) For rolling, without sliding, /rs JfcN but, N' - mg cos a-mco2/sina co2/ mg I sin a + cos a k( mgcos a - m co /sin a ) [Using C)] g Rearranging, we get, m co / ( cos a + fcsina) ss: (A: mg cos a - mg sin a ) Thus to 2S V g (it - tan a)/A + k tan a ) / = 2 rad/s 1.147 (a) Total kinetic energy in frame K' is This is minimum with respect to variation in V, when - 0, i.e. mt ( vT- V*J + m9 ( vT- v") - 0 or y» -v ml + m2 Hence, it is the frame of CM. in which kinetic energy of a system is minimum. (b) Linear momentum of the particle 1 in the K' or C frame ~^ *. w /n^ /712 w k. mxm2 or, px » M- ( vi ~ V2 )> wnere> M- - ■ reduced mass Similarly, p2 = fi ( v2 - vx So, |^|= |ft|-|>- |iv^ where, vrd = |v^-i^| C) Now the total kinetic energy of the system in the C frame is f-T+f- 2—+ -2- 1 2 2x 2m2 2 ji Hence

78 1.148 To find the relationship between the values of the mechanical energy of a system in the K and C reference frames, let us begin with the kinetic eneigy T of the system. The velocity of the i-th particle in the K frame may be represented as v^« vj+ v£. Now we can write Since in the C frame Y mi vi * 0, the previous expression takes the form J» J+~/wv£* J+ ~ m V (since according to the problem vc - V) A) 2 2m'V' Since the internal potential eneigy U of a system depends only on its configuration, the magnitude U is the same in all refrence frames. Adding U to the left and right hand sides of Eq. A), we obtain the sought relationship E- E + ^rmV2 1.149 As initially U - U * 0, so, E - J From the solution of 1.147 (b) As Thus J- | 1.150 Velocity of masses mx and m2, after r seconds are respectively. Hence the final momentum of the system, ^> (where, And radius vector, a>« vtt + zrwit » c c 2 c 1^ — a 2* = Vo^ + T^2» where v£* X

79 1.151 After releasing the bar 2 acquires the velocity v2, obtained by the energy, conservation : 1 2 2*K (i) Thus the sought velocity of CM. 1.152 Let us consider both blocks and spring as the physical system. The centre of mass of the system moves with acceleration a - towards right Let us work in the frame of nt + tn tn centre of mass. As this frame is a non-inertial frame (accelerated with respect to the ground) we have to apply a pseudo force ml a towards left on the block m1 and m2 a towards left on the block As the center of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The elongation of the spring will be maximum or minimum at this instant. Assume that the block m1 is displaced by the distance x1 and the block m2 through a distance x2 from the initial positions. From the energy equation in the frame of CM. A T+ U-A^ . (where Here, V77777777777777777777 also includes the work done by the pseudo forces) A7*0, U^^k(x1+x2J and W act mlF mx + ml & ( Xl + X2 or, So, 1., ,2 mibi+x^F -k(xx+x2) * — ^^ Or» Xl 2mlF k (/Wj + m2) Hence the maximum separation between the blocks equals 2miF 0 k (ml + m2) Obviously the minimum sepation corresponds to zero elongation and is equal to /0 1.153 (a) The initial compression in the spring A/ must be such that after burning of the thread, the upper cube rises to a height that produces a tension in the spring that is atleast equal to the weight of the lower cube. Actually, the spring will first go from its compressed

80 state to its natural length and then get elongated beyond this natural length. Let / be the maximum elongation produced under these circumstances. Then Kl-mg A) Now, from energy conservation, B) (Because at maximum elongation of the spring, the speed of upper cube becomes zero) From A) and B), -mg K 2 2mgM A/ - — K r K2 0 Or, A/ . , Therefore, acceptable solution of A/ equals K (b) Let v the velocity of upper cube at the position (say, at C ) when the low.er block breaks off the floor, then from energy conservation. or, (where / * mg/K and A / 32 ntv + 0 At the position C, the velocity of CM; vc (spring+ two cubes) further rises up to Now, from energy conservation, B) — —Let, the CM. of the system JL Bm)gArC2 7C1 2g 8g k But, uptil position C, the CM. of the system has already elevated by, (A/ + /) m + 0 4mg ^ ■B At k Hence, the net displacement of the CM. of the system, in Upward direction . 8m£ K 7 1.154 Due to ejection of mass from a moving system (which moves due to inertia) in a direction perpendicular to it, the velocity of moving system does not change. The momentum change being adjusted by the forces on the rails. Hence in our problem velocities of buggies change only due to the entrance of the man coming from the other buggy. From the

81 Solving A) and B), we get mv , Mv and 1 M-m 2 M-m As Vi 11 v and vT So, v, - /w v and 1 (M-m) °"u K2 (M-m) 1.155 From momentum conservation, for the system "rear buggy with man" From momentum conservation, for the system (front buggy + man coming from rear buggy) MvTh ~~ ~~ ^ So, v»= + 77 m M+m PuttingThe value of v£ from A), we get 1.156 (i) Let v^ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation of linear momentum, - 0 or, 77r M+2m A) (ii) Let v be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man + other man) from the conservation of linear momentum : 0- (M + m) v* + m E% v*) B) Let v£be the sought velocity of the buggy when the second man jump off the buggy; then from conservation of linear momentum of the system (buggy + one man) : C) Solving equations B) and C) we get mBM+3m)i/ (M From A) and D) V2 m — - 1 + Hence v2 > vx 1.157 The descending part of the chain is in free fall, it has speed v * \2gh at the instant, all its points have descended a distance y. The length of the chain which lands on the floor during the differential time interval dt following this instant is vdt.

82 For the incoming chain element on the floor : ■ From dpy « Fv dt (where y - axis is directed down) j 0-(kvdt)v- Fydt <k or Fy= -\v2= -2\gy g Hence, the force exerted on the falling chain equals X v and is directed upward. Therefore from third law the force exerted by the falling < chain on the table at the same instant of QH.~~^ time becomes X v2 and is directed downward. ^ Since a length of chain of weight (kyg) already lies on the table the total force on the floor is Bkyg) + (kyg) m CXyg) or the weight of a length 3y of chain. 1.158 Velocity of the ball, with which it hits the slab, v « v2gh v 1 After first impact, v' * ev (upward) but according to the problem v'« -,so e* - A) and momentum, imparted to the slab, -» mv - (- mv') m mv A + e) Similarly, velocity of the ball after second impact, v" * ev' = e v And momentum imparted « m (v\+ v" ) * m A + e) ev Again, momentum imparted during third impact, « m A + e) e v, and so on, Hence, net momentum, imparted * mv A + e) + mve A + e) + mve A + e) + ... - mv ~ £, (from summation of G.P.) A-e) - V2g*7 -J-- m y/2gh I fa + 1)/ (ri - 1) (Using Eq. 1) 1-- = 0*2 kg m/s. (On substitution) 1.159 (a) Since the resistance of water is negligibly small, the resultant of all external forces acting on the system "a man and a raft" is equal to zero. This means that the position of the CM. of the given system does not change in the process of motion. i.e. r£~ constant or, Ar£- 0 i.e. "V mi Arp- 0 or, m^A^+Ar^+MAr£ = 0 Thus, m(r+/)+M/« 0, or, (b) As net external force on "man-raft" system is equal to zero, therefore the momentum of this system does not change, So, 0- m[v*(r) + i£(r)] £

83 1.159 (a) Since the resistance of water is negligibly small, the resultant of all external forces acting on the system "a man and a raft" is equal to aero. This means that the position of the CM. of the given system does not change in the process of motion. i.e. rc « constant or, Arc » 0 i.e. mi or, Thus, m m , or, m +M (b) As net external force on "man-raft" system is equal to zero, therefore the momentum of this system does not change, So, 0 * m [ v*' @ + v%(t) ] + m v\t) or, m+M A) As v (t) or v2 (t) is along horizontal direction, thus the sought force on the raft Mm dv* (t) m + M dt Note : we may get the result of part (a), if we integrate Eq. A) over the time of motion of man or raft. dt 1.160 In the refrence frame fixed to the pulley axis the location of CM. of the given system is described by the radius vector m 2M But and Thus A r, ml" 2M Note : one may also solve this problem using momentum conservation. 1.161 Velocity of cannon as well as that of shell equals v 2 gl sin a down the inclined plane taken as the positive x - axis. From the linear impulse momentum theorem in projection form along x - axis for the system (connon + shell) i.e. Apx - p cos a - M Vzg/sina « Mg sin a A t (as mass of the shell is neligible) pcosa -A/V 2 gl sin a or, At- * ——: Mg sin a 1.162 From conservation of momentum, for the system (bullet + body) along the initial direction of bullet mv0 * (m + M) v, or, v mv. o m+Af

84 1.163 When the disc breaks off the body My its velocity towards right (along jc-axis) equals the velocity of the body M, and let the disc's velocity'in upward direction (along y-axis) at that moment be v' From conservation of momentum, along jc-axis for the system (disc + body) mv wv- (m+M)v* or v' * T7 A) * x m+M v 7 And from energy conservation, for the same system in the field of gravity : -mv2 = -(m+M) v'l + ~mvt2y + mgh! , where h' is the height of break off point from initial level. So, -mv - -(m+M) rM + m\+2mvy*mgt using A) Also, if h" is the height of the disc, from the break-off point, then, v'2y - 2 gh" Hence, the total height, raised from the initial level Mv2 2g(M+m) 1.164 (a) When the disc slides and comes to a plank, it has a velocity equal to v = ^2gh> to friction between the disc and the plank the disc slows down and after some time the disc moves in one piece with the plank with velocity v' (say). From the momentum conservation for the system (disc + plank) along horizontal towards right : mv mv- (m+M)vf or v' = — m +M Now from the equation of the increment of total mechanical energy of a system : - (M + m) v'2 - ^mv2 - Afr 1 m2v2 1 2 or, —(M+m) T-—mv - Air (m+MJ 2 fr so, Hence, 2 m M + m - m A ■fr where u* rr» reduced mass I m+M

85 (b) We look at the problem from a frame in which the hill is moving (together with the disc on it) to the right with speed il Then in this frame the speed of the disc when it just gets onto the plank is, by the law of addition of velocities, v = w + y/lgh. Similarly the common speed of the plank and the disc when they move together is m + M Then as above A *. « — (m + M)v2 - — mvz - ^rMu Z* JL JL 1 u y2izh + — 2sh m+M 3 (m+MJ 2v« . .,., u2-~m2uV2g7T-mg/i We see that A- is independent of u and is in fact just - \i g h as in (a). Thus the result obtained does not depend on the choice of reference frame. Do note however that it will be in correct to apply "conservation of enegy" formula in the frame in which the hill is moving. The energy carried by the hill is not negligible in this frame. See also the next problem. 1.165 In a frame moving relative to the earth, one has to include the kinetic energy of the earth as well as earth's acceleration to be able to apply conservation of energy to the problem. In a reference frame falling to the earth with velocity v0, the stone is initially going up with velocity v0 and so is the earth. The final velocity of the stone is 0 = vo - gt and that of the earth is vo + — gt (M is the mass of the earth), from Newton's third law, where t = time of fall. From conservation of energy 2 1 2 1 ..r 2 , 1 ..r/ rn -mvQ + ~Mv0 + mgh = -M v0 + u 1 2 Hence ~v0 m + I u mgh Negecting — in comparison with 1, we get 2 v20 « 2gh or vo « W2gh The point is this in earth's rest frame the effect of earth's accleration is of order — and M can be neglected but in a frame moving with respect to the earth the effect of earth's acceleration must be kept because it is of order one (i.e. large). 1.166 From conservation of momentum, for the closed system "both colliding particles" /WjVj + m2v2 = (m1 + m2) v -* !? 2? lCit2/) + 2D7t6*) r r r or, v* ~ LJ-z—1-j£ -= i + 2j-4k Hence | v] = VI +4 + 16 m/s = 4*6 m/s

86 1.167 For perfectly inelastic collision, in the CM. frame, final kinetic energy of the colliding system (both spheres) becomes zero. Hence initial kinetic energy of the system in CM. frame completely turns into the internal energy @ of the formed body. Hence - 1 ._► -+.2 2i 1 ._* _*,2 Now from energy conservation AJ«-j2«-~|iv1-v2 , In lab frame the same result is obtained as 1.168 (a) Let the initial and final velocities of ml and m2 are u± , u2 and v, v2 respectively. Then from conservation of momentum along horizontal and vertical directions, we get : m m2 v2 cos 9 A) and m1v1 = m2v2 sin 8 B) Squaring (i) and B) and then adding them, Now, from kinetic eneigy conservation, 1 .2 1 .2 . 1 .2 C) or, m (ul - v\) i(«2 + v2) [Using C)] or, or, Z n *t ft (VA kj ■ vl i / 2 So, fraction of kinetic energy lost by the particle 1, 1 2 1 IM 2 2 x x -m. 1- (b) When the collision occurs head on, [Using D)] m1v1 + and from conservation of kinetic energy, D) E) A)

87 1 1 2 L vj + 2 1 2 mv2 + -m («i - ViJ 1 /Wo 2 n L1 [Using E)] or, -1 m or, Fraction of kinetic energy, lost F) 1- m1-m2 4/WJ/W2 [Using F)] 1.169 (a) When the particles fly apart in opposite direction with equal velocities (say v), then from conservatin of momentum, m1 u + 0 * (m2 - m^) v A) and from conservation of kinetic energy, 1 2 1 7 or, From £q. A) and B), m2-m1 or, m22-3 /W2 ■ 0 B) Hence —- -- as m. k 0 m2 3 2 (b) When they fly apart symmetrically relative to the initial motion direction with the angle of divergence 0 » 60°, From conservation of momentum, along horizontal and vertical direction, mi ui * mi vicos @/2)+ m2 V2 cos @/2) (x) mx vx sin @/2) ~ m2 v2 sin @/2) and or, Now, from conservation of kinetic energy, m2 v2 B) C) From A) and B), mx ux« cos @/2) /W2 2 wx V! cos @/2)

88 ffln TH So, ux - 2vx cos @/2) From B), C), and D) 4 mx cos2 F/2) v\ - /»! vj or, 4 cos2 F/2) * 1 + — m2 D) 2v2 ivi m or, and putting the value of 6, we get, — « 2 1.170 If (v^jV^) are the instantaneous velocity components of the incident ball and ( v^, v2 ) are the velocity components of the struck ball at the same moment, then since there are no external impulsive forces (i.e. other than the mutual interaction of the balls) We have u sina * v-,.. , v~ ■ 0 m u cosa * m vu + m The impulsive force of mutual interaction satisfies d g . F (JF is along the x axis as the balls are smooth. Thus Y component of momentum is not transferred.) Since loss of K.E. is stored as deformation energy D, we have 2 1 -mu - ^ -mw2cos2a- 2 " 9— m w cos a - m v^ - (mucosa- mv^) — F 2m2wcosavljK - 1 * m ( We see that D is maximum when w2cos2a w cosa cosa and D mu2cos2a max D, rj^. max 1 2 1 Then r\ * -—— « — cos a * ~ 1^ 4 On substiuting a » 45* x:

89 1.171 From the conservation of linear momentum of the shell just before and after its fragmentation 3^«i£+i£+i£ A) where v1, v£ and v^ are the velocities of its fragments. From the eneigy conservation 3r|v2 * v2 + v2 + v2 B) Now i£or v^. ■ v^- v£ « v^- v* C) where v£ ■ v*= velocity of the CM. of the fragments the velocity of the shell. Obviously in the CM. frame the linear momentum of a system is equal to zero, so vT+v2 + v3*0 D) Using C) and D) in B), we get 3r|v2 * (v + v^J + (v + v£J + (v*- v[- v^f « 3V2 + 2v 2 + 2v 2 + 2 v v£ or, 2v2 + 2v1v2cos6 + 2v2 + 3(l-T|)v2«0 E) If we have had used v2 * - vx - v3, then Eq. 5 were contain v3 instead of v2 and so on. The problem being symmetrical we can look for the maximum of any one. Obviously it will be the same for each. For v^to be real in Eq. E) 4 vIcos26 *8Bv?, + 3 A -ti) v2) or 6(r\ - ljv2 * D - cos29)v\ So, v2i v A/ ' vi v or Hence v2(max)« 1^1^ - v + V2(ri - 1) v - v(l + ^Uy\ - 1)) - 1 km/s Thus owing to the symmetry ￼ Vl(max) " V2(max) = V3(max) s V f1 + V2(t| - 1)) l(max) " V2(max) = V 1.172 Since, the collision is head on, the particle 1 will continue moving along the same line as before the collision, but there will be a change in the magnitude of it's velocity vector. Let it starts moving with velocity vx and particle 2 with v2 after collision, then from the conservation of momentum mu * mvl + mv2 or, u « vx + v2 A) And from the condition, given, T, 1 or, vJ + v^-d-T,)^ B) From A) and B), 2 2 2 or, v\ + «2 - 2 uvj + vj » A - tj) tf

90 or, So, 0 Vi-2ii± - 8ti u Positive sign gives the velocity of the 2nd particle which lies ahead. The negative sign is correct for vl . So, vx « — u A - VI -2t] ) * 5 m/s will continue moving in the same direction. Note that vx ■ 0 if r| = 0 as it must. 1.173 Since, no external impulsive force is effective on the system "M + m", its total momentum along any direction will remain conserved. So from px ■ const. mu « Af vx cos 9 or, vx ■ — Mcos 8 and from p - const mv M x sin G or, v2- — vt sin G = u tan G, [using A)] Final kinetic energy of the system And initial kinetic energy of the system- — mu Tf - J. So, % change = -^— x 100 -mw2tan29 + -M 2 2 2 cos2 9 1 2 — mu 1 12 xlOO 2M xlOO / I X and putting the values of 9 and — , we get % of change in kinetic energy* - 40 % M 1.174 (a) Let the particles m^ and m2 move with velocities vx and v2 respectively. On the basis of solution of problem 1.147 (b)

91 As v1lv2 So, p « [i Vvj + v\ where \x (b) Again from 1.147 (b) ttl\ + fifty So, 1.175 From conservation of momentum Pi = Pi + so (pi "Pi ) ■ V / From conservation of energy ' cosOj + px El 2m, Pi ,2 2/Wi P2_ 2m, Eliminating p2' we get 0 1 + m, - 2/?1'jp1cos01 1- L This quadratic equation for/?]/ has a real solution in terms ofp1 and cos Bx only if 1- m 2> or sin2 or sin 0t ^ + 1 or sin 0, ^ L This clearly implies (since only + sign makes sense) that . Q m2 ml 1.176 From the symmetry of the problem, the velocity of the disc A will be directed either in the initial direction or opposite to it just after the impact Let the velocity of the disc A after the collision be v' and be directed towards right after the collision. It is also clear from the symmetry of problem that the discs B and C have equal speed (say v") in the directions, shown. From the condition of the problem, d ' ^ so, sin0« V4-ti2 12 cos 0 so> sin 0 V4ti 12 A) For the three discs, system, from the conservation of linear momentum in the symmetry direction (towards right) 2m v"sin0 + mv' or, v* 2v"sin0 + v' B)

92 From the definition of the coefficeint of restitution, we have for the discs A and B (or C) = v" - v' sin 0 v sin 0 - 0 But e = 1, for perfectly elastic collision, So, v sin 0 = v ' - v' sin 0 C) From B) and C), in2 v = v A - 2 sin2 0) in2 A+2 sin2 0) 6-rf Hence we have, {using A)} - 2) 6-n Therefore, the disc A will recoil if r\ < V2 and stop if r\ = vT. Note : One caw write the equations of momentum conservation along the direction per- perpendicular to the initial direction of disc A and the consevation of kinetic energy instead of the equation of restitution. 1.177 (a) Let a molecule comes with velocity v^to strike another stationary molecule and just ^ ^i after collision their velocities become v x and v 2 respectively. As the mass of the each molecule is same, conservation of linear momentum and conservation of kinetic energy for the system (both molecules) respectively gives : and 2 A' From the property of vector addition it is obvious from the obtained Eqs. that JL v —> v 2 = 0 ill (b) Due to the loss of kinetic energy in inelastic collision vx > v\ + v'2 ^» —>' so, v x • v 2 > 0 and therefore angle of divergence < 90°. 1.178 Suppose that at time t, the rocket has the mass m and the velocity v*, relative to the reference frame, employed. Now consider the inertial frame moving with the velocity that the rocket has at the given moment. In this reference frame, the momentum increament that the rocket & ejected gas system acquires during time dt is, —*» * — > —> -■* dp * md v + \kdtu= F dt —>• dv or, m dt F- [xu or, mw = F - \xu

93 1.179 According to the question,/7* Oandfx * - dm/dt so the equation for this system becomes, dv* dm -» m ~~~r~m ~~r~ u dt dt As dv*f i u* so, mdv= -udm. Integrating within the limits : m */*■-/ dm v — or — « In — m u m 0 m mQ Thus, v= win — m As dvrL u, so in vector form v • - u In — m 1.180 According to the question, F (external force) = 0 o dv dm -> So, m~7~» —7~w As so, in scalar form, mdv** -udm wdt dm 9 u m Integrating within the limits for m (i) m wt I dm v , m — = - I — or, — - -In — u J m u m0 mo Hence, m ■ mQ e " (m*/m ^ 1.181 As i7 = 0, from the equation of dynamics of a body with variable mass; dv -+dm ,-> -*dm ... m-_« w _-. or dv* u — A) dt dt m v ' Now £/v*t |w*and since ifX vTwe must have | dv*\ = voda (because v0 is constant) Where d a is the angle by which the spaceship .turns in time dt. - dm . . u dm So, - w — « vn a a or, a a - v0 m u C dm u . or, a * I — * — In v0 J m v0 m mQ m

94 1.182 We have — = - \i or, at - \idt Integrating I dm« - ji I dt or, m « m0 - 0 As w*= 0 so, from the equation of variable mass system : (m0 - \it) — m F or, — - hT» or, o 0 Hence — In 'o mo-\it 1.183 Let the car be moving in a reference frame to which the hopper is fixed and at any instant of time, let its mass be m and velocity v. Then from the general equation, for variable mass system. dv* 7T -*dm m —7"* F + u —7- dt dt We write the equation, for our system as, dv -^ -*dm -> -h m -7- = F - v -7- as, u * - v dt dt ' So and But so, dt (mv) - I? Ft v * — on integration, m m * m 0 A) Thus the sought acceleration, w dt m 0 1 + m 0 1.184 Let the length of the chain inside the smooth horizontal tube at an arbitrary instant is x. From the equation, dm mw - F + u dt

95 as IT- 0, T\ T w> for thc chain inside tbe m T where X* y Similarly for the overhanging part, IT« 0 Thus mw» F or \hw= "khg-T From A) and B), or, B) or, dv [As the length of the chain inside the tube decreases with time, ds . dx -dx.] or, -gh x + h o Integrating, r Lf & J vdv~ -8hJ xTh or, 1.185 Force moment relative to point 0 ; r at lot Let the angle between M and N, 45° at Then M-N* 72 So, 2 *2 r04 « a2 + b2 r04 or, r0 - yf±{* h cannot be negative) It is also obvious from the figure that the angle a is equal to 45* at the moment ,0, when a = i.e. f 0 and JV-

96 1.186 = mvogt sin -mv0gt2 cos a (-k ): Thus M @ mv0 g t cos a Thus angular momentum at maximum height i.e. at t- T v0 sin a M\\\ Alternate : 2 / 3 g . 2 sin a cos a * 37 kg - m /s ////// /////////VT M@)«0so, M(t) j (rxmg) o 0 ±?<2 00/7 1.187 (a) The disc experiences gravity, the force of reaction of the horizontal surface, and the force R of reaction of the wall at the moment of the impact against it. The first two forces counter-balance each other, leaving only the force R. It's moment relative to any point of the line along which the vector R acts or along normal to the wall is equal to zero and therefore the angular momentum of the disc relative to any of these points does not change in the given process. (b) During the course of collision with wall the position of disc is same and is equal to r , Obviously the increment in linear momentum of the ball Ap * 2mv cos a n Here, AAf - r^.xAp^ 2mvl cos a n and directed normally emerging from the plane of figure Thus | AM|= 2mv/cosa 1.188 (a) The ball is under the influence of forces T and m g*at all the moments of time, while moving along a horizontal circle. Obviously the vertical component of T balance m g and

97 so the net moment of these two about any point becoems zero. The horizontal component of T, which provides the centripetal acceleration to ball is already directed toward the centre (C) of the horizontal circle, thus its moment about the point C equals zero at all the moments of time. Hence the net moment of the force acting on the ball about point C equals zero and that's why the angular mommetum of the ball is conserved about the horizontal circle. (b) Let a be the angle which the thread forms with the vertical. Now from equation of particle dynamics : Tcosa* mg and Tsina* moo /sina Hence on solving cos a « -4- A) CO / As | M | is constant in magnitude so from figure. | AM |* 2Mcos a where M- \M,\- \Mf\ - - |r£xmv*|« mvl/as i^l Thus| A M | = 2mvl cos a = 2 mco / sin a cos a CO (using 1). 1.189 During the free fall time t» x = y — , the reference point O moves in hoizontal direction > (say towards right) by the distance Vx. In the translating frame as M (O) * 0, so (-Vxi + hj)xm[gxj-Vi] -mVg%2'h + mVh(+'k) 1(9) -mVg mVh(+k) * - Hence 1.190 The Coriolis force is.B/n v* x S Here co is along the z-axis (vertical). The moving disc is moving with velocity v0 which is constant. The motion is along the jc-axis say. Then the Coriolis force is along y-axis and has the magnitude 2m v0 to. At time t> the distance of the centre of moving disc from O is VqT (along jc-axis). Thus the torque N due to the coriolis force is N * 2m v0 o)*VqT along the z-axis.

98 Hence equating this to —7- dM _ 2 «*- 22 "T"" 2/wv0cof or Af = mvQ(ot + constant. The constant is irrelevant and may be put equal to zero if the disc is originally set in motion from the point O. This discussion is approximate. The Coriolis force will cause the disc to swerve from straight line motion and thus cause deviation from the above formula which will be substantial for large t. 1.191 If r = radial velocity of the particle then the total energy of the particle at any instant is where the second term is the kinetic energy of angular motion about the centre O. Then the extreme values of r are determined by r - 0 and solving the resulting quadratic equation we get ? m Ik From this we see that E - Hr\ + r2) B) where r1 is the minimum distance from O and r2 is the maximum distance. Then 2jfcr2 Hence, m « —j~ Note : Eq. A) can be derived from the standard expression for kinetic energy and angular momentum in plane poler coordinates : 2 * M = angular momentum = mr 8 1.192 The swinging sphere experiences two forces : The gravitational force and the tension of the thread. Now, it is clear from the condition, given in the problem, that the moment of these forces about the vertical axis, passing through the point of suspension Nz - 0. Con- Consequently, the angular momentum Mz of the sphere relative to the given axis (z) is constant. Thus mv0 (/ sin 0)« mvl A) where m is the mass of the sphere and v is it s velocity in the position, when the thread forms an angle — with the vertical. Mechanical energy is also conserved, as the sphere is

99 under the influence if only one other force, i.e. tension, which does not perform any work, as it is always perpendicular to the velocity. So, From A) and B), we get, j mvjj + mg I cos 9 = j mv2 B) vo= V2g//cos 9 1.193 Forces, acting on the mass m are shown in the figure. A&N** mg* the net torque of these two forces about any fixed point must be equal to zero. Tension T, acting on the mass m is a central force, which is always directed towards the centre O. Hence the moment of force T is also zero about the point O and therefore the angular momentum of the particle m is conserved about O. Let, the angular velocity of the particle be co, when the separation between hole and particle m is r, then from the conservation of momentum about the point 0, : or co = Now, from the second law of motion for m, J= F= m(o2r Hence the sought tension; m cog r*r m tog r04 r r 1.194 On the given system the weight of the body m is the only force whose moment is effective about the axis of pulley. Let us take the sense of co of the pulley at an arbitrary instant as the positive sense of axis of rotation (z-axis) (t)-fNzdt As Mz@)-0, so, Mz So, Mz (t) m J mgRdt** mgRt o 1.195 Let the point of contact of sphere at initial moment (f= 0) be at 0. At an arbitrary moment, the forces acting on the sphere are shown in the figure. We have normal reaction Mr - mg sin a and both pass through same line and the force of static friction passes through the point O, thus the moment about point O becomes zero. Hence mg sin a is the only force which has effectivejorque about point O, and is given by |^|« mgRsina normally emerging from the plane of figure. AsM(f= 0)- 0, so, AM= M(t) = JNdt Hence, M (t) ** Nt= mgR sin at

100 Let position vectors of the particles of the system be r^and r* with respect to the points O and O' respectively. Then we have, where f£ is the radius vector of O' with respect to O. Now, the angular momentum of the system relative to the point O can be written as follows; or, M - M + /r^x/A, where, p*= T j?* B) From B), if the total linear momentum of the system, />*= 0, then its angular momen- momentum does not depend on the choice of the point O. Note that in the CM. frame, the system of particles, as a whole is at rest 1.197 On the basis of solution of problem 1.196, we have concluded that; "in the CM. frame, the angular momentum of system of particles is independent of the choice of the point, relative to which it is determined" and in accordance with the problem, this is denoted We denote the angular momentum of the system of particles, relative to the point O, by M. Since the internal and proper angular momentum M, in the CM. frame, does not depend on the choice of the point O\ this point may be taken coincident with the point O of the £-frame, at a given moment of time. Then at that moment, the radius vectors of all the particles, in both reference frames, are equal (rt = r~) and the velocities are related by the equation, -* ** -* where v^ is the velocity of CM. frame, relative to the £-frame. Consequently, we may write, or, M- M+m/r^xv^V as Y mJ\- mK> wncre m== A mi- or, M- , 1.198 From conservation of linear momentum along the direction of incident ball for the system consists with colliding ball and phhere mv0 = mv' + j v1 A) where v' and v1 are the velocities of ball and sphere 1 respectively after collision. (Remember that the collision is head on). As the collision is perfectly elastic, from the definition of co-efficeint of restitution, i or, v'- vx- -v0 B) 0-v0

101 Solving A) and B), we get, 4v, -, directed towards right. In the CM. frame of spheres 1 and 2 (Fig.) Also, r lc As rlc JL so, , thus # - / m/2 4v0 A" 2 2 (where n is the unit vector in the sense of r\c xjp^) mvol Hence M = —-— L199 In the CM. frame of the system (both the discs + spring), the linear momentum of the discs are related by the relation, p1» -p^ at all the moments of time, where, p« « p0 « p = u v^t And the total kinetic energy of the system, J* i u vi, [See solution of 1.147 (b)] Bearing in mind that at the moment of maximum deformation of the spring, the projection of v^j along the length of the spring becomes zero, i.e. v^^ * 0. The conservation of mechanical energy of the considered system in the CM. frame gives. Now from the conservation of angular momentum of the system about the CM., 2 2 m or, Vo -1 - V, i 0 , as x « I o B) Using B) in A), o or, 1- ^ 2r x2) 1 "■ "I + TT or, , [neglecting As x * 0, thus jc ^ kL

102 1.4 UNIVERSAL GRAVITATION 1.200 We have Mv2 yMmA or r Thus v CD* - ym v2 (Here m, is the mass of the Sun.) So T 2jtx6-67xl01xl-97xl0 30 1-94 x 10' sec - 225 days. v3 C4-9 x 103K (The answer is incorrectly written in terms of the planetary mass M) 1.201 For any planet ,, yMm. or @ So, (a) Thus So — 00 rv 3/2 (b) 5-24. 2/3 2* So (Y ms r9 B .273 or, Bnyms 2/3 where T* 12 years, m^- mass of ths Sun. Putting the values we get Vj * 12-97 km/s vj Bnyms) Acceleration ■ —-■ \fh 2-15 xlO km/s2

103 1.202 Semi-major axis* (r + R)/2 r + R It is sufficient to consider the motion be along a circle of semi-major axis —~— for T does not depend on eccentricity. 3/2 r + R 2n\ 2 \ Hence T* \ —- nV(r + RK/2vms Vym, s (again ms is the mass of the Sun) 1.203 We can think of the body as moving in a very elongated orbit of maximum distance R and minimum distance 0 so semi major axis « R/2. Hence if x is the time of fall then or x * Tl 4V2 * 365 / 4V2 = 64-5 days. 1.204 y^ If the distances are scaled down, R decreases by a factor r\ and so does ms . Hence T does not change. 1.205 The double star can be replaced by a single star of mass moving about the centre of mass subjected to the force y m1 m21 r . Then _. 2nr3/2 2nr3/2 1 = ym1m2/ — 1 z ml-¥m. So r3/2 or, 2/3 73 = VyA/(J/2 jiJ 1.206 (a) The gravitational potential due to m1 at the point of location of m2 : 00 00 So, G,, « m9 V, 21 " ' r 2 Similarly JJ ~ - -

104 Hence Un-U21- u ym1m2 ^^•* ^^mmm^mm *^m *^ ^^m (b) Choose the location of the point mass as the origin. Then the potential erfeigy dU of M an element of mass dM = -fdx of the rod in the field of the point mass is dU=-ym -ydx — I x where x is the distance between the element and the point (Note that the rod and the point mass are on a straight line.) If then a is the distance of the nearer end of the rod from the point mass. t J X a + l mM\ dx M, (. I) — m- ym— In 1 + - x I \ a The force of interaction is F = - da mM 1 1 ymM Minus sign means attraction. 1.207 As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the focii of the ellipse) So, mv1rl- mv2r2 or, A) From the conservation of mechanical eneigy of the system (Sun + planet), ymsm i 2 ymsm 1 2 or, Thus, ^ [Using A)] v2- V2 y ms rx / r2 (rx B) Hence M * mv2r2= m V 2 2r2

105 1.208 From the previous problem, if rx , r2 are the maximum and minimum distances from the sun to the planet and vx , v2 are the corresponding velocities, then, say, 1 2 ymms 2 2 r2 ymms ymm. T1~ r2 Y"™ rl+ r2 of L where 2d = major axis * rt + r2. The same result can also be obtained directly by writing an equation analogous to Eq A) of problem 1.191. 1\J2 vnttti (Here M is angular momentum of the planet and m is its mass). For extreme position r m 0 and we get the quadratic The sum of the two roots of this equation are ymms Thus ymms £ m - -—— « constant la 1209 From the conservtion of angular momentum about the Sun. /wvorosina» mv1r1^ /wv2r2 or, v1r1 From conservation of mechanical energy, ymsm _ _1 2 1 v2 r2 * v0 r2 sin a or, or, yms (Using 1) vo- '2 + 2YmJr1-v2r2sina- 0 So, -2ym.± ^ sin2 W VQ^sin2a ro[l±Vl-B-r])r]sin2a] b where "H = v§ fo / Y m,s» (wjs the mass of the Sun). A)

106 1.210 At the minimum separation with the Sun, the cosmic body's velocity is perpendicular to its position vector relative to the Sun. If rain be the sought minimum distance, from con- conservation of angular momentum about the Sun (C). mvo/« mvr^ or, v- min (i) From conservation of mechanical energy of the system (sun + cosmic body), 1 2 ymsm 1 2 + 2 So, or, So, 2 min ..2 'mm (using 1) 2^2 2 ,2 min - 2y m2 + 4v02 v02 I2 Hence, taking positive root 1.211 Suppose that the sphere has a radius equal to a. We may imagine that the sphere is made up of concentric thin spherical shells (layers) with radii ranging from 0 to a\ and each spherical layer is made up of elementry bands (rings). Let us first calculate potential due to an elementry band of a spherjcal layer at the point of location of the point mass m (say point P) (Fig.). As all the points of the band are located at the distance / from the point P, so, .„_ UK (where mass of the band) < JM Bx a sin 8) (a dB) B) And tdM\ . Q .. —r- sin0 d& I2- a2 + r2-2areosQ C) I Differentiating Eq. C), we get arsinQdQ D) Hence using above equations E)

107 Now integrating this Eq. over the whole spherical layer J r-a So ydM r F) Equation F) demonstrates that the potential produced by a thin uniform spherical layer outside the layer is such as if the whole mass of the layer were concentrated at it's centre; Hence the potential due to the sphere at point P\ yM ,- *-J— G) r This expression is similar to that of Eq. F) Hence thte sought potential energy of gravitational interaction of the particle m and the sphere, yMm (b) Using the Eq., G'--*r (using Eq. 7) So G* --^r and F-/wG*--L-irr (8) r r 1.212 (The problem has already a dear hint in the answer sheet of the problem book). Here we adopt a different method. Let m be the mass of the spherical layer, wich is imagined to be made up of rings. At a point inside the spherical layer at distance r from the centre, the gravitational potential due to a ring element of radius a equals, - ^p- dl (see Eq. E) of solution of 1.211) s», ,.;«,._£(•„._= a-r Hence Gr - - Hence gravitational field strength as well as field force becomes zero, inside a thin sphereical layer. 1.213 One can imagine that the uniform hemisphere is made up of thin hemispherical layers of radii ranging from 0 to R. Let us consider such a layer (Fig.). Potential at point O, due to this layer is,

108 , ydm 3yM , , d cp = - J = - ' a rdry where T D ^ 4ji;r dr (This is because all points of each hemispherical shell are equidistant from O.) Hence, cp« fd<p=* ~ R o 2R M Hence, the work done by the gravitational field force on the particle of mass m, to remove it to infinity is given by the formula **, A = mqp, since <p - 0 at infinity. Hence the sought work, 3ymM (The work done by the external agent is -A) 1.214 In the solution of problem 1.211, we have obtained cp and G due to a uniform shpere, at a distance r from it's centre outside it We have from Eqs. G) and (8) of 1.211, <P - s— and G « - ±-r- r r r3 (A) Accordance with the Eq. A) of the solution of 1.212, potential due to a spherical shell of radius a, at any point, inside it becomes yM qp a Const, and Gr 0 (B) For a point (say P) which lies inside the uniform solid sphere, the potential (p at that point may be represented as a sum. inside where qpt is the potential of a solid sphere having radius r and qp2 is the potential of the layer of radii r and R. In accordance with equation (A) Y / M £ 3\ yM 2 tPl="r(D/3)ni?33ICr )"~ R3' The potential qp2 produced by the layer (thick shell) is the same at all points inside it. The potential cp2 *s easiest to calculate, for the point positioned at the layer's centre. Using Eq. (B) R <P2= " where dM - M D/3) nR 4jir dr CM R rdr is the mass of a thin layer between the radii r and r + dr. Thus . .. inside (C)

109 From the Eq. dr n Cr r or yMr 4 M (where p = , is the density of the sphere) 4 n^ (D) The plots qp (r) and G (r) for a uniform sphere of radius R are shown in figure of answersheet. Alternate : Like Gauss's theorem of electrostatics, one can derive Gauss's theorem for gravitation in the form § GdS- - . For calculation of G at a point inside the sphere at a distance r from its centre, let us consider a Gaussian surface of radius r, Then, R R Hence, G» - as p D/3) nR 00 So, cp = r r R Integrating and summing up, we get, 3- And from Gauss's theorem for outside it : -4xyM or 00 Thus 1.215 Treating the cavity as negative mass of density - p in a uniform sphere density + p and using the superposition principle, the sought field strength is : or jT + -jY*(-p) FT (where r^ and r_ are the position vectors of an orbitrary point P inside the cavity with respect to centre of sphere and cavity respectively.) 4 -^ -♦ 4 Thus

110 1.216 We partition the solid sphere into thin spherical layers and consider a layer of thickness dr lying at a distance r from the centre of the ball. Each spherical layer presses on the layers within it The considered layer is attracted to the part of the sphere lying within it (the outer part does not act on the layer). Hence for the considered layer r2* dF nr2drp) or, dPAnr (where p is the mean density of sphere) or, Thus (The pressure.must vanish at r = R.) or, p - |/l - (r*/K2)) yM2/nR4, Putting p - M/D/3) **3 Putting r « 0, we have the pressure at sphere's centre, and treating it as the Earth where 3 3 2 mean density is equal to p - 5*5 x 10 kg/m and R ■ 64 x 10 km we have, p - 1-73 x 1011 Pa or 1-72 x 106 atms. 1.217 (a) Since the potential at each point of a spherical surface (shell) is constant and is equal to qp« - ^r-, [as we have in Eq. A) of solution of problem 1.212] We obtain in accordance with the equation 17 = -J^mcp* -<pf dm HI) R m ym2 " 22? 1 . (The factor — is needed otherwise contribution of different mass elements is counted twice.) (b) In this case the potential inside the sphere depends only on r (see Eq. (C) of the solution of problem 1.214) -2 \ <P 3 ym 2R 1- 3R Here dm is the mass of an elementry spherical layer confined between the radii r and r + dr: dm** Djtr2drp)» r2dr

Ill U~ 2J Cm R3 r2dr « 3 ym 2R .2 \ 1- 3R After integrating, we get 5 R 1,218 Let co V r • circular frequency of the satellite in the outer oibit, co, -v yM, (r-Ar) - m circular frequency of the satellite in the inner orbit So, relative angular velocity « o>0 ± eo where - sign is to be taken when the satellites are moving in the same sense and + sign if they are moving in opposite sense. Hence, time between closest approaches 2n 2n coo± co 3/2 2r 4-5 days F * 0) 0-80 hour F - 2) where 6 is 0 in the first case and 2 in the second case. 219 yM 6-67x10 nx 5-96 xlO24 IT F-37 x 106J R 2x22 24 x 3600 x 7 6-37 x 106 - 0-034 m/s2 VMs 6-67 xlO^ 1-97 xlO30 .. lft.3 /2 and @3- —2—- 7 t*z— - 5-9^x10 m/s A. Then A49-50 x 106 x 103J : co2 : co3 - 1: 0-0034 : 0-0006 1.220 Let h be the sought height in the first case, so 99 g yM ioo ass yM

112 99 / h\~2 From the statement of the problem, it is obvious that in this case h«R „ 99 (. 2h\ . R /6400 „. 111115 Too" (x-*j or h' So" (W1™- 32km (j ( In the other case if h' be the sought height, than j) ot2m[l+i) From the language of the problem, in this case ti is not very small in comparision with R. Therefore in this case we cannot use the approximation adopted in the previous case. Here, A + j) -2* So, ~«±V2-1 As -ve sign is not acceptable h'= (V2 -1)*- (V2 -1N400km * 2650km 1.221 Let the mass of the body be m and let it go upto a height h. From conservation of mechanical energy of the system 1 2 Using *—j- * g, in above equation and on solving we get, 1.222 Gravitational pull provides the required centripetal acceleration to the satelile. Thus if h be the sought distance, we have mv2 ymM ,D .. 2 w so, ——tt« -1 =■ or, (/? + /*) v = vM or, Rv2 + hv2= gR29 as g 2R2-n 2 Hence «• - ^ v2 1.223 A satellite that hovers above the earth's equator and corotates with it moving from the west to east with the diurnal angular velocity of the earth appears stationary to an observer on the earth. It is called geostationary. For this calculation we may neglect the annual motion of the earth as well as all other influences. Then, by Newton's law, 2 yMm I 2jt

113 where M = mass of the earth, T = 86400 seconds = period of daily rotation of the earth and r = distance of the satellite from the centre of the earth. Then Substitution of M = 5-96 x 1024 kg gives r - 4-220 x 104 km The instantaneous velocity with respect to an inertial frame fixed to the centre of the earth at that moment will be Y )r- 3-07 km/s and the acceleration will be the centripetal acceleration. ^ r» 0-223 m/s2 1.224 We know from the previous problem that a satellite moving west to east at a distance R « 2-00 x 104 km from the centre of the earth will be fevolving round the earth with an angular velocity faster than the earth's diurnal angualr velocity. Let (o = angular velocity of the satellite 2n o>0 « -=• = anuglar velocity of the earth. Then 2k GO - O>n - U X as the relative angular velocity with respect to earth. Now by Newton's law _. 2*-2 So, Substitution gives M* 6-27 xlO24 kg 1225 Tne velocity of the satellite in the inertial space fixed frame is y ±— east to west. With respect to the Earth fixed frame, from the v^ * v^- (h^ x rf the velocity is * 7-03 km/s Here M is the mass of the earth and T is its period of rotation about its own axis.

114 It would be - 2?i R T * R To find the acceleration we note the formula , if the satellite were moving from west to east. Here i7 * - nt * -v —» -v —» /? and v 1 co and v x a) is directed towards the centre of the Earth. Thus 2nR 2n_ T 2k T toward the earth's rotation axis lit 4-94 m/s on substitution. 1226 From the well known relationship between the velocities of a particle w.r.t a space fixed frame (iQ rotating frame (IC) v-v +(wxr) v 1 * v "" Thus kinetic energy of the satellite in the earth's frame 2jti?2 Obviously when the s^Uite moves in opposite sense comared to the rotation of the Earth its velocity relative to the same frame would be 2iC And kinetic energy 2 - 2m*'*m 2m 2jiR B) From A) and B) T' +—j V- 231R C) Now from Newton's second law yMm m v2 Using D) and C) R or v VgR D) T2 231R •27 nearly (Using Appendices)

115 1.227 For a satellite in a circular orbit about any massive body, the following relation holds between kinetic, potential & total energy : J«-£, U = 1E A) Thus since total mechanical energy must decrease due to resistance of the cosmic dust, the kintetic energy will increase and the satellite will 'fall', We see then, by work energy theorm o j. 2 j otrfr dv So, mvav - av vat or, -=- m v2 Now from Netow's law at an arbitray radius r from the moon's centre. v2 vM — si-^- or v r - /vM V— v r (M is the mass of the moon.) Then V-- where £= moon's radius. So 7 x ax m 0 f()^^^ where g is moon's gravity. The averaging implied by Eq. A) (for nondrcular orbits) makes the result approximate. 1.228 From Newton's second law 2 or vo* V jT " 1*67km/s mv0 From conservation of mechanical energy 0 or, ve« Y ^5. » 2-37 km/s2 B)

116 In Eq. A) and B), M and R are the mass of the moon and its radius. In Eq. A) if M and R represent the mass of the earth and its radius, then, using appendices, we can easily get v0* 7-9 km/s and vc= 11-2 km/s. 1.229 In a parabolic orbit, E - 0 o 1 2 yMm n r^^ /yM ￼ So 2mV' J?~* °r V" ^V where M - mass of the Moon, R « its radius. (This is just the escape velocity.) On the other hand in orbit 2d yMm mvf R = ' o or vf f R2 f Thus Av= A-V2) V3^ " -°*70 km/s- 1.230 From 1.228 for the Earth surface and ve Thus the sought additional velocity XV This 'kick' in velocity must be given along the direction of motion of the satellite in its orbit 1,231 Let r be the sought distance, then yM 2 or Tlr or (nR-ry r nR yfnr- (nR-r) orr« ™ « 3-8 x 104 km. V T| + 1 1.232 Between the earth and the moon, the potential energy of the spaceship will have a maximum at the point where the attractions of the earth and the moon balance each other. This maximum RE. is approximately zero. We can also neglect the contribution of either body to the p.E. of the spaceship sufficiently near the other body. Then the minimum energy that must be imparted to the spaceship to cross the maximum of the RE. is clearly (using E to denote the earth)

117 With this eneigy the spaceship will cross over the hump in the P.E. and coast down the hill of p.E. towards the moon and crashland on it. What the problem seeks is the minimum eneigy reguired for softlanding. That reguies the use of rockets to loving about the braking of the spaceship and since the kinetic energy of the gases ejected from the rocket will always be positive, the total eneigy required for softlanding is greater than that required for crashlanding. To calculate this eneigy we assume that the rockets are used fairly close to the moon when the spaceship has nealy attained its terminal velocity on the moon 2yMo — where Mo is the mass of the moon and Ro is its radius. In general dE = vdp and since the speed of the ejected gases is not less than the speed of the rocket, and momentum transfered to the ejected gases must equal the momentum of the spaceship the energy E of the gass ejected is not less than the kinetic energy of spaceship Addding the two we get the minimum work done on the ejected gases to bring about the softlanding. « ym On substitution we get 1-3 x 108 kJ. 1.233 Assume first that the attraction of the earth can be neglected. Then the minimum velocity, that must be imparted to the body to escape from the Sun's pull, is, as in 1-230, equal to where v\ « yMs/rtr** radius of the earth's orbit, Ms « mass of the Sun. In the actual case near the earth, the pull of the Sun is small and does not change much over distances, which are several times the radius of the Earth. The velocity v3 in question is that which overcomes the earth's pull with sufficient velocity to escape the Sun's pull. Thus where R « radius of the earth, ME * mass of the earth. Writing vx2 « yME /R, we get 16-6 km/s

118 1.5 DYNAMICS OF A SOLID BODY 1.234 Since, motion of the rod is purely translational, net torque about the CM. of the rod should be equal to zero. For the translational motion of rod. F mwe B) From A) and B) 2oF2 qj i 1/2 F2 * mwc 1.235 Sought momentN* FxF- (ai + bj)x(Ai and arm of the force /« -=• 1.236 Relative to point 0, the net moment of force : - abT+AB{-T)~ (ab-AB)T A) Resultant of the external force F-F^ + ^« AJ+bV B) As N'F = 0 (as N±F) so the sought arm / of the force F ab-AB /- N/F- 1.237 por coplanar forces, about any point in the same plane, \ rL x Fi (where Ftut * Y Fi * resultant force) or, N^ * r x Thus length of the arm, / * Here obviously [F^ | = IF and it is directed toward right along AC. Take the origin at C. Then about C, N~ I yfl aF + -j^F- VJaF 1 directed normally into the plane of figure. (Here a = side of the square.) Thus N= F-j= directed into the plane of the figure. Hence /- V '■ Jjf - j Thus the point of application of force is at the mid point of the side BC.

119 1.238 (a) Consider a strip of length dx at a perpendicular distance x from the axis about which we have to find the moment of inertia of the rod. The elemental mass of the rod equals dm* -rdx Moment of inertia of this element about the axis dmx2** -y2 Thus, moment of inertia of the rod, as a whole about the given axis J J m (b) Let us imagine the plane of plate as xy plane taking the origin at the intersection point of the sides of the plate (Fig.). Obviously Ix- J dmy2 o X dx ma Similarly mb Hence from perpendicular axis theorem which is the sought moment of inertia. 1.239 (a) Consider an elementry disc of thickness dx. Moment of inertia of this element about the 2 -axis, passing through its CM. dl. (dm)R pSdx R 2 ■ 2 where p = density of the material of the plate and S = area of cross section of the plate. Thus the sought moment of inertia b d2 dx* ^?Sb nR2\

120 putting all the vallues we get, Iz» 2- gnvm (b) Consider an element disc of radius r and thickness dx at a distance x from the point 0. Then r-x tana and volume of the disc « nx2lan2adx Hence, its mass dm « n x tana dx-p (where A 1 2 i p * density of the cone - m/— nR h) p Moment of inertia of this element, about the axis OAy r2 dl= dm — ( n x tan a dx ) 2 2 x tan a h Thus the sought moment of inertia / tan a I x dx. dx 0 n pR A-h 4 as tana = — h Hence 3m R 10 / 2 / putting p h ! .240 (a) Let us consider a lamina of an arbitrary shape and indicate by 1,2 and 3, three axes coinciding with jc, y and z - axes and the plane of lamina as x -y plane. Now, moment of inertia of a point mass about x - axis, dlx » dm y Thus moment of inertia of the lamina about this axis, lx- f dmy Similarly,' Iy - Cdmx and Iz = J dmr 2 = fdm(x2+y2) as r- Vx7Ty7 Thus, Iz - Ix + Iy or, I3 « (b) Let us take the plane of the disc as x -y plane and origin to the centre of the disc (Fig.) From the symmetry Ix « Iy. Let us consider a ring element of radius r and thickness dry then the moment of inertia of the ring element about the y - axis. xCO

121 dlz= dmr2 ^ Thus the moment of inertia of the disc about z- axis 2m f 3 . ~~T I r dr R J mR o But we have Thus 2 mR dx 1.241 For simplicity let us use a mathematical trick. We consider the portion of the given disc as the superposition of two- complete discs (without holes), one of positive density and radius R and other of negative density but of same magnitude and radius R/2. As (area) a (mass), the respective masses of the considered discs are Dm/3) and (-ml3) respectively, and these masses can be imagined to be situated at their respective centers (CM). Let us take point O as origin and point x - axis towards right Obviously the CM. of the shaded position of given shape lies on the x - axis. Hence the CM. (C) of the shaded portion is given by (-m/3)(-£/2)+Dm/3H R * * ~6 Thus CM. of the shape is at a distance R/6 from point O toward x - axis Using parallel axis theorem and bearing in mind that the moment of inertia of a complete homogeneous disc of radius m0 and radius r0 equals — m0 r^. The moment of inetia of the small disc of mass (- m / 3) and radius R/2 about the axis passing through point C and perpendicular to the plane of the disc 2 *x Similarly \mR mR' 27 Thus the sought moment of inertia, 15 he 2-'

122 1.242 Moment of inertia of the shaded portion, about the axis passing through it's certre, /4 3 \ 3 7*r P r 24 Now, \fR « r + dr> the shaded portion becomes a shell, which is the required shape to calculate the moment of inertia. Now, _ 5 3 7™ n p / r +5r ar Neglecting higher terms. 2 2 r - — 1.243 (a) Net force which is effective on the system (cylinder M + body m ) is the weight of the body m in a uniform gravitational field, which is a constant. Thus the initial acceleration of the body m is also constant. From the conservation of mechanical energy of the said system in the uniform field of gravity at time t« A t : AT+ AG « 0 or or. — ( 2m + M) v - mg A/i« 0 [ as v « co/? at all times ] But v Hence using it in Eq. A), we get — ( 2m + M ) 2w ish - mg tJt 0 or w From the kinematical relationship, 2mg R Bm+M)R Thus the sought angular velocity of the cylinder gt (b) Sought kinetic energy. 2 - — i A)

123 1244 For equilibrium of the disc and axle 2J« mg or J« mg/2 As the disc unwinds, it has an angular acceleration C given by 27> mgr / * / /p« 27> or The corresponding linear acceleration is mgr Since the disc remains stationary under the combined action of this acceleration and the acceleration (-w) of the bar which is transmitted to the axle, we must have mgr2 wrm o 1245 Let the rod be deviated through an angle qp'from its initial position at an arbitrary instant of time, measured relative to the initial position in the positive direction. From the equation of the increment of the mechanical energy of the system. AT- A or, or, 2 3 co = I jP/ coscp dcp = jF7 sincp Thus, co V6F sincp Ml 1.246 First of all, let us sketch free body diagram of each body. Since the cylinder is rotating and massive, the tension will be different in both the sections' of threads. From Newton's law in projection form for the bodies mx and m2 and noting that wx - w2 ■ w = $R> (as no thread slipping), we have (m1 > m2) ■ mxw= mx and T2-m2g= m2w A) Now from the equation of rotational dynamics of a solid about stationary axis of rotation, i.e. Nz- I Pz, for the cylinder. or, (T^T^R- /p - mtf2p/2 B) Simultaneous solution of the above equations yields (mx - m2 ) g Ti mi (m + 4m, and 2 m R I m1 + m2 + y T2 m2(m + 4m1

124 1.247 As the systenr(m + ml + m2) is under constant forces, the acceleration of body m1 an m2 is constant In addition to it the velocities and accelerations of bodies ml and m^ ai equal in magnitude (say v and w) because the length of the thread is constant From the equation of increament of mechanical energy i.e. AJ + AC/* Afrf at time t whe block m1 is distance h below from initial position corresponding to t« 0, 2 1 2 2\ V2 R -m2gh~ - (as angular velocity co = v/R for no slipping of thread.) But So using it in A), we get 2wh w 2(m2-kml)g m + 2 (m m B. Thus the work done by the friction force on ml 11 A/r* -hngh* -kmgl -wt km^m^km^g t . - " m + 2(m + m ) (USmg 2*' 1.248 In the problem, the rigid body is in translation equlibnum but there is an angular retardation. We first sketch the free body diagram of the cylinder. Obviously the friction forces, acting on the cylinder, are kinetic. From the condition of translational equlibrium for the cylinder, + JW2; N2~ Hence, AT, mg . mg For pure rotation *of the cylinder about its rotation axis, Nz « /Pz or, -kNiR-kN^- mR 2' kmgR A + k) mR or, - or, Now, from the kinematical equation, 0) « @0 + 2p2 A qp we have, y////// Acp because a) - 0

125 Hence, the sought number of turns, n 2k" %nk(l+k)g 1.249 It is the moment of friction force which brings the disc to rest The force of friction is applied to each section of the disc, and since these sections lie at different distances from the axis, the moments of the forces of friction differ from section to section. To find Nz> where z is the axis of rotation of the disc let us partition the disc into thin rings (Fig.). The force of friction acting on the considered element dfr = kBnrdro)g> (where a is the density of the disc) The moment of this force of friction is m -rdfr* -2nkogr2dr Integrating with respect to r from zero to Ry we get R Nz= -2nkog \r2dr= --nkogR*. o For the rotation of the disc about the stationary axis z, from the equation N = -jKkogR3 or . 2 rz " rz 3R Thus from the angular kinematical equation 1.250 According to the question, Integrating, I^" -b/co or, Z*-^:* -kdt dt V(o kt or, @ = k2t2 4/2 - —9— + (o0, (Noting that at t = 0, (o = Let the flywheel stops at t * tQ then from Eq. A), t0 = Hence sought average angular velocity < @ > 0 'k2t2 4/: + CO, dt CO, 0

126 dM. 1.251 Let us use the equation , z « ^relative to the axis through O A) For this purpose, let us find the angular momentum of the system Mz about the given rotation axis and the corresponding torque N^ The angular momentum is Mz * Tec + mvR /W/i 2 m 0^2 [where /= —R and v* coi? (no cord slipping)] 4* So, dMz IT fMR mR B) The downward pull of gravity on the overhanging part is the only external force, which exerts a torque about the z -axis, passing through O and is given by, Hence from the equation dt (MR z* JX8R Thus, 2mgx >0 Note : We may solve this problem using conservation of mechanical energy of the system (cylinder + thread) in the uniform field of gravity. 1.252 (a) Let us indicate the forces acting on the sphere and their points of application. Choose positive direction of x and cp (rotation angle) along the incline in downward direction and in the sense of oT(for undirectional rotation) respectively. Now from equations of dynamics of rigid body i.e. Fx - mwa and Ncz « Ic Pz we get : mg sin a -/r« mw and But /r* kmg cosa In addition, the absence of slipping provides the kinematical realtionship between the accelerations : w= $R D) The simultaneous solution of all the four equations yields : 2 2 k cos a ^ — sin a, or k ^ — tan a (b) Solving Eqs. A) and B) [of part (a)], we get : N A) B) C)

127 As the sphere starts at t axis, for pure rolling wc« -gsina. 0 along positive x vc@ E) Hence the sought kinetic energy as a) * v/R) 1.253 (a) Let us indicate the forces and their points of application for the cylinder. Choosing the positive direction for x and cp as shown in the figure, we write the equation of motion of the cylinder axis and the equation of moments in the CM. frame relative to that axis i.e. from equation Fx - mwc and Nz « Ic pr n 2 mg-2J- mwc; 2TR - As there is no slipping of thread on the cylinder we- ptf From these three equations T« ^L« 13N, p« ~£« 5xl02rad/s2 (b) we have So, wc * —g > 0 or, in vector form '2 3 1.254 Let us depict the forces and their points of application corresponding to the cylinder attached with the elevator. Newton's second law for solid in vector form in the frame of elevator, gives : 2T + /ng*+ m (- m^) * mw A) The equation of moment in the CM. frame relative to the cylinder axis i.e. from 2TR mR P mR2 W 2 r 2 R [as thread does not slip on the cylinder, w'

128 or, As A) so in vector form mw mw B) Solving Eqs. A) and B), w / ^ w — (g - w>0) and sought force 1.255 Let us depict the forces and their points of application for the spool. Choosing the positive direction for x and cp as shown in the fig., we apply Fx = mw^ and Nc*« Ic §z and get mg sin a - T= mw; Tr « /p "Notice that if a point of a solid in plane motion is connected with a thread, the projection of velocity vector of the solid's point of contact along the length of the thread equals the velocity of the other end of the thread (if it is not slacked)" Thus in our problem, v ■ v0 but v0 = 0, hence point P is the instantaneous centre of rotation of zero velocity for the spool. Therefore vc = cor and subsequently wc « Pr. Solving the equations simultaneously, we get g sin a 2 w = — — = 1*6 m/s 1 + mr 1.256 Let us sketch the force diagram for solid cylinder and apply Newton's second law in projection form along x and y axes (Fig.) : i + fc = rnwc A) and 0 or N1+N2= mg + F B) Now choosing positive direction of qp as shown in the figure and using N( we get mR2 cz » Ic [as for pure rolling wc ]. In addition to, D) S2E7

129 Solving the Eqs., we get 3Jcmg_ B-3*)' 3*mg 2-3* and w c(max) m mg 2-3* 2-3* 1.257 (a) Let us choose the positive direction of the rotation angle cp, such that wa and pz have identical signs (Fig.). Equation of motion, Fx - mw^ and Ncz = Ic P2 gives : Fcos a-fr= mw^ifrR-Fr* /CP2 = ymR2 p2 In the absence of the slipping of the spool w = p R From the three equations wcx (b) As static friction (fr) does not work on the spool, from the equation of the increment of mechanical energy A^ = AT. .2 . •■■ .. .„ r> 2 R •cx- Tm(l+yJ>v It-wc lC0SCC " J F [cos a - (r/R) 1 , r vv = —' ——\ f J , where cos a > — c (l) ' /? A) 2 m A + y) Note\that at cos a - r/R, there is no rolling and for cos a < r/R , wcx < 0, *'.e f/*e spool will move towards negative x-axis and rotate in anticlockwise sense. 1.258 For the cylinder from the equation N ~ /p about its stationary axis of rotation. „ r or p = — A) 2 v mr K ' For the rotation of the lower cylinder from the equation Ncz - /c P2 mr2 , , 4T 2 mr Now for the translational motion of lower cylinder from the Eq. Fx - mwa : As there is no slipping of threads on the cylinders : C)

130 Simultaneous solution of A), B) and C) yields 1.259 L*t us depict the forces acting on the pulley and weight A, and indicate positive direction for x and qp as shown in the figure. For the cylinder from the equation Fx= m $ and Ncz= TA-2T-Mwc and For the weight A from the equation A) B) mWx C) As there is no slipping of the threads on the pulleys. wa wc+ 2 P* " wc+ 2 wc " 3 Simultaneous solutions of above four equations gives : \M+9m 1.260 (a) For the translational motion of the system (mx+ m^)y from the equation : Fx -T ■" V./Wi + "*2\ "'a "'> ^c -* ' \"*1 **" "*2' \ / Now for the rotational motion of cylinder from the equation : m.r1 2F Fr= -4—P or pr« — 2 K K mi B) But c + P r» Ifly tfl^ (/Hj + C) (b) From the equation of increment of mechanical energy : AT = Here AT- 7@, so, T(t)~ A^ As force F is constant and is directed along Jt-axis the sought work done. A^ = Fx (where x is the displacement of the point of application of the force F during time interval t)

131 (I F\ — I 2 2m1(m1 (using Eq. C) Alternate : T(t) = TtnuulMM (t) + T (t) T(t) 2 m2) Ft m2) 2 mj m2) 1,261 Choosing the positive direction for x and qp as shown in Fig, let us we write the equation of motion for the sphere Fx « mwcx and Ncz * Ic jt ■*/Wo Wo i //•/'■t/Wo r p (vv2 is the acceleration of the CM. of sphere.) For the plank from the Eq. Fx ■ mwx In addition, the condition for the absence of slipping of the sphere yields the kinematical relation between the accelerations : Simultaneous solution of the four equations yields : F . 2 l and w- 1.262 (a) Let us depict the forces acting on the cylinder and their point of applications for the cylinder and indicate positive direction of x and qp as shown in the figure. From the equations for the plane motion of a solid Fx « mwcx and kmg** mw^ or wCJC= kg 2 -kmgR mR or pz« - R A) B) Let the cylinder starts pure rolling at t * tQ after releasing on the horizontal floor at r - 0. From the angular kinematical equation co « co ■+ or CO C) From the equation of the linear kinematics, /77//77j ex ocx or D)

132 But at the moment t = tQy when pure rolling starts vc = so, lo Thus 3kg (b) As the cylinder pick, up speed till it starts rolling, the point of contact has a purely 1 2 translatory movement equal to — wc fjj in the forward directions but there is also a backward Jit 1 2 movement of the point of contact of magnitude (co0 x0 - — p t0) R. Because of slipping 4br the net displacement is backwards. The total work done is then, Ab = kmg = kmg kmg 3kg (OqR 6 The same result can also be obtained by the work-energy theorem, A J. 1.263 Let us write the equation of motion for the centre of the sphere at the moment of breaking-off: mv /(R + r)« mg cos 0, where v is the velocity of the centre of the sphere at that moment, and 6 is the corresponding angle (Fig.). The velocity v can be found from the energy conservation law : m gh= - mv2 + - /co2, where / is the moment of inertia of the sphere relative to the axis passing through the sphere's 2 2 centre, i.e. / = ~ mr . In addition, v = cor; h = (/* + r) A - cos 0). From these four equations we obtain co» Vl0g(/UrI7r2. 1.264 Since the cylinder moves without sliding, the centre of the cylinder rotates about the point Oy while passing through the common edge of the planes. In other words, the point O becomes the foot of the instantaneous axis of rotation of the cylinder. It at any instant during this motion the velocity of the CM. is v1 when the angle (shown in the figure) is p, we have mv. mg cos p -N,

133 where N is the normal reaction of the edge or, gtf cos p - m . A) From the energy conservation law, But mR4 mR2 « |mK2, 77//////////0 (from the parallel axis theorem) Thus, From A) and B) N aC gi?(l-cosp) B) The angle p in this equation is clearly smaller than or equal to a so putting p = a we get where No is the corresponding reaction. Note that N this turning if No > 0. Hence, v0 must be less than o. No jumping occurs during max G cos a - 4) 1.265 Clearly the tendency of bouncing of the hoop will be maximum when the small body A, will be at the highest point of the hoop during its rolling motion. Let the velocity of CM. of the hoop equal v at this position. The static friction does no work on the hoop, so from conservation of mechanical energy; Ex « E2 2 2 -mgR= imBvJ + JW + imW~| +mgR or, or, 3v2« \%-2gR A) From the equation Fn « mwn for body A at final position 2 : ~| R B)

134 As the hoop has no acceleration in vertical direction, so for the hoop, N + JST-mg C) From Eqs. B) and C), tn v ?f D) As the hoop does not bounce, N 2t 0 E) So from Eqs. A), D) and E), or 3R * Hence vo£ 1.266 Since the lower part of the belt is in contact with the rigid floor, velocity of this part becomes zero. The crawler moves with velocity v, hence the velocity of upper part of the belt becomes 2v by the rolling condition and kinetic energy of upper part 1 lm\ 2 2 = — \ — \ Bv) = mv , which is also the sought kinetic energy, assuming that the length of the belt is much larger than the radius of the wheels. 1.267 The sphere has two types of motion, one is the rotation about its own axis and the other is motion in a circle of radius R. Hence the sought kinetic energy \l\<»l (i) where /x is the moment of inertia about its own axis, and I2 is the moment of inertia about the vertical axis, passing through 0, But, /x * -^mr and 72 ■ -^mr + mR (using parallel axis theorem,) B) In addition to <*>!-■: and <°2 » J Using B) and C) in A), we get T' « j- mv2 10 1 + 1R 2 1.268 For a point mass of mass dm, looked at from C rotating frame, the equation is ■pi —> n y/ —>' ► dmw = f + dmu> r +2dm(v xco) y where r = radius vector in the rotating frame with respect to rotation axis and v * velocity in the same frame. The total centrifugal force is clearly Rc is the radius vector of the CM. of the body with respect to rotation axis, also where we have used the definitions mRc= y. dmr and mvc - \ dmv

135 1.269 Consider a small element of length dx at a distance x from the point C, which is rotating in a circle of radius r ■ x sin 0 rm Now, mass of the element ■ y \dx So, centrifugal force acting on this element - I -r] dx oJ x sin 0 and moment of this force about C, \dN\ = (j\ dx (o2x sinQ-X cose 2/ and hence, total moment 1/2 J 0 1.270 Let us consider the system in a frame rotating with the rod. In this frame, the rod is at rest and experiences not only the gravitational force m g and the reaction force R, but also the centrifugal force Fcf. In the considered frame, from the condition of equilibrium i.e. NOz = 0 or, NCf « mg — sin 0 where Ncf is the moment of centrifugal force about O. To calculate Ncf , let us consider an element of length dx, situated at a distance x from the point O. This element is subjected to a horizontal pseudo force ■ 9 dx(o x sin 0. The moment of this pseudo force about the axis of rotation through the point O is dN cf -y I dx eo2 x sin 0 x cos 0 —-— sin 0 cos 0 x dx So xr Tmco 7vcyr ■ I —j— sin 0 cos mco2/2 sin 0 cos 0 0 It follows from Eqs. A) and B) that, cos 0 2o>2/ or 0 = cos 3g 2 oo2/ B) C)

136 1.271 When the cube is given an initial velocity on the table in some direction (as shown) it acquires an angular momentum about an axis on the table perpendicular to the initial velocity and (say) just below the C.G.. This angular momentum will disappear when the cube stops and this can only by due to a torque. Frictional forces cannot do this by themselves because they act in the plain containing the axis. But if the force of normal reaction act eccentrically (as shown), their torque can bring about the vanishing of the angular momentum. We can calculate the distance Ax between the point of application of the normal reaction and the C.G. of the cube as follows. Take the moment about C.G. of all the forces. This must vanish because the cube does not turn or tumble on the table. Then if the force of friction is fr /r |=JVAjc But N = mg and fr « kmgy so Ax= ka/2 1.272 In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary. Hence, it follows that Initial angular momentum Initial velocity Axis ± to the initial velocity on the table 1M 1 2,2 l 1M/ CO (co is the final angular velocity of the rod) and Ml Ml co, co From these equations we obtain co co0/|l + 3M) M and v'« co0// Vl+3m/M 1.273 Due to hitting of the ball, the angular impulse received by the rod about the CM. is equal (i) to p —. If co is the angular velocity acquired by the rod, we have ml2 pi 6p Hm2 OXWml In the frame of CM., the rod is rotating about an axis passing through its mid point with the angular velocity co. Hence the force exerted by one half on the other = mass of one half x acceleration of CM. of that part, in the frame of CM. m 8 2ml = 9N

137 1.274 (a) In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary. Hence it follows that 1 2 1 ,2 1 —mv - — mv +-r fMl2) CO and I fl Mi co From these equations we obtain 3m -AM 3m* AM As V t t v^so in vector form v v. and co 4v /(l+4m/3A/) 3m-4M 3m + AM (b) Obviously the sought force provides the centripetal acceleration to the CM. of the rod and is Afto2i- 8Mv 2 / A + 4M/3m f 1.275 (a) About the axis of rotation of the rod, the angular momentum of the system is conserved. Thus if the velocity of the flying bullet is v. mvl ml1* Ml 2 N CO CO mv M 3 3mv Ml as m « M Now from the conservation of mechanical energy of-the system (rod with bullet) in the uniform field of gravity If ,2 2\ml Ml 2 ^ a) — A-cosa ) B) [because CM. of rod raises by the height — A - cosa) ] Solving A) and B), we get M\^ II m . a m2" (b) Sought Ap m (o>/) + M f Ml -mv I where. co/ is the velocdty of the bullet and o> — equals the velocity of CM. of the rod after the impact. Putting the value of v and co we get Ap — — mv « M This is caused by the reaction at the hinge on the upper end.

138 (c) Let the rod starts swinging with angular velocity g>', in this case. Then, like part (a) <Ml2 mvx- 7MX I a) or o) « , 3mvjc Ml Final momentum is mjceo , r ,M, M ,, 3 x 0 f L Li So, This vanishes for Pf-Pi-mvl—-! 1.276 (a) As force F on the body is radial so its angular momentum about the axis becomes zero and the angular momentum of the system about the given axis is conserved. Thus MR G>0 + m oH R MR go or oo0i 1 + 2m (b) From the equation of the increment of the mechanical energy of the system : AJ-A 1MR2 MR act Putting the value of go from part (a) and solving we get 1 + 2m M 1.277 (a) Let z be the rotation axis of disc and qp be its rotation angle in accordance with right-hand screw rule (Fig.), (qp and qp' are to be measured in the same sense algebraically.) As Mz of the system (disc + man) is conserved and Mz^ initUd ^« 0, we have at any instant, or, m2/2 dtp' On integrating or, 0 m m2/2 dip' A) This gives the total angle of rotation of the disc.

139 (b) Fron dtp dt i Eq (i) «i ) m2 dtp dt ml v'(r) J? \ / \ / Differentiating with respect to time d2tp dt2 If1+ 1 dv'(t) R Thus the sought force moment from the Eq. Nz= I $a 1 rf v; (r) dt /W2 Hence R dt m1m2R dv'(t) 2m1 + m dt 1.278 (a) Frome the law of conservation of angular momentum of the system relative to vertical axis z, it follows that: Hence A) Not that for (oz > 0, the corresponding vector to coincides with the poitive direction to the z axis, and vice versa. As both discs rotates about the same vertical axis z, thus in vector form. 11 x x / > 1 2. However, the problem makes sense only if o>1 11 &>2 or a^ t A o>2 (b) From the equation of increment of mechanical energy of a system: A,r i 1 r i 1 AT. 2 Using Eq. A) 1.279 For the closed system (disc + rod), the angular momentum is conserved about any axis. Thus from the conservation of angular momentum of the system about the rotation axis of rod passing through its CM. gives : I , Tim/ U A)

140 (v' is the final velocity of the disc and to angular velocity of the rod) For the closed system linear momentum is also conserved. Hence mv - mv' + Y] mvc B) (where vc is the velocity of CM. of the rod) From Eqs A) and B) we get vc * — and v - v' * r\vc Applying conservation of kinetic energy, as the collision is elastic 1 2 1 ,2 1 2 1 y\fnl 2 2 1 +- C) or Then 0 0 0 v - V » 4t|vc and hence v + v' » 4vc , 4-Y] 4 + T1 v and a) 12 v 7 D 1 Vectorially, noting that we have taken v parallel to v So, J7* « 0 for T| « 4 and w* 1 f v for r| > 4 c 1.280 See the\ diagram in the book (Fig. 1.72) (a) When the shaft BB' is turned through 90° the platform must start turning with angular velocity Q so that the angular momentum remains constant. Here 70(o0 or, The work performed by the motor is therefore /OoH 2 2 If the shaft is turned through 180°, angular velocity of the sphere changes sign. Thus from conservation of angular momentum, /Q-/Oo>0» /Oo>0 (Here - /0 co0 is the complete angular momentum of the sphere i. e. we assume that the angular velocity of the sphere is just - co0). Then and the work done must be, 1^ 2 2 1 2 2_I 0 2 2 2

141 (b) In the case (a), first part, the angular momentum vector of the sphere is precessing with angular velocity Q. Thus a torque, co 0 i is needed. o 1.281 The total centrifugal force can be calculated by, 2 » 1 t co xdx= -rmlQ@ o Then for equilibrium, / and, = ^m/nco2 /0 Thus rx vanishes, when CO o = 6 rad/s Then T2~ mgj - 25 N 72* 0 A 0 B Tng cr 1.282 See the diagram in the book (Fig. 1.71). (a) The angular velocity co*about OO' can be resolved into a component parallel to the rod and a component co sin0 perpendicular to the rod through C. The component parallel to the rod does not contribute so the angular momentum 1 7 M * / co sin0 » — m I co sin0 12 Also, Mz= Msin0= — m/2cosin20 This can be obtained directly also, (b) The modulus of M does not change but —> —> the modulus of the change of M is | A M \. AM| - 2Msin(9O-0)« ^- m 12 co sin20 (c) Here Mx = McosB = /co sin0 cos0 Now dM dt = 7cosin0cos0 CO dt —r dt 1 24 1 ,2 2.2^ = ^rm/ co sin 0 24 as A/ precesses with angular velocity co.

142 1.283 Here M= /co is along the symmetry axis. It has two components, the part /cocosG constant and the part M1« Jco sin0 presesses, then dM dt I a) sinB to' » mg/ sinB or, eo'« precession frequency 0-7 jad/s (b) This force is the centripetal force due to precession. It acts inward and has the magnitude ma)'2/sin6 - 12mN. p* is the distance of the i th element from the axis. This is the force that the table will exert on the top. See the diagram in the answer sheet t 1.284 See the diagram in the book (Fig. 1.73). 1 2 The moment of inertia of the disc about its symmentry axis is —■ m R . If the angulai velocity of the disc is to then the angular momentum is —mR co. The precession frequency being 2k n, we have dM dt 1 2 — mR eo x 2nn 2 This must equal m (g + w) /, the effective gravitational torques (g being replaced by g + w in the elevator). Thus,

143 1.285 The effective g is Vg + w inclined at angle tan" — with the vertical. Then with reference to the new " vertical" we proceed as in problem 1-283. Thus , ml\g2 + w2 a)' * -T « 0-8 rad/s. 7@ The vector o> forms an angle 0 « tan" — « 6° with the normal vertical. g 2 2 1.286 The moment of inertia of the sphere is -=mR and hence the value of angular momentum 2 2 is —mR a). Since it precesses at speed co' the torque required is So, F'« |mK2G>o>'// - 300N (The force F' must be vertical.) 1 2 1 2 1.287 The moment of inertia is — mr and angular momentum is — mr co. The axle oscillates about a horizontal axis making an instantaneous angle. qp= qpmsin — This means that there is a variable precession with a rate of precession ~. The maximum qpm value of this is —=—. When the angle between the axle and the axis is at its maximum value, a torque /to Q 1 2 ^m pm - -mr a) —^r~= ^ acts on it. The corresponding gyroscopic force will be -— = 90 N 1.288 The revolutions per minute of the flywheel being «, the angular momentum of the flywheel v is / x 2jw. The rate of precession is — R. ThxxsN « 2nINV/R - 5-97 kN. m. 1.289 As in the previous problem a couple 2nInv/R must come in play. This can be done if a force, ——— acts on the rails in opposite directions in addition to the centrifugal and other Rl forces. The force on the outer rail is increased and that on the inner rail decreased. The additional force in this case has the magnitude 1*4 kN. m.

144 1.6 ELASTIC DEFORMATIONS OF A SOLID BODY ) or —« a At A) 1.290 Variation of length with temperature is given by 1,-/0A+ But e-f, Thus a ■ aAtEy which is the sought stress of pressure. Putting the value of a and E from Appendix and taking Ar = 100°C, we get a * 2-2 x 103 atm. 1.291 (a) Consider a transverse section of the tube and concentrate on an element which subtends an angle Aqp at the centre. The forces acting on a portion of length A/ on the element are A) tensile forces side ways of magnitude oArAl. The resultant of these is 2oArAlsin oArAlAq) radially towards the cente. B) The force due to fluid pressure « prAtpAl Ar Since these balance, we get Pmvxm crm — where a_ is the maximum tensile force. Putting the values we get p max 19-7 atmos. (b) Consider an element of area dS « n(r A8/2 ) about z -axis chosen arbitrarily. There are tangential tensile forces all around the ring of the cap. Their resultant is o 2n\ r A9 Ar . A9 Hence in the limit rAG 2 or 2omAr m 39-5 atmos. 1.292 Let us consider an element of rod at a distance x from its rotation axis (Fig.). From Newton's second law in projection form directed towards the rotation axis -dT = On integrating - T j(x>2xdx moJjc2 , x — — + C (constant)

145 But at Thus Hence Thus ±-or free end, 0 WOJ/ 8 _ -*max 8 . Condition required for the problem is Hence the sought number of r p s nm SL = J_ A/— [using the table n - 0-8 x 102rps ] 2k nl v p 1293 Let us consider an element of the ring (Fig.). From Newton's law Fn element, we get, TdQ = ( zr- dQ \(o 2 r [see solution of 1.93or 1.92} mwn for this So, Condition for the problem is T m m(j>2r m or @ ** —r~ ~T ' "" Jir2BJtrp) pr Thus sought number of rps n Usinc the table of appendices n - 23rps 1.294 LI! point O desenS by the disUnce x (Fig.). From the condition of equihbnum of pa*. O. , + x2 A) Now, £ o- c£ or J- c£jt — 2 ( a here is stress and e is strain.)

146 In addition to it, V(//2) e = 1/2 From Eqs. A), B) and C) x x- as x < < I Vuff So, mgl —°- 2/ 1/3 or, jc= / mg 2nEd = 2-5 cm 1.295 Let us consider an element of the rod at a distance x from the free end (Fig.). For the considered element T - T' are internal restoring forces which produce elongation and dT provides the acceleration to the element. For the element from Newton's law : m dT « (dm) w « As free end has zero tension, on integrating the above expression, T<& — - -r I Jm 4/* Elongation in the considered element of lenght dx : a - Thus total elengation Hence the sought strain SEl a I E-L 2SE 1.296 Let us consider an element of the rod at a distance r from it's rotation axis. As the element rotates in a horizontal circle of radius r, we have from Newton's second law in projection form directed toward the axis of rotation : - (dm)co2r or, -dT fm , \ 2 m 2 j ydr (o r* -r-co rdr

147 At the free end tension becomes zero. Integrating the above experession we get, thus Thus m>?(l2-r2) I nnx> I \-r- l2 Elongation in elemental length dr is given by : (where S is the cross sectional area of the rod and T is the tension in the rod at the considered element) or, 2SE I Thus the sought elongation dr I 2SE o \-r- 1 dr dr or, ma?I 21 (SI p) 2 2SE 3 3SE 1 D CO / @ 3 E (where p is the density of the copper.) 1.297 Volume of a solid cylinder r21 ^r x2rArl rcr^A/ 2Ar A/ SO, -7^ ■ But longitudinal strain A/// and accompanying lateral strain A r/r are related as Ar A/ Using B) in A), we get : AV A/ V * / A/ - / * E (Because the increment in the length of cylinder A/ is negative) AV -F V * I But So, A - 2 |x) A) B) C)

148 Thus, AV - —^ A - 2 \x) E Negative sign means that the volume of the cylinder has decreased. 1.298 (a) As free end has zero tension, thus the tension in the rod at a vestical distance y from its lower end T- ^gy (l) Let dl be the elongation of the element of length dy> then T Tnsydy - T^dy= z\ _/ - pgydy/E (where p is the density of the copper) SE SIE Thus the sought elongation a/ Ja/- pgf^f- \pgi2/E B) (b) If the longitudinal (tensile) strain is e * —, the accompanying lateral (compressive) strain is given by e' - — - - |A e C) Then since V * Jir/we have AV 2A r A/ « A - 2|*) j [Using C)] where — is given in part (a), \x is the Poisson ratio for copper. 1.299 Consider a cube of unit length before pressure is applied. The pressure acts on each face. The pressures on the opposite faces constitute a tensile stress producing longitudianl com- compression and lateral extension. The compressions is ^ and the lateral extension is \i ^ E t The net result is a compression *z A - 2[x) in each side. E tt &V ty,. * x ^ r AV oAl Hence — = - -±r A - 2 \x) because from symmetry — = 3 —

149 (b) Let us consider a cube under an equal compressive stress a, acting on all its faces. CO Then, volume strain = - ky where A: is the bulk modulus of elasticity. So 5L ^ ° km E or, [i * — if £ and P are both to remain positive. 1300 A beam clamped at one end and supporting an applied load at the free end is called a cantilever. The theory of cantilevers is discussed in advanced text book on mechanics. The key result is that elastic forces in the beam generate a couple, whose moment, called the moment of resistances, balances the external bending moment due to weight of the beam, load etc. The moment of resistance, also called internal bending moment (I.B.M) is given by I.B.M. = EI/R Here R is the radius of curvature of the beam at the representative point (jc, y). I is called the geometrical moment of inertia of the cross section relative to the axis passing through the netural layer which remains unstretched. (Fig.l.). The section of the beam beyond P exerts the bending moment N(x) and we have, If there is no load other than that due to the weight of the beam, then where p = density of steel. Hence, at x « 0 L 2 Z I 'ds 2EI Here b = width of the beam perpendicular to paper. A/2 Also, / -A/2 HCnCe>l«|0- E# 12* @.121 km)

150 1.301 We use the equation given above and use the result that when y is small Thus m Thus _ R dx2 ' dx2 EJ (a) Here N(x)= N0\s a constant. Then integration gives, dy dx" But \a\ " 0 f°r ^ " 0> so Ci * 0- Integrating again, 7 2EI where we have used y * 0 for x « 0 to set the constant of integration at zero. This is the equation of a parabola. The sag of the free end is " y{x" '" 2EI (b) In this case N(x)- F(l- x) because the load F at the extremity is balanced by a similar force at F directed upward and they constitute a couple. Then dll.l dx2" El ■ t , .. dy F(lx-x2/2) _ Integrating, ^ y—^ '- + C1 As before Cx = 0. Integrating again, using y » 0 for x « 0 llx2 x3\ - here X Ft El 3 El Here for a square cross section a/2 =\ -a/2 = a4/12. 1302 One can think of it as analogous to the previous case but with a beam of length 1/2 loaded upward by a force F/2. ^ F I 1 On using the last result of the previous problem. " 1 2 1.303 (a) In this case N(x)= —pgbh(l-x) where b = width of the girder. Also / - b h3/12. Then,

151 Integrating, using Thus d y pgbh 2 12 ^" 2 (/ dx 0 fpr jc * 0. Again integrating 6pg Eh2 fl2x2 Ixl j[_ 3 +12 X = 4a_i ±) 2 3 + 12 6pg/4 3 3pg/4 12* (b) As before, £/ —f-« JV (jc) where JV (jc) is the bending moment due to section PB. dx1 This bending moment is clearly IX AT J w 2lz-2xU w (Here w = p g b h is weight of the beam per unit length) dy Now integrating, El -£-~ w x2i or since -*-= 0 for x « /, cn = w/3/3 rjc jc r 0 Integrating again, Ely** w 24 6 As y = 0 for jc * 0, cx * 0. From this we find 24 1304 The deflection of the plate can be noticed by going to a co- rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance § from the axis is a = § p and the moment of the forces exerted by the section between x and / is

152 N From the fundamental equation jp/*P(/3-*3). A/2 The moment of inertia / lh: Idz neutral surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and z is perpendicular to it. dx Eh - A Integrating Eh Since dy -f- = 0 , for x * 0, c2 * 0. Integrating again, 4pp I)'+ ^ c2 * 0 because y « 0 for jc * 0 Thus 1305 (a) Consider a hollow cylinder of length /, outer radius r + Ar inner radius r, fixed at one end and twisted at the other by means of a couple of moment N. The angular displacement cp, at a distance / from the fixed end, is proportional to both / and jY. Consider an element of length dx at the twisted end. It is moved by an angle cp as shown. A vertical section is also shown and the twisting of the parallelopipe of length / and area Ar dx under the action of the twisting couple can be discussed by elementary means. If / is the tangential force generated then shearing stress is f/Ardx and this must equal G0 Hence, 9 since 0 /» GArdx~-*-. The force / has moment fr about the axis and so the total moment is

153 (b) For a solid cylinder we must integrate over r. Thus xt \ 2nr3 drq>G nrAGqp J / 2/ o 1.306 Clearly JV /In? dr I <pG 327 d./2 using G« 81GPa = 84x10 10 N m = 5x 10m, ^= 3x 10~2m 71 71 cp * 2*0° * — radians, / = 3 m 31 X 84 X 31 32x3x90 <625-81)xlON-m = 0-5033 x 103 N-m « 0-5 k N-m 1.307 The maximum power that can be transmitted by means of a shaft rotating about its axis is clearly N co where N is the moment of the couple producing the maximum permissible torsion, (p. Thus ~ . (o =* 16-9 kw 2/ 1308 Consider an elementary ring of width dr at a distant r from the axis. The part outside exerts a couple N + -r- dr on this ring while the part inside exerts a couple N in the opposite direction. We have for equilibrium dN^ dr

154 where dl is the moment of inertia of the elementary ring, p is the angular acceleration and minus sign is needed because the couple N (r) decreases, with distance vanshing at the outer radius, N (r^ « 0. Now dl m Inrdr r4 Thus dN r5dr or, N * T / 2 2x (r2 " r )> on integration dr 1309 We assume that the deformation is wholly due to external load, neglecting the effect of the weight of the rod (see next problem). Then a well known formula says, elastic energy per unit volume - —stress x strain »-o£ This gives ——Ez ~ 0-04 kJ for the total deformation energy. 2 p 1310 When a rod is deformed by its own weight the stress increases as one moves up, the stretching force being the weight of the portion below the element considered. The stress on the element dx is pnr2(l-x)g/nr2~ pg(l-x) The extension of the element is kdx= d&x= p g(l-x) dx/E jq 1 2 Integrating A/ * -pg/ /E is the extension of the whole rod. The elastic energy of the element is Integrating AC/- l l-x m

155 1.311 The work done to make a loop out of a steel band appears as the elastic energy of the loop and may be calculated from the same. If the length of the band is £ the radius of the loop R - -—. Now consider an element 2 jc ABCD of the loop. The elastic energy of this element can be calculated by the same sort of arguments as used to derive the formula for internal bending moment. Consider a fibre at a distance z from the neutral surface PQ. This fibre experiences a force/? and undergoes ds Z an extension ds where ds = Zdyy while PQ * s * R d (p. Thus strain — = —. If a is the s R cross sectional area of the fibre, the elastic energy associated with it is _. 2 2 Rdq> a Summing over all the fibres we get Ely v 72 E Idy YiTZaZ= For the whole loop this gives, ing I dcp- 2 ji, B using EIk _ 2EIn R I 6/2 Now So the energy is -6/2 1 K2Ehb3 12 = 0-08 kJ 1.312 When the rod is twisted through an angle 8, a couple 21 9 appears to resist this. Work done in twisting the rod by an angle cp is then N(Q) nr G ——— cp2 = 7 J on putting the values. o 3 i 1.313 The energy between radii r and r + dr is, iJy differentiation, —-t—G cp Its density is k r dr G cp2 _ 1 2nrdrl I ,2 2 1.314 The energy density is as usual 1/2 stress x strain. Stress is the pressure p g h. Strain is x p gh by defination of p. Thus 1 2 3 u = — p (p g/i) - 23-5 kJ/m on putting the values.

156 1.7 HYDRODYNAMICS 1315 Between 1 and 2 fluid particles are in nearly circular motion and therefore have centripetal acceleration. The force for this acceleration, like for any other situation in an ideal fluid, can only come from the pressure variation along the line joining 1 and 2. This requires that pressure at 1 should be greater than the pressure at 2 i.e. Pl>P2 so that the fluid particles can have required acceleration. If there is no turbulence, the motion can be taken as irrotational. Then by considering 0 along the circuit shown we infer that V2>V1 (The portion of the circuit near 1 and 2 are streamlines while the other two arms are at right angle to streamlines) In an incompressible liquid we also have div v - 0 By electrostatic analogy we then find that the density of streamlines is proportional to the velocity at that point. 1316 From the conservation of mass Ml - V2*2 But Sx < S2 as shown in the figure of the problem, therefore vi > vi As every streamline is horizontal between 1 & 2, Bernoulli theorem becomes constant, which gives A) P + 2 Py2 P\ < Pi as vl > V2 As the difference in height of the water column is M, therefore Pi -Pi " B) From Bernoulli theorem between points 1 and 2 of a streamline Pi + />2 + 1 or, 2 .2 2 /^2 " /^l " j P (V1 " ^ or p g A /i - - p (vx - using A) in C), we get C) (using Eq. 2) 12 V C-2 ^2 J2 ~ °1 Hence the sought volume of water flowing per see

157 1.317 Applying Bernoulli's theorem for the point A and B, Pa = Pb + 2 p y2 as> Va = ° or> ^" P * So, B -V 2A/ip0g Thus, rate of flow of gas, Q « Sv« S W2A/*p0g P The gas flows over the tube past it at B. But at A the gas becomes stationary as the gas will move into the tube which already contains gas. P 1 2 In applying Bernoulli's theorem we should remember that — + — v + gz is constant along P ^ a streamline. In the present case, we are really applying Bernoulli's theorem somewhat indirectly. The streamline at A is not the streamline at B. Nevertheless the result is correct. To be convinced of this, we need only apply Bernoull's theorem to the streamline that goes through A by comparing the situation at A with that above B on the same level. In steady conditions, this agrees with the result derived because there cannot be a transverse pressure differential. 1.318 Since, the density of water is greater than that of kerosene oil, it will collect at the bottom. Now, pressure due to water level equals hlp1g and pressure due to kerosene oil level equals h^ p2 g. So, net pressure becomes hx px g + h2 p2 g. From Bernoulli's theorem, this pressure energy will be converted into kinetic energy while flowing through the whole A. i.e. hx px g + h2 p2 g = - Pi v Hence v h± + h2 £2 Pi g « 3 m/s f\ ' — '~~ ~~ — — — ~ 1.319 Let, H be the total height of water column and the hole is made at a height h from the bottom. Then from Bernoulli's theorem ~* ~ 1 -pv (H-h)pg or, v= \(H-h) 2g, which is directed horizontally. For the horizontal range, / * v t

158 xt c • t d (Hh-A) n Now, for maximum /, —— - « 0 an which yields h * -t" * 25 cm. 1*320 Let the velocity of the water jet, near the orifice be v', then applying BemouUis theorem, 1 2 . 1 2 APS + Pv or, v'«Vv2-2g/i0 A) Here the pressure term on both sides is the same and equal to atmospheric pressure. (In the problem book Fig. should be more clear.) Now, if it rises upto a height h, then at this height, whole of its kinetic* energy will be converted into potential energy. So, pgh A or - 20 cm, [using Eq. A)] 1*321 Water flows through the small clearance into the orifice. Let d be the clearance. Then from the equation of continuity or v^pvr.v^ A) where vt, v2 and v are respectively the inward radial velocities of the fluid at 1, 2 and 3. Now by Bernoulli's theorem just before 2 and just after it in the clearance //////////////A///////////////S Applying the same theorem at 3 and 1 we find that this also equals p+ - C) (since the pressure in the orifice is p0 ) B) Z$V From Eqs. B) and C) we also hence D) and P - Po + TPvi 1 - [Using A) and D)]

159 \z\ \ht forct, *cfcft& «a fot piston Y>t F and foe lengfo oi foe cylinder be I. Then, work done - Fl A) £^WVjv&% ^tmnXVV* ^tottm lot points A and B,p~ -pv2 where p is the density and v is the velocity at point B. Now, force on the piston, B) where A is the cross section area of piston. Also, discharge through the orifice during time interval t - Svt and this is equal to the volume of the cylinder, i.e., 5 V- Svt or v From Eq. A), B) and C) work done — St C) (as Aim V) 1323 Let at any moment of time, water level in the vessel be H then speed of flow of water through the orifice, at that moment will be v» y/lgH In the time interval dt, the volume of water ejected through orifice, dV- svdt On the other hand, the volume of water in the vessel at time t equal! V-SH Differentiating C) with respect to time, A) B) D) Eqs. B) and D) C All SdH* svdt or rfr* —==r,frOmr2) o Integrating, V2h Thus, s g 1324 In a rotating frame (with constant angular velocity) the Eulerian equation is 2p(v*' x 3) + poJr*« p dt In the frame of rotating tube the liquid in the "column^ is practically static because the orifice is sufficiently small. Thus the Eulerian Eq. in projection form along 7* (which is

160 the position vector reduces to of an arbitrary liquid element of lenth dr relative to the rotation axis) or, so, dr dp " p<o2 rdr p r f dp = pio2 / 0 t A 0 *— r i r >l Thus A) Hence the pressure at the end B just before the orifice i.e. applying BemouU's theorem at the orifice for the points just inside and outside of the end B />„ + 5 P «>2 B / * - *2) So, 1.325 The Euler's equation is p -j- - wherC V • • y whcrc z Now But ar > sr we consider the steady (i.e. dV7dt - 0)jQow of an incompressible fluid then p and as the motion is irrotational Curl v*- 0 So from A) and B) p V f \ v2\ - - V (p + p B) constant. or, Hence -pv2 - 0 constant. 1 326 Let the velocity of water, flowing through A be vA and that through B be v, then discharging rate through A - QA - 5 vA and similarly through B - S v, Now, force of reaction at A,

161 Hence, the net force, F= PS(v2A-v2B) as F^UK A) Applying Bernoulli's theorem to the liquid flowing out of A we get Po and similarly at B Po + 2 Po + PS Po Hence Thus A/tpg = 2pgSA/t - 0-50 N 1.327 Consider an element of height dy at a distance y from the top. The velocity of the fluid coming out of the element is v=* Vlgy The force of reaction dF due to this is dF« p dA v , as in the previous problem, = p(bdyJgy Integrating F= pgb J lydy h-l 2 pgblBh-l) (The slit runs from a depth h-l to a depth A from the top.) 1328 Let the velocity of water flowing through the tube at a certain instant of time be u, thtn u * n 2 -^r, where Q is the rate of flow of water and ji r is the cross section area of the tube. nr -I x 0 From impulse momentum theorem, for the stream of water striking the tube corner, in jc-direction in the time interval dt, ~ Fxdt= -pQudt or Fx« - p Qu *^~ and similarly, F ** pQu 2Z Therefore, the force exerted on the water stream ^. by the tube, ^ F* -pQui + pQuj pi According to third law, the reaction forcj on —;i the tube's wall by the stream equals (-F) -y - pQui-pQuj. ^7 Hence, the sought moment of force about 0 IT becomes ~H \ N- l(-i)*(pQui-pQuj)- pQulk nr Ik and \N\ 0-70 N-m nr

162 1329 Suppose the radius at A is R and it decreases uniformaly to r at B where S s * nr . Assume also that the semi vertical angle at 0 is a. Then R r y wmmm mm —— mm %**m nR and So JR-r (x - xy2u where y is the radius at the point P distant x from the vertex O. Suppose the velocity with which the liquid flows out is V at A, v at B and u at P. Then by the equation of continuity The velocity v of efflux is given by v * VJgh and Bernoulli's theorem gives .2 1 1 Pn + T A) + where /? is the pressure at P and /?0 is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then "~-l A B «— I j 0 = J Lz X Lt We have subtracted p0 which is the force due to atmosphenic pressure the factor sin 6 gives horizontal component of the force and ds is the length of the element of nozzle surface, ds-dx sec 6 and R-r tan 6 Thus R JCp 1 - npv 2 1 R — r dx R jt(i?2 - r2J) R * pgh (S - s)r/S * 6-02 N on putting the values. Note : If we try to calculate F from the momentum change of the liquid flowing out w< will be wrong even as regards the sign of the force. There is of course the effect of pressure at S and s but quantitative derivation of F fron Newton's law is difficult.

163 (fa) —• —♦ —» —4 1330 The Euler's equation is p —* f-Vp in the space fixed frame where /* -pgk downward. We assume incompressible fluid so p is constant. Then / = - V (pg z) where z is the height vertically upwards from some fixed origin. We go to rotating frame where the equation becomes p —* -V(p + pgz) + pco r + 2p(v the additional terms on the right are the well known coriolis and centrifugal forces. In the frame rotating with the liquid v * 0 so = 0 or p + ojez-T-oco^r** constant On the free surface p = constant, thus 0J JZ «. * — r^ + constant 2 If we choose the origin at point r ■ 0 (i.e. the axis) of the free surface then "cosntant" = 0 and CO 7 z= —r (The paraboloid of revolution) At the bottom z = constant So ^= -pco r + constant lfp=p0 on the axis at the bottom, then P- 1331 When the disc rotates the fiiild in contact with, corotates but the fluid in contact with the walls of the cavity does not rotate. A velocity gradient is then setup leading to viscous forces. At a distance r from the axis the linear velocity is co r so there is a velocity gradient -r- both in the upper and lower clearance. The corresponding force on the element whose radial width is dr is tj 2jirdr — (from the formular F ■ J\A — ) The torque due to this force is * . or r\ Ircrdr -7- r h 2nd the net torque considering both the upper and lower clearance is R 2 J r\2ju*drj- * Jl/?4COT|//j So power developed is P * nRWr\/h * 9 • 05 W (on putting the values). (As instructed end effects i.e. rotation of fluid in the clearance r > R has been neglected.)

164 1332 Let us consider a coaxial cylinder of radius r and thickness dry then force of friction or dv viscous force on this elemental layer, F» 2nrly\—. dr This force must be constant from layer to layer so that steady motion may be possible. FJr r Integrating, or, 2 7i I r\dv. A) dv 0 or, Putting F\n S S 2nlv\v B) r = Rv we get Fin — * 2 7t lr\ v0 R2 ijt__I-LLJi \ From B) by C) we get, V= Vf In r/R2 0 In R/R2 Note : The force F is supplied by the agency which tries to carry the inner cylinder with velocity v0 . 1.333 (a) Let us consider an elemental cylinder of radius r and thickness dr then from Newton's formula , d(o „ , 2 dw> xrlnr — ^ 2tzIr\r — 1 dr * dr and moment of this force acting on the element, or, dr 2 ji / t^ do * N -j r dr B) As in the previous problem N is constant when conditions are steady (O Integrating, Till] I dlD« N I — o or, Putting i / N C) co « co2 , we get 2 Jl / T] C02 * -Z" D)

165 From C) and D), 2R2 @ « @, (b) From Eq. D), N R\ r2 R\R2 1334 (a) Let dV be the volume flowing per second through the cylindrical shell of thickness dr then, dV= - Bnrdr) v0 and the total volume, / 1 \ R2 f r - R 1 r3' r - — K ji 4 * ° 2 (b) Let, dE be the kinetic energy, within the above cylindrical shell. Then dT= - 2 Ulxrldrp)v2 L, 1±L r- R2 dr Hence, total energy of the fluid, R T- 0 2r3 ri' R2 RA. nR2plv20 (c) Here frictional force is the shearing force on the tube, exerted by the fluid, which 1 C^V equals - r] S —. Given, So, v= v, o And at dv

166 Then, viscous force is given by, F « - t| BnRl) | — / (d) Taking a cylindrical shell of thickness dr and radius r viscous force, 2 dv Let Ap be the pressure difference, then net force on the element = Ap Jt r + 2 ji tj / r — But, since the flow is steady, F^ * 0 a dr or, A * = nr1 1.335 The loss of pressure head in travelling a distance / is seen from the middle section to be h2- /*! - 10 cm. Since h2 - hx « hx in our problem and /f3 - /^ ■ 15 cm « 5 + /i2 - ^i > we see that a pressure head of 5 cm remains incompensated and must be converted into kinetic energy, the liquid flowing out. Thus v2 = pg A h where A h « /i3 - Thus v« \2gAh m 1 m/s 1.336 We know that, Reynold's number (Re) is defined as, Re* pv l/x\y where v is the velocity / is the characteristic length and r\ the coefficient of viscosity. In the case of circular cross section the chracteristic length is the diameter of cross-section dy and v is taken as average velocity of flow of liquid. pd1v1 N R (Rld' b t f th i d) where vx is the velocity at distance jcx and similarly, Re « - From equation of continuity, A or, ji 7^ Vi — Ji / TJ 1V1" P2V2 SO A2 or €\ V2 d2V2 rlm <hV1r2 -a Ax e"aA)C (as ^2-^- Ax) Thus tt~« e « 5 1.337 We know that Reynold's number for turbulent flow is greater than that on laminar flow ,_, pvd 2p1v1r1 2p2v2r2 Now, (/Qz = fjj- « —— and (Re)t -

167 But, (/g, * (/Q, Pi vlrl ^2 so v2 - * 5 |i m/s on putting the values. — p2 r2 Ti vp0 d 1338 We have f? « and v is given by ti (p * density of lead, p0 * density of glycerine.) 2 , 7 1 / v ' 18^(p~Po)g v ,1/3 and d * [9 T|vp0 (p - Pq) g ] * 5-2 mm on putting the values. 1339 m—- /ng-6jiTirv 6 k 11 r or T" + dt m f 6 jc ti r +kVmgk- ktdv . ht ht d ia ft or e -p + lce v*ge or -z-e v«ge or vefe« £-eh + C or v-7 + Ce"fe (where C is const.) A: A: x Since v- 0 for r- 0, 0 - f + C So C~ - f k Thus The steady state velocity is ~. v differs from f- by n where e""fa* n k or 1 k 4 3 6 n » n r« Tir n 4^p 18 ti Thus . , ^8ti 18 We have neglected buoyancy in olive oil.

168 1.8 RELATIVISTIC MECHANICS 1.340 From the formula for length contraction c2 , So, v_ c4 v- cVtiB-ti) 1.341 (a) In the frame in which the triangle is at rest the space coordinates of the vertices are @00), a a 2 ' 29 a a 2 y 2> , all measured at the same time t. In the moving frame the corresponding coordinates at time t' are A : « 0, 0), B : '2' 2' The perimeter P is then 1/2 ij = fl/l+V4-3p2 (b) The coordinates in the first frame are shown at time u The coordinates in the moving frame are, @,0,0) A:(vf',0,0),B:(|Vl-p2+vt', a The perimeter P is then 1 /O , C : (aVl - p2 + W , O,o x 2* a(Vl-p2 +V4-p2 j here p = V c 1342 In the rest frame, the coordinates of the ends of the rod in terms of proper length /0 A : @,0,0) B : (/0 cos0o , lQ sin0o , 0) at time u In the laboratory frame the coordinates at time /' are A : (vtf, 0,0), B : f/0 cos0o V1 - p2 + vt\ lo sin0o, 0 j

169 Therefore we can write, / cos 60 - /0 cos60 VI - p2 and / sin 6 * /0 sin60 Hence /02« (/2) or, v i-p2 1343 In the frame K in which the cone is at rest the coordinates of A are @,0,0) and of B are (/*, h tan 0, 0). In the frame ICt which is moving with velocity v along the axis of the cone, the coordinates of A and B at time t' are A : (- vt\ 0,0), B : (Wl-p2 - vt\ h tan 6, o' Thus the taper angle in the frame K is S M Ml ♦ a> tan 0 tan0 « B and the lateral surface area is, 5» n h'2 sec0; tan0; 1-P' Here So~ nh sec6 tanG is the lateral surface area in the rest frame and h'- ZtVl-p2 , p- v/c. 1344 Because of time dilation, a moving clock reads less time. We write, fVl-p2, Thus, or, 1345 In the frame K the length / of the rod is related to the time of flight A/ by /« vAf In the reference frame fixed to the rod (frame AT')the proper length /0 of the rod is given by But Vi_62 Vi_82>

170 Thus, VA/ Vl- p So At' or and /0« cV(A/'J-(AO 1346 The distance travelled ii} the laboratory frame of reference is vA t where v is the velocity of the particle. But by time dilation Af- \ So v- cVl-(AVA02 VI - vVc2 Thus the distance traversed is 1347 (a) If x0 is the proper life time of the muon the life time in the moving frame is VI - vVc and hence / VI - vVc Thus xo«fvnv7? (The words "from the muon's stand point" are not part of any standard terminology) 1348 In the frame K in which the particles are at rest, their positions are A and B whose coordinates may be taken as, A : @,0,0),B- (Io,0,0) In the frame K' with respect to which K is moving with a velocity v the coordinates of A and B at time t' in the moving frame are A - « 0,0) B - (l0 Vl-P2 + vt\ 0,0l Suppose B hits a stationary taiget in K after t'B while A hits it after time tB + A/. Then, So, vAf Zo 1349 In the reference frame fixed to the ruler the rod is moving with a velocity v and suffers Lorentz contraction. If /0 is the proper length of the rod, its measured length will be !- /0Vl-p2, c

171 In the reference frame fixed to the rod the ruler suffers Lorentz contraction and we must have Vl - P2 - /0 thus /0 and 1-p Ax, or v«c Vl- Ax, 1350 The coordinates of the ends of the rods in the frame fixed to the left rod are shown. The points B and D coincides when * cx-vt0 or ro« -^ The points A and E coincide when 0 /oVl-p2 -vh> Thus A/ 8 D or -1 *o = 1-p2- 1- (O-\>tfi0) From this 2 c2 At/I o 1351 In Kq, the rest frame of the particles, the events corresponding to the decay of the particles are, A : @,0,0,0) and @,/q, 0,0) - B In the reference frame Ky the corresponding coordintes are by Lorentz transformation A: @,0,0,0), B: lo :2Vl-P2'Vl-p2 ,0,0 Now by Lorentz Fitzgerald contraction formula. Thus the time lag of the decay time of B is vl vl * 3. .2 B decays later (B is the forward particle in the direction of motion) 1352 (a) In the reference frame K with respect to which the rod is moving with velocity v, the coordinates of A and B are

172 Thus l-xA-xB-v(rA-rB) So /, Vl - vVc2 ± /o- (since jca - xB can be either + /„ or - Thus v (tA - tB) - (± 1 -Vl-vVc2)/ i.e. tA -tR= — A /I y 1- c f A -+■ or tB-tA= -^11 h 1353 At the instant the picture is taken the coordintes of A, B, A', B' in the rest frame of AB are A : @,0,0,0) fl B' O O- B': @,0,0, 0) :O .0,0) In this frame the coordinates of B' at other times are B': (t, vr, 0, 0). So B' is opposite to 'o B at time t (B) - —. In the frame in which B', A' is at rest the time corresponding this is by Lorentz tranfonnation. t° (B' c2 Similarly in the rest frame of A,B, te coordinates of A at other times are - —2 + Vty 0, 0 c A' is opposite to A at time f (A) The corresponding time in the frame in which A', B1 are at rest is L 1.354 By Lorentz transformation t' t- vx

173 So at time If x > 01' < 0, if x < 0, f' > 0 and we get the diagram given below "in terms of theJC-clock". The situation in terms of the K! clock is reversed. 1355 Suppose x (t) is the locus of points in the frame K at which the readings of the clocks of both reference system are permanently identical, then by Lorentz transformation t' VI - V2/c2 t- Vx (t)) So differentiating x (t) Let tan/i8, 0^ 8<oo, Then x(t) tan/t6 (l-Vl-tan/t28 cos/t8 sin/t 8 1- cos h 6 cos /i 8 - 1 - / V cos /i 8 - 1 c tan/t — • (tan /t 8 is a monotonically increasing function of 8) 1356 We can take the coordinates of the two events to be A : @,0,0, 0) B : (Ar, a, 0, 0) For B to be the effect and A to be cause we must have Ar > In the moving frame the coordinates of A and B become A:@,0,0,0),fl: L Y(a-7Ar),0,0 where y Since (AT'J-4 c At- we must have A/' >

174 1357 (a) The four-dimensional interval between A and B (assuming Ay * Az » 0) is : 52-32* 16 units Therefore the time interval between these two events in the reference frame in which the events occurred at the same place is c (''« -fA)m vTo7 = 4 m or (b) The four dimensional interval between A and C is (assuming Ay * Az * 0) 32-52= -16 Ct 7 6 5 4 3 Z So the distance between the two events in the frame * in which they are simultaneous is 4 units = 4m. 0 1.358 By the velocity addition formula 4 i a| I 3 r 4 B 1 5 t C v -V x 1- Vv ' r vx vyVl-V2/c2 1- v V x r X and v + v 1- v 7 1359 (a) By definition the velocity of apporach is dx2 ^preach" in the reference frame K . (b) The relative velocity is obtained by the transformation law l - (- v2) 1360 The velocity of one of the rods in the reference frame fixed to the other rod is Vag v + v 2v 1 + The length of the moving rod in this frame is 0 2J + P2) 1361 The approach velocity is defined by -* dr? dr7 v approach in the laboratory frame. So Vapproach- vv2 + w

175 On the other hand, the relative velocity can be obtained by using the velocity addition formula and has the components r A so vxv2 1362 The components of the velocity of the unstable particle in the frame K are c / so the velocity relative to K is 2T/2 The life time in this frame dilates to c2 c2 and the distance traversed is v'2-(v'2V2)/c: ¥' 0Vi-v7c2Vi-v'2/c2 1363 In the frame K the components of the velocity of the particle are v cos 8 - V 1- vVcos 8 v sin 8 Vl - V2/c2 vV 1 - —y cos 8 Hence, tan 8 = —f- - —— v' v cos 8 - V - V2 i9 v 1364 In K! the coordinates of A and B are A : (f\ 0, - v' f', 0); B : (f\ /, - v' t'y 0) After performing Lorentz transformation to the frame K we get ', Vl\ V' t' - V' tf 0 z- 0 By translating t' -* tf j-, we can write the coordinates of B as B :t= y t'

176 c2 Vt'y c ,z- 0 Thus Ar* / V- (v2) v'W Hence tan 8' v'V /y dt 1.365 v v + M>d!f In ^ the velocities at time f and t + dt are respectively v and v + M>*fr along x - axis which is parallel to the vector V. In the frame K' moving with velocity V with respect to Ky the velocities are respectively, v-V . v + wdt-V and vV a- 1 - (V + V r - The latter velocity is written as v-V wdt v-V wV v-V l_v' l-v~ 11-^ c c Also by Lorentz transformation 1- J dt f ^-^^:/c2 , 1 - vV/c Vi - y2/c2 Vi - Thus the acceleration in the K' frame is w dt' '-7' f V2) 3/2 (b) In the K frame the velocities of the particle at the time / and t + di are repectively @, v, 0) and @, v + wdt> 0) where V is along jc-axis. In the IC frame the velocities are M and (- V, (v + wdi) Vl - V2/c2 , (A respectively

177 Thus the acceleration w dt' » w We have used dt' dt along the y-axis. Vl - 1366 In the instantaneous rest frame v * V and w w 3/2 (from 1.365a) c2 So, 3/2 w'df is constant by assumption. Thus integration gives W t Integrating once again x W 1367 The boost time x0 in the reference frame fixed to the rocket is related to the time x elapsed on the earth by dt 0 0 1 + 1/2 dt (Wx)/c dt o >'tV ./^ - —7 In ViTF w 1368 m m, Vl-p: For 1369 We define the density p in the frame K in such a way that pdxdydz is the rest mass dmQ of the element That is pdxdydz- p0 dx0 dyQ dz0 , where p0 is the proper density > ^0 are tne dimensions of the element in the rest frame Kq. Now dy- dy0, dz- dz0, dxm dx

178 if the frame K is moving with velocity, v relative to the frame KQ. Thus Po Defining r\ by p = p0 A + y\) We get 1 + r\ or 1370 We have V.-4 or, -=■ - 1 2 mov or, = J A+r,J A+rj) lo or A 2 2 moc +p 2 2 p +moc or So C - V ■ 1- / 1 + K p c\ / 2 / -1/2 x!00% 1371 By definition of r\> mov v2 1 T]mov or 1 - ■ ""T or v* c xl00% 1372 The work done is equal to change in kinetic energy which is different in the two cases Classically i.e. in nonrelativistic mechanics, the change in kinetic energy is 2 i-m0 c2 (@-8J - @-6J \- ^moc20-28 - 044 m0 c: Relativistically it is, mQc moc Vl - @-8J Vl - @-6J °*6 °*8 = 0-416 m0c2= 0-42 m0c2 m0c2 A-666-1-250)

178 1.373 2moc4 or or *— as — f»r 1 — —— as — •2 2 Ot r2 4 c V3 v VF . — « —— • i.e. v « c — c 2 2 1374 Relativistically So Thus But Classically Hence if , fic/ * y 27 SO 3 7 4^^ 7 4 <7£ moc 2 3 the velocity f) is given by the classical formula with an error less than £. 1.375 From the formula mQC > P Vl-v2/c2 we find E2 - c2p2 + m^ c4 or (/w0 c2 + 7J = c2/?2 + /Wq c4 or c2/ i. 1376 Let the total force exerted by the beam on the target surface be F and the power liberated there be P. Then, using the result of the previous problem we see / since / Np being the number of particles striking the target per second. Also, P= N moc Vl-v2/c2 -/woc = -T These will be, respectively, equal to the pressure and power developed per unit area of the target if / is current density.

180 1.377 In the frame fixed to the sphere :- The momentum transferred to the eastically scattered particle is 2mv 1- The density of the moving element is, from 1.369, n and the momentum transferred per unit time per unit area is ,, 2 mv 1 p * the pressure * —z== n —z= 2mnv2 c2 In the frame fixed to the gas :- When the sphere hits a stationary particle, the latter recoils with a velocity v +v 2v i V cr m-2v The momentum transferred is 1 + v2/? 2mv and the pressure is . 2mv n • v 1- l- 1.378 The equation of motion is mov Integrating v/c 1- Vi-p3 moc ~, using v* 0 for t» 0 or, Frt 2 ot> v or x fJ Vi2 + kCJ F c + constant or using jc » 0 at t« 0, we get, x / 2\2 m0c

161 1379 x « Va2 + c f, SO X * V = c2t or, V.-4 ^- Thus d c2 > moc 1.380 jr_ £- dt OT0V vr: /w, 0 Vi- m0 -j v • v Thus * w vi-p- , IV* V, IV, V _ p2) 2K/2 1381 By definition, c2 mQ c E = m0 vx cmodx where ds Thus, 2 2 c dt dx is the invariant interval dx' (dx - Vdt) d!z- 0) Vl - V2/c2 F, 1382 For a photon moving in the x direction > Py In the moving frame, e' Note that 1383 As before e 1 1-8 lf "T7 or 3c E = dt dx_ ds

182 Similarly Then Ql py=m»cds> dz - m2c4 (c ds2 - dy2-d?) 24.. . , -« moc is invariant 1384 (b) & (a) In the CM frame, the total momentum is zero, Thus V c Vj(r+2moc2) 2mQc' V where we have used the result of problem A.375) Then Vl - V2/c2 VT- 2m0c2 Total energy in the CM frame is 2 2m0c Vl - V2/c2 = 2m, 0c2V 1— 2/w0 C V2m0c2G + 2m0c2) m T+2mQc: So 2m -1 0 Also 2 :4 = V2m0 c2 G + 2m0 c2) , 2moc27, or p* \ h VBm0 c2 + 7J - rB/w0 c2 + 7) = V2m0 c2 Bm0 c2 + 7) = c V2m0 Bm0 c2 + 7) Also + 2m0 c2 1386 Let 7* = kinetic energy of a proton striking another stationary particle of the same rest mass. Then, combined kinetic energy in the CM frame 2moc4 r 2m0c2 -1 2Ty f T moc + 1 1 + 2m0c2 moc

183 1387 We have m0c2y 0 Hence (m0 cz - Erf - cLp\ « (E2 + E3¥ - l c4 m2) c - 2m0 c2 Ex The L.H.S. = (ml c4 - £xJ - c The R.H.S. is an invariant We can evaluate it in any frame. Choose the CM frame of the particles 2 and 3. In this frame R.H.S. = (E'2 + E\f = (m2 ^ m^f c4 Thus or 2mQc /n3Jc4 f O K. or 2 2 2 + ml - (m2 -r m3) 2m0 1388 The velocity of ejected gases is u realtive to the rocket. In an earth centred frame it is v - u 1- vu 7 in the direction of the rocket The momentum conservation equation then reads (m + dm) (v + dv) + v -u (- dm) - mv or mdv- v - u c2 - V = 0 Here - dm is the mass of the ejected gases, so - u + mdv - = 0, or mdv + u ' 2^ V dm= 0 (neglecting 1 - —z since w is non- relativistic.) Integrating (p= -|, _p u I dm n t 1-t I —- 0, In -g m 1-P The constant * — In m0 since 0 initially. u/c Thus i- fe m o .u/c m « constant m o

PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS 2.1 EQUATION OF THE GAS STATE • PROCESSES 2.1 Let rnl and m2 be the masses of the gas in the vessel before and after the gas is released* Hence mass of the gas released, Now from ideal gas equation R R T dV as V and T are same before and after the release of the gas. R R so, (px -p2) V= (m1-m2) jjT0= Am — To Am ^ j^r- A) We also know p « p — T so, —-=- - — B) M RTo Po (where p0 « standard atmospheric pressure and TQ - 273 K) From Eqs. A) and B) we get Am= pV-^-= l-3x30x^L 30g Po 1 2.2 Let m1 be the mass of the gas enclosed. Then, P\V= vx R Tx When heated, some gas, passes into the evacuated vessel till pressure difference becomes Ap. Let p\ and p'2 be the pressure on the two sides of the valve. Then p\V= v\RT2 and

185 fPlV p\V\ RT2 or p2 Pi T, T2 But, So, P'l-P'2 Pl or, Pl 2 PlT2 (PlT2 -p'2-Ap 0-08 atm 2.3 Let the mixture contain vx and v2 moles of H2 and He respectively. If molecular weights of H2 and Hc are Mx and M2, then respective masses in the mixture are equal to Therefore, for the total mass of the mixture we get, m =* m1 + m2 or m » vx Jl/j + v2 M2 A) Also, if v is the total number of moles of the mixture in the vessels, then we know, B) v = v Solving A) and B) for vx and v2, we get, (vM2-m) m- l= M2-M1 ' M2-Mx Therefore, we get m, 1 (v M2 - m) 1- M2-Mt and m (m - v Mx) -y^ —— M2-Mx or, x (v M2 - m) (m - v Mx One can also express the above result in terms of the effective molecular weight M of the mixture, defined as, m= RT v Thus, Using the data and table, we get : M2-M 1-M/M2 3-0 g and, — = 0-50

186 2.4 We know, for the mixture, N2 and CO2 (being regarded as ideal gases, their mixture too behaves like an ideal gas) pV=vRT, so pQV~ vRT where, v is the total number of moles of the gases (mixture) present and V is the volume of the vessel. If vx and v2 are number of moles of N2 and CO2 respectively present in the mixture, then v - vx + v2 Now number of moles of N2 and CO2 is, by definition, given by /Wj m2 v^-and, v2-_ where, m1 is the mass of N2 (Moleculer weight = Mx) in the mixture and m2 is the mass of CO2 (Molecular weight = M^) in the mixture. Therefore density of the mixture is given by V (vRT/P0) pQ mx + m2 pQ (ml + m^ Mx M2 RT vx + v2 ■ 1-5 kg/m on substitution (a) The mixture contains vx, v2 and v3 moles of O2> N2 and CO2 respectively. Then the totai number of moles of the mixture v - v1 + v2 + v3 We know, ideal gas equation for the mixture pV- vRT or A 2 3) or, p - — » 1-968 atm on substitution (b) Mass of oxygen (O2) present in the mixture : m1 * VjMj^ Mass of nitrogen (N2) present in the mixture : m2 » v2 M2 Mass of carbon dioxide (CO2) present in the mixture : m3 - v3 M3 So, mass of the mixture m = m1 + m2 + m3= vx Mr + v2 M2 + v ., . . -Al_ . ,, mass of the mixture Moleculer mass of the mixture : M - Vl + V2 + V3 total number of moles 36-7 g/mol. on substitution

187 2.6 Let px and p2 be the pressure in the upper and lower part of the cylinder respectively at temperature Jq. At the equilibrium position for the piston : + pxS But So, p2S or, p2 (m is the mass of the piston.) RTn Pi * —v" (where Vq is the initial volume of the lower part) RT0 0 or, mg Let T be the sought temperature and at this temperature the volume of the lower part becomes V, then according to the problem the volume of the upper part becomes r\f V Hence, From A) and B). RTC mg RT S * V 1 ~Zj --T or, T o 1--1V vM-6 11' As, the total volume must be constant, V0(l+tl)« V(l+T]') or, V Putting the value of V in Eq. C), we get A+tT) T0(r\2-l)r]' = 0-42 kK 2.7 Let pj be the density after the first stroke. The the mass remains constant , or, Similarly, if p2 is the density after second stroke or, p2 Pi Po In this way after nth stroke. n p« V+AV Po Since pressure a density, B)

188 f V \" ~y—77? Po (because temperature is constant) It is required by — to be — Po i\ i ( v \n tt lnri Hence n = — ' lnll+^y m RT 2.8 From the ideal gas equation p » — — d£_ RT dm dt^ MV dt In each stroke, volume v of the gas is ejected, where v is given by V r In case of continuous ejection, if (tfi^j) corresponds to mass of gas in the vessel at time /, then mN is the mass at time t + At, where A/, is the time in which volume v of the gas has come out. The rate of evacuation is therefore — i.e. At _ v V m{t + At) - m(t) m(t + At) In the limit At - 0, we get i/ rim B) m From A) and B) dp C mRT C dp C , -JL- = — s — — n or -*- = — — a/ rfr FMF Vp p V o /dp C f p VJ Integrating | ^= -^ | <* or In f-. --I Thus p - /?£f a/v 2.9 Let p be the instantaneous density, then instantaneous mass = V . In a short interval dt the volume is increased by Cdt. So, " Vp= (V + Cdt) (p + dp) (because mass remains constant in a short interval dt)

189 so, dp C , —*-» - —dt p V Since pressure a density -*- - -irzdt p V or f- J V*' or r-£ln — In— 1-0 min C ri 2.10 The physical system consists of one mole of gas confined in the smooth vertical tube. Let m1 and m2 be the masses of upper and lower pistons and Sx and S2 are their respective areas. For the lower piston or, Similarly for the upper piston ■ Ps\> A) / / JLLU. or, T« (p-po)S1-m1g From A) and B) ~Po) B) i+ m2> or, so, h \r //7T A / y / mg p - -t"c" +Po ~ constant From the gas law, p V- vRT vR AT (because p is constant) So, Hence, AT AS 1= RAT, 0-9K 2.11 (a) ~ po-a(-j) (as, V= RT/p for one mole of gas) For T —v dp must be zero

190 which yields, Po B) Hence, max SO In —, and J P P 3\R max dT For T the condition is — - 0 , which yields Po * Po In — P A) Hence using this value of p in Eq. A), we get max 2.12 J» o R2T2 (as, V» RT/p for one mole of gas) So, For ' dT 'Which givcS T - 2T0 From A) and B), we get, 2RVETT; A) B) 2.13 Consider a thin layer at a height h and thickness dh. Lttp and dp +p be the pressure on the two sides of the layer. The mass of the layer is Sdhp. Equating vertical downward force to the upward force acting on the layer. pS So, dh A) But, p* -£r/*r,wehaverf/?« ^rr M M or, ~MdTss PgM So, ah That means, temperature of air drops by 34°C at a height of 1 km above bottom.

191 2.14 We have, ^« - pg (See 2.13) A) But, from p - Cpn (where C is, a const) -J- - Crip" B) We have from gas low p « p — J, so using B) P " P^T> °r T" ~RCp n -1 Thus, But, = dh dp dp dh Thus ^ /? RT Integrating, we get C) rc("-1)p 7^Fi M 2.15 We have, dp - - pgdh and from gas law p « -=^= /? A) (where p0 is the pressure at the surface of the Earth.) P Poe > • [Under standard condition, p0 « 1 atm, T ■ 273 K Pressure at a height of 5 atm « 1 x e^*9****™^4*273 m Q>5 atm Pressure in a mine at a depth of 5 km = 1 x e" M x 9'81 x (" 5000>/8314 x m = 2 atm.] 2.16 We have dp « -pgdh but from gas law /? = 77^^ Thus djP = $RT at const, temperature M So, p Integrating within limits J P J RT

192 in- p0 RT So, m Po Pn (a) Given T» 273% — « e Thus h « -T-r-lne l * 8 km. (b) J * 273° K and 0.01 or ■£- - 0.99 Po Po Thus h - - 77- In -^ - 0.09 km on substitution -Mfc Po 2.17 From the Barometric formula, we have P m Poe and from gas law p « *-— RF So, at constant temperature from these two Eqs. J*MH1 A) £q. A) shows that density varies with height in the same manner as pressure. Let us consider the mass element of the gas contained in the coltimn. Hence the sought mass, h Mpo» | -Mrh/RT » i'O** /^ _ -M**/Rr\ j e 0 2.18 As the gravitational field is constant the centre of gravity and the centre of mass are same. The location of CM. 00 00 \ hdm I hpdh CO 00 J dm I pdh 0 0 But from Barometric formula and gas law p « p0 e" 8

193 a J h(e-Mgh/RT)dh So, o a C RT Mg dh o 2.19 (a) We know that the variation of pressure with height of a fluid is given by But from gas law p = -t^RT or, r D_ M Kl From these two Eqs. or, A) -Mgdh p " J?ro(l-«A) Integrating, /d^ -M£ C dh p * RT0J A-ahY we get 0 Hence, Po p = p0 A - a/i) ». Obvionsly /i<: — (b) Proceed up to Eq. A) of part (a), and then put T in the same fashion to get Po TQ A + a h) and proceed further f ahf*/aRTo 2.20 Let us consider the mass element of the gas (thin layer) in the cylinder at a distance r from its open end as shown in the figure. Using Newton's second law for the element n dp)S-pS= (pSdr)co2r or, dp** p co2rdr = *~:co2rdr RT

194 So, dp 7 Mco rdr or' I dp J 7 0 P M(O 0 Thus, ?.21 For an ideal gas law *'M*T So, p- 0082x 300 x -~~-atms - 279-5 atmosphere For Vander Waal gas Eq. p + —j\ (V-vb)' vRT, where V or, av2 mRT/M am2 V-vb mb V2M M-pb 2.22 (a) r RT a (The pressure is less for a Vander Waal gas than for an ideal gas) or, <* A + RT or, , (here V^ is the molar volume.) 1-35 x 1-1 x A-0-039) 0-082 x @-139) (b) The corresponding pressure is a RT JL a a 1-35 0-961

195 So, or. Pi'Pi R(T2-TX) P2-P1 V-b or, b=V- P2-P1 Also, P2"Pl_ _£ 1 ~ V: 1 T - i2 or, Tr2 T —T V i2 ^1 a -Pi T2-Tx Using Tx - 300 K, px - 90 atms, J2 - 350 K, p2 « 110 atm, 7« 0-250 litre a * 1-87atm. litreVmole2, b* 0045 litre/mole 2.24 P RT a 7 or, 7 W; RTVd-2a(V-by V2(V-b) -1 \RTV3-2a(V-bJl 2.25 For an ideal gas k0 RT Now k RTV 21 -1 2 %- RTV Now 1- K>K 2a 1 ., to leading order in a, 2a 2b a RTV>T °r T<hR If a, b do not vary much with temperature, then the effect at high temperature is clearly determined by b and its effect is repulsive so compressibility is less.

196 2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 2.26 Internal energy of air, treating as an ideal gas M v My-1 y- R co Using Cv « 7, since Cn - Cv - R and -rf- * y K y - 1 p v Cy Thus at constant pressure U » constant, because the volume of the room is a constant. Puting the value ofp =*patm and V in Eq. A), we get 17* 10 MJ. 2.27 From energy conservation Ut + i(vM)v2 - £7, or, At/ - \vMv2 A) But from U = v r, AC/ = AT (trom the previous problem) B) Hence from Eqs. A) and B). 2.28 On opening the valve, the air will flow from the vessel at heigher pressure to the vessel at lower pressure till both vessels have the same air pressure. If this air pressure is p, the total volume of the air in the two vessels will be (V1 + V^). Also if v1 and v2 be the number of moles of air initially in the two vessels, we have Pi Vi = vi *ri and Pi V2 - V2 RT2 (!) After the air is mixed up, the total number of moles are (v1 + v2) and the mixture is at •temperature T. Hence P^ + VJ* (v1+v2)/?J B) Let us look at the two portions of air as one single system. Since this system is contained in a thermally insulated vessel, no heat exchange is involved in the process. That is, total heat transfer for the combined system Q-0 Moreover, this combined system does not perform mechanical work either. The walls of the containers are rigid and there are no pistons etc to be pushed, looking at the total system, we know A = 0. Hence, internal energy of the combined system does not change in the process. Initially energy of the combined system is equal to the sum of internal energies of the two portions of air :

197 Final internal energy of (nx + n2) moles of air at temperature T is given by (v1 + v2)/?r Therefore, Ui «■ £/* implies : Vi + v2 * (px V^i) + (ft V2/T2) = ' From B), therefore, final pressure is given by : This process in an example of free adiabatic expansion of ideal gas. 2.29 By the first law of thermodynamics, G.« &U + A Here A » 0, as the volume remains constant, So, Q= AU= ~-AT From gas law, p0 V= vRT0 VAT Hence amount of heat lost = - AU = 0.25 kJ 2.30 By the first law of thermodynamics Q = AC/ + A p AF A But At/ » *-—— * ——- (as p is constant) y1 y-lv 7 2.31 Under isobaric process A = pAF= i? AT (as v = 1) = 0*6 kJ From the first law of thermodynamics = lkJ D A T Again increment in internal energy AU = , for v = 1 y-1 Tin. fi-^AT-^: or Y- ^-|^= 1-6 2.32 Let v » 2 moles of the gas. In the first phase, under isochoric process, Ax« 0, therefore from gas law if pressure is reduced n times so that temperature i.e. new temperature becomes Now from first law of thermodynamics = 'JIT

198 vR (To T\ v^ro(l-H) "y-iU"°r «(y-i) \ / During the second phase (under isobaric process), A2«/>A7- vR&T Thus from first law of thermodynamics : G2- AC/2+A2-^^+vi?Ar Y -1 vRT0(n- Y-l n(Y-D Hence the total amount of heat absorbed vRTQ A - n) vRT0(n -l)y "(Y-l) 2.33 Total no. of moles of the mixture v - vx + v2 At a certain temperature, U» Ux + U2 or v Cv Thus V Similarly v + v, Y2 Thus ZL R v, r + pttVio\is problem R Y2 15-23/mo\e.K

199 and Yi - Y2 ~ 23-85 J/mole. K Now molar mass of the mixture (M) Total mass 20 + 7 Total number of moles 1 1 36 Hence cv * -~r * 0-42 J/g -K and cp M 0-66 J/g-K 2.35 Let S be the area of the piston and F be the force exerted by the external agent Then, F +pS « p0S (Fig.) at an arbitrary instant of time. Here p is the pressure at the instant the volume is V. (Initially the pressure inside is A (Work done by the agent)» \ F dx »/(Po-/>)S<&-/(Po-/>)^ o ^o -fp*iy -/ = vRT(r\ -l-\nr))= RT(r\-l-\ny\) (For v * 1 mole) 2.36 Let the agent move the piston to the right by r. In equilibirium position, PlS+F0gm,mP2S> Or> Fagent* &2 Work done by the agent in an infinitesmal change dx is (p2-Pi)Sdx- ip2-p^ By applying pV» constant, for the two parts, P i (Vo + Sx) * p0 Vo a nd p2 G0 - Sx) * /?0 Vo So, (wherc When the volume of the left end is r\ times the volume of the right end (Vo + V) - ri G0 - V), or,

200 o 0 0 P0 Vo [ In (V2 - V2) - In V02 ] lnJ (r\-l > + 1 V, 0 -In V2 . - in k0 In 4ti + 1)' In 1)' 2.37 In the isothermal process, heat transfer to the gas is given by 2 /! For r\ * — - — ^ Pi In the isochoric process, A « 0 Thus heat transfer to the gas is given by vR But P2 ^o Pi' T vCvAJ , or, AT for C, for yi * — or, Thus, net heat transfer to the gas so, vR vR Y-l -DJn or, or, or, o vRZ y « 1 + 0 6-1 -lnt] 80 x 103 1-4 3 x 8-314 x 273 2.38 (a) From ideal gas law p fyR kT where k -In 6 vR' V For isochoric process, obviously k « constant, thus/7 ■ JfcT, represents a straight line passing through the origin and its slope becomes k. For isobaric process/? - constant, thus on/? - T curve, it is a horizontal straight line parallel to T- axis, if T is along horizontal (or x - axis) For isothermal process, J* constant, thus on/? - T curve, it represents a vertical straight line if T is taken along horizontal (or x - axis) For adiabatic process Typx'y- constant After diffrentiating, we get (l-y)p~ydp- Ty + yp1~y * Ty~x • dT« 0

201 dT 1-y -Y The approximate plots of isochoric, isobaric, isothermal, and adiabatic processess are drawn in the answersheet. (b) As p is not considered as variable, we have from ideal gas law y« —T* ifrfwhereifc'-: — P \ P On V - T co-ordinate system let us, take T along x - axis. For isochoric process F = constant, thus k'« constant and F« k'T obviously represents a straight line pa sing through the origin of the co-ordinate system and kf is its slope. For isothermal process J« constant. Thus on the stated co- ordinate system it represents a straight line parallel to the V-axis. For adiabatic process TVy~ « constant After differentiating, we get (y - 1) VydV- J+ VY dT- 0 dV (_±_) V dT' " The approximate plots of isochoric, isobaric, isothermal and adiabatic processess are drawn in the answer sheet 2.39 According to T-p relation in adiabatic process, T1'- kpy~ (where k*= constant) and So, -•o P2\ for ti ■ — Pi Hence 290xl0A4"iyi4- 0-56kK (b) Using the solution of part (a), sought work done y-1 y-1 \ 5.61 kJ (on substitution) 2.40 Let (p0, Vq, Tq) be the initial state of the gas. -vRAT We know Y-1 (work done by the gas) But from the equation TVy'1 * constant, we get AT- To fr\y~ - 1 \ Thus Y-1 On the other hand, we know A^ ■ vRT0 In j—j = - vR To In r\ (work done by the gas) Thus ^adia (Y-l)lnT] 0-4 x In 5 1-4

202 2.41 Since here the piston is conducting and it is moved slowly the temperature on the two sides increases and maintained at the same value. Elementary work done by the agent = Work done in compression - Work done in expansion i.e. dA» p2 dV-p1 dV» (p2 -px) dV where p1 and p2 are pressures at any instant of the gas on expansion and compression side respectively. From the gas law px (Vo + Sx) » vRT and p2 (VQ - Sx)« vRTy for each section (x is the displacement of the piston towards section 2) So, P2 "Pi m vRT 2Sx vRT 2V V02 - V2 (as Sx - V) So vRT 2V - v2 Also, from the first law of thermodynamics = -dU= -iv-^—dT (as dQ - 0) y- 1 D So, work done on the gas = - dA » 2v dT Thus R or, Y-1 dT vRT 2V-dV V2-V2> VdV When the left end is r\ times the volume of the right end. (Vo + V) = T) (Vo - V) or ^ On integrating dT T VdV o o or \n4r- (Y-1) 0 Y-1 In Vo - In V02 < 1- Hence r= 1- fa-1) "h + i 2 V / f(r, + IJ ) 4n m 1 2 o

203 2.42 From energy conservation as in the derivation of Bernoullis theorem it reads P 1 2 + tv +gz + u + Qd » constant A) P 2 In the Eq. A) u is the internal energy per unit mass and in this case is the thermal energy per unit mass of the gas. As the gas vessel is thermally insulated Qd ■ 0, also in our case. Just inside the vessel u ■ —— « -rrz -7 also *- » -rr * Inside the vessel v « 0 also. Just M M(y-l) p M outside />«0, and u « 0. Ingeneral gz is not very significant for gases. Thus applying Eq. A) just inside and outside the hole, we get P RT RT m yRT M M (y-1) „ 2 2y RT A / 2yRT m u t Hence v * —J — or, v * y -77-7 77 * 3.22 km/s. M(y- Note : The velocity here is the velocity of hydrodynamic flow of the gas into vaccum. This requires that the diameter of the hole is not too small (D > mean free path /). In the opposite case (D <<l) the flow is called effusion. Then the above result does not apply and kinetic theory methods are needed. 2.43 The differential work done by the gas dA- pdV« ^-f--^dJ- -vRdT as pV- v/?Jand V- j\ t+at So, A - - J vRdT = - \RAT T From the first law of thermodynamics y-1 R AT^- (for v = 1 mole) y1 R AT^ y-1 y-1 2.44 According to the problem : AaU or dA = aU (where a is proportionality constant) avRdT /1X or, /k/K= A) From ideal gas law, pV *= vRT, on differentiating pdV +Vdp= v i? JT B)

204 Thus from A) and B) pdV (pdV+ Vdp) or, pdV(k - 1) + kVdp « 0 (where k Y-l another constant) or, 0 or, pdVn + Vic^p * 0 (where k-1 ratio) Dividing both the sides by pV dV dp n V p On integrating n In V + In />« In C (where C is constant) or, In (pV") - In C or, pV* - C (const.) 2.45 In the polytropic process work done by the gas vR[r,-7>] (where Tt and T* are initial and final temperature of the gas like in adiabatic process) vR and Ac/* — By the first law of thermodynamics Q « AJ7 + A (Tf-T)vR 1 Y-l «- v/? [n - According to definition of molar heat capacity when number of moles v = 1 and AT= 1 then Q - Molar heat capacity. Here, Cn« , R(?~Y\, <0 for l<n<y n (/i - 1) (y - 1) 2.46 j^t the process be polytropic according to the law pVn = constant Thus, So, or, Pfi \ J / p a or In P = n In a or n In a In the polytropic process molar heat capacity is given by

205 R(n-y) _R R_ R Ring , lnB 7-7-^—: , where n* 7-^ Y - 1 In P - In a In a c ^ 8-314 8-314 In 4 .„_, n 1*66 - 1 In 8 - In 4 2.47 (a) Increment of internal energy for A7*, becomes ATT vRAT RAT ~, T/ , , x AC/= 7-= 7= -324J(as v= 1 mole) Y-l Y-l From first law of thermodynamics ATT A RAT RAT HUM * A U + A* 7- 7* Oil kJ Y-1 n-1 (b) Sought work done, An - f pdV= I — (where /jV"- k= PiV?- pf 1-n 1-n * 0-43 kJ (as v « 1 mole) 7 7 n- 1 «-1 2.48 LaW of thc process is /? * a V or /?V~ * * a so the process is polytropic of index n - - 1 As /? = aV so, ^ =^ a Vq and py = a tj Vq (a) Increment of the internal energy is given by y-l (b) Work done by the gas is given by . PjVj -PfVf a Vg - ail Vo • 1] Vo «-l " -1-1 1 - 2 2a o(y] " J (c) Molar heat capacity is given by (-1-Y) (n-l)(Y-l) (- 1 - 1) (y - 1) 2y-l

206 2.49 (a) AU- -^yAJ and Q- vCnAT where Cn is the molar heat capacity in the process. It is given that Q - - AU So, CAT- -2- AT, or C R y-1—' ~» y-l (b) By the first law of thermodynamics, dQ » dU + dA, or, 2vCndT = pdV, or, ^-^+ /wJY = 0 2RV vRT So- f^^^-O, or, ^-^2- constant (c) ^^g^^ But from part (a), we have Cn - - Y-l __ R (nj)R ... ... Thus . ' . '——which yields Y-l (n - 1) (y - 1) 1+Y "- 2 From part (b); we know TV™~ ^ - constant So, -^r * \T7"\ * ti^^2 (where T is the final temperature) t \yoj Work done by the gas for one mole is given by 2RT0[l-i\ll-iyi] n-1 Y-l 2.50 Given p** aTa (for one mole of gas) /nlA"a So, pT ■ a or, p or, p1~av~a^ aR~a or, pV^*'^* constant polytropic exponent n ■ (a) In the polytropic process for one mole of gas : . RtsI RAT 1 - n I. a RAT (l-a) a-1 (b) Molar heat capacity is given by R R R R R Y-l _^_! 7? (l-a)

207 2.51 Given LT« aVa or, aVa, or, aV or, or s — Cv pV Ra pV 1-a Ra constant * a (y - 1) as y-l So polytropric index n ■ 1 - a. (a) Work done by the gas is given by Hence 4 -vfl AT , ArT vRAT A « — and Ac/ ■ r- n-1 y-l -ALT (y-l) ALT (y-l) n-1 a 'By the first law of thermodynamics, Q * ALT + A (as n * 1 - a) = ALT + ALT(y-l) a = ALT (b) Molar heat capacity is given by R R R 1 + y-1 R y-1 n-1 y-1 1-a-l R R — (as n * 1 - a) y-1 a 2.52 (a) By the first law .of thermodynamics dQ= dU+dA- vCvdT+pdV Molar specific heat according to definition dQ CvdT+pdV vdT vdT vRT _ We have T=Toe aV aV After differentiating, we get dT * aToe -dV So, Hence t a Toe RToe Cv+ C, v aVTQeav ] (b) Process is p= poe aV RT

208 So, C= C 2.53 Using 2.52 (a) C = We havep Therefore RTdV V dT or, r u a v xr je -V R a RT R poea A+aV) (for onc molc of ^y or, /?I=jp0Vr+a RdT-podV, So, Hence P0 1 + a R ' R R + aR yR aR (b) Work done is given by = C R {P2X2 Pill R R (for one mole) Y-l (Po + - (p0 + ^ By the first law of thermodynamics Q « AC/+A Y-l Y-l . 2 + a In — 2.54 (a) Heat capacity is given by RT dV C m Cv + — — (see solution of 2.52) v + We have or, T T -- — a a After differentiating, we get, dV J^ dT~ a

209 Hence RT 1 a R y - 1 (b) Given J * Jo + aV As T* *~ for one mole of gas a = oR aV Now (for one mole) 2 7 1 aCv(V2-V0 By the first law of thermodynamics Q cxR Y- AU 1 + 1 Y-l Y2 2.55 Heat capacity is given by C * CK+ ~77~~pp (a) Given C - CK + So, Cv+aJ-Cv+ —— or, jdT= Integrating both sides, we get — J - In V + In Co - In VC0, Co is a constant Or, <b)C V-Co- ear//e or V-eaT/R constant

210 . „ _ RTdV _ RTdV and C=CV + —— so, Cv—— T^-pVor' ^-^T or> v Integrating both sides, we get — 3—- - In T + In Co « In PP-1 So, lnJ.C0«--g- rC0«e-^v or, Je^^- -^-- constant pV u Co TD'TT (c) C« Cv + flp and C» c So, Cv+flp«Cv+ —— so, <*P~-ydf RT RTdV, RT f , * x or, <i — * — — (as p - — for one mole of gas) or, ^- a or, <*V« oiiJ or, dT"- — y So,r « — + constant or V - cJ « constant a 2.56 (a) By the first law of thermodynamics A « Q - AG or, - CdJ - CvdJ * (C - Cv) dl (for one mole) Given C - ~ So, a In tj-CVTO fa-1) - alnt) + rOl-1) »-?£ ♦»-?£♦* Given or, or, or, Integrating both sides, C* i a we a y R Y-l dV V get so C 1 d RT* 1 a ,7 N ai RTdV^ / ' RJ2 1 dT Y — 1 T [y — 1) a a r ir r

211 or, or, or, or, 2.57 The work done is if RT RK pV RK constant *t V2-b RT\n T7 , -h a Vb 2.58 (a) The increment in the internal energy is Atf- dV But from second law dV On the other hand P" T RT a -P V-b or, 7 So, AU« a (b) From the first law RT , C and \ a fJL AU* RT\n V2-b 2.59 (a) From the first law for an adiabatic dQ= dU+pdV = From the previous problem 'dU\ 0 dU ar dV dV- a So, 0

212 This equation can be integrated if we assume that Cv and b are constant then R dV dT . /? , or, lnJ+—ln(V- v constant or, (b) We use Jl/C. rftf Now, So along constantp, Thus On differentiating, 0 - constant v2 Cp- Cv+ ^ or, df (V-bf V2 RT/V-b RT la (V-bf V3 RT V-b R dT V-b and C - R 1_2*(V-&)' 2.60 From the first law Q - Uf-Ui+A** 0, as the vessels are themally insulated. As this is free expansion, A » 0, so, C/^ » C/,. tix2 But #XV So, -aV2v or, Substitution gives AT- - 3 K 2.61 Q= Uf-U.+A** Uf-Uiy (as A » 0 in free expansion). So at constant temperature. -«v2 0'33 kJ from the given data.

213 23 KINETIC THEORY OF GASES. BOLTZMANN'S LAW AND MAXWELL'S DISTRIBUTION 2.62 From the formula nkT n 4xlQ-15xl-01xl05 1-38 xKT^x 300 per m* lx 1011 perm3 105 per ex Mean distance between molecules A0~5c.c.I/3- - 0-2 mm. 2.63 After dissociation each N2 molecule becomes two A^-atoms and so contributes, 2x3 degrees of freedom. Thus the number of moles becomes m ,* x j mRT ,* v -A+ti) and p - — A + T!) Here M is the molecular weight in grams of NT 2.64 Let nx » number density of He atoms, n2 - number density of N2 molecules Then p « nx mi + n2 m2 where m1« mass of He atom, m2 « mass of N2 molecule also p « (nx + n2) kT From these two equations we get 2--SL kT m* i 2.65 p nvx2mv cos QxdA cos 0 dA 2 mnv* cos 0 2.66 From the formula V p p If i = number of degrees of freedom of the gas then = CV + RT and ■ j-RT .2 .2 1 + — or t = r i Y-l 2.67 yRT M , and SO, sound rms

214 fa) For monoatomic gases i - 3 sound v ▼ 9 rms * Vf- °-75 (b) For rigid diatomic molecules i - 5 sound V rms 2.68 For a general noncollinear, nonplanar molecule 3 3 mean energy - - k T (translational) + - kT (rotational) + C N - 6) kT (vibrational) ■ CiV^ - 3) kT per molecule 3 For linear molecules, mean energy » —kT (translational) + kT (rotational) + CN - 5) kT (vibrational) 3 N - — I kT per molecule Translational energy is a fraction 1 and 1 in the two cases. 2.69 (a) A diatomic molecule has 2 translational, 2 rotational and one vibrational degrees of freedom. The corresponding energy per mole is 3 1 ~J?7, (for translational) +2x —RTy (for rotational) + 1 xRT, (for vibrational) » -RT Thus, 2*' and Y* C 9 7 (b) For linear N- atomic molecules energy per mole 3N - - \RT as before So, 3N-j\R 6AT-3 (c) For noncollinear N- atomic molecules 3N Cv - 3 (N - 1) R as betore B68) y 3AT- 2_ N-2/3 3* AT-1 2.70 In the isobaric process, work done is A « />^/w « RdT per mole. On the other hand heat transferred Q - C Now C =* CiV - 2) i? for non-coil inear molecules and C — IR for linear molecules

Thus non collinear A Q 215 1 linear For monoatomic gases, cp - — and — ■ -z 2.71 Given specific heats cp, cv (per unit mass) R M(c-cv)~ R or, M cp 2 2 2c Also y ■ " — + h os> *m Cv 20-7 29 7 C » =—-R v* -=—. 1-4 . — v 8-3 ' Y 20-7 5 i- 5 (b) In the process pZ» const - const, So 2 — - — - 0 Thus Cd7- CvdT+pdV or C« CK+2rt« \ — \R So Hence i - 3 (monoatomic) 2.73 Obviously 3 5 (Since a monoatomic gas has Cv» —/? and a diatomic gas has Cv» —/?. [The diatomic molecule is rigid so no vibration]) lc 3 5 R p~ 2 HenCC Y"i 3Yl + 5Y2 2.74 The internal energy of the molecules are

216 where v - velocity of the vessel, N = number of molecules, each of mass m. When the vessel is stopped, internal energy becomes — mN <u > 1 2 So there is an increase in internal energy of Af/= —mNv. This will give rise to a rise in temperature of A7- i' mNv2 iR there being no flow of heat This change of temperature will lead to an excess pressure flA7 mNv2 V - iV and finally ^ = ^~ - 2-2 % p iRT where M = molecular weight of N2, i ■ number of degrees of freedom of ^2 2.75 (a) From the equipartition theorem 6 x 1(T21 J; and vrmy« \^^ - V£^L - 0-47km/s (b) In equilibrium the mean kinetic energy of the droplet will be equal to that of a molecule. / 2 k T or v_-3V-£4£- -015m/s 5 7 2,76 Here i ■ 5, Cv- —R, y « •=■ given 2 D 3RT Now in an adiabatic process TVy -1 _ j yM m constant or 7 T^2 - constant M — 7 2 ^ or Vrj"- K or V 2.77 Here Cv- ^^i?(i- 5 here) The gas must be expanded r\ times, i.e 7-6 times. 5 2M m= mass of the gas, M= molecular weight If vrms increases rj times, the temperature will have increased r\ times. This will require (neglecting expansion of the vessels) a heat flow of amount ~R{ri2-l)T. 10 kJ.

217 2.78 The root mean square angular velocity is given by 1 2 i —/co - 2x- kT B degrees of rotations) or co - V^ - 6-3 x 1012 rad/s 2.79 Under compression, the temperature will rise TVy~1= constant, 7V27';* constant or, r (if * V^'- To VJJ" or, T-xf2" So mean kinetic energy of rotation per molecule in the compressed state kT - k 70 ri^1' - 0-72 x 100 J 2.80 No. of collisions ■ -n <v> « v 4 vT v' «' <v'> 1 Now, —* ■ — V « <V> T| (When the gas is expanded r| times, n decreases by a factor r|). Also mV) -TV or r»Ti F so, — ■ —ri - ti < v ^ i.e. collisions decrease by a factor y\ ~7~ , i - 5 here. In a polytropic process pVn » constant., where n is called the polytropic index. For this process pVn m constant or TVn "x » constant Then dQ= CdT= dU+pdV= CvdT+pdV - -RdT + — dV= -RdT-——RdT= T-—T 2 K Z /I — 1 Z W- 1 / • * Now C- R so 7--^ 1 i , i-2 or, -—7 - r- - 1 - ~7r~ or v' Now — 1 i~l 1 /1V-2 /1V-2 dzl PI PI ti «-2 times - —~r times 2*52

218 2.82 If a is the polytropic index then pVa » constant, TVa Now Hence Then v " n <v> " V ▼ 7 1 1 t 7 - - — or a » -1 a-1 2 constant -1/2 -1/2 2 + 2 2.83 v. 45 km/s, 51 km/s and v__- v rms Y p km/s 2.84 (a) The formula is 4 2 -«2j . V i/ e dut where u * — Now Prob J rf/(«) (b) Prob v- v. rmr <6r| 0-0166 Prob Prob rms <5r| u -vr 4 e 2 2 12 V3 . 00185

219 2. T-T Av - 384K (b) Clearly v is the most probable speed at this temperature. So V m v or 7 mv 342 K 2.86 (a) We have, vl 2 / 2 V2 2/2 v: v: 2 2/ 2 o OkT 1/ — V / V 2 £*r\l i 2' p Of 2 2 V (In So 7- ———~- - 330K vi 2A:ln-~ (b) F (v) x — I — comes from F (v) dv - J / (w), £iv Thus P2 m m T| now mv or -r-Tlnri o Thus m 2.87 nty N o Av = N 363 K 1 — 2 2 2 -v/v o } 222 2 V -v/v v 3/2 « lnm^v. /mir vl ~ 3it7 ^-—- , Putting the values we get v - 1-60 km/s

220 2.89 dN(v) N4v2dv _vV For a given range v to v + dv (i.e. given v and dv ) tbis is maximum when 0 -4 2 2 - V /V or, o .. Thus . 2 p m 3 2.90 3/2 Thus dn(v)=N 2.91 <v > = 0 by symmetry 00 00 00 |vj ^ ~ 2«F dvxj I e ut dvx = I vx e ikt dvx J I - mv 2kT dv. — 00 0 0 0 00 00 du/Je du o 0 oo 00 Vf !> o o 2Vjc 2.92 00 A)/ 'V 2 mv oo v** o 00 00 m xe -X Jl 2VJC -x dx 2Vjc m m 2.93 Here vdA = No. of molecules bitting an area dA of the wall per second 00 -JdN(vx) vxdA

221 or, 00 1/2 mv m 00 1/2 o n BkT\ -u . hn where <v> « 2.94 Let, dn (vx) » n 1/2 m \ -mv~/2MTi *' " ' 2 ji *7 ' " * be the number of molecules per unit volume with x component of velocity in the range vx to Then 00 I 2mvx'vxdn(vx) o 00 1/2 2mvtn 00 o 1 2*7 f 2 -«2 - 2mn—r= I ire yfn m J o 00 4=nkT- \ -x dx 00 2.95 3/2 2nkT mv e~?kT Anv dv — 0 00 3/2 m 1 2kT C ^T I 2 m J -X e~xdx o m \ I 2m \ J16 m?t\ n2 MTJ n<v>

222 3/2 2.96 dN(v)-N m 2nkT or, e 4 ji v dv 3/2 v dN(t) J dt iTtkT dz Now, or, 1 2 dv 1 e« -mv so — ■ — 2 ae mv 3/2 dt 2jtifc7J /n VJt i.e. VJl The most probable kinetic energy is given from i -1/2 dt _ > 2 kT The corresponding velocity is v "/r V 2.97 The mean kinetic energy is 00 0 Thus o kT E/2) C/2) 2 VJt -3/2 3/2 26 If 6 ti - 1% this gives 0-9 % 00 2.98 -3/2 oo e (e0 » Jt7)

223 (In evaluating the integral, we have taken out Ve" as Ve^ since the integral is dominated by the lower limit.) 2.99 (a) F(v)-Av3e"mv2/2tr For the most probable value of the velocity '2kT This should be compared with the value v » "^ for the Maxwellian distribution. 1 2 (b) In terms of energy, 8 - — mv 3 ^-mv AB* F (e) - Av* e 3/2 =mAe V2me m2 From this the probable energy comes out as follows : F (e) - 0 implies . o or 2.100 The number of molecules reaching a unit area of wall at angle between 6 and 0 + dQ to its normal per unit time is v <■ oo dv « I dn (v) —^ v cos 0 J 4jt v- 0 00 3/2 c-mv/2*r v3 dvsin G cos 9 d 9 x 2 n 0 N 7 /2tr \1/2 n\ sin0cos0d0 I mn J 2.101 Similarly the number of molecules reaching the wall (per unit area of the wall with velocities in the interval v to v + dv per unit time is 6- x/2 dv » I J/i (v) —— v cos 0 J 4ji e. o

224 6- n/2 2 jt e- o 3/2 2 iinO cos9 dQ x 2;t 3/2 iTikTj 2.102 If the force exerted is F then the law of variation of concentration with height reads -FZ/kT « „ Fbh/kT _ t? Wlnii .^20 n(Z)-noe-FZ/kT So, ^ eFAMT or F- ^^ - 9 x 100 N 2.103 Here F- f ^Apg» ^^ or Na= 6*Tlnl\ 6 #* a fA In the problem, -*- = 1*39 here mg 9-8 m/s2A p = 0-2 x 103kg/m3 5m, J = 4 x 10m,g = 9-8 m/s2, A p = 0-2 x 103kg/m and /? = 8-31 J/k ^ xr 6x8-31 x 290 x In 2 1A26 r-^ in23 , -l Hence, Nn = ——ttttzxr—t x 10 = 6-36 x 10 mole * jt; x 64 x 9-8 200 x 4 concetration of TTF^f Ho M concentration ofJV2 u e~Mn So more ^2 at ^e bottom, 1.39 here / 21O^ /# \ -mtgh/kt ,,\ -m.gh/kT ^* They are equal at a height h where — - e*hlmi-mJ'kT tr In n, - In «, or n = g mi - m 2.106 At a temperature J the concentration n(s) varies with height according to ~mgz/kT nQe 00 This means that the cylinder contains I 11B ) dz o 00 «oitr dz = mg o particles per unit area of the base. Clearly this cannot change. Thus n0 kT** po= pressure at the bottom of the cylinder must not change with change of temperature.

225 00 oo 2.107 mgze -mgz/kt dz xe"xdx o 00 kT o 00 -mgz/kT -X e ----dz I e~ dx o o When there are many kinds of molecules, this formula holds for each kind and the average energy kT if, where /f. a. fractional concentration of each kind at the ground level. 2.108 The constant acceleration is equivalent to a pseudo force wherein a concentration gradient is set up. Then = 1-11 v\KT "'- MAl "Mj 2.109 In a centrifuge rotating with angular velocity co about an axis, there is a centrifugal acceleration co r where r is the radial distance from the axis. In a fluid if there are suspended colloidal particles they experience an additional force. If m is the mass of each particle then its volume is — and the excess force on this particle is P — (p - p0) co r outward corresponding to a potential energy - ^^(p - p0) to r This gives rise to a concentration variation m n (r) - n0 exp + Thus where Thus 2pkT M m M X£ k RyM M* m is the molecular weight 2pRT\ny\ (P - Po) 2.110 The potential energy associated with each molecule is : - — m co2 r2 and there is a concentration variation Thus n (r) = «0 exp r|« exp mco 2fcT = «oexp 2RT 2RT or co v 2RT Ml2

226 Using M= 12 + 32= 44gm, /= 100cm, R - 8-31 x 107 we get go = 280 radians per second. 300, 2.111 Here n (r) = «0 exp f 2\ ar ' kT (a) The number of molecules located at the distance between r and r + dr is 2 / "^ 4nr drn(r)» 4nn0exp r2dr • • 1S n or> If (c) The fraction of molecules lying between r and r + dr is (raP'/kT) 00 J 4 7ir2drn0exp (- ar2/kT) o 00 00 J 47ir2dr exp AT T n( exp (- x) Thus 3/2 3/2 a So n(r)« N a nkT 2 1/2 drexp exp kT (-a ar2\ kT 1 2\ / 2\ r / 3/2 .. 2.112 When J decreases r| times «@) « n0 will increase T| times. Write l/= ar or so A U 5. so exp a 2VU iVaU The most probable value of U is given by l/2\ ( \ dU\dU 0 U 2VJ7 exp or, pr From 2.111 (b), the potential energy at the most probable distance is kT.

227 2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 2.113 The efficiency is given by T >Ti Now in the two cases the efficiencies are facrcascd , T2 decreased Thus 2.114 For H2 , y - j vi " P2 V2> p, v;, Pi Define n by V3 ■ n V2 Pa v* Pa vj Thenp$= so 17 » Pi V? so V^-t- 711-T»1-T or V^ Also 2 /-v 3 1- Finally p2 (b) Define nbyp3- — 1-7 0-242 -^n or So we get the formulae here by n -* n y in the previous case Tl-l-rt177" -l-»~7-0-18 2.115 Used as a refrigerator, the refrigerating efficiency of a heat engine is given by 1- where T| is the efficiency of the heat engine.

228 2.116 Given V2 - n Vx, V4 = n V3 Qx = Heat taken at the upper temperature - RTX In n + R T2 In n - /? G\ + T2) In « 1 /T* \ v — 1 Ji Now or V, Similarly V5 Thus B2 * neat ejected at the lower temperature « - /? J3 In — In in rr^ 1 n v'2/ Thus 1- 2J, 2.117 fi ' (T2 - T3) (p2 - ThUS T] = 1 - jtj^ ^1 (Pi - On the other hand, V2Y, p3 VI - p4 V? also V2 - n VI Thus P2»\ P* and T| * 1 - n1, with y » —for N2 this is r| 0-602 2.118 Ql - -fPl (V2- P2 So Nowpj * np2, P2 pi <y2 - or Vs * n * 1 so ri = 1 • « i-i

229 2.119 Since the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion Pi " nP2 aiu* ^2 " n ^i • Ci - Gii + fin where Qn = Cp (« - 1) 7\ and j212 - Cv Tx Heat rejected is G'2 = Q'21 + e'22 where Thus T| C'9 Gi 1- /1 - 1 + v [l - — 1 « - 1- 777j > 77 n ny 2.120 (a) Here p2 />0 -1 But VJ or, Q' RTt 0 Y-l Inn In Temp n To AV, adiahaJUc P1.V1 Pojh Qa isofhermat _ In n _ Thus ti - 1 , on using Cv n1 v y-l (b) Here V2 - «VX, Pl Vx - and />! (n K^ - p0 Vj or or Thus — lnw 7o adiabcttic isothermat •To

230 2.121 Here the isothermal process proceeds at the maximum temperature instead of at the minimum temperature of the cycle as in 2.120. adubatk adiabatic V 0?) (Po,Vo,To) isothermal (PhV,Jo) (a) Here pl Vx Pi vi ~ Po vo or Pi vi i.e. Or V, « VnrtY-l 1- Yony V, RT« T0U-± CVTO Inn- Thus C'2 - (b) Here V2 - -^,/>„ Vo -/>i ^i rt-?7Y „ VS-1*-^7 or Vx- «(Y/Y-1}70 7, 0 n n« C Thus 1- n\nn 2.122 The section from (pl9 Vv To) to (p2, 72, Tq/w) is a polytropic process of index a. We shall assume that the conesponding speciflc heat C is + ve. Here, dQ = CdT= CvdT+pdV Now/?Va« constant or TV01" constant. sopdV= Then C We have px Vx a-l a-l Y-l a-l RT0 PlVx n n PO V0 = P\ Vl * nPi V2> PO V0 isothermal \ L Poly tropic of \ V

231 PiV?~PiV? or n or 1 1 or m fty-l" a-lVt 0 Now<2'2 In n Thus tj- 1- nlnn 2.123 adiabatic To, Vo,To (a) Here Q'2 = C. Along the adiabatic line n so oo n or, ^-i). Thus n- i- 1 T o n s ,7-1 Y(") .7-1 n^ - Jo), fix - C^ • 7\ (n - 1) 7 Along the adiabatic line TV1" » constant i0 y0 T n Thus or nY-l 2.124 PofVo,To 77' vo'rr Vo,Tot) Po,Vo,To (O.) (b)

232 00 Q' i|, Q'\- RT0\nn, Q\= CVTO c. \ 1-- n \ So r n 1- 1 + - 1- win ft n- 1 + (y - l)n\nn So 2.125 We have !'!- x/?Tolnv, Cl r0 (n 1- q '» in n " - 1 + (y - 1) In n 2= C^IoCx-lJG^ G'i + G"i a^ > cvr0(x-i) as well as Q1 = g/ + Q" and e2' - e2" SO Tl- 1 Cv(x-l)+/?lnv 1- - 1 (x- l)lnv X-1 . . X-1 + x ln v x In v + Y-l 2.126 Here g/' = C^Iq (x - 1), Q'l " and in addition to we have i - Ci' + &" and 2' - e2"+q2'" So n = 1 - -r- = 1 - Ci X-1 + = 1 — 1-- T-1 + T-1 xln hvn 7o (x - 1) ln n 1 - —|x ln« xlnn Y(t-D Y-l

233 2.127 Because of the linearity of the section B C whose equation is Po vo We have ~» v or v» Vt v Here Q- CVTO(Vt - 1), Q nt Vr" Qi Thus Q'2 - g + Q' T 7^1 Along BC, the specific heat C is given by CdT= 1_ 2 dT Thus y-I Vx" Finally Vx +1 l)(Vx +1) 2.128 We write Claussius inequality in the form 0 where <f B is the heat transefened to the system but d2 Q is heat rejected by the system, both are +ve and this explains the minus sign before <f2 Q, In this inequality ' > T> Trmn and we can write IPS •/ ■* mar •/ T IT max *^ nun < o Thus _Cl_< _Cj_ ^ ?^n< T T T max min max 2 or max • carnot 2.129 We consider an infinitesimal carnot cycle with isothermal process at temperatures T + dT and T. Let bA be the work done in the cycle and 6B, be the heat received at the higher temperature. Then by Carnot *s theorem

234 6A dT " T On the other hand 6A » dpdV« |^| dTdV while dUn +pdV fdU) dV +P dV Hence 2.130 (a) In an isochoric process the entropy change will be V+dv CvdT For carbon dioxide y - 1*30 so, AS= 19-2 Joule/°K-mole (b) For an isobaric process, Inn Cplnf » 25 Joule/'IC-mole 2.131 In an isothermal expansion AS- vR\n -f- SO, eLS/VR m 2.0 tjmes \ \ \ t isotherm \ \ \ 2.132 2.132 The entropy change depends on the final & initial states only, so we can calculate it directly along the isotherm, it is AS = 2/? Inn = 20J/°K (assuming that the final volume is n times the inital volume) 2.133 If the initial temperature is Jo and volume is Vo then in adiabatic expansion. so, Tx where n* — Vx being the volume at the end of the adiabatic process. There is no entropy change in this process. Next the gas is compressed isobarically and the net entropy change is AS m M cMai

235 But So * Vo T —m T n~y or or In — « - 77 C_ In n « - 77 n M * My-1 lnw « -9-7J/K 2.134 The entropy change depends on the initial and final state only so can be calculated for any process whatsoever. We choose to evaluate the entropy change along the pair of lines shown above. Then / dT ■ h Po,Va]b ^ = —Y (y In a - In p)« - 11 -5^- 2.135 To calculate the required entropy difference we only have to calculate the entropy difference for a process in which the state of the gas in vessel 1 is changed to that in vessel 2. A5- v f T. c vR lna- Y-l With y= f > «= 2 and p= 1-5, v= 1-2, this gives A5 - 0-85 Joule/°K ■Mr 2.136 For the polytropic process with index n p V « constant Along this process (See 2.122) 1 1 \ n-y R XT. So

236 2.137 The process in question may be written as P V Po vo where a is a constant and /?0, Vo are some reference values. For this process (see 2.127) the specific heat is Along the line volume increases a times then so does the pressure. The temperature must then increase a times. Thus yR Y+1 7iT if 46-1 Joule/°K 2.138 Let (pv VJ be a reference point on the line P - Po - aV and let (p, V) be any other point. The entropy difference AS« S(p,V)- For an exetremum of A5 dV or or V - 0 or YPo m This gives a maximum of AS because ^ AS dV2 + 1) <0 (Note :- a maximum of AS is a maximum of S (p> V)) 2.139 Along the process line : S » a7+ Cvln T dS or the specific heat is : C » 7-^= « aJ+ On the other hand : CdT* CvdT + pdV for an ideal gas. Thus, dTdT

237 or Using — — = dT or, —In 7+constant- T a V a R V J- Jo when V- Vo, we get, J- TQ + —In — 0 2.140 For a Vander Waal gas (V-b)- RT The entropy change along an isotherm can be calculated from AS as} dV dV It follows from B.129) that V~b assuming a, !> to be known constants. Thus AS- R\n Vx-b T ,i2 2.141 We use, AS as' dV 'T- -6 T assuming Cv, a,b to be known constants. 2.142 We can take S -* 0 as T-* 0 Then T T f f f 0 0 2.143 AS T mJ T T r yx-6

238 2.144 Here aSn or S = 1 i-1 n Then Clearly 2.145 We know, n a"n n C < 0 if n < 0. 5-5 o CdT T o assuming C to be a known constant. !S - 5,v Then 2.146 (a) C= T T (b) (c) W= AQ-AU Since for an ideal gas Cv is constant and At/- (U does not depend on V) 2.147 (a) We have from the definition B-1 TdS = area under the curve To (Sj - So) z> Thus, using T!= 1- 12. n' 1+1 21, 1 — n- 1 0 2 In (b) HcrcQi- ^-E1-50)(ri + r0) 1 1- 2Ti 7, + Jo 70-7! n + 1 T OO So,7b

239 2.148 In this case, called free expansion no work is done and no heat is exchanged. So internal energy must remain unchanged U* « £/,-. For an ideal gas this implies constant temperature Tf ■ Tp The process is irreversible but the entropy change can be calculated by considering a reversible isothermal process. Then, as before J ^» I - vRlnn - 20-1 J/K 2.149 The process consists of two parts. The first part is free expansion in which U* ■ t/(-. The second part is adiabatic compression in which work done results in change of internal energy. Obviously, vo 0= UF-Uf+fpdV, Vr 2V0 vf Now in the first part p* ■ — p^ V*» 2VQ, because there is no change of temperature, In the second part, p V- -pQ BV0f - 27" 1p0 Vj 4mt V pdV 2 V, o ■1-7 Y ■ ~ 2V. o R T Thus A 17- Up-Ui = —~-BY-l) The entropy change AS = AS'/ + AS/7 tSj = /? In 2 and ZiuS1^ = 0 as the process is reversible adiabatic. Thus AS = /? In 2. 2.150 In all adiabatic processes by virtue of the first law of thermodynamics. Thus, For a slow process, A' = I pdV where for a quasistatic adiabatic process pVy = constant. On the other hand for a fast process the external work done is A" < A'. In fact A" » 0 for free expansion. Thus U'* (slow) <U"* (fast) Since U depends on temperature only, T* < T'* Consequently, p", >p'f (From the ideal gas equation pV = RT)

240 2.151 Let Vx - Vo, V2 - n Vo Since the temperature is the same, the required entropy change can be calculated by con- considering isothermal expansion of the gas in either parts into trie whole vessel. Vl + ^2 Vl + ^2 Thus AS- AS7+AS/7« vtR\n \ + v2mn \ 2 - vxRIn A + n) + v2l* In ^-^ - 51 J/K 2.152 Let c1« speciGc heat of copper specific heat of water = c2 To 97 + 273 Then AS- I ^ I n^c2 In — -mlcx In — 7 + 273 rQ Tq is found from 280 m«> Co + 370 mi c, o - 280) - mx q C70 - Jo) or To using cx - 0-39 J/g °K, c2 - 448 J/g °K, To ~ 300°K and A5 - 28-4 - 24-5 « 3-9 J / °K 2.153 For an ideal gas the internal energy depends on temperature only. We can consider the process in question to be one of simultaneous free expansion. Then the total energy 17= UX^U2. Since T, +72 Ux= CvTlf U2= CVT29 U« l£v and iJ^ + T^ll is the final temperature. The entropy change is obtained by considering isochoric processes because in effect, the gas remains confined to its vessel. / ' T AS- I -i=- - I C=£-« Cvln Since (Tx + TJ2 * (Tx -72J + 4T1T2, AS > 0 2.154 (a) Each atom has a probability — to be in either campartment Thus (b) Typical atomic velocity at room temperature is^ 10 cm/s so it takes an atom 10" sec to cross the vessel. This is the relevant time scale for our problem. Let T = 10 ~ sec, then in time t there will be t/T crossing or arrangements of the atoms. will be large enough to produce the given arrangement if - 2 ^ 1 or JV^ x 2 ^ 1 or JV^ r^r x In 2

241 2.155 The statistical weight is . m 10x9x8x7x6^ C"n" n i —» " 8x4x3x2 The probability distribution is Nc 2~"« 252x20» 24-6% 2.156 The probabilites that the half A contains n molecules is Nl 1~N 2.157 The probability of one molecule being confined to the marked volume is V We can choose this molecule in many (Nc ) ways. The probability that n molecules get confined to the marked volume is cearly 2.158 In a sphere of diameter d there are N** -—£- Wq molecules where n0 * Loschmidt's number = No. of molecules per unit volume A cc) under NTP. The relative fluctuation in this number is w Vn i " N 6 or —r * -r *r «« or ar ■ r or \ / = 0-41 The average number of molecules in this sphere is —y » 10 2.159 For a monoatomic gas CK» —R per mole The entropy change in the process is AS- S-So ' " A7^ ^0 Now from the Boltzmann equation 5- JHnQ 3x6 2 T— x 10 300 ■ 1013 * 1021 Thus the statistical weight increases by this factor.

242 2.5 LIQUIDS. CAPILLARY EFFECTS 2.160 (a) '3 4x490xl0'3 N 1-5 x 10"* (b) The soap bubble has two surfaces , ^ <n6N ^ 1*307x10 —r- 13 atmosphere m2 m2 so 8a 8x45 — 3 -t »•> .. -t n — 3 3x10 -3 x 10 « 1-2 x 10 "^ atomsphere. 2.161 The pressure just inside the hole will be less than the outside pressure by 4 a/d. This can support a height h of Hg where , 4a , 4a pg*.T or *- — 4 x 490 x 10 200 13-6 x 103x 9-8x 70x 10 136 x 70 - -21 m of Hg 2.162 By Boyle's law or d) 3 U 1- 3B f ( 8a -1) Thus 2.163 The pressure has terms due to hydrostatic pressure and capillarity and they add P ' Po + PS* + /. 5x9-8xlO3 4x-73xl0 1 + 4x10 -6 x 10 " 51 atoms » 2-22 atom. 2.164 By Boyle's law or 4a\ n 4a ji 6 or g p » 4-98 meter of water

243 2.165 Clearly p g » 4 a | cos 9 An x d2 pg 11 mm 2.166 In a capillary with diameter d = 0-5 mm water will rise to a height 2a 4a 1 ~mm mm— ~—m I 4 x 73 x 10 -3 ;r 103x9-8x0-5xl0 '3 59'6 mm Since this is greater than the height ( = 25 mm) of the tube, a meniscus of radius R will be formed at the top of the tube, where R 2a 2 x 73 x 10 -3 103 x 9-8 x 25 x 10 -3 «* 0'6mm 2.167 Initially the pressure of air in the cppillary is p0 and it's length is /. When submerged under water, the pressure of air in the portion above water must be p0 + 4-r, since the level of water inside the capillary is the same as the level outside. Thus by Boyle's law 4a> Pol or 4a (/-.*)= pox or I 1 + 4a 2.168 We have by Boyle's law \Po - 4 a cos 0' . f or, Hence, 4 a cos 9 . /^o 3 P8* a pg/i /-A l-h d 4 cos 0 2.169 Suppose the liquid rises to a height A. Then the total energy of the liquid in the capillary is h x P^ x T" - di)ah Minimising E we get h 4a - di) 6 cm.

244 2.170 Let h be the height of the water level at a distance x from the edge. Then the total eneigy of water in the wedge above the level outside is. E= I xbydx-h- pg^-2| dxh-acosO J 2 J r i J 2 2 ~ 2 a cos 9 , -2 —h xpgby 2 2 a cos 0' *Pgocp rp,. . . . . . 2 a cos 0 Inis is minimum when h « xpgby 2.171 From the equation of continuity / 1_ 2 !*- 4 a2 cos2 8 JC -ar-v or We then apply Bernoulli's theorem v + constant P 2 The pressure p differs from the atmospheric pressure by capillary effects. At the upper section 2a neglecting ihc curvature in the vertical plane. Thus, 2a 2 no. i* 1 4 2 V or n4-l Finally, the liquid coming out per second is, 2.172 The radius of curvature of the drop is R1 at the upper end of the drop and R2 at the lower end. Then the pressure inside the drop is pQ + — at the top end and pQ + — at the bottom /<! R2 end. Hence 2a 2a or 2a(R2-R{) h 1 To a first approximation/?!- /?2- — so R2-R1& — pg/t3/a. » 0-20 mm if /t - 2-3 mm, a = 73 mN/m

245 2.173 We must first calculate the pressure difference inside the film from that outside. This is Here 2 rx |cos 9 |« h and r2 „ - /? the radius of the tablet and can be neglected. Thus the total force exerted by mercury drop on the upper glass plate is 2 Jt/?2 a [cos 8 | h typically We should put hi n for h because the tablet is compresed n times. Then since Hg is nearly, incompressible, tvR h = constants so R -» /tv^T. Thus, , , ir 2jii?2a|cos9| 2 total force * r-* n h Part of the force is needed to keep the Hg in the shape of a table rather than in the shape of infinitely thin sheet. This part can be calculated being putting n » 1 above. Thus 2jt/?2a|cosG| 2 nR2 a | cos = or wg + m _ 2 jc /? a jcos 91, 2 ,v n«, = —' l (rr - 1) « 0-7 kg hg 2.174 The pressure inside the film is less than that outside by an amount a (— + — I where rx and r2 are the principal radii of curvature of the meniscus. One of these is small being given by /*» 2^ cos 9 while the other is large and will be ignored. Then p » . a where ^ « area of the water film between the plates. h XT A Ml _ Now A = — so F when 9 (the angle of contact) = 0 2.175 This is analogous to the previous problem except that : A = nR" 2nR2a nA - « 0*6 kN So h 2.176 The energy of the liquid between the plates is 2 1 2a 2 a2/ 7 pgd This energy is minimum when, h 2a and 2 a2/ the minimum potential energy is then Exmn = - The force of attraction between the plates can be obtained from this as min 3d 2 a2/ Pg (minus sign means the force is attractive.) Thus Fm _ 13N

246 2.177 Suppose the radius of the bubble is x at some instant. Then the pressure inside is 4a p0 + —. The flow through the capillary is by Poiseuille's equation, Cnr4 4a . idx %x\l x dt Jir4a Integrating —z—j-t« K (R - x ) where we have used the fact thaij t» 0 where x » R. This gives t - — 4 as the life time of the bubble corresponding to x = 0 ar 2.178 If the liquid rises to a height ti, the energy of the liquid column becomes rr 2» h - , 1 /i^a\ 2?i a £« pgnrh- — -2nrha** — pgn\rh-2— This is minimum when rh « and that is relevant height to which water must rise. 98 At this point, 2 jia 2?ia2 Since .E « 0 in the absence of surface tension a heat Q « must have been liberated. Pg 2.179 (a) The free energy per unit area being a, iiad2 - (b) P» 2nad because the soap bubble has two surfaces. Substitution gives F » 10 jaJ 2.180 When two mercury drops each of diameter d merge, the resulting drop has diameter dx where ^d^^dsx2 or, dx- The increase in free energy is AF- n22/sd2a-2nd2a« 2nd2a B"y3- 1) - -1-43 2.181 Work must be done to stretch the soap Him and compress the air inside. The former is simply 2ax4jtl* « SnR a, there being two sides of the film. To get the latter we note that the compression is isothermal and work done is v ■ v 4a) pdV where VQp0« />o + ~£ V m V. i 0 •v; 4 Jt 3 4a Po R and minus sign is needed becaue we are calculating work done on the system. Thus since pV remains constants, the work done is Va £ pVln V Po So A'- ZnR2a+pV\rv£- Po

247 2.182 When heat is given to a soap bubble the temperature of the air inside rises and the bubble expands but unless the bubble bursts, the amount of air inside does not change. Further we shall neglect the variation of the surface tension with temperature. Then from the gas equations Po 4a\ 4 Differentiating or Now from the first law or — r3 « vR 7, v = Constant 3r 4nr dr = vRdT vRdT Po 8a 3r vCvdT+ vRdT 3r C= C 4a r 8a P037 4aN r using 2.183 Consider an infinitesimal Carnot cycle with isothenns at 7- dT and 7. Let A be the work da done during the cycle. Then A« [aG-</7)-aG)]6a« - Where 6a is the change in the area of film (we are considering only one surface). T-dT Then tj = dT fir by Carnot therom. or qoo or q= -T T dT 2.184 As before we can calculate the heat required. It, is taking into account two sides of the soap film da 6ax2 dT Thus da da* Now AF= 2a6a so, At/- AF + TA5- 2|a- r-j=l 5o

248 2.6 PHASE TRANSFORMATIONS 2.185 The condensation takes place at constant pressure and temperature and the work done is pAV where AV is the volume of the condensed vapour in the vapour phase. It is pAV= ^RT = 120-6 J M where M« 18 gm is the molecular weight of water. 2.186 The specific volume of water (the liquid) will be written as Vv Since Vv> > Vlt most of the weight is due to water. Thus if rnl is mass of the liquid and mv that of the vapour then or V- mV, = mv (Vv - V,) So m = — 77T** 20 gm in the present case. Its volume is m V ■ 1*01 V 1 2.187 The volume of the condensed vapour was originally Vo- V at temperature J= 373 K. Its mass will be given by m Mp (Vo - V) P (^o - ^0 ■ Tjr^ or m" Wt ™ ^ ^m w^ere P * atmospheric pressure 2.188 We let Vt« specific volume of liquid. Vv = NVt** specific volume of vapour. Let V= Original volume of the vapour. Then pV V V So (Ar-l)mfV> V 1-- - -(w-1) or T| - --y-1' ^7-7 In the case when the final volume of the substance corresponds to the midpoint of a horizontal portion of the isothermal line in the py v diagram, the final volume must be Yi A + N) — per unit mass of the substance. Of this the volume of the liquid is V^/2 per unit total mass of the substance. Thus T| = — 2.189 From the first law of thermodynamics A17 + j4= Q= mq where q is the specific latent heat of vaporization RT Now A = jp(Vv-V/)/w= m^r M I RT\ Thus A£/= m #- — 1 M t For water this gives « 2*08 x 106 Joules.

249 2.190 Some of the heat used in heating water to the boiling temperature J= 100°C = 373 K The remaining heat » g-mcAJ (c » specific heat of water, AJ« 100 K) is used to create vapour. If the piston rises to a height h then the volume of vapour will be - ^(neglecting water). Its mass will be p0 sh p0 sh Mq x M and heat of vapourization will be —~—. To this must b^ added the work Kl RT done in creating the saturated vapour - posh. Thus ^ RT ( o) quantity of saturated vapour must condense to heat the water to boiling point J= 373°K (Here c » specific heat of water, To « 295 K « initial water temperature). The work done in lowering the piston will then be mc(T-T0) RT q * M RT since work done per unit mass of the condensed vapour is p V » — M —RT 2.192 ^- AD Pv2a Pv 4a M Given AP- --- -x_ - x t,^. r.^ 4aiii or For water a» 73 dynes/cm, M« l»8gm, p,« gm/cc, J* 300K, and with r\~ 001, we get dm 0-2J 2.193 In equilibrium the number of "liquid" molecules evaporoting must equals the number of "vapour" molecules condensing. By kinetic theory, this number is 1 -/8 kT 1 4 Its mass is 1 4 4 ￼v Jt m VkT , „+/ m » r\nkT y -—77 ' °*35 where/?0 is atmospheric pressure and T= 373 K and M = molecular weight of water.

250 2.194 Here we must assume that \l is also the rate at which the tungsten filament loses mass when in an atmosphere of its own vapour at this temperature and that r\ (of the previous problem) •» 1. Then RT - 0-9 nPa M from the previous problem where p = pressure of the saturated vapour. 2.195 From the Vander Waals equation RT a pm V-b" v2 where V=* Volume of one gm mole of the substances. For water V« 18 c.c. per mole « 1-8 x 10" litre per mole a =■ 5'47 atmos litre2 mole2 If molecular attraction vanished the equation will be w RT for the same specific volume. Thus a 5*47 a a A/? = —j = "T5—TIT x 10 atmos •» VI x 10 atmos 2.196 The internal pressure being —j, the work done in condensation is v £L JL This by assumption is Mq, M being the molecular weight and Vx, V being the molar volumes of the liquid and gas. Thus p, = —~ « ^r?-« p q V2 Vi where p is the density of the liquid. For water p» « 3*3 x 10 atm 2.197 The Vandar Waal's equation can be written as (for one mole) RT a At the critical point Mh and BVt T dV 2 vanish. Thus T n RT 2a RT la 0 « n + —r or (V-b? V* (V-bf V 2RT 6 a RT 3a " (V-bK VA °r (V-bK" V4

251 Solving these simultaneously we get on division MCr This is the critical molar volume. Putting this back 2a _ 8a Finally Per From these we see that 4 f b2 lTCr Pc 27 b3 a b v2 VMCr > Vmo> 0T 1Cr 4a " 27b2 a/9b 27 a 9b 3 bR 2 w a 27 b2 RZ Cr Sa/27b 8 2.198 Cr a/27 R?Cr Thus and 8 a/27 b Sb = R k2m 64a 27 Cr 0^082 x 304 73x8 = 0 043 litre/mol Per or a 64 3-59 atm-litre" 2.199 Specific volume is molar volume divided by molecular weight. Thus yMCr 3RTCr 3 x -082x562 litre 471£c Cr~ M " SMpc ' 8x78x47 g * g 2.200 a M (Vm -b)-RT a or ym~b 8 T V, or a Per Vl MCr V - 8 3T' where or When Jt = m , X AfCr Cr 27 fc i 3 8 3 8 12 and v » -, x » - x 24 x - » - 2.201 (a) The ciritical Volume VMCr is the maximum volume in the liquid phase and the mimmum volume in the gaseous. Thus 1000 rpax 18 x 3 x -030 litre - 5 litre

252 (b) The critical pressure is the maximum possible pressure in the vapour phase in equilibrium with liquid phase. Thus 5-47 /W" JiT2~ 27x-03x-03- 22&toM»l*« 2.202 'cr- 21 bR 8 3-62 27 -043 x -082 M 44 Pcr - 36 - 3 x 43 - 304K gm/c.c. » 0-34gm/c.c. 2.203 The vessel is such that either vapour or liquid of mass m occupies it at critical point. Then its volume will be VCr M g The corresponding volume in liquid phase at room temperature is where p = density of liquid ether at room tmeperature. Thus V using the given data (and p ■ 720gm per litre) 2.204 We apply the relation (J» constant) pdV «/ to the cycle 1234531. Here & dS=& dU« 0 So £ pdV= 0 This implies that the areas I and II are equal. This reasoning is inapplicable to the cycle 1231, for example. This cycle is irreversible because it involves the irreversible transition from a single phase to a two-phase state at the point 3. v 2.205 When a portion of supercool water turns into ice some heat is liberated, which should heat it upto ice point. Neglecting the variation of specific heat of water, the fraction of water turning inot ice is clearly where c » specific heat of water and q •» latent heat of fusion of ice, Clearly - 1 at t » - 80°C

253 2.206 From the Claussius-Clapeyron (C-C)equations q12 qui& the specific latent heat absorbed in 1 -* 2 A = solid, 2 = liquid) 273 x -091 . atmxcm3xK _ x I - 333 joule 3 x latmxcm 2L_ « lO, AJ= --0075K Joule Joule 2.207 Here 1 = liquid, 2 = Steam T/f fe 2250 0-9 2.208 From C-C equations dp TV2 Assuming the saturated vapour to be ideal gas jr- g. n- and />« /?0 11 + —^ AJU 1-04 atmosphere 2.209 From C-C equation, neglecting the voolume of the liquid dp <?i2 Mq p Mq dT Now pV= -r;RT or m « ^L for a perfect gas So —» - "zr (V is Const = specific volume)

254 2.210 From C-C equation dp rv Mq Integrating lnp - constant - KT So Po exP This is reasonable for \T - Jo| « To, and far below critical temperature. 2.211 As before B.206) the lowering of melting point is given by -—— p The superheated ice will then melt in part. The fraction that will melt is - —p ~ -03 2.212 (a) The equations of the transition lines are log/? = 9-05 - : Solid gas 1310 6-78 - : Liquid gas At the triple point they intersect Thus 490 or 1 490 2-27 - 216 K corresponding p^ is 5.14 atmosphere. In the formula logp = a-- we compare 6 with the corresponding term in the equation in 2.210. Then ^^ So, 2-303- or, ^sublimation "liquid-gas ~" 2-303 x 1800 x 8-31 44 2-303 x 1310 x 8-31 44 R 783 J/gm = 570J/gm Finally - liquid ~ 213 J/gm on subtraction 2.213 f J m

255 qm 2.214 AS- ^ •H+4-181nif~8-56J/°K 2.215 c « specific heat of copper « 0-39 ^Suppose all ice does not melt, then heat rejected = 90 x 0-39 (90 - 0) - 3159 J heat gained by ice * 50 x 209 x 3 + x x 333 Thus x= 8-5 gm The hypothesis is correct and final temperature will be 7= 273K. Hence change in entropy of copper piece 273 - me In ^ « - 10 J/K 363 2.216 (a) Here t2 = 60°C. Suppose the final temperature is t °C. Then heat lost by water = m2 c (t2 - t) heat gained by ice * mlqm^mlc{t- t^ , if all ice melts In this case ml qm « m2 x 448 F0 - t)t for ml * m2 So the final temperature will be 0°C and only some ice will melt. Then 100 x 448 F0) - m\ x 333 m\ » 75-3 gm « amount of ice that will melt Finally A5- 75- AS = * m2 " m2C 3^ 2/ T2 ~+100x + nt2 c Ti m T2 4- In - 273 181nSf T T2 8-8 J/K (b) If m2ct2>m1 qm then all ice will melt as one can check and the final temperature can be obtained like this m2c G2 - T) = m1 qm^mlc G- Tx) (/W2 72 + m1 Tx) c-m1qm« (m^+ /»2) cT nuqm * 2 + lfIlil" - or 7- — « 280 K m1q ( 7 and AS- -zr- + c rn^n —-m2ln 7 19 J/K

256 2.217 AS- - . where 1 1 « 0-2245 + 0-2564 - 0-48 J/K 2.218 When heat dQ is given to the vapour its temperature will change by dT> pressure by dp and volume by dV, it being assumed that the vapour remains saturated. Then by C-C equation on the other hand, pY « rr M vapour »VUq), or dp RT So pdV + Vdp-^f, Hence ^"^ s (m " f) finally dQ= CdT= dU+pdV' (Cp, Cv refer to unit mass here). Thus For water Cp - —*- • -^ with y * 1-32 and M - 18 So Cp« l-90J/gmK and C» -443J/gm°K- -74J/moleK 2.219 The required entropy change can be calculated along a process in which the water is heated from Tx to T2 and then allowed to evaporate. The entropy change for this is where q « specific latent heat of vaporization.

257 2.1 TRANSPORT PHENOMENA 2.220 (a) The fraction of gas molecules which traverses distances exceeding the mean free path without collision is just the probability to traverse the distance s « X without collision. Thus (b) This probability is 2.221 From the formula P- e -1- I- 0-37 0-23 or X A/ lnr] 2.222 (a) Let P (t) ■ probability of no collision in the interval @,0- Then - P(t)(l-adt) or dt - -a.P(t) or P(t)m e -at where we have used JP @) » 1 (b) The mean interval between collision is also the mean interval of no collision. Then 00 -at te~aidt oo i rB) a T A) a -at dt 0 1223 , x . 1 kT (a) X» -7= 5— - -==— yflnd2n Sin 1-38 273 V2?i@-37xl0-9Jxl05 6-2xlO"8m X 6-2xlO~8 t« a — s= -136 ns > 454 X - 6.2 x 106 m (b) T] ■ 1-36 x 104 s - 3-8 hours 2.224 The mean distance between molecules is of the order 22-4 x 10 -3 6-0 xlO23 x 10 "9 meters » 3-34 x 10 ~9 meters This is about 18.5 times smaller than the mean free path calculated in 2.223 (a) above. 2.225 We know that the Vander Waal's constant b is four times the molecular volume. Thus / 3b Hence ^-^ or d- 6 f kT N A 2/3

258 2.226 The volodty of sound in N2 is so, 2.227 (a) \>l if p< Now <-fc7, for O, ofOis 0-7 Pa. (b) The corresponding n is obtained by dividing by JfcT and is 1-84 x 1020 per 3 14 ^ / m » 1*84 per c.c. and the corresponding mean distance is "TTj. n lO @484I/3 x 10 < - 1*8 x 10" m ~ 048 urn. 5 2^28 (a) v - - -V2xd2n<v>= -74x 1010s"l (see 2.223) (b) Total number of collisions is -nv « l-OxlO^scm Note, the factor — • When two molecules collide we must not count it twice. 1.129 (a) d is a constant and n is a constant for an isochoric process so X is constant for an isochoric process. (b) X» ~t= ^ — aTfor an isobaric process. y/ld2P <v> Vf 1 . v « —r— a -zr ■ = for an isobaric process. 2.230 (a) In an isochoric process X is constant and v a Vf aVpV aVp a Vn kT (b) X» -7= 5— must decrease n times in an isothermal process and v must increase V2d2 n times because <v > is constant in an isothermal process.

2.231 (a) XaJ Thus XaVand But in an adiabatic process |y - —here 7VY " * « constant so TV2''5 •■ constant or (b) Xa- P — ap~Uy But of—I - constant or — ap~Uy or \P) P Thus M. M. xJL ill (c) XaV But TV^5- constant or VaT'5/2 Thus Xa7/2 2.232 In the polytropic process of index n pVH « constant, TV* "l - constant and /?x" R T* - constant (a) KaV 1/2 1-w -»♦! va * V T - V 2 7- 7 2 (b) Xa-, TnapH~l or Tapx'n w p> so Xa/j7* <v> /> i-1+.L 111 —r— a-4-»a/7 2 2 T (c) Xaj, "nl 259

260 2.233 (a) The number of collisions between the ;nolecules in a unit volume is 1 1 A2 2 2 V2 V2 This remains constant in the poly process pV~d « constant Using B.122) the molar specific heat for the polytropic process pVa« constant, is ^l 4j ^2 4 It can also be written as —R A+20 where i - 5 Vr i (b) In this case -jt- - constant and so pV » constant It can also be written as — (i + 1) 2.234 We can assume that all molecules, incident on the hole, leak out. Then, , dt dt or dn- -n A ,„ « -n — 4v/S <v> x Integrating n « w0 e~t/x. Hence <v > = 2.235 If the temperature of the compartment 2 is rj times more than that of compartment 1, it must contain — times less number of molecules since pressure must be the same when the big hole is open. If M » mass of the gas in 1 than the mass of the gas in 2 must be —. So immediately after the big hole is closed. M o M ni o m o ' n mVx\ where m - mass of each molecule and n\, n2 are concentrations in 1 and 2. After the big hole is closed the pressures will differ and concentration will become nl and n2 where M 1 * m Vr\ On the other hand nl <V1 > " n2 <V2 > ^e* nl = ^H n2

261 Thus So ru** ru i 2.236 We know Thus r] changing a times implies T changing a2times. On the other hand TT kT 3 Vjtw V2xd2p Thus 2) changing P times means changing p times So p must change — times 2.237 Da — n (a) Z) will increase n times r) will remain constant if T is constant 3/2 1/2 Thus D will increase n times, t] will increase n times, if/? is constant 2.238 In an adiabatic process TVy~ « constant, or 7« V "^ Now V is decreased — times. Thus H1 --4/5 So D decreases «4/ times and r) increase n times. 2.239 (a) Thus D remains constant in the process pV3 = constant So polytropic index n «■ 3 (b) r\aVfaVpV

262 So r] remains constant in the isothermal process pV= constant, «« 1, here (c) Heat conductivity k ■ r\ Cv and Cy is a constant for the ideal gas Thus n » 1 here also, 2.240 mkT 1 or Jl* 3 x 18-9 xlOc 4 x 8-31 x 273 x 10~3l 1/4 10 2*241 k « — <v > X p cv 3x18 4 x 834 x 273' Jt3 x -36 x 36 x 10 - 0178nm 46 8kT mn yf2nd2n mn v Cvis the specific heat capacity which is — I. Now Cv is the same for all monoatomic M gases such as He and A. Thus tea SMd or 8-7 4 1-658- 1-7 2.242 In this case A-A or 2RAR or To decrease Nv n times T] must be decreased n times. Nqw v\ does not depend on pressure until the pressure is so low that the mean free path equals, say, — A& Then the mean free path is fixed and r\ decreases with pressure. The mean free path equals — A/? when V2nd2n0 (nn ■ concentration)

263 Corresponding pressure is p0 yflkT nd'&R The sought pressure is n times less p- ^ » 707 x 10 -23 nd nAR 10-18xl0 071 Pa The answer is qualitative and depends on the choice — A/? for the mean free path. 2.243 We neglect the moment of inertia of the gas in a shell. Then the moment of friction forces on a unit length of the cylinder must be a constant as a function of r. So, - 3 d(ti t x 2 Jt r ri -7- » iVi or co (r) ■ - 1 1^ and co = 4 JlTj A 1) or T] 2.244 We consider two adjoining layers. The angular velocity gradient is —. So the moment of the frictional force is co h C r -2Krdr"r\r-r nr\a co 2h o 2.245 In the ultrararefied gas we must determine r\ by taking X « — h. Then I 3 r8itr 2 *T 3 2M so, 2.246 Take an infinitesimal section of length dfc and apply Poiseuilles equation to this. Then dV_ -n a4 dp From the formula or dt dt pV- pdV- 8t] 5x RT — 'M RT . naA M pdp 8r)RT dx This equation implies that if the flow is isothermal then p -*- must be a constant and so equals Thus, d2-d2 P2 Pi 2/ in magnitude. naM \p\-p\ \6i\RT T~

264 2.247 Let 7* temperature of the interface. Then heat flowing from left = heat flowing into right in equilibrium. Thus, Ki-V T-T2 U or 7 / rp 'l I * J h \ 2,248 We have or h T 'l >us result Ki 'i M l2 7 n K K2 i + 1 7 r-r2 '2 • K2J2 '2 7,-7, or k /l^2 Kt -K- 2.249 By definition the heat flux (per unit area) is — = -a-T-ln7= constant = ax dx In 71/72 ;— Integrating In where 7X « temperature at the end x * 0 x/7 So 7= 7\ J2] and a In 7/72 2.250 Suppose the chunks have temperatures Tv 72 at time t and 7X - dTx, 72 + aT2 at time dt + t. Thus xdT^ C2dT2 2) where A7« 7^7 Hence A7- where — x KS I

268 2.251 Q « k: dx a* ~3A dx , (A « constant) Thus or using « constant - y ( 1^ - i2 r- r, = 7\ at jc= 0 or r r 3/2 3/2 7 ^ T \3/2 v y -1 J 2/3 2.252 R3/2jT3/2 -1 Then from the previous problem 2//?3/2(J23/2-J13/2) 9x3/2d2VMNAl ' 3 here. 2.253 ^ this pressure and average temparature « 27°C ■ 300K « T ♦ r2) 2330 x 10 ~5m = 23-3mm > > 5-0mm = / The gas is ultrathin and we write X = — / here Then dT / where 1 - 3 1 - 2 MP R x and where < v > . We have used J2 - 7\ « —-— here.

266 dT A 2.254 In equilibrium 2jitk—« -A- constant So 7» B--—In r n dr 2jik But 7=7^ when r« /?x and 7« J2. when r 2i r From this we find 7 « 7< + -— In — 2.255 In equilibrium 4jir k—« -A» constant 4jtk r Using T- T± when r- i?j and T- T2 w^en r T2-Ti R2 2.256 The heat flux vector is - k grad T and its divergence equals w. Thus id/ dJ\ w. t. .. . .. # or r-—=-—-in cylmdncal coordinates. r dr\ dr j k j or i-2>+Alnr- —r 2 Since I is finite at r - 0, A » 0. Also J* Jo at r - R so 5= ro + -^- u 4k Thus J- T0 + -£- u 4k r here is the distance from the axis of wire (axial radius). 2.257 Here again v2r« -~ K So in spherical polar coordinates, t r ■— = -— or r —- -t—r +A d J k dr 3k or 75rr r ok Again A ^= 0 and B = Jo + —R ° u 6k so finally 7= To+ ^-

PART THREE ELECTRODYNAMICS 3.1 CONSTANT ELECTRIC FIELD IN VACUUM 3.1 Fd (for electorns) Thus Similarly 4 K e0 r r (for electrons) 4 n e0 y nC ~19CJ A-602 x 10 "X'C) 4x10 9x10' x 6-67 x 10 " n m3 / (kg • s2) x (9-11 x 10 " 31 kg) — (for proton) 9CJ A-602 x 10-1VC) 1 x 1036 9 x 10' - 11 -27 x 6-67 x 10 "lL mV(kg • sz) x A-672 x 10 " z/ kg) For F^ » F. <L r2 or ■**■■ V4 3X eft y 9x10" 3.2 Total number of atotns in the sphere of mass 1 gm 86 x 10 ~10 C/kg 1 x 6023 x 1023 63-54 ,23 So the total nuclear charge X » —T^Ta x 1*6 x 10~19 x 29 Now the charge on the sphere - Total nuclear charge - Total electronic charge

268 Hence force of interaction between these two spheres, 1 [4-398 x 102]2 9 x 19.348 N . 4jie0 i2 33 Let the balls be deviated by an angle 6, from the vertical when separtion between them equals x. Applying Newton's second law of motion for any one of the sphere, we get, T cos 0 « mg A) B) and JsinG- Fe From the Eqs. A) and B) Fe tanG- — mg But from the figure x x tan 6 2/ as x«l D) or l_ 2/ From Eqs. C) and D) mgx 11 Thus Differentiating Eqn. E) with respect to time 2 Jt £0 mg jt E) dt I dx dt According to the problem — - v - a/Vx (approach velocity is — ) B n e0 mg J so, jr I a dq 3 3.4 Let us choose coordinate axes as shown in the figure and fix three charges, qt, q2 and q3 having position vectors rl9 r2 and r3 respectively. Now, for the equilibrium of q3 - 0

269 or, ft because or, or, Also for the equilibrium of r2""r3 rl""r3 I r2 - r3 I lrl-r3l 0 or, ft = -ft r2""rl Substituting the value of rj we get, -ftft 3.5 When the charge <?0 is placed at the centre of the ring, the wire get stretched and the extra tension, produced in the wire, will balance the electric force due to the charge q0. Let the tension produced in the wire, after placing the charge qo> be J. From Newton's second law in projection form Fl 1 mw n. TdQ- 4ne 0 r 2nr or, 3.6 Sought field strength dQ - (dm)Oy J- 8 jt2 e 0 « 4'5 kV/m on putting the values. 3.7 Let us fix the coordinate system by taking the point of intersection of the diagonals as the origin and let k be directed normally, emerging from the plane of figure. h fild Hence the sought field strength :

270 li + xk -<? t -q lj+xk | q (j) 4 Jl 60 (I2 + X2f/2 4jt60 A2+X2K/2 -4 A,00) Thus £ = 3.8 From the symmetry of the problem the sought field. E= \ dEx W where the projection of field strength along x - axis due to an elemental charge is da cos § qRcosQdQ dE^ - ^ - 4 Jt je/2 Hence £ = jt/2 3.9 From the symmetry of the condition, it is clear that, the field along the normal will be zero i.e. En = 0 and E - El da Now cos 0 But Hence dq = p dx and cos 0 = £ = 0 2 otE = (/2+/?2K/2 and for /» R, the ring behaves like a point charge, reducing the field to the value, 1 q E ~ 4jte0 I2

271 dE For £maY, we should have —rr = 0 max' £1 0 or 0 Thus / = -= and £ V2 max 3.10 The electric potential at a distance x from the given ring is given by, ( \ 4 4 Hence, the field strength along jc-axis (which is the net field strength in *our case), _ 4 1 qx dx 4 n e0 (R 2 + x2 ) 3/2 4?ie 0 1 + — 2 X 3/2 -1 4 Jie 0 3 R2 3 R* Z v O y • ] Neglecting the higher power of R/x, as jc » jR. 3qR2 8?180jc4' Note : Instead of cp (jc), we may write E (x) directly using 3.9 3.11 From the solution of 3.9, the electric field strength due to ring at a point on its axis (say is) at distance x from the centre of the ring is given by : qx E(x)- 2 + x2f/2 And from symmetry E at every point on the axis is directed along the x-axis (Fig.). Let us consider an element (dx) on thread which carries the charge (kdx). The electric force experienced by the element in the field of ring. = 4jieo(/*2+x2K/2 Thus the sought interaction Xqxdx 3/2 On integrating we get, F =

272 3.12 (a) The given charge distribution is shown in Fig. The symmetry of this distribution implies that vector E at the point O is directed to the right, and its magnitude is equal to —> —> the sum of the projection onto the direction of E of vectors dE from elementary charges dq. The projection of vector dE onto vector E is dE cos qp = cos qp, where dq = KR dtp - Xq R cos cp dcp. Integrating A) over cp between 0 and 2 jc we find the magnitude of the vector E: ^0 , P = — I 4 x E0RJ 0 2 j cos cp a cp It should be noted that this integral is evaluated in the most simple way if we take into account that <cos2 qp > = 1/2. Then 2k 1 V cos cp d cp «■ <cos cp> 2 n « jl 0 (b) Take an element S at an azimuthal angle cp from the jc-axis, the element subtending an angle d cp at the centre. The elementary field at P due to the element is Xq cos opdcpR ^ z— alone SP with components 4mo(x2+R2) 6 cos cp d cp R where x { cos 0 along OP, sin 0 along OS } cos 0 (x2 + R 2 The component along OP vanishes on integration as I cos cp d cp « 0 o The component alon OS can be broken into the parts along OX and OY with Xo R 2 cos qp d qp 2 2 3/2 x { cos ^ a^onS OX, sin cp along On integration, the part along OY vanishes. Finally s For x»R E = 4e 0 2 4 Jt e0 x: where p =

273 3.13 (a) It is clear from symmetry considerations that vector E must be directed as shown in the figure. This shows the way of solving this problem : we must find the component dEr of the field created by the element dl of the rod, having the charge dq and then integrate the result over all the elements of the rod. In this case 1 Xdl dE cosa nEo r\ cosa, whereX = ~- is the linear charge density. Let us reduce this equation of the fonn convenient for integration. Figure shows that dl cos a « r0 d a and r0 cos n Consequently, dE X r0 d a 4 jc eor cos a d a r 4 n eo r\ This expression can be easily integrated : dl E = -i 2 I cos a d a = 2 sin an 4 jc tor J 4 ji e0 r u o where a0 is the maximum value of the angle a, sin a Thus [sole o- a , E~ that i NaK? q/2a 2 4 ji e0 r - in this case a Ja2 + r also E 2 4?ti q 4 jceo r2 q '^a2 + r2 for r »t for r » a as of the field of a point charge. (b) Let, us consider the element of length dl at a distance / from the centre of the rod, as shown in the figure. fy Then field at P, due to this element. f i va \ y \dl ^l^ V P 4mo(r-l f if the element lies on the side, shown in the diagram, and dE other side. 4 jc £0 (r + /)' if it lies on Hence E= I dE« I ^-h J J 4jieo(r-/J o o On integrating and putting X » ^-, we get, £ r 4 n e 0 For r» a,

274 3.14 The problem is reduced to finding Ex and E viz. the projections of E in Fig, where it is assumed that X > 0. Let us start with Ex. The contribution to Ex from the charge element of the segment dx is 1 "kdx . n 4jte 0 A) w Let us reduce this expression to the form convenient for integration. In our case, dx= r da/cos a, r = y/cosa. Then X dE. sin a d a. * 4;ieo;y Integrating this expression over a between 0 and Jt/2, we find Ex= X/4jte0y. In order to find the projection Ey it is sufficient to recall that dEv differs from dEx in that sin a in A) is simply replaced by cos a. This gives dE « (X cos a d a )/4 Jt e0 y and E = X/4 Jt e0 y. We have obtained an interesting result : Ex = E independently of y, i.e. E is oriented at the angle of 45° to the rod. The modulus of E is 3.15 (a) Using the solution of 3.14, the net electric field strength at the point O due to straight parts of the thread equals zero. For the curved part (arc) let us derive a general expression i.e. let us calculate the field strength at the centre of arc of radius R and linear charge density X and which subtends angle 90 at the centre. From the symmetry the sought field strength will be directed along the bisector of the angle 0O and is given by V2 -e/2 X (R dQ) 4m0R2 a 2 In our problem 0O = Jt/2, thus the field strength due to the turned part at the point V2 k f . f . _ = — which is also the sought result. dE (b) Using the solution of 3.14 (a), net field strength at O due to stright parts equals V2 2 n tQR and is directed vertically down. Now using the solution of 3.15

275 (a), field strength due to the given curved part (semi-circle) at the point O becomes — and is directed vertically upward. Hence the sought net field strengh becomes zero. 3.16 Given charge distribution on the surface_o = a • r* is shown in the figure. Symmetry of this distribution implies that the sought E at the centre O of the sphere is opposite to a dq= o B x r sin 9) r d 9 = (?• r*) 2 ji r2 sin 9 d 9 - 2 ji a r3 sin 9 co^ d 9 Again frgm symmetry, field strength due to any ring element dE is also opposite to a i.e. dE \\ a. Hence dq r cos 9 4 jt e0 (r sin 9 + r cos 9) B Jt a r3 sin 9 cos 9 d 9) r cos 9 (- a*) sing the result of 3.9) 4 ji e0 r" a - a r 2e sin 9 cos2 d 9 o Thus j ^r f si e0 J o sin 9 cos2 9dQ 7?" a r 2 a r Integrating, we get E * - — j « - — 3.17 We start from two charged spherical balls eadj of radius R with equal and opposite charge densities + p and - p. The centre of the balls are at + — and- — respectively so the z* z* equation of their surfaces are -* a a to be small. The distance between * R or r - -cos9* /? and r + -^089* R, considering the two surfaces in the radial direction at angle 9 is acos9 | and does not depend on the azimuthal angle. It is seen from the diagram that the surface of the sphere has in effect a surface density o = o0 cos9 when o0 = pa. Inside any uniformly charged spherical ball, the field is radial and has the magnitude given by Gauss's theorm 4ji 3 -j-r p/e0 or In vector notation, using the fact the V must be measured from the centre of the ball, we get, for the present case E - P 3e o -*■ a r+2

276 "Or* 0 = 3eo* When E*is the unit vector along the polar axis from which 0 is measured. 3.18 Let us consider an elemental spherical shell of thickness dr. Thus surface charge density of the shell a = p dr = (a*- r*) dr. Thus using the solution of 3.16, field strength due to this sperical shell a r 3 e dr o Hence the sought field strength E= - 3e rdr = - 0 o 6e0* 3.19 From the solution of 3.14 field strength at a perpendicular distance r < R from its left end 4 Jt eor 4 Jt eor Here er is a unit vector along radial direction. Let us consider an elemental surface, dS = dy dz flux of E (r) over the element dS is given by C) dz(r dQ) a figure. Thus E-dS 4 jieor 4 Jt eor (r 4jte drdQ( 0 2x The sought flux, $ = - 7-^— \ dr \ d 9 = - ^. & 4jt£0J J 2e0 0 0 If we have taken dS | T (- i )> then <E> were XR 2e0 Hence 1*1 XR 2e 0 3.20 Let us consider an elemental surface area as shown in the figure. Then flux of the vector E through the elemental area, EdS= EdS dS (r d 9) dr 2ql r dr dQ 4jteo(r2 3/2

277 where Eo= ^ r— is magnitude of 4jteo(/ +r ) field strength due to any point charge at the point of location of considered elemental area. Thus <£> ^J {/+l2f/2 J o o R 2flL 4 lx 2 Jt I rdr ■ / 2 12\ '0 o 1- It can also be solved by considering a ring element or by using solid angle. 3.21 Let us consider a ring element of radius x and thickness dxy as shown in the figure. Now, flux over the considered element, d <£>=£*•dS** E_dS cos 9 ^ ^ But Er = or 0 from Gauss's theorem, r0 and dS * 2nxdx , cos 9 = — r Thus 3e o r P ri J7 0 Hence sought flux V J?2 J2 ■ — I xdx 3e0 J 2;ipro(tf2-r2) p r 0 3e o 3e o 2 -r0) o 3.22 The field at P due to the threads at A and B are both of magnitude and directed along AP and BP. The resultant is along OP with 2 X cos 9 \x n2 /4) '0 2Vx +/ 0 2Vx This is maximum when x = 1/2 and then £ = £, max JC 8 ft

278 3.23 Take a section of the cylinder perpendicular to its axis through the point where the electric field is to be calculated. (All points on the axis are equivalent.) Consider an element S with azimuthal angle cp. The length of the element is R^d(pyR being the radius of cross section of the cylinder. The element itself is a section of an infinite strip. The electric field at O due to this strip is ^- ^^ R ACO o0 cos cp (R dtp) 2ne0R along SO This can be resolved into a0 cos cp d cp 2 Jt e o cos cp along OX towards O sincp along YO On integration the component along YO vanishes. What remains is a0cos a 2;te o o '0 along XO i.e. along the direction cp = jl 3.24 Since the field is axisymmetric (as the field ^\H^&YS&$ ^fe^fcfi lament), we conclude that the flux through the sphere of radius R is equal to the flux through the lateral surface of a cylinder having the same radius and the height 2Rt as arranged in the figure. Now, But Thus E-dS- ErS E -- tr R -5- %2itR-2R- 4naR R R 3.25 (a) Let us consider a sphere of radius r < R then charge, inclosed by the considered sphere, 4nr2 Po(l-j)dr A) o o Now, applying Gauss' theorem, E 4nr2= q'wclosed y (Where E is the projection of electric field along the radial line.) f c or, Po --bUr o 3e o r - 4R

279 And for a point, outside the sphere r > R. R I 4;tr dr p0 11 - — (as there is no charge outside the ball) o \ ' inclosed Again from Gauss' theorem, Er4nr* 4nrdr r_ R 0 or, Po r2e 0 *} R_ 4R Po*: 12 r (b) As-magnitude of electric field decreases with increasing r for r > Ry field will bo maximum for r < R. Now, for Er to be maximum, dr 4R = 0 or 3r or r= rm= — Hence Po* max '0 3.26 Let the charge carried by the sphere be qy then using Gauss' theorem for a spherical surface having radius r > Ry we can write. 2 ^inclosed losed q 1 I a 0 e0 zQJr On integrating we get, 2 fcQ Z, fcQ The intensity E does not depend on r when the experession in the parentheses is equal to zero. Hence 2 q = 2no.R and 2e 0 3.27 Let us consider a spherical layer of radius r and thickness dry having its centre coinciding with the centre of the system. Then using Gauss' theorem for this surface, r2= pdV 0 o ar'4 71 r2dr f o 0

280 After integration 4nr4 3 eoa or, 3 ea r Now when a r < < 1, Er« -— And wnen ar > > 1, Er** I 3 e0 a rA 3.28 Using Gauss theorem we can easily show that the electric field strength within a uniformly charged sphere is E = -^— 13 e0 The cavity, in our problem, may be considered as the superposition of two balls, one with the charge density p and the other with - p. Let P be a point inside the cavity such that its position vector with respect to the centre of cavity be r_ and with respect to the centre of the ball r\. Then from the principle of superposition, field inside the cavity, at an arbitrary point P, 3e, a Note : Obtained expression for E shows that it is valid regardless of the ratio between the radii of the sphere and the distance between their centres. 3.29 Let us consider a cylinderical Gaussian surface of radius r and height h inside an infinitely long charged cylinder with charge density p. Now from Gauss theorem : T? i j ^inclosed (where Er is the field inside the cylinder at a distance r from its axis.) - . pJir h or, Er2nrh= or Er 2e 0 Now, using the method of 3.28 field at a point ', inside the cavity, is F- o a o

281 3.30 The arrangement of the rings are as shown in the figure. Now, potential at the point 1, - potential at 1 due to the ring 1 + potential at 1 due to the ring 2. 4ke0R 4 K eo Similarly, the potential at point 2, CP2" 4^R+^T Hence, the sought potential difference, Q . -q - cp2 * Acp * 2 4ke0R 4 ji e0 (/?2 + a2) 1/2 1- Vl + (a/RJ 3.31 We know from Gauss theorem that the electric field due to an infinietly long straight wire, at a perpendicular distance r from it equals, Er « . So, the work done is / Erdr r (where x is perpendicular distance from the thread by which point 1 is removed from it.) A(Pi2 Hence 2jte o 3.32 Let us consider a ring element as shown in the figure. Then the charge, carried by the element, dq = B Jt R sin 0) R d 0 a, Hence, the potential due to the considered element at the centre of the hemisphere, 1 dq 2koR sin 0a*0 oR . Q ,Q -_» » -—sin 0a 0 2e0 4 Jie o So potential due to the whole hemisphere a/2 oR Ro C . *-2T0J S1 0 sin 0 d 0 2e o Now from the symmetry of the problem, net electric field of the hemisphere is directed towards the negative y-axis. We have a ,_ 1 dqcosQ at - y 4 Jt e0 2 2e sin 0 cos 0 dQ o re/2 re/2 Thus E — fsi 2e0 J Sl o sin 0 cos 6 d a C . = -— I si 4e0J a sin 2 0 d 0 = -^--, along YO 4e o

282 3.33 Let us consider an elementary ring of thickness dy and radius y as shown in the figure. Then potential at a point P, at distance / from the centre of the disc, is olnydy dtp 1/2 Hence potential due to the whole disc, R /olnydy 4*eo(y2W2 From symmetry 2e 0 (VTT {R/iy - E>--dl o 2e o 11 o 0 1- V., (R/l) when / 0, <p « -—, E = -— and when l»R , <P 4e0/ 3.34 By definition, the potential in the case of a surface charge distribution is defined by integral qp ^ I jn order to simplify integration, we shall choose the area element dS 4ji£0J r in the form of a part of the ring of radius r and width dr in (Fig.). Then dS = 29 r dry r - 2R cos 8 and dr ■ - 2/? sin 0 d 0. After substituting these expressions into integral ^p 4 jt e o odS , we obtain the expression for qp at the point O: o » _°A fesin k e0 J We integrate by parts, denoting 0 « a and sin 9 J 9 ) sin 0 d 8 « - 8 cos 0 +/cos 0 J 0 « - 0 cos 0 + sin 0 which gives -1 after substituting the limits of integration. As a result, we obtain cp = oR/ti £o.

233 3.35 In accordance with the problem <p * a - r Thus from the equation : E • - V ~[axi + a j + zk ] = -< -^(axx)i + -(ayy)j + -{axz)k 3.36 (a) Given, <p - a (x2 - So, The sought shape of field lines is as shown in the figure (a) of answersheet assuming a>0: (b) Since qp « axy So, £"= -Vqp* -ayi-axj Plot as shown in the figure (b) of answersheet 3.37 Given, qp - a (x2 + /) + fez2 So, £*« -Vqp= -[2axT+2ayf+2bzT] Hence Shape of the equipotential surface : ^ ^ ^ ry 0 0 Put p* xi+yj or p = jc +>^ Then the equipotential surface has the equation a p + b z ■ constant = qp If a > 0, b > 0 then qp > 0 and the equation of the equipotential surface is qp/a qp/6 which is an ellipse in p, z coordinates. In three dimensions the surface is an ellipsoid of revolution with semi- axis Vcp/a , Vqp/a , Vqp/£. If a > 0, ft < 0 then qp can be fc 0. If qp > 0 then the equation is This is a single cavity hyperboloid of revolution about z axis. If cp = 0 then 0 or is the equation of a right circular cone. If qp < 0 then the equation can be written as - « p2 or |cp|/a This is a two cavity hyperboloid of revolution about z-axis.

284 3.38 From Gauss' theorem intensity at a point, inside the sphere at a distance r from the centre is given by, Er ■^— and outside it, is given by Er » ~. 3 e0 . 4 n e0 r (a) Potential at the centre of the sphere, R 00 00 <Po E-d7 3e o o 0 4 ji e0 r >dr=3s2 '0 as as p 3_q_\ 4nR (b) Now, potential at any point, inside the sphere, at a distance r from it s centre. R 00 i-rdr 3e0 J 4?ie0 r2 On integration : cp (r) 1- <Po 1- 3.39 Let two charges +# and -^ be separated by a distance /. Then electric potential at a point at distance r>> I from this dipole, +£. 4 ji e0 r+ 4 ji e0 r_ 4 ji e( 0 r+ But r _ - r + - From Eqs. A) and B), / cos 9 p cos 9 0 r_ / cos 9 and r+ r_ '0 - r 2 z 4 Jt e0 r 4 Jt e0 r 3 4?te0r where p is magnitude of electric moment vector. xt zr dW 2/? COS 9 Now, Er« - - ^ 4 ji e0 r and dcp /? sin 9 8 4jie0r3 So £= 4?ie0r V4cos29 +sin29 3.40 From the results, obtained in the previous problem, p sin 9 2/? cos 9 —^- 5- and £fi 431 e0r3 e 4 31 From the given figure, it is clear that, Ez = cos 9 - EQ sin 9 4 n e0 r C cos2 9-1) A)

285 and When So jr c • a r a 3/? sin 9 cos 9 EL ■ Er sin 9 + £0 cos 9 = —— x— 4 jc e0 r ELp, EL and 3 cos 9=1 and cos 9 = 9 1 V3 Thus ELp at the points located on the lateral surface of the cone, having its axis, coinciding with the direction of z-axis and semi vertex angle 9 = cos 1 / V3. 3.41 Let us assume that the dipole is at the centre of the one equipotential surface which is spherical (Fig.). On an equipotential surface the net electric field strength along the tangent of it becomes zero. Thus psin 9 - En sin 9 + Ea = 9 or - 2sn sin 9 + '0 o 4 jc e0 r3 = 9 1/3 Hence Alternate : Potential at the point, near the dipole is given by, cp = —-*--—r - EQ • F+ constant, 4 n e0 r '0 cos 9 + Const For cp to be constant, 4 ji e0 r j-E0= 0 or, 4 jc e0 r .1/3 Thus 3.42 Let P be a point, at distace r » I and at an angle to 9 the vector / (Fig.). Thus E at P K r + 2 X 2 jc e0 F+//2 r+2 2JC8 0 2 i ^ r + ■—- + r I cos 9 /•+---/•/ cos 9 4 4 2 jce o 2/r r2 r3 Hence E 2 jc e cos 9 , r »/ 0

286 Also, <P 2 m, r/cos8 X/cosO 2xe0r 3.43 The potential can be calculated by superposition. Choose the plane of the upper ring as x * 1/2 and that of the lower ring as x = - 1/2. Then <P 4*eo[# 4 Jt e 0 (x + //2J V1 4 ji e0 [i?2 + x2 - Ir]172 4 ji e0 [R2 + jc2 + 2 1- bc qlx 4 n e0 (R2 + x2) J72 For The electric field is E - - 4 ■f T- 4jteo(*2+*2K/2 2(i?2+;c2M/2 2 572 For M»/?f ,£:- 4 ji e0 (i?2 + jc2) The plot is as given in the book. 3.44 The field of a pair of oppositely charged sheets with holes can by superposition be reduced to that of a pair of unifosm opposite charged sheets and discs with opposite charges. Now the charged sheets do not contribute any field outside them. Thus using the result of the previous problem (-q) \2nrdrx 4*zo(r2+x2) 2K/2 «2 2 R +x oxl 4 en y3/2 oxl 2e, dcp a I ~ dx" ~ 2 e o (R2 + 3/2 olR The plot is as shown in the answersheet.

287 3.45 For x> 0 we can use the result as given above and write ~ ± al 2 En 1- 1/2 for the solution that vanishes at a. There is a discontinuity in potential for | x\ * 0. The solution for negative x is obtained by o -+ -o. Thus olx 1/2 + constant Hence ignoring the jump olR ►2 ^ v2x3/2 for large [*| cp <■ ± to 2e0 (/T + xl) P j r P and £«- j (where p ** nR o I) 3.46 HereE 0 and a/ (a) /Talong the thread. E does not change as the point of observation is moved along the thread. F- 0 (b) p along rT On 3ieor (c) /Talong 2;te0r2 8 2jte0r2' 3.47 Force on a dipole of moment p is given by, a? 9 a/ In our problem, field, due to a dipole at a distance /, where a dipole is placed, P Hence, the force of interaction, 2jie0/3 ,- 24xlO6N 3.48 -dcp» £• Jr^ a(ydx+xdy)** ad(xy) On integrating, cp = - a xy + C 3.49 - F ^ or, 2axydx-¥a(j^ -y2)dy ad (j^ y) - ay2 dy

288 On integrating, we get, / 2 3 3.50 Given, again - dcp * E • dr=* (ayi + (ax + bz)j + byk)* (dx i + dyj + = a (y dx + ax Jy) + b (z dy + y dz) * ad (x y) + bd (yz) On integrating, cp = - (a xy + b yz) + C 3.51 Field intensity along jc-axis. *" 3 ax2 A) dx Then using Gauss's theorem in differential from e0 so, p (x) = 6a e0 jc. 3.52 In the space between the plates we have the Poisson equation d2q>_ _Po dx2 ~ e0 or, cp« --—jc2+Ajc+5 where p0 is the constant space charge density between the plates. We can choose cp @) = 0 so B = 0 Po ^2 A q> Po Then w (d) = A cp = AJ - -^^ or, A = —-r- + ^ 2g ^ 2 Now £---1^*—^-^-° forx-0 dx e0 if A- ^ + ^-- 0 d 2e0 2 e0 Acp then p0 = eo Also E {d) = 3.53 Field intensity is along radial line and is £ §JE. _ 2 a r A) From the Gauss' theorem, eo, where dq is the charge contained between the sphere of radii r and r + dr. r Hence 4;tr2£r» 4jit2 x (-2ar) - — JV2 p (r')dr' B) 0 o Differentiating B) p = - 6 £0 a

289 3.2 CONDUCTORS AND DIELECTRICS IN AN ELECTRIC FIELD 3.54 When the ball is charged, for the equilibrium of ball, electric force on it must counter balance the excess spring force, exerted, on the ball due to the extension in the spring. or. " ~ ■ k jc, (The force on the charge 4*eoB/J iK q might be considered as arised from attraction by the electrical image) or,# = 4/ Vji e0 k x , sought charge on the sphere. /777T777 im \ ©- 3.55 By definition, the work of this force done upon an elementry displacement dx (Fig.) is given by dA~ 4 Jt £0 Bxy where the expression for the force is obtained with the help of the image method. Integrating this equation over x between / and oo, we find 00 A = - 16 0 dx jc2 l_ 16 n e0 / * 3.56 (a) Using the concept of electrical image, it is clear that the magnitude of the force acting on each charge, 4 n e0 / 4 n e0 • 8jte0/2 (b) Also, from the figure, magnitude of electrical field strength at P 1 \_q_ T 1- n 5V5 I Jteo/2 3.57 Using the concept of electrical image, it is easily seen that the force on the charge q is, _ ^ , hs£ 4jie0 B/J 4 n e0 B V2 Q2 BVI-l)^2 f 0 I I f/77T///// 32 Jt 60 / (It is attractive) -i \L J 1 *

290 3.58 Using the concept of electrical image, force on the dipole p* dE dE F » p —r, where E is field at the location of j?*due to (- or, \F as, \E\ dE_ dl 32*e0/4 z 3*59 To find the surface charge density, we must know the electric field at the point P (Fig.) which is at a distance r from the point O . Using the image mirror method, the field at P, E- 2£cosct« 2 2—5-« 3~—5 4ke0x x 2keoA +r Now from Gauss' theorem the surface charge density on conductor is connected with the electric field near its surface (in vaccum) through the relation a « e0 Eny where En is the projection of E onto the outward normal n^with respect to the conductor). As our field strength E 11 «Tso 3M (a) The force F1 on unit length of the thread is given by where Ex is the field at the thread due to image charge : E, - B1) Thus 4m0l minus singn means that the force is one of attraction. (b) There is an image thread with charge density- X behind the conducting plane. We calculate the electric field on the conductor. It is E(x)-EAx) I I on considering the thread and its image. Thus

291 3.61 (a) AtO, EAO) « 2 00 2ktol So 2*/ /"Kdx 4*eo(*2 + r2) (** + (x2 + r2) (x2 + r2I72 27te0 00 00 4 > on putting .2 2 I Hence a(r)- b0E, 3.62 It can be easily seen that in accordance with the image method, a charge -q must be located on a similar ring but on the other side of the conducting plane. (Fig.) at the same perpendicular distance. From the solution of 3.9 net electric field at O, (- /0 where «*is outward normal with respect to the conducting plane. rt Now Hence a where minus sign indicates that the induced carge is opposite in sign to that of charge q>0. 3/2 -4 3.63 Potential qp is the same for all the points of the sphere. Thus we calculate its value at the centre O of the sphere. Thus we can calculate its value at the centre O of the sphere, because only for this point, it can be calculated in the most simple way.

292 where the first term is the potential of the charge qy while the second is the potential due to the charges induced on the surface of the sphere. But since all induced charges are at the same distance equal to the radius of the circle from the point C and the total induced charge is equal to zero, cp' * 0, as well. Thus equation A) is reduced to the form, <P 4JI£O / 3.64 As the sphere has conducting layers, charge -q is induced on the inner surface of the sphere q and consequently charge + q is induced on the outer layer as the sphere as a whole is uncharged. Hence, the potential at O is given by, <Po = +4isU 4 n e0 r 4 n e0 R1 4m0R2 It should be noticed that the potential can be found in such a simple way only at 0, since all the induced charges are at the same distance from this point, and their distribution, (which is unknown to us), does not play any role. 3.65 Potential at the inside sphere, V a 4 ji e0 a 4m0b Obviously When %m 4,i And when r £ b = 0 for b_ a by = 1 - - / r, using Eq. A). 3ie0r 4?ie0l aj j ^ iw 4jie0r 4 Jt 60 1 r a 3.66 (a) As the metallic plates 1 and 4 are isolated and conncted by means of a conductor, (Pi = cp4. Plates 2 and 3 have the same amount of positive and negative charges and due to induction, plates 1 and 4 are respectively negatively and positively charged and in addition to it all the four plates are located a small but at equal distance d relative to each

293 other, the magnitude of electric field strength between 1-2 and 3 - 4 are both equal in magnitude and direction (say E). Let E' be the field strength between the plates 2 and 3, which is directed form 2 to 3. Hence E \ \ E (Fig.). According to the problem E'd- Acp- cp2-cp3 A) o In addition to T 1 T4 or, 0- - Ed + Acp -Ed or, Acp* 2Ed or E Hence Acp 2d B) (b) Since E a a, we can state that according to equation B) for part (a) the charge on the pla^te 2 is divided into two parts; such that 1 / 3 rd of it lies on the upper side and 2 / 3 rd on its lower face. Thus charge density of upper face of plate 2 or of plate 1 or plate 4 and lower face of e0 Acp - and charge density of lower face of 2 or upper face of 3 3a = eo£ a' » e0 E » e( Acp 3 e0 Acp Hence the net charge density of plate 2 or 3 becomes o + o'' - ^r~> which is obvious 2d from the argument 3.67 The problem of point charge between two conducting planes is more easily tackled (if we want only the total charge induced on the planes) if we replace the point charge by a uniformly charged plane sheet Let a be the charge density on this sheet and Ex, E2 outward electric field on the two sides of this sheet. i lien i-t ■% ~ "*^9 The conducting planes will be assumed to be grounded. Then Ex x = E2(l- *). Hence — (/ - x), E2 = -—x ( ), 2 e0 / e0 This means that the induced charge density on the plane conductors are (f) - -jx Hence - ^(/ -x\ q2** - y x

294 3.68 Near the conductor E - £« — This field can be written as the sum of two parts Ex and E2. Ex is the electric field due to an infinitesimal area dS. Very near it a 27 The remaining part contributes E2 o 27 on o both sides. In calculating the force on the element dS we drop E1 (because it is a self-force.) Thus .2 dF_ dS a 27 a 0 3.69 The total force on the hemisphere is n/2 2 S O 2e cosQ-lnRs'mQRdQ o o n/2 *R2o2 C 2e0 J cos 9 sin 8 d G o X2X 4__\ 2* 3.70 We know that the force acting on the area element dS of a conductor is, d?= \oldS A) It follows from symmetry considerations that the resultant force F is directed along thez-axis, and hence it can be represented as the sum (integral) of the projection of elementary forces A) onto the z-axis : dFz - dF cos 9 B) For simplicity let us consider an element area dS = 2nRsinQRdG(Fig.). Now considering that E * a/e0. Equation B) takes the from 71 O2R2 . f » sin t '0 cos3 G J cos G

295 Integrating this expression over the half sphere (i.e. with respect to cos 0 between 1 and 0), we obtain F' ■ F = —~— 4e0 nop 3.71 The total polarization isP ■= (e - 1) e0 JE. This must equals —rrr- where nois the concemtation of water molecules. Thus 2-93 x 10 on putting the values 3.72 From the general formula E « (-, where r * / and r* f f 4jte0 /3> This will cause the induction of a dipole moment. Thus the force, /3 a/4^e0 /3 " 4jt2e0/7 3.73 The electric field E at distance x from the centre of the ring is, 4 Jt e0 (FT + jO The induced dipole moment is p** fizQE - 4 Jt (tf2 + jc2) The force on this molecule is ± R This vanishes for x ■ "T^ (apart from * ■ 0, jc « oo) It is maximum when = 0 or, (/?2-2jc2)(/?2+x2)-4x!2(^2 + x2)-8jc2(/?2-2jc2)- 0 or, /*4-13*2/*2 + 10x4= 0 or, *2= ^-(l3± Vl29 \ ^—V13 ± Vl29 (on either side), Plot of Fx{x) is as shown in the answersheet.

296 3.74 Inside the ball Also Also, 3.75 *= D or q = - fn ~p si c* conductor e - e j or, 1 JL 4jc / e - 1 q r e-1 a o e-1 a a » - P a This is the surface density of bound charges. 3.76 From the solution of the previous problem q'- charge on the interior surface of the conductor fodS= -5_! Since the dielectric as a whole is neutral there must be a total charge equal to e-1 outer = + q on the outer surface of the dielectric. 3.77 (a) Positive extraneous charge is distributed uniformly over the internal surface layer. Let a0 be the surface density of the charge. Clearly, E = 0, for r<a For a < r e0 E x 4 Jt r2 = 4 Jt a2 a0 by Gauss theorem. eoe I r or, For r > by similarly r> b Now, So by integration from infinity where qp (oo) = 0, eor

297 a <r <b cp eer , B is a constant a0a2/i i\ a0a2 or by continuity, cp« 1 — — —I + tot [r b) For r < a. cp * A * Constant , a<r<b By continuity, cp o0a eoe [a b) eob (b) Positive extraneous charge is distributed uniformly over the internal volume of the dielectric Let p0 « Volume density of the charge in the dielectric, for a < r < b. £-0, r<a e0 e 4 — (r3 - a3) p0, ( a < r < b ) or, Po — 3 e0 e or. By integration, r , " a ) Po c E » =— for r > 3eor2 -« JPo 3e0r for r or, qp» J5- Po 3e0e By continuity or. b3-a po=B- Po 3efte o 62 a3^ B Po 3e0e -«3) a 3V Finally Po 3 e0 e I 2 2 e0 e ,r<a On the basis of obtained expressions E (r) and (cp) (r) can be plotted as shown in the answer-sheet.

298 3.78 Let the field in the dielectric be E making an angle a with nTihen we have the boundary conditions, Eo cos a0 = e E cos a and Eo sin c0 = £ sin a So E = Eo V sin a0 + -^ cos2 a0 and tan a « e tan a0 e In the dielectric the normal component of the induction vector is Dn = a= e0£0cosa0 e0 Eo cos a0 or, a = e0£0cosa0 e-1 3.79 From the previous problem,a' = e0 -—— Eo cos 8 n Bo ^ Length I (a) 6 (b) D (eo£osin0 - e 60JE0sin9) = - (e - 1) eoEo /sin 9 — dDx 3.80 (a) divZ)= —— = p and D = p/ dx e e , l<d and o and <p(x) 0 constant for l> d , / > ^ then cp (jc) 0 0 2e by continuity. On the basis of obtained expressions Ex (x) and cp (jc) can be plotted as shown in the figure of answersheet.

(b) p' - - div P = - div (e -1) eo£ = - p (e-D o' - Pin-p2n> wnere n *s the normal from 1 to 2. - Pln, (P2 - 0 as 2 is vacuum.) 299 3.81 I d 2 '/■" P A = 0 as Dr rt oo at r - 0 /Thus, 3 e e o For By continuity of Dr at r = /? ; B so, 3e0r> C * + and cp = - 2 6 e e o 5+ r 3 e0 6 e e0 , by continuity of op. See answer sheet for graphs of E (r) and cp (r) -♦ 1 ^ f r3 0>) p' 3.82 Because there is a discontinuity in polarization at the boundary of the dielectric disc, a bound surface charge appears, which is the source of the electric field inside and outside the disc. We have for the electric field at the origin. Jo'dS ^ 4 71 e0 r3 where r « radius vector to the origin from the element dS.

300 a' « P » P cos 8 on the curved surface n (Pn =0on the flat surface.) Here 0 « angle between r and P By symmetry, E will be parallel to P. Thus 2* Em - Pcos8i? 8 o where, r » /? if d « /?. So, Em - and Pd 4e0R 3.83. Since there are no free extraneous charges anywhere -> dDx divD« —-- 0 or, £> - Constant dx x But D' - 0 at oo , so, D » 0, every where. Thus, £-3 1 d2 So, <P + constant Hence, cp (+ d) - cp (- d) 2PQd 4P0d 3.84 (a) We have Dx * D2> or, e Also, or> Hence, e + 1 and El o and e + 1 (b) Dj » D2> or, e o o Thus, — and eo£o

301 3.85 (a) Constant voltage acros the plates; 3.86 (b) Constant charge across the plates; Ex(l 0 or El At the interface of the dielectric and vacuum, Elt The electric field must be radial and A Elm E2 ,a<r<b eoer Now, q 4 ^ e or, cD o e + 1 3.87 In air the forces are as shown. In AT-oil, F'-* F'« F/z and mg->mg 1- Po Since the inclinations do not change Po — si — e p or, or, Po P P- Po T where p0 is the density of K-oil and p that of the material of which the balls are made. 3.88 Within the ball the electric field can be resolved into normal and tangential components. En m Eco&B,Et* E sin 9 Then, Dn ■ tt0E cos 9 and e0£cos8 or, a'» (e - 1) e0E cos 9 so> and total charge of one sign,

302 J (e- tzR2e0 (e - \)E o (Since we are interested in the total charge of one sign we must intergrate co§ 6 from 0 to 1 only). 3.89 The charge is at A in the medium 1 and has an image point at A' in the medium 2. The electric field in the medium 1 is due to the actual charge q at A and the image charge q' at A'. The electric field in 2 is due to a corrected charge c[ at A. Thus on the boundary between 1 and 2, In COS 0 - 4 neor COS 0 '2n A 2 4 Jieor it , 2 4 Jt eor cos 0 sin 0 + 4?ie sul 0 sin 0 The boundary conditions are and E u *q"m q-q' q"-q + q' So, .it e-1 (a) The surface density of the bound charge on the surface of the dielectric 2/i e-1 cos 0 » - 2;ir 00 Total bound charge . e-1 f I 1S>-—[?J 23C(/2 + o 3/2 £ + l 3.90 The force on the pdint charge q is due to the bound charges. This can be calculated from the field at this charge after extracting out the self field. This image field is e -1 q E: imagc + l4;ieoB0: Thus, e-1 16jte0/2

303 3.91 E. q'r2 4ttp r 4 ir »• * 4 * , P in 2 where —-* e + 1 '» a" -a In the limit / -> 0 4 jt e0 r 2 ji e0 A + e) r j, in either part Thus, £. 2 Jt e0 A + e) r2 2 Jt e0 A + e) r D 2 Jt e0 A + e) r2 x « 1 in vacuum e in dielectric 3.92 E 4 jt e0 e r2 4 Jt e0 r{ i-^; P in 2 p 4 Jt e0 r23 ; P in 1 Using the boundary conditions, 2t This implies q-E q' = q" and g + e q' » e So, Then, as earlier', 1 e +1 e 3.93 To calculate the electric field, first we note that an image charge will be needed to ensure that the electric field on the metal boundary is normal to the surface.

304 The image charge must have magnitude - so that the tangential component of the electric field may vanish. Now, '0 _2J 2 cos 6 2 n e0 e r3 Then P. This is the density of bound charge on the surface. 3.94 Since the condenser plates are connected, E1h+E2{d-h)= 0 3.95 3.96 and or, o Ph 1 = £0E2 &2 Thus, E2d- —« 0, or, E2 8 Ph o 1 i d Given P« ar^ where r^ distance from the axis. The space density of charges is given by, p' = - div P m - 2a On using. div 7*= — —(r^r^1" 2 In a uniformly charged sphere, 7* Po -+ 3e or, o 3e o The total electric field is 3e o Por~ p0 where p 6 r « - P (dipole moment is defined with its direction being from the -ve charge to +ve charge.) The potential outside is _+ cp = 4 ji e0 I r - 6r| I' 4 ji e0 r3 ' 4jc ^ » where p0 » - — /? p0 dr is the total dipole moment.

305 3.97 The electric field Eo in a spherical cavity ki a uniform dielectric of permittivity e is related to the far away field E, in the following mannerjmagine the cavity to be filled up with the dielectric. Then there will be a uniform field E everywhere and a polarization P, given by, Now take out the sphere making the cavity, the_£lectric field inside the sphere will be P 3e 0 By superposition. Eo - -— « E 3e0 '///777, 3.98 By superposition the field E inside the baj^ is given by -* -* P 3.99 0 On the other hand, if the sphere is not too small, the macroscopic equation P = (e - 1) ZqE must hold. Thus, £|l + |(e-l)|-£0 or, 3E, Also This is to be handled by the same trick as in 3.96. We have effectively a two dimensional situation. For a uniform cylinder full of charge with charge density p0 (charge per unit volume), the electric field E at an inside point is along the (cylindrical) radius vector 7 and equal to, 7* 1 — o hence, Er - -— r Therefore the polarized cylinder can be thought of as two equal and opposite charge dis- distributions displaced with respect to each other 1 ^1 ,-+ ^ 1 ,— ? Since P = - p 6 r. (direction of electric dipole moment vector being from the negative charge to positive charge .) 3.100 As in 3.98, we write F= £^ - — 2e0 using here the result of the foregoing problem. Also So, E\—r— I- £0, or, £ and P-

306 3.3 ELECTRIC CAPACITANCE ENERGY OF AN ELECTRIC FIELD 3.101 Let us mentally impart a charge q on the conductor, then 00 H 2dr+ \ 1 2dr 0er J 4jie0r R, 4 Jieoe 4 Jteoe Hence the sought capacitance, q q 4 n e0 e 4xz0R2 C = 4 n e0 e R 3.102 From the symmetry of the problem, the voltage across each capacitor, Aqp « \/2 and charge on each capacitor q « C § / 2 in the absence of dielectric. Now when the dielectric is Glled up in one of the capacitors, the equivalent capacitance of the system, and the potential difference across the capacitor, which is filled with dielectric, Ce A(p' But m eC A + e) Ce A + e) tpaE 1 So, as cp decreases t- A + e) times, the field strength also decreases by the same factor and flow of charge,Aq « q' - q A + e) _£ l 25 2 (e + 1) 3.103 (a) As it is series combination of two capacitors, or, C eo5 K* Cq Gj J On Co (b) Let, a be the initial surface charge density, then density of bound charge on the boundary plane. o'-all--=-|-o 1 E2 / » a

307 But, a- cv s So, 3.104 (a) We point the jt-axis lowards right and place the origin on the left hand side plate. The left plate is assumed to be positively charged. Since e varies linearly, we can write, z(x) = a + bx where a and b can be determined from the boundary condition. We have e = ex at x = 0 and e = e2 at x = dy Thus, e2- 1 Now potential difference between the plates d d qp. — qp_ = I E-dr= I —dx o I a od o e o e2- In — Hence, the sought capacitance,C = - SL _ oS + — cp__ In and the space density of bound charges is Sde2(x) 3.105 Let, us mentally impart a charge q to the conductor. Now potential difference between the plates, cp+ - cp_ * I E - dr _ I ._— m £X IT P Hence, the sought capacitance, ^ In 4nz0a In /

300 3.106 Let X be the linear chaige density then, E m At Jt a /\i £i and, '2m (i) B) t oq z\2 c2 The breakdown in either case will occur at the smaller value of r for a simultaneous breakdown of both dielectrics. From A) and B) 2> fe the sought relationship. 3.107 Let, X be the linear charge density then, the sought potential difference, /X f X 2 x e0 e2 r J 2 3i e0 e2 r ih Now, as, E1R1e< E2 R2 e2> so + — In R3/R2 82 is the maximum acceptable value, and for values greater than E1 Rx will take place, Henctf, the maximum potential difference between the plates, EXRX l " ~ ~ 1 34QS Let us suppose that linear charge density of the wires be X then, the potential difference, V+ - "V- " <P " (" V)" 2 cp. The intensity of the electric field created by one of the wires at a distance x from its axis can be easily found with the help of the Gauss's theorem, X X f b-a In b-a Then, cp - I E dx a Hence, capacitance, per unit length, a lnfe/a dielectric breakdowB

309 3.109 The field in the region between the conducting plane and the wire can ht obtained by using an oppositely charged wire as an image on the other side. Then the potential difference between the wire and the plane, Acp « I E - dr b I 2 ji e0 r 2 n e0 Bb - r) dr 2jte0 2 Jt e0 In a 2 Jt e, 26-a In o 26 -a a In —, as b» a 2 Jt e0 a Hence, the sought mutual capacitance of the system per unit length of the wire 2 Tit o Acp In 26/a 3.110 When 6 » a, the charge distribution on each spherical conductor is practically unaffected by the presence of the other conductor. Then, the potential cp+ (cp_ ) on the positive (respectively negative) charged conductor is q I Thus cp+ - and C 2jte0ea. Note : if we require terms which depend on —, we have to take account of distribution of charge on the conductors. 3.111 As in 3.109 we apply the method of image. Then the potentical difference between the +vely charged sphere and the conducting plane is one half the nominal potential difference between the sphere and its image and is 1 Thus C = Acp . for a.

310 3.112 <=> (a) Since <px = <pB and cp2 - <pA The arrangement of capacitors shown in the problem is equivalent to the arrangement shown in the Fig. c c c and hence the capacitance between A and B is, C= (B) From the symmetry of the problem, there is no P.d. between D and E.. So, the combination reduces to a simple arrangement shown in the Fig and hence the net capacitance, '0 C C y+2- c 3.113 (a) In the given arrangement, we have three E0S capacitors of equal capacitance C = —r~ and the first and third plates are at the same potential. Hence, we can resolve the network into a simple form using series and parallel grouping of capacitors, as shown in the figure. Thus the equivalent capacitance 0 2 ~ 3

311 (b) Let us mentally impart the charges +q and -q to the plates 1 and 2 and then distribute them to other plates using charge conservation and electric induction. (Fig.)* As the potential difference between the plates 1 and 2 is zero, ft ft (wherc or, ?2 = 2ql9 The potential difference between A and B, Hence the sought capacitane, 3eo5 0 cp q2/C 2qx/C 2 2d 3.114 Amount of charge, that the capacitor of capacitance Cx can withstand, q1 = Cx V1 and similarly the charge, that the capacitor of capacitance C2 can withstand, q2 = C2 V2. But in series combination, charge on both the capacitors will be same, so, qm&x> that the combination can withstand = Cx Vv as C1 V1 < C2 V2, from the numerical data, given. Now, net capacitance of the system, and hence, V. r "max max •-» / ft Co- 1 L J c c C1U2 ~=r = 9kV 3.115 Let us distribute the charges, as shown in the figure. Now, we know that in a closed circuit, - Acp * 0 -So, in the loop, DCFED, ft or, Again in the loop DGHED, A 1^ W ^2 A) B) D

312 Using Eqs. A) and B), we get Now, 1 2 or, 3C + 1 T| + 3 T| + 1 10 V 3.11* The infinite circuit, may be reduced to the circuit, shown in the Fig. where, Co is the net capacitance of the combination. 1 1 1 C C + Cn C C 0 Solving the quadratic, we get, o C, taking only +ve value as Co can not be negative. 3.117 Let, us make the charge distribution, as shown in the figure. C2 Now, or, Hence, voltage across the capacitor and voltage ao-oss the capacitor, C2 C,= 5 V 3.118 Let |j > \ly then using - Acp - 0 in the closed circuit, (Fig.) -ff - ff Cx C2 or, D

313 Hence the RD. accross the left and right plates of capacitors, q and similarly .d + C 3.119 Taking benefit of trie foregoing problem, the amount of charge on each capacitor 3.120 Make the charge distribution, as shown in the figure. In the circuit, 12561. - Acp = 0 yields - 4 c c -S- 0 or, and in the circuit 13461, or, o C, Now - cpB _ fc c2 c3 - x + C2) (C3 + C4) It becomes zero, when (C2C3-C1C4)= 0. or 3.121 Let, the charge ^ flows through the connecting wires, then at the state of equilibrium, charge distribution will be as shown in the Fig. In the closed circuit 12341, using -Aqp= 0 -HCV-4) or, q = = 006 mC C, (l/Cx + 1/C2 + 3.122 Initially, charge on the capacitor Cl or C2, —, as they are in series combination (Fig.-a) 12

314 when the switch is closed, in the circuit CDEFC from - Acp * 0, /(Fig. b ) -pr- 0 or And in the closed loop BCFAB from - Acp = 0 Cx C2 A) B) J+1o Cr I From A) and B) qx = 0 Now, charge flown through section 1 « and charge flown through section 2= - q1- q = - 3.123 When the switch is open, (Fig-a) ,1 St- ■in I and when the switch is closed, a, = £ C, and q0 = ^ C, Hence, the flow of charge, due to the shortening of switch, \ Ci - C2 through section 1 = qx- qo= ? C} ^ — = - 24 u C through the section 2 = - q2- (%) = - C = - 36 and through the section 3 = q2-(q2- qj - 0 = ^ (C2 - Cx) = - 60 lH, ll - =

315 3.124 First of all, make the charge distribution, as shown in the figure. In the loop 12341, using - Acp » 0 -0 A) Similarly, in the loop 61456, using - Acp - 0 C -I2-O B) From Eqs. A) and B) we have — +1 Hence, 3.125 In the loop ABDEA, using - Aqp = + Qi - 0 0 A) Similarly in the loop ODEF, O -=r- 0 B) f Solving Eqs. A) and B), we get, C2 " Cl C1 Now, (Pi - cp0 = <Pi - - —-p, , as (cp0 = 0) 4 8 1 f 3 0 And using the symmetry, cp2 = and The answers have wrong sign in the book. 3.126 Taking the advantage of symmetry of the problem charge distribution may be made, as shown in the figure. In the loop, 12561, - Acp - 0

316 or C C or <7i_ C1 (C3 + C2) q2 ^ 2 v 1 3/ A) Now, capacitance of the network, q2 C B) From Eqs. A) and B) 1 2 Cx C2 + C3 (Cx + C2) Ci + C2 + 2 C3 3.127 (a) Interaction energy of any two point charges qx and q2 is given by is the separation between the charges. 4 jc e0 r where r n a L I i i i H Hence, interaction energy of the system, n2 n2 2 4 ji e0 a 4 ji e0 4 Jc e0 a 4 jc e0 (V2 and - 2 3teoa 4jieo« 4xeo(V2a) Vlq2 4 jc e0 a

317 3.128 As the chain is of infinite length any two charge of same sign will occur symmetrically to any other charge of opposite sign. So, interaction energy of each charge with all the others, .111 U= -2 But and putting x - 1 we get In 2 up to up to oo 00 A) From Eqs. A) and B), U 1 1 1 l-2 + l --2<72ln2 4 Jieoa up to 00 B) 3.129 Using electrical image method, interaction energy of the charge q with those induced on the plane. U 4 Jt e0 B1) 8 Jt e0 / 3.130 Consider the interaction energy of one of the balls (say 1) and thin spherical shell of the other. This interaction energy can be written as fdyq - f . ^ J 4jie J (r) q1 r2 sin 0 d 0 dr 2eo(/ 2lr cos 0) 1/2 Ur ^—jdr \ dxp2(r) l-r :4nr2drp2(r) Jt e0 / ri Hence finally integrating 00 0 J 4n 3.131 Charge contained in the capacitor of capacitance C1 is q in it : x qp and the energy, stored 2C1 2 Now, when the capacitors are connected in parallel, equivalent capacitance of the system, C = Cx + C2 and hence, energy stored in the system :

318 U. c,V 2 (C + C \ V 1 1/ So, increment in the energy, ci «P2 / 1 , as charge remains conserved during the process. AU 1 ^ c l 2 (Cx + C2) - 0-03 mJ 3.132 The charge on the condensers in position 1 are as shown. Here q and Hence, + %) o o c + c 1} c or, o C0 + 2C o CO) After the switch is thrown to position 2, the charges change as shown in (Fig-b). A charge q0 has flown in the right loop through the two condensers and a charge q0 through the cell, Because of the symmetry of the problem there is no change in the energy stored in the condensers. Thus H (Heat produced) = Energy delivered by the cell C0 + 2C 3.133 Initially, the charge on the right plate orthe capacitor, q= C (^ -%2) and finally, when switched to the position, 2. charge on the same plate of capacitor ; So, ^ Now, from energy conservation, AU + Heat liberated » A ^^ where AC/ is the electrical energy.

319 + Heat liberated = as only the cell with e.m.f. ^ is responsible for redistribution of the charge. So, 1 o C 6, So - - C &; + Heat liberated - C 1 2 Hence heat liberated » -C^ 3.134 Self energy of each shell is given by ^p, where cp is the potential of the shell, created only by the charge q, on it. Hence, self energy of the shells 1 and 2 are : The interaction energy between the charged shells equals charge q of one shell, multiplied by the potential cp, created by other shell, at the point of location of charge q. So, - fli Hence, total enegy of the system, 4 Jie o 2R, 3.135 Electric fields inside and outside the sphere with the help of Gauss theorem : Sought self energy of the ball 00 0 Hence, U 4nto5R and 3.136 (a) By the expression I — e0sE2dV= I — se0E24nr2dr, for a spherical layer. To find the electrostatic energy inside the dielectric layer, we have to integrate the upper expression in the limit [a, b] b .2 u 1 f 2e°eJ 4 n £0 e 4 Jir2dr i 8 Jt e0 e 1.1 a b 27 mJ

320 3.137 As the field is conservative total work done by the field force, 2 4jte 0 1 R 8jte O 3.138 Initially, energy of the system, Ui s ^ + W12 where, Wx is the self energy and W12 is the mutual energy. So, «'-\ <l% and on expansion, energy of the system, Z, ^r JC £q ■t\<j ^r JC £i Now, work done by the field force, A equals the decrement in the electrical energy, q (q0 + q/2) ( \ \\ i.e. A- 4 jc e 0 Alternate : The work of electric forces is equal to the decrease in electric energy of the system, A = U--Uf In order to find the difference Ui - Up we note that upon expansion of the shell, the electric field and hence the energy localized in it, changed only in the hatched spherical layer consequently (Fig.)» h- vf 2 where Ex and E2 are the field intensities (in the hatched region at a distance r from the centre of the system) before and after the expansion of the shell. By using Gauss' theorem, we find 1 4 jc e< and o r As a result of integration, we obtain 4 jc e0 r2 4 jc e 0

321 3.139 Energy of the charged sphere of radius r, from the equation Um i !„__£ q2 If the radius of the shell changes by dr then work done is 4Tcr2Fudr = - dU = ^/SnE0r2 Thus sought force per unit area, D n M2 . a 4 Jt r (8 ji e0 r ) 4;irx8jte0r 2 e0 3.140 Initially, there will be induced charges of magnitude -q and +# on the inner and outer surface of the spherical layer respectively. Hence, the total electrical energy of the system is the sum of self energies of spherical shells, having radii a and b, and their mutual energies including the point charge q. -qq qq -qq or 8 ;o 1 1 Finally, charge q is at infinity hence, Ur= 0 Now, work done by the agent = increment in the energy urut 8jie o n a 3.141 (a) Sought work is equivalent to the work performed against the electric field created by one plate, holding at rest and to bring the other plate away. Therefore the required work, A agent where E ■ -— is the intensity of the field created by one plate at the location of other. Z E 0 So, a Alternate : A ext = AU (as field is potential) 2eo5 £* Cr» kJ ^O Crv kJ (b) When voltage is kept const., the force acing on each plate of capacitor will depend on the distance between the plates. So, elementary work done by agent, in its displacement over a distance dxy relative to the other, dA= -F^dx But, S o(x) and o(x) = z0V Hence, 2 OX — x1 x2

322 Alternate : From energy Conservation, or 1 2 0 0 V2+A agent So eo5V2 agent *2 ■ m 3.142 (a) When metal plate of thickness r\d is inserted inside the capacitor, capacitance of the system becomes Cn - 0 Now, initially, charge on the capacitor, qQ « Co V - 7; r a A - r|) Finally, capacitance of the capacitor, C « As the source is disconnected, charge on the plates will remain same during the process. Now, from energy conservation, - U: = A . (as cell does no work) 2 0 or, ~ ^_ *~ ~ ^— ■* a agent 2 2 C 2C 0 agent Hence A agent n _q-T1)i c c 2 (i - mJ (b) Initially, capacitance of the system is given by, Ce Ce r (this is the capacitance of two capacitors in series) -e) + ev r r ' So, charge on the plate, q0 ■ C0V Capacitance of the capacitor, after the glass plate has been removed equals C From energy conservation, c c, 1 C V2 e ti (e - 1) 2[e-r1(e-l)]2 0-8 mJ 3.143 When the capactior which is immersed in water is connected to a constant voltage source, it gets charged. Suppose a0 is the free charge density on the condenser plates. Because water is a dielectric, bound charges also appear in it Let a' be the surface density of bound charges. (Because of homogeneity of the medium and uniformity of the field when we ignore edge effects no volume density of bound charges exists.) The electric field due II fT to free charges only —; that due to bound charges is — and the total electric field is a 0 ee, . Recalling that the sign of bound charges is opposite of the free charges, we have

323 ao ao a' , / e -1 \ — = or, a = a0 ee0 eo eo I e / Because of the field that exists due to the free charges (not the total field; the field due to the bound charges must be excluded for this purpose as they only give rise to self energy effects), there is a force attracting the bound charges to the near by plates. This force is 1 ,ao (e - 1) a02 ttO — =* -^ u per unit area. 2 e0 2ee0 The factor — needs an explanation. Normally the force on a test charge is qE in an electric field E. But if the charge itself is produced by the electric filed then the force must be constructed bit by bit and is ■I q{ E') dE o if q(E') <xE ' then we get This factor of — is well known. For example the energy of a dipole of moment p in an electric field Eo is -p -EQ while the energy per unit volume of a linear dielectric in an electric field is - — P • Eo where P is the polarization vector (i.e. dipole moment per unit volume). Now the force per unit area manifests ifself as excess pressure of the liquid. V °o Noting that -r = — d 660 eoe(e - 1)V2 We get A p = * 2? Substitution, using e = 81 for water, gives Ap - 7-17 k Pa ■ 0-07 atm. 3.144 One way of doing this problem will be exactly as in the previous case so let us try an alternative method based on energy. Suppose the liquid rises by a distance h. Then let us calculate the extra energy of the liquid as a sum of polarization energy and the ordinary gravitiational energy. The latter is If a is the free charge surface density on the plate, the bound charge density is, from the previous problem, a = a This is also the volume density of induced dipole moment i.e. Polarization. Then the energy is, as before 1 -1 , a -(e-l)a2 2 ° £° 2 ° e0" 2e0e

324 and the total polarization energy is 0 Then, total energy is The actual height to which the liquid rises is determined from the formula § This gives h « -^—i-L2_ 2eoepg 3.145 We know that energy of a capacitor,U » Hence, from .F = — dx we have, dc I dx I Now, since </« iJ, the capacitance of the given capacitor can be calculated by the formula of a parallel plate capacitor. Therefore, if the dielectric is introduced upto a depth x and the length of the capacitor is /, we have, j-qERx 2xRto(l-x) + d From A) and B), we get, 3.146 When the capacitor is kept at a constant potential difference V, the work performed by the moment of electrostatic forces between the plates when the inner moveable plate is rotated by an angle rfcp equals the increase in the potential energy of the system. This comes about because when charges are made, charges flow from the battery to keep the potential constant and the amount of the work done by these charges is twice in magnitude and opposite in sign to the change in the energy of the capacitor Thus Nz • a- 5cp Now the capacitor can be thought of as made up two parts (with and without the dielectric) in parallel. Thus . ^J . 2rf 2rf as the area of a sector of angle qp is -K2 ep. Differentiation then gives (e-1) The negative sign of Nz indicates that the moment of the force is acting clockwise (i.e. trying to suck in the* dielectric).

325 3.4 ELECTRIC CURRENT 3.147 The convection current is dt here, dq** Xdxy where X is the linear charge density. But, from the Gauss' theorem, electric field at the surface of the cylinder, X Hence, substituting the value of X and subsequently of dq in Eqs. A), we get 2E Jt Eft adx 1 T dt - 2ne0Eav, as -7*- v 3.148 Since d « r, the capacitance of the given capacitor can be calculated using the formula for a parallel plate capacitor. Therefore if the water (permittivity e) is introduced up to a height x and the capacitor is of length /, we have, eeo-2 Jt rx e0 (/ - x) 2 Jtr e0 2 nr C + - d d Hence charge on the plate at that instant, q = CV Again we know that the electric current intensity, dq_ d(CV) dt dt Veo2Krd(sx + l-x) V2nre0 " d dt " d ^"^ dt dx But, --v. 2jtreo(e-lO So, ' ° 3.149 We have, Rt - /?0 A + o/),(l) where /?, and i?0 are resistances at f C and 0° C respectively and a is the mean temperature coefficient of resistance. So, /?! - Ro A + ax r) and /?2 - r\ Ro A + a2 r) (a) In case of series combination, R = 2/?, so R= R1 + R2= Ro [A + T|) + (ax + T] a2) t] A) B) 1 +T]

326 Comparing Eqs. A) and B), we conclude that temperature co-efficient of resistance of the circuit, a2 + r\ a2 a = 1 +T| (b) In parallel combination = R' A + a' r), where R' Now, neglecting the terms, proportional to the product of temperature coefficients, as being very small, we get, a'« 3.150 (a) The currents are as shown. From Ohm's law applied between 1 and 7 via 1487 (say) / J/+2J2 Thus, R (b) Between 1 and 2 from the loop 14321, Ix R « 2I2R+I3R or, /x » /3 + 2/2 From the loop 48734, 2 or, so Or A —j" Then, (/x + 2/2) R e =

or (c) Between 1 and 3 From the loop 15621 ^ ^R or, Then, *hReq I2R+I2R Hence, (o 3.151 Total resistance of the circuit will be independent of the number of cells, 327 R B Rx B if R. (RX + 2R)R R+2R+R or, R2 + 2RRX-2R2 = 0 On solving and rejecting the negative root of the quadratic equation, we have, Rx-R 3.152 Let Ro be the resistance of the network, iA 'VWW Ri Ro <=> **> B 0 2 ^ then, i?0 = — — or Rq-RQR1-R1R2 0 On solving we get, VP 1+4-^ 6Q

328 3.153 Suppose that the voltage V is applied between the points A and B then = 7fl=7 where R is resistance of whole the grid, /, the current through the grid and 70, the current through the segment AS. Now from symmetry, 1/4 is the part of the current, flowing through all the four wire segments, meeting at the point A and similarly the amount of current flowing through the wires, meeting at B is also 7/4. Thus a current 7/2 flows through the conductor AS, i.e. Hence, 3.154 Let us mentally isolate a thin cylindrical layer of inner and outer radii r and r + dr respectively. As lines of current at all the points of this layer are perpendicular to it, such a layer can be treated as a cylindrical conductor of thickness dr and cross-sectional area 2 ji rl. So, we have, dr dr dR Sir) - P rl and integrating between the limits, we get, R " ■ b I S + ■i «• I 3.155 Let us mentally isolate a thin spherical layer of inner and outer radii r and r + dr. Lines of current at all the points of the this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness dr and cross sectional area 4 ji r . So dr A) And integrating A) between the limits [a, ft], we get, a ft Now, for ft -* oo, we have 4 3.156 In our system, resistance of the medium R where p is the resistivity of the medium 4* a i ft The current 4* 9 a 1" ft

329 a i . -dq d(C w) ^dq> Also, i m —*■ = - v T' » - C -f-, Jr A J* ' So, equating A) and B) we get, dw j = j» - C j , as capacitance is constant. at at at B) ±- a or, At 4ji a or, in Hence, resistivity of the medium, At 4 nab Cp(b-a) 4 Jt At ab 3.157 Let us mentally impart the charge +^ and -q to the balls respectively. The electric field strength at the surface of a ball will be determined only by its own charge and the charge can be considered to be uniformly distributed over the surface, because the other ball is at infinite distance. Magnitude of the field strength is given by, So, current density j » — ^ P 4 ji e0 a2 and electric current 4na = 2 £_ p4m0a4 But, potential difference between the balls, Peo Hence, the sought resistance, qp+ - qp_ 2#/4 Jt e0 a 0 3.158 (a) The potential in the unshaded region beyond the conductor as the potential of the given charge and its image and has the form cp = A rl r2 where rv r2 are \he distances of the point from the charge and its image. The potential has been taken to be zero on the conducting plane and on the ball --^ -7 <S"^ 1' '

330 So A m Va> In this calculation the conditions a « I is used to ignore the variation of cp over the ball. The electric field at P can be calculated similarly. The charge on the ball is Q - and Va r cosO 2aW Then i = — E * —5- normal to the plane. P pr3 (b) The total current flowing into the conducting plane is (On putting >» = Inxdxj ■ 0 0 +/2) 2alV 77 2 ao _ 2nalV C p~ J dy 4mV f R » -h Hence 3.159 (a) The wires themselves will be assumed to be perfect conductors so the resistance is entirely due to the medium. If the wire is of length Ly the resistance R of the medium is a ■=- because different sections of the wire are connected in parallel (by the medium) rather than in series. Thus if /^ is the resistance per unit length of the wire then R « R^/L. Unit of Rl is ohm-meter. The potential at a point P is by symmetry and superposition (for / » a) A , ri ~r In — 2 a A 2 a A *i -=- In — V A a Then qpx = — = — In — (for the potential of 1)

331 or, and A - - VAn - a ln We then calculate the field at a point P which is equidistant from 1 & 2 and at a distance r from both : Then —j— f-lx 2sin0 2 ln//fl I r VI 1 2 In I/a and (b) Near either wire and Then Which gives 3.160 Let us mentally impart the charges Then capacitance of the network, oE - - p 21n//a 1 21n//a a V V R K In I/a and -q to the plates of the capacitor. 1 ee 0 <P A) Now, electric current, Hence, using A) in B), we get, ee, as 7t f ^ B) Cq? = —^ 1-5 ji A -o P eeo 3.161 Let us mentally impart charges +9 and -q to the conductors. As the medium is poorly conducting, the surfaces of the conductors are equipotential and the field configuration is same as in the absence of the medium. Let us surround, for example, the positively charged conductor, by a closed surface 5, just containing the conductor, ; as ; f t then, 1 / jdS JoEndS and So, <P <P 0 RC = — = p ee0 a u

332 3.162 The dielectric ends in a conductor. It is given that on one side (the dielectric side) the electric displacement D is as shown. Within the conductor, at any point A, there can be no normal component of electric field. For if there were such a field, a current will flow towards depositing charge there which in turn will set up countering electric field causing the normal component to vanish. Then by Gauss theorem, we easily derive a « Dn « D cos a where a is the surface charge density at A. The tangential component is determined from the circulation theorem E-dr= 0 It must be continuous across the surface of the conductor. Thus, inside the conductor there is a tangential electric field of magnitude, d)t)eCtV\C 77 Dsina eoe at A . This implies a current, by Ohm's law, of JD since ^(Conductor) 3.163 The resistance of a layer of the medium, of thickness dx and at a distance x from the first plate of the capacitor is given by, o(x) S K) Now, since a varies linearly with the distance from the plate. It may be represented as, x , at a distance x from any one of the plate. From Eq. A) dR a _dx x S / or, R. 1 f dx S J o2 - ax 0 a, + ;— °2 In — r In S (a2 - a,) o Hence, . V l~ R- °2 din — a. 3.164 By charge conservation, current j, leaving the medium A) must enter the medium B). Thus j cos ax = j2 cos 0L2 m F Another relation follows from

333 which is a consequence of/ E dr- 0 Thus—jjSincij- —j2 sin ai a2 B) or, tan ax tan a1 tan (t) or, o J, 3.165 The electric field in conductor 1 is and that in 2 is Applying Gauss' theorem to a small cylindrical pill-box at the boundary. Thus, nR °m eo (P2"Pi) '0 and charge at the boundary- e0 (p2 - 3.166 We hawtJE1 dx+E <idi- V and by current conservation 1 Pi 1 - i-£ " P2 2 Thus, Pi ^1 + p2 di 2 Pi d\ + P2 ^2 At the boundary between the two dielectrics, o » 3.167 By current conservation E(x) E(x) + dE(x) + p2d: (x) (e2 p2 - dx p (x) p(x) + dp(x) dp (x) This has the solution, £(*)- Cp(jc)- +v I

334 Hence charge induced in the slice per unit area da = zo-[{z(x) + dz(x)}{p(x) + dp(x)}-t (x) p (x) ] = e0-d[e (x) p (x) ] Thus, dQ= told[e(x)p(x)] Hence total charge induced, is by integration, 3.168 As in the previous problem - Cp(jc)- C(p0+Plx) frl - 1) Po where p0 + px d = r] p0 or, d By integration V= J Cp(x)dx = o 711118 C - Thus volume density of charge present in the medium dQ Sdx = tQdE (x)/dx 2e0 V (r\ - 1) Po X - 3.169 (a) Consider a cylinder of unit length and divide it into shells of radius r and thickness dr Different sections are in parallel. For a typical section, / 1 \ 1 2jc r dr 2 jc r dr (a/r2) \ / Integrating, i<12a2jca _ 2 jca t ^ _2 or, R1 = —2~", where 5= jc/c (b) Suppose the electric filed inside is Ez = Eo ( Z axis is along the axiz of the conductor). This electric field cannot depend on r in steady conditions when other components of E are absent, otherwise one violates the circulation theorem § E-dr = 0 The current through a section between radii (r + dr, r) is 1 3 E Inrdr ^ E « 2jtr d oc/r Thus / = - - ^ £ 0 2 oc Hence E = —x— when S = jlR2 2

335 3.170 The formula is, or. or, C or, 1 -e -t/RC 0 -t/RC Hence, t = RC In Thus t = 0.6 \aS. o Vo-V In 10, if V= 0-9 3.171 The charge decays according to the foumula Here, RC « mean life = Half-life/ln 2 So, half life= J= RC In2 But, Hei\ce, p = 13 7-x = 1.4xlO"Q-m In 2 3.172 Suppose ^ is the charge at time t. Initially q = C §, at Then at time f, 0. But,i So = - -p (- si (- sign because charge decreases) Z -IR 0 RC or, or, dt ae t1]/*c- l R q= +Ae -tr\/RC , from gr = c | at r = 0 Hence, Finally, »■ ^ -t^/RC 3.173 Let r * internal resistance of the battery. We shall take the resistance of the ammeter to be = 0 and that of voltmeter to be G. Initially.^- |_/r,/-

336 After the voltmeter is shunted r + RG R + G (Voltmeter) B) and r + RG r + G (Ammeter) C) R + G From B) and C) we have From A) and D) r\ r + G R — = r + G - r\r or G = rir Then A) gives the required reading r\ r\ +1 3.174 Assume the current flow, as shown. Then potentials are as shown. Thus, or, And So, qp2 i Rf = -4V t ' D) 3.175 Let, us consider the current i, flowing through the circuit, as shown in the figure. Applying loop rule for the circuit, - A cp = 0 « 0 or, i{Rx+R{+R) or, Now, if i = - cp2 = 0 0 2 ^ or, and R2+R+Rx or, /? * R1 - /?2, which is not possible as R2 > Thus, cp2 - cp3 * -

337 or, So, 3.176 (a) Current, i R « R2-Rv which is the required resistance. oc, as |- aR (given) w^-^ii?= naR - naR= 0 3.177 As the capacitor is fully charged, no current flows through it So, current L(as And hence, t -12 + iR T A -0-5V 3.178 Let us make the current distribution, as shown in the figure. Current i * |—— (using loop rule) So, current through the resistor Rv _D pp - 1-2 A R Aj + R /V2 + i\j R2 and siinilary, current through the resistor R2, 0-8 A xR o 3.179 Total resistance' /-x 0 /-x 0 IR+xR 0 R o IR+xR 0 Vo V

338 Then V xR °IR+xR o xR 0 '0Rx/< IR+Rox\l / For R»R0}Vm Voy 3.180 Let us connect a load of resistance K between the points A and B (Fig.) From the loop rule, A qp = 0, we obtain - 'i*i A) and i/t -12 - (i-ii or i(«+^- xa Solving Eqs. A) and B), we get %l^l + ?2*2 *l+*2 R + ^1^2 Rx+R2 So R + R C) o where I Rx + R and R o Rx + R 1 T JV2 xxl T JV2 Thus one can replace the given arrangement of the cells by a single cell having the emf £n and internal resistance Rn. 3.181 Make the current distribution, as shown in the diagram. Now, in the loop 12341, applying - Acp = 0 iR + ixRx + %x- 0 A) and in the loop 23562, tt-fe + fi-i!)/^- 0 B) Solving A) and B), we obtain current through the resistance Rt 1. -ll-f- 2/ 3 RRl+RR 0-02 A and it is directed from left to the right 3.182 At first indicate the currents in the branches using charge conservation (which also includes the point rule). In the loops 1BA 61 and B34AB from the & . A i loop rule, - A cp » 0, we get, respectively til r* 0 B) On solving Eqs A) and B), we obtain « 006A A-S Ri Thus cpA - qp5 0-9 V

339 3.183 Indicate the currents in all the branches using charge conservation as shown in the figure. Applying loop rule, - A<p = 0 in the loops 1A781, 1B681 and B456B, respectively, we get R2 k 6 , 5 '3*3 +'1*2 -lo = 0 (i) 6,—e B) and C) £*4= Solving Eqs. A), B) and C), we get the sought current 1- -ir) R2 R3 to A B 3 T% 3.184 Indicate the currents in all the branches using charge conservation as shown in the figure. Applying the loop rule (- A (p - 0) in the loops 12341 and 15781, we get -0 A) and Solving Eqs. A) and B), we get B) 8 t (/t R 1 Hence, the sought p.d. Si = -1 V 3.185 Let us distribute the currents in the paths as shown in the figure. Now, (Pi - <p2 ■ iR\ + *i^2 A) B) and ^ - cp3 » iRx + Simplifying Eqs. A) and B) we get + R2 0-2 A 3.186 Current is as shown. From Kirchhoff's Second law ?2 V3 k+h

340 Eliminating i2 Hence R R l-3 ■MB 3 /?3(/?1+/?2)-JR1(/?3+/?4) On substitution we get i3 «1*OA from C to D 3.187 From the symmetry of the problem, cunent flow is indicated, as shown in the figure. Now, cpA - cpa - i\ r + (i - ix)R A) In the loop 12561, from - A cp « 0 - - L r » 0 or, B) Equivalent resistance between the terminals A and B using A) and B), o 3.188 Let, at any moment of time, charge on the plates be across the capacitor, cp « q/C A) Now, from charge conservation, 4 and -q respectively, then voltage where B) In the loop 65146, using - A cp • 0. C) [using A) and B)] In the loop 25632, using — A <p — 0 - 0 or, ijJt- -t- D)

341 From A) and B), dt or, da E) On integrating the expression E) between suitable limits, 9 / C or> -2 o Thus - e 3.189 (a) As current i is linear function of time, and at t = 0 and At, it equals i0 and zero respectively, it may be represented as, Thus o o So, Hence, The heat generated. Af Af A/ o 0 ,2 (b) Obviously the current through the coil is given by 00 00 Then charge 0 0 So, At And hence, heat generated in the circuit in the time interval t [0, 00 00 0 0 2At

142 M90 The equivalent circuit may be drawn as in the figure. Resistance of the network * RQ + \R/3) Let, us assume that e.m.f. of the cell is §, then current Ro + (R/3) Now, thermal power, generated in the circuit P= i2R/3 o + (R/3) dP (R/3) For P to be maximum, ^r- * 0, which yields = 3R0 S>Rt 3.191 We assume current conservation but not KirchhofTs second law. Then thermal power dissipated is o 22 u - Rx+R The resistances being positive we see that the power dissipated is minimum when R, This corresponds to usual distribution of currents over resistance joined is parallel, 3.192 Let, internal resistance of the cell be r, then -ir A) where i is the current in the circuit. We know that thermal power generated in the battery. Q = ?' B) Putting r from A) in B), we obtain, Q = (| - V) i = 0-6 W In a battery work is done by electric forces (whose origin lies in the chemical processes going on inside the cell). The work so done is stored and used in the electric circuit outside. Its magnitude just equals the power used in the electric circuit We can say that net power developed by the electric forces is =-2-0W 2 R Minus sign means that this is generated not consumed.

343 3.193 As far as motor is concerned the power delivered is dissipated and can.be represented by a load, Ro . Thus V 2 and P- I2R0 ^ (*„+*) This is maximum when Ro = R and the current / is then The maximum power delivered is 4R V2 V2 The power input is -=—— and its value when P is maximum is — R + Rq 2a The efficiency then is — « 50% 3.194 If the wire diameter decreases by 6 then by the information given V2 P * Power input * — * heat lost through the surface, H. R Now, H <* A - 6) like the surface area and V2 So, ■£- A - 6J - A A - 6) or, V2 A - 6) - constant But V«l+T] so A+ T]J A - 6) - Const - 1 Thus 6 - 2 T| - 2% 3.195 The equation of heat balance is Put So, or, or, where A C 9 + fc § « — d is a constant. Clearly |- 0 at r= 0, r2 ,^ SO or, i A ,,2 V2 CR' kt/c + A V2 V '" C^R and hence,

344 3.196 Let, WA-<pB** <p Now, thermal power generated in the resistance Rxf R, i'2R. ^ Rn R^ Rn + R. For P to be independent of R3 dP R x> A—* « dR. R. - 0, which yeilds 12 Q 3.197 Indicate the currents in the circuit as shown in the figure. Appying loop rule in the closed loop 12561, - Acp = Owe get ii*-!! + «!-<> A) and in the loop 23452, (i-i1)JR2 + %2-»1JR- 0 B) 6 Solving A) and B), we get, So, thermal power, generated in the resistance R, P- L2R For P to be maximum, R R^ + R-^ R2 + RR dP R 0, which fields R "T& Hence, 3.198 Let, there are x number of cells, connected in series in each of the n parallel groups then, n x = N or, x * — Now, for any one of the loop, consisting of x cells and the resistor R> from loop rule Soyi iR + —jcr- n N n 0 xr R + — n Nr n2 , using A)

345 Heat generated in the resistor Ry •2 „ n2R+NR R and for Q to be maximum.-j6-» 0, which yields 3.199 When switch 1 is closed, maximum charge accumulated on the capacitor, and when switch 2 is closed, at any arbitrary instant of time, because capacitor is discharging. or, I —dq- - 9. o Integrating, we get In nax or, (max Differentiating with respect to time, max or, i (r) = Negative sign is ignored, as we are not interested in the direction of the current, thus, B) B) C) When the switch (Sw) is at the position 1, the charge (maximum) accumalated on the capacitor is, When the Sw is thrown to position 2, the capacitor starts discharging and as a result the electric energy stored in the capacitor totally turns into heat energy tho' the resistors R1 and R2 (during a very long interval of time). Thus from the energy conservation, the total heat liberated tho' the resistors. H - U: - = i 2C 2

346 During the process of discharging of the capacitor, the current tho' the resistors R1 and R2 is the same at all the moments of time, thus Ht « Rt and H2 oc R2 So, H*x 3.200 When the plate is absent the capacity of the condenser is c d When it is present, the capacity is Cm (a) The energy increment is clearly. 2 2 2(l-rO (b) The charge on the plate is CV q.« initially and fy- CV finally. CVr\ CV2r\ A charge l has flown through the battery charging it and withdrawing — l units of energy from the system into the battery. The energy of the capacitor has decreased by 1 CV2t\ just half of this. The remaining half i.e. —— l must be the work done by the external agent in withdrawing the plate. This ensures conservation of energy. 3.201 Initially, capacitance of the system = Ce. 1 2 So, initial energy of the system : Ut ■ ~ (C e) V 1 2 and finally, energy of the capacitor : Uf** -^C V Hence capacitance eneigy increment, V2- -\ CV1 (e - 1)- -0-5mJ From energy conservation AU' Acdl+Aagmt (as there is no heat liberation) But A^= (Cf-C^V2- (C-Ce)V2-

347 A - e) V- 0-5 m J 3.202 If Co is the initial capacitance of the condenser before water rises in it then 1 2 zo2lnR ui " 2C°V ' where C° * ~ (R is the mean radius and / is the length of the capacitor plates.) Suppose the liquid rises to a height h in it. Then the capacitance of the condenser is + (e - 1) A) and energy of the capacitor and the liquid (including both gravitational and electrosatic contributions) is i Ea2nR u ±^ + pgBxRhd) | If the capacitor were not connected to a battery this energy would have to be minimized. But the capacitor is connected to the battery and, in effect, the potential energy of the whole system has to be minimized. Suppose we increase h by 5/t. Then the energy of the capacitor and the liquid increases by bh 2d (e-l)V2 + pgBnRd)h and that of the cell diminishes by the quantity Acdl which is the product of charge flown and V e0 BnR) ~ bh ° V, (e - 1) V2 d In equilibrium, the two must balance; so e0 (e - 1) V2 2d 0 (e - l)V7 pgdh Hence h ■ 3.203 (a) Let us mentally islolate a thin spherical layer with inner and outer radii r and r + dr respectively. Lines of current at all the points of this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness dr ry and cross sectional area 4 ji r . Now, we know that resistance, dr dr S(r) r4nr: Integrating expression A) between the limits,

348 o b or, _£_ a Capacitance of the network,C 111 a b and also, q~ Cqp ' where g is the charge at any arbitrary moment Ry as capacitor is discharging. From Eqs. B), C), D) and E) we get, 1 1 P 4 jieoe a'b a or. Integrating ■ MM* *»#• ^kJ 0 dt pee0 B) C) D) E) Hence q « #0 e p eo8 (b) From energy conservation heat generated, during the spreading of the charge, H » £7. - Uf [because A^ « 0] 2 _ _ _2 2 4 ft eoe 8 ji eoe 3.204 (a) Let, at any moment of time, charge on the plates be (qQ - q) then current through the resistor, i = - or, dt , because the capacitor is discharging. Now, applying loop rule in the circuit, or, or, dq 1 — —

349 At t** 0, q** 0 and at f« x, q** q So, In -x Thus ? = qo(\-e~x/RC\- 018mC (b) Amount of heat generated = decrement in capacitance energy 2 C 2 2 C RC 82 mJ 3.205 Let, at any moment of time, charge flown be q then current i = Applying loop rule in the circuit, - Aq> - 0, we get • 0 dqm- dt or, dq R CV0-2q RC dt RC *c CV0-2q So'ln-cvf-=-2 or, Hence, /- §« 2 Now, heat lib era led, for 0 £ t * t dt 2 RC —— R 00 = J i2Rdt= o RC 3.206 jn a rotating frame, to first order in co, the main effect is a coriolis force This unbalanced force will cause electrons to react by setting up a magnetic field B so that the magnetic force e v*x B balances the coriolis force. Thus The flux associated with this is £ ^* —* ^ 2/n —* r— B = co or, B = - — co 2m e Nxr2B= Nxr2 — co

350 where N ■ is the number of turns of the ring. If co changes (and there is time for the electron to rearrange) then B also changes and so does <3>. An emf will be induced and a current will flow. This is —co/jR e The total charge flowing through the ballastic galvanometer, as the ring is stopped, is 2m e Nnr2/ — <o/R e_ INnrO) /car ' m~ qR = qR 3.207 Let, n0 be the total number of electorns then, total momentum of electorns, A) Now, /- PSxvd- ^-Sxvd. ^vd B) Here Sx = Cross sectional area, p = electron charge density, V = volume of sample From A) and B) p- — IU 0-40 jaNs e 3.208 By definition nevd= j (where vd is the drift velocity, n is number density of electrons.) ml tk l Then x - — vd J So distance actually travelled nel < v > Sm < v>x- : (<v> = mean velocity of thermal motion of an electron) 3.209 Let, n be the volume density of electrons, then from / - p Sx vd, /» neSx | <v*> | « neSx- neSJ So, t~ —j—- 3 pis. (b) Sum of electric forces _* ~-* (nv) eE\= \nS lepjl where p is resistivity of the material. - nSlep-** nelpl* 1-0|iN

351 3.210 From Gauss theorem field strength at a surface of a cylindrical shape equals, X , where X is the linear charge density. Now, Also, or, 1 2 2mcv or» v dq-Xdx so, dx /= Xv or, V using A) Hence 2*7 32V/m (b) For the point, inside the solid charged cylinder, applying Gauss' theorem, or, 2nrhE _2 _r« So,from or, Hence, r * f ^ I -aq)« I J J 2m0R <Pi 0 rdr r2 2 0 4 JC8 0 - 0-80 V 3.211 Between the plates <p - a je 4/3 or, dx 4 3 4 ax - - p/e0 or, p- - -2/3 Let the charge on the electron be - e,

352 1 2 then -zmv -ecp= Const. = 0, as the electron is initially emitted with neligible energy. 2 qp V « ^ V m m ▼ m -2/3 so, ;= _pv= (/ is measued from the anode to cathode, so the - ve sign.) V 3.212 E = - So by the definition of the mobility + V and (The negative ions move towards the anode and the positive ion towards the cathode and the total current is the sum of the currents due to them.) On the other hand, in equilibrium n+ - n_ I So, n+= n_= / (wJ + m0) d — * 2-3xl08cm~3 3.213 Velocity = mobility x field or, v = m — sin co r, which is positive for 0 & co t £ ji So, maximum displacement in one direction is 2uV0 0 20 At co= coo, *„„- /, so, /m - / 2«V0 Thus m- 3.214 When the current is saturated, all the ions, produced, reach the plate. —~ • sat , <n9 -3 -1 Then, n- - — - 6 x 10 cm s ev (Both positive ions and negative ions are counted here) The equation of balance is, -r- ■ ni - rn

353 The first term on the right is the production rate and the second term is the recombination ry rate which by the usual statistical arguments is proportional to n (= no of positive ions x no. of -ve ion). In equilibrium, dt so, neq - V — - 6 x 107 cm 3.215 Initially n~ n0- Vn. /r Since we can assume that the long exposure to the ionizer has caused equilibrium to be set up. Afer the ionizer is switched off, dn 2 dt dn 1 or rdt= -—r, or, rt- — +constant n2 n nil But n - nft at t - 0, so, rt n n0 The concentration will decrease by a factor r) when n0 iil or, /a= = ■ 13 ms 3.216 Ions produced will cause charge to decay. Clearly, r\CV~ decrease of charge* n^Adt" —j q) or, f * « " 4-6 days n^ed Note, that niy here, is the number of ion pairs produced. 3.217 If v « number of electrons moving to the anode at distance x, then dv ax -r~-av or v - vne dx u Assuming saturation, /= evoead 3.218 Since the electrons are produced unifonnly through the volume, the total current attaining saturation is clearly, d o Thus, j= — = en. I ■ (ead-l A Ma

354 3.5 CONSTANT MAGNETIC FIELD. MAGNETICS 3.219 (a) From the Biot - Savart law, dB Mo .dlxr -— i ——, so 4jt r3 77 Mo . (RdQ)R From the symmetry dlLr) B -f J dB- R Mo 6-3 o (b) From Biot-Savart's law : B — tj -^ (here r-J? So, 4jc <Ux R+f dlxx Since Jc*is a constant vector and \R | is also constant So, <P dlxx*~ Uf dl > C -* —* £ — and <Q dlxR** <f Rdln 0 [because (P V J Here n isa unit vector perpendicular to the plane containing the current loop (Fig.) and in the direction of x Thus we get Mo 2nR2i 3/2 n 3JE20 As L AOB * ±1-, OC or perpendicular distance of any segment from centre equals R cos ~ . Now magnetic induction at O, due to the right current carrying element AB n 4k 2sin (From Biot- Savart's law, the magnetic field at O due to any section such as Ad is perpendicular to the plane of the figure and has the magnitude.)

355 B f . dx 4jt 4n jRcos - n cosG - cos— sec 8 n Rcos— n — 2sin— Jt n As there are n number of sides and magnetic induction vectors, due to each side at 0, are equal in magnitude and direction. So, B ni o 4?t jRcos 2 sin — • n n n Mo m A n , c -—r- tan — and for n -» «> 2ji R n B° 2RLJ tan n 2 it 3.221 We know that magnetic induction due to a straight current carrying wire at any point, at a perpendicular distance from it is given by : An r l 2 f where r is the perpendicular distance of the wire from the point, considered, and 6X is the angle between the line, joining the upper point of straight wire to the considered point and the perpendicular drawn to the wire and 62 that from the lower point of the straight wire. Here, An (d/2)sin| P COS t" + COS T and BA An (d/2) cos sin *• + sin Hence, the magnitude of total magnetic induction at O, Ai d/2 cos -J* sin ^ + . q> & sm2 "*2 Jt d sin cp 010 mT

358 3.222 Magnetic induction due to the arc segment at O, and magnetic induction due to the line segment at O, Ho i r^ . , So, total magnetic induction at O, * — [jt - cp + tan cp] « 28 [i T I 3.223 (a) From the Biot-Savart law, So, magnetic field induction due to the segment 1 at Oy also B2 = J34» 0, as d/tf and Hence, i?0» B1+B2+B3+ B4 ' 2 ji-(p (b) Here, ^ Ho i . .co * - — — sin 45 , 3 4% b d Ho 1 . -co BA« — 7- sin 45°, 4 4 6 4jt z, 5 /(a? 4 / b '/ f and So, Bo 0 1 Anal A fsin45o+-^-tsin45° + 0 Anb Anb 2a + b

357 3.224 The thin walled tube with a longitudinal slit can be considered equivalent to a full tube and a strip carrying the same current density in the opposite direction. Inside the tube, the former does not contribute so the total magnetic field is simply that due to the strip. It is B (I/2nR)h r where r is the distance of the field point from the strip. 3.225 First of all let us find out the direction of vector B at point O. For this purpose, we divide the entire conductor into elementary fragments with current di It is obvious that the sum of any two symmetric fragments gives a resultant along JSTshown in the figure and con- consequently, vector B will also be directed as shown ~" j dB sin cp A) 0 di sincp So, B 2nR Ho 2k2 R o Hence i sin op a cp, as di= — aw \i0i/n2R 3.226 (a) From symmetry - 0+ A D „ 4jiJ? (b) From symmetry 4 R Bx+B2+B3 Ho i Ho / 3ji /? 2 Ho i 1 + (c) From symmetry 3.227 B Q 1 or, 2-0 ji T, (using 3.221)

358 3.228 (a) B1+B2+B3 So, (b) 5 Mo < So, 0-34 (c) Here using the law of parallel resistances So, Hence Thus B o Thus, -1, and *!« -i An # 011 ^i T 3.229 (a) We apply circulation theorem as shown. The current is vertically upwards in the plane and the magnetic field is horizontal and parallel to the plane. IV T dfc 2Bl or, B (b) Each plane contributes \iQ — between the planes and outside the plane that cancel. Thus f \iQi between the plane 0 outside.

359 3.230 We assume that the current flows perpendicular to the plane of the paper, by circulation theorem, IBdl- \L0Bxdl)j or, \ioxjy Outside, 2Bdl- \ioBddl)j or, B » \i0 dj | x\ st d. 3.231 It is easy to convince oneself that both in the regions. 1 and 2, there can only be a circuital magnetic field (i.e. the component B^. Any radial field in region 1 or any Bz away from the current plane will imply a violation of Gauss' law of magnetostatics, B must obviously be symmetrical about the straight wire. Then in 1, or, In 2, B. 2nr B9-2nr- 0, or 3.232 On the axis,/? 0 i?, along the axis. 1 00 00 Thus, I B-dr* I Bxdx= r — 00 Jdx (R2 + x2K/2 — 00 Ho/tf x/2 ; -a/2 sec t 5 f-f-, on putting x 5 sec 0 fltanG 3C/2 -K/2 00 The physical interpretation of this result is that J Bx dx can be thought of as the circulation — 00 of B over a closed loop by imaging that the two ends of the axis are connected, by a line at infinity (e.g. a semicircle of infinite radius). 3.233 By circulation theorem inside the conductor r2 or, 1 -r i.e., Similarly outside the conductor,

360 - \iojz7tR2 or, R So, 2.234 We can think of the given current which will be assumed uniform, as arising due to a negative current, flowing in the cavity, superimposed on the true current, everywhere including the cavity. Then from the previous problem, by superposition. If / vanishes so that the cavity is concentric with the conductor, there is no magnetic field in the cavity. 3,235 By Circulation theorem, By • 2 ji r - \i0 j j (r') x 2 ji r' dr' or using B - fcrainside the stream, <p bra + 1- VL0fj(r')r'dr' So by differentiation, Hence, (a + 1) bra - (i0; (r) r ■ 3.236 On the surface of the solenoid there is a surface current density —♦ A » nle A ^, Then, k o where r0 is the vector from the current element to the field point, which is the centre of the solenoid, Now, - ew x rZ« R ez -1/2 Thus, B- B. Vo*! -» D2 I dz —A—x2jc^ I —=—7T 4* J (R +2T) -1/2 3/2

361 tan 2R cos ad a (on putting z = R tan a) -tan -l / 2R 1/2 ina= \ionl 3.237 We proceed exactly as in the previous problem. Then (a) the magnetic induction on the axis at a distance x from one end is clearly, 00 00 V*nl _ d2 f dz 1 td2 C dz 4» J [«2 + (z-jcJ]3/2 2^° J (z2 + /? 0 x a/2 1 f 2^onI J 1 -\ionl tan * 1- x> 0 menas that the field point is outside the solenoid. B then falls with x. x < 0 means that the field point gets more and more inside the solenoid. B then increases with (x) and eventually becomes constant, equal to \ionl. The B -x graph is as given in the answer script. (b) We have, B o 1- VF+ 0 or, VR2+x2 1-2ti Since r| is small (~ 1%), x0 must be negative. Thus xQ « - |jc0 | and 1-2ti or, 2Vt|A-ti) 3.238 If the strip is tightly wound, it must have a pitch of h. This means that the current will flow obliquely, partly along e and partly along e^ Obviously, the surface current density is, h h

362 By comparision with the case of a solenoid and a hollow straight conductor, we see that field inside the coil (Cf. B m \xonl). Outside, only the other term contributes, so or, Note - Surface current density is defined as current flowing normally acrpss a unit length over a surface. 3.239 Suppose a is the radius of cross section of the core. The winding has a pitch 27iR/Ny so the surface current density is J*m 2itR/N *1+ where e± is a unit vector along the cross section of the core and e2 is a unit vector along its length. The magnetic field inside the cross section of the core is due to first term above, and is given by B^2jtR = \i0NI (NI is total current due to the above surface current (first term.)) Thus, B^ = \i0NI/2nR. The magnetic field at the centre of the core can be obtained from the basic formula. dB dS and is due to the second term. 0 So, or, M^ J_ C ± 4 ji 2tui J ^: Rd(p x 2na B. 2R The ratio of the two magnetic field, is ■ — Jt 3.240 We need the flux through the shaded area. Now by Ampere's theorem; B = \x0 3t/v or, B The flux through the shaded region is,

363 R o R R2 4* 3.241 Using 3.237, the magnetic field is given by, 1- At the end,£« —\ionl= "zBq, where Bo is the field deep inside the solenoid. Thus, $ = — [i0 nfS = %/2y where ^> = [i0 nIS is the flux of the vector B through the cross section deep inside the solenid. 3.242 Byljzr* \x0NI or, B <p Then, rss 4 jc In T], where t] - b/a 3.243 Magnetic moment of a current loop is given by pm - «i S (where « is the number of j- 2 ^0 i turns and 5, the cross sectional area.) In our problem, n = 1, 5 = jri? and 5 = T"B" L K So, m 2kBR 3,244 Take an element of length rd 6 containing — • rdQ turns. Its magnetic moment is normal to the plane of cross section. We resolve it along OA and OB. The moment along OA integrates to n jvd o A 4 cos - 0 0 while that along OB gives 31 /Nd2I 1 i — srnedQ--Nd I B o

364 3.245 (a) From Biot-Savart's law, the magnetic induction due to a circular current carrying wire loop at its centre is given by, The plane spiral is made up of concentric circular loops, having different radii, varying from a to b. Therefore, the total magnetic induction at the centre, (i) where — i is the contribution of one turn of radius r and dN is the number of turns in 2r the interval (r, r + dr) N J 2r i.e. b-a Substituting in equation A) and integrating the result over r between a and b, we obtain, s In — ^nr dr — _ .,r In 2r (b-a) 2(b-a) a (b) The magnetic moment of a turn of radius r is pm = inr and of all turns, b C au C • N 7iiN(bsa2 2 N A 7iiN(bs-a2) r dr = ——7 :—L b dr 7: b-a 3 (b-a) a $.246 (a) Let us take a ring element of radius r and thickness dr, then charge on the ring element., dq- a 271 rdr and current, due to this element, at - •* — - a co r dr 2 ji II st' So, magnetic induction at the centre, due to this element : dB ■ — r C \i0o<x>rdr \i0 and hence, from symmetry : B - J dB - I — o 0 (b) Magnetic moment of the element, considered, dpm = (di) Ttr2 = a co rfr ji r2 = o 71 (o r3 dr Hence, the sought magnetic moment, ajicor3dr-acoji~ 4 4~ 0

365 3.247 As only the outer surface of the sphere is charged, consider the element as a ring, as shown in the figure. The equivalent current due to the ring element, 7T- 2jc B nr sin 9 rd 9) a and magnetic induction due to this loop element at the centre of the sphere, O, .„ l*o ,. 2 n r sin 6 r sin 6 cus m — ai 4 „. sin2 9 di r° 4ji r [Using 3.219 (b) ] Hence, the total magnetic induction due to the sphere at the centre, Oy n/2 TSivQ J B S'-f in2 2 n r2 sin 9 d 9 sin2 9 a 2jt [using A)] 0 Hence, B- 1*0 a cor 4* - 29pT o 3.248 The magnetic moment must clearly be along the axis of rotation. Consider a volume element dV It contains a charge 4n/3R to revolve around the axis and constitute a current. dV. The rotation of the sphere causes this charge rffx CO Its magnetic moment will be So the total magnetic moment is R n A-88 I I —^rsinOdQx 0 0 The mechanical moment is cor2sin29 co 4 , So, § 3.249 Because of polarization a space charge is present within the cylinder. It's density is p « - div P^« - 2a Since the cylinder as a whole is neutral a surface charge density a must be present on the surface of the cylinder also. This has the magnitude (algebraically)

366 a x 2tlR - 2 a nR 2 or, a When the cylinder rotates, currents are set up which give rise to magnetic fields. The contribution of p and a can be calculated separately and then added. For the surface chaige the current is (for a particular element) CO ? aRx 2nRdxx — - aR (odx 2k Its contribution to the magnetic field at the centre is \i0R2(aR2<x)dx) 2(x2+R2)s/2 and the total magnetic field is - 00 00 /VL0R2(aR2a>dx) n0a*4co C dx m _.________— I ___________ _ , 9 _»9v 3/9 <"> I , 9 00 — 00 As for the volume chaige density consider a circle of radius r, radial thickness dr and length dx. The current is- 2a x 2 k r dr dx x — • -2ardra)dx 2n The total magnetic field due to the volume charge distribution is 2+ r2f/2 R oo R oo ^r I dx 2 n r co ^^—r-^r « - I a ja0 co r dr I dx (jc + r ) J 2(jr + r) ^ J 0-oo v/ 0 -oo R /9 a ji0 cor dr x 2 * - ji0 a co R so, B * 2^ + Bv - 0 o 3.250 Force of magnetic interaction,/7,^ « e (io< J5 ) Where, D ^ ^ So, ___ _£ 4?i r3 1 er And Feu= &= e \Fmag\ 2 M « v „ \mag\ 2 Hence, -» — « - v un eft IF I lc - 1-OOxlO

367 3.251 (a) The magnetic field at O is only due to the curved path, as for the line element, dl ]*\ rJ Hencc> Thus = 0-20 N/m (b) In this part, magnetic induction B at O will be effective only due to the two semi infinite segments of wire. Hence B= 2- Jt/ 4*1- (-*) nsm2("*) Thus force per unit length, Til 3.252 Each element of length dl experiences a force BIdL This causes a tension T in the wire. For equilibrium, Tda= BIdl, where da is the angle subtended by the element at the centre. BIdl Then, 7= do. = BIR The wire experiences a stress BIR nd2/4 This must equals the breaking stress om for rupture. Thus, T B 7i d2 o m max 4IR 3.253 The Ampere forces on the sides OP and O' P are directed along the same line, in opposite directions and have equal values, hence the net force as well as the net torque of these forces about the axis OO' is zero. The Ampere-force on the segment PP and the cor- corresponding moment of this force about the axis OO' is effective and is deflecting in nature.

368 In equilibrium (in the dotted position) the deflecting torque must be equal to the restoring torque, developed due to the weight of the shape. Let, the length of each side be / and p be the density of the material then, UB (/ cos G) = (SI p) gtrsin G + {S I p) g --sin 0 + (S/p)g/sin8 or, Hence, B 2Spg/2sin9 tan 9 3.254 We know that the torque acting on a magnetic dipole. But, pm « i 5 n, where n is the normal on the plane of the loop and is directed in the direction of advancement of a right handed screw, if we rotate the screw in the sense of current in the loop. On passing a current through the coil, this torque /////// acting on the magnetic dipol, is counterbalanced kj by the moment of additional weight, about 0. Hence, the direction of current in the loop must '*^ I it) be in the direction, shown in the figure. p^xB - - / x Amg or, NiSB** A/wg/ kmgl \ I o s So, B NiS 0-4 T on putting the values. 3.255 (a) As is clear from the condition, Ampere's forces on the sides B) and D) are equal in magnitude but opposite in direction. Hence the net effective force on the frame is the resultant of the forces, experienced by the sides A) and C). Now, the Ampere force on A), Ho "' F, - -r- o 2k r and that on C), «o* 2it ■1 + 21 So, the resultant force on the frame « Fl - F3, (as they are opposite in nature.)

369 0-40 >iN. (b) Work done in turning the frame through some angle, A * fi d 4> ■ i C^ - <$,), where <$f is the flux through the frame in final position, and <£-, that in the the initial position. Here, I ^/l " I 3>J - <$ and <bi - - so, A<I> - 2 <I> and A - i 2 <£> Hence, A » 2i I ; a In 2ji r x [ 2r\ -1 3.256 There are excess surface charges on each wire (irrespective of whether the cunent is flowing through them or not). Hence in addition to the magnetic force Fm, we must take into account the electric force F^ Suppose that an excess charge X corresponds to a unit length of the wire, then electric force exerted per unit length of the wire by other wire can be found with the help of Gauss's theorem. 4 n e0 / 4 n e0 where / is the distance between the axes of the wires. The magnetic force acting per unit length of the wire can be found with the help of the theorem on circulation of vector B •p m 4* 2i2 \*~ Fe where i is the current in the wire. B) * Now, from the relation, X * C <p, where C is the capacitance of the wires per unit lengths and is given in problem 3.108 and qp = iR or, i-JEL C) rj X mR v ' Dividing B) by A) and then substuting the value of — from C), we get, A. Up (In TlJ The resultant force of interaction vanishes when this ratio equals unity. This is possible when R= Rq, where

370 0 3.257 Use 3.225 The magnetic field due to the conductor with semicircular cross section is B = n2R Then 3.258 We know that Ampere's force per unit length on a wire element in a magnetic field is given by. —* . A A A dFn = i(nxB) where n is the unit vector along the direction of current. A) Now, let us take an element of the conductor i2, as shown in the figure. This wire element is in the magnetic field, produced by the current ii9 which is directed normally into the sheet of the paper and its magnitude is given by, Ho A B 2nr B) From Eqs. A) and B) A . .* - . A dFn = — dr{nxB\ (because the current through the element equals — dr Ho A A dr r Hence the magnetic force on the conductor : a + b So, dFn = r±J-l?L towards left (as n lrB ). n 2jt b - v n -«i In I rlr ^T1 I — (towards left) . —7— in — 2k b a h (towards left). Then according to the Newton's third law the magnitude of sought magnetic interaction force In + b a 3.259 By the circulation theorem B = \x0 i, where i = current per unit length flowing along the plane perpendicular to the paper. Currents flow in the opposite sense in the two planes and produce the given field B by superposition. The field due to one of the plates is just —B. The force on the plate is, 1 B? — B x ix Length x Breadth - -— per unit area. 2 2\i0 (Recall the formula F = Bit on a straight wire)

371 B lf1 +n2 3.260 (a) The external field must be —-—, which when superposed with the internal field —— (of opposite sign on the two sides of the plate) must give actual field. Now 2 *l-*2 1 • or, Thus, F- (b) Here, the external field must be Bx-B2 Br Pl+B2 upward with an internal field, —-—, upward on the left and downward on the right Thus, i = t • D D ^ D d and F= — (c) Our boundary condition following from Gauss' law is, Bx cos 9j « B2 cos 92. where i - current per unit length. 2 The external field parallel to the plate must be - (The perpendicular component Bx cos Bj, does not matter since the corresponding force is tangential) B* sin2 Ql - B2 sin2 92 Thus, F = ^ per unit area 0 per unit area. The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing. 3.261 The Current density is —, where L is the length of the section. The difference in pressure produced must be, 1 IB = —7- x B x (abL)/ab - — aL a sin '\ >\ /1 - B2 sin 92 a

372 3.2B Let t - thickness of the wall of the cylinder. Then, J= 1/2 nRt along z axis. The magnetic field due to this at a distance r III-- < r < R + -I is given by, r2-[R-' or> Now, F= I JxBdV I *♦* RL~ 2nRL 3 8n2R2t2r 2nRL x 2 71 rLdr 8 M2 f 3.243 When self-forces are involved, a typical factor of — comes into play. For example, the £* force on a current carrying straight wire in a magnetic induction B is BIL If the magnetic induction B is due to the current itself then the force can be written as, 0 If B (/') a /', then this becomes, F - | B (I) 11 In the present case, B(T)= \ionl and this acts on nl ampere turns per unit length, so, F 1. Area 2 Ixnlxlxl 1 2r' Ho" 7 pressure /? 3.244 The magnetic induction B in the solenoid is given by B - |i0 n/. The force on an element dl of the current carrying conductor is, This is radially outwards. The factor — is explained above.

373 To relate dF to the tensile strength F{im we proceed as follows. Consider the equilibrium of the element dl. The longitudinal forces F have a radial component equal to, ^-- FdQ 1 2 Thus using dl** RdQ, F = -r\ionl R This equals Flim when ,/- /h-.- V lim * \ionR Note that Flim, here, is actually a force and not a stress. 3.265 Resistance of the liquid between the plates Voltage between the plates = Ed = v Bd, Current through the plates - pd Power, generated, in the external resistance R, v2B2d2R v2B2d2 v2B2d2 + sVr R 1/4 1/2 This is maximum when R and P. max sVr v2B2Sd 4p + 2 3.266 The electrons in the conductor are drifting with a speed of, xR2ne' where e = magnitude of the charge on the electron, n = concentration of the conduction electorns. The magnetic field inside the conductor due to this current is given by, B*r) - *r It mo or> V ^ A radial electric field vB must come into being in equilibrium. Its P.D. is, R ■2 Acp - I =— « dr - r J xR2ne2xR2 KR2 ne mo 431 4nR ne o

374 3.267 Here,^ = — and ; = ne vd — JD 200xl04j~-xlT 2 jB m2 so, n= •<-=■ = 7z « 2-5 x 1028 per m3 - 2-5 x 1022 per c.c. Atomic weight of Na being 23 and its density « 1, molar volume is 23 c.c. Thus number c 4. -4.1 .6x10 ~ * + r»22 of atoms per unit volume is ——— = 2*6 x 10 per c.c. Thus there is almost one conduction electron per atom.' „ . ,o o , f. ... ..... dirft velocity v 3.268 By definition, mobility = —J—: or ji - — Electric field component causing this drift EL On other hand, EL 1 3 2 £ = v5 - —, as given so, \x « —— « 3-2 x 10 m /(V • s) 3.269 Due to the straight conductor, Bm = We use the formula, F = (p^- V) B (a) The vector jj£ is parallel to the straight conductor. because neither the direction nor the magnitude of B depends on z (b) The vector pm is oriented along the radius vector r The direction of B at r + dr is parallel to the direction at r. Thus only the cp component of F will survive. * Pm dr 2jir (c) The vector/*^ coincides in direction with the magnetic field, produced by the conductor carrying current / 2jt

375 3.270 Fx = But, cft °°> 3.271 Vm dx x X ~ A — F — Ho 'f J 1-2 Rdl (J + Ri) Jt A J 6 Jt R Ipm x * R Y2 T7 — P r " ^2ot = ^2^ 3/2 ZJ a/ a dl 2 • Mo n Pirn /3 rp 2 — - K/2 r*-» —» 2 ' r5 - 3 H0i'lmi'2m O 14 3.272 From 3.270, for x » R, -5 x 2x3 -1mK or, i « -1 3.273 ^_= 2x3xl(rJrx(l(rim) 2" l-26xl0-6xA0-2mL 5'^ = 5 cos a, //', = —B sin a, 1 Ho - B't= B' H' SO, 3.114 (a)/ 9 nN X B \x sin a + cos a <iS= -(P J-dS, since Vaccum £-d£= 0 Now / is nonvanishing only in the bottom naif of the sphere. d!,H>

376 Here, 5' « B cos G, H' - — 5sin0, 5' - \iBsinQ, H'~ — cos G 71 Mo 1--) and Mo Only Jn contributes the surface integral and -S n lower lower x (b) ^ F-rfrt. {Bt-B't)l= r 3.275 Inside the cylindrical wire there is an external current of density —=-. This gives a magnetic txR field H^ with —7T or, R2 ' MMo u-1 /r Y/r From this Btn = T and / = -^ =■ = ——T = Magnetization. Hence total volume molecular current is, r- R The surface current is obtained by using the equivalence of the surface current density to Jx nf this gives rise to a surface current density in the z-direction of - The total molecular surface current is, The two currents have opposite signs. 3.276 We can obtain the form of the curves, required here, by qualitative arguments. From y H-dh* /, we get H (x » 0) - H (x « 0) - nl Then B (x » 0) - \i\ionl B(x -<0)= \ionl Also, B(x<0)= \i0H(x<0) / (jc < 0) - 0 B is continuous at x = 0, if is not These give the required curves as shown in the answer- sheet.

377 3.277 The lines of the B as well as H field are circles around the wire. Thus nr Hx xr+H2TLr= I or, Also Thus 1= \i2H2[i0= = B2= B H, \i2 nr and B 3.278 The medium I is vacuum and contains a circular current carrying coil with current /. The medium II is a magnetic with permeability \i. The boundary is the plane z- 0 and the coil is in the plane z = /.To find the magnetic induction, we note that the effect of the magnetic medium can be written as due to an image coil in II as far as the medium I is concerned. On the other hand, the induction in II can be written as due to the coil in I, carrying a different current. It is sufficient to consider the far away fields and ensure that the boundary conditions are satisfied there. Now for actual coil in medium /, Bx Br X I 2p cos 0' W PmSm so, B2 = where B cos 0 - sin 0 ) and 2T (- 3 sin 0' cos 0') pm = / (jt or), a = radius of the coil. Similarly due to the image coil, Bz = B cos2 0' - sin2 0'), Bx = C sin 0' cos 0'), p'm = /' (* a2 ) As far as the medium II is concerned, we write similarly - B cos2 0' - sin2 0'), Bx = = I"{na2)

378 The boundary conditions are, pm +p'm - p"m (from Bu ~Pm+P'm= -\p"m (from Hlt = 7^) Thus, /" /, 7 /, r f / Ji + 1 |A + 1 In the limit, when the coil is on the boundary, the magnetic field enverywhere can be obtained by taking the current to be —^—/. Thus, B = —+ Bn 3.279 We use the fact that within an isolated uniformly magnetized ball, Hr - - J/3, , where J is the magnetization vector. Then in a uniform magnetic field with induction BQy we have by superposition, or, ^ + 2^ 3^ also, 3.280 The coercive force Hc is just the magnetic field within the cylinder. This is by circulation theorem, Hc = f = 6 Wm (from (y H-d7*= /, total current, considering a rectangular contour.) 3.281 We use 0 Neglecting the fringing of the lines of force, we write this as + —b= 0 or, 101 A/m The sense of H is opposite to B 3.282 Here,* H-dl= NI or, J tt Hence, pi = 3.283 One has to draw the graph of \i = \i0NI-Bb JD , so, = 3700 NI\iQ-Bb versus if from the given graph. The \x - H graph starts out horizontally, and then rises steeply at about H = 0-04 k A/m before falling agian. It is easy to check that \i « 10,000.

379 3.284 From the theorem on circulation of vector H . B b d + — - \i0 or, A-51-0-987)i/, where B is in Tesla and H in kA/m. Besides, S and H are interrelated as in the Fig. 3.76 of the text. Thus we have to solve for B, H graphically by simultaneously drawing the two curves (the hysterisis curve and the straight line, given above) and find the point of intersection. It is at H~ 0-26 kA/m, B - 1-25 T Then, 3.285 From the formula, \x - 4000. Thus -£- f (IW)B dV or since B is predominantly along the jc-axis, \x\x0 3.286 The force in question is, dx since B is essentiatlly in the x-direction. So, x dx d dxK This is maximum when its derivative vanishes 2 x2 i.e. 16 a2 x2 - 4 a « 0, or, jc ot V4a The maximum force is, So, X VB2 - 3-6 x 10 3.287 dB \x0 This force is attractive and an equal force must be applied for balance. The work done by applied forces is, X'L . i. y VB2 X' 0

380 3.6 ELECTROMAGNETIC INDUCTION. MAXWELL'S EQUATIONS 3.288 Obviously, from Lenz's law, the induced current and hence the induced e.m.f. in the loop is anticlockwise. From Faraday's law of electromagnetic indcution, Here, and from Hence, dt = BdS = -2Bxdyy a using t at wy 3.289 Let us assume, B is directed into the plane of the loop. Then the motional e.m.f. I dl = vBl and directed in the same of (v x B) (Fig.) So, Bvl L2 As J?! and R2 are in parallel connections. R1R2 RrtRz 3.290 (a) As the metal disc rotates, any free electron also rotates with it with same angular velocity co, and that's why an electron must have an acceleration co2r directed towards the disc's centre, where t is separation of the electron from the centre of the disc. We know from Newton's second law that if a particle has some acceleration then there must be a net effecetive force on it in the direction of acceleration. We also know that a charged particle can be influenced by two fields electric and magnetic. In our problem magnetic field is absent hence we reach at the conslusion that there is an electric field near any electron and is directed opposite to the acceleration of the electron. If E be the electric field strength at a distance r from the centre of the disc, we have from Newton's second law. = mw n „ 2 „ m co2 r eE= m r co , or, E = and the potential difference, as 0 0

381 Thus - A <p - — y = 30 IlV (b) When field B is present, by definition, of motional e.m.f. : 2 dr Hence the sought potential difference, J ~ J - > (aS V= COT) 0 Thus <Prim-<Pcen= <P = o 1 a2= 20 mV eB (In general co < — so we can neglect the effect discussed in A) here). 3.291 By definition, So, A A 0 But, v = cor, where r is the perpendicular distance of the point from A. 10 mV Hence, I E-dr*= I -<&Brdr= ---(nBd2= - a o This result can be generalized to a wire AC of arbitary planar shape. We have c c c J F- d?= - J ( ?x F) • dr = - f ((co x O x iT) • dT --/ d being AC and r being measured from A. 3.292 Flux at any moment of time, where cp is the sector angle, enclosed by the field. Now, magnitude of induced e.m.f. is given by,

382 ~dt BR2dq> BR co, 2 dt where co is the angular velocity of the disc. But as it starts rotating from rest at t - 0 with an angular acceleration p its angular velocity co (t) = fit. So, in 2 According to Lenz law the first half cycle current in the loop is in anticlockwise sense, and in subsequent half cycle it is in clockwise sense. 2 Thus in general, |^ « (- 1) nBR t9 where n in number of half revolutions. The plot ^ (f), where tn = V2 k n/$ is shown in the answer sheet 3.293 Field, due to the current canying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by, B = -— where r is the perpendicular distance from the wire. 2k r r r As B is same along the length of the rod thus motional e.m.f. 2 and it is directed in the sense of (v*x B ) So, current (induced) in the loop, m R 2 nRr 3.294 Field, due to the current carrying wire, at a perpendicular distance x from it is given by, Motional e.m.f is given by dl There will be no induced e.m.f. in the segments B) and D) as, v*^ t d I and magnitude of e.m.f. induced in 1 and 3, will be (a + x)} respectively, and their sense will be in the direction of (v*x B). So, e.m.f., induced in the network = 61-69 fas 5i > av\ioi 2k 1 1 x a+x 2kx(ci+x)

383 3.295 As the rod rotates, an emf. dtl is induced in it The net current in the conductor is then R A magnetic force will then act on the conductor of magnitude BI per unit length. Its direction will be normal to B and the rod and its torque will be R dxBx Obviously both magnetic and mechanical torque acting on the CM. of the rod must be equal but opposite in sense. Then for equilibrium at constant co or, R mgR D Or 2 2 mga sin cor sin cor (a3B2 co + 2 mgR sin co r) 2 aB 2 aB (The answer given in the book is incorrect dimensionally.) 3.296 From Lenz's law, the current through the connector is directed form A to B. Here |^ = vBl between A and B where v is the velocity of the rod at any moment. For the rod, from Fx = mwx or, mg sin a - HB - mw For steady state, acceleration of the rod must be equal to zero. Hence, mg sin a = i IB A) But, R From A) and B) v - vBl R mg sin a R B212 3.297 From Lenz *s 1 aw, the current through the copper bar is directed from 1 to 2 or in other words, the induced crrrent in the circuit is in clockwise sense. Potential difference across the capacitor plates, C or> <1 " C

384 Hence, the induced current in the loop, dt ~ dt But the variation of magnetic flux through the loop is caused by the movement of the bar. So, the induced e.m.f. ^ = B Iv and, Hence, dt dt dt = CBlw Now, the forces acting on the bars are the weight and the Ampere's force, where Famp= HB(CBlw) B= Cl2B2w. From Newton's second law, for the rod, Fx = mwx or, mg sin a - C I2B2 w = mw Hence w = mgsina Cl2B2 + gsina m 1.298 Flux of By at an arbitrary moment of time t : B -S « B —— cos co ty From Faraday's law, induced e.m.f., §^ dt / 2 \ Bn — cos co r \ / dt B na co . z sm co t. and induced current, in Bit a 2R Now thermal power, generated in the circuit, at the moment t = t: 2 ' 2 \ B%a co — sin co t R and mean thermal power generated, 1 r 2 — I sin co t dt R J 0 dt 2R a2 B 0 Note : The claculation of ^n which can also be checked by using motional emf is correct even though the conductor is not a closed semicircle , for the flux linked to the rectangular part containing the resistance R is not changing. The answer given in the book is off by a factor 1/4.

385 3.299 The flux through the coil changes sign. Initially it is BS per turn. Finally it is - BS per turn. Now if flux is <$ at an intermediate state then the current at that moment will be -N dt R So charge that flows during a sudden turning of the coil is q= [idt= -^[<D-(-<D)]« 2NBS/R J R Hence, 1 qR 2NS 0-5 T on putting the values. 3.300 According to Ohm's law and Faraday's law of induction, the current i0 appearing in the frame, during its rotation, is determined by the formula, Ldl o ^ dt dt Hence, the required amount of electricity (charge) is, fio dt- - 3>2- Since the frame has been stopped after rotation, the current in it vanishes, and hence A /0 = 0. It remains for us to find the increment of the flux A<$> through the frame Let us choose the nonnal n to the plane of the frame, for instance, so that in the final position, n*is directed behind the plane of the figure (along B ). * Then it can be easily seen that in the final position, <$>2 > 0> while in the initial position, $x < 0 (the normal is opposite to B ), and A <£ turns out to be simply equal to the fulx through the surface bounded by the final and initial positions of the frame : b + a A <£> <£>2 + jl - J B a dr, b-a where B is a function of r, whose form can be easily found with the help of the theorem of circulation. Finally omitting the minus sign, we obtain, A<X> Mo a 11 b + a 3301 As By due to the straight current carrying wire, varies along the rod (connector) and enters linerarly so, to make the calculations simple, B is made constant by taking its average value in the range [a, b].

386 b b r D . r h0 h . I Bdr I dr J J 2n r b b fdr Jdr or, lo t b 2 ji (b - fl) fl (a) The flux of B changes through the loop due to the movement of the connector. According to Lenz's law, the current in the loop will be anticlockwise. The magnitude of motional e.m.f., ,- v<B>(b-a) 2it (b-a) So, induced current b ,, a n r* 2ji • IT In Ho . 2 7t (b) The force required to maintain the constant velocity of the connector must be the magnitude equal to that of Ampere's acting on the connector, but in opposite direction. So, o:/ "in lQ 2xR a (b-a) «b - a) a v H . ln~ a , and will be directed as shown in the (Fig.) 3302 (a) The flux through the loop changes due to the movement of the rod AB. Recording to Lenz's law current should be anticlockwise in sense as we have assumed B is directed into the plane of the loop. The motion e.m.f |^(r) * Blv and induced current L in vBl R From Newton's law in projection form Fx= mwx vdv But So, Off^ vB2/2 dv —-—*- mv — R dx

387 o or, /. mR C . wR ax « - o o I dv or, jc - —r— 52/2J 52/ 'o (b) From equation of energy conservation; Ef - Ei + Heat liberated 1. 2 o + Heat liberated -0 + 0 So, heat liberated = — m v0 3.303 With the help of the calculation, done in the previous problem, Ampere's force on the connector, -* vB2l2 Famn * —n— directed towards left. amp fl Now from Newton's second law, A dv 2l2 So, vB2l —- R dv — dt or, I dt- m I dv B ® ^- ^ , R > n 0 0 F- or, Thus m B2l2 1-e In R -tB2l2\ RF Rm B2l2 3304 According to Lenz, the sense of induced e.m.f. is such that it opposes the cause of change of flux. In our problem, magnetic field is directed away from the reader and is diminishing. (a) (b) So, in figure (a), in the round conductor, it is clockwise and there is no current in the connector In figure (b) in the outside conductor, clockwise.

388 In figure (c) in both the conductor, clockwise; and there is no current in the connector to obey the charge conservation. In figure (d) in the left side of the figure, clockwise. 3.305 The loops are connected in such a way that if the current is clockwise in one, it is anticlockwise in the other. Hence the e.m.f. in loop b opposes the e.m.f. in loop a. d 2 id e.m.f. in loop a ■ — (a B) - a -~ (Bo sin cor) Similarly, e.m.f. in loop b = b Bo co cos cor. Hence, net e.m.f. in the circuit «■ (a - b ) Bo co cos cor, as both the e.m.f *s are in opposite sense, and resistance of the circuit = 4 (a + b) p Therefore, the amplitude of the current (a2 - b2) Bo co 4 (a + b) p 0-5 A. 3.306 The flat shape is made up of concentric loops, having different radii, varying from 0 to a. Let us consider an elementary loop of radius r, then e.m.f. induced due to this loop = n r J5 co cos cor. ° dt and the total induced e.m.f., a / (*r2B0<x>coso)t)dN, A) o where ji r 2 co cos cor is the contribution of one turn of radius r and dN is the number of turns in the interval [r, r + dr\ So, dN- @)dr B) a 2 /2 tf tzBq(o cos co tN a - (jt r B0<n cos cor) — dr « r 0 1 2 Maximum value of e.m.f. amplitude ^^ = — nB0(&Na 3.307 The flux through the loop changes due to the variation in B with time and also due to the movement of the corinector. So, d(B-S) dt dips) dt as 5 and B are coll iniear 1 2 But, 5, after r sec. of beginning of motion « J5r, and S becomes = l — wt , as connector starts moving from rest with a constant acceleration w. So,

389 3.308 We use B - \xQ n I Then, from the law of electromagnetic induction / dt So, for r < a 2 1 E 2 % r - - % r \ionl or, E - - — |i0 n r /. (where / - dl/dt) For r>a Jt # ™" Jl U pin f* J ^*> CD *i) The meaning of minus sign can be deduced from Lenz's law. 3.309 The e.m.f. induced in the turn is \ionln — m_ . . . 71 d The resistance is -— p. \xon!Sd So, the current is = 2 m A, where p is the resistivity of copper. 4p 3.310 The changing magnetic field will induce an e.m.f. in the ring, which is obviously equal, in the two parts by symmetry (the e.m.f. induced by electromagnetic induction does not depend on resistance). The current, that will flow due to this, will be different in the two parts. This will cause an acceleration of charge, leading to the setting up of an electric field E which has opposite sign in the two parts. Thus, ^= rl and, ~ + 7ia where % is the total induced e.m.f. From this, and 2%a 2 But by Faraday's law,^ - Tea b so, E= -rah 2 rahr 2 rj + 1 3.311 Go to the rotating frame with an instantaneous angular velocity aT(f). In this frame, a Coriofis force, 2m v x co (t) acts which must be balanced by the magnetic force, e v*x B (t\ Thus, c^W38 -7T~B(t) . 2 m (It is assumed that a?is small and varies slowly, so co and co can be neglected.)

390 3.312 The solenoid has an inductance, L = \i0 n2 x b 2 /, where n = number of turns of the solenoid per unit length. When the solenoid is connected to the source an e.m.f. is set up, which, because of the inductance and resistance, rises slowly, according to the equation, This has the well known solution, /-jfd-e-'*1). Corresponding to this current, an e.m.f. is induced in the ring. Its magnetic field dF dl -tR/L ,4 -tR/L dF , 2 2 2 B - fi0 n I in the solenoid, produces a force per unit length, -jf = B i = \i0 n Ka 11/r 2*a2V2 RL acting on each segment of the ring. This force is zero initially and zero for large t. Its maximum value is for some finite t. The maximum value of N2 - tR/L \ L I -L _ - tR/L So dF. max 2 T72 a V dl 4rRlb 3.313 The amount of heat generated in the loop during a small time interval dt, So, f/Rdt, but, dQ- K and hence, the amount of heat, generated in the loop during the time interval 0 to t. ; Bat-ax) R a2x3 0 3.314 Take an elementary ring of radius r and width dr. ' 2 The e.m.f. induced in this elementary ring is n r p. Now the conductance of this ring is. R hdr p2nr so hrdr 2p Integrating we get the total current, b J hrdr h p (b2 - a2) T^P= 4p

391 2 11 3.315 Given Z, * \xQn V~ \xon lonR , where R is the radius of the solenoid. Thus, So, length of the wire required is, v 4nLl0 /= nlo2nR = V - 010km. 3316 From the previous problem, we know that, VL14K /' = length of the wire needed- V , where / ■ length of solenoid here. Po'' Now, R = -—, (where S = area of crossection of the wire. Also m = p S V) , RS Rm Thus, Po PPo where p0 = resistivity of copper and p = its density. c ,. Rm LI Equating, = —7-— PPo H-o'4 ^ ^0 mR or, 4 jr pp0 / 3.317 The current at time t is given by, The steady state value is, 70 - —• R and ^ r-L-lnr=^ or> '•-itar^-1'491 3.318 The time constant x is given by 1 1 x - —- — P05 where, p0 - resistivity, /0 - length of the winding wire, S - cross section of the wire. But m « / p0 5 Z, So eliminating 5,x = ^ '0 ppo/o m/p/0

392 From problem 3.315 l0 = (note the interchange of / and /0 because of difference in notation here.) mL , m Thus, x = \io4x PPo —L PPo/ 0-7 ms, 3.319 Between the cables, where a<r<b> the magnetic field H satisfies So lf,2*r-J or, B. r- The associated flux per unit length is,<I> = I — J 2nr x 1 x dr = r= a In — Hence, the inductance per unit length L1 = — = ——In r\9 Where r] = — We get L1 - 0-26 3.320 Within the solenoid,//-2 nr= NI or ff B - and the flux,^ = = N — JS/J I 2n J r Finally, «■ flln 1 + r 6 3.321 Neglecting end effects the magnetic field B, between the plates, which is mainly parallel to the plates, is B = ^ — (For a derivation see 3.229 b) Thus, the associated flux per unit length of the plates is, <$>= |io~x/ixl = b -1 x /. \ j^ = inductance per unit length = \i0 — « 25 nH/m.

393 3.322 For a single current carrying wire,/? - ~— (r > a). For the double line cable, with current, flowing in opposite directions, in the two conductors, , between the cables, by superposition. The associated flux is, d-a C I J drxl M. d Ho. T .fl . In — = — In ri x /, per unit length an n r nan Ho Hence, JL = —\nr\ 1 n is the inductance per unit length. 3.323 In a superconductor there is no resistance, Hence, T dl L — = + dt dt So integrating, na2B because A<$> = Of - ^, Of = n a2 B, Ot = 0 Also, the work done is, A = J '%Idt = I I dt dt - l T T2- l \T C V H 3.324 In a solenoid, the inductance Z, = HM-qW V= |i Ho—-—, where S - area of cross section of the solenoid, / - its length, V - 57, N = nl ■ total number of turns. When the length of the solenoid is increased, for example, by pulling it, its inductance will decrease. If the current remains unchanged, the flux, linked to the solenoid, will also decrease. An induced e.m.f. will then come into play, which by Lenz's law will try to oppose the decrease of flux, for example, by increasing the current. In the superconducting state the flux will not change and so, = constant Hence, - = j-, or, /= Io - = Io A +11) 3.325 The flux linked to the ring can not change on transition to the superconduction state, for reasons, similar to that given above. Thus a current / must be induced in the ring, where, 3> na2B naB L /. 8a_^ / ft Sa ^ In —- r b /

394 3326 We write the equation of the circuit as, r\ dt for t £ 0. The current at t - 0 just after inductance is changed, is i - "H d ' so ^at ^e ^ux through the inductance is unchanged. We look for a solution of the above equation in the form i= A+Be -t/c L Substituting C - — ,J5- t| - 1,A - ^ Thus, - lj e -t)Rt/L di 3.327 Clearly, Lj(= R(I-i) =?-/?/ So, ZLj This equation has the solution (as in 3.312) 1 <^> R T-i 7 R 3.328 The equations are, Li Then, jt (I1i1-Z2i2)= 0 or, Lli1-L2i2SSi constant But initially at t - 0, ix - ^ ™ ^ so constant must be zero and at all times, L2 h = i 5 In the final steady atate, current must obviously be ix + i2 - ■?■ . Thus in steady state 3.329 Here,J5 at a distance r from the wire. The flux through the frame is obtained as, 2 it In a Thus, Z,12 = * 12 a

395 3.330 Here also, B » -— and 2 nr Thus, / fhdr N [i\iohN a 3.331 The direct calculation of the flux 3>2 is a rather complicated problem, since the configuration of the field itself is complicated. However, the application of the reciprocity theorem simplifies the solution of the problem. Indeed, let the same current / flow through loop 2. Then the magnetic flux created by this current through loop 1 can be easily found. Magnetic induction at the centre of the loop, : B = 2b So, flux throug loop 1, : <£>12 = 2b and from reciprocity theorem, <Pl2 " *21 > *21 = '21 1 2- \iQjia i 2b So, Mo 3332 Let pm be the magnetic moment of the magnet M. Then the magnetic field due to this magnet is, n / trtn r r The flux associated with this, when the magnet is along the axis at a distance x from the centre, is 3 (pf- r) f 4 Ji. - ^o r 2ji where,<l>2= -—p I —■=— *n *J (x + 0 and 3 1*0A HP, m f l So, a' 2 (x2 + a2f/2 / 2 2V (* +P) 3/2 PjSf 2n n J (x + o A _ 1

396 When the flux changes, an e.m.f. - N ^~r- is induced and a current - — ZL-r- flows. The dt R dt total charge qy flowing, as the magnet is removed to infinity from x = 0 is, or, 3.333 If a current / flows in one of the coils, the magnetic field at the centre of the other coil is, 2aqR B \xoa2l 3/2 2/ , as /» a. The flux associated with the second coil is then approximately \x0 jc a 1/2 I Hence, \x0 na dl, 3.334 When the current in one of the loop is 7: = at, an e.m.f. L12—7—= L12a, is induced i the other loop. Then if the current in the other loop is 72 we must have, This familiar equation has the solution, -tR L12 ex / L \ I2 = —-— 1 - e 2 which is the required current R 3.335 Initially, after a steady current is set up, the current is flowing as shown. In steady condition i20 = ^ , i10 = -~ • a Ro When the switch is disconnected, the current through Ro changes from i10 to the right, to i20 to the left. (The current in the inductance cannot change suddenly.). We then have the equation, dL In == U. dt This equation has the solution / The heat dissipated in the coil is, L I Sw R. 0 0 + KoyLdt Rl™ *2(R + R0)~ 2R(R+R0)

397 3.336 To find the magnetic field energy we recall that the flux varies linearly with current Thus, when the flux is 3> for current i, we can write <(> = A L The total energy inclosed in the field, when the current is /, is [\idt= f J * J W [\idt f N^ J * J dt N A i di o The characteristic factor — appears in this way. 3.337 We apply circulation theorem, H-2nb- NIy or, H» NI/2nb. Thus the total energy, %2a2 b BE. Given N,/, b we know Hy and can find out B from the B -H curve. Then W can be calculated. 3.338 From 6 H-dr*= NIy B_ Mo NI Also, B ■ u,iinH. Thus, H = r ^ ^u nd+\xb Since B is continuous across the gap, B is given by, NI B in the magnetic and the gap. «2 ^—xSxb (b) The flux is So, Energy wise; total energy B2 M- j 2 u

398 The Ly found in the one way, agrees with that, found in the other way. Note that, in calculating the flux, we do not consider the field in the gap, since it is not linked to the winding. But the total energy includes that of the gap. 3.339 When the cylinder with a linear charge density X rotates with a circular frequency co, a surface current density (charge / length x time) of . is set up. The direction of the surface current is normal to the plane of paper at Q and the contribution of this current to the magnetic field at P is dB dS where e is the direction of the current. In magnitude, |i^<r^= r, since £*is normal to r*and the direction of dB is as shown. It's component, dB» cancels out by cylindrical symmetry. The component that survives is, dBn ; cos 0 J dQ- where we have used dS cos 0 d Q and I d Q = 4 jt, the total solid angle around any point. The magnetic field vanishes outside the cylinder by similar argument. The total energy per unit length of the cylinder is, 8% a 3.340 wE = -£0£ , for the electric field, l w B for the magnetic field. B Thus, when 3.341 The electric field at P is, B 3 x 108 V/m ql .2 3/2

399 To get the magnetic field, note that the rotating ring constitutes a current i the corresponding magnetic field at P is, q co/2 ji, and B. a2 i 3/2 • Thus, W M f qlx2 ) 4 Jt e0 ji0 a i 1 f I a2 co w or, M fE 2 4 /i e0 \x0 co a // P 3342 The total energy of the magnetic field is, BB dV-irl J-BdV. The second term can be interpreted as the energy of magnetization, and has the density -\rt 3.343 (a) In series, the current I flows through both coils, and the total e.m.f. induced, when the current changes is, f dt dt or, (b) In parallel, the current flowing through either coil is, — and the e.m.f. induced is Equating this to -Z/ —, we find V = —L at 2 3344 We use So, 3345 The interaction energy is 1 C ,-*,2 1 C 2 H-o J II 2 \i0 J 1 f-*-*„„ _ ■ D . D Ji/ ■ ■I *■ ^ Here, if 5X is the magnetic field produced by the first of the current carrying loops, and B2y that of the second one, then the magnetic field due to both the loops will be Bx + B2-

400 3.346 We can think of the smaller coil as constituting a magnet of dipole moment, Its direction is normal to the loop and makes an angle 6 with the direction of the magnetic field, due to the bigger loop. This magnetic Geld is, = 2b The interaction energy has the magnitude, I tx/ I ^oA^2 2 Q ' I" ~~Th—n cos Its sign depends on the sense of the currents. 3.347 (a) There is a radial outward conduction current Let Q be the instantaneous charge on the inner sphere, then, •a 2 dQ r* 1 dQ a / x 4 n r = - -f- or, / = z -r~ r- J dt J 4nr2 dt dD d On the other hand Ad = -— = -r x/d dt dt ( Q A r - -j n a (b) At the given moment, E - 3 r r 4 ^c e0 e r and by Ohm's lawj ^ P 4 3t e0 e p r Then, 4 jc e0 e , *r\ ir, ~i m -~ -— COS V7 and r eoe P The surface integral must be -ve because jd, being opposite of;, is inward. 3.348 Here also we see that neglecting edge effects, jd~ -;. Thus Maxwell's equations reduce to,div 1?= 0, Curl H** 0, 5*- \iH A general solution of this equation is B - constant » Bo - Bo can be thought of as an extraneous magnetic field. If it is zero, B = 0. 3.349 Given / = Im sin cor. We see that I Im or, D - —- cos cor, so, Em = is the amplitude of the electric field and is 7V/cm

401 3350 The electric field between the plates can be written as, Vm . Vm E « Re —£ el w \ instead of —j- cos <ot a a This gives rise to a conduction current, and a displacement current, 3D Jd~ IF" Reeoel>(°-f el<of The total current is, jT » -j-vo2 + (e0 e coJ cos (co r + a) where, tan a a co on taking the real part of the resultant. The corresponding magnetic field is obtained by using circulation theorem, r=* nr2jT or, H - #m cos (cor + a), where, #m * ^ltV?+ (e0 e co): 3351 Inside the solenoid, there is a magnetic field, B n Im sin Since this varies in time there is an associated electric field. This is obtained by using, Edr dt J BdS For r< R,2%rE= -B-nr2, or, Br 2 For r> 2r The associated displacement current density is, dE \-*oBr/2 The answer, given in the book, is dimensionally incorrect without the factor 3352 In the non-relativistic limit 4 k e0 r" (a) On a straight line coinciding with the charge path, Jdm e0 BE q dt " 4% r3 , fusing, V dt - v.

402 But in this case, r » - v and v - » v? so, y^ « —^- 4 3T (b) In this case,r » 0, as, r*L v^Thus, 4;ir 3353 We have, £^ 4 ji e0 then j4m — . e0 —- ——f—5^j(a -2r) This is maximum, when x « jc^ « 0, and minimum at some other value. The maximum displacement current density is 4]tfl3 To check this we calculate •—— ; dx This vanishes for x - 0 and for * » y -• a. The latter is easily shown to be a smaller local minimum (negative maximum). 3354 We use Maxwell's equations in the form, when the conduction current vanishes at the site. We know that, —T~ V JQ» Ti£—2 ji A - cos 6), 4 Jt e where, 2ji A - cos 9) is the solid angle, formed by the disc like surface, at the charge. X —* —*■ 1 Thus, G> B-dr= 2xaB~ j\ioq-smQ-d On the other hand,* * a cot 0 differentiating and using — = - v, v = a cosec 0 0 \i0 q v r sin 9 Thus, 2*« , 43i r3

This can be written as, B 403 —"W 9 and if - -j1 — (The sense has to be checked independently.) 3355 (a) If F= lB{t\ then, Curl £ = Z^E- * 0. So, E cannot vanish. (b) Here also, curl E * 0, so E cannot be uniform. (c) Suppose for instance, E » a*f (t) where 5* is spatially and temporally fixed vector. Then - — = curl E = 0. Generally speaking this contradicts the other equation curl H'- —- * 0 for in this case the left hand side is time independent but RHS. depends on time. The only exception is when / (t) is linear function. Then a uniform field E can be time dependent. 3356 From the equation Curl H - ; at We get on taking divergence of both sides ^ "-* _ -— div D = div/ But div Z) = p and hence div ; + ^ ■ 0 ot 3357 From Vx£ = - ot we get on taking divergence 0= - — di dt This is compatible with div B « 0 3358 A rotating magnetic field can be represented by, Bx = 2?0 cos co t; 5^ ■ Bo sin cor and Bz « 5^ Then curl, /?■- So, - (Curl E )x = - co ^0 sin cor = - y - (Curl E )y * co J50 cos cor « cofi^ and - (Curl £ )z * 0 Hence, Curl E » - ccTx ^ , where, co » e3 co .

404 3359 Consider a particle with charge ey moving with velocity v? in frame K. It experiences a force F = ev*x It In the frame K*> moving with velocity irrelative to Kt the particle is at rest. This means that there must be an electric field E in K, so that the particle experinces a force, Thus, F- ^x? 3360 Within the plate, there will appear a (v*x B) force, which will cause charges inside the plate to drift, until a countervailing electric field is set up. This electric field is related to B, by E ** eBt since vScB are mutually perpendicular, and E is perpendicular to both. The charge density ± a, on the force of the plate, producing this electric field, is given by E* ~ or o«e0v5« 0-40pC/m2 3361 Choose ccTtT B along the z-axis, and choose rj*as the cylindrical polar radius vector of a reference point (perpendicular distance from the axis). This point has the velocity, v - co x r, and experiences a (vxB) force, which must be counterbalanced by an electric field, E » - (ccTx r"} x B » - (oT- B ) rt There must appear a space charge density, p » e0 div E » - 2 e0 <o- B » - 8 pC/m Since the cylinder, as a whole is electrically neutral, the surface of the cylinder must acquire a positive charge of surface density, 2e0(co-2?)jia _*-♦ 2 a = + » eft a co • B » + -2 pC/m 2«a ° r 3362 In the reference frame K, moving with the particle, XD — £f u B-vxE/c => 0. Here, v^« velocity of K\ relative to the K frame, in which the particle has velocity Clearly, Vq " vTFrom the second equation, C 2 jteo ,3

405 3363 Suppose, there is only electric field Ey inX. Then in K'y considering nonrelativistic velocity v xE E'B'- 0 So, In the relativistic case, E1 \\" E\\ El V 1 - vVc2 B B1 - v*x E/c Vl - Now, ' • 5' - £'„ • 5'n + £'x • B\ - 0, since \-B\" - Et • (yx 3.364 lnK,B- b^f^r, b - vVc2) constant 0 A A In A A The electric field is radial (r^ xi+yj ). 3365 Z^ A A = a-^r*= (xi+yj r In The magnetic lines are circular. v xE arxv c2 ' c2? 2 , 2 3366 In the non relativistic limit, we neglect v /c and write, EL m EL + v x B '± * 5± - vxE/cA These two equations can be combined to give, 3367 Choose E in the direction of the z-axis, E ■ @,0, E). The frame K! is moving with velocity v * (v sin a, 0, v cos a), in the x - z plane. Then in the frame K'y E n m Ew B Km 0 The vector along v is e is, Vi - vw x vr^ (sin a, 0, cos a) and the perpendicular vector in the x-z plane /= (- cos a, 0, sin a),

406 (a) Thus using E = E cos a e + E sin a /, E\\ - E cos a and E „ - , =, " " Vl - vVc2 So £'-£ Vi= cos2 a 1-P and tana' tan a Vl - vVc2 (b) fi'n = 0, 2?'x - Vi- B' pjgsina c Vl - P2 3368 Choose 5 in the z direction, and the velocity v^» (v sin a, 0, v cos a) in the x - z plane, then in the K' frame, vxB 1 V 1 - vVc2 B 1 Vl - vVc2 We find similarly, E' c p B sin a Vi - p Br= B A/1 - V i-P tan a' tana 3.369 (a) We see that, '„ • B\ + £'± • 5'± v xE 2 1 " 2 C 1 ~* ~~* /—' £■. • 5, — (v' xB 2 - V X XJ ) • (v x 7c2 E xl 2 -7 But, - A CB -D -A DBCy 2\ 1- so, (b) £'2 - | - c2 B'\ + ^'i - c2 B'

407 2 -2 n2 ^ - C D\\ + V2 2 vx£ c2 ' 2 ^ J - \ (?*x £x c 1- 1- .2\ I = E2-c2B2, since, 3370 In this case, E * B ** 0, as the fields are mutually perpendicular. Also, 2 c252 20 x 108 f E2 - - 20 x 108 f—| is - ve. Thus, we can find a frame, in which E' = 0, and B'= -\c2B2-E2 » fi c A/ S2 -%/ Vl--f7 = 0-20 VI- 4xlO4 3xl08x2xl0 = 015 mT 3371 Suppose the charge q moves in the positive direction of the jc-axis of the frame K. Let us go over to the moving frame K\ at whose origin the charge is at rest. We take the x and x' axes of the two frames to be coincident, and the y &.yf axes, to be parallel. In the K' frame, E 4;ie0 r'3 ' and this has the following components, 4 ^c e 4 % e 0 r y 4 % e0 r' Now let us go back to the frame K. At the moment, when the origins of the two frames coincide, we take t * 0. Then, x = r cos 0 jc' y 1 - -jT , y = r sin 0 = y' Also, E=E'E=E'Hl-v2/c2 From these equations, r ,2 r2 A - P2 sin2 0) E = 2 . 2^3/2 4*eor3(l-p2sin26) P2K/2 f A 4 Jt e0 r3 A - p2 sin2 0K/2 J

408 3.7 MOTION OF CHARGED PARTICLES IN ELECTRIC AND MAGNETIC FIELDS 3372 Let the electron leave the negative plate of the capacitor at time t = 0 AS * dxy i r and, therefore, the acceleration of the electron, eE eat dv_ eat_ m ml ' dt ml or, o o But, from s = J vdt. /, ea C » 1 ea 2 ^\ ml J y y 2 ml v y , 1 ea C 7 j ea? I ■ ~ —T \ t dt= -—r or, f 2 m/ J 6 m/ 6m/2N ea o Putting the value of t in A), leaFml2Y (9 aleV 2 1 2 ml[ ea J 12 m ™ 16km/s. 3.373 The electric field inside the capacitor varies with time as, E= at. Hence, electric force on the proton, F" eat and subsequently, acceleration of the proton, eat m Now, if t is the time elapsed during the motion of the proton between the plates, then t = —, as no acceleration is effective in this direction. (Here v is velocity along the length Vii " of the plate.) dvL From kinematics, —r— - w j dvL « J wdt, so, 0 0 (as initially, the component of velocity in the direction, X to plates, was zero.)

409 /ea t ml 2 eal2 or „ - 2m 2m o vl eal2 Now, tan a vn 2mvi i eal2 ■ BeV\* , as vM - I 1 , from energy conservation. m 2 m - A/ ~ 4 3.374 The equation of motion is, dv dv q d m Integrating \ 2 Q 1 2 — v - -1- (£ft jc - — ojc ) = constant. 2 m v 2 But initially v - 0 when x - 0, so "constant" = 0 Thus, v m ' 1 2 \ Thus,v = 0, again for x m The corresponding acceleration is, m 3.375 From the law of relativistic conservation cf energy 2 = mnc . as the electron is at rest (v - 0 for x - 0) initially. Thus clearly T= eEx. 2 or, or, l-(vz/c') = v V (m0 ( c ct-fcdt m moc c2 + e m0c2 -/ oc + eEx ExJ - ml cA + eEx \l( 2 F \2 2 4

410 The "constant" - 0, at r - 0, for x « 0, So, cf- Finally, using J= it ceE As before, T** eEx Now in linear motion, q c4 . +2m0c) or, V7(T+2/w0c2) mow v dt Vl - Vl - v2/c2 A - m 0 A - vVc2) moc w So, 3.377 The equations are, i dt Hence, m o 1 + r ) -3 = 0 and Vl - v2/c2 eE VW/c2 Also, by energy conservation, = constant Vl - (v^/c2) Vl- + e£y Dividing voeo TO0C o Vl - Vl - Also, Thus, "constant" = 0 as vy - 0 at f m 0 c eEt + constant. 0. Integrating again, 1 2 1 2 2 + constant. - m0 c + constani

411 "constant" = 0, as y = 0, at t = 0. e £ tf - (ey EJ + 2e0 or, or, Hence, and q ~ eo ceEt = V (e0 + eEyf - 6q also, v tan9 v. 3.378 From the figure, sma= — m v /n v As radius of the arc R - —— > where v is the qB velocity of the particle, when it enteres into the field. From initial condition of the problem, qV= —mv2 or, v Hence, sin a = dgB m = dB 2mV m and a s sin -l dB v 2mV 30°, on putting the values. 3.379 (a) For motion along a circle, the magnetic force acted on the particle, will provide the centripetal force, necessary for its circular motion. i.e. mv ~R and the period of revolution,J evB or, v 2ji CO eBR m 2%m eB (b) Generally, dj?_ d_ ' <* = <*'Vl - (v2/c2)" Vl-(v2/c2)' For transverse motion, v • v = 0 so, mov mov m o 3/2 mov m o Vl - (v2 / c2) here.

412 Thus, m0v2 rVl-(v2/c2) B ev or, v/c Ber Vl - (v2 / c2) mo or, c Ber Finally,, 2 Jim o eB 2 -2 . __J2 J2 r2 + mo 3.380 (a) As before,/? - B qr. (b) (c) using the result for v from the previous problem. 3.381 From C.279), j_ Jt e (relativistic), T. 2 Jim, 0 c2eB (nonrelativistic), Here, Thus, Now, m0c2/Vl-v2/c2 - E 67 — > W-K.E.) , so, 3.382 T= (The given potential difference is not large enough to cause significant deviations from the nonrelativistic formula). Thus, m So, Now, and Pitch m cos a, v± sin a m m p- V«T" Be .Bev, or, r= 2 Jim Be cos a - 2 mv cos a

413 3383 The charged particles will traverse a helical trajectory and will be focussed on the axis after traversing a number of turms. Thus / 2 ji/w ( 1 v 2 Tun vA qB1 qB, So, Hence, n n +1 i / B2-Bx lion v0 q (B2 - Bx or, m or, 3.384 Let us take the point A as the origin O and the axis of the solenoid as z-axis. At an arbitrary moment of time let us resolve the velocity of electron into its two rectangular components, v.i along the axis and v£ to the axis of solenoid. We know the magnetic force does no work, so the kinetic energy as well as the speed of the electron | v± | will remain constant in the x-y plane. Thus v£ can change only its direction as shown in the Fig.. Vii will remain constant as it is parallel to B. Thus at t = t vx = vL cos cor = v sin a cos cor, and v± sin cor « v sin a sin co r u eB vr = v cos a , where co = — z m As at r = 0, we have x = y = z = 0, so the motion law of the electron is. z = vcosar vsina CO vsina co smco r (cos cor - 1) (The equation of the helix) On the screen, /, so r = vcosa Then, 2 2 2 2 v^ sin* a /. co/ r « x + v - ^ 1 - cos co vcos a r = 2 v sin a co sin co/ 2 v cos a eB sin a sm 2 mv cos a

414 3.385 Choose the wire along the z-axis, and the initial direction of the electron, along the jc-axis. Then the magnetic field in' the x - z plane is along the y-axis and outside the wire it is, -0, if y-0) The motion must be confined to the x-z plane. Then the equations of motion are, d Multiplying the first equation by vx and the second by vz and then adding, v dvr dt z dt or, v2 + v2 - v2 , say, or, vz - V vg - v\ dx m u x 2nx _ dx °r> /~r J 2 Jim a: V2 on usingjV^ = v0 , if x = a (i.e. initially). Now, v^ = 0, when x = jcm, so, xM= aev° , where 6 = m 2iun 3.386 Inside the capacitor, the electric field follows a — law, and so the potential can be written as _ V lnr/a -V 1 ^ \nb/a ' Inb/ar' Here r is the distance from the axis of the capacitor. A1 mv2 qV 1 2 Also, = 7—^- or mv r \nbIar r = 7^or mv \ r \nbIar r \nbIa On the other hand, mv - 4 5 r in the magnetic field. Thus, v = ——:—7-7— and -^ = tt- = BrXnbla m Br :77 and tt ^1 BrXnbla m Br B2r2ln(b/a)

415 dvx dv. dv. 3.387 The equations of motion are, dvy t —f-« qE and These equations can be solved easily. First, vy - Then, v2x + v2z = constant - v\ as before. In fact, vx = v0 cos cor and vz « v0 sin cor as one can check. Integrating again and using x=z=0, atf=O m x » — sin cor, z = — A - cos cor) co co Thus. 2 = 0 for n At that instant, yn = ^- x —=-.— x n2 x Also, 2n co 2K2mEn2 tan an = —, (vz = 0 at this moment) 0 gB 1 X -*—X 'o ^f^rn <?£ m 2jtn 2nEn 3.388 The equation of the trajectory is, )t), y = ^r^2 as before see C.384). Now on the screen x = /, so y y x = — sin cor, z = — A - cos coi CO CO co/ . _ i co/ sin cor« — or, cor- sin — At that moment, qE I . _ico/ 'sin — so, and CO/ 0 sui . -v/2mco2y ui V =r^- qE vo0 . 2 cor —2 sin -— 2 - sin . \/2 q B2y in V —*;—z- Em 2 For small CO ' . - 1 CO/ sm — f cor /tan--- 2 or, ^2y_ 2^~ T" / 2mE z2 qB2i . z is a parabola.

416 3389 In crossed field, eE = evBy so v= - B Then^F = force exerted on the plate = - x m — * —— F e B eB 3*390 When the electric field is switched off, the path followed by the particle will be helical, and pitch, A/ = v J, (where v is the velocity of the particle, parallel to B> and J, the time period of revolution.) = v cos (90 - cp) T ■ v sin cp T 2 Jim / _ 2ji ■= v sin cp —— as i» —— v qB \ qB A) Now, when both the fields were present, qE - qvB sin (90 - qp), as no net force was effective on the system. or, v = ■—— B) B cos qp From A) and B), A/ ■ •=• —— tan cp « 6 cm. 3 qB 3.391 When there is no deviation, _^ _+ -qE** q(v*xB ) or, in scalar from, £■ vfi(as vl5) or, v « — A) Now, when the magnetic field is switched on, let the deviation in the field be *. Then, where t is the time required to pass through this region. also, f= — v Thus x lfqvB) 2 m 2 r>2 £\ _ 1 q a'B 2m B) For the region where the field is absent, velocity in upward direction J m J m 32* ? Now, Ax - x m q qB b . i b bB /A. - ^ —=— when /'-—-—- D) m E v r From B) and D), Ax-^ or, 2 m E m E 2Ebx

3.392 (a) The equation of motion is, m dt ■^ ^ a f j k . Now, v x B 0 0 So, the equation becomes, qBvy dv iBy-jBx m vv, and ~= 0 dt mm* dt 417 Here, vx - i , v• - y, vz - z. The last equation is easy to integrate; vz - constant - 0, since vz is zero initially. Thus integrating again, z« constant" 0, and motion is confined to the x - y plane. We now multiply the second equation by i and add to the first equation. we get the equation, dt . E B m This equation after being multiplied by e'^can be rewritten as, ~(§e )= icoe B and integrated at once to give, where C and a are two real constants. Taking real and imaginary parts. vxss "^ + C cos (co r + a) and vy « - C sin (cor + a)^ £ Since vy = 0, when r = 0, we can take a = 0, then vx = 0 at r = 0 gives, C = - — and we get, is £ vx « -- A - cos cor) and v• ■ — sin cor. Integrating again and using x - El sin ( 0, at r » 0, we get E This is the equation of a cycloid. (b) The velocity is zero, when cor « 2 n Jt. We see that - ff) B-2coscor)

418 or, ds dt IE —— B sin cor —— 2 The quantity inside the modulus is positive for 0 < cor < 2 ji. Thus we can drop the modulus and write for the distance traversed between two successive zeroes of velocity. 4£ 1 - cos cor Putting cof= 2ji, we get SmE coB (c) The drift velocity is in the Jt-direction and has the magnitude, B% - A - cos cor) 3.393 When a current / flows along the axis, a magnetic field B = is set up where 2 2 2 p - x + y . In terms of components, \iolx 9 2 Jt p' and B«0 Suppose a p.d. V is set up between the inner cathode and the outer anode. This means a potential function of the form Inp/fr as one can check by solving Laplace equation. The electric field corresponding to this is, Vx „ Vy p2\na/b 0. The equations of motion are, d_ dt Jtmvy \e\Vz .MM... p2\na/b 2 ji p2 \e\Vy xz p2\na/b and d — tnv dt z i -\e 2 Jt p" -|e|y--lnp z 7i at (- | e |) is the charge on the electron. Integrating the last equation, mv. -\e | e | —— In p / a « mz. 2k

419 since vz « 0 where p « a. We now substitute this z in the other two equations to get e\V In a /b m \2n e\V \e m Inp/b xx + yy P2 a b e\V \e\ mmmmm m v / < J p Integrating and using v ■ 0, at p = b> we get, 1^ 2 e\V a b 2m In £l The RHS must be positive, for all a > p > b. The condition for this is, 2 i a lnb 2 m 2jt and the magnetic field is along the 3.394 This differs from the previous problem in (a z-direction. Thus Bx = By « 0, Bz = B Assuming as usual the charge of the electron to be - | e \, we write the equation of motion dt \e\Vx d Ifi^ . |n. p"ln a and dt z 0 The motion is confined to the plane z = 0. Eliminating B from the first two equations, d A i\ \e\V xx+yy dt[2 ) \nb/a p2 or, i-»- I In p/a In6/fl so, as expected, since magnetic forces do not work, ^: m , at p

420 On the other hand, eliminating V, we also get, Jtm(xvy-yvx)= \e\B(xx + yy) t \ \e\B 2 ♦ * i.e. m> - yvj « 1-rJ— p + constant y x 2m 2m The constant is easily evaluated, since v is zero at p - a. Thus, At p- bt (xvy-yvjs vb Thus, vb * L|L^ (b2 - a2) or, or, B z 3395 The equations are as in 3.392. = v cos cor -:L- v^. and — dt m y dt m m x dt with co « -L—, % - v + iv , we get, ? = 1 — co cos (at - 1 co E dt B * or multiplying by ela>tf or integrating, or, since S-Oatr-0, C--^. Thus, 1- i^J sincor + ^'0 m Em • m or, v^ = — cor sin cor and vy * — sin cor + — cor cos cor Integrating again, x - —r (sin cor - cor cos cor), y « ^— r sin cor. 2co2 2co

421 qEm where a - , and we have used x « y - 0, at t« 0. The trajectory is an unwinding spiral. 3396 We know that for a charged particle (proton) in a magnetic field, mv « B 4v or mv » Ber r But, co - m Thus E** —mv2= - So, AE = roco2rAr = 4 On the other hand AE = 2 eV, where V is the effective acceleration voltage, across the Dees, there being two crossings per revolution. So, 2jt2v2mrAr/e tnv 3.397 (a) From = Bev, or, mv = and (b) From 2jt = CO 2nr V 7 2m 12MeV we get, /_j- t^ —V?" - 15 MHz 3.398 (a) The total time of acceleration is, where n is the number of passages of the Dees. But, 7= neV= D2 2 2 Ber 2m or, n B2er2 nBr2 k2 mv r2 So, t - -r-— x - - —— - 30 \i s e5/m 2mV 2V eV (b) The distance covered is, s = V v« * 9— But, v n So, s=V-^t 7 v» = v^-t |v«(fti-v^!»3/2 2mv2 Z^ v 2rov2 ^ v 2wv2 3

422 But, n B2e2r2 _ 2jt2mvV 2eVm" eV rn 4 ji3 v2 mr2 Thus, 5- r-jT - 1-24 km Z JT T ft 3.399 In the nth orbit, = n 7n= -• We ignore the rest mass of the electron and write vn v vrt« c. Also W« cp= cBern. Thus, rt c. Also W cp cBern. 2kW n Bee2 2jiWv _ or, n - r— - 9 2 3.400 The basic condition is the relativistic equation, mv2 _ wov D = 5ov, or, /wv = . = = Bar. Vl - vV2 Or calling, co = —" , m o we get, co = — -, con c2 is the radius of the instantaneous orbit. The time of acceleration is, N N N t ￼ 2vb" Zj ton" Zj qBc2' rt- 1 rt» 1 n N is the number of crossing of either Dee. 2 nAW But, T^ = m0 c + —~—, there being two crossings of the Dees per revolution. So, t- > —4 + V 7imo< Also, rss rN — m "" TIF* ^—d~ " (oN 7i dN 2qBc

423 Hence finally, co co0 AW2 N< y A^B2c' m co0 co0 V (AW? m qBAW Jl 7W0 C 3.401 When the magnetic field is being set up in the solenoid, and electric field will be induced in it, this will accelerate the charged particle. If B is the rate, at which the magnetic field is increasing, then. 2 ' 1 ' ji r f? = 2nrE or E = ~^r^ Thus, m~Ts" 9*r^7> or v* After the field is set up, the particle will execute a circular motion of radius p, where mv = Bqp, or p = — r 3.402 The increment in energy per revolution is e <t>, so the number of revolutions is, W The distance traversed is, s = 2 nrN 3.403 On the one hand, meE dt 2xr dt 2xr dt o On the other , p = B (r) er, r - constant. so, &L=erftB(r)merB(r) Hence, erB (r) - -—it r2~r<B> ' K 2nr dt So, i(|| This equations is most easily satisfied by taking B (r0) - y < 5 >. ro 1 1 f 3.404 The condition, B (r0) - ~ < 5 > * ~J ,2 0 0

424 or, B (r^ « -j f Brdr This gives r0. In the present case, ro °o ro« 3.405 The induced electric field (or eddy current field) is given by, r 0 Hence, r dE 1_£ dt This vanishes for r = r0 by the betatron condition, where r0 is the radius of the equilibrium orbit 3.406 From the betatron condition, dB Thus, 4<B> r dt Ar d<B> d& z and "" «r d dt dt At So, energy increment per revolution is, 2n?eB e dt - Ar 3.407 (a) Even in the relatjvistic case, we know that: p = Ber Thus, W= y/c2p2 + ml c4 - m0c2 - m0 c2 Nl + iB (b) The distance traversed is, W WAt 2nr2eB/At on using the result of the previous problem.