Cengage MECHANICS 1

7.160 Physics for IIT-JEE: Mechanics I 58. a. Usc concept of function force for drawing the graph. +al = G2 a3 59. d. t = (M = dv Equation of moti~m (i) m)- T - N = inial (ii) ~ .N =1112([\\ (iii) 2 dt ~n2g - T = fn2(l2 (iv) m3g - T = m3G3 11 12 Using the above equation we can calculate the values. f dv=f 2(M t+ m) .dt For Problems 69~70 o0 69. d., 70. b. v = 6 mls After 12 second, up to 15 seconds, for 3 seconds the acceleration Sol. Free' body diagram Fig. 7.731: of plank will be ~~f L ___J-F a= 0.2 x 2 x 10 =lm/s2 Fig. 7.731 4 v' = 6+ 3 Vi = 9 111/8 +sa,,, = VI V = 135 = 0.6 m/s2 For Problems 60~62 Let both the blocks moves together 60. d., 61. b., 62. a. F Sol. Figure 7.729 Acceleration of the blocks, a = -;--= Force in spring can't change abruptly whereas the tension in the (m+M) string can change. f=m (_F) m+M R If both the blocks moves together, f :5 lung mF +-:---= < /l,mg (m M)- ~ SON F :5 /L (m + M) g , a If the cube begins to slide then, Fig. 7.729 F = f.I.(m +M)g For Problems 63~65 (towards +x direction) 63. d., 64. a., 65; a. am = -f = f.lg Sol. (towards +x direction) m mgsin30° - T = rna (I\" -IUnl?) F~f.l.mg ~~- aM= - - _..- M a4 m, At = a4 m - a4 M = Vg - 50-5=IOa f.I.(m +M)g - F a = 4.5 m/s2 M 0+45 r21If the cube falls from the plank it will cover a distance I -2-' = 2.25 m/s =} 1-,/- V ([M,I/I For Problems 66-68 !2---~ 66. b., 67. c., 68. d. 2iM Sol. Free body diagrams Fig. 7.730: +t = - - - - Fig. 7.730 F -It(m M)g Constraint relations: For Problems 71~74 71. a., 72. b., 73. c., 74. c. +Xl = X2 X3 Sol. Let the acceleration of block I and the pullcy P2 relative to ground is x in the directions shown in Fig. 7.732. Similarly, the acceleration of block 4 and pulley p) relative to the pulley P2 is y and the acceleration of block 2 and block 3 relat.ive to the pulley p) is z. Then, absolute acceleration of 1 is a2 = x

absolute acceleration of 2 is a2 = Y + z - x Newton's Law of Motion 7.161 absolute acceleration of 3 is a3 = - y + z - x ~al and absolute acceleration of 4 is ([4 = -x - y at t I Fig. 7.733 Fig. 7.732 Velocity of block at any time: v\" = v, + a,l +::::} Vb = VI j.tgt Now given that. Velocity of plank at any time: vp = V2 + al a2J = a2 - (11 = -1 mjs2 At t = to. both the velocities are same: or, y + z - 2x = - I + + V2 - Vj (131 = (l::~ - al = -5 m/s2 v, /Lglo = V2 alo =? 10 = - - - /Lg - a 80. a. For I < 10: Velocity of block at any time: v\" = v, + !LI;t slope of graph is /Lg Velocity ofbloek at 1= 10 : VbI) = v, + /Lg/o Fort> 10 (Fig. 7.719): Velocity of bloek at any time: Vb = VbO + aU - to) + +=? Vb = V, (/Lg - a )/0 al slope of graph is a. So slope will decrease. Hence the correet graph is (a) (i) 81. b. For t < 10: (see Fig. 7.734) ~=p=n=,g=- or, y-z-2x=-5 _ _-,(ii) -Ii> F II _ _M ----' (134 = 03 - G4 = 0 (iii) - -Ii> a or, 2y - Z'= 0 Fig. 7.734 Solving these three equations, we find x = 2 m/s2 y = I m/s2, and Let F be the forward force, then z = 2 m/s2 +F - /Llrtg = Ma =? F = /LillI! Ma or, (I, = 2 m/s2 (upwards) '--------'_IIFort > to: :==ma=-_-,----+F a2 = I m/s2 (upwards) IAI (I) = -3 m/s2 (downwards) {/4 = -3 m/s2 (downwards) For Problems 75-76 Fig. 7.735 75. c., 76, a. Let F be the forward force, then Sol, +F - rna = M a =? I\" = (m M)a 75. c. h = ~/2 - ,.2 = I m, For Problems 82-84 If f!.oTime period: To = 2\" = 2\" = ~s 82. b., 83. d. 84. c. Sol. Let x is the compression in spring at any time Fig. 7.736 76. a. w'=g/h Fig. 7.736 =l' IIlW2[ = 500 x ~ x./2 = S./2N -=Fl - kx mlaO, F2 - kx = m2aO 1000 h Solve to get: ao = 1\", - F'z For Problems 77-78 77. d\" 78. b. -'---'- Sol. 77. d. Tease = mg, l' sin I! = Inv2/r Squaring and adding both, we get the answer. tn! -m2 78. h. Dividing the above equations: Tnl/<2 - F]m2 and kx = -'-=---'--=- v = ,Jrg tan e ml- nI2 For Problems 79-81 79. c., 80. a., 81. b. 84. c. Just after n12 is removed: Sol. kx F2 - m2aO F2 Q2=-= =--ao nl2 nt2 m2 79. c. Figure 7.733 For Problems 85-88 85. e., 86. a., 87. b., 88. d. For I < to : .f = /Lmg. a, = -f = /Lg Sol. m

7.162 Physics for lIT-JEE: Mechanics I 85. c. The forces acting on the block arc F1, F2, mg, normal Fig. 7.738 contact force, and frictional force. Here the frictional force to = - _2__- = -3-0- = 13.33 s won't act along the vertical direction as the component of 2,25/15 2,25 the resultant force along the smface acting on the body is 91. c. Once the block comes to rest, kinetic friction disap- pears and static friction comes into the existence and then not along vertical direction and the direction of the friction situation would be identical to that of question no 40. force is either opposite to the motion of block (direction 92. a. of acceleration of block) if it is moving or opposite to component of resultant force along the surface if it. is not 93. b. Sol. iL, = 24 N [Limiting friction force between 10 kg moving, block and incline Fig, 7,739] N, = 300N So, IL = I\"N, = Q,6x 300 = ]80 N Resultant of 4g and F, is 107,7 N making an angle of ~)tan- I ( with the horizontal. As the force applied along the surface is h\" so the block doesn't move and the friction is static in nature. (~)F = 107,7 N making an angle of tan ,1 with the horizontal in upward direction. ,1 l' 4g Fig. 7.737 4g 86. 3. As Fnpplicd < fL, so the block remains at rest. Fig. 7.739 87. b. For F, = ISO N, h. = 0,6 x ISO =.90 N FL2 = 3 N (Limiting friction force between 5 g block and incline) If we ignore the friction for the time being, then the system has As the component of resultant force along the sur- the tendency to move down the incline as shown in Fig. 7.739 face is 107,7 N and it is greater than ji\", so above. So we can say that the friction force is acting opposite to the kinetic friction comes into existence, i.e., fric- direction of this tending motion. As the system is not moving, 11 and /2 are static in nature, tion force acquires the value I = I\"kN, = 0,5 x 150 For equilibrium of both the blocks, 61? = - T + II and = 75 N. Its direction is opposite to component of resultant 4g + j, = T, force along the surface. Other condition arc II < 24 Nand h < 3 N, F·rom hit and trial (better substitute 12 first) we can draw some 88. d. Acceleration of block = 107,7 - 75 2 4 = conclusions. 8,175 m/s If h = 0, T = 40 N, II = 20 N For Problems 89-93 If h = 3 N, T = 43 N, f, = 17 N 89. b., 90, a., 91. c., 92. a., 93. b. So, h should lies between () to 3 N Sol. II should lie between 20 to 17 N 89. b. As acceleration is coming -ye for both the possible T should lie bctween 40 to 43 N direction of motion, so it means that the net applied force For Problems 94-96 is not enough to cause the motion of system or to overcome .the limiting friction force. 94. d. 95. c., 96. b. 90. a. As it is given that the system is moving initially in a Sol. From the dat.a given we can find the limiting friction force spscific direction, so it means that the direction offdetian for the two surfaces Fig. 7.740. is known and it is kinetic in nature (Fig, 7,738), '3 F _l(-,)~~,,_si_n_37_0_-_5\",g:-;s;---in_5_3_'_-\"\"\"fl - h I (1= Fig. 7.740 15 . II = I\"kl X 109 cos 37\" = 20,0 ILl =0,5 x 3x 1.0= 15N h = I\"k2 X 5g cos 53° = 2,25 a = - -2,-25m / s 2 15 So, the required time,

Newton's Law of Motion 7.163 fu = 0.2 x 5 x 10 ~ ]() N 3 So, FOlin = -rng For F<fL2 4 Both the blocks remain at rest and fl = F, fz = fl' .and 101. c. If both the blocks are stationary (Fig. 7.742). al = a2 ~ O. tV, For F > fLz and F < a certain value say FI , the motion starts at a lower surface but both the blocks continue to move with the same acceleration. The friction on lower surface becomes kinetic in nature. Fig. 7.742 F - fk2 2 Here, a = al = a2 = 5 m/s F - fl = 3a and fl - fk = 2a '* eBalancing the forces along x-direction .. 2F +3fk2 All these equatIOns glve fl = 5 F = N sin 0 N = F / sin For relative motion to start between two bodies, fl ?: fLI Balancing the forces along y-direction F :::: 30 N, So, minimum value of F to cause relative motion between blocks is 30 N. Ny=mg+NcosO For Problems 97-99 97. a., 98. a., 99. c. (--.!---)= mg + cosO = mg + F cote Sol. sme 97. a. From expo 1, fL ~ 450 N Ny=mg+ 54F '*I',N = 450 1'., = 0.45 102. a. To keep a regular contact see Fig. 7.743 [c. N = mg = 1000 NJ 98. a. In 1st set the force applied is less than fL, so the block bsin $ b can't move, b cos 0 99. C. From 2nd and 3rd set of expo 2. aSj~ 2.0m = ~ [600 - fk] (2 2m and 3.0m =. -2I [750-fk] t2 a cos e m Fig. 7.743 '*fk = 300N I'k = 0.3 asine = bcose For 2nd .set, a = 600 - 300 b= a tan 0. = 3 100 = 3 m/s2 -4a For Problems 100-102 For Problems 103-104 103. a., 104. d. 100. a., 101. c., 102. a. Sol. Sol. 103. a. Figure 7.744 100. a. For equilibrium of block (A) (Fig. 7.741) y a - (-M\"t'o-t ) g B Ncos 8 Fig. 7.744 AN 104. d; T = 2M,M,g 2m (MIn. - m) g mg . '1 =2 (mM.o1 - m2») g .1 M tot For Problems 105-107 Fig. 7.741 105. d., 106. a., 107. b. F = Nsine Sol. N = F/ sine To lift block B from the ground N eose 2: mg => -F- eose > mg sine - (~)F?: mgtanO = mg

7.164 Physics for IIT-JEE: Mechanics I F(N) 105. d. If F 20 N, 10 kg block will not move and it will not 4.5 press 5 kg block. So N = O. (Fig. 7.745) 4 44.5 f--+-+-+---\"t'-+-r-t>- ((s) !1.523 15 -8 -------------- Breaking Breaking - 13.5 strength strength Fig. 7.747 JlX lOxg=:40N /lx5 x g\"\"20N Fig. 7.745 For Problems 114-116 = =106. a. If F 50 N, force on 5 kg block IO N (Fig. 7.746) 114. a., 115. c., 116. d. 50 ~ • [N~~Ql , IN:1°1 Sol. 40N f,..•. ION 114. a. Let m I and m2 do not accelerate up or down, then FI :;::; mig, F2 = m2g· But In! -I m2 so FJ f=. Fl. Fig. 7.746 Hence net horizontal force on M is FJ - f2. So M cannot So friction force = ION be in equilibrium. If M accelerates horizontally, then tnl 107. b. Until thc 10 kg block is sticked with ground (F = 40 and m2 also accelerate horizontally. N), no force will be felt by 5 kg block. After F = 40 N, the 115. c. friction force on 5 kg increases, till F = 60 N, and after =Net force FI - F2 = In I g - m2g that, the kinetic friction starts acting on 5 kg block, which a= will be constant (20 N). net mass M+m!+m2 M+ml+m2 For PI'ohlems 108-110 116. d. In the horizontal direction, both tn land tn2 will have the 108. b., 109. b., 110. b. same' acceleration for any case, In vertical direction also. Sol. they should have same acceleration. Let it be a upwards =2 for both, then 108. b. OJ = - 2\" 4 radls FJ -mig = mla (I) \" r2 -rn2g = m2Cl (2) T =mult =m(4)21= 16ml FI -F2. From equations (1) and (2) - e e l6t109. b. cos = wg2[ =g16[ =} = cos··1 [ g ] = e11O. b. v = (,Ir = \"I[ sin m! m2 C~J2=wrv'l-cos2 e -41/1- For Problems 117-119 117. a., 118. c., 119. d. Sol. For Problems 111-113 117. a. Figure 7.748 111. d., 112. b., 113. b. T=Mg Sol. 111. d. Initial velocity: u = 0 1', f l'I = Fdt =} m(v - u) = (6t - 2t2)dl AB =} 2v = 3t2 - 2 3 _/ 3 Put v = 0, we get t = 4.5 s l' Mg 112. h. If we draw 1\"-1 graph (Fig. 7.747). We can see that upto t .::: 3 S, force is +ve, so velocity Fig. 7.748 increases uptD 3 s and after that velocity starts decreasing. So at 3 s velocity is maximum. T, = T + Mg = 2Mg 113. b. We can see from the figure above that when force be- 2Tl 4Mg comes maximum (at t = 1.5 s) for the first time, velocity is not maximum at that instant. Hence (a) is incorrect Kx=2TI or x =K- = -K- (b) is correct as explained in the above solution. (c) is incorrect because when force becomes zero, velocity 118. c. Total tension on lower support: (Fig. 7.749) does 110t become zero. TI + T = 2Mg + Mg = 3Mg 119. d. TI = 3Mg

'Newton's law of Motion 7.165 11 For Problems 123-125 lower support 123. c., 124. b., 125. c. Fig. 7.749 Sol. 123. c. Since there is no friction betwcen A and B. so B will remain at rest (Fig. 7.752). ' -_ _ _--\"-A_ _ _ _..JI --l> a T, Fig. 7.752 fk = /1 (2 + 2) g = 0.1(2 + 2)g = 4N Mg Mg T'\" /'1!g J,Acceleration of A; a :::; - = -4 = 2 m/s2 Fig. 7.750 m2 Net tension at lower support (Fig. 7.749) I T+Tl =Mg+3Mg=4Mg For 13 to fall off A: IS = ul + 2al2 For Problems 120-122 =} 4 = 0 X 1+ 2212 =} t = 2 s 120. c., 121. a., 122. d. 124. b,VA = +U al = 0 + 2 x 2 = 4 m/s 120. c. Figure 7.749 125. c. JlIst when B falls of A, take this instant to be I = 0 (Fig, 7.753) jj = fk = 0.5 x 2g = IO N Fig. 7.753 -\" att = 0: velocity of A: UA = 4 m/s, velocity of B : UlJ = 0 <-,- acc. of A: al = ILlg = 0.1 x 10 = 1 m/s2 acc. of B : (/, = fl2g =0.4 x 10 = 4 m/s' I, Let us see when A comes to rest (w.r.!. belt): Fig. 7.751 +For this: VA = lolA all Initially, F = 20 N > jj, so the block will start accelerating =} 8=4+Jxl=}I=4s immediately. Let us see when B comes to rest (w.r.t. belt): At anytime t : F=20-21 +For this: VB = Uu a2l Acceleration: a = 20-2t - fk IO-2t =} 8=O+41=} 1=2s = --- (I) So B comes to rest earlier, and til1 that A continues to move with mm (2) acceleration 1 m/s2. 10 - 2t So we have to hnd separation between the blocks at t = 2 s. At For a = 0, 2 = 0 =} I = 5 s t = 2 s: dv IO - 2t SA = 4 x 2 + 2I x J x (2)2 = 10m From equation (I) - = - - - = 5 - I SII = 0 x 2 + 21 x 4 X (2)2 = 8 ill dl 2 separation = SA - Sl! = JO - 8 = 2 m r10 dv = 1{0' (5 - I)dt =} v = 5t - ~2 Matching . Let liS see when the velocity becomes zero. For this: t2 Column TYJ:!.e 51 - - = 0 =} 1 = 10 s 1. 3. -7 q., r., s.; b. -+ S; C. -+ p.Sj d. -+ p., s. 2 2. i. -> a.,e., ii. -+ b., c., We see that at I = lOs, also F = O. So the block has no tendency to iii. -+ b., c., iv. --;,. b., d. move. Hence the acceleration is zero at this time. Now the block Acceleration of the whole system towards right: will not move from I = 10 s to 15 s because for this magnitude F of F < 10 N. So block will remain at rest from t = lOs to 15 s a= ---. or acceleration is zero from] 0 s to 15 s. e eM+m 121. a. From equation (2), first time acceleration becomes zero at 1 = 5 s. Velocity at this time: F - mg sin = rna cos 12 52 V = 51 - - = 5 x 5 - - = 12.5 m/s 22 122. d. From the above discussion clearly, velocity is zero at 1= 12 s.

7.166 Physics for IIT-JEE: Mechanics I at Pj P2 ~a )<;{F t mgsinil ~gCOSO OJ 4 ~a'l mg Fig. 7.754 F F - In!? sine = m - - - cosO eM=} ~ M +I\"n =< F = -:::+~-m=-)=mcg::s.ine . M+m-m.cosO pseudo force on m as seen from the frame of M: mF So acceleration of 4 is downwards Hence (ii) -> (a,d) F:~1 = rna = m + M Acceleration of 2 w.r.t. 3: e( e))= mg sin M+m (1111-cos < mg sin 0 = =a2/3 ([2 - a3 02 - al = 2(a - (1) < 0 Hence (a) --+ (P. r) This is downwards. pseudo force on M as seen from the frame of 111: Hence (iii) --+ (d) Acceleration of 2 w.r.t. 4: F,,-,=Ma= mM+FM (>~) a'/4 = a2 - (-a4) = 4a > 0 This is upwards. m+M Hence (iv) --+ (e) =mg sin e ( M +m(M1- cosO) ) < mg sin 0 4. i. -0- a., c. ii. -0- b., d. iii. -0- a., b., c., d. iv. -0- c., d. Hence (b) --+ ('1. r) (i) Force of friction is zero in (a) and (c) because the block has no tendency to move. Now mg cos e - N = ma sin e (il) Forccoffriction is 2.5 N in (b) and (d) because the applied force in horizontal direction in both is 2.5 N. -> N..,..,... rngcosO - rna sinO (iii) Acceleration is zero in all the cases. (iv) Normal force is not equal to 2 g in (c) and (d) because e.Hence N is less than tIlg cos Hence it will also be less some extra vertical force is also acting. ethan mg sine, because = 45°. 5. i. -+ b., d., ii. --;,. c. Applying equation on m in horizontal direction: iii. -)- a., d., iv. -7 b., d. FcosO - Nsine = ma =< F FcosO-NsinO=m---· M+m =< N=~\"(M+m)COSe-m) +I-i - M +m mS1I18 ~F lOON put 0 = 45\" =< N=~(M+m-v0-m) mF +-- >--- j, M+m m m+M (a) (b) (e) Hence (c)--+ (q,r) Fig. 7.756 Normal force between ground and M will be (M + m)g. it, = 0.2 X 2g = 4 N e. mF ii, = 0.1 X 515 = 5 N It is greater. than m,g sin It is also greater than - M ' it, = 0.1 X 109 = lON +m Friction on 3 kg block is towards left and non-zero. rnF . Hence (i) ..... b, d e.because - - - is less than mg sm fl, < fl, hence 5 kg block will not move. So the net M+m Iriction on 5 kg will be zero. Hence (d) --+ (q. s) Hence (ii) ..... c 3. i. --? b., c., ii. --+ 3., d., iii. -+ d., iv. --+ c, Let accelerations of various blocks are as shown in Fig. 7.755. Pulley P, will have downward acceleration a. 6. i. -7 d., ii. -+ c. iii. -+ b iv -0- n. Now a = -a! -+a-2 =< a, = 2a - a, > 0 From Fig. 7.757 F=(mt+m,)a 2 T sine = nl2a So .acceleration of 2 is upwards Hence (i) ..... (b, cJ (i) (ii) and a = +--G-I' - -G-4 2

1111 T cos () T Newton's Law of Motion 7.167 F / [\"=e __a e T Fig. 7.758 mig D---+TsinO --+0 Fig. 7.757 T = mIg sece (iii) eFrom equations (ii) and (iii) -0 = a = g tan r=:\".'../I.,·' <~\"-'t.e.•_.,.•~~<,..,. I· Put in 1- I\" = (m., +m2)gtane mlF --=--Net force acting on ml = 1n2a = ml +m2 Force acting on In I by wire: mlg+TcosO=mlg+m2g Fig. 7.759 7. i. -> c., d., ii. -+ a., c. F -.Ii = (2 -I- 3 + S)a iii. -\": a., c., iv. -+ c., d. For (i), it is not mentioned whether the object is accel- a= 12 - 10 2 mjsw, erated or moving with a constant velocity. So nothing can be predicted with surcty. --- = --- ,If no net: force is acting along east; then also it can move with a constant velocity, and if no force is acting at all, then 10 10 also it can move with a constant velocity. 2 For (ii) and (iii): As the object is accelerated (whether f,=2a=-N uniform or non-uniform) a force must act on the object in such a manner that a component'or whole of the force would S bc along east, and also thc net force must be towards east. h = 12 - f, - 3a = 11 N For (iv) : It is moving with a constant velocity, so the net force must be zero that implies no force may act on the l',=ION object. For F = 15 N, the situation is similar. 8. i. -+ b., ii. -+ ,a., c. iii. -+ b., iv. -+ a., c. For relative motion to start between Band C. ,h ::: fL2 For (i) and (iii): If I\" > f it means objcct is moving. and F -.Ii = lOa and I\" - 12 = 5a friction i,s kinetic in nature, ]h = F - Sa = F - 5 [F ~/l = F ~ J3 For (ii) and (iv): If I\" > f, then object is at rest and F+ 10 friction is static in nature. Another possibility is that F = .fL. -~- > IS =} F > 20 N 9. i. -+ a., ii. -+ c. 2 111. -+ c., iv. -+ c. Just after spring 2 is cut. fCondition for relative motion to start between B- and C) The net force acting on thc block D changes and it is For relative motion to start between A and E, acting in upward direction and hence D accelerates. As there f, 2: fLi = 8 N is no change in the elongation of spring 1, the equilibrium of F - f, - 12 = 3 a and f, = 2 a A and B wouldn't be disturbed. 15J/,=2 [ -F--S- >8 Similarly. we call find reasons for D. I\" > 3S N 10. i. -+ c., ii. -+ c. iii. --* b., c. iv. -+ a., b., c. rcondition for relative motion between A and Bl Let f\" /2, h represent the friction forces between 3 11. i. -+ b., ii. -+ c. iii. --'\" b., iv. -+ c. Sol. (i) Leta be the acceleration of two block systems towards contact sUlfaecs A - B, B - C, and C - Ground, respectively. Limiting values of friction forces at 3 surfaces are 8 N. 15 N, right (Fig. 7.760), then a = ~F2-- -1\", and 10 N, respectively (Fig. 7.758). Inl +m2 For relative motion between C and Ground, the mini- Fig, 7.760 mum force needed is F = 10 N. F2 - T = nl2' a For F = 12 N, all the 3 blocks mOve together with same acceleration (Fig. 7.759), i.e., at = (/2 = (/, = a. Solv\"mg.1, = I1'/ltn2 (F-2 + -F--l) m! +m2 m2 In! Oi) Replace F, by - F t is result of A

7.168 Physics for IJT-JEE: Mechanics I (iii) Let a be the acceleration of two block system towards CD left, then a = \"-' - F, Equilibrium 1111 +m2 - - - - -position .....\"- -~D~h Fig. 7.761 Fig. 7.762 Fl - N = mza Hence (iii) - a., d, , .Solvmg, N = tnjm2 (F- I + -F-2) But if the block is at position '2', then the velocity is ml +m2 In! 1112 zero and the acceleration in upward direction. (iv) Replace F, by - F, in result of C Hence (i) - a. mlm2 (1\"2 PJ) When the block is bctwcen position '3' and' I', then +N = m! tn2 m2 - ml mg > kx. So the net force in downward direction, hence ac- 12. i. -+ c., ii. -+ h. iii. -+ d., iv. -+ b. celeration is in downward direction, But velocity may be ei- (A), (B) After spring 2 is cut, tension in string AB will ther in upward or downward direction. not change. Hence (iv) - b., d, (1(:0); =4mg 14. i. -)- c. ii. -+ c. iii. -+ c. iv. -+ h., d. Let the maximum downward displacement of m is Xo, (TeD)! = mDg + mf)' mA +m[/ -me -Inn . g '*then 1- mA+mB+mC+mD -kxF; = mgxo Xo = 2 mg/k 2 (1=2mg +~) = 2.4mg To lift the block (M): ho = Mg .5 '* 2mg = MM '* mg = Mg/2 Hence (i) - (e) Hence TeD decreases. (e), (D) After string between C and pulley is cut tension (ii) When 111 is in equilibrium (Fig. 7.763) in string AB wi II become zero. ,:l, T +(Teo); = (mf) mE) g = 4mg ~ Aeeelcration of C and D blocks is kx mg kx (me + mf)) g + mEg = (mc + mf))a Fig. 7,763 6mg 3 kx = mg, T = 2kx +mg = 3mg = T3Mil a=--=-g Hence (ii) -+ (a). 4mg 2 (iii) Figure 7.764 (Tco); + meg = mea N +kx = Mg 3 ~.x ;1;1 (Tef))/ = 2m'2g - 2mg = mg Mg tN The tension decrease. 'Fig,7,764 l1L~L, ~~~d,~~~d~~~~~ N = Mg -kx = Mg- 2Mg ='M2 g In Fig. 7.762, 3 is the equilibrium position where ve- locity is maximum and acceleration is zero. 1 and 2 are the Hence (iii) -+ c. extreme positions where velocity is zero and acceleration is (iv) Tension ~ kxo = Mg = 2mg maximum. I is the unstretched position. Hence (iv) -+ (b\" d.) When the block is at position '3', then mg = kx. So net 15. -t -+ c., ii. -+ c. iii. --+ h., d., iv. -+ n., d. force is zero, hence acceleration is zero. But velocity may be either in upward or downward direction. Hence (ii) - c., d. When the block is between position '3' and '2', then kx > ing, So the net force is in upward direction, hence ac- celeration is in upward direction. But the velodty may be either in upward or downward direction.

Newton's Law of Motion 7.169 (i) If VI = V2 = 0, then there is no force on M in hori- zontal direction, So IV! does not accelerate Hence (i) -+ (e) (ii) If 1\"1 = It2 oj 0 Fig, 7,768 Fig, 7,765 fit '= 0..2 x 4g = 8 N 11 <1\"1 mg, h<V2 mg ii, = 0.4 x 6g = 24 N ./)3 = 0.5 x 12g = 60 N 11 and h will be of same magnitude at any time whether the blocks slip on larger block or not, so the net force on M is zero (Fig, 7.765). Hence M does not accelerate. so (ii) -+ (e) (iii) 1\"1 > V2, here f, > .h hence (iii) -+ (h,d) (iv) VI < V2 here II < .h hence (iv) --+ (a, d) 16. i........,. 3., d., ii. -+ a., d. iii. -+ b., c., d., iv. -7 b., c., d. •(>SN Maximum possible acclerate of m: ao = jJ-g =:: 0.5 g So (d) matches with all (i), (ii), (iii) and (iv) - J o o fl;, = 8 N Let us assume that Tn and 2 m move together with accel- ~-- . a: a = ml!?- 6 kg cratlOn 3m+mj •~ J~~8N if a = ao ';:;::} 111-1 =:: 0,5 g 3m +111-1 =>- 1111 = 3 ill Fig, 7.769 So 3 m is the maximum value of nt I so that both move Here only 4 kg will accelerate, 2 kg and 6 kg will remain at rest (Fig. 7.769). together. 20. i. -> b., d., ii. ->- c., iii. --+ a., c. iv. ->- d. (i) 1111 = 2m < 3m hence (i) --+ (a), (e1) (ii) In I = 3 m hence (ii) --+ (a), (d) From Fig. 7.770 (iii) tnl = 4 m > 3 m hence (iii) -> (b), (c), (d) 2T, cos 37' = 120 (iv) 1111 = 6 m > 3m hence (iv) --+ (b), (c), (d) 4 17. i. --)- b., ii. -7 a. iii. -7 c., d. iv. -+ c., d. => 2T, X = 120 => T, = 75 N - .5 The direction of accelerations of various blocks are as ,, shown in Fig. 7.766. i l la 4--- , ! 53° ~--.a T !J Fig, 7,766 ,31\": ace. of B is towards right, Hence 0) -+ b ,120 T, ace. of C w.r.L II is towards left, Hence (ii) --+ a Fig, 7,770 aeacc. of' A w,r.t~ C : 0'1/<: = Ali-A - TI cos 37\" = 73 cos 53' = \"-aj - (-ai) = ai - aj as shown below 75 x :45 = 7'1 X :35 => T3 = 100 N T2 + 1', sin 37° = 7'3 sin 53' T2 + 75 x 3 = 100 x 4 5- -5 a 1'2 = 35 N Fig, 7,767 t Hence (iii) --+ (c,d), Similarly (iv) --+ (c,d) 18. i. -? b., c., ii.~:> a., d. iii. --+ b., d., iv. -~ a. 'Archives ' ,,,, ',:' 19. i. ~~ b., d., ii. --+ c. iii. --+ a., d., iv. --+ c. \\' From Fig. 7.768 Fill in the Blanks Type 1. Method I The maximum frictional force that can act oli the mass of one kg will be = I1N =mg

7.170 Physics for IIT-JEE: Mechanics I 'mmn~;'~h•11I111IliAII11I11I1 = 0.6 x I x 9.8 = 5.88 N The truck is accelerating at 5 m/g2, The pseudo force acting on the mass Fig. 7.771 as seen by the observer on the truck is mxa=lx5=5N J Fig. 7.773 Fig. 7.771 mv2 T - mgcose =-- The frictional force will try io oppose the movement of the mass by this force, Therefore the frictional force acting r will be 5 N. At extreme end v = 0 T = mgcose Method II As seen by the observer on the ground, the fric- T tional force is responsible to more the mass with an acceler- ation of 5 m/s2. Therefore the rrictional force would be mg sine mg mg cose =mxa=lx5=5N Fig. 7.774 2. .Let A he the area ofcross-scct.ion orthe rod Fig. 7,772, There- 3. False fore m For mass m T -mg =ma p = - => m = pilL = mass of rod For mass 2m (Fig. 7.775) AL F (i) (ii) Fig. 7.772 The rod is moving with an acceleration a under the action of F F = pALa (i) Consider the half rod of length ~. Let the force acting on the mid portion be f. Applying f~\"t = ma for the dOlled position fIIg 2mg F - .f = pA2L a Oi) Fig. 7.775 From equations (i) and (ii), 2mg - T = 2ma From equations (i) and (ii) => f = pilL --a a =g/3 2 T - lng = mal 2mg -mg =ma' => Stress at . . Af =. pLa -2- T T=2mg rmd pomt a't .j.\" B True or False mg F 1. False Concept: Friction force opposes the relative motion of the Fig. 7.776 surfaces of contact. When a person walks on a rough surface, the foot is the surface of contact Fig. 7.773. When he pushes the foot backward, the motion of surface of contact is back- ward. Therefore, the frictional force will act forward (in the direction of motion of the person). 2. False When the angular displacement is 20°. the mass is at extreme end (Fig. 7.774)

Newton's Law of Motion 7.171 .. a' =g mg 4. Since no external force is acting on the two particle systelll Fig. 7.779 .. a em = 0 =? Vcm = Constant. The statement is false. Multiple Choice Questions with One Correct Answer mv2 mg+N =-- F 5 x ]04 5 -3 2 1. c. a = -m = 3 x 107 = -3 x 10 mls r Tfthe surface is smooth then on applying conservation of v=~2as = /2 x ~ x 10-3 X 3 =0.1 mls mechanical energy, the velocity of the body is always same at the top most point. Hence Nand r have inverse relationship. e e2. Since, /-,mg cos > mg sin From the figure it is clear that /' is min for first figure therefore e.. force of friction is f = mg sin N will be maximum. 7. a. The two forces acting at the insect are mg and N. Let us 3. a. We give power to rear wheel, so friction is rear wheel acts resolve mg into two components (Fig. 7.780) in forward direction. Front wheel is a free wheel. On this mg sin a balances N mgsina is balanced by the frictional force. friction acts in backward direction. Net friction is in forward direction, due to which cycle accelerates. P\" 113 4. f ,,,8 T 8 Fig. 7.780 mg N=mgcosa Fig. 7.777 f = mg sin Q' But f = /-'N = /-,mgcasa FBD of bob . 7.777) is Tsm. il = Imvi2 (Fig. f-Img cos ex = mg sin ex and Tcose = mg. I =} cola = -- =} cota = 3 e v2 (10)2 /-' tan = -- = -'---'- 8. For equilibrium in vertical direction for body B we have Rg (10)(10) (Fig. 7.781) tanil = I ,, or e = 45\" +T cosO T cosO 5. The magnitude of the frictional force f has to balance the weight 0.98 N acting downwards (Fig. 7.778). Ii- - 0.5 e', 8 f T TsinO T 5N +-5N A lJ Tsin8 B 5N mg J2mg mg 0.1 ><>.8 - O.98N Fig. 7.781 Fig. 7.778 ../img = 2Tcase Therefore, the frictional force is 0.98 N. Hence option (b) is ../img = 2(mg)cos@ the ,~orrect option. 6. a. Since the body presses the surface with a force N hence T = mg (at equilibrium) according to Newton's third law the surface presses the body with a force N (Fig. 7.779). The other force acting on the e eI = 45\" body is its weight mg. cos = - For circular motion to take place, a centripetal force is =} required which is provided by (mg + N). ../i 9. d. Forces on the pulley are (Fig. 7.782)

7.172 Physics for IIT-JEE: ,~echanics I F) (m + lvf)g Assertion and Reasoning Fig. 7.782 1. b. The cloth can be pulled out without dislodging the dishes 10. a. The forces acting on the block are shown in Fig. 7.783. from the table due to the law of inertia, which is Newton's first law. While, the statement II is true, but it is Newton's Since the block is not moving forward for the maximum force third law. F applied, thereforc (i) 2. b. and c. Both statements are correct. But statement II does (ii) not explain correctly statement L F cos61l\" = f = ILN, and Correct explanation is : There is increase in normal re- action when the object is pushed and there is decrease in F sin 60° + mg = N normal reaction when the object is pulled (but strictly not horizontally). From equations (i) and (ii) Multiple Choice with More than One Correct Answer F cos 60° 1. b. Sol. From b. MNO, using pythagorous theorem (Fig. 7.785) M N2+ N02 = M02 N AB F Fsin 60° Fig. 7.783 F cos 60\" = {.LIF sin 600 + mg] uu F = ___,.1 mg_.____ pQ cos 60° - IJ., sin 600 ._I-.- x ~ x 10 S Fig. 7,785 M' 2y3 =.:.. =20 N ~ Here x is a constant Differentiating the above equation -1 -2-1~- -I 2 x- 4 by t 2 dz dl 11. b. 21' cos 0 = F 0+2z- = 21- d/ dt =? ZVM = IV F =? v'\" .= ~U = ~ =.!!--.- ('\" cose = .':1) case z zjl 2. a. is wrong since earth is an accelerated frame and hence cannot be an ineltial frame. h. is correct. c. is incorrect strictly speaking as Earth is accelerated ref- erence frame [earth is treat.ed as a reference' frame for practical examples and Newton's laws are applicable to it Fig. 7.784 only as a limiting casel. d. is correct. T = 'F2 sccO 3. a. Since the body is moving in a circular path (Fig. 7.786) Acceleration ofparlicle (Fig. 7.(84) TsinO FtanO 2 = therefore, it needs centripetal force ( MtV ) 111 2fn g sine g g cosO Fx Fig. 7.786 2m ,Ja2 - x 2 e12. b. For sliding tan 0 2: ./3(= 1.732) For toppling tan 2: 2 -(= 0.(7) 3

Newton's Law of Motion 7,173 MV2 of P as shown at any intermediate position, the horizontal T-lngcosO =---- velocity first increases (due to N sin e), reaches a max value I at 0 and then decreases. Thus it always remains greater than Also the tangential acceleration acting on the mass is g sin /) Fig. 7.787. y. T~ercfore, tp <'tQ: • e A fir B T 0,4 f Neose f N p Nsin8 v gsinO geosO mg 0 mg Fig. 7.787 Fig. 7.788 (b) and (c) are correct options. 4. a; At A the horizontal speeds of both the masses is the same (Fig. 7.788). The velocity of Q remains the same in horizon· tal as,no force is acting on the horizontal direction. But in case



Appendix Solutions to Concept Application Exercises Chapter 1 3. v = u. + at => v = 25 - 21 Exercise 1.1 v(rnls) 1. (1+x)-2= 1+(-2)x+ (-2)(-2-1)x 2 + 25 15 Il Fig. 8·1.4 (-2)(-2 -1)(··2 - 2) x3 1-2x+3x2 -4x3 + ... Exercise 1.3 ~ 1. a.y=x'-8x.a= l,b=-8,c=0 [( Ax) 3 Vertex X=-:a=-C~J=4 2. Q 1+~ -1] = Q [13+l1-;x--1] = -3Qxl-Ix. At x = 4, Y =42 - 8 x 4 =-16, Atx= 0, y=O Exercise 1.2 At x =8,y =0 1. i. 3x + 2y = 0 y y --'k--i--:.2:_ _ x Fig. 8·1.1 16 • . . . . . • • . ii. x- 3y+ 6 =0 => .Y= -3x+2 . Fig. 8-1.5 h. y = _2x2 + 3, a = - 2, b = 0, c = 3 Vertex b Fig, 8·1.2 X= -2-a =0' atx=0,y=3 2. v = 1 + 2t v(m/s) At x= I,y= I, alx=-I,y= I 5 y 3 4 3 2 .1 --/-j-+----t----~I(S) I 1:2 Fig. S-1,3 Fig, S-1,6 c. y = x 2 - 6x + 4, a = 1, b = - 6, c = 4

A-2 Physics for IIT-JEE: Mechanics I Vertex: x=-h-=--(--6=) 3 dy = (x2+3x)d(2X+7) +[-<!.Y+3X)] At dx dx dx 2a 2x 1 (2x+ 7) = (x2 + 3x) 2 + (2x + 3) (2x + 7) X= 3,)'=-5; alx= 0,y=4; alx= 6,y= 4 = 6x2 + 26x + 21 ---j-\\--+-f-',~-~x b. Let y = (3x2 + 2) (4x - 3x3) 6 dy = (3x2 +2)~(4x-3xJ) Fig. S·1.7 dx dx 2. x=1+12,a= I,b= l,c=O + [~(3h2)](4X-3x3) \\ = (3x2 + 2) (4 - 9x2) + 6x (4x _ 3x3) = -45x\" + 18x2 + 8 c.Lety= v'X(X'J +x2 ~3x) :::: X712 +x512 ~ 3x312 \\ \\ \\ \\ \\ \\ = -2.JJ-x (73x- +5x2-9x) \\ \\1 \\I ,\\ 2 I .... .....::. / - 1 d. Let y = sin x log x 4 - Fig. S·1.8 dy = sinx~(IOgx)+[~(sinX)]IOgX dx dx dx Vertex: t= b = ---1 , '= --1-, = -si-n+x cosx logx x -- 2a 2xl 2 3. a. Let y = tan 3x = (tan x)3, ~~ = 3(tan 2 x)(scc2 x) 11 a.tt= --2,x=--4 Exercise 1.4 :tz.b. Let y =tan(x2), = sec2 (x2)2x dx 1. a. ~ (9) = 0, since differentiation of constant is zero. (.Jx)c. Let y = sin2 = (sin.Jxt dx idd~l' = 2(s'm\"'x)cos ('\\I'X) 1 X-\\12 h. \".(£(1l4) = 0, since differentiation of constant is zero. dx = 2~sin(2.Jx) (2ec. ~ 3 ) = 0, since differentiation of constant is zero. d. .<i-.(x2 +5) = 2x+O= 2x 4. x = a(O+sinO) => dx = a(l+cosO) (1) dO dx e. -d[( x+5).'12J = --J( x+5 )--312 y = a(l+sinO) => dy = a(O+oosO) (2) dx 2 dO f. ~(5x3/2) = 5~c£(x3/2) = 5 X ~X1I2 = _~x!/2 D.lV.ld. mg(2)and(J) -dy'l-dO- = -el-y = -s-in-O , = Ian (012) dyldO dx J+cosO dx dx 2 2 g. ~(.JX+3) = ~[(X+3(2J = .!c(x+3r'12 dx dx 2 (j)= dO :::: -2-,1-'\" +.1- :::: ..,I-,- + -J h. ~[(2x2+9)'] = 3(2X2 +9)2 (4x) = 12(2x2 +9)'x 12 I dx 5.0= -+- 20 5' dl 20 5 10 5 2•. a. Let y = (x2 + 3x)(2x + 7) -4- + _J. = 0.6 mel/s JO 5

Appendix: Solutions to Concept Application Exercises Aw 3 6. A=5t2 +4t+8 3• x= 2 +51+ 7t 2 :::> dx =v=5+ 14t Rate of inc. of area: - dA = 10t+4 dt dt (dA) = 10x3+4 = 34m2/s a. vt=o =5 + 14 x 0 = 5 111/s dt /'ds b. V/:;,o 4 S = 5 + 14 x 4 = 61 m/s Exercise 1.5 c.a=-e=lv 14m/s 2 ff 3 dt 2 . 1. a. f (x 2 + 2) dx = x2 dx+ 2dx = ~+2x+c d,x'.5, =2+ 5 x 5 +7 x 5 =202m .3 4. a = t3 _ 3t2 + 5 b. f (x\" - x1 + 3x) dx = f x3dx - flx dx + f 3x dx = t4 3t3 Ic. (X 3/2 - Xl/2 + Sx) dx v= fadt+Cl -- - + 5 t + C1 4 3 = -2x-5/-2 -2-x+31' -5-x2 +c 5 32 At t = 1 s. v = 6.25 mig J f f 2x t 5 t 4 5t 2 x= fvdt+C, =-- - + - + 2 t + C2 d. (e2x + 5)dx = e'x{h+ 5dx = \"-2-+ 5x + c - 20 4 2 15 At t= 1 g.x=8.30m 4 8.30= 15 ]4 5 ---+-(1)2+2xl+4 =>C2 =4 20 4 2 x == _t 5 _ _t 4 5 +2t+4. nowput/=2.s +_[2 20 4 2 24 V = --- - 23 + 5 x 2 + 2 = 8 mig 4 25 24 5 2 X= ·------+-(2) +2x 2+4= 15.6m 20 4 2 f\"] (C. log, )4 = log.,(413) 5. x = 2t3 _ 3t2 + 1 v = dx = 6t' - 6t 3 -;dx = x log,4-]og,3 = 3 dt a. Put v = 0 => 0 = 6t2 -- 6t => t = 1 g, t = 0 s x312 ) 9 b. x = 0 => 0 = 2t3 - 3t2 + 1 => (t - 1)2(2t + 1) = (J f l--)d. -2 (9312 - ) 9 .[;dx = ( = = 38/3 => 1 == 1 S, V/::::: 1 s == 6 x 12 - 6 x l == 0 rnls 4312 4 312 4 3 6. x=t[t-I] [t-2]=(t 2 -t)(1-2) [Sinffl4 2X]lf/4 == 13 _ 2t2 _ [2 + 2t == t3 _ 3P + 2t e. Io cos 2xdx == - 2 - 0 = ~[gin[2[7r/4lJ-sin[4xoJ] = ~ v= -(h=312 - 6t+2 dl Exercise 1.6 a. v1=0 = 2 m/s 1. Y =6t2 + 31 + 4 => v = -dy = 121 + 3 m/s dt a,.b. a = ddvt = 6t - 6 , 0 = - 6 mls' 2. v= 12+3(t+7I')= 12+31+2112 c. x = 0 => t = 0 s, t = 1 S, t = 2 S a = -dv = 3+421 d. v = 0 => 3t 2 - 61 + 2 = 0 dl

Aw 4 Physics for IIT-JEE: Mechanics I =1+_1 = .)3±1 -1Inv2 =50-0 => -1x-1v5, = 500 -.)3 .)3 2 3 2 10 3 x=l Jl -IJl(.)3'--11(~ 1(.)3-1 1 .)3 .)3 .)3 -2) v = lOOO x 2=1OJ20. mls 45 3 2 = 20 J5 m!s 3 = 373 10. x = a + bt + eF x= l Jl -IJl(.)3+Ji(.)3+1 1(.)3+1 1 And .)3 .)3 .)3 -2) -2 ds 11. v= - = 15-0.8t =>7= 15-0.8t =>1= lOs 3.)3 dl 12. s = 13 - 61' + 31 + 4 => v = 3t2 - 121 + 3 a=61-12=0 =>1=2s v = 3(2)' - 12 x 2 + 3 = - 9 m!s 13. x = et2 => vx = -dx ;:;: 2et dl dy =>v.y = -=2bt dl 7. a. Distance = area = .!. x 2x 10 = 10 m Speed at any time = v = 2 14.1=.Jx+3 =>x=(t-3)2 b. Distance = area = 2 x 10 = 20 m c. Total distance = 10 + 20 = 30 m v= \"(b':; '=2( t-3) =0 =>1=3s 8. a. Maximum speed of the car will be at I = 2 s, because dt area is. + ive till that time, xl\"3, = (3_3)' = 0 Change in vel. = area of a - t graph => vmax - 0 = 2 x 2 => VmllX = 4 m/s. b. Since finally the car comes to rest, so both area should be same, 4 = 10 x 4 => to = 1 s xo-2 4-xo 10 Chapter 2 9. a. - - = - - =>xo= -m 100 50 3 Exercise 2.1 F 1. Totall,or-cc )F = 2\"i+\"3j\"+k,+,i+, ,j+\"k = 3\"i+\"4j+A2k 100 FI I or F = )32 + 4' + 22 = .Ji9 N o i'-+--+-ft'-l--l-.,l--_X 2. The required vector is 50 Fig. S-1.9 w= 11 =100J = lsi +20) -xoxlOO--(6-xo)50 22 .!. mv' = 100 =>.!. x !.? v' = 100 3. S1.I1 IJ' =C- => sinlJ' = B12 R 2 2 10 B ii~- v= Pf~ = ~ m!s ~ __ ._.~c A' ,Fig. S-2.1

Appendix: Solutions to Concept Application Exercises A-5- Now, 0+0'= 180' =>0+30'= 180°=>0= 150° Exercise 2.2 4. tan j3 = :: or j3 = tan -I (::) Jii x811. Area of parallelogram = 33 2. Their dot product should be zero. Y~-7:x (4[ + J-3k).(zm! +6mJ+k) = 0 => 8m + 6m - 3 = 0 3 Fig. 8-2.2 => m=3114 IAx813. Given =13A.8 =>ABsinB= 13ABcosO \"\"\"* ~) ._-} -7 __>~) \"\" => tan 0= 13 => 0= 60° 4. If 0 is the required angle, then 5. A +B +C =Oor C = - (A +B) = -(3i+4k) MF6. A=2F,B= ..fiF,/?= /?2= A2+B2 = 2ABcosO ...... 10 F2 = 4p2 + 2F2 + 2 (2P) . .fiF cosO AB cosO=-- cos 0= 1/..fi => 0= 45' AB 7. Let the two forces beA and B, then given A - B = 10, A2 + B2 = 502 s. ij k Solve to get A = 40 N, B = 30 N v = I -Z 2 8. Resultant of two forces, )(F / 2)2 + (F / 2)2 = F /..fi The third force should be opposite to the resultant and of 0 4 -3 ;ame magnitude, Hence its magnitude is PI.J2. = [[6-8]+J[0+3J+k[4-01 9. Let P be the unknown force = -27+3J+4,C Ivl=v=,/4+9+16 20N P = 59 units. 30° 20 ~3 N -t\" \",,-7 Fig. S-2.3 6. /? = 3i-2j+k, A = 12[+3J-4k P'= (2013)2 +202 -2(20)(Z013)cos30~ 0) L. or I\" = 1600 - IZOO = 400 or P = 20 N (ii) a .pA IO.P+Q=18 Fig. S-2.4 p2 + Q2+ 2PQeos 0= 144 -> -> QsinO ~\"'-c_---, = tan 900 = 00 Component of /? along A : P+Q cosO Chapter 3 => P + Q cos 0= 0 or Q cos 0= - I' Exercise 3.1 From equation Oil, 1'2 + Q' + 2(1' (-1')) = 144 1. No, it is a unit of length. or Q2_ 1'2 = 144 or (Q -1') (Q + 1') = 144 2. No, the numerical value may change. but not the magnitude. 3. It is not true, constants may have dimensions. For cxample, or (Q - 1') 18 = 144 [From equation (i) I G, universal gravitational constant has dimensions. or Q-P=8 (iii) 4. nm stands for nano-meter, a unit of length, mN stands for Adding equations (i) and (iii), 2Q = 26 or Q = 13 milli-newton, a unit of force, and Nm stands for ncwton- IDeter, a unit of torque. Now, I' + 13 = 18 5. Work, energy and torque all have same dimcnsions as M I } r 2• or 1'= 5

A-6 Physics for IIT~JEE: Mechanics I ilV = ± 0.0895 x 3.45 = ± 0.3 6. When we add length into length or subtract length from (rounded off to one place of decimal) length, we obtain another length. V = (3.45 ± 0.3) ms-I 7. Here KI is dimensionless. Hence [K] = 1111] I Percentage error = ilV So unit of K :::: sec-I:::: H. ~xlOO 8. Unit of a:::: unit of P x unit of V 2. V :::: dynexcm6::::dynexcm4 em2 = 0.0895 x 100 = 8.95% E 6. Volume V = 43 =>ilV= -41C3r 2Llr f9. Planck's constant, h = -1[1' [h] = [ML2r2] = [Ml}r l ] 33 [rl] ilV x 100 = 3 Llr x 100 =3xl% d% 10.v=al+b - Vr t+c RIR, 5.0x 10.0 50 [al] = [vi = [LT\" I ] 7. InparalIel, Rl'= - - ' - = -_·_··-=--=3.30 = [Lr2] RI + R2 5.0+ 10.0 15 Dimensions of c = [t] = [T] (we can add quantities of same dimensions only). Also ilRI' x 100 = LlRI x 100+ ilR2 X lOa RI' RI R2 :::: 7% [F11. I¢] = [BA] = A] ~ [ML~2I3-] = [ML2T -2A-I] Rl' = 3.3Q±7% qv ATLT-I 12, alT] = [MLT-2] => a = [MLr'] 8. The final result should contain three significant figures. 9. Length 1= 2.53 + 1.27 = 3.80 em biT] = [MLr2] or b = [MLT--'J Lll=O.OI +0.01 =0.02 13.1 N= GJ I kg x 1 kg orG= I Nm2kg-2 (Most probable errors of both the rods arc added) Hence true value = (3.80 ± 0.(2) em 2 1m S·mce 10• p:::: In -2- 14. 1 N = 6.67 x 10_ll~lx-l, = 6.67 x 10-17 N ffr I (10) 1( -il-p-')\"XIOO= (i-lmx'2M~+t-,L) xlOO 15. 1 J = I N x J km \\p m r L = =6.67 x lO-17 N x J0' m 6.67 x 10-14 J = (2c003 + 2 x O.OO~ + 2.06) x 100 0.3 0.5 6 Exercise 3.2 = (O.OJ + 0.02 + 0.01) x 100 = 4 1. c. 2.000 em, because it contains maximum number of significant figures, 4. 11. Maximum percentage error in p:::: 4% + 2 x 2%:::: 8% 12. Maximum error in density:::: 3 + 3 x 2:::: 9%. 2. Not significant In a number without decimal, the zeros on Exercise 3.3 the right of non-zero digits are not significant. 3. Probable error reduces to 115 as the number of observations 1. a. Main scale division:::: x:::: 1 mm is made 5 times. Lctybetheverm.erscalc dIV\"ISl'onh,t eny:::: ~9:0::: 0.9mm 4. Quantity having higher powers, because errors get multiplied 100 by powers. Vernier constant: x ~ )':::: I - 0.9 :::: 0.1 mm S. Here S = (13.8 ± 0.2) em; I = (4.0 ± 0.3) s 2. c. 19x = 20y and x - y = 0.1 V = 13.8 = 3.45 ms-I Solve to getx = 2 mm 4.0 3. a. Maximum error is taken to be half of least count. il: = ±(~~ +~t)=±C%~+~~~) . Pitch :::: I lllin X IO~ mm = ±O.0895 4. a. Least count:::: N . . . . --:::: 5 0.01 dlvlslOn 200

Appendix: Solutions to Concept Application Exercises A~7 Chapter 4 8, AB = 10 x 10 = 100 m, BC = 20 x 10 = 200 m A Exercise 4.1 ~ 1. a. True, because a revolving object is under at;:cdcration. b. True. because for a constant velocity, acceleration is zero BC c. True, when a plane takes offit has varying velocity, hence under acceleration. Fig, 5-4,2 d. False, a person in circular motion is under acceleration. e. False, here velocity is varying so the guard is under Displacement, AC = ~ AB' + BC' = 1001:5 m acceleration. Av. veL =AC = 1001:5 = 51:5 m/s 2. a. No, if velocity of a body is zero, there may be acceleration t 20 in the body. For example, a body at its highest point during Av. speed = .A-B.+-~B-C- = -10-0-+-2-0-0 = 15m/, vertical motion. t 20 h. No, if acceleration of a body is zero, its velocity may be constant or zero. 9, 1j=3i+2), 1;=71+6}, d~r,-i;=4i+4] c. Yes, change in velocity = acceleration x time, 10, u = 108 kmlh ~ 30 mis, v = 36 km/h = 10 mls and average acceleration::::;; change in velocity/time v2 = u' + 2as => 10' = 30' + 2a x 200 => a = -2 mis' \"2 =\"2d. True, here u = (), s = ut + 1 at 2 1 at 2 => s oc [2 v = u + at => 10 = 30 - 2t => t = 10 s e. Twe, D\" = u + '\" (2n - I) => D\" ex; (2n -1) 11, sat' => s =kt2 => v=d~, =2kt 2 dt 3. a. Yes, if its velocity is slowing down towards north, b. Yes, if velocity and acceleration are in opposite direction. a =d.v- = 2k -+ (constant) c. Yes, if an accelerating object starts decelerating. d. No, because average speed can also be greater than dt average ve~ocity. 12, 8 =0 + al x 10 => aI =0.8 mis' e. No, change in velocity takes place in the direction of acceleration. 0' = 8' + 2a, x 64 => \", = - OS mis' 0= 8 + (12 t3 =? t3 = 16 s 4. a. Yes, for example, a freely falling particle at its highest point has zero velocity but accelerating downward. al 1=\\0 '264m h. No, if distance is zero then displacement is also zero. IIfI c. Yes, if velocity is also in negative direction. d. Yes, if motion takes place continuously in one direction, A sl B s2 C °2'3 D 5. Distance:::: 5 = Jrr:;:> r:::: 5/Jr m u=O ---8m/s ---8m/s v=O Displacement = 2,. = 1011C m ~'ig, 8-4.3 SI = -Ialt , ~-I(0.8)10, =40m 22 Distance 5 II S2 = 584 - 40 - 64 = 480 m, t2 = -S2 = 480 c::-:--:--- = - ~ - -= 60s Displacement lOl1C 7 v8 6. AB=RO => 10= -150 => 0=-21-C- Total time taken = 10 + 60 + 16 = 86 s 1C 3 13. lO~ u+\"(2x2-1) and 25=u+\"'(2x5-1) o 22 e Solving, we get u = ~ mis, a = 5 m/s2 A ';'----7'B 2 D7= -5+5- [2x7-11=35m 22 Fig, S-4,l 14.25= 1<+\"'(2x5-1) and33= u+(/.(2x7-1) .22 AB=2(OB).s11. 160o= 2 x1-5 x1-3=-1-51e3m 7f 2 Solving, we get u = 7 mis, a = 4 m/s2 J[ 15, Vwl = 50 + 40 = 90 km/h = 25 m/s awl = 30 + 20 = 50 emls2 = 0.5 m/s2 7. v,w = -s = 100x10+200x20 :::: 500 I , => 100 + 100 = 25 t + I .2 10+20 0.5 t t -~-'-m/s Srel = Urel t + - arel t . _. 22 3 Solving, we get t = 10m -55 s

A~8 Physics for IIT-JEE: Mechanics I Exercise 4.2 I ., -1lOt 2 =;1., '-21-8=0 3. S=ul+ -af' =; -40= 101- 2 2• 1. x:::::t 3 +3t 2 +2t dx :::::3t 2 +6t+2 =>V= =; t=4',-28 - dl Ignoring -·ve value, we get t :::: 4 s dv 4, v=u+al =; 0=20-gsinOx4, =; a= - =61+6 ,I dl 2. x=3+81+712 dx =; Sill 0= - =; 0= 30' v=-=8+141 2 dl v 2 ::::u 2 +2as Vt =2, =8+14x2 = 36 m/s 0 2 = (20)2 + 2(-g sinO) (hlsinO) =; h = 20 m a :::::d-\"v\"-::::: 14 m/s2 5. a, s=ul+-1al2 =; -200=OxI+-I(-IO)t2 dl 22 \"2 =; 1= J4ij 8 3. fdv= fadl b, -200 = 101--IlOt2 =; I'' -2t-40= 0 2 20 r313 2t3' ]2 =; 1= l+v'4i s v-2=l-+--+2t =; v=18m/s ,32 0 C. -200=-IOI-.l1012 =; 12 +21-40=0 2 4. x::::: ~t+_':_2_t2 => d 2x 2a, 23 =; 1=-I+v'4is a=-,~:::::-------=- tit 2 3 5. s= 13 -61 2 +31+4, v=-ti=s 31 2 -121+3 6. H u2 I +12 ), 1 dl =-, u =-g(l, 11=2-gl,12 2 2g dv I (1= -=61-12 H dl Put a = 0 =; 61 - 12 = 0 =; 1= 2 s v, =2,3(2)2 - 12 x 2 + 3 = - 9 mls Exercise 4.3 1. i. a. False, time of ascent:::: time of descent. 1u b,True, D,, = u+-(I(2n-1 ) = 0+1-0(2x3-1) =25m Fig, 8·4,4 22 '2GI,ven -I, 1, h =~3 H c. True, because at the time of dropping the velocity of = - . solvmg the packet is upward and same as that of balloon. 3' , we get 4 d. True, acceleration due to gravity is same for all bodies. 7. u=~2xfH => u=1i[ ii. a. only velocity is zero b. time of descent c. second half ,,2 _ _ 2, a. H = 20 = - =; u = /2g x 20 = 20 m/s 2g , , u 20 =2s b.1 = '- =1-0 g a c, On hitting the ground, the velocity will be equal to the initial Fig. S·4,5 velocity but in opposite direction. Hence, answer is - u -H = ul--1 gl2 => gl 2 -2111-2H =0 = 20 m/, 2 I, gl2_fiji 1-2H=0 d, S, = 20 x 0,5 -\" 10 (0.5)', => g12_ 2 fiji I+fiji 1-2H =0 2 !iti[iI-2JH1+JH r!iI-2JH1=0 S, = 20 x 2,5 .. .l 10 (3,5)' => (!iI+JH)(!it-2JH) =0 => 1=2v{iiif 2 Required displacement = S2 - S, = 10 m e, 15 = 201 - .l 101 2 =; 12 - 4t + 3 = 0 =; I = I s, 3 s 2

Appendix: Solutions to Concept Application Exercises A-9 8. Velocity after 50 m fall. ul = )2gl1 =)100 g b. a =-2 mIs', 11= 8 mls 32 =uf +2x(2)h =;. 9 = 100 g -4h V = U + at = 8 - 2t, S:::: ut + -I- at 2 = 8f - t 2 100x9.8-9 rm/S2 2 sCm) h = 242.75 = 243 m ) I(s) 16 '~I(s) 4 -2 Net height = - 243 + 50 = 293 m 9. h= -1gn,-h,-=-I g(u -1) 2,· where . the total tI. me 9 2 22 IllS taken. Solving, we get 4 ~9 - - - -4- - - 8: 1(5) -- I(S) ../2 ( ../2 V../2+11 r;:: Fig. 8-4.8 u= f2-1 =;. n=lJ2=i)lJ2;i) =2+v2s c. a = 2 mis', u =-8 mls 10, Every ball will travel a distance of 5 m in the last second of V :::: U + at :::: - 8 + 2t, S:::: ut + -Ia,t :::: -8t +. t 2 going up to its highest point, provided its velocity is > 10 m/ 2 s initially (It at least should travel a maximum height of 5 m). rl11/S2 ~(mlS) sCm) Exercise 4.4 I(S) 4 9 )• 1. a, We know that the slope of position-time graph is equal to 1(5) velocity. So (i) as the slope is zero, it is zero velocity -2 1(5) -8 (ii) as the slope is positive and constant, it is constant -16 ---- positive velocity Fig. 8-4,9 (iii) as the slope is infinite it is infinite velocity d, a = -2 mIS', u = -8 mls (iv) as the slope is negative and constant, it is constant r 8~'(m/S)~v=u+al=-8-21,S=ut+ -Ial2 =-8t-12 2 negative velocity (v) as the slope is positive and increasing, it is positive increasing velocity lll S2 sCm) 16 (vi) as the slope is positive and decreasing, it is positive / ) I(S) decreasing velocity I(S) (vii) as the slope is positive and increasing. it is positive ~2 increasing velocity 4 (viii) as the slope is positive and decreasing, it is positive Fig. S-4.10 decreasing velocity h. True, as dvldt = acceleration 4. Let us first find the change in velocity. Change in velocity is the area under accelcration--time graph. For first 20 s, c. Change in velocity d. No, because it wil indicate infinite acceleration which is Area = tw = -1 x 10 x 20 + 10 x 20 = 300 mls not possible 2 c. Zero, because acceleration will be zero in uniform motion. 2, P--I . t;v 300 ., Av. acceleratIOn = - = ~ = 15 m/s velocity t;1 20 a S, From 0 to 4 s, at I = O. a = 5 mls 2 VI = U == at == 0 + 5 x 4 = 20 m/s b~1 ~I •U\"\"'O VI v2 I 1= 0 I 1=4 s t= 8 s Fig. 8-4.11 Fig. 8-4.6 From 4 to 8 s, a = -5 mis' 3. a, a = 2 mIs', u = 8 mls v2 :::: v! + at => v2 = 20 - 5 x,4 == 0 mls v = u + at = 8 + 2t, S = ut + ~ at2 = 8t + (2 v(m/s) 2 20 ------- ~'(m/s) . sCm) 10 10 - -- 20l2t----: \"--'----''--~-'>-_ I(s) 2 4 68 8 ,: 9- : . o 1~ls I(s) Fig, S-4.12 I(S) 12 Fig, 8-4,7

. A~10 Physics for IIT~JEE: Mechanics I f. True, sin a = ': =0.25 =! =:> a = 30° v 0.5 2 Maximum velocity is 20 m/s at t:::: 4 s, Displacement from 2 to 6 s = area from 2 to 6 s B= 90°+ a = 120° = lOx4+(l/2)x4x 1O=60m Since the motion takes place along the same direction only, II Net velocity so distance:::: displacement = 60 m 6. No, As shown, at a given instant of time, the body is at two Fig. 8-5.4 different positions which is not possible. 2. Distance to be covered = 1000 + 200 = 1200 m ,--mntTime Velocity = 72 km/h = 20 mig L_ _ _-_' _ _\"-.l_ _ _~) Position t = Distance = 1200 = 60 s Velocity 20 Fig. S-4.13 3. Required separat.ion \"'\" (v2 - vj ) t Chapter 5 = (72-36) -20. = 12km Exercise 5.1 60 1. a. True, as the ball gains horizontal velocity due to t.he 4. t = ~:~tallength = 110+ 90 =8s motion of the train and also it moves downward with constant acceleration resulting the net path to be Relative velocity (36+54)x(5118) parabolic w,r,t, an observer on the ground, Howeverw.r,t, an observer in train the path of the ball will be a straight 5. Required separation:::: (v2 - v1) t line vertically downward, = (l0-6)3 = 12 km h. False, the path is a straight line vertically downward, 6. VA =54. VB =-90 c. True, let the boat moves at an angle Bas shown in Fig, W ~'------~E Fig. 8-5.5 _____'-S-5.\\ \"L2 v - River flow Fig. 5-5.1 Relative velocity of B w.r,t, A, vBI A :::: VB -VA -90-54 = -144 km/h = - 40 m/s = 40 m/s. west Then time taken to cross the river, t ::co -\"-'- Relativ? velocity of ground w,r,t, B VsinB vCIlJ = Vc - VB = 0 - (-90) = 90 km/h = 25 mis, east Fortmin , sinB\"\"'1=>B=90° 7. 6 mis due south. Because relative velocity of B W,Lt. A will Hence, boat needs to move perpendicularly to the direction of flow of river. be equal and opposite to the relative velocity of A w,r.t. B. d. False. Bshould be > 90° -) ,.. --) A 8. a. vl\\=5i, vB=?) Relative velocity of B w.r.t. A /I Net velocity -}--+--+ AA AA v vilA = VB-VA. =7 j-5i \"\",-Si+7j VBA = ~52 + 72 =.J74 m/s Fig. 8-5.2 N c. True, vrlli! \"'\" v,. - vII! 7 Let rain be falling vertically e tan () v ,0 = ta(n l-!v!!...')\\ W--~~--~-------E ::::...!!J... => ·5 VI\" Vr e F\\i~:-v,,, S , e :~'\") Fig. 8-5.6 . Vc/\", (%)tanti=% =:>B=tan··t WofN V 'r ______________ ~ Fig. 8·5.3

Appendix: Solutions to Concept Application Exercises A-ll ~, l ' = -2u-sgi-nil, as 0~< l'!0°,SO 'It WI'II take,1ess ti'mo. 9.v,,=~15j 5. The gun is lined slightly above the target This is to N~, accommodate the effect of gravity, as due to gravity the bullet Vb wi!! descend some height by the time it reaches the target. 6. True, acceleration during projectile motion is same at all points and is equal to g, the acceleration due to gravity, It remains constant both in magnitude and direction. 7. At highest point, velocity is in horizontal direction and acceleration (due to gravity, g) is in vertical direction. So angle between them is 90°. => H I : =t a n 2a: as Hoc sin2 a _ H2 S 9. (i) Highest point, height is maximum. Fig. S-5.7 (ii) Point of projection, velocity is ma~imum, hence K.E. is maximum. (iii) Same at all points, sum of K.E. + P.E.:= total mechanical energy, it always remains same. 10. Since their time of flight is same, so maximum height attained by them will also be same. It is because both depend only upon vertical component of velocity. tan iI = ~~ = 25 iI = tan -, (25) S of E sf', 10 ' s 10. i. When she walks in same direction, relative velocity of 2 woman W.f.t. ground: R= ueos iI T=25cos60\"x sf2', 125f', =--m 4 v,VI :::: 1+ 1 5 = 2 5 m/s tlInc taken ::::: -s = -3S-- = 14 s ~, , 25 12. a. v = u cos iJi + (u sin iJ\" gt)j ii. When she walks in opposite direction, relative velocity of = 100cos 30\" i + (l OOsin 30° ~ lOx 2)] = sof', i + 30 J woman W.f.t. ground: _, s 3S b. ~\": :=ucos6i+usineJ :=lOOcos30oi+lOOsin30o] v') ::::: 1.5 - 1 ::::: 0.) mIs, tune taken::::: - = - ::::: 70 s = Sf',i+50j • v2 0,5 11. a. t = ---d'- 58-00~ ~--- = 1330 s 2.5 x\"':\"\" x 810 l20° evsin 18 ,5 Fig. S-S.8 b. x = (u + v cosil) t = (1.5 + 2.5cosI200) x ~ x 1330 -'!o -} -, ~ 18 U· v = 92.4 m Angle between u and v', cos a := , - - \" S 100 _ UV 12. v,< =40x-=-km/h, v\" =20m/s 18 9 sof', x50f', +30x50 (5)~ u=ta n _,l(ev\",)'\\ = _, 9 tan => a = ~,[ 2-J921 ] Exercise 5.2 cos 1. No, they do not depend upon mass, c. H = 2 = 502 2. All quantities will increase, because g is less on moon in --=125m ~L 2 x 10 comparison to that on earth. 2g 3. Ca) Highest point 13, a. ~I5=20sin30t~-~IOt2 => 12~2t~3=0 (b) Point of projection and point of return, 2 4. Letthe other angle is iJ, then iI+ 60° = 90° => iJ= 30° => (t-,,3)(I+I)=0=>t=3s

AM12 Physics for IIT-JEE: Mechanics I 2 -10-0 100 Ii = 20 m/s 4. G, = ( -221T) =9.87m1s 2 < h= 15 In tv Chapter 7 ~---x---+ Exercise 7.1 I<'ig. S-5.9 1. a. The horse cart will not have any reaction on it in empty space, ~b. x = 20 cos 30° x t = 20 x 3 = 30.[:i m b. Because of inertia. c. v2 = ,,2 + 2gh = 20' + 2 x 10 x j 5 = 700 => v = j o.fi mls c. As F = I!.P, so if the player moves his hands backward, I!.t Let the angle made by v with the horizontal is a Then vcosa=ucosB => lO.ficosa = 20cos30° /),t will increase and hence F will decrease. 2. a. Force on a body acts in a direction of change in its => cos a=-V\"f37 momentum. In both cases change in momentum of ball is d.H= -,u,,--\"s.:,e'\"2,--B,- =2-0'2-'sCiCn-' .3_0_ = 5 m along XO. Hence, in both the cases force will be along x-direction. 2g 2 x 10 y Maximum height attained above ground =h+H= 15+5=20m oLx 14. Both will reach at the same time, because initial vertical 0.\"2.4 velocity of both is zero and the time of flight depends upon the vCltical component of velocity only. (a) 15. On reaching the ground: Fig. S-7.1 VA=~u2+2gh, VB =,j2gh b. Case (a) Clearly, Momentum of the ball before reflection =- mv (along OX) Exercise 5.3 Momentum after reflecfion =- mg (along XO) = -nJV (along XO) 21T IT 1. a. (0 = -- = -rad/s Now, / =- change in momentum 60 30 =- -mv - mv =- -2mv (along OX) _ _ mvcos f) b. m= -2-1T- = --ITrad/s v 3600 1800 e C. (iJ =- 21T 1[ tmv sin a = - - - rad/s 12x3600 21600 mv cos e-+--- e (w -)'10x2,,- x 2 ~ 19.74 2 v 2. a, = air = m/s jm mv sin e ( 'I' (b) l3. a= 30-) +22 = 2.7 mis' 500 Fig. S-7.2 l((15/2)2)2 Case (b) !' =- ~mv cos B- InV cos B=- -2mv cos () a= ~-) +(0.5)' = 0.86 mis' tana=-G,=- 0.5 x 4 x 80' =-3-2::::::>a=-t1a(n32-) I =- ----2-~m~~v'-; -/ = G, 15 x 15 45 45 Ratio = - /' 2mvcose /' cosB

3. Just before t = 2 s. the velocity of the particle is Appendix: Solutions to Concept Application Exercises A-13 2-0 e u = - - = I em/s = 0.01 m/s Fig. 8-7.3 2-0 Magnitude of initial momentum = Pi = (Am) V Just after t:::;; 2 s, the velocity of the particle is Magnitude of final momentum = Pj= (ilm)V v = .2... _-0- = - I em/s = -0.01 m/s Change in momentum =: b..P =: ~r - 1; 2-0 t..P can be calculated by the vector subtraction, geometrically. The magnitude of impulse.T =Im(v - u)1 AsP;=P! => ()=45° = 10.04 (- 0.01 - 0.01) 1= 8 X 10.4 Ns I1P= j(p;2+p/) = )2(I1m)2V2 = J211mV The givenx- 1 graph (see Fig. 7.11. Chapter 7) may represent the repeated rebounding of a particle between two elastic =- - ---- =\"l.'.orce exerted on the h.ql.lld = -!:J' J211mV r2;;V pAV, walls at x ::: 0 and x :.:: 2 em. The particle will get an 111 111 impulse of 8 x IO\"A Ns after every 2 s. Hence, the pipe must be pushed at the corners with force 4. Velocity of the ball just before collision: v2 = 0 + 28h J2p AV 2 at an angle of 45°. v=v1 = ,f2gh = \"/2x10x5 =lOm/s Momentum of the ball before collision, ~ ::: mVi = O.o50x(-IO)] Ns = -0.50]Ns Velocity of the ball just after collision, vf = ,f2gh x x.£l.7. Impulse,! = 2 mu eos30° = 2 3 x 10 = 30jl Ns = ../2 x 10 x 1.25 = 5.0 m/s 2 Momentum of the ball just after collision, m JPf = mVf = 0.050 x (5.0]) = 0.25 Ns u Now impulse imparted by the ground on the ball 60° I = I1P =Pr -f; =0.25] -(-0.50]) =O.75]Ns Required force, F = !:J' = 0.75 = 7.5 N I1t 0.1 5. After 3 s of pouring, the bucket contains (3 s)(0.25 Lis) = 0.75 L of water, with mass 0.75 Lx (l kg/] L) = 0.75 kg, and feeling gravitational force 0.75 kg (9.S m/s2) = 7.35 1I N. The scale through the bucket must exert 7.35 N upward Fig. S-7.4 on this stationary water to support its weight. The scale must F = ~ = 30jl = 150jl N \"111 0.2 exert another 7.35 N to support the 0.75 kg bucket itself. 8. a. Theimpulse is to the right and equal to the area under the Water is entering the bucket with the speed given by F - t graph (see Fig. 7.15, Chapter 7): 1 = ([5 + lJI2) x mgytop :::;; (1I2)mvimpact2 4 = 12 Ns \"imp\", = (2gY,op)'12 = [2(9.S m/s2 )2.6 mJ112 = 7.14 m/s mvb. mVi + if =: f downward. ' => 2.5 x 0 + 12 = (2.5)v => v = 4.8 m/s c. From the same equation, The scale exerts an extra upward force to stop the downward => 2.5 x (-2) + 12 = (2.5)v => v = 2.8 m/s d. F,og \"\" = 12.0 => F\"g = 2.40 N motion of this additional water, as described by Exerdse 7,2 1. False. F is not acting on M. m ¥impact + f~xtra t = m ~f 2. No force is acting in horizontal direction on m. The rate of change of momentum is the force itself (dmldt)limpuct + Fcxtra = 0 F;,,,,,\" = -(dm/dt) \";mp,\" = -(0.25 kg/s)(-7.14 m/s) = + 1.78 N = 16.5 N Altogether the scale must exert 7.35 N + 7.35 N + 1.78 N = 16.5 N 6. Consider a mass Am of liquid flowing across the corner in time At. We will apply Newton's second law of motion to the mass Am.

A~14 Physics for IIT-JEE: Mechanics I TJ N ~ 37\" ~ON T= 400 N ¢ mg Fig. S-7.9 Fig. 8-7.5 As the system is in equilibrium, net sum of all the forces at No relative motion of m. Acceleration of In is zero. the junction must be zero. For this we resolve the tensions in 3. Weight shown by machine is N == Mg cos B. So the mass the horizontal and the perpendicular direction. In horizontal shown by the machine will be Nig = M cosO direction, we get f /iv cos53°1\"2 - cos37°1; == 0 (i) Mgcos8 '~ In vertical direction, we get _ _ _+-sze=~Mg sin e sin 37°1; + 8in53°T2 -400 ::::: {) (ii) Mg On solving the above equations, we get Fig. 8-7.6 T = 240 Nand T, = 320 N 4.10+ 1O=m!?=>m=2kg J 10 N 10 N 7. A horizontal string cannot balance a vertical weight. IOn N\". \".<on N 45°\"~ / 45° 10 N 10 N mg lOON Fig. 8-7.7 FBD orthe block 5. Let the man exerts force F on string. Fig. 8-7.10 2F=F+N+mg 8. The free body diagram is drawn so that only the string T4 is F+N= Mg cut, as shown in Fig. S-7.11 (I) (2) F IlOON N Fig. S-7.11 mg L Fr = 0 => T4 cos 60° = 100 => T, = 200 N Fig. S-7.8 9. Pseudo force == ma in backward direction. 10. In all the three cases the spring balance reads 10 kg. Fromequat.lOns(1)and(2):F= ~+~g (M-~g 2. an.dN=· 2 - Let us cut a section inside the spring as shown in Fig. 8-7.12 6. From the FBD of the block it is clear that tension in lower r---------------~r---------------~ cord is 400 N. Figure S-7.9 shows the junction where the r ::,,, ~'F:\"\"\" :~-',,, ~---------- I~______ _____ - - - - - - - - - - - Fig. 8-7.12 three cords meet.

Appendix: Solutions to Concept Ap,plication Exercises A~15 As each part of the spring is at rest, so F = T. As the block is By Newton's second law for 1 kg block, stationary, so T = F = 100 N. we have F2 = I a = I x I = 1 N 11. Let T be the reading of the spring balance. Then 14. By Newton's second law, we have for 20 kg block: 20g - T = 20a (i) for 10 kg block: T - 109 = lOa (ii) Ig-T,=la (i) (ii) Solving equations 0) and Oil, we get a =Ii , T = 40 g (T, + 2g) - 1', = 2a (iii) 33 and 7; -2g = 2a 40 . So the spring balance reading is - kg. Solving equatioIls (i), Oil, and (iii), we get 3 12, The FBD for the two cases is showIl in Fig. 7.13. a = 2 mis', T, = 8 N, In first case, let the force exerted by the man on the floor is and 7;=24N N j • Consider the forces inside the dotted box, We have, N, = l' + 50g Block is to be raised without acceleration, so T = 25R N, = 25g + 50g = 75g = 75 x 10 = 750 N In second case, let the force exerted by the man on the floor is N2. Consider the forces inside the dotted box. we have, =N2 50g - T, and T= 25g N2 = 50g - 25g = 25g = 25 x lO = 250 N As the floor yields to a downward force of700 N, so the man should adopt the second case. SOg SOg Ig Fig, S·7,J3 Fig. S-7,16 13. Since both the blocks are in contact, therefore they will move together with an acceleration By shortcut method: [(2+1)-I]g =2m/s' a = F;lel = -3- 0) 0+2+2) 2+1 x M{Olal 1', =m(g-a)= I (10-2)=8N 15. ~~ \\\\\\\\\\\\ \\ \\ \\ \\ '\\ Note: As the motion of the block is on s'mooth horizontal surface, so no need to mark the forces along vertical Fig, S·7,14 direction. Let the force of interaction between them is FJ. FBD Fig, 8·7.17 By Newton's second law for 2 kg block, we have F, = 2a = 2 x I = 2 N The same result can also be obtained from I kg block 3 - F, = la => F, = 2 N Let the force of interaction between thenris .F2 ,---------, £~!J~J:~;:~\\ \\ \\ \\ \\ \\\\ '-._--------, Fig. S·7.15

A-16 Physics for IIT-JEE: Mechanics I By Newton's second law, we have (i) T - (lOg + 8g) = 10 x 2 + 8 x 0 (ii) or T= 18g+20=200N T, - Ig = la (iii) T2 - T, = 3a 18. When the student takes a step'on the platform of the balance, there is a sudden downward change in motion in the and 2g - 72 = 2a beginning, so there is downward acceleration. Also, there is a sudden upward charge at the end of the step, so the Solving above equations, we get acceleration is in upward direction. g, 7g and 5g 19. Suppose 1', > 1\",. Then the motion of the rod will take a=6 m/s , T'=6N, T'=3N place in the direction of the force F2 with acceleration, say, a. Considering the motion of the entire rod, we have ForceonpulleyP\"F,= )r,2+r,' =.fir, F2 - F! :::: net force on the rod = mL x a where In is the mass of the rod per unit length of it. Force on pulley P2, 1\"2 = )T; + T; = .fi T, By shortcut method: Unbalanced load (2-1)g a= (1+3+2) Total mass = 1L mfs2 Fig, S-7.20 6 eConsidering the motion of the length from the first end T,= n~,p(g+a)=1(g+~)=7: N F - F j :::: rn€a. Dividing, we get 2(g 5:T, = m\"owo (g - a) = F-F, e . (F,-r,)e + -~) = N --- :::: FI ~~- or F :::: -----\"-- F2 -F, L L 16, By Newtons' second law 20. Let acceleration of mass I'n relative to wedgc down the plane T T is ar. Its absolute acceleration in horizontal directi9n is ar cos 60 0 - a (towards right). Hence, let N he the normal T reaction between the mass and the wedge. Then, N sin 0 = Ma :::: In (a r co-s60° - a) T (M+m)a (4+1)(2) 2 lkg or, a = - =-.----. = 20 mls , I11c0560' (1)(112) . 19 21. The point to this problem is that the monkey and the bananas Fig, 8-7,18 have the same weight, and the tension in the string is the same at the point where the bananas are suspended and where the 4g sin30 - T= 4a (i) monkey is pulling. In all the cases, the monkey and the and T-lg=1a (ii) bananas will have the same net force and hence the same Solving equations (i) and (ii), we get acceleration, direction, and magnitude. a. The bananas move up. (I = 2 mis' and T = 12 N b. The monkey and bananas always move at the same velocity, so the distance between them stays the same. -'--====-:::.:.::.By shortcut method: a = Unbalanced load c. Both the monkey and the bananas are in free fall, and as Total mass - they have the same initial velocity, the distance be~ween thcm docs not change. = 4g sin 30° - Jg 0 d. The bananas will slow down at the same rate as the monkey; if the monkey comes to a stop, so will the 4+1 ::::2m/s\" bananas. r<9, 22. From t:::: 0 to 2 s, the lift has a uniform acceleration a, = Ll.v = ~~_::() = 1.8 mls2 Ll.I 2-0 Fig, 8-7,19 From 1 '= 2 to lOs, the lift has a constant speed and hence T= M\"p(g+a) = 1(10+2)= J2N acceleration a2 during the period is () m/s 2 Force on the pulley, F = )T2 + T2 + 2 IT cos 60° = -f3 T . 17. Let tension in the rope is T. By Newton's second law, we From t:::: lOs to 12 s, the lift has a uniform acceleration. If a3 have is the acceleration during the period, then !\\v V2-Vj 0-3.6 a = - =--- = - - - -1.8 mis' -' 111 12 -1, 12-10

Appendix: Solutions to Concept Application Exercises Aw17 a. here m = 1500 kg, g = 9.8 m/s2 forces on the mass along x are T and a fictitious force i. At t == 1 S, a == Cli == 1.8 m/s2 (- Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the xwdirection, :. tension T, =m(g+a)= 1500(10+ I.S)= 17,700N ii. At t == 6 S, a == (12 = 0 Exercise 7.3 :. tension T, = meg + a,) = mg = 1500 x 10 = 15,000 N ii. At t = 11 sa = a3 = -1.8 mis2 1. Let in a given time dispiacementA isx l•B isx2' and that ofy :. tension T, = meg + a3 ) = 1500 (10 - 1.S) = 12,300 N b. Height reached by the lift willis x3 in downward direction, == Area of enclosed by v- t curve and t-axis (sec Fig. 7.86, As length of string remain constant, so we can write Chapter 7) Xl + (x j -x3) +x3 +(X2 -X3) +X2 = 0 = Area of quadrilateral OABC = 36 m .:::::> 2x +2x2 - X3 =0 J Average velocity = Total diplacment 26 -=~\"\"\"\"'= . = - = 3 mls Total tIme 12 Average acceleration = Net change in velocity . Total lIme = -o = 0 mis' , 12 23. a. When the lift moves upward with an acceleration, then Z the apparent weight is Aj X2 Xl Fig. S-7.21 W' == W + ma or W' == W + W a v,a. Differentiating. we get 2v, + 2v2 - vJ = 0 => 2v,+2(-1)-(-2)=0 => =0 g b. Again differentiating. we get 2a I + 2a2 - {lJ = 0 ( a1 => 2a 1 +2(-3)-1 =0 =>a, =7/2m/s2 = Wl1+g) here W' = 50 N, {l = 2.45 m/s2, g = 9.8 mis' 2. a.vp::: vA+vs :::::>5 +10 => VA =0 --- 2 2 50 = 1 +2-.4..5· ) =W x-5=> W=4-x 50 ( 9.8 4 5 = 40N b.5= vA -20 => VA = 30 mls b. Again if a is upward acceleration, then W' = W 2 ( a1. 3. a. Length of the string between A and D will remain the ll+g) gIves same. For this, (al - a4) + (a2 - (4) +a2 ::: 0 (1) =>a=30_ 1 = g 40 4 a= g =>-,,-=30 _I=_~ (1= _.1£ a, j D a, 4 g 40 4 4 • )\"'C la, t (Negativc sign shown that lift descends) a'i J; la2 ja; Le., lift moves downward with acceleration g 4 c. When the elevator cable breaks, the lift falls freely wilh acceleration 1. (I-~):. apparent weight W' = W =0 N LF,24. a. Fig. S-7.22 =Ma Length of the string from E to C will remain the same. a= T = IS.ON = 3.60 m/s2 to the right. a4 + a3 = 0 => a, = -a3 (ij) - --- M 5.00kg From equations (i) and (U) at + 2a2 -r 2a3 = 0 b. If v == constant, a = 0, so T == 0 (This is also known as an equilibrium situation). b. (a! +(2)+(a! +a3)+03 ::: 0 c. Someone in the car (non-inertial observer) claims-that the .:::::> 2al +a2 +203 =0

A-18 Physics for IIT-JEE: Mechanics I 4 ..!!M 4. Components of velocities along the string should be the same. a So v:::: U cose Fig. 8-7.26 8u u cose Hence, (lBA =- 2a Fig. S-7.23 ii. (lJlG::::a--aBA =-a-2a::::-a Alternatively: b. i. '(/BA =- 3a Length of string, e:::: y + ~1~2 + x2 B o ely 2x dx -a Differentiating we get 0:::: -dt + 1.22 dt Fig. 8-7.27 2;Jh~ +x~ O=~v+(cos(})u =>v::::ucosO H. (lBC:::: ~a2+a~A +2aaBAcosB = a~10+6cosO 5. let the speed of the ring is ll, then u cose=- v 8. ma -? pseudo force; ma cosB= mg sinO a::::gtanB e II cose Fig. S-7.24 mgcos8 :::::> u:::: v/cosB=- v secB 6. 3. Length of (l) and (3) will increase at the cost of part e 4(Fig. S-7.25). Hence decrease in length of (4) will be Fig. 8-7.28 .twice of increase in lengths of either 1 or 3, F = (M + lI1)a = (M + m)g tanO So aA:::: 2a +)9. i. Length of the string: e :::: X h2 + y2 4 ~A(w.r.t wedge) Differentiating, we get -d:f::: -d,y- + 2---,2\"x---,d-x- A dt dt 2)11' + x 2 dt _________ !~J j<'ig. 8-7.29 2y aA \"\" 2a e 180·e-_ a Fig. S-7.25 b. a l :::: ~a2 +a~ +2a~ C~~S(180-0) =- aJ5-4cosB 7. a. i. Length of (5) will decrease at the cost of increase in length of (I) and (3).

Appendix: Solutions to'Concept Application Exercises A-19 +)ii. Length of string. I! = 2x /]2 + y2 For this a2 = al sine (i) (ij) Again differentiate to get u = vlcosO N sin e= IV/a! (iii) 10. i. For block 2M. 2T= 2 Ma (i) mg eosO~ N = ma2 For block M. Mg - T= M2a (ii) Solving equations (i) and (ii). we get a = g13. Solving equations (i), (ii), and (iii). wc get ii. For block 2 M. (i) emg sin Beos mg sin 2 eeos () 2Mg sinB- T' = 2Ma al= M+msin 2 B ' a2 = M+msio\"-'}'() r 12.7F-N-Mg=Ma (1) r F+N-mg=ma (ii) r____ Movable 2F pulley F TJJ j' ~ 2a Mg Fig. 8·7.30 For movable pullcy, T' - 2T= 0 x a => T' = 2T (ii) For blnck M, T - Mg = M2a (iii) Solving equations (i), (ii), and (iii), we get a = g[sinB-IJ Fig. 5-7.33 3 This is negative, hence directions of accelerations will be opposite to those shown in Fig.S-7 JO. Note: Here we assume that the block 2M is moving From equations (i) and (ii), we get N = F [7m - M J\" down the plane, You may assume that it is moving up M+m the plane. Ihe magnitude afthe acceleration/aund will remain the same. Now for both to move together,'N;? 0 ::::;:. mlM z 117 13. Ace. of wedge should be a l = g tan 0 iii. For block 2M, (i) ~n M ~ 2M!? sinO- T= 2M(2a) N -;;- e e Fig. 8-7.34 T NsinO= 2Ma l=2MgtanO =>N= 2Mg T cose Fig. 8-7.31 N/:::=\\==+1==:;;m;::]1 ja, mg Fig. 7.35 For block M, 2T - Mg = Ma (ii) Solving, we get Tn = -2=M~ l-tan 20 Solving equations (i) and (ii), we get a = g[4sin 0-1] 9 14. VI = 1 mis, v2 :::: 3 m/s 11. Components of acceleration of M and m perpendicular to the 3 m!s incline should be same as both always remain in cpntact. ~ 4 - 3C h I mfs B Fig. 8-7.32 Fig. 5·7.36

A~20 Physics for IIT·JEE: Mechanics'r c. Pm1n = Fmax = PsN= 8 N d. When the block is in motion, F = JlkN = 0.2 x 20 = 4 N Let the velocity of cis V3 downward W.r.t. B. Then, Pm'\" = Fk,oo\"c = Jl..N = (0.20) (20 N) = 4 N (vI - v2) + (VI - v2) + v3 = 0 =? v3 = 2(v2 - v,) = 2(3 - 1) = 4 mls c. Since in this case P > J.1sN, the block accelerates and the friction force will be kinetic. Net velocity of C = )vi + vi = )32 + 42 = 5 mls 2. a. 15. Case I: T-N=40a, 20g - T= 20a N= 20a =? a =g14 Psin8 p Psine T 40 kg +Img f'C-~8~p cos8 8 mg a~ (b) 20g Fig. S-7.41 Fig. S-7.37 In first case: N! = mg - PsinB In second case: N2 = mg + PsinB Acceleration of B = .fia = g 12.fi Case II: T= 40a, 20g - T= 20a =? a = g/3 Since normal force in second case is greater so the friction will be greater in second case when the block impends the T motion. Hence in first case, less effort is required. b. P cos e= pN, = p(mg - P sine) ~ I\" T P = _-LJl::,:\"'£g__ cosB+)lsinB a~ A c. For pushing, P cosO> f.1 Nz =? P cose> p (mg + P sini1) 20g P > _~fc:l:.:m=\"g_-c case - Jlsine Fig. S-7.38 As Bincreases, denominator cosf)- J.1 sin Bdecreases and Ratio: g3 = - 3--n.--- =? n = 1 .' hence P increases. Let at f) = 00' denominator becomes --x - eozero. So coseo -p sineo= 0 =? = coc'(p). e eo,At = P> 00 which is not possible. . 2.fi g 2.fi 3. Ie = Jl.,N = Jl.,mg = 0.3 x I x 10 = 3 N Exercise 7.4 I/ = Jl.,N = Jl.,mg = 0.25 x I x 10 = 2.5 N a. F = I N, F < Ir, so the motion will not start and the 1. a. When F cxt = 0 I Ffriction = 0 friction will be static. So frictionf= F = I N. mg b. F = 2 N, F <.II' so the motion will not start and the Fig. S-7.39 b. First, we calculate maximum friction force. friction will be static. So frictionI = F = 2 N. F me> = Jl.,N = (0.40 X 20) = 8 N e. I = 3 N, F = If' Here the motion is just about to sta,t. Hence the friction force will be limiting. So friction f mg = Ir = 3 N. Fig. 8-7.40 d. F=4 N, F >fI, so the motion will start and friction will be Since in this case, P < F max' the block is in static kinetic. Hence friction will be I= IK = 2.5 N. equilibrium. Le., e. F = 20 N, F > fI, so motion will start and friction will be F=P=5N kinetic. Hence friction isf=fK= 2.5 N. 4. Here Ir = 10 N =? Jl,mg = 10 10 10 J1,= - = - - =0.2 . mg 5xl0 fK = 8 N =? JlKmg = 8 88 JlK = - = - - =0.16 mg 5x 10

5. Since the body is not moving. Appendix: Solutions to Concept Apptication Exercises A~21 8. Fig. S-7.42 mgsin (} mgcos 8 So I~ p andN~mg Net force applied by surface on the'body: Fig. 8-7.45 F~ )/2 +N' ~ )p2 +m2g2 N = mg cosO IK ~ JiK N = JiK mg cose 6mHence, required force is )lK 2 + N2 ~ N 6. i. I\",,, = ,uN ~ (0.2~(IOOl= 20 N 9. Applya~gsin600-PKgcos60° Since mg> Imax therefore, friction force is equal to.hnax and get JiK= J3-l i.e.,f =I\",,, ~ 20 N 10. From FBD friction,f ~ mg sina ii./\",,,~JIN~O,2x500~ lOON LN=mgcosa = tana 'N Since mg > [max., therefore, friction force is equal to mg. / This means that l ~ mg = 50 N N tOO N 5 0 0 N ]1,' 300 --.~-- --~--N SOON mg SON mg\"~50N } ~ mg\"\"'50N mgcosa Fig. S-7.43 Fig. 8-7.46 ~JlN ~ ~iii. f\",,, For the maximum possible value of a, the friction becomes =p(lOOcos300) =(0.2)100 1OJ3 N limiting friction, sO\"Jl=tana => cota= 3 Since the vertical component of the external force can 11. balance the weight of the block, therefore l ~ O. IN--:f5l ~! T ~8N 7. //1 I/2 30 sin 30° 30N Fig. 8-7.47 f<C-+-'---- 30 cos30° Itl = p,rn,g ~ 0.lx2x1O =2N h 11111111 NI \\\\\\\\\\'i\\ liZ ~rn2rn2g=0.2x3x 10~6N mg T+I\" =8 =>T~2N Fig. 8-7.44 Il+l~T =>f,~T-l~2-I~lN N~ mg-30sin30° ,As the speed is constant, so the acceleration is zero. 12. ff ~ pH ~ 0.4 x 2 g ~ 8 N 30cos30° = IK ~ JiKN => 30cos30° ~ PK (mg - 30 sin 30\") F~ 2.5 N Solving, we get fiK:;:: 3J3/7 Since F </r. So the block will not move and friction will be static and equal to 2.5 N. 13. Here the weights of the blocks will balance each other so that no friction force is required. 14. Push is a force directed inward (towards the centre) and pull is a force directed away from the centre, Let P be the force required to move the roller by pulling and Q be the force required to move it by pushing applied at the same inclination Bto the horizontal. Considering free body diagrams in the two cases.

A~22 Physics for IIT-JEE: Mechanics I (;iP Q'r;:\\ 17. T= Mg (T = tension in the rope) N::::: 60 g - T sin 60° (N is normal reaction between the man ~~ and the ground) and Teos60° == pN :-·~\"fL >tN'-.~;'n Solving these three equations with proper substitutions, we .e Ji e getM = 32.15 kg im . P 18. For an angle of 45.0°, the tensions in the horizontal and vertical wires will be the same. Mg a. The tension in the vertical wire will be equal to the weight w = 12.0 N. This must be the tension in the horizontal Mg wire, and hence the friction force on block A is also 12.0 N. Fig. S-7.48 b. The maximum frictional force is /l, >VA = (0.25)(60.0 N) = 15 N, this will be the tension in N + PsinB::::: mg N::'J-QsinB~mg both the horizontal and vertical parts of the wire, so the p cosO~.tiim ~ fl\"N maximum weight is 15 N. Q cosO:::::f'lim ::::: j.ls N' 19. The block has a tendency to slide down. A block placed on Hence P ~ _-,/l~,,-'_n_,,g__ Hence Q::::: /l mg an inclined plane has a t.endency Jo 'slide down the inclined s. plane due to its weight if the inclination angle of plane is ecos B+ f.'s sin cos B- fls sm B greater than the angle of repose, P<Q This is why it is easier to pull than to push a lawn roller. (~)Anglc of repose, ¢ = tan 1 = 30° 15. No. Let P be the pull of the engine. Considering the first four Here 0> 1/1. hence the block has a tendency to slide down compartments P-4f-T= 4 ma TT F m*gmsg~c\"ose Fig. 8:7.50 ~a 'F f where.f= frictional force on each compartment, m::::: mass of Fig. S-7.51 each compartment, a ~ acceleration of the train, T ~ tension. MgcosO+N=N' N' = Mg cosO+ mg cosO Considering the free body diagrams of the 21 st compartment For the block to remain stationary P- 21f- T' = 21 ma f= Mg sinO -Q If f is static in nature;I< fmax (i) (ii) ;-0-----~P Mg sin 0< pN' Mg sinO< p(Mg cosO+ mg cosO) ~ ~ Mg sinB-pMg cosO< p mg eose f f Fig. S-7.50 Hence T= P-4(f+ rna) and T'=P-21(f+ma) T' < T 16. a. Uno horizontal forci..'- is applied, no friction force is needed 20. Here also angle of inclination of the plane> angle of repose to keep the box in equilibrium. Hence, the block has a tendency to slide down the plane, b. The maximum static friction force is, Free body diagram of the block as seen from the frame of reference of the wedge. fl,N =I', W = (0.40)(40.0) = 16.0 N Considering the equilibrium of M in the plane of wedge. so the box will not move and the friction force balances the applied force of 6.0 N. c. The maximum friction force found in part (b). 16.0 N. d. Fromi, = !1,N = (0.20)(40.0 N) = 8.0 N. c. The applied force is enough either to start the box moving or to keep it moving. The answer to part (d) is independent of speed (as long as the box is moving), so the friction forcc is 8.0 N. The acceleration is (F\"- f,)lm = 2.45 mis'

Fig. S·7.52 Appendix: Solutions to Concept Application Exercises AM23 mg sin 6=/+ mamin cosO (i) For the string on the right, the vertical component of the strings tension T' is equal to 2mg (if the mass docs not move ef= mg sin O~ mamin cos down). The tension of the string itself, however, is T' > 2 mg. Therefore. the system is not in equilibrium. The right hand For friction 1.0 be static nature mass will have a greater pull. 2. No, force cannot determine the direction of motion. Yes, f <!,,,,, => f < J.1N force determines the direction of acceleration. F()r example, in circular motion force is directed towards the centre but the e-mg sin !namin cos B< (mamill sin 6 + mg cos B) motion is along the tangent, but acceleration is definitely la in along the normal which is the direction of the force. IJ tanIJ-p 3. a. The normal force is always perpendicular to the surface em > g sinli-J.1eos ::::> > g [ ,utan B+ I l that applies the force. Because your car maintains its ,usin B+cos J(lnlin orientation at all points on the ride, the normal force is [ J always upward. b. Your centripetal acceleration is downward toward the center [ J3-··J·3·~1=> amill > 10 -~l-'..\"- :::> amin > -\"1'r:::'0m/.s2 of the circle, so the net force on you must be downward. J3 J3+1 '/3 4. (a) Because the speed is constant, the only direction the force can have is that of the centripetal acceleration. The , force is larger at (C) than at (A) because the radius at (C) ,, iI; smaller. There is no force at (B) because the wire is straight. (b) In addition to the forces in the centripetal N direction in (a), there are now tangential forces to provide the tangential acceleration. The tangential force is the same at all the three points because the tangential acceleration is constant. B ©(a) emg sin Fig. S-7.53 Cd) F.,, 4 ,--) mamax cos f)= I/: + mg sin () ®,~~) J\"~. F N;:::; mamax sin B F, ' -, , .., I,,F -4 = p(mamax sin ff+ 111g cos 0)+ 111g sin B ,,, },~ 111g sin f} + pmg cos 0 l'~ (llmtx ::::: mcosB-,wnsinO ©(b) g(tanIJ+J.1) Fig. S-7.54 = (-11' tan g) 5. Outer wall will exert non-zero normal force. It is because the 1O(J3 +-.1.)1JJ333_amax ::::: block requires inward force on it which will provide the . - ' - - c -:.c: = 00 centripetal force to the block. Inward force can be given by outer wall only. 1__ mv2 6. a.N~ R , f - mg = O,f=J.1,N, F= J3 T So there is no limit on maximum acceleration. Exercise 7.5 1. Considering the string over the left nail, the vertical m components of the forces of tension T acting on t.he weights are mg if the string is secured on the naiL From Newton's -, third law the knot (point 0) is acted upon by the same forces T, whose resultant is 2 mg. mg Fig. S-7.55

A-24 Physics for IIT-JEE: Mechanics I Solving, we get T = ~/4;rg'Rp\" Mv 2 b, T= 2.54 s b.AtB,N-Mg= --- R The maximum speed at B corresponds to N::::: 0 rev = Irev (60S) =23.6 .r:ev -Mg::::: -m-v~2 ~ min 2.545 min min R c. The gravitational and the frictional forces remain constant. Vm\" = fRi = JiS.O(IO.O) =5.J(5 mls The normal force increases. The person remains in 9. a. Since the object of mass Jn2 is in equilibrium, motion with the walL d. The gravitational force remains constant. The normal and L;F,. =T-m2g =OOrT=m2g the frictional forces decrease. The person slides relative b. The tension in the string provides the required centripetal to the wall and downward into the pit. acceleration of the puck. 7. Let the tension at the lowest point be T. Thus, Fc::::: T= nt2g. LF ;:: mv2 (m, I ma; T-mg;::: mac=: r l-')gR In, fiG~f' d. The puck will spiral inward, gaining speed as it does so. It gains speed because the extra-large string tension ~ Forces produces forward tangential acceleration as well as inward radial acceleration of the puck, pulling at an angle mg~ of less than 90° to the direction of the inward spiraling ~'ig, S-7.56 velocity. e.'The puck will spiral outward, slowing down as it docs so. 10. r = R sine, Neose= mg (i) .0m/s2+(-,8_.0_0_m_/s_2'.'-)] moil' 16.0m x[l = 1.26 kN > 1000 N 8 mg N He does not make it across the river because the vine breaks. 8. a. v =: 20.0 mIs, N =: force of track on roller coaster, and Fig. 8-7.58 R = 1.00 m. Nsin 0::::: mo},. B N=:mm 2R ,:\"C \"c ~--~ ,, , ,, (ii) ,, 10m 15 m , eFrom equations (i) and (ii), we get cos = -g- ~ Ii OJ'/{ , ,, , ,, A ~----- Fig. 8-7.57 Mv 2 11, L;F = -- =N-Mg R From this, we find n= Mg+Mv-'= (500kg)(lOm/s2) R (500 kg)(2.0 m/ s') v + Fig. 8-7.59 lO.Om N = 5000 N + 20,000 N = 2.50 x 104 N

Appendix: Solutions to Concept Application Exercises A-25 2 Ftangential = 111 a L (i) (ii) eA~celeration at pointA, aJ = g sinB, at point B: a2 = ~ where v~ ~2gh = ~2g(e-fcosO) Net force Fo\" = ~(maL)' + (mOJ' L)2 => a, = 2g(1 - cosO) Friction force f = Fnet Given a, = Q 2 => g sinO= 2g(l - cosiJ) Squaring both sides and solving, we get f= ~(maLJ' +(mOJ' L)' cosO= -3 => 0= 53° 5 12. v = ~2gh = ~2gecosO , at = g tan 0 ~, a,. = f = 2g cosO 15, a. rim b. h ft, : __ ---' v \\ a, ' Fig. 8-7.60 mg sine + m ro2,. Net acceleration will be horizontal if the vertical components Fig. 8-7.61 of ac and a l cancel each other. f'? mg sinB+ In oil' For this, ac cos B= at =!> 2g colB= g sin2 B j1mgcosB '? mgsinB +mro2,. tan 0= .fi fl dan O+ -a-ir- = - I + =4-f'f-':x.:.1:0:=89--x-;3,;0. 13. i. Force on the particle when the fan is at full speed g cos 0 .J3 36 x 1000 x .J3 (_I)F = maiR = (2fff)2 R - -I+ 4ff2 x 363 n 1000 60 -.J3 4.J3 x 1000 = (_I)(2ffX 1500)' (~) 1000 60 100 I ff 2 = . J3 (I + x .363) = ,:l,ff2 N I => I r,;- + (I + 35) r,;- (45) = 2.6 2 -../3 -../3 ii. The friction force exerts this force. '2\". , j11east = 2.6 ,3 , 16. III. Fparticle = FII = 1r 14. a. If the block does not slip f ,; liN N mo,' L,; flmg => OJ'; (fl: )'\" b. It is the case of non-uniform circular motion. F- mdd-vt=1'dn(Od-Jt-L)=m.LddO-tJ= maL Fig. 8-7.62 tangential -

A-26 PhyskS\"'fo-r IIT-JEE: Mechanics r In the frame of roel, a. N-mg= -m-v-2 Also, mg sin 0 = pN + mra/ r Here N= mg cos-O To feel weightless at top, N = 0 mg sin B = J.ang cos (J ~ n11\"(J)2 b. r = 0.45 m, 0= 50' t1IIv2/2 sin 50\" = 0.766 and cos 50' = 0.64 mg sinB-mrro2 sin (J- roi Fig. 8-7.63 JI = mg cosO g cosB mv2 = 0.55 N-mg= -- 18. If v be the speed of the particle when radius vector makes an r angle Owith vertical, then (cos lJ)mg-N= 1111} m2 +2gr(l .. coslJ)1 -- =-[u rr 17. At the time of slipping, maximum friction acts on the body. N mrw2 + 11N mg sin e mg mgcos e Fig. 8-7.65 e m- Fig. 8-7.64 mg cos 0= -[(7)+2)gr-2grcosO] r when the particle loses contact N = 0 cos Ii= (7)+2)-2cosO=;.cos Ii= ('7;2)

... c



UT-JEE 2010 Solved Paper Physics Objective Type h. 11lultipie choice questio1ts :with olle correct answer R2 1. Incandescent bulbs are designed by keeping in mind v that the resistance of their filament increases with the increase in temperature, If at room temperature, 100 c. W, 60 Wand 40 W bulbs have filament resistances R lOo' R60 and R40; respectively, the relation between these resistances is a•....... _ ..L + ..L R2 R,oo - Ii,o R60 v c. Rwo > R60 > R40 d. G,~~2 Sol. d. WWv G, Power\" 11R IiT 2. To verify Olun's law, a student is provided with a test R, resistor Rr a high resistance R I' a small resistance R , V 2 two identical galvanometers G, and G , and a variable 2 voltage source V. The correct circuit to carry out the experiment is a. Sol. c. Ii, G is acting as voltmeter and G is acting as ammeter. 12 v 3.· An AC voltage source of variable angUlar frequency OJ and fixed amplitude Vo is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero), When (J) is increased

(!iJ .IIT-JEE 2010 SQlved Paper the critical temperature T (B) is a function of the magnetic 16. If the total energy of the particle is E, it will perform periodic motion only if field strength B. The dep~ndence of T (E)' on E is shown in the figure. <.: a. E<O b. E>O In the grapll1i below, the resistanceR ofasnperCOllductor c. Lv '0 >]-'>0 is shov.'l1 as a function of its temperature T for two different magnetic fields E,'(solid line) and E, (dashed Sol. c. line). IfIi, is larger than H, which of the followillg graphs shows the correct variation of R with T in these fields? Energy must be less than Vo' 17. For periodic motion of small amplitude .4, the time period T of this particle is proportional to a. b. R a. AVCf'1lT b·HVl c• ..htfJaji o fl, 1 [if T d. j'iilT d. c. Sol. h. R R [a]=MC'r' fl, Only option (b) has JimelLSioll of time. Alternatively, o T o .T i1/ll(d(IXi )' + kx4 = kA\" Sol. a. Larger the magnetic field, smaller the critical (eI\")''dt = 2k (.14 _ 4) 111 A X temperature. Substituting, we get x :::; Au. 15. A superconductor has Tc (0) = ] 00 K. When a magnetic 18. The acceleration of this particle for Ixl > Xo is field of 7.5 T is applied. its Tc decreases to 75 K. For a. proportional to Vo h. proportional to VimXo this material, one can definitely say that when c. proportional to~J..);}l:?~~d.zero a. D = 5 T. 7;, (D) = 80 K Sol. d. b. E=5T, 75K<Tc (D)<]OOK c. B=]OT,75K<7~.<]OOK As potential energy is conshmt for !x! > Xo' the force d. B = ] 0 T, Tc = 70 K on the particle is zero. Hence, acceleration is zero. Sol. b. Selicexplainatory Integer type Problems 16-18: This section contains ten questions. The ans\\ycr to each When a particle of mass 111 moves on the x-axis ina potential question is a single-digit integer ranging from 0 to 9. The of the form V(x)::::: /0;2 it performs simple harmonic motion. correct digit below the question number in the ORS is to be bubbled. The corresponding time period is proportional to y/;,f\", as can be seen easily using dimensional analysis. However, the motion of a particle call be periodic even when its potential energy increases on both sides of x = 0 in a way different from kY'2 and its total energy is such that the pm1iclc does Hot escape to infinity. Consider a particle of mass 111 moving on the x-axis. Its potential energy is Vex) = ax' (a> 0) for Ixl near the origin and becomes a constant equal to J..-~ for !x! > Xo (see fig.ure). Ft') 19. Gravitational acceleration on the surface of a planet is \\',(;~/llg, where g is the gravitational acceleration <. on the surface of the earth. The average mass density of the planet is 2/3' times that of the earth. If the es- cape speed on the surface of the eat1h is taken 10 be 11 kms-1, the escape speed on the surface of the planet in kms -1 will bo____

[!I)IIT-JEE 201()Soived Paper Sol. (3) [g''( =nvi-;; p' 2 24. When two identical batteries of internal resistance P = T 1 Q each are connected 'in series across a resistor R, the rate of heat produced in R is -J]. When the same lIenee, IIIf' =-3:[;f26 batteries are connected in parallel across R, the rate is J . IfJ ~ 2.25 J, then the value of Ii in Q is- --------,,-... 21 2 Vi iR'2p\"~\"7 I .,esc '1 Sol. (4) \" ~ =,':: iT J,~(i!~JII R'f, v' = 3 km/s J ~ (_II..J+o'.·1·1·~2··)Ii since 'JII' f ~ 2 25 esc ,2 2 . 20. A pieccofiee (heateapacity ~ 2100 J kg- 1 oC- 1 and II~4Q latent heat = 3.36 x lOs J kg-l) of mass 111 granL'> is at -SoC at atmospheric pressure. It is given 420 J of 25. Two spherical bodies A (radius 6 eni) and 13 (radius heat so that the ice starts melting. Finally, when the 18 cm) are at temperature T1 and.?'2, respectively. . The icc-\\vatcr mixture is in equilibrium, it is found that 1 maximum intensity in the emission spectrum of ,A is g of ice has melted. Assuming there is no other heat at 500 nm and in that of B is at 1500 nm. Considering exchange in the process, the value of m .............. . them to be black bodies, what will be the ratio of the Sol. (8) mte of total energy radiated by A to that of H? 420 ~ (m x 2100 x 5 + 1 x 3.36 x 10j ) x 10-3 Sol. (9) where m is in gIll. AT::::: constant 21. A stationary source is emitting sound at a fixed A'\"T ~A T A A 8 lJ frequency1.-, which is reflected by two cars approaching source. oThe f•requenci.es f Rate of total energy radiated\" AT' the difference between the 0' sound reflected from the cars is 1.2% 01'/. What is the 26. When two progressive waves)\\ ::::: 4 sin(2Y - 6/) and IJ (2x -Y2::::: 3sin 6t - }.) are superimposed, the ampli- difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound ,,/hich is tude of the resultant wave is_ _ 330 IDS ! Sol. (5) 1\\\\'0 vvavcs have phase difference lil2. Sol. (7) .~pp :::::,10 cc-.-+:- vv' 2!,c df~'-\"• '(J ~idv (c - v) where c is spcee,l of sound 4 df~ l6~j;, 27. A 0.1 kg mass is suspended from a wire of negligible m~lss. The lengthofthe wire is lmand its cross-sectional lIence, dv \"\" 7 km/hr. area is 4.9 x 10-7 m2 If the mass is pulled a little in the vertically downward direction and released, it performs 22. The It)eal length of a thill biconvex lells is 20 em. simple harmonic m01ion of angular frequency 140 rad S~l. I1'thc Youn~'s modulus of the material ofrhe wire When an object is moved from a distance of 25 em is 11 x 109 Nm~2, the value 01\"11 is_..\"\".._. in front of it to 50 em, the magnification of its image Sol. (4) {]-;1 chanbocs from lJl 2 5to)1'1(1_). The ratio m .Ill! _ '\" _~ __ .~~... (j) ~ Vi/IT 2) )0 28. A binary star e011<ists of two stars A (mass 2.2M) and ·Sol. (6) B (mass 11M), where Ai, is the mass of the sun:'They are separated by distance d and arc rotating ahout their 111=j+'-\"[u centre of mass, \\vrnchis stationary. The ratio of the total 23. An a- particle and. a proton arc accelcrHted from rest angulM momentum of the binary star to the angular by a potential difference of 100 V. Ailer this, their de momentum of star B about the centre of mass is Broglie wavelengths are A and A respectively. The ra- up' tio ApfA.a, to the nearest integ'e.r,_is\"_ ,...._-\"\". Sol. (3) Sol. (6)

(A.!J JIT-JEE20) o~()(\"edP\"per Paper 2 Objective Type Sol. d. Alultiple choice questions witlt one correct answer 1nR3pg= qE = 6P'lRv T 1. A block of mass 2 kg is free to move along the x-axis. It is at rest and fro111 t ::::: 0 onwards it is subjected to a :.q = 8.0 X 10- 19 C time-dependent force F(I) in lhe x direction. The force F(t) varies with t as shown in the figure. The kinetic 4. A vernier callipers has 1 mIll marks on the main scale. energy oflbe block after 4.5 seconds is It has 20 equal divisions on the vernier scale which match \\\"ith 16 main scale divisions. For this vernier F(/) callipers, the least count is 4N a. 0.02 mm b. 0.05 111m 4.5s c. O.IImn d. 0.2 lIun o 3s Sol. d. L.c. = I M.S.D - I V.S.D (I -1~) M.S.D a. 4.50 J b. 7.50 J )(1( I -} mm) = 0.2 mm c. 5.06 J d.14.06J 5. A hieonvex lens of focal length 15 em is in front of a So!. c. plane mirror. The distance between the lens and the Area under 1\"1 curve = 4.5 kg-m/see mirror is 10 em. A small ohjeet is kept at a distance of 30 em from the lens. The final image is 1(2)K.E. = ('Ii} )' =5.06 J a. virtual and at a distance of 16 em from the mirror h. real and at a distance of 16 cm from the mirror 2. A uniformly charged thin spherical shell of radius R c. virtual and at a distance of 20 em from the mirror carries uniform surHtce charge density of a per unit d. real and at a distance of 20 em from the mirror arca. It is made of two hemispherical shells, held to- gether by pressing them with force F. F is proportional Sol. b. to FF a. -Lo\"R' b·ILifR o '0 Eo ,~ ~ ~~ >~....... ................... 6cm IOcm c. I 02 d. lif 6. A hollow pipe of length 0.8 m is closed at one end. At t~lr r;;jl' its open end a 0.5 m long uniform string is vibrating So!. a. in its second harmonic and it resonates with the Pressure = ~ and force = . ri. x nR2 ilmdamental frequency of the pipe. If the tension in 2Eo 2Eo 3. A tiny spherical oil drop carrying a net charge q is bal- the wire is 50 N and the speed'ofsound is 320 Ins-I. anced in still air with a vertical uniform electric field of the mass of the string is strength(8Inl7) x 10'Ym-1 Whenthefieldisswitehed ofl'. the drop is observed to fall with terminal velocity 2 a.5g b.lOg X ]()-3 ms- I. Giveng = 9.8 ms -2, viscosity of the air = 1.8 X 10-5 Ns m -2 and the density of oil = 900 kg c. 20 g d. 40 g m- 3, the magnitude of q is Sol. b. a. 1.6 X 10- 19 C b. 3.2 X 10- 19 C c. 4.8 X 10-19 C d. 8.0 X 10- 19 C

Integer type 10. A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of This section contains five questions. The answer to each question is a single-digit integer, ranging from 0 to 9. the gas is 1',. (in kelvin) and the final temperature is aT,, The correct digit below the question number in the ORS is to be bubbled. the value of a is 7. A large glass slab (p = 5/3) of thickness 8 em is placed Sol. (8) over a point source qf light on a plane surface. It is TV,,-1 = constant seen that light emerges out of the top surface of the 71/,/5-1:=; ar(j~r/5-1 slab from a circular area of radius R em. What is the :.a =4. value of R? R 11. At time t = 0, a battery of 10 V is connected across Sol. (6) Scm points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) sin (J, = 3/5 does the voltage across them becomes 4 volt? [take In :.R=6cm 5 = 1.6, In3 = 1.1] 8. Image of an object approaching a convex mirror Sol. (2) of radius of curvature 20 In along its optical axis is 4= 10(1- e-';4) obscrved to move from 25/3 111 to 5017 m in30seconds. What is the speed of the object in km per hour ? :. t = 2 sec Sol. (3) Linked compreheltsion type For VI =~7Qm, \"I = -25 m Problems 12-14: When liquid medicine of density p is to be put in the eye, v = 235 m 11 = -50 m it is done with the help of a dropper. As the bulb on the top 2 '2 = ~g. X .~Ii = 3 kmph. Speed of object 19. To determine the balf-life of a radioactive element, a of the dropper is pressed, a drop forms at the opening of dN(t) versus t. l' .d_Nd-t(t-) the dropper. We wish to efostrimmeadteatthtehesiozpeemo.fntg.heIS dsrpohpe. n.Weael studcnt pIots a graph 0 fl 1{ -d~t first assume that the drop ~ere is the rate of radioactive decay at time t. If the number because that requires a minimum increase in its surface of radioactive nuclei of tlus element decreases by a energy, To deteruLine the size, we calculate the net vertical factor ofp after 4.16 years, the value ofpis_ _. force due to the surface tension Twhen the mdius of the drop is R. When the force becomes smaller than the weight of the 6 -----r----T----~----------~----l drop, the drop gets detached from the dropper. II I !I 5 -----II -----Ir----TI ----'I-----rI ----TI 12. If the radius of the opening of the dropper is 1', the III I I vertical force due to the surface tension 011 the drop of radius R (assuming I' <i( R) is_ 4 _____ I I I I I -dN-(t)] L___ - - - - 4 - _ _ _ ~-_---~_---~ dt In III I I [ I I! I 3 -----:-----t-----:---- : ----~-----1 \", a. 2.nrT b. )\",RT III I I 2 -----r----T----~-----r----r----~ 2.nR2T IIII I I d. --r-~ II I I I I I~--~'----~--~----~----7_--~ 2345678 Years Sol. c. Sol. (8) Surface tension force =2.nr7'R = 2JrtT N=Noe-AI 13. Ifr=5 X 1O-4m,p=103 kgm- 3,g=10m/s2,T=0.1l InldNldl1 = In(Ni) -At Nm- 1 the radius of the drop when it detaches from the , iFrom graph, A= per ycar dropper is approximately_____. 1112 = Q'j~~J = 1.386 year a. 1.4 X 10-3 III b. 3.3 X 10-3 m 4.16 yrs = 31 c. 2.0 X 10-3 m d. 4.1 X 1O-3 m 112 Sol. a. :.p = 8 ~bRu~T=mg=}4nR3 pg

~Ur-J~E201OS9IvedPaper 14. After the drop.detaches. its surface encrgy is_ _. 1=11Ir 2 +1111).. a. 1.4 X 10-6 J b. 2.7 X 10-6 J 1J 2 2 c. 5.4 X 10-6 J :.d= 1.3x 10- 10 111 Sol. b. ,nI, •<•:;=========~.~)==( -:;$I) 1112 Surface energy ~ T(4\"R') ~ 2.7 X 10-6 J r r2 Problems 15-17: Match the column type The key feature of Bohr's theory of spcetmm of hydrogen Match the entries in Column I with appropriate options atom is the quantization of angular momentum ,,,,hOll an electron is revolving around a proton. We will extend this in Column n. to a general rotational motion to fmd quantized rotational energy of a diatomic molecule assum,ing it to be rigid. The 18. Two transparent media of refractive indices /l' and f1 mle to be applied is Bohr '8 quantization condition. I3 15. A diatomic molecule has moment of inertia 1. By have a solid lens shaped transparent material of re- Bohr's quantization condition its rotational energy in fractive index/'2 between them as shown in ilgures in the n1h level (n :::: 0 is not allowed) i8_ _, Column II. A ray traversing these media is also shown in the figures. In Column I different relationships be- twceu,ul',uz al1d,u3 arc given. Match them to the my diagram shown in Column II. a. ;:1; ( 8:hrr'2[ ) c. 11 (8~221) d. n,(-/,7'-) 8n 1 Sol. d. 1.• = 1b11r1 b. )1-K.I.,.~ z.::' ~( 1111 2\" 21 21 16. It is fonnd that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to 41n X IO!l Hz. Then the moment of inertia of CO molecule about its centre of lllass is close to__ (Take 11 ~ 2\" X 10-34 Js) a. 2.76 X 10-46 kg m' b. 1.87 X 10-46 kg m' c. 4.67 X 10-47 kg m' d. 1.17 X 10-47 kg m2 Sol. b. hv ::: k.E \"'2 - kE~Fl II 1 = 1.87 X 10-46 kg m2 17. In a CO molecule. the distance between C (mass = 12 Sol. a.-» p., r.; b.--.)- q.,'s., t.; C.-7 p., r., t.; d. -;,. q., s. a.m.u) and 0 (mass::: 16 a.m.u.), where I a.m.ll. i 19. You are given many resistances, capacitors and inductors. These are connected to a variable DC X 10-27 kg. is close to_ _ _. 3 voltage source (the first two circuits) or anAC voltage source of 50 Hz frequency (the next three circuits) in a. 2.4 X 10- 10 m b. 1.9x 10- 10 111 diffcrent ways as shown in Column II. When a current 1(steady state for DC or nns for AC) flows through the C. 1.3 X 10- 10 m d. 4.4x 10- 11 III circuit, the corresponding voltage VI and V (indicated' Sol. e. 'm,d md 2 rl 111 +m andr =111 ;111 in circuits) are relatcd as shown in Column 1. Match I2 2 12 the lwo

I IIT·JEE201 O$olved Paper.~ii-I a. I ¢ 0, V] is propoltional to I II I¢ O, V2>\"V1 d. J Sol. a.~ r., s., to; b.~ q., r., s., to; C.~ p., q.; d....,. q., r., s., t.