Tiêu đề 03B.SHM-PHY LEVEL - V KEY ( 115-130 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 115 JEE ADVANCED - VOL - III JEE MAINS - CW - VOL - I SHM NISHITH Multimedia India (Pvt.) Ltd., K E Y LEVEL-V SINGLE ANSWER TYPE 01) B 02) B 03) C 04) B 05) A 06) D 07) D 08) C 09) C 10) D 11) C 12) C 13) B 14) C 15) A 16) D 17) B 18) B 19) D 20) B 21) D 22) B 23) C 24) A 25) B 26) D 27) A 28) A 29) A 30) A 31) B 32) C 33) B 34) A 35) A 36) A 37) C 38) B 39) B 40) B 41) A 42) C 43) D 44) C 45) A 46) D 47) D 48) B 49) D MULTIPLE ANSWER TYPE 50) B,D 51) B,C 52) B,C 53) A,C,D 54) A,B,D 55) A,C,D 56) D 57) A,C COMPREHENSION QUESTIONS 58) C 59) A 60) D 61) D 62) B 63) D 64) A 65) A 66) A 67) B 68) A 69) C 70) A 71) A 72) A 73) B 74) A MATRIX MATCH TYPE 75) A-q ; B-t ; C-p ; D-t 76) A - r ; B - p ; C - p ; D - s 77) A - q,r ; B - q,r,s ; C - p ; D - p 78) A - r, s ; B - r, s; C - p, q, s ; D - r,s 79) A - s; B - q,r ; C -p; D - s 80) A - p,q,r; B - p,q,r ; C - r; D - s 81) A-p,q,r,s; B-r, s ; C-p,q,r,s ; D-p,q,r,s 82) A - r, s; B - q,r,s ; C - p,r ; D - p,q,r,s ASSERTION AND REASON TYPE 83) D 84) B 85) A 86) D 87) D 88) A 89) D 90) A 91) D INTEGER TYPE QUESTIONS 92) 8 93) 1 94) 1 95) 2 96) 1 97) 2 98) 1 99) 1 100) 2 HINTS LEVEL -V SINGLE ANSWER TYPE 1. (B) y t t   sin 3 cos   2sin 3 t           2 2 d y a g dt   t  ? 2. (B) 2 l T g a    if a  0 , then T T  1 1 if 2 2 , d y a dt  then T T  2 2 2 1 ? T T   3. (C) Time taken to collide on left wall and get back to the mass attached with spring is v L t 2 1  . Time to get the spring compressed once and comes back is, K m K T m t      2 2 2 2  Average time between two successive collisions = 1 2 2 t t t   collision frequency = 1 ? t  4. (B) From L.C. linear momentum : M m v mu     1 From L.C. energy :   1 1 2 2 2 2 M m v KA   5. (A) We know that time period is directly proportional to the square root of its length. Two
SHM 116 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., different pendulae can be in phase at the earliest with a difference in the number of oscillations being a maximum of one.  NTshorter N TLonger  ( 1) N 1  (N 1) 16 N  (N 1)4  4N  4 3N  4 ; 3 4 N  i.e., when 4 oscillations of the shorter and three of the longer pendulae are over they will be in phase. So, choice (a) is correct and rest of the choices are wrong. 6. (D) Given: 3  T  second, l 1m As the car accelerates it will given a pseudoforce on bob and the bob will tilt itself from the vertical by an angle           g 1 a  tan and the time period of oscillation is 2 2 2 g a l T    squaring both sides with T and L 2 2 2 2 1 4 3 g  a           36 2 2 g  a  2 2 2 a  36 10 2 2 2 36 10 34.5  a    ms so, choice (d) is correct. 7. (D) Time period of a rod oscillating about a horizontal axis with center of mass at a distance x is Mgx I Mx T 2 2    For least T,  0 dx dT 0 (1 ) .2 ( ) 2 1 . 2 2 2 1 2                   Mgx Mgx Mx Mx I Mx dx dT   2 2 2 2 1 Mx I Mx I Mx Mgx    2 3 2 Mgx.Mx  (I  Mx ) Solve to get x and substitute back in T to get 12 l xleast  8. (C) Using momentum conservation we get, 3 . u mu  Mv  m  M mu v 3 2  K.E. with the bob = 2 2 2 2 9 4 2 1 2 1 M m u Mv  M = M m u 9 2 2 2 Comparing with 1 2 2 . . 2 K E M A   at mean position, We get 2 2 2 2 2 9 4 M A m u   Or MA mu 3 2   1 2 2 2 3 mu v v MA             Frequency (C) is correct. 9. (C) Since total energy can be the maximum P.E., we get, 2 3 3 2 1 mv  Ky  KA  2 2 1 mv = K  3 3 A y  ; m K A y V 2 ( ) 3 3   Velocity maximum is therefore m KA V 3 max 2   3 / 2 Vmax  A so choice (C) is correct. 10. (C) As the rod is displaced by, to the spring will get
NISHITH Multimedia India (Pvt.) Ltd., 117 JEE ADVANCED - VOL - III JEE MAINS - CW - VOL - I SHM NISHITH Multimedia India (Pvt.) Ltd., compressed by y and the angular shift, tan  x y     2 2 2 2 1 2 1 2 1 I Ky mv Constant           2 2 2 2 2 2 1 2 1 2 3 1 mL Ky mv ML  Constant           2 2 2 2 1 2 1 2 3 1 m v Ky mv M Constant Differentiating w.r.t. time , we get,           m va K yv m va M 2 2 1 2 2 1 2 2 3 1 0 a = -        m  m M Ky 3          m M K 2 3  So, choice (c) is correct and choices (a) and (b) are incorrect. Choice (d) is not possible since, it is incorrect dimensionally. 11. (C) free body diagram is 12. (C) If the string is displaced slightly downward by x , we can write, the net (restoring) force on the peg        x x g xg 2 2 ; 2     Force gx 2    M a gx . 2    5 2   l a gx  2    a x   2 5    g . or 2 5 2 2 l T g      . 13. (B) Time period of spring         k m T 2 k, being the force constant of spring. For first spring.           1 1 k m t 2 ..... (1) For second spring           2 2 k m t 2 ..... (2) The effective force constant in their series combination is 1 2 1 2 k k k k k   Therefore, time period of combination            1 2 1 2 k k m k k T 2    1 2 1 2 2 2 k k 4 m k k T    ..... (3) From equations (1) and (2), we obtain             1 2 2 2 2 2 1 k m k m t t 4  2 2 2 2 t 1  t  T [from equation (3)] 14. (C) Let T be the tension in the cord and xa and xb the displacements of pulleys A and B respectively. Now assume that pulley B is fixed ; then extension of spring b b or x 2x 2 x x   . Similarly if we imagine that pulley A is fixed, . a x  2x But neither pulley B nor pulley A is fixed.
SHM 118 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd.,  a b x  2x  2x .... (1) From free body diagram of pulleys, b b 2T  k x .... (2) or a a 2T  k x .... (3) If eq k denotes equivalent spring constant, a b eq x 2x 2x k T    From equations (2) and (3) b b a a k 2T and x k 2T x   i.e.,           a b eq k 1 k 1 4 1 k Hence   a b eq a b 4 m k k k k m k     k xb b T T B A T T k xb b 2nd Method 15. ( A) When the plank is situated symmetrically on the drums, the reactions on the plank from the drums will be equal and so the force of friction will be equal in magnitude but opposite in direction and hence, the plank will be in equilibrium along vertical as well as in horizontal direction. Now if the plank is displaced by x to the right, the reaction will not be equal. For vertical equilibrium of the plank R R mg A B   ...(i) And for rotational of plank, taking moment about center of mass we have R L x R L x A B ( ) ( )    ...(ii) Solving Eqns. (i) and (ii), we get 2         A L x R mg L and 2         B L x R mg L Now as f R   , so friction at B will be more than at A and will bring the plank back, i.e., restoring force here         ( ) ( ) B A B A mg F f f R R x L As the restoring force is linear, the motion will be simple harmonic motion with force constant   mg k L So that     2 2  m L T k g . 16. (D) If    , the ball collides with the wall and rebounds with same speed. The motion of ball from A to Q is one part of a simple pendulum. Time period of ball  2( ) AQ t . Consider A as the starting point ( 0) t  . Equation of motion is x t A t ( ) cos   x t t ( ) cos ,     because amplitude    A  time from A to Q is the time t when x becomes   .        cost 1 1/ cos AQ t t              The return path from Q to A will involve the same time interval. Hence time period of ball  2 AQ t 2 1 1 cos 2 cos l g                       1 2 2 cos               g g .