Title 04. MOtion in a plane Easy Ans.pdf ✅

1. (c) v = r  r v  = = constant [As v and r are constant] 2. (c) As time periods are equal therefore ratio of angular speeds will be same. T   2 = 3. (b) r mv F 2 =  2 F  v . If v becomes double then F (tendency to overturn) will become four times. 4. (b) Work done by centripetal force is always zero. 5. (c) It is always directed in a direction of tangent to circle. 6. (c) Stone flies in the direction of instantaneous velocity due to inertia 7. (c) Centripetal acceleration r v 2 = = constant. Direction keeps changing. 8. (c) Linear velocity, acceleration and force varies in direction. 9. (b) Angular velocity of particle P about point A, r v r v AB A 2  = = Angular velocity of particle P about point C, r v r v BC C = = Ratio . 2 1 / /2 = = v r v r C A   10. (a) . 2 r mv F = If m and v are constants then r F 1           = 1 2 2 1 r r F F 11. (a) In uniform circular motion (constant angular velocity) kinetic energy remains constant but due to change in velocity of particle its momentum varies. 12. (c) 13. (a) When speed is constant in circular motion, it means work done by centripetal force is zero. 14. (d) 15. (a) This horizontal inward component provides required centripetal force. 16. (a) Thrust at the lowest point of concave bridge = r mv mg 2 + 17. (d) 18. (a) Because the reaction on inner wheel decreases and becomes zero. So it leaves the ground first. 19. (b) 20. (a) r R r R T T r R a a R r r R r R =  =   = 2 2 2 2   [As Tr = TR] 21. (c) min Rad 60 2 min   = and min Rad hr 12 60 2  =    2 /12 60 min 2 /60  =     hr 22. (d) The particle performing circular motion flies off tangentially. 23. (a) The angle of banking, rg v 2 tan =  10 (150 ) tan 12 2   = r  r 10.6 10 m 10.6 km 3 =  = 24. (c) K.E. 2 2 1 = mv . Which is scalar, so it remains constant. 25. (b) v = 72 km / hour = 20 m / sec tan (2) 20 10 20 20 tan tan 1 1 2 −1 − −  =        =         = rg v  26. (a) 27. (d) / sec 4 / sec 60 2 120 rev / min 120 rad  rad  =  = 28. (c) In uniform circular motion, acceleration causes due to change in direction and is directed radially towards centre. 29. (b) Reaction on inner wheel         = − ra v h R M g 2 1 2 1 Reaction on outer wheel         = + ra v h R M g 2 2 2 1 where, r = radius of circular path, 2a = distance between two wheels and h = height of centre of gravity of car. A 2r r v C B P
30. (d) Maximum tension m r 2 =  = m   n  r 2 2 4 By substituting the values we get Tmax = 87.64 N 31. (d) l h rg v = 2  l rgh v = 8.57 m / s 10 50 1.5 9.8 =   = 32. (b) 2 2 2 2 2 3 a =  r = 4 n r = 4 1  20 10  a= 5 2 8 10 m/sec 33. (c) 34. (d) In 15 second's hand rotate through 90°. Change in velocity v = 2v sin( / 2) = 2(r)sin(90 / 2) 2 2 1 = 2 1   T  30 sec 2 60 2 4  cm = = [As T = 60 sec] 35. (c) Since n = 2 , 2  = 2  2 = 4 rad / s So acceleration r 2 = 2 2 2 / 4 100 25 = (4)  m s =  36. (b) r n r 2 2 2  = 4 = 30 60 1200 4 3 2         2 = 4740 m /s 37. (a) 38. (c) Particles of cream are lighter so they get deposited near the centre of circular path. 39. (d) Radial force mr p m p r m r mv 2 2 2  =      = = [As p = mv] 40. (b) r K r mv  2  v  r° i.e. speed of the particle is independent of r. 41. (b) If the both mass are revolving about the axis yy' and tension in both the threads are equal then ( ) 2 2 M x = m l − x  Mx = m(l − x)  M m ml x + = 42. (b) 20 9.8 400 tan 2  = = rg v    = 63.9 43. (d) In complete revolution change in velocity becomes zero so average acceleration will be zero. 44. (a) We know that Rg v 2 tan = and b h tan = Hence Rg v b h 2 =  Rg v b h 2 = 45. (b) 46. (a) r mv R mg 2 = cos − when  decreases cos increases i.e., R increases. 47. (d) Tension in the string T m r n mr 2 2 2 =  = 4  2 T  n  1 2 1 2 T T n n =  rpm T T n 7 2 2 = 5 = 48. (b) 49. (a) T m r 2 =   10 0.25 0.1 2 =     = 20 rad /s 50. (c) s m h km v = 36 = 10  1000 . 50 500 100 2 N r mv F =  = = 51. (a) r mv T 2 =  1.96 0.25 25 2  v =  v =14 m / s 52. (b) Centripetal force mr 5 1 (2) 20 N 2 2 =  =   = 53. (a) 2 2 r k r mv =  r k mv = 2  K.E.= r k mv 2 2 1 2 = P.E.  = F dr r k dr r k = = −  2  Total energy = K.E. + P.E. r k r k r k 2 2 = − = − 54. (d) Maximum tension = N r mv 16 2 =  16 144 16 2 =  v  v = 12 m/s v2 → v1 → 90° M x m y l y' R  mg cos mg v  Convex bridge
55. (a) The maximum velocity for a banked road with friction,         − + =     1 tan 2 tan v gr        −  + =   1 0.5 1 0.5 1 9.8 1000 2 v  v = 172 m / s 56. (d) mm s T r v r 6.28 / 60 2 0.06 2 =  =  = =    Magnitude of change in velocity = | | 2 1 v − v v v 8.88 mm / s 2 2 2 = 1 + = (Asv v 6.28mm /s) 1 = 2 = 57. (a) Work done by centripetal force in uniform circular motion is always equal to zero. 58. (b) v = r = 20 10 cm /s = 2m /s 59. (d) vmax = rg = 0.2 100 9.8 = 14m / s 60. (d) r mv F mg 2 = − 61. (a) rad s T / 60 30 2 2   = = = 62. (b) n 10.47 rad /s 60 2 100 2 =  = =    63. (d) Work done in circular motion is always zero. 64. (d) In complete revolution total displacement is zero so average velocity is zero 65. (c) vmax = rg = 0.75  60  9.8 = 21 m /s 66. (a) Distance covered in ‘n’ revolution = n 2 r= n D  2000  D = 9500 [As n = 2000, distance = 9500 m]  D 1.5m 2000 9500 =  =  67. (c) Centripetal acceleration = 4 2n 2 r=4 2×(1)×0.4=1.6 2 68. (a) 69. (b) Due to centrifugal force. 70. (d) As momentum is vector quantity  change in momentum P = 2mv sin(/ 2) = 2mv sin(90) = 2mv But kinetic energy remains always constant so change in kinetic energy is zero. 71. (a) rad s r v 1 / 100 100  = = = 72. (c) = = 0 dt d  (As  = constant) 73. (b) i j k i j k v r ˆ 2 ˆ 13 ˆ 18 5 6 6 3 4 1 ˆ ˆ ˆ = − − + − =   = − 74. (a) 2 2 2 2 2 50 493 / 2 1 a 4 n r 4   = cm s      =  =  75. (c) Maximum force of friction = centripetal force N r mv 270 30 100 (9) 2 2 =  = 76. (a) v = rg = 0.4  30  9.8 = 10.84 m / s 77. (b) v = r = 0.5 70 = 35 m /s 78. (a) 2r = 34.3  2 34.3 r = and 22 2 2 r T r v   = = Angle of binding =          = − tan 45 2 1 rg v  79. (c) 80. (a) T m r 2 =     T  5 rpm 4 2 1 1 2 1 2 =  = =     81. (d) tan [ 3] 20 3 9.8 (14 3) tan tan 1 2 1 2 −1 − − =          =         = rg v  = 60 82. (c) Centripetal acceleration 2 2 2 2 2 4 4 2 1 4 4   =       = n r = 83. (b) Centripetal force = breaking force  m r = 2  breaking stress  cross sectional area  m r = p  A 2   10 0.3 4.8 10 10 7 6    =  = − mr p A    = 4 rad / sec 180° mv mv
84. (a) Because velocity is always tangential and centripetal acceleration is radial. 85. (c) T = tension, W = weight and F = centrifugal force. 86. (c) 0.61 4 9.8 (4.9) 2 2 =  = = rg v  87. (d) As body covers equal angle in equal time intervals. its angular velocity and hence magnitude of linear velocity is constant. 88. (b) rad s r v 0.1 / 100 10  = = = 89. (a) r mv F 2 =  m rF v = 90. (d) Electrostatic force provides necessary centripetal force for circular motion of electron. 91. (a) Acceleration v T v r v r    2 2 2 = = = = 92. (b) v = rg = 0.6 150 10 = 30 m/s 93. (b) 94. (c) r mv F 2 =  2 F  v i.e. force will become 4 times. 95. (d) v = rg = 0.25  40 10 = 10 m / s 96. (d) Time period = 40 sec No. of revolution Time period Total time = 3.5 Rev 40 sec 140 sec = = . So, distance = 3.5  2R = 3.5  2 10 = 220 m . 97. (a) 2 2 13 4 4 10 − m  n r =   0.08 10 /sec. 8 n =  cycles 98. (b) Momentum changes by 2mv but kinetic energy remains same. 99. (c) L = I. In U.C.M. =constant  L = constant. 100.(c)  W = FS cos   = 90° 101. (b) 102.(c) In uniform circular motion tangential acceleration remains zero but magnitude of radial acceleration remains constant. 103.(d) The inclination of person from vertical is given by, 5 1 50 10 (10) tan 2 2 =  = = rg v   tan (1 / 5) −1  = 104.(d) The centripetal force, r mv F 2 =  F mv r 2 =  2 r  v or v  r (If m and F are constant),  2 1 2 1 2 1 = = r r v v 105.(b) As the speed is constant throughout the circular motion therefore its average speed is equal to instantaneous speed. 106.(a) Linear velocity, v =r = 2nr = 23.14 3 0.1 = 1.88 m /s Acceleration, 2 2 2 a = r = (6) 0.1 = 35.5m /s Tension in string, T m r 1 (6 ) 0.1 35.5 N 2 2 =  =    = 107.(a) 3 2 2 2 2 10 / 1 / 160 (400 ) m s k m s r v a = = = = 108.(b) vmax . = rg = 0.5  40 9.8 = 14 m / s 109.(b) N r mv F 3 2 10 50 500 100 =  = = 110.(b) R T F m         = 2 2 4 . If masses and time periods are same then F  R  1 2 1 2 F / F = R / R 111.(b) It is a vector quantity. 112.(a) v r v a = = 2  a v  = a       = 2 (2 )  i.e. remains constant. 113.(d) Tension in the string 2 T0 = mR0 In the second case 2 0 2 T = m(2R)(40 ) = 8mR 0 = 8T 114.(b) Average velocity 1 2 1 2 Total time Total displaceme nt − = = = ms s m 115.(d) Let  is the angular speed of revolution O T1 A T2 B T3 C l l l