Title 31. Rotational Motion Easy Ans(1).pdf ✅

ROTATIONAL MOTION 1. B 2. C 3. C 4. B 5. C 6. A 7. A 8. A 9. C 10. C 11. B 12. A 13. C 14. C 15. A 16. A 17. A 18. B 19. C 20. D 21. A 22. D 23. C 24. B 25. B 26. C 27. D 28. C 29. A 30. B 31. D 32. C 33. C 34. C 35. C 36. D 37. B 38. C 39. B 40. C 41. B 42. A 43. B 44. A 45. A 46. D 47. A 48. B 49. A 50. D 51. B 52. D 53. D 54. B 55. A 56. D 57. B 58. C 59. D 60. A
61. B 62. D 63. A 64. B 65. C 66. D 67. B 68. C 69. B 70. C 71. D 72. B 73. C 74. C 75. B 76. B 77. B 78. C 79. A 80. A 81. C 82. C 83. A 84. D 85. C 86. B 87. C 88. D 89. B 90. A 91. C 92. C 93. B 94. B 95. A 96. D 97. B 98. D 99. D 100. A 101. A 102. C 103. D 104. D 105. C 106. D 107. C 108. B 109. C 110. A 111. C 112. C 113. C 114. A 115. A 116. B 117. D 118. A 119. A 120. A 121. B
122.D 123.B 124.D 125.D 126.A 127.C 128.B 129.C 130.C 131.A 132.C 133.B 134.B 135.B 136.A 137.C 138.B 139.C 140.B 141.D 142.B 143.A 144.D 145.A 146.D 147.D 148.B 149.A 150.A 151.A 152.A 153.A 154.(c) Let carbon atom is at the origin and the oxygen atom is placed at x-axis m1 = 12 , m2 = 16 , r i j r i j ˆ 0 ˆ and 1.1 ˆ 0 ˆ 1 = 0 + 2 = + → → 1 2 1 1 2 2 m m m r m r r + + = → → → i ˆ 28 16  1.1 = r i ˆ = 0.63 → i.e. 0.63 Å from carbon atom. 155.(a) Velocity of centre of mass 1 2 3 1 1 2 2 3 3 m m m m v m v m v vcm + + + + = 100 ˆ 50 10 ˆ 30 10 ˆ 20  10i +  j +  k = i j k ˆ 5 ˆ 3 ˆ = 2 + + . 156.(b) Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem) Diagonal of square d = a 2 = 80 cm  a = 40 2cm 8 , 1 m = kg 2 , 2 m = kg 4 , 3 m = kg m4 = 2kg Let 1 2 3 4 r , r , r , r are the position vectors of respective masses r i j ˆ 0 ˆ 0 1 = + , r ai j ˆ 0 ˆ 2 = + , r ai aj ˆ ˆ 3 = + , r i aj ˆ ˆ 4 = 0 + From the formula of centre of mass 1 2 3 4 1 1 2 2 3 3 4 4 m m m m m r m r m r m r r + + + + + + = i j ˆ = 15 2 + 15 2  co-ordinates of centre of mass = (15 2,15 2) and co-ordination of the corner = (0, 0) From the formula of distance between two points ( , ) 1 1 x y and ( , ) 2 2 x y distance 2 2 1 2 2 1 = (x − x ) +(y − y ) = 2 2 (15 2 − 0) + (15 2 − 0) = 900 = 30cm y m1 m2 C O x
157.(c) m1 = 7gm , m 2 = 4 gm , m3 = 10 gm and ), (2 5 7 ), ˆ 3 ˆ 5 ˆ ( 1 2 r = i + j − k r = i + j + k ) ˆ ˆ 3 ˆ r3 = (3i + j −k Position vector of center mass 7 4 10 ) ˆ ˆ 3 ˆ ) 10(3 ˆ 7 ˆ 5 ˆ ) 4(2 ˆ 3 ˆ 5 ˆ 7( + + + − + + + + + − = i j k i j k i j k r 21 ) ˆ 3 ˆ 85 ˆ (45i + j − k =  r i j k ˆ 7 1 ˆ 21 85 ˆ 7 15 = + − . So coordinates of centre of mass       − 7 1 , 21 85 , 7 15 . 158.(b) We know that second's hand completes its revolution (2) in 60 sec  rad sec t / 60 30  2   = = = 159.(d) Angular acceleration () = rate of change of angular speed t 2 (n n ) 2 − 1 =  10 60 4500 1200 2       − =  2 2 360 10 60 3300 2 sec degree   =  2 = 1980 degree / sec . 160.(d) Angular acceleration ( ) 2 3 2 2 2 2 at bt ct dt d dt d = = + +   = 2b + 6ct 161.(a) Distance covered by wheel in 1 rotation = 2r = D (Where D= 2r = diameter of wheel)  Distance covered in 2000 rotation = 2000 D = m 3 9.5 10 (given)  D = 1.5 meter 162.(d) Angular acceleration 2 1 2 12 / 5 60 0 rad sec t = − = − =    Now from 2 1 2 1  =  t +  t = (12)(5) 150 . 2 1 0 2 + = rad 163.(c) Angular displacement in first one second 2 (1) 2 1 2 1   =  = ......(i) [From 2 1 2 1  =  t +  t ] Now again we will consider motion from the rest and angular displacement in total two seconds    (2) 2 2 1 2 1 + 2 = = ......(ii) Solving (i) and (ii) we get 2 1   = and 2 3 2   =  3 1 2 =   . 164.(d) Number of revolution = Area between the graph and time axis = Area of trapezium = (2.5 5) 3000 2 1  +  = 11250 revolution. 165.(a) v = r = ) ˆ 2 ˆ ˆ ) (0 ˆ 0 ˆ 4 ˆ (3i + j + k  i + j + k i j k i j k ˆ 3 ˆ 6 ˆ 8 0 1 2 3 4 0 ˆ ˆ ˆ = = − + 166.(b)We will not consider the moment of inertia of disc because it doesn't have any mass so moment of inertia of five particle system 2 2 I = 5 mr = 5  2 (0.1) 2 = 0.1 kg-m . 167.(a) Moment of Inertia of disc I = 2 2 1 MR 2 2 ( ) 2 1 = R t R 4 2 1 = t R [As M = V   = R t 2 where t = thickness,  = density]  4         = x y x y x y R R t t I I [If  = constant]  (4) 64 4 1 4 = = x y I I [Given Ry = 4Rx , 4 x y t t = ] y B 8kg x 2kg 2kg 4kg D C A (0, 0) (a, 0) (0, a) (a, a) 402 C.M.