Title AITS 02 SOLUTION (1).pdf ✅

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[1] ANSWER KEY PHYSICS 1. (2) 2. (3) 3. (4) 4. (2) 5. (4) 6. (2) 7. (3) 8. (4) 9. (1) 10. (4) 11. (3) 12. (3) 13. (4) 14. (1) 15. (2) 16. (3) 17. (4) 18. (2) 19. (1) 20. (4) 21. (2) 22. (1) 23. (2) 24. (4) 25. (1) 26. (1) 27. (2) 28. (1) 29. (1) 30. (3) 31. (3) 32. (3) 33. (2) 34. (1) 35. (1) 36. (2) 37. (1) 38. (3) 39. (1) 40. (2) 41. (4) 42. (2) 43. (1) 44. (1) 45. (2) 46. (3) 47. (1) 48. (1) 49. (1) 50. (1) CHEMISTRY 51. (3) 52. (1) 53. (4) 54. (4) 55. (3) 56. (3) 57. (2) 58. (3) 59. (2) 60. (4) 61. (1) 62. (4) 63. (3) 64. (1) 65. (3) 66. (3) 67. (2) 68. (1) 69. (3) 70. (4) 71. (3) 72. (1) 73. (4) 74. (2) 75. (4) 76. (4) 77. (1) 78. (3) 79. (3) 80. (3) 81. (3) 82. (1) 83. (2) 84. (1) 85. (2) 86. (2) 87. (3) 88. (1) 89. (1) 90. (3) 91. (2) 92. (3) 93. (1) 94. (3) 95. (2) 96. (1) 97. (2) 98. (4) 99. (2) 100. (3) BOTANY 101. (4) 102. (2) 103. (3) 104. (3) 105. (4) 106. (2) 107. (4) 108. (1) 109. (4) 110. (4) 111. (2) 112. (3) 113. (2) 114. (4) 115. (1) 116. (1) 117. (4) 118. (1) 119. (3) 120. (3) 121. (3) 122. (1) 123. (3) 124. (3) 125. (4) 126. (3) 127 (2) 128. (2) 129. (4) 130. (3) 131. (1) 132. (2) 133. (4) 134. (2) 135. (2) 136. (2) 137. (2) 138. (4) 139. (2) 140. (2) 141. (1) 142. (2) 143. (2) 144. (1) 145. (1) 146. (2) 147. (2) 148. (3) 149. (4) 150. (3) ZOOLOGY 151. (3) 152. (1) 153. (1) 154. (4) 155. (3) 156. (2) 157. (4) 158. (1) 159. (1) 160. (1) 161. (1) 162. (1) 163. (4) 164. (4) 165. (2) 166. (1) 167. (3) 168. (4) 169. (3) 170. (4) 171. (1) 172. (2) 173. (2) 174. (4) 175. (4) 176. (1) 177. (2) 178. (2) 179. (1) 180. (4) 181. (1) 182. (1) 183. (4) 184. (2) 185. (2) 186. (3) 187. (1) 188. (1) 189. (1) 190. (4) 191. (2) 192. (4) 193. (1) 194. (1) 195. (1) 196. (4) 197. (2) 198. (1) 199. (2) 200. (1) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 12/02/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 02 Dropper M
[2] SECTION – I (PHYSICS) 1. (2) Centre of mass of the rod moves a height, 2 = l h Work done, W mgh =  2 = mgl W 2. (3) From work-energy theorem ( ) 2 2 2 1 1 2 W m v v = − For first condition v1 = 0 v2 = 10 m/s ( ) 1 2 10 2 W m = W = 50 m For second condition ' v2 = 20 m/s ' 1 v =10 m/s W' = ? ( ) ( ) 2 2 ' ' 2 1 1 ' 2   = −     W m v v ( ) 1 ' 400 100 2 W m = − 300 150 m 2 = = m W' = 150 m W' = 3 × 50 m W ' 3 W = 3. (4) 1 1 2 2 I I  =  2 2 2 2 2 4 ML ML T T    =   2 4 T T = 4. (2) Total potential energy of spring = Total kinetic energy of block 1 1 1 2 2 2 2 2 2 kx kx mv + = 2 2 1 2 kx mv =  2 = k v x m 5. (4) −1/3 F S  or −1/3 a S      =   vdv a dS  −1/3  vdv S dS  −1/3 vdv S dS  −1/3   vdv S dS  2 2/3 v S  1/3 v S  Now, power, P F v   −1/3 1/3 F S S   0 P S  6. (2) For a given body, mass is same, so it will depend only on the distribution of mass about the axis The mass is farthest from axis BC, so I2 is maximum. Mass is nearest to axis AC, so I3 is minimum. Hence, the correct sequence will be I2 > I1 > I3 7. (3) I4 = 0 For 2 1 2 3 mL I I = = I3 = mL2 I = I1 + I2 + I3 + I4 2 2 2 5 2 3 3 mL mL = + = mL
[3] 8. (4) Power of the machine gun Total work done time = 1 2 2 = n mv t 1 2 , 1 sec 2 k mv t = =  The power of the machine gun = nk 9. (1) W F r r =  − ( 2 1) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ = + +  + + 4 3 11 11 15 i j k i j k = 44 + 11 + 45 W = 100 J 10. (4) From FBD, Normal reaction force on the block is  N = m(g + a) 2   = +     g m g 3 2 = g m  3 2 = mg N Displacement of the block 1 2 2 2   = +     g S ut t (u = 0) 2 4 = g S t Work done = FN × S 3 2 2 4 =  mg g t 2 2 3 Work done = 8 mg t 11. (3) ( ) F i j = − 6 8 ˆ ˆ N ( ) S i j = + 4 6 ˆ ˆ m ( ) ( ) W F S i j i j =  = −  + 6 8 4 6 ˆ ˆ ˆ ˆ = 24 – 48 = – 24 J 12. (3) A1 = area of complete circle 2 A a 1 =  A2 = area of small circle 2 2 2 2 4    =  =     a a A 2 2 4  = a A (X1, Y1) = coordinates of center of mass of large circle = (0, 0) (X2, Y2) = coordinates of center of mass of small circle = ,0 2       a Using 1 1 2 2 com 1 2 − = − A X A X X A A 2 2 2 0 4 2 4    −     =   − a a a a 3 2 8 3 6 4 − − = =  a a a com 1 2 0 0 0 − = = − Y A A . . . ,0 6   − =     a C O M 13. (4) vcm = 0, because internal force cannot change the velocity of centre of mass. 14. (1) Before collision After collision Momentum conservation mv + 0 = m × 0 + 2mv1  v = 2v1  1 2 = v v Coefficient of restitution (e) Velocity of sepration Velocity of approch = 0 2 0 − = − v e v 1 0.5 2 e = =