Tiêu đề AITS 05 SOL.pdf ✅

[1] ANSWER KEY PHYSICS 1. (4) 2. (3) 3. (1) 4. (1) 5. (3) 6. (3) 7. (1) 8. (2) 9. (4) 10. (1) 11. (1) 12. (3) 13. (3) 14. (2) 15. (1) 16. (2) 17. (1) 18. (1) 19. (3) 20. (2) 21. (3) 22. (2) 23. (2) 24. (4) 25. (4) 26. (2) 27. (2) 28. (3) 29. (2) 30. (2) 31. (3) 32. (4) 33. (3) 34. (1) 35. (4) 36. (4) 37. (2) 38. (1) 39. (2) 40. (4) 41. (1) 42. (4) 43. (1) 44. (1) 45. (2) 46. (4) 47. (3) 48. (2) 49. (3) 50. (3) CHEMISTRY 51. (4) 52. (2) 53. (2) 54. (2) 55. (2) 56. (1) 57. (3) 58. (3) 59. (3) 60. (1) 61. (4) 62. (1) 63. (2) 64. (1) 65. (2) 66. (3) 67. (4) 68. (1) 69. (1) 70. (2) 71. (3) 72. (2) 73. (4) 74. (4) 75. (3) 76. (3) 77. (2) 78. (2) 79. (4) 80. (1) 81. (1) 82. (3) 83. (2) 84. (3) 85. (3) 86. (1) 87. (3) 88. (2) 89. (4) 90. (1) 91. (2) 92. (2) 93. (4) 94. (2) 95. (3) 96. (3) 97. (3) 98. (3) 99. (1) 100. (2) BOTANY 101. (3) 102. (3) 103. (3) 104. (2) 105. (3) 106. (4) 107. (4) 108. (3) 109. (1) 110. (2) 111. (2) 112. (1) 113. (1) 114. (4) 115. (4) 116. (3) 117. (3) 118. (4) 119. (4) 120. (3) 121. (4) 122. (3) 123. (3) 124. (4) 125. (4) 126. (2) 127. (1) 128. (1) 129. (1) 130. (3) 131. (1) 132. (1) 133. (3) 134. (2) 135. (4) 136. (4) 137. (2) 138. (2) 139. (3) 140. (2) 141. (1) 142. (3) 143. (3) 144. (3) 145. (1) 146. (2) 147. (1) 148. (2) 149. (2) 150. (1) ZOOLOGY 151. (1) 152. (2) 153. (3) 154. (2) 155. (2) 156. (2) 157. (4) 158. (1) 159. (2) 160. (1) 161. (4) 162. (3) 163. (4) 164. (2) 165. (1) 166. (3) 167. (4) 168. (4) 169. (1) 170. (2) 171. (4) 172. (2) 173. (2) 174. (2) 175. (4) 176. (2) 177. (3) 178. (1) 179. (1) 180. (1) 181. (1) 182. (4) 183. (2) 184. (4) 185. (3) 186. (3) 187. (4) 188. (3) 189. (2) 190. (3) 191. (1) 192. (4) 193. (3) 194. (3) 195. (2) 196. (2) 197. (4) 198. (2) 199. (1) 200. (4) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 26/03/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 05 Dropper M
[2] SECTION – I (PHYSICS) 1. (4) Motion of centre of mass is not affected by internal forces so path of the centre of mass will remain unchanged. 2. (3) 3 < 2 < 1. The steamlined shaped body experiences less air resistance. 3. (1) Fl Y A l =   YA l F l  = (3 ) ( / 3) Y A l F l  = = 9 F 4. (1) No external force in horizontal direction so the centre of mass will fall vertically downward. 5. (3) Acceleration 2 v r towards centre  2 4 16 | | 8 2 2 a = = = towards centre i.e. = a j = − 8( ) ˆ 6. (3) As per founding off rule and significant figure. 0.00274 becomes 0.0027. 7. (1) 1 2 max 1 2 2 3 10 2 3 a a v t a a  = =  + + = 12 m/s 8. (2) As the ant can move on a plane, it has 2 degrees of freedom. 9. (4) Q = 35 J W = –15 J Q = W + U 35 = –15 + U U = 50 J 10. (1) Maximum velocity in SHM = A =  Maximum acceleration in SHM =  2A =   2 2 T T     = =  =   11. (1) Adiabatic relation between T and V is 1 1 1 TV − = Constant  1 1 1 2 1 2 TV T V − − = 12. (3)  N = 30 N 13. (3) 1 2 T v m =  To double v, the tension T must be made 4 times the original tension. 14. (2) For any gas, 3 B rms k T v v m = =  2 2 v v T T   = Constant 15. (1) When K.E. = P.E. 4 2 2 2 2 a y === cm 16. (2) 2 2 1 dx x dx x − =   2 1 1 2 1 x c x c − + − = + = − + − + 17. (1) V = g xh y LT–1 = [LT–2 ] x [L] y –2x = –1 1 1 , 2 2 x y = =
[3] 18. (1) L = 0.5 m, b = 0.2 m, h = 0.25 m Volume = length × breadth × 2.5 × 10–2 m 3 So, the order of magnitude is 10–2m 3 . 19. (3) v = u + at 7.5 = 2.5 + 0.50 t t = 10 sec 20. (2) Loss in P.E. = Gain in K.E. 1 2 2 2 =  L mg I  2 1 2 2 2 3   =        L mL mg 3  = . g L 21. (3)  = I 1000 = 200  = 5 rad/s2  = 0 + t  = 0 + 5 × 3  = 15 rad/s 22. (2) Use direct formula, Loss in K.E., 1 2 2 2 1 2 1 2 1 ( ) (1 ) 2 m m K u u e m m  = − − + 2 2 0 1 2 ( 0) [1 (1/ 2) ] 2 2 m m K v m m   = − − + 2 0 1 4 = m v 23. (2) 2 2 Whole disc 1 (2 ) 2 I M r Mr = =  2 Half disc 1 2 I Mr = 24. (4) Factual 25. (4) 2 2 5 I mR = As ext  = 0 so, ext 0 dL dt  = =  L = constant 26. (2) As V = Al ... (i) Where A is the area of cross-section of the wire. Young’s modulus, ( / ) ( / ) F A Fl Y l l A l = =   2 Fl Fl l YA YV  = = [Using (i)] l  l 2 Hence, the graph between l and l 2 is a straight line. 27. (2) Pressure amplitude, 2  =  =   0 Ak v vA As, v k   = =  Intensity, 2 2 2 2 n a I v  =  0 p va =   0p a  When ' 0p a a , then ' 3 = I' = 9I  800% I I  = 28. (3) 2 2 H O 2 2 2 kT f E E kT f N       = =       29. (2) y a t kx 1 =  − sin , ( ) 2 sin 2 y a t kx    =  − +      Phase difference is /2.  2 2 1 2 1 2 A a a a a = + +  2 cos 2 2 2 2 cos 2 2 a a a a  = + + = 30. (2) Time period of simple pendulum = 2 l T g