Tiêu đề AITS 01 SOLUTION.pdf ✅

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[1] ANSWER KEY PHYSICS 1. (3) 2. (4) 3. (1) 4. (1) 5. (2) 6. (2) 7. (1) 8. (3) 9. (2) 10. (2) 11. (2) 12. (4) 13. (3) 14. (1) 15. (1) 16. (1) 17. (4) 18. (2) 19. (2) 20. (4) 21. (2) 22. (1) 23. (2) 24. (1) 25. (2) 26. (3) 27. (2) 28. (2) 29. (2) 30. (1) 31. (4) 32. (2) 33. (1) 34. (2) 35. (2) 36. (2) 37. (3) 38. (3) 39. (1) 40. (1) 41. (2) 42. (2) 43. (3) 44. (3) 45. (1) 46. (2) 47. (4) 48. (3) 49. (3) 50. (4) CHEMISTRY 51. (2) 52. (2) 53. (4) 54. (4) 55. (1) 56. (4) 57. (4) 58. (3) 59. (4) 60. (3) 61. (3) 62. (1) 63. (1) 64. (3) 65. (1) 66. (3) 67. (1) 68. (1) 69. (3) 70. (4) 71. (1) 72. (2) 73. (1) 74. (3) 75. (1) 76. (1) 77. (4) 78. (1) 79. (4) 80. (3) 81. (2) 82. (1) 83. (2) 84. (4) 85. (2) 86. (4) 87. (1) 88. (1) 89. (1) 90. (4) 91. (4) 92. (3) 93. (2) 94. (3) 95. (1) 96. (1) 97. (2) 98. (1) 99. (3) 100. (3) BOTANY 101. (1) 102. (4) 103. (2) 104. (3) 105. (2) 106. (1) 107. (2) 108. (3) 109. (2) 110. (3) 111. (3) 112. (4) 113. (1) 114. (2) 115. (3) 116. (4) 117. (2) 118. (3) 119. (4) 120. (1) 121. (4) 122. (2) 123. (2) 124. (3) 125. (4) 126. (2) 127. (2) 128. (4) 129. (1) 130. (2) 131. (1) 132. (2) 133. (3) 134. (2) 135. (4) 136. (1) 137. (2) 138. (3) 139. (1) 140. (4) 141. (2) 142. (4) 143. (2) 144. (1) 145. (3) 146. (2) 147. (1) 148. (2) 149. (3) 150. (4) ZOOLOGY 151. (4) 152. (4) 153. (3) 154. (3) 155. (2) 156. (2) 157. (3) 158. (4) 159. (2) 160. (2) 161. (2) 162. (4) 163. (1) 164. (1) 165. (3) 166. (2) 167. (4) 168. (1) 169. (3) 170. (4) 171. (1) 172. (4) 173. (4) 174. (2) 175. (1) 176. (1) 177. (2) 178. (1) 179. (2) 180. (1) 181. (2) 182. (3) 183. (4) 184. (2) 185. (4) 186. (3) 187. (1) 188. (1) 189. (1) 190. (4) 191. (1) 192. (2) 193. (2) 194. (2) 195. (1) 196. (1) 197. (3) 198. (4) 199. (1) 200. (2) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 29/01/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 01 Dropper E
[2] SECTION – I (PHYSICS) 1. (3) a = 2 V r = constant 2. (4) 2 180 u g =  u = 1800 30 2 = = 30 × 1.4 = 42 = 42 m/s 3. (1) 2F cos 60 40 3 2      =    3 2 40 3 2 F =  F = 40 N 4. (1) Substituting the values of ‘a’ and ‘b’, we get: x(t) = (t 2 + 1) x(3) = 9 + 1 = 10 m x(5) = 25 + 1 = 26 m avg (5) (3) 5 3 x x V − = − 26 10 2 − = = 8 m/s 5. (2) For  and 90° –  range will be same. 6. (2) t = 50 50 100 10 15 25 + = + = 4 s 7. (1) tan  = 100 500 y x V V =   = tan–1 1 5       8. (3) At highest point 2 V V  = 9. (2) V = dr dt 10. (2) 2 2 av (6 2) (8 4) 4 2 2 2 2 2 V − + − = = = m/s tan  = 4 1 4 =   = 45° 11. (2) 5 4 x v t = − 10 y v = ax =−4 m/s2 0 y a = anet = – 4 m/s2 12. (4) T = 2 sin u g  = same for all because u sin  is same for all. 13. (3) 14. (1) V = 81 36 117 + = 15. (1) Vrel = cos30 cos60 3 u  +  u = 2 2 u u + = u  t = x u 16. (1) 2 P Q 0 a 2a 3 0 a 3  =  − − =  = 17. (4) As seen from the figure the displacement is 2 2 (AF) (FD) 7 2m + = 18. (2) Let nˆ1 and nˆ 2 are the two unit vectors, then the sum is n⃗ s = nˆ1 + nˆ 2 or ns 2 = n1 2 + n2 2 + 2n1n2cos θ =
[3] 1 + 1 + 2cos θ Since it is given that ns is also a unit vector, therefore 1 = 1 + 1 + 2cos θ cos θ = − 1 2 ∖ θ = 120∘ Now the difference vector is nˆd = nˆ1 − nˆ 2 or nd 2 = n1 2 + n2 2 − 2n1n2cos θ = 1 + 1 − 2cos(120∘ ) nd 2 = 2 − 2(−1/2) = 2 + 1 = 3 ⇔ nd = √3 19. (2) A 2B 3C − + ˆ ˆ ˆ ˆ ˆ ˆ = + − − + − (2i j) 2(3j k) 3(6i 2k) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = + − + + − = − − 2i j 6j 2k 18i 6k 20i 5j 4k 20. (4) If two vectors A⃗⃗ and B⃗ are given then the resultant Rmax = A + B = 7N and Rmin = 4 – 3 = 1N i.e. net force on the particle is between 1 N and 7 N. 21. (2) Horizontal range 2 u sin 2 R ...(1) g  = Maximum height 2 2 u sin H ...(2) 2g  = According to the problem R = H 2 2 2 u sin 2 u sin g 2g   = 22. (1) Speed along the shortest path 1 4km / hr 15 / 60 = = Speed of water 2 2 v 5 4 3km / hr = − = 23. (2) Given m = 2 kg u = 4 m/s v = 0 m/s (body comes to rset) t = 2s Now, by Newton’s equations: v – u = at 0 – 4 = a(2) Thus, a = – 2 m/s2 So, Force, F = ma = 2 × 2 = 4 N 24. (1) By definition dx v dt = Where x, t are distance and time respectively. As slope represents velocity, for constant velocity the slope should be constant. In uniform acceleration, velocity keeps uniformly increasing which is the slope in a distance time graph hence a parabolic path is observed. 25. (2) Both will reach on the ground with same speed 2 v u 2gh 100 2400 50m / s = + = + = 26. (3) Both wind and car speed will force the flag to point towards S – E 27. (2) It is clear from the diagram that the shortest distance PQ Hence sin 45 OQ  =1 PQ 100 50 2m 2  =  = Also, 2 2 10 10 AB v = + 10 2 km h = 5 AB PQ t h v = = 28. (2) The train is going to stop and hence the acceleration is negative. When the boy drops the apple, the apple will not have any retardation along the horizontal direction. Whereas the boy sitting vertically below due to his contact with the train will have a retardation.