Tiêu đề AITS 03 SOLUTION.pdf ✅

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[1] ANSWER KEY PHYSICS 1. (2) 2. (3) 3. (3) 4. (4) 5. (1) 6. (3) 7. (3) 8. (2) 9. (3) 10. (3) 11. (3) 12. (3) 13. (3) 14. (4) 15. (3) 16. (3) 17. (4) 18. (3) 19. (2) 20. (1) 21. (1) 22. (2) 23. (2) 24. (3) 25. (4) 26. (1) 27. (4) 28. (4) 29. (2) 30. (3) 31. (2) 32. (3) 33. (1) 34. (3) 35. (1) 36. (3) 37. (1) 38. (1) 39. (2) 40. (3) 41. (2) 42. (2) 43. (3) 44. (4) 45. (4) 46. (4) 47. (1) 48. (1) 49. (1) 50. (1) CHEMISTRY 51. (2) 52. (4) 53. (4) 54. (2) 55. (3) 56. (2) 57. (3) 58. (4) 59. (3) 60. (4) 61. (1) 62. (4) 63. (4) 64. (1) 65. (3) 66. (3) 67. (2) 68. (1) 69. (1) 70. (2) 71. (3) 72. (3) 73. (4) 74. (3) 75. (2) 76. (2) 77. (4) 78. (2) 79. (1) 80. (3) 81. (2) 82. (4) 83. (1) 84. (3) 85. (4) 86. (1) 87. (2) 88. (3) 89. (3) 90. (4) 91. (2) 92. (3) 93. (3) 94. (2) 95. (2) 96. (3) 97. (3) 98. (1) 99. (3) 100. (1) BOTANY 101. (4) 102. (2) 103. (4) 104. (2) 105. (3) 106. (2) 107. (3) 108. (3) 109. (4) 110. (1) 111. (1) 112. (1) 113. (2) 114. (1) 115. (4) 116. (4) 117. (1) 118. (3) 119. (2) 120. (3) 121. (1) 122. (3) 123. (2) 124. (3) 125. (1) 126. (1) 127. (3) 128. (2) 129. (2) 130. (4) 131. (1) 132. (4) 133. (4) 134. (1) 135. (2) 136. (1) 137. (4) 138. (3) 139. (4) 140. (2) 141. (3) 142. (2) 143. (3) 144. (2) 145. (1) 146. (1) 147. (1) 148. (3) 149. (2) 150. (4) ZOOLOGY 151. (4) 152. (4) 153. (2) 154. (3) 155. (3) 156. (2) 157. (2) 158. (2) 159. (3) 160. (3) 161. (3) 162. (1) 163. (2) 164. (2) 165. (3) 166. (3) 167. (2) 168. (2) 169. (2) 170. (2) 171. (2) 172. (4) 173. (3) 174. (4) 175. (3) 176. (3) 177. (4) 178. (4) 179. (3) 180. (4) 181. (3) 182. (2) 183. (2) 184. (3) 185. (3) 186. (3) 187. (4) 188. (4) 189. (1) 190. (3) 191. (3) 192. (4) 193. (1) 194. (1) 195. (2) 196. (2) 197. (1) 198. (1) 199. (4) 200. (4) DDUURRAATTIIOONN::29000MMiinnuutteess DATE : 26/02/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 03 Dropper M
[2] SECTION – I (PHYSICS) 1. (2) g  R 1 1 2 2 2 3    g R g R 2. (3) Polar satellites are low altitude satellites. 3. (3) Value of K changes continuously  Value of U is negative always  Total energy of planet is also negative  Since there is no torque. So angular momentum remains conserved. 4. (4) 2 2 0    R cos 45  g 2R 0 2 2 R        g g 5. (1) Energy stored per unit volume 1 Stress Strain 2 u    1 Stress Strain Strain 2 Strain            Y Y 1 2 Young modulus ( Strain) 2    6. (3) Young’s modulus of wire doesn’t depends on the length and diameter of the wire. It depends only on the material of the wire. 7. (3) If contact angle for the given liquid solid pair is less than 90°, liquid rise in the capillary tube. 8. (2) M  B g F Vtotaliceberg  insidewater g V g total inside total 900 9 1000 10    V V V inside total 100  90% V V 9. (3) 1 2 1 1 2 2 2 1 1 2 2                   H H H t t H H 10. (3) mg  ΔP  A 4 3 8 10 10 Δ 180 Δ 4.44 10 pascal 4.44kPa         mg P A P 11. (3)     9 9 Δ 80 40 40 5 5 72 F      F 12. (3) eq 1 2 1 1 1   R R R eq 2    A B K A K A K A L L L eq 2 5 7 2 2 2       KA KB K K K K 13. (3) Let junction temperature is  ,       3 200 100 2 Then, 0 0          KA KA L L KA L  3  600  2  200    0  6  800 400 °C 3    14. (4) 3 2 2 . Δ 3 3 F I F I I AY AY   Δ 8 3 l F l AY  15. (3) 1 B P V  
[3] 16. (3) Forces acting on both wires would be equal to T1 and T2 respectively by free body diagram Let equivalent force constant of wire  K K1  K2  K [Cross-sectional area will double when both wires taken together] 1 2 2 1 2 2 AY AY AY Y Y Y L L L      17. (4) normal 1 Stress Area Area F F   cosθ cosθ F A  2 cos θ F A  Cross-sectional area of PO  A Cross-sectional area of cosθ cosθ PO A PQ   18. (3) mg P A  mg V B A V   DV 3 V V R   19. (2) When coin falls into water block has to displace lesser volume to stay afloat. Implies that block will go up and water will go down. Hence both l and h will decrease. 20. (1) Pressure at 1 and 2 will be same ρ ρ ρ 2 X X Y Y Hg gH  gH  g  3.36 10 8 13.6 2    Y    Solving this we get 1 ρ 0.8 gcc Y   21. (1) upthurst  volume inside object ρ g 22. (2) A1 1 2 2 3 3 v  A v  A v 4 A 1 2A v3A 2 m / s 3 v  23. (2) A1 1 2 2 v  A v 1 2 1 2 A v v A   2 2 4 2 m / s 4  v    24. (3) 2 3 T  R 3/2 24 7 3.5         h R T R 24 6 2 2 2    h T h 25. (4) 4        GMm GMm GMm R R R h 4 3    R R h 3   R h 26. (1) Angular momentum is conserved and r is minimum at point A and maximum at point C . So, K A  KB  KC 27. (4) Total energy of bound system is negative.