Tiêu đề AITS 07 SOLUTION .pdf ✅

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[1] ANSWER KEY PHYSICS 1. (2) 2. (1) 3. (1) 4. (3) 5. (1) 6. (1) 7. (4) 8. (1) 9. (1) 10. (1) 11. (2) 12. (2) 13. (1) 14. (1) 15. (3) 16. (3) 17. (4) 18. (3) 19. (1) 20. (3) 21. (3) 22. (3) 23. (2) 24. (3) 25. (4) 26. (2) 27. (1) 28. (4) 29. (1) 30. (4) 31. (2) 32. (1) 33. (2) 34. (1) 35. (4) 36. (2) 37. (2) 38. (4) 39. (3) 40. (3) 41. (4) 42. (3) 43. (1) 44. (4) 45. (1) 46. (2) 47. (1) 48. (2) 49. (4) 50. (2) CHEMISTRY 51. (4) 52. (3) 53. (2) 54. (1) 55. (4) 56. (2) 57. (2) 58. (1) 59. (3) 60. (3) 61. (4) 62. (4) 63. (4) 64. (3) 65. (3) 66. (1) 67. (3) 68. (2) 69. (1) 70. (4) 71. (3) 72. (2) 73. (2) 74. (2) 75. (1) 76. (3) 77. (4) 78. (3) 79. (1) 80. (3) 81. (3) 82. (3) 83. (4) 84. (4) 85. (1) 86. (1) 87. (4) 88. (1) 89. (3) 90. (2) 91. (2) 92. (1) 93. (1) 94. (2) 95. (1) 96. (1) 97. (4) 98. (1) 99. (3) 100. (4) BOTANY 101. (4) 102. (3) 103. (1) 104. (2) 105. (1) 106. (2) 107. (3) 108. (1) 109. (1) 110. (3) 111. (2) 112. (3) 113. (4) 114. (1) 115. (2) 116. (1) 117. (2) 118. (4) 119. (4) 120. (3) 121. (2) 122. (1) 123. (3) 124. (1) 125. (4) 126. (1) 127. (1) 128. (4) 129. (3) 130. (1) 131. (2) 132. (2) 133. (2) 134. (3) 135. (4) 136. (2) 137. (1) 138. (2) 139. (3) 140. (1) 141. (1) 142. (1) 143. (4) 144. (3) 145. (4) 146. (1) 147. (2) 148. (1) 149. (4) 150. (4) ZOOLOGY 151. (2) 152. (4) 153. (2) 154. (4) 155. (2) 156. (4) 157. (3) 158. (4) 159. (4) 160. (4) 161. (3) 162. (4) 163. (4) 164. (1) 165. (1) 166. (4) 167. (3) 168. (4) 169. (3) 170. (4) 171. (2) 172. (3) 173. (1) 174. (2) 175. (3) 176. (3) 177. (1) 178. (2) 179. (2) 180. (3) 181. (4) 182. (1) 183. (3) 184. (3) 185. (2) 186. (4) 187. (4) 188. (4) 189. (2) 190. (1) 191. (2) 192. (4) 193. (4) 194. (3) 195. (3) 196. (4) 197. (3) 198. (4) 199. (2) 200. (3) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 09/04/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 07 Dropper M
[2] SECTION – I (PHYSICS) 1. (2) 2 2 M M M net =  +  x y Mx = M cos 60° – M sin 60° Mx = M and My = M sin 60° + M sin 60° M M y = 3 Mnet 2 2 = + M M( 3) 2 = 4M = 2 M 2. (1) [cos0 cos60 ] 2 2 =  −  = = W MB MB W MB and sin60 3 2 2  =   = MB W  = 3W 3. (1) tan cos tan   =  tan tan cos  =    1 3 tan 2 −    =       4. (3) 2 2 2 1 2 2 2 2 cot cot cot cot 30 cot 45 cot cot 2  +  =   +  =   = 1 cos (2) −  = 5. (1) 6. (1) = = C x T C xT1/ = x C T y = mx m C =  = tan 3 0.4 57 7 10 C − = =  7. (4) At point P net magnetic field 2 2 B B B net 1 2 = + where, 0 0 1 2 3 3 2 . and . 4 4   = =   M M B B d d 0 net 3 5 . 4   =  M B d 8. (1) 1 2 3 1 2 3 1 2 3 1 1 1 + + = + + V V V R R R V R R R 30 20 10 3 2 1 1 1 1 3 2 1 + +  = + + V (10 10 10) 2 3 6 6 + + =   + +     30 6 11  = 180 11 = V 9. (1) 1 2 90 1 1 4 0.5 80     = − = − =          l r R l
[3] 10. (1) 10 10 15 100 15 =  = − AD DB R l R l l = 40 cm 11. (2) 1200 400 300 1200 400  = =  + Rp Reading of voltmeter 300 300 100 300 600 =  = + V 12. (2) 10 0 2 5 I − = = A 13. (1) The current in the circuit are assumed as shown in the figure. Applying KVL along the loop ABDA, we get – 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 .....(i) Applying KVL along the loop BCDB, we get – 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 .....(ii) Solving equation (i) and (ii) for i2, we get i2 = 5 A 14. (1) Resistivity of some material is its intrinsic property and is constant at particular temperature. Resistivity does not depend upon shape. 15. (3) Current through 6 resistance in parallel with 3 resistance = 0.4 A So total current = 0.8 + 0.4 = 1.2 A Potential drop across 4 1.2 4 4.8  =  = V 16. (3) The given circuit can be redrawn as follows Equivalent resistance between A and B is R and current V i R , current through AFCEB. i = i/2 = 2 V R 17. (4) 6  and 6  are in series, so effective resistance is 12  which is in parallel with 3 , so 1 1 1 15 R 3 12 36  36 15 R  4.8 1 5 2 36 V I A R  = = = 18. (3) When a charged particle enters a transverse magnetic field it traverses a circular path. Its kinetic energy remains constant. 19. (1) 0 2 i B r    =      = 0 2 4 i B r    =     20. (3) Angle between v and B is zero. 21. (3) B ni =0 1 0 0 ( ) (2 ) 2 n B i ni B   =  =  =      B1 = B 22. (3) 0 0 0 2 1 2 1 2 2 4 ( ) 4 ( / 2) 4 ( / 2) 4 i i B i i r r r    = − = −    0 0 4 (5 2.5) . 4 5 2   = − =   23. (2) According to Ampere's circuit law ∮ B dI I A A . 2 1 3 =  =  + =  0 enclosed 0 0 ( )