Tiêu đề AITS 08 SOL EVENING.pdf ✅

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[1] ANSWER KEY PHYSICS 1. (1) 2. (4) 3. (1) 4. (3) 5. (1) 6. (1) 7. (4) 8. (4) 9. (4) 10. (4) 11. (2) 12. (2) 13. (2) 14. (1) 15. (3) 16. (1) 17. (1) 18. (2) 19. (3) 20. (1) 21. (4) 22. (2) 23. (3) 24. (1) 25. (2) 26. (1) 27. (2) 28. (1) 29. (4) 30. (3) 31. (4) 32. (4) 33. (2) 34. (1) 35. (3) 36. (4) 37. (3) 38. (2) 39. (4) 40. (1) 41. (2) 42. (4) 43. (4) 44. (3) 45. (3) 46. (2) 47. (1) 48. (2) 49. (3) 50. (1) CHEMISTRY 51. (2) 52. (1) 53. (2) 54. (4) 55. (3) 56. (1) 57. (3) 58. (4) 59. (4) 60. (3) 61. (4) 62. (2) 63. (1) 64. (4) 65. (1) 66. (2) 67. (4) 68. (1) 69. (2) 70. (2) 71. (3) 72. (3) 73. (2) 74. (2) 75. (2) 76. (4) 77. (1) 78. (1) 79. (3) 80. (4) 81. (2) 82. (1) 83. (2) 84. (4) 85. (1) 86. (3) 87. (1) 88. (4) 89. (2) 90. (3) 91. (4) 92. (4) 93. (3) 94. (1) 95. (3) 96. (3) 97. (3) 98. (3) 99. (3) 100. (4) BOTANY 101. (3) 102. (2) 103. (3) 104. (1) 105. (4) 106. (2) 107. (2) 108. (2) 109. (3) 110. (4) 111. (3) 112. (4) 113. (1) 114. (3) 115. (4) 116. (1) 117. (3) 118. (1) 119. (2) 120. (3) 121. (1) 122. (3) 123. (3) 124. (3) 125. (3) 126. (4) 127. (1) 128. (1) 129. (2) 130. (1) 131. (3) 132. (2) 133. (2) 134. (2) 135. (1) 136. (3) 137. (3) 138. (4) 139. (2) 140. (3) 141. (3) 142. (1) 143. (3) 144. (3) 145. (1) 146. (1) 147. (3) 148. (3) 149. (3) 150. (4) ZOOLOGY 151. (1) 152. (3) 153. (4) 154. (1) 155. (2) 156. (3) 157. (2) 158. (3) 159. (1) 160. (1) 161. (1) 162. (4) 163. (4) 164. (1) 165. (4) 166. (1) 167. (3) 168. (1) 169. (2) 170. (4) 171. (2) 172. (4) 173. (2) 174. (2) 175. (4) 176. (1) 177. (4) 178. (4) 179. (3) 180. (3) 181. (2) 182. (2) 183. (2) 184. (3) 185. (1) 186. (4) 187. (1) 188. (1) 189. (2) 190. (3) 191. (1) 192. (4) 193. (3) 194. (3) 195. (1) 196. (2) 197. (4) 198. (2) 199. (4) 200. (2) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 12/04/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 08 Dropper E
[2] SECTION – I (PHYSICS) 1. (1) 1 2 2 E Li = ( )( ) 1 2 5 10 2 = 200 10 20 i A   = =     E = 250 J 2. (4) d e dt  = 4 500 100 10 1 B NA dt  − = − =    = 5V 3. (1) wg sin sin90 c  =  sin w g c  =  4/3 sin 5/3 = c 1 4 sin 5 c −    =     4. (3) 1 3 sin 4 c = =  1 3 sin 4 c −    = =      5. (1) Factual. 6. (1) 0 0 , E c B = also speed of wave c k  = 0 0 E B k   = E0k = B0ω. 7. (4) Let I = I0 sinωt, Where I0 = 10, ω = 100 π Then dI M dt  = = 0 sin d M I t dt =  = MI0ω cos ωt max = MI0ω M = 5 mH 8. (4) ( ) 6 2 y 2.5 cos 2 10 10 N E t x C   − =  −    Direction -X axis wavelength 2 2 200m k 10−    = = =   6 2 10 6 10 Hz 2 2 f   = = =   9. (4) If XL = XC VL = VC  Reading of voltmeter = |VL – VC| = 0 10. (4) 11. (2) c =30° 1 sin30 vm c = =  8 3 10 8 1.5 10 m/s 2 vm  = =  12. (2) tan 8 R C = .....(i) 5 3 sin C = 1 sin 90o sin C = 3 5 ⇒ C = 37o 3 4 8 R = ⇒ R = 6 cm.
[3] 13. (2) 2 0 2 0 avg i dt i dt =   Areaof graph between 0to 2 2 i t t t − = = = 1 2 10 2 5 2 A   = = 14. (1) 1 2 ( 1) f R   = −     2( 1) R f = − ' 2( 1) R f = − " 1 R f = −  f ' = f & f " = 2f 15. (3) 16. (1) At any time t, the side of the square a = (a0 – αt), (where a0 = side) At t = 0, flux through the square φ = BA cos 0° = B (a0 – αt)2  emf induced d E dt  = −  E = – B.2 (a0 – αt) (0 – α) = + 2aαB 17. (1) , ,and v B are coplanar 18. (2) 1 1000 XC c = =  6 1 200 5 10 1000 −  = =   100 2 v  = =   cycle/s 19. (3) By using V = V0 sin ωt 0 0 2 1 2 sin sin 2 2 V t V t T T      =  =     2 sin sin 4 t T      =     2 sec. 4 8 T t t T    =  = 20. (1) The speed of light 0 0 1 1 1 0.25 2 8 4 C = = = =    21. (4) φ = BA cos θ = 2.0 × 0.5 × cos 60° 2.0 0.5 0.5weber. 2  = = 22. (2) As, dI M dt  = d M dt = (I0 sin ωt) = MI0ωt  max = 0.005 × 10 × 100π × 1 = 5π 23. (3) 0 0 ; ; N Ni L BA B ni i l   =  = =  = 2 0 2 0 N L A n Al l  = =  Where n is the number of turns per unit length L n 2 24. (1) Frequency remains constant during refraction vmed = 0 0 1 4 2 c =    med med air air /2 1 2 c c   = = =   wavelength is halved and frequency remains unchanged. 25. (2) 26. (1) ( ) 2 2 V V V V net = + − R L C Vnet = 5V