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[1] ANSWER KEY PHYSICS 1. (1) 2. (4) 3. (2) 4. (3) 5. (2) 6. (4) 7. (2) 8. (3) 9. (2) 10. (1) 11. (1) 12. (2) 13. (2) 14. (1) 15. (3) 16. (1) 17. (3) 18. (1) 19. (2) 20. (1) 21. (4) 22. (2) 23. (3) 24. (1) 25. (1) 26. (4) 27. (1) 28. (3) 29. (3) 30. (3) 31. (4) 32. (3) 33. (2) 34. (3) 35. (1) 36. (2) 37. (2) 38. (2) 39. (1) 40. (1) 41. (1) 42. (4) 43. (1) 44. (3) 45. (1) 46. (2) 47. (2) 48. (1) 49. (2) 50. (4) CHEMISTRY 51. (3) 52. (4) 53. (2) 54. (3) 55. (3) 56. (3) 57. (1) 58. (1) 59. (4) 60. (3) 61. (3) 62. (2) 63. (2) 64. (3) 65. (1) 66. (1) 67. (2) 68. (2) 69. (3) 70. (2) 71. (2) 72. (1) 73. (2) 74. (4) 75. (2) 76. (4) 77. (4) 78. (3) 79. (3) 80. (1) 81. (2) 82. (3) 83. (2) 84. (4) 85. (1) 86. (1) 87. (4) 88. (3) 89. (4) 90. (2) 91. (4) 92. (2) 93. (3) 94. (4) 95. (2) 96. (3) 97. (4) 98. (4) 99. (4) 100. (1) BOTANY 101. (4) 102. (3) 103. (4) 104. (4) 105. (3) 106. (2) 107. (3) 108. (1) 109. (1) 110. (3) 111. (2) 112. (4) 113. (2) 114. (3) 115. (4) 116. (3) 117. (2) 118. (3) 119. (2) 120. (2) 121. (4) 122. (4) 123. (2) 124. (4) 125. (2) 126. (1) 127. (3) 128. (2) 129. (1) 130. (2) 131. (4) 132. (1) 133. (3) 134. (1) 135. (4) 136. (3) 137. (4) 138. (1) 139. (4) 140. (4) 141. (1) 142. (1) 143. (2) 144. (1) 145. (4) 146. (1) 147. (2) 148. (3) 149. (4) 150. (1) ZOOLOGY 151. (3) 152. (4) 153. (2) 154. (3) 155. (4) 156. (4) 157. (2) 158. (3) 159. (3) 160. (2) 161. (1) 162. (1) 163. (3) 164. (3) 165. (2) 166. (3) 167. (3) 168. (2) 169. (2) 170. (2) 171. (3) 172. (4) 173. (4) 174. (3) 175. (4) 176. (2) 177. (2) 178. (4) 179. (2) 180. (3) 181. (3) 182. (3) 183. (2) 184. (1) 185. (3) 186. (3) 187. (3) 188. (3) 189. (2) 190. (3) 191. (4) 192. (3) 193. (3) 194. (3) 195. (2) 196. (3) 197. (1) 198. (2) 199. (2) 200. (1) DURATION : 90 Minutes DURATION : 200 Minutes DATE : 16/04/2023 M. MARKS : 720 All India Test Series (NEET-2023) Part Test – 09 Dropper E
[2] SECTION – I (PHYSICS) 1. (1) 2r = n   = 2r n Å 2. (4) rn  n 2 2 4 2 1 (4) (1) = r r r4 = 16 × 0.53 r4 = 8.48 Å 3. (2) n11 = n22 or n11 = n22 or 12 × 600 = n2 × 400  n2 = 18 4. (3) In  - particle scattering experiment electrostatic force acts. 5. (2) Relation between V0 – , V0 = h e – 0 h e Put it in the form of y = mx – c, here V0 = y,  = x, 0 h e = c  y =       h e x – c  m = h e 6. (4) 7. (2) Intrinsic semiconductor. 8. (3)  = = h h P mv , so with the increase in velocity of electron, wavelength decrease, and so fringe width decreases. 9. (2) max K1 = h1 –  = 1 – 0.5 = 0.5 eV max K2 = 2.5 – 0.5 = 2.0 eV Thus max K1 : max K2 = 0.5 : 2 = 1 : 4 10. (1) The energy of each photon = 20 4 10 200  = 5 × 10–19 J Wavelength =  = hc E = 19 34 8 5 10 (6.63 10 ) (3 10 ) − −       = 4.0 × 10–7 = 400 nm 11. (1)  0 =  hc  (0 )Na= 19 34 8 2 1.6 10 6.6 10 3 10 − −      = 6188 Å  0   1  0 copper 0 sodium ( ) ( )   = sodium copper ( ) ( )    (0 )Cu= 4 2 × 6188 = 3094 Å To eject photo-electrons from sodium the longest wavelength is 6188 Å and that for copper is 3094 Å. Hence for light of wavelength 4000 Å, sodium is suitable. 12. (2) Wave nature of light. 13. (2) P = 10 × 103 watt  = 300 m P =  nhc t 104 = 34 8 6.62 10 3 10 300 1 −      n n = 34 8 4 6.62 10 10 300 10    − = 1.5 × 1031 14. (1) Collision ionisation. 15. (3) h = h0 + Ek 6.6 × 10–34 × 3 × 1015 = 4 × 1.6 × 10–19 + Ek 19.8 × 10–19 – 6.4 × 10–19 = Ek Ek = 13.4 × 10–19 J  2 1 mv2 max = 13.4 × 10–19
[3] vmax = 19 2 13.4 10−   m = 31 19 9 10 2 13.4 10 − −    = 1.73 × 106 m/s 16. (1)  = = rms h h P mv Also 3 rms = kT v m 3  = h mkT 17. (3)  = h mv  v =  h m v = 31 10 34 9.1 10 10 10 6.6 10 − − −     = 7.2 × 105 m/s 18. (1) p = 2 p p h m e V  d = 2 d d h m e V    d p = p p d d m e m e  md = 2mp , ed = ep    d p = 2 p p p p m e m e = 2 1 19. (2) The de-broglie wavelength is 2r2 = 2  = r2 r2 = 22r1 = 4r1 = 4 × 0.53 = 2.12 Å  = 3.14 × 2.12 = 6.66 Å 20. (1) IE = IB + IC IB = IE – IC = 1000 × 10–6 – 0.96 × 10–3 = 1 mA – 0.96 mA = 0.04 mA 21. (4) Let the energy of one photon = hc/,  Energy of n photons E = nhc/  10–7 = 34 8 10 6.6 10 3 10 5000 10 − −      n n = 26 10 7 19.8 10 5000 10 10 − − −    = 0.25 × 1012 n = 2.5 × 1011 22. (2) The maximum kinetic energy is Kmax=  hc – = 280 1242 eV nm − nm – 2.5 eV = 4.4 eV – 2.5 eV = 1.9 eV Stopping potential V is given by eV = Kmax V = Kmax e = e 1.9 eV = 1.9 V 23. (3) The gap between conduction band and valence band is near about 1 eV. 24. (1) Fermi level of energy of an intrinsic semiconductor lies in the middle of forbidden gap. 25. (1)  = D d . When d → 0,  → , and so fringes will not be seen over the screen. 26. (4) 1 3 R R A = 0 1 15 3 1.2 10 (64) − =   R = 4.8 × 10–15 27. (1) In forward biasing, resistance of p−n junction diode is zero, so whole voltage appears across the resistance. 28. (3) 29. (3) The energy of 10 eV means that E = eV = 10 e Volt V = 10 Volt The electron was accelerated through a p.d. of 10V Now,  = V 12.27 Å = 10 12.27 Å = 3.9 Å 30. (3) A full-wave rectifier rectifies both the half cycles of the AC input i.e., it conducts twice during a cycle. Output frequency is double to that of input frequency. 31. (4)  = D d and (2 ) / ' 4 ( / 2)   = =  D d