NARAYANACODE A GROUP OF EDUCATIONAL INSTITUTIONS CBSE-PMT (PRELIMS) : 2008 ANSWER KEY Q. CODE No. A B C D Q. CODE No. A B C 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. (4) (3) (4) (1) (3) (4) (2) (2) (2) (2) (3) (3) (2) (4) (3) (1) (3) (4) (1) (3) (4) (1) (1) (2) (4) (3) (2) (2) (3) (1) (2) (1) (4) (3) (2) (1) (2) (4) (4) (4) (1) (2) (4) (4) (1) (2) (1) (1) (4) (3) (2) (1) (4) (3) (2) (4) (1) (1) (1) (3) (3) (3) (3) (3) (2) (4) (1) (3) (2) (4) (3) (3) (2) (4) (1) (1) (4) (3) (2) (1) (2) (4) (2) (4) (1) (1) (4) (4) (3) (1) (3) (2) (1) (2) (2) (3) (1) (1) (2) (3) (3) (3) (4) (3) (4) (3) (4) (3) (2) (1) (3) (1) (3) (4) (3) (4) (4) (1) (2) (3) (3) (2) (3) (1) (2) (2) (4) (3) (3) (4) (4) (4) (3) (2) (2) (1) (4) (4) (3) (4) (4) (3) (2) (4) (1) (2) (2) (4) (2) (4) (4) (2) (2) (4) (4) (2) (4) (4) (3) (2) (4) (1) (1) (1) (1) (3) (3) (4) (1) (3) (2) (4) (2) (2) (4) (4) (1) (1) (1) (3) (3) (3) (2) (2) (2) (3) (3) (3) (4) (1) (2) (3) (3) (1) (4) (2) (4) (1) (2) (2) (2) (4) (3) (2) (4) (2) (1) (2) (4) (4) (2) (3) (4) (4) (3) (1) (4) (3) (3) (3) (2) (1) (4) (4) (2) (4) (1) (4) (1) (4) (1) (3) (1) (3) (3) (3) (4) (2) (2) (2) (1) (3) (2) (4) (2) (1) (2) (1) (3) (2) (2) (2) (2) (4) (2) (1) (1) (4) (4) (4) (3) (4) (1) (2) (3) (3) (3) (1) (2) (1) (3) (2) (4) (3) (2) (1) (3) (2) (2) (4) (1) (2) (1) (4) (3) (2) (2) (1) (2) (2) (1) (4) (2) (2) (2) (4) (1) (4) (1) (1) (3) (4) (2) (4) (1) (4) (2) (1) (1) (2) (3) (4) (1) (4) (4) (2) (1) (3) (3) (3) (2) (4) (4) (2) (1) (3) (3) (1) (1) (1) (2) (4) (1) (1) (3) (3) (2) (4) (3) (4) (4) (2) (3) (1) (4) (4) (1) (4) (3) (3) D (2) (1) (3) (3) (4) (2) (4) (3) (4) (3) (2) (2) (2) (2) (3) (4) (2) (1) (4) (1) (3) (2) (1) (4) (1) (2) (3) (2) (1) (4) (1) (4) (1) (2) (4) (4) (4) (1) (3) (2) (3) (4) (2) (1) (4) (4) (1) (2) (2) (1) Q. CODE No. A B C D Q. No. CODE A B C D 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. 118. 119. 120. 121. 122. 123. 124. 125. 126. 127. 128. 129. 130. 131. 132. 133. 134. 135. 136. 137. 138. 139. 140. 141. 142. 143. 144. 145. 146. 147. 148. 149. 150. 151. 152. 153. 154. 155. 156. 157. 158. 159. 160. 161. 162. 163. 164. 165. 166. 167. 168. 169. 170. 171. 172. 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. 185. 186. 187. 188. 189. 190. 191. 192. 193. 194. 195. 196. 197. 198. 199. 200. (2) (4) (1) (2) (1) (2) (1) (3) (4) (1) (3) (2) (2) (1) (3) (1) (4) (1) (2) (2) (4) (4) (1) (3) (2) (2) (4) (4) (3) (1) (2) (3) (2) (2) (3) (4) (1) (4) (3) (2) (1) (3) (2) (1) (1) (4) (2) (3) (2) (4) (3) (2) (3) (3) (3) (2) (1) (1) (1) (3) (2) (1) (2) (1) (4) (1) (2) (2) (4) (3) (4) (4) (4) (4) (4) (2) (4) (1) (1) (2) (1) (1) (3) (1) (4) (2) (4) (2) (3) (4) (2) (2) (1) (4) (4) (2) (4) (3) (2) (3) (3) (2) (3) (1) (4) (1) (2) (3) (4) (2) (3) (2) (2) (3) (3) (3) (3) (4) (3) (1) (1) (2) (1) (2) (2) (3) (2) (4) (2) (3) (4) (3) (3) (2) (3) (3) (3) (4) (4) (2) (3) (2) (4) (3) (1) (3) (1) (4) (2) (1) (3) (4) (1) (1) (1) (4) (4) (3) (3) (2) (1) (2) (2) (1) (1) (2) (1) (1) (4) (2) (3) (3) (4) (1) (1) (2) (1) (1) (3) (3) (4) (3) (2) (4) (4) (1) (2) (4) (3) (3) (3) (1) (2) (2) (2) (3) (1) (1) (4) (1) (1) (1) (1) (1) (3) (3) (4) (2) (3) (2) (4) (3) (1) (1) (4) (4) (4) (2) (1) (2) (4) (1) (3) (1) (4) (4) (2) (3) (1) (3) (4) (4) (3) (1) (1) (2) (2) (3) (4) (3) (2) (1) (2) (2) (2) (2) (3) (3) (4) (3) (4) (3) (4) (4) (1) (2) (4) (3) (4) (2) (1) (2) (1) (3) (4) (4) (1) (4) (2) (3) (4) (2) (4) (1) (1) (1) (4) (4) (3) (1) (1) (1) (3) (3) (1) (3) (1) (1) (3) (3) (4) (1) (4) (4) (1) (1) (4) (3) (1) (4) (4) (2) (1) (4) (3) (1) (2) (4) (3) (4) (2) (1) (2) (3) (2) (2) (3) (4) (3) (3) (3) (2) (1) (3) (1) (2) (1) (1) (1) (4) (1) (2) (3) (4) (3) (1) (3) (4) (4) (2) (1) (3) (2) (4) (1) (3) (1) (4) (3) (4) (1) (1) (4) (3) (1) (1) (4) (3) (3) (2) (4) (4) (3) (3) (1) (3) (1) (2) (2) (3) (3) (1) (1) (1) (4) (2) (4) (3) (2) (4) (4) (1) (3) (1) (4) (3) (2) (4) (2) (4) (3) (1) (2) (2) (4) (3) (2) (1) (1) (3) Janak Puri: A-1/171 A, Janakpuri, N.D. 58,. Ph.: 41576123/24/25: South Extn.: F-38, South Extension-I, New Delhi-49, Ph.: 46052731/32; Rohini Centre : D-11/141, Sector - 8, Rohini, New Delhi - 110085, Phone : 47016398/99: Rohtak Centre : SCF-1, HUDA Complex, Rohtak, Haryana: Tel.: 9255321118/1119: Head Office.: Vittalwadi, Narayanaguda, Hyderabad-500 029, Ph.: 23220400, 23226800 Narayana Institute of Correspondence Courses : 63 FNS House, K.S. Market, Sarvpriya Vihar, New Delhi-110016; Ph.:011-46080617/18 www.narayanadelhi.com, www.narayanaicc.com, www.narayanaonline.com, e-mail :
[email protected] We have taken the best efforts to provide the right answers, but in case of any controversy, Narayana Institute cannot be held liable. [1] CBSE PMT - 2008 Prelims Paper (Code - A) with Hints & Solutions CODE NARAYANA A GROUP OF EDUCATIONAL INSTITUTIONS CBSE-PMT : 2008 (PRELIMS) PAPER WITH HINTS & SOLUTIONS PHYSICS Which two of the following five physical parameters have the same dimensions ? (a) Energy density (b) Refractive index (c) Dielectric constant (d) Young’s modulus (e) magnetic field (1) (a) and (e) (2) (b) and (d) (3) (c) and (e) (4) (a) and (d) [W] Sol.(4) [Energy density] = 2. [F].[L] [L3 ][L] = (2) 10 (3) 1.8 (4) 12 [F] [L3 ] ∴ = [P] Sol.(1) Using v2 – u2 = 2as 202 – 102 = 2 × a × 135 ∆V 3∆R = V R 19 m 3 S 3rd Using v – u = at 20 – 10 = 10 ×t 9 t = 9 sec A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point S D C A (2) 6 m B t (3) 4 m Sol.(4) 300 10 =a = 270 9 5. V ∝ R3 The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms–2, in the third second is (1) A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms–1 to 20 ms–1 while passing through a distance 135 m in t second. The value of t is (1) 9 = [Y] If the error in the measurement of radius of a sphere is 2 %, then the error in the determination of volume of the sphere will be (1) 2 % (2) 4 % (3) 6 % (4) 8 % Sol.(3) 3. L3 = 4. Distance 1. (4) =u+ 10 m 3 1 a(2n − 1) 2 1 4 10 = 0 + × × (6 − 1) = m 2 3 3 Time (1) A (2) B (3) C (4) D Sol.(3) V maximum velocity point means point at which dx = slope is maximum dt ∴ ‘C’ NARAYANA INSTITUTE : A-1/171 A, Janak Puri, N.D.58; Ph.: 011-41576123/24/25; F-38, South Extn.-I, N.D.-49; D-11/141, Sec.-8, Rohini, N.D.-85; NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House, K.S. Market, Sarvpriya Vihar, New Delhi-110016; Ph.:011-46080617/18 Head Office.:Vittalwadi, Narayanaguda, Hyderabad-29 www.narayanadelhi.com, www.narayanaicc.com, www.narayanaonline.com, e-mail :
[email protected] (4) ∆p = m(v − u) (4) mv 2 ! ˆ − mv(cos 45iˆ + sin 45j) ˆ | | ∆p |=| mv[cos 45iˆ − sin 45j] = 2 mv sin 45° = 7. The ratio of the radii of gyration of a circular disc to that of a circular ring. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is (1) (3) 2ML2 24 ML2 12 (2) ML2 24 (4) ML2 6 NARAYANA INSTITUTE : A-1/171 A.2008 Prelims Paper (Code . When the particle (3) 10.com.com . The force necessary to keep the belt moving with a constant velocity of v m /s will be (1) Zero (2) Mv newton (3) 2 Mv newton (4) Mv / 2 newton Sol. A particle of mass m is projected with velocity v making (1) 7. e-mail : info@narayanaicc. 2 mv F= d(mv) dm = v dt dt as V = constant ∴ F = Mv Three forces acting on a body are shown in the figure.-85.S. Market.(3) 2N (1) 3N (2) 0.(2) 8. (1) 1 1 × 4 × V2 + × 0. A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1. (2) Solving (1) and (2) v = 100 m/s 11. Narayanaguda.2 kW (4) 12.narayanaonline. To have the resultant force only along the y-direction.:Vittalwadi.narayanadelhi.A) with Hints & Solutions 6. N. Hyderabad-29 www.3 kW land on the level ground the magnitude of the change 90 in its momentum will be Sol..5 N 3 N 4 (4) Sol. D-11/141. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.0 kW (2) 8.com.com.58. force] ! ˆ now.D.(2) Frequired = [–x component of res.-8. www.D. The losses due to frictional forces are 10 % of energy.1 KW 10.2 v = 0 . Ph.(2) 4V + 0.1 kW an angle of 45° with the horizontal. www. N. N. K. Rohini.... Sec.2 × v2 = 1050 2 2 . The initial velocity of the shell is (1) 120 ms–1 (2) 100 ms–1 (3) 80 ms–1 (4) 40 ms–1 Sol..5 N (3) 1. Ph. Sand is being dropped on a conveyor belt at the rate of M kg/s... Fres = (4 – 2)(cos30jˆ – sin 30i) ˆ +1(cos60iˆ + sin 60j) ∴ 9. New Delhi-110016.:011-46080617/18 Head Office. around their respective axes is (1) 2 : 3 (2) 3 : (4) 2 :1 2 30° (3) 1 : 60° x Sol. Thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. How much power is generated by the turbine ? (g = 10 m/s2) ∴ 2 Iring K disc = = K ring Idisc K ring K disc = mr 2 1 2 mr 2 1 2 12. F-38. each of same mass and radius.[2] CBSE PMT .05 kJ of energy. the magnitude of the minimum additional force needed is y 1N 4N = Mgh 90 × t 100 = 15 × 10 × 60 90 × 1 100 = 8. Janak Puri.: 011-41576123/24/25. ! 1 1 Freqd = − − ˆi + ˆi = ˆi 2 2 Water falls from a height of 60 m at the rate of 15 kg/ s to operate a turbine.-49.-I. Sarvpriya Vihar.narayanaicc. South Extn..D.(2) Pgenerated = Pinput × 100 (1) Zero (2) 2 mv (3) mv / 2 ! ! ! Sol. 14 m/s 14.-I.(4) = 0. At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x.: 011-41576123/24/25. www.S. e-mail :
[email protected] m and wavelength λ = 0. change in internal energy and the work done in a closed cycle process. F-38. Ph. If Q. D-11/141.narayanaonline. The speed of the car at the top of the hill is between (1) 13 m/s and 14 m/s (2) 14 m/s and 15 m/s (3) 15 m/s and 16 m/s (4) 16 m/s and 17 m/s MV2 = Mg R PV 1 = RT m M0 Sol. Sarvpriya Vihar.com . Sec. South Extn. www.58.narayanaicc.(2) 10 x 110 a max = ω2 A ω2 = 12 (∵ A is same) a max 2 ω2 a max1 (100) 2 = (1000) 2 = 1 102 18. New Delhi-110016.D. On a new scale of temperature (which is linear) and called the W scale. where x and y are in meters and t in seconds.-8.[3] Sol.(4) For a cyclic process ∆U = 0 or E = 0 15. Janak Puri.:011-46080617/18 Head Office.-85.L 2 2 M (1) 283 x 383 (2) x (3) 383 x 283 (4) 2 axis V = Rg N=0 MV2/R Mg V = 20 × 10 ∴ Sol.L 2 2 Sol. Narayanaguda. Rohini.(3) CBSE PMT . is a wave travelling along the (1) Negative x direction with amplitude 0.-49.narayanadelhi.(3) ρ 1 ∝ P T P RT = ρ M0 13.2 m (2) Negative x direction with frequency 1 Hz (3) Positive x direction with frequency π Hz and wavelength λ = 0. corresponding to a temperature of 39°C on the Celsius scale ? (1) 139°W (2) 78°W (3) 117°W (4) 200°W t − LFP t − LFP = UFP − LFP true UFP − LFP faulty 39°C − 0°C t − 39°W = 100°C − 0° 239°W − 39° W t = 117°W 16.2008 Prelims Paper (Code . E and W denote respectively the heat added. the freezing and boiling points of water are 39°W and 239°W respectively.:Vittalwadi. N.com. Two Simple Harmonic Motions of angular frequency 100 and 1000 rad s–1 have the same displacement amplitude. as 2 πt = ωt f = 1Hz ∴ (4) is correct NARAYANA INSTITUTE : A-1/171 A.(3) = 14. K.A) with Hints & Solutions I = 2× 1 ML 3 2 2 ML 2 = 12 L M .(1) PV = nRT M. then (1) Q = 0 (2) W = 0 (3) Q = W = 0 (4) E = 0 Sol. Market. N.2 m (4) Positive x direction with frequency 1 Hz and wavelength λ = 0. N.25 sin (10πx − 2πt) .com.D.25sin(10πx − 2πt) as ωt and kx have opposite sign wave travels along positive x. Hyderabad-29 www.2 m Sol.com. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Ph. The ratio of their maximum acceleration is (1) 1 : 104 (2) 1 : 10 2 (3) 1 : 10 (4) 1 : 103 Sol. The wave described by y = 0.D. What will be the temperature on the new scale. At 110°C this ratio is ρ1 / P1 T2 = ρ2 / P2 T1 x 383K = (ρ2 / P2 ) 283K ρ2 283 = x P2 383 17. A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is 20 m. Sec. Ph. Two thin lenses of focal lengths f1 and f2 are in contact and coaxial.: 011-41576123/24/25. Janak Puri. Narayanaguda.(1) x = a sin ωt + 2 6 I1 + I2 (3) dx π V= = aω cos ωt + 2 dt 6 I1 − I2 (4) aω π = aω cos ωt + 2 6 Sol. Hyderabad-29 www. N.(3) π (2) 3 2π 3 v = 300 m/s ∴λ = v 300 = = 15m v 20 ∆x ∆φ 5 ∆φ 2π = ⇒ = ⇒ ∆φ = 3 λ 2π 15 2π 21.05 sec and the velocity of the wave is 300 m/sec.5 × 1011 m putting I = 9. N.05 sec ∴ (4) π ( 1 1 1 = + Feq f1 f 2 f +f Peq = 1 2 f1f 2 24. Two periodic waves of intensities I1 and I2 pass of the time period the velocity of the point will be through a region at the same time in the same direction.(2) 1 v = O u O = 1.com . What is the phase difference between the oscillation of two points ? ωt + π (1) 6 (3) Sol.05 ) I max + I min = ∆X = 15m – 10m = 5m T = 0. K.:Vittalwadi.D.(4) v = µ ε a standard formula 0 0 x = a sin(wt + π6) . Ph.58.2 × 10–4 m NARAYANA INSTITUTE : A-1/171 A. After the elapse of what fraction 22.-85.D.4 × 10–4 m (2) 9.D.com. South Extn. www. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.39 × 109 m v = 0.(1) 1 π = cos ωt + 2 6 π π π = ωt = 6 3 6 2π π T t= ∴ t= 12 T 6 20. The power of the combination is (1) f1 + f 2 f1f 2 (2) f1 f2 (3) f2 f1 (4) f1 + f 2 2 Sol.-I.narayanaonline. What is the diameter of the sun’s image on the paper ? (1) 12.narayanadelhi. www.com. New Delhi-110016.:011-46080617/18 Head Office.5 × 1011 m.A) with Hints & Solutions 19. The velocity of electromagnetic radiation in a medium of permittivity ε0 and permeability µ0 is given by (1) µ0 ε0 (2) (3) µ0ε0 (4) ε0 µ0 1 µ0 ε0 ( ) ( I min = ( I max = ) 2 I2 ) I1 + I 2 I1 − ( 2 I1 + I 2 ) +( 2 I1 − I 2 ) 2 = 2(I1 + I2) 23.narayanaicc.39 × 109 m and its mean distance from the earth is 1. D-11/141. Two points are located at a distance of 10 m and 15 m from the source of oscillation. e-mail :
[email protected]. Market. Sarvpriya Vihar. A point performs simple harmonic oscillation of period 1 T and the equation of motion is given by Sol.2008 Prelims Paper (Code .S. The period of oscillation is 0. The diameter of the sun is 1.(1) 1 ∴v = = 20Hz 0.2 × 10–4 m (3) 6. equal to half of its maximum velocity The sum of the maximum and minimum intensities is (1) T/12 (2) T/8 (1) 2(I1 + I2) (3) T/6 (4) T/3 (2) I1 + I2 π Sol.[4] CBSE PMT .5 × 10–4 m (4) 6. F-38. N.com. Rohini. A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm.-8.1m u = 1.5 × 10–5 m Sol. N.5 ohm (4) 2.D.2008 Prelims Paper (Code .com. A Thin conducting ring of radius R is given a charge +Q. Sarvpriya Vihar. 1.0 ohm Sol.D. Narayanaguda. The electric field cross-section A such that the uniform electric field at the point is between the plates is E.narayanadelhi. N. Market. www.21 times. NARAYANA INSTITUTE : A-1/171 A. Janak Puri.(2) (4) ε0 E 2 Ad Sol.[5] CBSE PMT . N. The energy required to charge a parallel plate 27.1 times 1.-8.-I. A cell can be balanced against 110 cm and 100 cm of potentiometer wire.S.2 times.0 ohm (3) 0. K.com.1 R r = 1 ohm 29.:Vittalwadi.narayanaicc. D-11/141. Ph. New Delhi-110016. The electric potential at a point in free space due to a condenser of plate separation d and plate area of charge Q coulomb is Q × 1011 volts.:011-46080617/18 Head Office.com . Hyderabad-29 www.com. is (1) 12πε0 Q × 1022 volt / m 1 2 (1) ε0 E Ad (2) 4 πε0 Q × 1022 volt / m 2 (2) (3) 12 πε0 Q × 10 20 volt / m 1 ε0 E 2 / A. same Both remain the same Specific resistance remain same while resistance change.d 2 (4) 4 πε0 Q × 1020 volt / m (3) ε0 E / Ad 2 Sol. e-mail : info@narayanaicc.(2) ER ∝ l2 r+R r + R l1 = R l2 D (1) 3E along OK (2) 3E along KO (3) E along OK (4) E along KO Sol.-49. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E.1 times 1.D.-85. www.: 011-41576123/24/25. A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively (1) (2) (3) (4) Sol. Rohini. Its internal resistance is (1) Zero (2) 1. respectively with and without being short circuited through a resistance of 10 ohm. F-38.(4) Energy of charged capacitor = 1 CV2 2 Energy given by cell = CV2 V= 1 Q × 4πε0 r E= 1 Q × 2 4πε0 r 4πε 0 V 2 Q 2 × 1022 = 4πε0 × Q Q Aε0 2 × ( Ed ) = d E= = Aε 0 E 2 d E = 4πε0 Q × 1022 volt / m 26.58. South Extn.narayanaonline.1 times. Sec. Ph. 1. The electric field at the centre due to the charge on the part ACDB of the ring is A K C B O 28.(3) ! E total = 0 ! ! E AKB + E ACDB = 0 ! ! E ACDB = −E AKB ! E ACDB = E along OK E ∝ l1 r + 1 = 1.A) with Hints & Solutions 25.(3) 1. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. (1) voltage across 2 ohm.: 011-41576123/24/25. SR and RQ are F 1 .(2) φ = BScos θ = 1 × π(0. N.2 volt 1Ω + 0. The potential B .(2) V I t = ms∆t ⇒ t= 32.D.(4) Magnetic field can never increase the energy of 3Ω a charge particle so its kinetic energy will remain T. charge Q and kinetic energy T 30. the force on the segment QP is (1) 3.narayanaonline. N.06 ωb (4) 0. Narayanaguda. F-38. P M 34.2 meter is placed in a 1 ωb in such π m 2 ! a way that its axis makes an angle of 60° With B .4 min (4) 12. In the circuit shown.2)2 cos 60° = 0. How much time will it take to boil 1 kg water from temperature 20°C ? The temperature of boiling water is 100°C (1) 4. www.5Ω respectively and are in the plane of the paper and 0. Market. Janak Puri.6 min Sol.5Ω along the directions shown.narayanadelhi.:Vittalwadi. A closed loop PQRS carrying a current is placed in a uniform magnetic field.3 min VI 220 × 4 F1 A current of 3 amp flows through the 2 ohm resistor shown in the circuit.01 ωb (2) 0. F 2 and F 3 N 1Ω 0.58. A circular disc of radius 0.-85.A) with Hints & Solutions 33.02 ωb (3) 0. Ph.-49. After 3 seconds the kinetic energy of the particle difference between the points M and N is will be (1) 4 T (2) 3 T 4Ω (3) 2 T (4) T Sol. If the magnetic forces on segments PS. K. D-11/141.-I.08 ωb Sol. Ph. Hyderabad-29 www. N.5 volt Q (3) 1. Sarvpriya Vihar. www.(1) VPM = 1 A × 4 ohm = 4 volt So VMN 31.D. Rohini.(3) Force on QP will be equal and opposite to sum of forces on other 3 sides ms∆t 1× 4200 × 80 = = 6. South Extn. the current through the 4 ohm enters a transverse uniform magnetic field of induction resistor is 1 amp when the points P and M are ! connected to a d.com.D.5 volt Sol.0 volt (4) 0. e-mail :
[email protected] min (2) 6.25Ω F3 F1 R S F3 (1) F3 – F1 + F2 (3) (2) F3 – F1 – F2 (F3 − F1 ) 2 + F22 (4) (F3 − F1 ) 2 − F22 Sol.The magnetic flux linked with the disc is uniform magnetic field of induction (1) 5 watt (2) 4 watt (3) 2 watt (4) 1 watt Sol.3 min (3) 8. voltage source. P 1Ω = × 4 volt = 3.narayanaicc. The power dissipated in the 5 ohm resistor is F3 F2 2Ω So FQP = (F3 − F1 ) 2 + F22 4Ω 1Ω 5Ω 35. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.:011-46080617/18 Head Office.[6] CBSE PMT . A particle of mass m. Sec. New Delhi-110016.com.02ωb π NARAYANA INSTITUTE : A-1/171 A.2 volt (2) 1.com . v = 3 × 2 = 6 volt 6v So current across 5 ohm = = 1A 1Ω + 5Ω Power across 5 ohm = P = I2R = (1)2 × 5 ohm = 5 watt (1) 0.c. An electric kettle takes 4A current at 220 V.com.S.2008 Prelims Paper (Code . In an a.-8.0 henry Sol.2008 Prelims Paper (Code .m. K.f.-49.(4) Pav = E 0 I0 cos φ 2 10 –3 2 × 20 = × 10 –3 Amp Current for 20 division = 40. In the phenomenon of electric discharge through 30 3 gases at low pressure. susceptisicitry of (1) X-ray region ferromagnetic substance varies with absolute (2) Ultraviolet region 1 (3) Visible region temperature as X n ∝ (T − T ) .com . (e) and the current (i) at 36.D. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Sarvpriya Vihar. Rohini.A) with Hints & Solutions 39.2 ev = 11.(4) Net flux through solenoid φ = 500 × 4 × 10 –3 = 2ωb φ = LI ⇒ 2ωb = L × 2Amp ⇒ L = 1 henry Sol. The self-inductance of the solenoid is (1) 4. Ph. circuit the e.narayanaonline. is (3) 5550 Ω (4) 6050 Ω E 0 I0 Sol. A galvanometer of resistance 50 Ω is connected to a any instant are given respectively by battery of 3V along with a resistance of 2950 Ω in series. N. New Delhi-110016.S.(2) meve = mv v= me ve m NARAYANA INSTITUTE : A-1/171 A. A full scale deflection of 30 divisions is obtained e = E 0 sin ωt in the galvanometer.7 × 10–21 ms– 1 (2) 2.-I.2 eV.58.(1) KEmax = hv − φ or hv = KE max + φ = 5 ev + 6. but for C (4) Infrared region 1 .com. Narayanaguda.5 henry (3) 2.0 henry (2) 2.-85.(2) According to curie-weisis law. e-mail :
[email protected]. Sec. F-38. N.1 × 10–31 kg) Sol.(1) Current through galvanometer (2) (1) E0I0 2 3V –3 I= = 10 Amp E 0 I0 E 0 I0 sin φ cos φ 50Ω + 2950Ω (3) (4) 2 2 Current for 30 division = 10–3 Amp 1 Sol. A long solenoid has 500 turns.D. So after Tc i.D. N. Ph. T curie temperature every ferromagnetic substance behaves like paramagnetic. Curie temperature is the temperature above which (2) Excitation of electrons in the atoms (1) Ferromagnetic material becomes diamagnetic (3) Collision between the atoms of the gas material (4) Collisions between the charged particles emitted (2) Ferromagnetic material becomes paramagnetic from the cathode and the atoms of the gas material Sol. Market. the resulting magnetic flux linked with each turn of the solenoid is 4 × 10–3 ωb . The velocity of the particle is (1) 2. the colored glow in the tube a pears as a result of 2 3 ⇒ ×10−3 = ⇒ R = 4450Ω (1) Collision between different electrons of the atoms 3 50Ω + R of the gas 37.c. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 106 ms–1.0 henry (4) 1. The wavelength of the incident (4) Paramagnetic material becomes ferromagnetic radiation for which the stopping potential is 5 V lies in material the Sol.7 × 10–18 ms– 1 (3) 9 × 10–2 ms– 1 (4) 3 × 10–31 ms– 1 (Mass of electron = 9.narayanaicc.narayanadelhi.:Vittalwadi.[7] CBSE PMT .c.(4) Theoretical (3) Paramagnetic material becomes diamagnetic 41.com. the resistance in series should be The average power in the circuit over one cycle of (2) 5050 Ω (1) 4450 Ω a. South Extn.:011-46080617/18 Head Office. Hyderabad-29 www.: 011-41576123/24/25. paramagnetic substance X n ∝ 38. The work function of a surface of a photosensitive material material is 6. In order to reduced this deflection i = I0 sin(ωt − φ) to 20 divisions. www. D-11/141.2 ev (lies in x rays) 42. Janak Puri. When a current of 2 ampere is passed through it. www.e. N. Ph.D. K. The circuit NOR NAND NOT is equivalent to (1) OR gate (2) AND gate (3) NAND gate (4) NOR gate Sol.8 eV Av (4) 10. The minimum frequency of the radiation that can be absorbed by the material is nearly (1) 20 × 1014 Hz (2) 10 × 1014 Hz (3) 5 × 1014 Hz (4) 1 × 1014 Hz Sol.(2) Theoretical Av = 10 = 100 0. Z) = ZMp + (A – Z) Mn – BE Sol.(4) Theoretical 50. New Delhi-110016. proton and neutron respectively in units of u (1u = 931.D.-8. Two radioactive materials X1 and X2 have decay constants 5λ and λ repetitively.[8] CBSE PMT .2 eV Sol. Market.D. Rohini. www.(3) 2 ev = hv ∴ ν= 2ev h NARAYANA INSTITUTE : A-1/171 A. The ground state energy of hydrogen atom is –13. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.5 MeV/C2) and BE represents its bonding energy in MeV. Mp and Mn denote the masses of the nucleus A Z X .6 feedback is 10.2008 Prelims Paper (Code .1 48.com.com. N.(4) E1 = – 13.10 Å (4) 2. If M (A.:011-46080617/18 Head Office. D-11/141.27Å (2) 1. Hyderabad-29 www.4 eV (3) 10 (4) 1. Two nuclei have their mass numbers in the ratio of 1 : 3. then the atomic radius in fcc Crystal is (1) 1. Then ratio of their nuclear densities would be (1) 1 : 1 (2) 1 : 3 (3) 3 : 1 (4) (3)1/3 : 1 Sol.com . Sec.narayanaicc. The voltage gain of an amplifier with 9 % negative 43. Z) = ZMp + (A – Z) Mn – BE / C2 (3) M(A.narayanadelhi. 46.:Vittalwadi.A) with Hints & Solutions 47.-85.6 ev 4 ∆E = E 2 – E1 44. South Extn. Sarvpriya Vihar. If initially they have the same number of nuclei. The voltage gain without feedback eV.(4) 1 4λ N1 1 e –5λt = = = e –4λt N 2 e e – λt ∴ 1 = 4λ t 1 or t = 4λ 45.(1) Voltage gain = 1 + β A v Sol. When its electron is in the first excited state. Janak Puri.6 eV E2 = Av 10 = 1 + 9 /100A v –13.com. www.(1) Density is independent of mass number.0 eV. Z) = ZMp + (A – Z) Mn + BE / C2 (2) M(A.S. If the lattice parameter for a crystalline structure is 3.-49. F-38. its will be excision energy is (1) 0 (1) 100 (2) 90 (2) 3. N.(1) 4 r = 2a 2a 4 ∴ r= 49. Ph.: 011-41576123/24/25.narayanaonline.25 (3) 6. A p-n photodiode is made of a material with a band gap of 2. then (1) M(A. Z).92Å Sol.81Å (3) 2. Z) = ZMp + (A – Z) Mn + BE (4) M(A.-I. Narayanaguda. then the ratio of the number of nuclei of X1 to that X2 will be 1/e after a time (1) e λ (2) λ (3) 1 λ 2 (4) Sol.6 Å.58. e-mail : info@narayanaicc. E.-8..com.D. If ‘a’ stands for the edge length of the cubic systems simple cubic. it indicates that: (1) Number of the molecules of gas increases (2) Kinetic energy of molecules decreases (3) Pressure of the gas increases (4) Kinetic energy of molecules remains the same 1 8 1 (3) 64 1 1 a: 3a : a 2 2 B. = 1 1 ##$ # HI(g) %## # H 2 (g) + I 2 (g) is 8. each ion makes definite contribution to equivalent conductance of an electrolyte. K.-85. Ph.A) with Hints & Solutions CHEMISTRY 51.0 2 2 The equilibrium constant of the reaction ##$ # H 2 (g) + I 2 (g) %## # 2HI(g) will be: (2) 1 16 (4) 16 1 1 ##$ # Sol : (3) HI %## # H 2 + I2 2 2 K=8 then. If a gas expands at constant temperature. each ion makes definite contribution to equivalent conductance of an electrolyte.1 × 109 cm / s 9 NARAYANA INSTITUTE : A-1/171 A. Ph. Sarvpriya Vihar. N. Kohlrausch’s law states that at: (1) Infinite dilution.com.com . r = (2) 1 3 1 a: a: a 2 4 2 2 (4) 1 3 2 a: a: a 2 2 2 a 2 a 2 2 3a 4 rSCC : rBCC : rFCC 1 h ∆v = 2m π 52. www.narayanaonline. e-mail : info@narayanaicc. body centred cubic and face centred cubic. South Extn.narayanadelhi. each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte Sol : (4) According to Kohlransch law. The measurement of the electron position is associated with an uncertainty in momentum. New Delhi-110016. Rohini. N. K. each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte 56.. at infinite dilution. remains constant 53. Sec.-I.narayanaicc.58. then the ratio of radii of the spheres in these systems will be respectively.C.. which is equal to 1 × 10–18g cms–1. D-11/141. a 3a 1 : : a 2 4 2 2 55.E.D.D. Market. Hyderabad-29 www.C. www.[9] CBSE PMT .C. If uncertainty in position and momentum are equal. N. The value of equilibrium constant of the reaction Sol : (4) Avg.-49.C. whatever be the nature of the other ion of the electrolyte (2) finite dilution. The uncertainty in electron velocity is. r = 1 h 2 π (1) (3) F. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. const.: 011-41576123/24/25.S. temperature K. Narayanaguda.C. then uncertainty in velocity is : (1) h π (2) 1 h 2m π (3) h 2π (4) 1 h m π Sol : (2) ∆x ⋅ ∆p = h 4π 3 RT 2 At.:Vittalwadi.C.:011-46080617/18 Head Office.com. H 2 + I 2 %## ##$ # # 2HI 2 Sol : (2) S. Janak Puri.2008 Prelims Paper (Code . (mass of an electron is 9 × 10–28g) (2) 1 × 109 cms–1 (1) 1 × 1011 cms–1 (3) 1 × 106 cms–1 (4) 1 × 105 cms–1 Sol : (2) ∆P = m∆V 1 × 10–18 = 9 ×10–28 × ∆V ∆V = 1010 = 1. F-38. r = h ∆p2 = 4π ∆p = (1) 1a: 3a : 2a 1 1 K= = 8 64 54. whatever be the nature of the other ion of the electrolyte (3) Infinite dilution each icon makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte (4) Infinite dilution. 58.S. 242 and 431 kJ mol–1 respectively. Rohini.10 0.05 0.11 × 10–4 M (3) 3. q gas (C3H8) measured under the same conditions ? III. III and IV (4) I.30 0.1 × 10–4 Based on these data.com.-8.10 0.30 0.40 0.05 0. D-11/141. Market. H-TS (1) 10 L (2) 7 L (1) II and III (2) I and IV (3) 6 L (4) 5 L (3) II. (1) & (2) Br2 5.-I.48 (3) Molecular solids are generally volatile (4) The number of carbon atoms in an unit cell of Diamond is 4 Sol : (2) The fraction of total volume occupied in primitive cell is 0.7 × 10−5 Br2 b b 0.11] 3 1.7 × 10−5 + = 1.30 0. Sarvpriya Vihar. Ph. Sec.narayanaonline. Which of the following are not state functions ? and 1 atm. w IV. Hyderabad-29 www. Janak Puri. Which of the following statement is not correct ? (1) The number of Bravais lattices in which a crystal can be categorized is 14 (2) The fraction of the total volume occupied by the atoms in a primitive cell is 0. N.05 0.narayanaicc. (2) & (3) c H +2 5. Equal volumes of three acid solutions of pH 3. K. CH 3COCH 3 (aq) + Br2 (aq) → CH 3COCH 2 Br(aq) + H + (aq) + Br − (aq) These kinetic data were obtained for given reaction concentrations Initial concentrations. What will be the H+ ion concentration in the mixture ? (1) 1.4 ltr → 5 × 22.475] ∴b=a 22. c= 1 ∴ is (2) = 10−3 [1.05 0. 62. What volume of oxygen gas (O2) measured at 0°C 57.7 × 10–5 1. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. New Delhi-110016.20 Initial Rate. Bond dissociation enthalpy of H2.narayanadelhi.10 = [0. Ph.-85.A) with Hints & Solutions 59.[10] CBSE PMT .:011-46080617/18 Head Office. is needed to burn completely 1L of propane I. Enthalpy of formation of HCl is: (1) 245 kJ mol–1 (2) 93 kJ mol–1 (3) –245 kJmol–1 (4) –93 kJmol–1 Sol : (4) H 2 + Cl2 → 2HCl ∆H = [434 + 242] − [2 × 431] 2 676 − 862 = −93kJ / mol 2 61.7 × 10–3 M = 10 Sol : (3) H + = M = −3 × 1 + 10−4 × 1 + 10−5 × 1 3 10−3 1 + 10−1 + 10−2 = 3 c 1 2 = 0. e-mail :
[email protected] ltr → 5 ltr ∴ 1 ltr 60.52. Cl2 and HCl are 434.11 × 10−3 = 3.:Vittalwadi.: 011-41576123/24/25. Ms–1 5.10 = [1] From expt.com.-49. q + w II..com .7 × 10–4 M (4) 3.2 × 10−4 H3 c 0. South Extn.D. disappearance of Br2.D.7 × 10–5 5. www.2 × 10–4 3.7 × 10−5 = 5. F-38. 4 and 5 are mixed in a vessel.7 × 10−4 3 NARAYANA INSTITUTE : A-1/171 A. The bromination of acetone that occurs in acid solution 1 mol → 5 × 22.11 × 10–3 M (2) 1.10 0.2008 Prelims Paper (Code .com. II and III Sol : (1) Heat and work are not state function Sol : (4) C3 H8 + 5O 2 → 3CO 2 + 4H 2 O 58. Narayanaguda.4 ltr is represented by this equation. M [Br2] [H +] [CH3COCH3] 0.D.05 0.475 = approx. the rate equation is : (1) Rate = k[CH3COCH3][Br2][H+] (2) Rate = k[CH3COCH3][H+] (3) Rate = k[CH=COCH3][Br2] (4) Rate = k[CH3COCH3][Br2][H+]2 Sol : (2) From Expt. N. www. N.05 0. Market.com . kc depends only on temperature 3 10ie − − and pentane (g). N.2 for H2O(l).4 and –8.. Standard free energies of formation (in kJ/mol) at 298 K are –237.com.. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.. D-11/141...968V (3) 2. F-38. If degree of dissociation of X and A be equal.303 (4) 2000 K 2 P1= 36×P2 Sol : (1) K1=K2 P1 36 = P2 1 16 10 e 64...2008 Prelims Paper (Code .(1) and ##$ # PCl5 (g) %## # PCl3 (g) + Cl2 (g) ##$ # A %## # 2B..D. For the gas phase reaction. CO2(g) since.D.: 011-41576123/24/25. The values of Kp1 and Kp2 for the reactions ##$ # X %## # Y + Z.0968V (4) 1.narayanadelhi. The value of E °cell for the pentane-oxygen fuel cell is: (1) 0.D. N. Ph.com. respectively.58.) + 3OH (aq. then equilibrium concentration 4 of Fe3+ will increase by: (1) 4 times (2) 8 times (3) 16 times (4) 64 times decreased by +3 − ##$ # Sol : (4) Fe(OH)3 (s) %## # Fe (aq. Hyderabad-29 www.) K c = Fe +3 OH − 3 3 Fe+3 OH − = Fe+3 OH − i F 3 y x × y = x × 4 3 x = x′ × 1 1 64 x ′ = 64x 2000 T 2000 T = 10 ie 15 − 1000 T 1000 − =e T 1000 −2000 log e 10 + =− T T 2. then total pressure at equilibrium (1) and (2) are in the ratio : (1) 1 : 1 (2) 3 : 1 (3) 1 : 9 (4) 36 : 1 Sol : (4) Kp1 = Kp2 ##$ # X %## #Y+ Z 1 − − α α 1− α ##$ # A %## # 2B 1 − 1− α 2α n y in z P1 ∆ 1× n x y1 ng = 9× ( n B )2 P2 ∆ = 9× × ( n A ) y2 ng 4α P × 2 1− α 1+ α (1) ∆H > O and ∆S < O (2) ∆H = O and ∆S < O (3) ∆H > O and ∆S > O (4) ∆H < O and ∆S < O Sol : (3) For dissociation reaction enthalpy is positive and since no. If the concentration of OH– ions in the reaction 3+ − ##$ # Fe(OH)3 (s) %## # Fe (aq) + 3OH (aq) is 1 times.:Vittalwadi.S.-49.-I.2. The temperature at which k1 = k2 is: (1) 1000 K 2.narayanaonline.303 = T = 2000 1000 − T T 1000 K 2 . respectively. The rate constants k1 and k2 for two different reactions are 1016 i e–2000/T and 1015 i e–1000/T. www.0968V Sol : (4) C5 H12 + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l ) 1 → H 2O(l ). South Extn.3 0 3 67.. –394.0968V (2) 1. Sarvpriya Vihar.[11] CBSE PMT .(2) which of the following conditions are correct ? are in ratio of 9:1. Narayanaguda. Ph. and it is an example of thermal decomposition 66.. Sec.:011-46080617/18 Head Office.narayanaicc. Janak Puri. Rohini. (1) H 2 + O2 2 ∆G1 = −237. K.. www.A) with Hints & Solutions 65...2 kJ / mol NARAYANA INSTITUTE : A-1/171 A.com. New Delhi-110016. of gaseous moles are increasing the entropy also increases.303 (2) 1000 K (3) 2000 K 2.-85. 63... e-mail :
[email protected]. N.. S. Ti = 22.com.08767 36. of HCl = 3. Rohini. is: (1) Ti > V > Cr> Mn (2) Cr > Mn > V > Ti (3) V > Mn > Cr > Ti (4) Mn > Cr > Ti > V Sol : (2) Cr > Mn > V > Ti 72.narayanadelhi. ∆G 3 = −8.2 kJ / mol ∆G 4 = 6 × ∆G1 + 5× ∆G 2 + ( −∆G 3 ) = 6×(–23.com. Ph.[12] (2) C + O2 (g) → CO 2 (g). of PbO = 6. How many moles of lead (II) chloride will be formed from a reaction between 6. D-11/141. The dissociation equilibrium of gas AB2 can be represented as : ##$ # 2AB2 (g) %## # 2AB(g) + B2 (g) The degree of dissociation is ‘x’ and is small compared to 1. Percentage of free space in a body centred cubic unit cell is: (1) 28% (2) 30% (3) 32% (4) 34% Sol : (1) PbO + 2HCl → PbCl2 + H 2 O 223 +73 Given 6. of PbO = eq. V = 23. of PbCl2 0.2 = –3387 kJ/mol –nFE° = –3387 × 1000 E° = 3387 × 1000 96500 × 32 CBSE PMT . NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. nos.com.72)+5×(–394.D.5 Now.-85.058 = n ×z n= 0. New Delhi-110016. The sequence of ionic mobility in aqueous solution is: (1) Na+ > K+ > Rb+ > Cs+ (2) K+ > Na+ > Rb+ > Cs+ (3) Cs+ > Rb+ >K+ > Na+ (4) Rb+ >K+ > Cs+ > Na+ Sol : (3) Ionic mobility increases down the group as the hydrated ionic radii decreases down the group 70. Which of the following complexes exhibits the highest paramagnetic behaviour ? (1) [Ti(NH3)6]3+ (2) [V(gly)2(OH)2(NH3)2]+ (3) [Fe(en)(bpy)(NH3)2]2+ (4) [Co(OX)2(OH)2]– Where gly = glycine.333 (4) 0. eq.2008 Prelims Paper (Code . Sec.-I.D.029 (2) 0. Cr(24) and Mn(25). N.2) = –1423.4) (–8.058 = 0.011 = 1. e-mail : info@narayanaicc. Market. V(23).0968V 68.2 × 1 = 0.58. Narayanaguda.2g of HCl ? (1) 0.5gm 278gm 3.D.4 kJ / mol (3) 5C + 6H2 (g) → C5H12 (g). Co = 27) Sol : (4) [Ti(NH3)6]3+ Ti +3 → 4s1 lonpaired e– [V(Gly)2(OH)2(NCl3)2]+ V +5 → 3p6 o unpaired e– Fe+2 → 3d 6 but pairing takes place. Ph.058 (limiting reagent) 223 eq.A) with Hints & Solutions Sol : (3) Packing fraction in BCC is 68%.:Vittalwadi. ∆G 2 = −394. so free space = 32% 71. Sarvpriya Vihar.narayanaonline.5 × 2 = 0.com . www. www. Hyderabad-29 www. of unpaired e– is zero [Fe(en)(bpy)(NH3)2]+2 Co+5 → 3d 4 even after pairing their will be 2 unpaired e–. N. en = enthylenediamine and pby=bipyridyl moities) (at. The correct order of decreasing second ionisation enthalpy of Ti(22). South Extn. Fe = 26. Janak Puri. N. F-38.: 011-41576123/24/25. so no. K.5g of PbO and 3.2 –1972 + 8. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is: (1) (Kp/P) (2) (2Kp/P) 1/3 (3) (2Kp/P) (4) (2Kp/P)1/2 Sol : (3) 2AB2 %## ##$ # # 2AB + B2 1 1– α α 2 α 1 α P 2 i Kp = (1 − α )2 1 + α 2 α2 i 1/ 3 2K p α3 P Kp = ⇒ α= 2 P 69.2gm eq. S thus highest paramagnetism [Co(OX)2(OH)2]– NARAYANA INSTITUTE : A-1/171 A.029 2 73.-8.044 (3) 0.:011-46080617/18 Head Office.narayanaicc.-49. www. Narayanaguda. Hyderabad-29 www.-49. Volume occupied by one molecule of water (density the correct pictures of the trends indicated against it? = 1 g cm–3) is : –23 3 (1) 5. Market. In which of the following coordination entities the magnitudes of ∆0 (CFS E in octahedral field) will be maximum ? (1) [Co(C2O4)3]3– (2) [Co(H2O)6]3+ (3) [Co(NH3)6]3+ (4) [Co(CN)6]3– 81.-85.-I. Sarvpriya Vihar.A) with Hints & Solutions 78. N.narayanaonline.com.4 moles (4) 7. www. New Delhi-110016. The thermal stability of these hydrides decreases in which of the following orders? (1) LiH > NaH > KH > RbH > CsH (2) CsH > RbH > KH > NaH > LiH (3) KH > NaH > LiH > CsH > RbH (4) NaH > LiH > KH > RbH > CsH Sol : (1) Down the group the thermal stability of hydrides decreases because the difference in size of increases + Sol : (1) O O O 82.:Vittalwadi.023 × 1023 Cl2>Br2>F2 > I2 3 = 3 × 10 cm 75.D. of FeC2O4 n= 79.narayanaicc.2 moles (2) 0. The correct order of increasing bond angles in the following triatomic species is: (1) NO +2 < NO −2 < NO 2 (2) NO−2 < NO+2 < NO 2 (3) NO−2 < NO2 < NO+2 (4) NO +2 < NO 2 < NO 2− Sol : (3) NO+2 →180° NO 2 →132° NO2 →118° NARAYANA INSTITUTE : A-1/171 A. D-11/141.com. 77.:011-46080617/18 Head Office.35 V 3– (2) Germanium Sol : (4) CN– is the strongest ligand & thus CFSE is maximum n × 5 = 1 ×3 Fe 2+ → Fe3+ + e −1.2008 Prelims Paper (Code .77 V 4– (1) [Fe(CN)6] (2) [Fe(CN)6] 2+ (3) Fe (4) Fe3+ Sol : (4) Fe+3 has highest SRP & thus behaves as strongest oxidising agent.6 moles (3) 0. The angular shape of ozone molecule (O3) consists of: 3 = 0.com.-8. Sec.D.Co = 27) Eq. Ph.S. Ph.[13] CBSE PMT . (1) Boron (At. On the basis of the following E° values.6 moles 5 76. E° = –0. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Which one of the following arrangement does not give 74.58. of MnO −4 = Eq.D.023 × 10 cm (3) F2>Cl2>Br2>I2 : Electron gain enthalpy (4) 3. Janak Puri. N.com . the strongest oxidizing agent is: –1 [Fe(CN) 6 ]4− → [Fe(CN)6 ]3− + e .0 × 10–23 cm3 (4) F2>Cl2>Br2>I2 : Bond dissociation energy Sol : (4) Volume of one water molecules is Sol : (4) The correct bond dissociation energy is 18cm3 23 6. Rohini.0 × 10 cm (2) F2>Cl2>Br2>I2 : Oxidizing power –23 3 (3) 6.5 × 10 cm (1) F2>Cl2>Br2>I2 : Electronegativity –23 3 (2) 9. K.narayanadelhi. F-38. no.: 011-41576123/24/25. The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature.5 moles H+ Sol : (2) MnO − + FeC O → Fe +3 + Mn +2 4 2 4 (3) Arsenic (4) Selenium Sol : (1) Boron is added to form p type semiconductor 80. With which one of the following elements silicon should be doped so as to give p-tyupe of semiconductor ? (1) 2 sigma and 1 pi bonds (2) 1 sigma and 2 pi bonds (3) 2 sigma and 2 pi bonds (4) 1 sigma and 1 pi bonds E° = –0. South Extn. N. Number of moles of MnO −4 required to oxidize one mole of ferrous oxalate completely in acidic medium will be : (1) 0. e-mail : info@narayanaicc. N.2008 Prelims Paper (Code . sp2.: 011-41576123/24/25. F-38.O. of geometrical isomer = 2 (cis and trans) No. 91. = 3 NO B. Sec. sp (2) sp2. K.-8. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. = 0. www. Ph.5 He+2 B. In the hydrocarbon Sol : (2) No. sp. Ph. The stability of carbanions in the following : (a) RC=C (b) (c) R2C=CH (d) R3C–CH2 is the order of (1) (a) > (c) > (b) > (d) (2) (a) > (b) > (c) > (d) CH 3 − CH = CH − CH 2 − C ≡ CH (3) (b) > (c) > (d) > (a) 6 5 4 3 2 1 The state of hybridization of carbons 1. sp2.A) with Hints & Solutions 88.:Vittalwadi. Narayanaguda. better is the leaving group and more is rate of nucleophilic substitution.[14] CBSE PMT . New Delhi-110016. Rohini. Janak Puri. Four diatomic species are listed below in different sequences. N.-85. Equimolar solutions of the following were prepared in water separately. sp3.58. Sarvpriya Vihar. How many stereoisomers does this molecule have ? 83.O.-49.5 84.com. =2. sp2 (4) sp.com .narayanadelhi.com. = 1. Hyderabad-29 www.D.O. Market. e-mail :
[email protected]. Which one of the solutions will record the highest pH? (1) CaCl2 (2) SrCl2 (3) BaCl2 (4) MgCl2 Sol : (3) On going down the group basic strength increases thus pH increases 85. D-11/141. sp3 (3) sp. of active optical isomer = 2 (d and l) Total 4 89.:011-46080617/18 Head Office.5 O −2 B. N. A strong base can abstract an α -hydrogen from : (1) Alkane (2) Alkene (3) Amine (4) Ketone Sol : (4) α hydrogen of carbon of ketone is attached to electron withdrawing group (–I effect) Sol : (2) (a) > (b) > (c) > (d) 90. www.narayanaicc. South Extn. The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of : (1) Acyl chloride > Ester > Acid anhydride > Amide (2) Acyl chloride > Acid anhydride > Ester > Amide (3) Ester > Acyl chloride > Amide > Acid anhydride (4) Acid anhydride > Amide > Ester > Acyl chloride Sol : (2) Weaker is basicity. sp3 (4) (d) > (b) > (c) > (a) Sol : (3) CH3 − CH = CH − CH 2 − C ≡ CH ↓ ↓ ↓ sp 2 sp3 sp 86. Base strength of: (a) H 3CCH 2 (b) H2C=CH and (c) H–C–C is the order of : (1) (a) > (b) > (c) (2) (b) > (a) > (c) (3) (c) > (b) > (a) (4) (a) > (c) > (b) NARAYANA INSTITUTE : A-1/171 A.D.S.narayanaonline.D. Which of these presents the correct order CH3CH=CHCH2CHBrCH3 of their increasing bond order ? (1) 2 (1) He 2+ < O 2− < NO < C 22− (2) 4 (2) O 2− < NO < C 22− < He 2+ (3) 6 (3) NO < C 22− < O −2 < He +2 (4) 8 (4) C 22− < He +2 < NO < O −2 Sol : (1) C 22 − B.O. Green Chemistry means such reactions which : (1) Study the reactions in plants (2) Produce colour during reactions (3) Reduce the use and production of hazardous chemicals (4) Are related to the depletion of ozone layer Sol : (3) Green chemistry deals with the minimum use of toxic chemicals 87.com. 3 and 5 are in the following sequence: (1) sp3. S. 38. Sarvpriya Vihar. H3C − CH − CH = CH 2 + HBr → A | 94. An organic compound contains carbon.D. hydrogen and oxygen.-49.Thymine and Guanine . K. Ph.com. Rohini. In a SN2 substitution reaction of the type R − Br + Cl − → R − Cl + Br − . the complimentary bases are Sol : (1) Higher the electron density on carbanion more is the basic strength Adenine . D-11/141.A) with Hints & Solutions Sol : (2) In DNA. New Delhi-110016. Market. N.67 H 9.narayanaicc. Which one of the following is most reactive towards CH3 electrophilic attack ? A (Predominantly) is: (1) CH3 − CH − CH − CH3 | | CH3 Br (2) CH3 − CH − CH 2 − CH 2 Br | CH3 Br | CH − C − CH 2 CH3 3 (3) | CH3 (4) CH3 − CH − CH − CH3 | | Br CH3 H3C − CH − CH = CH 2 + H ⊕ | Sol : (3) CH3 Cl CH2OH (1) (2) NO2 (3) (4) Sol : (4) –OH group behaves as +R effect thereby increasing the electron density and increases the electrophilic substitution. the complimentary bases are: (1) Uracil and adenine. www. Narayanaguda.D. Ph.226 16 1 → 9. DMF which one of the following has the highest relative rate? (1) CH3CH2Br (2) CH3–CH2–CH2Br CH3 − CH − CH 2 Br | (3) CH3 CH3 | CH3 − C − CH 2 Br (4) | CH3 Sol : (1) Smaller is size of alkyl group faster is rate of SN2 reaction NARAYANA INSTITUTE : A-1/171 A.narayanaonline.71% and H. e-mail :
[email protected]%. Janak Puri.: 011-41576123/24/25.com .Cytosine 92.com. F-38.:011-46080617/18 Head Office. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. 9.62 O 51.58.[15] CBSE PMT .62 = 3. The empirical formula of the compound would be: (1) CH 4O (2) CH3O (3) CH 2O (4) CHO 38.67 =9. 95.2008 Prelims Paper (Code . Sec. guanine and cytosine (3) Adenine and thymine. Hyderabad-29 www.-I. www. N.67 1 3 → 38.narayanadelhi.-8. In DNA. cytosine and guanine (2) Adenine and thymine.com. Its elemental analysis gave C. South Extn. thymine and cytosine OH Thus empirical formula =CH3O 96.:Vittalwadi.-85.226 12 1 → 51.71 Sol : (2) C ⊕ CH3 − CH − C H − CH3 | CH3 (2°carbocation) 1. guanine and uracil (4) Adenine and guanine.71 =3. 2-hydride shift ⊕ CH3 − C− CH 2 − CH3 | CH3 3°carbocation (more stable) Br Br | CH3 − C − CH 2 − CH3 | CH3 93. N.D. D. Market. Acetophenone when reacted with a base. www.narayanadelhi. N. New Delhi-110016. Sec. In a reaction of aniline a coloured product C was obtained N NH2 B CH3 CH3 (2) In vulcanization.S. K. N.narayanaonline.com.com. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Sarvpriya Vihar.:Vittalwadi. Ph.com. F-38. D-11/141. 4-polymer of isoprene O || C6 H5 − C = CH − C − C6 H5 | CH3 HCl N 99. the formation of sulphur bridges between different chains make rubber harder and stronger (3) Natural rubber has the trans-configuration at every double bond (4) Buna-S is a copolymer of butadiene and styrene Sol : (3) Natural rubber has cis configuration at every double 100.-49. yields a stable compound which has the structure : N=N–CH2–N (2) CH3 CH–CH (1) OH OH CH3 CH3 C=CH–C (2) CH3 N=N (3) O CH3 CH–CH2C (3) CH3 O CH3 CH3 (4) (4) OH OH – NaNO HCl N O CH3 CH3 Cold C6H5–C–CH 2–C–C6H 5 H+ → CH3 CH3 N=N– OH O C6H5–C–CH2–C–C6H5 → 98.D. Janak Puri.D.2008 Prelims Paper (Code .A) with Hints & Solutions 97. Ph.: 011-41576123/24/25.-85. South Extn.58.narayanaicc. C2H5ONa.:011-46080617/18 Head Office.com .[16] CBSE PMT . www. N. Narayanaguda. Which one of the following statements is not true ? − H 2O CH3 NaNO2 CH3 2→ Sol : (1) O || C 2 H 5O − Na + Sol : (2) → C6 H5 − C − CH3 O N+2Cl NH2 CH – C N NH–NH N=N N CH3 (4) Insulin Sol : (2) Thyroxine is an amine hormone NARAYANA INSTITUTE : A-1/171 A. e-mail : info@narayanaicc. Rohini. Hyderabad-29 www.-I.-8. Which one of the following is an amine hormone ? C Cold (1) Progesterone (2) Thyroxine A (3) Oxypurin The structure of C would be : CH3 (1) CH3 (1) Natural rubber is a 1. Ph.[17] CBSE PMT . Methanococcus. Ph. e-mail : info@narayanaicc. N. Prawn. Sea urchin doesn’t have jointed appendages.: 011-41576123/24/25. www.:011-46080617/18 Head Office.narayanaonline.Ventral solid Cockroach central nervous system (4) Taenia solium doesn’t have metameric segmentation and sea cuccumber shows radial symmetry.-8. . Janak Puri.2008 Prelims Paper (Code . (3) 105. 108. thorax and abdomen and respiration by tracheae (3) Chordata – Notochord at some stage and separate anal and urinary openings to the outside (4) Echinodermata – Pentamerous radial symmetry and mostly internal fertilization (1) In all arthropods. Narayanaguda. Which one of the following groups of three animals each is correctly matched with their one characteristic morphological feature ? Sol. D-11/141. South Extn.-49. Hyderabad-29 www. . www. F-38. indicates their reptilian ancestry ? (1) Eggs with a calcareous shell (2) Scales on their hind limbs (3) Four-chambered heart (4) Two special chambers crop and gizzard in their digestive tract (2) Ascaris is characterized by: (1) Presence of true coelom and metamerism (metamerisation) (2) Absence of true coelom but presence of metamerism (3) Presence of neither true coelom nor metamerism (4) Presence of true coelom but absence of metamerism (3) Ascaris has pseudocoelom NARAYANA INSTITUTE : A-1/171 A. (3) Salvinia (4) Adiantum Sol. male and female gametophytes do not have free living independent existence ? (1) Cedrus Sol. (2) Archaebacteria have specific histone proteins which are similar to eukaryotic histone proteins involved in DNA coiling. Which one of the following in birds.S. Market. respiration is not by tracheae. . In which one of the following.D. from Cycas and Pinus and showing affinities with angiosperms: (1) Embryo development and apical meristem (2) Absence of resin duct and leaf venation (3) Presence of vessel elements and absence of archegonia (4) Perianth and two integuments Sol. Methanobacterium exemplify: and Sol. 103. 107. New Delhi-110016. Rohini. Select one of the following pairs of important features distinguishing Gnetum.Bilateral symmetry Sea anemone.-I. (3) 104. Sarvpriya Vihar. Animals Morphological feature (1) Cockroach.Jointed Sea urchin appendages (4) Scorpion. Which one of the following is hetcrosporous ? (1) Equisetum (2) Dryopteris Sol. K. Sea cucumber (3) Centipede.com.com.com . N.com. Spider.narayanaicc. Sec. (2) Pteris (3) Funaria (4) Polytrichum Sol. N. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. This is also used as an important evidence to show that early eucaryotic cells have evolved from archaebacteria. (3) 102. (3) Archaebacteria that lack any histones resembling those found in eukaryotes but whose DNA is negatively supercoiled (4) Bacteria whose DNA is relaxed or positively supercoiled but which have a cytoskeleton as well as mitochondria Sol. Locust . Which one of the following phyla is correctly matched with its two general characteristics ? (1) Mollusca – Normally oviparous and development through a trochophore or veliger larva (2) Arthropoda – Body divided into head.:Vittalwadi.narayanadelhi.-85. (1) Bacteria that contain a cytoskeleton and ribosomes (2) Archaebacteria that contain protein homologous to eukaryotic core histones 106.58.Metameric Taenia segmentation (2) Liver fluke.D. Thermococcus.D.A) with Hints & Solutions BIOLOGY 101. Carbohydrates are commonly found as starch in plant (3) Manganese (4) Magnesium storage organs.: 011-41576123/24/25.-49. Thorn of Bougainvillea and tendril of cucurbita are examples of: (b) Chemically non-reactive (1) Retrogressive evolution (c) Easily digested by animals (2) Analogous organs (d) Osmotically inactive (3) Homologous organs (e) Synthesized during photosynthesis (4) Vestigial organs The useful properties are: Sol. Keeping in view the ‘fluid mosaic model’ for the and lipids structure of cell membrane. (3) Sol.com.-I. (2) (1) Calcium (2) Copper 114. (3) (4) α -ketoglutarate 117. N.A) with Hints & Solutions 115. (4) Annelids have true coelom absent in other prokaryotes and eukaryotes 110. (2) Is membrane-bound and contains storage proteins 116. Narayanaguda. Janak Puri.2008 Prelims Paper (Code . D-11/141. nor proteins can flip-flop 112. F-38. Which of the following five properties of starch (a . (c) and (e) similar function Sol.com.e) make it useful as a storage material? Sol. The two sub-units of ribosome remain united at a (4) A ribosome with several subunits critical ion level of: Sol. which one of the following statements is (1) Ventral nerve cord true about archaea ? (2) Closed circulatory system (1) Archaea completely differ from prokaryotes (3) Segmentation (2) Archaea resemble eukarya in all respects (4) Pseudocoelom (3) Archaea have some novel features that are Sol.narayanaicc.-8. lipids can not (3) Oxaloacetate Sol.S. South Extn. Sarvpriya Vihar.narayanaonline. Hyderabad-29 www. (2) exclusively in the: 113. In the light of recent classification of living organisms 109.-85.com . Rohini.[18] CBSE PMT . e-mail : info@narayanaicc. archaea and of phylum Annelida ? eukarya). Polysome is formed by: (1) Mitochondria (1) Ribosomes attached to each other in a linear (2) Proplastids arrangement (3) Glyoxysomes (2) Several ribosomes attached to a single mRNA (4) Peroxisomes (3) Many ribosomes attached to a strand of Sol. Vacuole in a plant cell : different from both procaryotes and eucaryotes (1) Lacks membrane and contains water and (ii) The cell membrane of archae have single layered excretory substances branched lipid layer.D. Market.58. (4) (a) Easily translocated 119. which one of the (3) Is membrane-bound and contains water and following statements is correct with respect to the excretory substances movement of lipids and proteins from one lipid monolayer to the other (described as flip-flop (4) Lacks membrane and contains air movement) ? Sol. proteins can not (2) Malonate (4) While proteins can flip-flop. Which one of the following is NOT a characteristic into three domains of life (bacteria. (3) (1) Neither lipids.D. A competitive inhibitor of succinic dehydrogenase is: (2) Both lipids and proteins can flip-flop (1) Malate (3) While lipids can rarely flip-flop. N. (3) endoplasmic reticulum 118.:011-46080617/18 Head Office.:Vittalwadi.com. N. www. (2) (i) The nucleoticle sequence of rRNA of archae is 111. In germinating seeds fatty acids are degraded Sol.narayanadelhi. (3) Homologous organs have similar origin or basic (1) (a) and (e) (2) (b) and (c) fundamental structure and perform different or (3) (b) and (d) (4) (a). Ph. K. Ph. (3) NARAYANA INSTITUTE : A-1/171 A. www. Sec. Cellulose is the major component of cell walls of: (4) Archaea completely differ from both prokaryotes (1) Saccharomyces (2) Pythium and eukaryotes (3) Xanthomonas (4) Pseudomonas Sol.D. New Delhi-110016. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Thorn of Bougainvillea and (2) There is a decline in population as boys marry tendrils of Cucurbita are homologous organs. Which one of the following scientist’s name is all alleles whether dominant or recessive express correctly matched with the theory put forth by him? in the organism. Janak Puri. (2) Down’s syndrome is due to trisomy of (1) Connecting links (2) Adaptive radiation chromosome no. F-38. (1) and malpighian tubules organs 124. In the DNA molecule: 128.D. there will be diploids consangious marriages which will lead to (3) Mutagens penetrate in haploids more effectively increased expression of recessive traits.:Vittalwadi. Hyderabad-29 www. N.S. Uracil – Pyrimidines (4) de Vries – Natural selection (3) Uracil.narayanadelhi. D-11/141. Sarvpriya Vihar. (1) Mendel – Theory of Pangenesis 121. whether dominant or recessive are their life time pass this character on to their expressed in haploids progeny (2) Haploids are reproductively more stable than Sol. (4) Wings of honeybee & wings of crow are a large gene pool analogous organs. This is because: spread in the isolated population (4) Wrestlers who develop strong body muscles in (1) All mutations. Which one of the following pairs of nitrogenous bases (2) Weismann – Theory of continuity of of nucleic acids. (2) In the small tribal population. Colourblindess is x-lined and exythroblastosis foetalis is due to Rh-factor (3) Seasonal migration (4) Brood parasitism incompatibility. Market. Which one of the following is incorrect about the Sol.com.com. Thymine – Purines characters (2) Thymine.-49. Cytosine – Pyrimidines Sol. www. New Delhi-110016. What is true about the isolated small tribal populations? of Cockroach (1) There is no change in population size as they have Sol.: 011-41576123/24/25.com . Which one of the following pairs of items correctly belongs to the category of organs mentioned against (1) There are two strands which run antiparallel-one it ? in 5′ → 3′ direction and other in 3′ → 5′ (1) Wings of honey bee – Homologous (2) The total amount of purine nucleotides and and wings of crow organs pyrimidine nucleotides is not always equal (2) Thorn of Bougainvillea – Analogous (3) There are two strands which run parallel in the and tendrils of Cucurbita organs 5′ → 3′ direction (3) Nictitating membrane and – Vestigial (4) The proportion of Adenine in relation to thymine blind spot in human eye organs varies with the organism (4) Nephridia of earthworm – Excretory Sol. (2) 123.D. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/ (1) They could maintain an internal environment linkage ? (2) They were able to reproduce (1) Down syndrome – 44 autosomes + XO (3) They could separate combinations of molecules from the surroundings (2) Klinefelter’s syndrome – 44 autosomes +XXY (4) They were partially isolated from the surroundings (3) Colourblindness – Y-linked Sol. (2) (4) Guanine. is wrongly matched with the Germplasm category mentioned against it ? (3) Pasteur – Inheritance of acquired (1) Adenine. N. South Extn.58.-85. Sol. therefore decline in population. Haploids are more suitable for mutation studies than (3) Hereditary diseases like colour blindness do not the diploids. Adenine – Purines 126.2008 Prelims Paper (Code . (3) (4) Erythroblastosis foetalis – X-linked 127.[19] CBSE PMT . Sol. (1) Adenine and Guanine are purines where as characteristics of protobionts (coacervates and Thymine & Cytosine are pyrimidines microspheres) as envisaged in the abiogenic origin of life ? 122.D.:011-46080617/18 Head Office.com. www.A) with Hints & Solutions 120. The traits than is diploids which are recessive and lethal will therefore (4) Haploids are more abundant in nature than express in homozygous condition leading to more diploids deaths. Darwin’s Finches are an excellent example of: Sol. Narayanaguda. 21. K. Ph. Rohini. Ph. e-mail : info@narayanaicc. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.-I.-8.narayanaicc. (1) Haploids contain only one set of alleles therefore 125. N.narayanaonline. girls only from their own tribe Blind spot is not vestigeal in humans NARAYANA INSTITUTE : A-1/171 A. Sec. The energy-releasing process in which the substrate (1) Cremocarp (2) Caryopsis is oxidised without an external electron acceptor is (3) Cypsela (4) Berry called: Sol. which is necessary for Sol.D. www.A) with Hints & Solutions (2) Vessels and tracheid differentiation 129.:011-46080617/18 Head Office. The chemiosmotic coupling hypothesis of oxidative first shown in: phosphorylation proposes that adenosine triphosphate (1) Petunia (2) Letting (ATP) is formed because: (3) Tobacco (4) Cotton (1) There is a change in the permeability of the inner Sol. Ph.-I.: 011-41576123/24/25.com. N. N. developed from inferior ovary and has seeds with succulent testa in: (3) Leaf abscission (1) Cucumber (2) Pomegranate (4) Annual plants (3) Orange (4) Guava Sol.narayanadelhi. www. Market. (3) Cobesive and adhesive forces maintain the 142. (3) Sol.-8. Narayanaguda.S. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Janak Puri.D. (3) (1) Glycolysis (2) Fermentation 133. so CO2 generated during photorespiration can be used for Sol. (4) (3) Clostridium (4) Frankia 132. K.2008 Prelims Paper (Code . The rupture and fractionation do not usually occur (3) Photorespiration (4) Aerobic respiration in the water column in vessel/ tracheids during the Sol. (2) External electron accepter is oxygen which is not ascent of sap because of: used in fermentation (1) Transpiration pull 141. Vascular tissues in flowering plants develop from: than C3 plants because: (1) Dermatogen (2) Phellogen (1) They have more chloroplasts (3) Plerome (4) Periblem (2) The CO2 compensation point is more Sol. The C4 plants are photosynthetically more efficient 135. N. number of: 134.:Vittalwadi. (2) (1) Floral parts NARAYANA INSTITUTE : A-1/171 A.-85. Rohini. South Extn. In leaves of C4 plants malic acid synthesis during CO2 and recycled through PEP carboxylase fixation occurs in: (4) The CO2 efflux is not prevented (1) Guard cells (2) Epidermal cells Sol. Nitrogen fixation in root nodules of Alnus is brought (4) A proton gradient forms across the inner about by: membrane (1) Azorhizobium (2) Bradyrhizobium Sol. Replum is present in the ovary of flower of: (2) Lignified thick walls (1) Pea (2) Lemon (3) Cohesion and adhesion (3) Mustard (4) Sun flower (4) Weak gravitational pull Sol. Ph. Senescence as an active developmental cellular (1) Mericarps (2) Achenes process in the growth and functioning of a flowering (3) Samaras (4) Berries plant. (3) (3) CO2 generated during photorespiration is trapped 136.narayanaonline.narayanaicc. 137. Sec. The fleshy receptacle of syconus of fig encloses a continuity of water column in xylem vessles. Endosperm is consumed by developing embryo in the diphosphate (ADP) seed of: (2) High energy bonds are formed in mitochondrial (1) Maize (2) Coconut proteins (3) Castor (4) Pea (3) ADP is pumped out of the matrix into the Sol. (2) Vessels and tracheids are dead cells (components) of xylem. Importance of day length in flowering of plants was 131.com. (4) bicarpellary syncarpous inferior ovary is: 140. (3) Since in C4 cycle the second carboxylation (3) Mesophyll cells (4) Bundle sheath reaction occurs in bundle sheath cells.D.com. The fruit is chambered. Hyderabad-29 www. (3) CO2 fixation in the mesophyll cells.com . F-38. (4) Pea is an example of non-endospermic seeds intermembrane space 139. is indicated in: Sol. (2) the conduction of water 130. Sarvpriya Vihar. Dry indehiscent single-seeded fruit formed from Sol.[20] CBSE PMT .-49.58. D-11/141. (3) mitochondrial membrane toward adenosine 138. e-mail : info@narayanaicc. New Delhi-110016. head.-49.-I. Electrons from excited chlorophyll molecule of between Rod Cells and Cone Cells of our retina ? photosystem II are accepted first by: (1) Ferredoxin (2) Cytochrome-b Rod Cells Cone Cells (3) Cytochrome-f (4) Quinone (1) Distribution More Evenly Sol.S. gizzard and intestine function Vision in poor Colour vision light and detailed vision in bright light Sol. K. N. Given below is a diagrammatic cross section of a single loop of human cochlea: α Amylase Starch → Disaccharide (Maltose) Pepsin Amino Proteins → acids Sol. oesophagus. type I & type II Last 2 pair of ribs are floating ribs 150. the enzyme acting upon it and the end product ? (1) Stomach : Lipase micelles Fats → (2) Duodenum : Trypsin Triglycerides → monoglycerides (4) Small intestine: (4) Overall 10 pairs of spiracles (2 pairs on thorax and 8 pairs on abdomen) Sol. www.com . N.com. 147. (4) 149. Hyderabad-29 www. Ph.narayanadelhi. (3) In stomach protein digestion begins by pepsin. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.-85. 20 amino acids are protein amino acids. Diabetes is of 2 types.D. F-38.A) with Hints & Solutions 148. e-mail : info@narayanaicc. D-11/141. (4) concentrated distributed all 144. Narayanaguda. Sarvpriya Vihar.D.:Vittalwadi. Rohini. (2) Cervical vertebrae are 7 in humans.-8. Janak Puri. Market. Which type of white blood cells are concerned with in centre of over retina the release of histamine and the natural anti coagulant retina heparin ? (2) Visual acuity High Low (1) Monocytes (2) Neutrophils (3) Visual Iodopsin Rhodopsin (3) Basophils (4) Eosinophils pigment Sol. What is vital capacity of our lungs ? (1) Total lung capacity minus residual volume Which one of the following options correctly represents the names of three different parts ? (1) A: Perilymph. C: Perilymph.2008 Prelims Paper (Code .narayanaicc. A: Endolymph B: Tectorial membrane Sol.D.:011-46080617/18 Head Office.narayanaonline. D: Secretory cells (3) Total lung capacity minus expiratory reserve volume (3) C: Endolymph. www.com.[21] CBSE PMT . South Extn.com. New Delhi-110016. Ph. Which one of the following is the true description about an animal concerned ? (1) Cockroach – (2) Earthworm – (3) Frog (4) Rat – – The alimentary canal consists of a sequence of pharynx. N. stomach.58. A: Serum (4) Inspiratory reserve volume plus expiratory reserve volume Sol. (1) (4) D: Sensory hair cells. Micelles are formed in small intestine.16 146.3 Sol. D : Sensory hair cells. Which one of the following items gives its correct total number? Body divisible into three regions . (1) (3) Small intestine : contained (2) Floating ribs in humans . B: Tectorial membrane C: Endolymph (2) Inspiratory reserve volume plus tidal volume (2) B: Tectorial membrane. Sec. neck and trunk (1) Cervical vertebrae in humans .8 Left kidney is slightly higher in position than the right one (3) Amino acids found in proteins . Which one of the following is the correct difference 143. Which one of the following is the correct matching of the site of action on the given substrate. (1) NARAYANA INSTITUTE : A-1/171 A. (3) 145.: 011-41576123/24/25.4 (4) Types of diabetes . D. The blood calcium level is lowered by the deficiency (2) Shoot apical meristem of: (3) Position of axillary buds (1) Calcitonin (4) Size of leaf lamina at the node below each (2) Parathormone internode (3) Thyroxine (4) Both Calcitonin and Parathormone Sol.D) and their modes mobilising Ca2+ from bones to blood.-85. B-(d). Market. (3) 153.-I. In humans. (1) NARAYANA INSTITUTE : A-1/171 A. Which one of the following is resistant to enzyme potential results from the movement of: action ? (1) Na+ ions from extracellular fluid to intracellular (1) Leaf cuticle fluid (2) Cork (2) K+ ions from extracellular fluid to intracellular (3) Wood fibre fluid (4) Pollen exine (3) Na+ ions from intracellular fluid to extracellular Sol. the anterior end becomes turgid and acts as a D. New Delhi-110016. D-(b) Sol. Sarvpriya Vihar. Hyderabad-29 www. Earthworms have no skeleton but during burrowing. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Narayanaguda.com. D-(d) (3) Blood (4) Gut peristalsis (2) A-(c). blood passes from the post caval to the diastolic right atrium of heart due to: (4) A-(c).com. What will happen if the secretion of parietal cells of (2) Pushing open of the venous valves gastric glands is blocked with an inhibitor ? (3) Suction pull (1) Enterokinase will not be released from the (4) Stimulation of the sino auricular node duodenal mucosa and so trypsinogen is not Sol.-8. at the end of the first meiotic division. Select of this hormone will decrease blood Ca2+ level. the (2) Gastric juice will be deficient in chymosin male germ cells differentiate into the: (3) Gastric juice will be deficient in pepsinogen (1) Spermatozonia (4) In the absence of HCl secretion. their correct matching from the four options that follow: 155.com. The pill (a) Prevents sperms (3) Lymphocytes and macrophages reaching cervix (4) Eosinophils and lymphocytes B.: 011-41576123/24/25. (2) atrium 152. K. B-(c). Sec.2008 Prelims Paper (Code . (1) Action potentail is due to influx of Na . (4) Parietal cells or oxyntic cells secrete HCl which converts pepsinogen to pepsin Sol. (1) C. N. So far there fluid is known enzyme to degreade sporopollenin (4) K+ ions from intracellular fluid to extracellular 160. (1) Intercalary meristem 154. inactive (2) Primary spermatocytes pepsinogen is not converted into the active enzyme pepsin. D-11/141.58. C-(b).-49.D. Rohini. Condom (b) Prevents implantation Sol. Janak Puri. Vasectomy (c) Prevents ovulation 156.narayanaicc. The most active phagocytic white blood cells are: (1) Neutrophils and monocytes Method Mode of Action (2) Neutrophils and eosinophils A. www. D-(b) (1) Pressure difference between the post caval and Sol. In humans. (2) (3) A-(d). F-38. South Extn.com .S. N. the action 159. Ph. (4) Exine is composed of sporopollenin. Copper T (d) Semen contains no hydraulic skeleton. It is due to: sperms (1) Setae (2) Coelomic fluid (1) A-(b). (3) Secondary spermatocytes (4) Spermatids Sol. www. B-(a). D-(c) 157. Deficiency of action (a . N. (2) Parathormone increases plasma Ca2+ level by 151. The length of different internodes in a culm of fluid sugarcane is variable because of: + Sol. C-(d)..narayanaonline. Given below are four methods (A . B-(a). C-(a). Ph. During the propagation of a nerve impulse. e-mail : info@narayanaicc.[22] CBSE PMT . C-(a).D.d) in achieving contraception.A) with Hints & Solutions Sol. (1) converted to trypsin 158.:Vittalwadi.narayanadelhi.:011-46080617/18 Head Office. (2) 171. (1) 165. Rohini. N. Ph.: 011-41576123/24/25. 169.D. D-11/141. but not xenogamy (2) Is secreted by anterior pituitary Sol. (2) 170. It is released from posterior pituitary. (4) Oral contraceptive pills are to be taken daily 163.A) with Hints & Solutions 167. Market. F-38. d (4) It brings about opening of the pollen tube (4) a.D. there is especially abrupt increase in gonadotropic hormones Sol.com. Ph. Methane 20% (3) CO2 40%. Hyderabad-29 www. Sec. Unisexuality of flowers prevents: starting from 5th day of menstrual cycle. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.CFCs 30% (4) N2O 6%.narayanadelhi.2008 Prelims Paper (Code . The haemoglobin of a human foetus: (1) Has a higher affinity for oxygen than that of an adult (2) Has a lower affinity for oxygen than that of the adult (3) Its affinity for oxygen is the same as that of an adult (4) Has only 2 protein subunits instead of 4 Sol. Which one of the following pairs of plant structures contraception and answer as directed thereafter: has haploid number of chromosomes ? (a) Medical Termination of Pregnancy (MTP) during (1) Egg nucleus and secondary nucleus first trimester is generally safe (2) Megaspore mother cell and antipodal cells (b) Generally chances of conception are nil until (3) Egg cell and antipodal cells mother breast-feeds the infant upto two years (4) Nucellus and antipodal cells (c) Intrauterine devices like copper-T are effective Sol. Sarvpriya Vihar.:011-46080617/18 Head Office.narayanaonline.D. Narayanaguda.:Vittalwadi. e-mail : info@narayanaicc. (3) contraceptives 162. (4) NARAYANA INSTITUTE : A-1/171 A. Consider the statements given below regarding 161.-85.-8. CO2 86% Sol.-49. It is also called birth hormone because it causes uterine contractions & results in parturition. (1) Foetal haemoglobin has very high affinity for oxygen (3) Stimulates growth of mammary glands (4) Stimulates pituitary to secrete vasopressin Sol. N. New Delhi-110016.58. K. Quercus species are the dominant component in: (1) Tropical rain forests (2) Temperate deciduous forests (3) Alpine forests (4) Scrub forests Sol. In human adult females oxytocin: (2) Autogamy. South Extn. Which one of the following is the correct percentage of the two (out of the total of 4) green house gases that contribute to the total global warming ? (1) Methane 20%. www. but not geitonogamy (1) Causes strong uterine contractions during (3) Both geitonogamy and xenogamy parturition (4) Geitonogamy. Which one of the following statements is incorrect about menstruation ? (1) The beginning of the cycle of menstruation is called menarche (2) During normal menstruation about 40 ml blood is lost (3) The menstrual fluid can easily clot (4) At menopause in the female. (3) Menstrual fluid doesn’t clot 166. About 70% of total global carbon is found in: (1) Forests (2) Grasslands (3) Agroecosystems (4) Oceans Sol.com. (1) Autogamy and geitonogamy 168. b (3) It prevents entry of more than one pollen tube (2) b. N2O 18% (2) CFCs 14%. Janak Puri. (1) Oxytocin or pitocin is synthesized by hypothalamic neurons & stored in posterior pituitary. www. N.-I. What does the filiform apparatus do at the entrance (d) Contraception pills may be taken upto one week into ovule ? after coitus to prevent conception (1) It guides pollen tube from a synergid to egg Which two of the above statements are correct ? (2) It helps in the entry of pollen tube into a synergid (1) a. c Sol. c into the embryo sac (3) c.[23] CBSE PMT . Which extraembryonic membrane in humans prevents desiccation of the embryo inside the uterus ? (1) Amnion (2) Chorion (3) Allantois (4) Yolk sac Sol.com.S.narayanaicc. (2) 164.com . (2) Sol. (2) Sol. Which one of the following is not observed in of ten species (A . decomposes Which two of the above were the main causes of rapidly.d) about increased population of deers certain desert animals such as kangaroo rat. Consider the following four statements (a . N. (3) Condtions for high rate of decomposition for this ? A Climatic factors (a) Lots of urea and phosphate fertilizer were used (i) Temperature in the crops in the vicinity (ii) soil mositure (b) The area was sprayed with DDT by an aircraft (B) Chemical quality of detritus (c) The lake water turned green and stinky (i) Slow rate of decomposition if detritus is rich in (d) Phytoplankton populations in the lake declined substances such as lignin and chitin.2.0 or less (3) b and c (4) c and a Sol. Market. World Summit on Sustainable Development (2002) was held in : (1) South Africa (2) Brazil (3) Sweden (4) Argentina Sol. b (2) b. D-11/141.com. (3) NARAYANA INSTITUTE : A-1/171 A.-8. According to Central Pollution Control Board use water to regulate body temperature. Hyderabad-29 www. (4) 173. which particulate size in diameter (in Which two of the above statements for such animals micrometers) of the air pollutants is responsible for are true ? greatest harm to human health ? (1) a and b (2) c and d (1) 5. (2) (d) They excrete very concentrated urine and do not 176. A lake near a village suffered heavy mortality of fishes (4) Anaerobic environment around them within a few days. (4) (3) Poor nitrogen content 178. c (a) Removal of 80% tigers from an area resulted in (3) c. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. Rohini.D. Janak Puri. (c) The length of food chains is generally limited to (a) They have dark colour and high rate of 3 . Ph. Ph.:Vittalwadi.:011-46080617/18 Head Office. Sarvpriya Vihar. c greatly increased growth of vegetation Sol. F-38. (4) (b) Removal of most of the carnivores resulted in an 179. breathe at a slow rate trophic levels. South Extn.: 011-41576123/24/25. e-mail :
[email protected] .narayanaicc. Consider the following statements concerning food chains: (1) a.-85.2008 Prelims Paper (Code . Sec. Study the table and answer the question which (3) Accelerated species loss follows: (4) Lesser inter-specific competition Sol. Consider the following reasons Sol.narayanadelhi. initially thereby greatly reducing photosynthesis. b (2) b. N. c (c) They feed on dry seeds and do not require (3) c. www. to conserve water and have their body covered Which two of the above statements are correct ? with thick hairs (1) a.-49. N.5 (2) 2. K. New Delhi-110016. (1) Which area out of a to d shows maximum species 174.com . (CPCB).D. Narayanaguda.com.J) in four areas (a . fish mortality in the lake ? 175.com.S. The slow rate of decomposition of fallen logs in nature diversity ? is due to their: (1) a (2) b (1) Low cellulose content (3) c (4) d (2) Low moisture content Sol. d (4) a.d) consisting biodiversity hotspots ? of the number of habitats given within brackets against (1) Species richness (2) Endemism each. www.-I.A) with Hints & Solutions 177. d (4) a.5 or less (4) 1. (ii) Nirogen rich detritus.[24] CBSE PMT . The table below gives the populations (in thousands) 172. reproduction and excrete solid urine (d) The length of food chains may vary from 2 to 8 (b) They do not drink water.4 trophic levels due to energy loss. d drinking water Sol.D.5 or less (3) 1. UCA . K. N.D.:Vittalwadi. c-(iv).D. Consider the following statements about biomedical (2) Bacterial leaf blight of rice technologies: (3) Downy mildew of grapes (a) During open heart surgery blood is circulated in (4) Loose smut of wheat the heart-lung machine Sol.Stop 185. (4) NARAYANA INSTITUTE : A-1/171 A. d-(iii) 182. Which one of the following is the correct statement (3) Separation of DNA fragments according to their regarding the particular psychotropic drug specified? size (1) Barbiturates cause relaxation and temporary (4) Construction of recombinant DNA by joining with euphoria cloning vectors (2) Hashish causes after thought perceptions and Sol. New Delhi-110016.S.[25] CBSE PMT . Rohini. Gel electrophoresis is used for: (4) AUG. b-(iii).com.58.-I. Which one of the following is linked to the discovery of Bordeaux mixture as a popular fungicide ? (4) a-(ii).A) with Hints & Solutions 187. Sec. (4) Blockage in coronary artery is removed by (3) Reclamation of wastelands angioplasty. www. Market. (2) Barbiturates induce sleep and morphine is an analygesic Sol. c-(iii).narayanadelhi. South Extn.com.-8.Alanine altered by contaminants to its original condition. c-(iii). (2) (2) a-(i). D-11/141.Leucine enzymes to retum the natural envioronment (2) GUU. www.narayanaicc. A transgenic food crap which may help in solving the items (pathogen/prevention/ treatment) in Column II. (2) Bioremediation can be defined as a process that for the particular amino acid ? uses microrganisms. (2) organs like prostate glands and lungs 184. Janak Puri. d-(iv) (3) a-(ii). NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House.com . (1) (1) Black rust of wheat 188. Which one of the following is being tried in India as a biofuel substitute for fossil fuels ? (c) Computerised Axial Tomography (CAT) shows detailed internal structure as seen in a section of (1) Aegilops (2) Jatropha body (3) Azadirachta (4) Musa (d) X-ray provides clear and detailed images of Sol.D. d-(iv) Sol. (1) water 181. Hyderabad-29 www. N. (3) UAG. To which type of barriers under innate immunity. ACG . c-(i). do (3) Opium stimulates nervous system and causes the saliva in the mouth and the tears from the eyes.: 011-41576123/24/25. problem of night blindness in developing countries is: Column I Column II (1) Golden rice (2) Flavr Savr tomatoes (a) Amoebiasis (i) Treponema pallidum (3) Starlink maize (4) Bt Soybean (b) Diphtheria (ii) Use only sterilized food and Sol. hallucinations belong ? (4) Morphine leads to delusions and disturbed (1) Physical barriers (2) Cytokine barriers emotions (3) Cellular barriers (4) Physiological barriers Sol. Which one of the following pairs of codons is (4) Gene transfer in higher plants correctly matched with their function or the signal Sol. (3) (2) Cutting of DNA into fragments 190. Trichoderma harzianum has proved a useful Which two of the above statements are correct ? microorganism for: (1) a and b (2) b and d (1) Biological control of soil-borne plant pathogens (3) c and d (4) a and c (2) Bioremediation of contaminated soils Sol.2008 Prelims Paper (Code . b-(iv). Narayanaguda. (3) (b) Blockage in coronary arteries is removed by angiography 183. b-(i). e-mail : info@narayanaicc. 189. F-38.Start/Methionine (1) Isolation of DNA molecule Sol. N. Match the disease in Column I with the appropriate 180. d-(i) Sol. green plants or their (1) UUA.:011-46080617/18 Head Office.narayanaonline. UGA . Ph. Bacterial leaf blight of rice is caused by a species of: (c) Cholera (iii) DPT Vaccine (1) Envinia (2) Xanthomonas (d) Syphilis (iv) Use oval rehydration therapy (3) Pseudomonas (4) Alternaria (1) a-(ii). Ph. b-(ii).com. fungi.-49.-85. Sarvpriya Vihar. GCU . (3) hallucinations 186. Sarvpriya Vihar. What is antisense technology ? (1) RNA polymerase producing DNA (2) A cell displaying a foreign antigen used for synthesis of antigens (3) Production of somaclonal variants in tissue cultures (4) When a piece of RNA that is complementary in sequence is used to stop expression of a specific gene Sol.com . K. (2) Ovaries & pancreas are mixed tissues.:Vittalwadi.narayanaicc.com. N.-8. South Extn. e-mail : info@narayanaicc. Which one of the following proved effective for biological control of nematodal diseases in plants ? (1) Paecilomyces lilacinus (2) Pisolithus tinctorius (3) Pseudomonas cepacia (4) Gliocladium virens Sol. (1) 196.58.com.com. 199.[26] CBSE PMT .-I. Ph.: 011-41576123/24/25. Modern detergents contain enzyme preparations of: 197. Hyderabad-29 www. www.-49. Ph. testes produces sperm. Rohini.narayanaonline. 195. (1) Sol. 200. Human insulin is being commercially produced from (1) Thermophiles (2) Acidophiles a transgenic species of: (3) Alkaliphiles (4) Thermoacidophiles (1) Saccharomyces Sol. (2) 194. The linking of antibiotic resistance gene with the plasmid vector became possible with: (1) Exonucleases (2) DNA ligase (3) Endonucleases (4) DNA polymerase Sol.-85. Cornea transplant in humans is almost never rejected. These perform endocrine function as well as other functions also. N.narayanadelhi. www. (3) 193.2008 Prelims Paper (Code . Ovaries produce ova. Pancreas secrete enzymes as well as hormones. D-11/141.D. Which one of the following pairs of organs includes only the endocrine glands ? (1) Adrenal and Ovary (2) Parathyroid and Adrenal (3) Pancreas and Parathyroid (4) Thymus and Testes Sol. (1) Paecilonyces lilacinus infects nematodal larvae thereby helping in biocontrol of nematodal diseases. N. Main objective of production/use of herbicide resistant GM crops is to: (1) Reduce herbicide accumulation in food articles for health safety (2) Eliminate weeds from the field without the use of manual labour (3) Eliminate weeds from the field without the use of herbicides (4) Encourage eco-friendly herbicides Sol. Cry1 endotoxins obtained from Bacillus Which two of the above measures can control the Thuringiensis are effective against: disease ? (1) Boll worms (2) Mosquitoes (1) a and d (2) b and c (3) Flies (4) Nematodes (3) a and b (4) c and d Sol. New Delhi-110016. (4) NARAYANA INSTITUTE : A-1/171 A. Sec. Market.S. Janak Puri.d) that could be taken to successfully grow chickpea in an area where bacterial blight disease is common: (a) Spray with Bordeaux mixture (b) Control of the insect vector of the disease pathogen (c) Use of only disease-free seeds (d) Use of varieties resistant to the disease (2) Escherichia (3) Mycobacterium (4) Rhizobium Sol.D. Narayanaguda. (3) Cornea is avascular so there is no problem of immunological rejection. NARAYANA INSTITUTE OF CORRESPONDENCE COURSES : 63 FNS House. This is because: (1) It is a non-living layer (2) Its cells are least penetrable by bacteria (3) It has no blood supply (4) It is composed of enucleated cells Sol.:011-46080617/18 Head Office. (2) 198. Consider the following four measures (a . F-38.D.A) with Hints & Solutions 191. (2) 192.