CPT XII-NEW (REG) Batches (Date: ) ANSWERS PHYSICS CHEMISTRY MATHEMATICS 1. (D) 31. (D) 61. (D) 2. (B) 32. (D) 62. (B) 3. (B) 33.

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1 CPT-0/XII-S (NEW)/ NARAYANA I I T / P M T A C A D E M Y CPT - 0 XII-NEW (REG) Batches (Date: ) ANSWERS PHYSICS CHEMISTRY MATHEMATICS 1. (D) 31. (D) 61. (D). (B) 3. (D) 6. (B) 3. (B) 33. (C) 63. (A) 4. (C) 34. (D) 64. (A) 5. (B) 35. (C) 65. (A) 6. (D) 36. (D) 66. (B) 7. (B) 37. (C) 67. (D) 8. (B) 38. (A) 68. (A) 9. (C) 39. (A) 69. (A) 10. (C) 40. (B) 70. (D) 11. (B) 41. (A) 71. (C) 1. (C) 4. (C) 7. (C) 13. (B) 43. (A) 73. (A) 14. (B) 44. (D) 74. (C) 15. (A) 45. (A) 75. (A) 16. (D) 46. (B) 76. (D) 17. (C) 47. (B) 77. (D) 18. (B) 48. (B) 78. (D) 19. (C) 49. (C) 79. (D) 0. (A) 50. (A) 80. (A) 1. (C) 51. (C) 81. (A). (C) 5. (A) 8. (D) 3. (B) 53. (A) 83. (B) 4. (A) 54. (C) 84. (B) 5. (B) 55. (C) 85. (B) 6. (B) 56. (C) 86. (C) 7. (C) 57. (A) 87. (C) 8. (B) 58. (B) 88. (B) 9. (C) 59. (A) 89. (D) 30. (D) 60. (C) 90. (B) 1

2 Hints & Solution CPT-0/XII-S (NEW)/ PART A: PHYSICS 1. (D) 1 F r ; so when r is halved the force becomes four times.. (B) 1 q 1 q F1 F 1. and F πε 0 a 4πε0 a F13 3. (B) ( ) 4. (C) The comb gets charged and induces opposite charges in paper. The field due to the charges in the comb, polarises the atoms in the paper. Finally, it attracts the paper because opposite charges are induced 5. (B) Conceptual 6. (D) The resultant force at the centre will be 0 due to symmetry 7. (B) Charge at rest produces only electric field 8. (B) By using Q ne Q C 9. (C) 10. (C) In the LHS of q, the fields due to (-q) and (q) are opposite. 11. (B) 1. (C) qe KE mv m( at) m t m 13. (B) In the following fig, in equilibrium F e T sin 30 0, r 1m 1m m T cos µ C 9 Q T r r T T sin F e + 10µ C mg

3 6 ( ) (B) Charge is same mass is different. 15. (A) F k T T 1.8N. ( Q ) x x CPT-0/XII-S (NEW)/ d df 0 dx 16. (D) The net electrostatic force due to Cs ions at the centre (Cl ion) is zero, due to symmetry. 17. (C) Force on each charge is zero, But if any of the charge is displaced, the net force starts acting on all of them. 18. (B) Conceptual. 19. (C) Conceptual. 0. (A) Sol.Let q q be charges on smaller and bigger spheres. Then q ' 4q But q + q ' Q q ' q ' Q 4 + q q ' ( ) 4 R 4 R 1. (C) Conceptual.. (C) Conceptual. 3. (B) C tan θ ' 1 A 5 q ' Q 4 4 q ' Q 5 C B θ A π θ ' 45º 4 π 3 θ π π 4 4 θ 4. (A) Conceptual. 5. (B) Sol. Forces on charge q placed at the centre due to 18 charges will cancel out in pairs. Thus, force on q will be due to only one charge, ie., 3

4 1 qq F 4 0 a 6. (B) Conceptual. 7. (C) Conceptual. 8. (B) Conceptual. 9. (C) Conceptual. 30. (D) CPT-0/XII-S (NEW)/ C is the mid point of AB. The field is directed to the right (positive) in the region between A and C, and to the left (negative) between C and B. The magnitude of E will increase sharply for x 0 and x l. PART B: CHEMISTRY 31. (D) 3. (D) 33. (C) 34. (D) 35. (C) 5 10 urea mole glu cos e mole Solvent 85g TF (D) (C) Sol. M v 38. (A) Sol. M N ( relation between molarity and normality) 39. (A) Sol. The acidity of NH3 is 1 (normality) 40. Wt 1000 (B) Sol. molality GMW Wt of solvent in grams 41. (A) Sol. P K H. C i 1+ n 1 4. (C) Sol. ( ) 43. (A) Tb kb m 0.5 k b (D) Sol. Same osmotic pressure solutions are called isotonic solutions. 45. (A) Sol. no of mill moles M vol in ml (B) Sol. M M (B) 48. (B) 49. (C) Sol. No of milli equivalents normality x volume in ml 50. (A) 4

5 v Sol. STP v Given no. of molecule nh 51. (C) Sol. X H n + n H N 5. (A) Sol. According to Henry s law, PN K, H X N i.e., P x N N K 0 P P 53. (A) Sol. mole fraction of solute 0 P 54. (C) Sol. Elevation in boling point depends on number of solute particle number of particales are more in 0.1 BaCl highest boiling point 55. (C) Sol. Tb i kb m 56. (C) Sol. Ratio of depression in f.p are in the ratio of no. of particles 57. (A) 58. (B) Sol. CST 59. (A) Conceptual 60. (C) Sol. icrt 4.9 i (D) i n Percentage of ionization49% PART C: MATHEMATICS ( ( x )) log 1+ log CPT-0/XII-S (NEW)/ log 3 ( x 7) log 3 ( x 7) 3 1 x x x 4 6. (B) log(x + 1) + log(x 1) log 11 + log 3 log(x +1) + log (x 1) log 11 + log 3 log 11 + log 9 log (x 1) log 99 x 1 99 x ±10 x 10 is only the valid value. 63. (A) x 5x + 4 x 5x + 4 or x 5x + 4 x 5x 0 or x 5x (A) x + 3 x 5 6 case I: x < 3/ H 5

6 x 3 + x Case II: x < x x 5 6 4x 6 x 5 Case III: x x + 3 x x φ 65. (A) x x ( )( ) x φ CPT-0/XII-S (NEW)/ (B) x 3 1 Hence no solution 67. (D) x x + 4 Squaring both the sides, (x ) (x + 4) x 4x + 4 x + 8x (x + 1) 0 x [ 1,) 68. (A) x + x 1 5 Case I: x < 0, x + 1 x 5 x 4 x Case II: 0 x < 1, x + 1 x 5 (not possible) Case III: x 1, x + x 1 5 x 6 x 3 there are two real solutions 69. (A) x 1 x 1 > 0 3 x + 3x + x x x + 1 x + 1 > 0 ( )( ) 1 1 x (, 1 ), 0, ; Least positive integer (D) Using properly x + y < x + y xy < 0 x,0 0,5 x < x 5x + 5x 5x(x 5x) < 0 ( ) ( ) 71. (C) Using properly x + y x + y xy 0 (x 3 + 4x + 9)(x 3) 0 x, 7. (C) logab abcd logac abcd logad abcd logbc abcd logbd abcd logcd abcd loge ab loge ac loge ad loge bc loge bd loge cd log abcd log abcd log abcd log abcd log abcd log abcd e e e e e e 6

7 loge ab + loge ac + loge ad + loge bc + loge bd + loge cd log abcd 3 log(abcd) e 3log eabcd loge abcd loge abcd (A) x 8 x 8 log1.5 < 0 0 < < 1 x x 74. (C) x x + 65 (5 x) x x x 10x 40 8x x (A) Since A + B 90 o tan A tan B (D) f(x) is defined when x 4 0 and x 1 0 x 4 x 4 and (x 4 or x 1) (x > 4 or x 1) dom (f) (-, 1] (4, ) 77. (D) f(x) is defined only when log(x x ) 0 (x x ) 1 (1 + x x) 0 (1 x) 0 (1 x) 0 x 1 dom (f) {1} 78. (D) For domain of f ( x) e ( x ) x ( 4,6) log 3 x 3x + x 3x + 0 and x 3 > 0 x - 1, - and x > 3. f x 4 x (D) ( ) 1 sin x sin x (i) 4 x 0 x 4 - x (ii) sinx - sinx > 0 When sinx < 0, sinx - sinx - sinx > 0 sinx < 0 If sinx sinx then 0 > 0 not possible Domain is [-, 0). 1 x f x sin log 80. (A) To find domain of function ( ) CPT-0/XII-S (NEW)/ x For f(x) to be defined we should have 1 log 1 1 x 1 1 x 4 x 1 1 x x, 1 1, 81. (A) Factual 8. (D) y x y log ( x ) we must have x > 0 x < x < 1 or x (-, 1) 83. (B) f x 4x x or [ ] [ ] ( ) 7

8 84. (B) 1 1 { } { 4( 1) ( 1) } {( a ) ( a ) } a ( a) f a + f a a + a + a a ( ) ( ) ( ) ( ) x + x x + x + 1 x + x (B) 86. (C) Put cos x t, - 1 t 1 1 ( ( )) f f t y t + 3t 1, t + 3t 1 0 t 1 dy 4t dt t + 3t 1 4 f and f f ( 1) 0 Hence (C) is the correct answer. 1+ cos x 87. (C) f ( x) cos x Which will be injective in, (B) Case I : [cos -1 x] 1 cos -1 x π cos π x cos - 1 x cos x [-1, cos] 89. (D) cos (sinx) is defined for all real x log x {x} 0 Case I, 0 < x < 1 {x} 1; x (0, 1) Case II, x > 1 and {x} > 1 but 0 < {x} > 1 not possible 90. (B) CPT-0/XII-S (NEW)/ f(x) is defined if x for these values of x f(x) lies in the interval 0, ************************************************************************** 8