NARAYANA IIT/PMT ACADEMY

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1 NARAYANA IIT ACADEMY ANSWER KEY XI STUD (LJ) IITJEE MAINS MODEL Exam Date : physics Chemistry Mathematics 1. (D) 31. (D) 61. (B). (B) 3. (A) 6. (C) 3. (B) 33. (A) 63. (A) 4. (C) 34. (C) 64. (A) 5. (C) 35. (D) 65. (D) 6. (A) 36. (A) 66. (A) 7. (A) 37. (C) 67. (B) 8. (B) 38. (A) 68. (A) 9. (D) 39. (B) 69. (A) 10. (A) 40. (A) 70. (C) 11. (A) 41. (C) 71. (D) 1. (B) 4. (B) 7. (D) 13. (B) 43. (B) 73. (D) 14. (A) 44. (D) 74. (D) 15. (D) 45. (B) 75. (B) 16. (A) 46. (A) 76. (D) 17. (B) 47. (D) 77. (C) 18. (A) 48. (B) 78. (B) 19. (C) 49. (C) 79. (D) 0. (B) 50. (C) 80. (D) 1. (B) 51. (A) 81. (C). (A) 5. (C) 8. (D) 3. (C) 53. (B) 83. (C) 4. (B) 54. (D) 84. (B) 5. (A) 55. (C) 85. (A) 6. (D) 56. (A) 86. (C) 7. (B) 57. (A) 87. (A) 8. (A) 58. (D) 88. (B) 9. (B) 59. (A) 89. (D) 30. (D) 60. (A) 90. (C) Page No.1

2 HINTS & SOLUTION PHYSICS. (B) Centre of mass does not change. 3. (B) Distance from shore = 10 l d. L m ml 4. (C) X cm m 5. (C) m be the mass of each part 3m cc m cc1 Or ad X a Area of removed plate A Area of original plate d distance between centres. A a 6. (A) V cm parallel a cm. 7. (A) L L 3 4 kx L xdm dx kx L L dm dx; xcm L L 3 L kx L 4 dm dx L (B) Fractional length hanging, l l 0.5 l 10cm L (D) 10. (A) Let be the mass per unit length. The coordinates of 0 and 10 are (10,0) and (0,5) respectively from A. Distance of C.M. from A, r x y cm cm cm mass of removed part d shift here d 0cm. mass of remained part r a Shift of centre of mass x R r Where r = radius of removed disc R = radius of original disc a = distance between the centres Here shift must be R for exact approach to the solution. 11. (A) Acceleration of system, mg sin 60 mg sin 30 a m 3 1 ma1 ma a g, Now a cm 4 m Here, a 1 and a are 3 1 at right angles, 4 Page No.

3 a 3 1 Hence, acm g 4 1. (B) The masses of load, ladder and man are M, M-m and m respectively. Their velocities are v(upward), -v and v r respectively 13. (B) r mivi vcm m i v v v v M M m m m M M r h Range of C.M.= vcm g m1v 1 mv But vcm m1 m 14. (A) Maximum height attained by C.M u cm = initial height of C.M + g L 15. (D) d sin 4. m1 r1 m r 16. (A) r cm m1 m m v 17. (B) vb m M L 18. (A) rcm xcm ycm or rcm cos 4 m1 r1 m r 19. (C) r cm m m 0. (B) 1. (B). (A) shift R 1 3 r d r 3 3 r x y cm cm cm x, y 0,0 ; 0 cm cm m x m x m x m m m 1 3 m1 y1 m y m3 y3 0 m1 m m3 3. (C) m1x 1 mx m3x3 m4x4 xcm m m m m (B) md shift M m 5. (A) v cm is parallel to a cm. v r Page No.3

4 6. (D) CM coincides with point of intersection of diagonals. 7. (B) r a Distance of C.M. from centre of big disc x r R r - radius of small disc R - radius of big disc a - distance between the centres of discs 8. (A) r cm x cmi ycm j 9. (B) md shift M m 30. (D) Mass of ladder acts at.5m Mass of man acts at m m1 y1 m y y M m 31. D CHEMISTRY 3. A 33. A Page No.4

5 34. C 35. D 36. A 37. C Page No.5

6 38. A 39. B 40. A 41. C Page No.6

7 4. B 43. B 44. D 45. B D Page No.7

8 48. B : D 55. Page No.8

9 D 59. u H nrt J / mol 37.9 KJ / mole 60. A Heat of formation of water is 1 H O H g g O l Page No.9

10 61. B r1 r 6. C x y 30x 6y MATHEMATICS and g, f x, y Required tangent is 4x y 15x 4 3 y x y 15x 60 3y x y x y A Since 3x + y =0 is a tangent to the circle with centre at (, 1). Radius = length of the perpendicular from (, 1) on 3x + y = So, the equation of the circle is 5 5 x y 1 x y 4x y 0 The combined equation of the tangents drawn from the origin to the circle is x y 4x y x y 3x 8xy 3y 0 3x y 0 or x 3y A Let the required point be P(x 1, y 1 ). The equation of the chord of contact of P with respect to the given circle xx yy 3 x x y y 3 0 is The equation of the chord with mid-point (1, 1) is x y 3 x 1 y x y 3 Equating the ratios of the coefficients of x, y and the constant terms and solving for x, y we get x1 1, y D Required line x 9 y x 35y A QR is the chord of contact of the tangents to the circle x y a 0 (i) So, its equation will be 1 1 xx yy a 0.(ii) The circumcircle of PQR is a circle passing through the intersection of the circle (1) and the line () and the point P(x 1, y 1 ) Circle through the intersection of (1) and () is 1 1 x y a xx yy a 0.(iii) It will pass through 1 x 1, y 1, if x y a x y a 0 Hence equation of circle is 1 1 x y a xx yy a 0 or x y xx yy Page No.10

11 67. B Conceptual 68. A Now, AB = 5 = 7 = r r 1. Therefore, the two circles C 1 and C touch each other internally and hence they have only one common tangent. 69. A S = 0 and S' 0 cut orthogonally. This implies that , 70. C Two circles are x y 4x 6y 3 0 and x y x y 1 0 Centres : C1, 3 C 1, 1 ; radii:r1 4, r 1 We have C1C 5 r1 r, therefore there are 3 common tangents to the given circles. 71. D and x y 3 1 x y 3 1, 7. (D) Let radius of circle is R then R h a k b where centre of circle is, orthogonal condition 73. (D) Chord of contact This touches, d r r when given the st. line 1 We get 74. (D) The distance between the pair of parallel lines = diameter 75. (B) Common chord is x = 0,0,,0 c a c a lie on same side of x = 0 a1a has no real roots c 0 x y c 76. D (CM) = (CP) (PM) which is maximum when PM is minimum i.e. coincides with M (i.e. middle point of chord) Hence equation is x 3y + 4 = C The two circles are x + y ax + c = 0 and x + y by + c = 0 Centres : C 1 (a, 0), C (0, b) radii : r1 a c, r b c Since, the two circles touch each other externally, therefore, C 1 C = r 1 + r a b a c b c a + b = a c + b c + a c b c h k to Page No.11

12 c 4 = a b c (a + b ) + c 4 a b = c (a + b ) a b c 78. B Since x 3cos, y 3sin x y 9 centre = (, 0) and radius = D Given circle is, x + y = 4 equation of tangent is y x 1 1 x y 0 Hence, point of contact is, x 0 y x y sign gives, y mx a 1 m, and sign gives 80. D Equation of radical axis is (x + y x 4y 4) (x + y + x + 4y + 4) = 0 4x 8y 8 0 x + y + = 0 Now, centres of circle are (1, ) and 1, and radii of circle are 3 and 1. Since, C 1 C > r 1 + r So therefore, given circles neither touch no intersect each other. 81. C centre h, k be the image of 1, w.r to x y 3 h, k 5, x y and radius = D Let x y gx fy c 1 Put A t, on the circle t 4 3 t gt ct ft 1 0 pqrs C 84. B 0 be the circle x xy y x y x kxy y x 4y 1 Let two distinct chords from, Equation of chord A as a midpoint is s1 s11 It is passing P (a,b) represents a circle P a b are bisected on y axis at A (0, ) k 4 3 Page No.1

13 b 4a 85. A Center C(1,) r = equation is x y C Common chord x = k x y 1 x k 0 1 The family of circles x y and cuts orthogonally k 1 x y x 1 0 It is a real circle r A We have 3 1 i.e., i. e. sin i.e., 1 sin see fig. a a 1< a < 88. B x y 6 x 3y 1 Clearly 3 1 1, and, lie inside the smaller part D The given line subtends 10 at the centre. radius of the circle is C. ( d r)( d r) (13 5 5)( Page No.13

NARAYANA IIT ACADEMY

CPT-_XI (LJ) 7..7 NARAYANA IIT ACADEMY ANSWER KEY XI STUD (LJ) IITJEE MAINS MODEL Exam Date :7--7 physics Chemistry Mathematics. (C) 3. (B) 6. (C). (C) 3. (D) 6. (A) 3. (B) 33. (A) 63. (B) 4. (D) 34. (A)