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1 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) 64. (A) 5. (B) 35. (C) 65. (C) 6. (C) 36. (D) 66. (C) 7. (D) 37. (A) 67. (A) 8. (B) 38. (A) 68. (A) 9. (B) 39. (B) 69. (C). (A) 4. (A) 7. (C). (A) 4. (C) 7. (A). (B) 4. (B) 7. (A) 3. (C) 43. (C) 73. (B) 4. (C) 44. (C) 74. (D) 5. (A) 45. (D) 75. (A) 6. (C) 46. (C) 76. (C) 7. (B) 47. (B) 77. (A) 8. (A) 48. (A) 78. (D) 9. (B) 49. (B) 79. (B). (C) 5. (D) 8. (D). (A) 5. (A) 8. (C). (B) 5. (B) 8. (C) 3. (C) 53. (C) 83. (C) 4. (D) 54. (A) 84. (C) 5. (A) 55. (B) 85. (D) 6. (C) 56. (B) 86. (D) 7. (C) 57. (A) 87. (A) 8. (B) 58. (C) 88. (B) 9. (B) 59. (A) 89. (C) 3. (C) 6. (D) 9. (D) Page No.
2 CPT-_XI (LJ) 7..7 HINTS & SOLUTION PHYSICS 3. (B) l r l; r. 4. (D) m v vb. m M 5. (B) Coordinates of kg, kg, 3kg are (,), (,), (, 3 ) respectively m x m y X ; Y ; r X Y i i i i cm cm cm cm cm mi mi 6. (C) md shift M m 7. (D) md shift M m 8. (B) r a x R r 9. (B) md pd xcm m m p q md pd x ; shift x x m m p q. (A). (A) cm cm cm L r p m r v 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.. (B) F m m g T ext acm M m m mm g But T m m 3. (C) Theoretical. Page No.
3 CPT-_XI (LJ) (C) u cm hmax H g 5. (A) Acceleration of system, mg sin 6 mg sin 3 a m 3 a g, Now 4 a cm ma ma m 3 Here, a and a are at right angles, 4 a 3 Hence, acm g 4 6. (C) U U Kx K and F Kx x 7. (B) U Kx... U 5 x... Comparing eq.() and () to find K. 8. (A) mg h Gain in K.E. = Loss of P.E. = h 9. (B) mv m T mg mg gl cos r l. (C) mv T H where vh 3gr r. (A) Kx KE KE and mv mv. (B) By work energy theorem W mgh mu mv 3. (C) 4. (D) Height of fall, KEmax w Fdx. 4 8 H g g After on second, body is at a height of 3 5g H 8 g. g Page No.3
4 CPT-_XI (LJ) 7..7 H % of P.E. retained = % H 5. (A) K. E. mvbottom vtop 6. (C) T mg 3 cos 7. (C) If is angular amplitude, k W x K f 4 x. dx 8. (B) g x ; from conservation of mechanical energy; K 9. (B) E Potential energy of particle at x k 3. (C) v ac k rt or v krt r dv Tangential acceleration is, at kr dt Hence P F vcos ma v cos. 3. B We know t G.33RT log p t is zero x mgx Kx mv ; CHEMISTRY K and R 8.34JK mol, here m 5 KE E or mv E G JK mol 98K log Therefore kg. =63 J mol or 63 kj mol 3. D vaporisation We can represent the process of evaporation as H O H O ; n l g v E H pv vaph nrt vap vap (Assuming stema behaving as an ideal gas) 4.66kJ mol E vap 8.34 KJ k mol 373K 37.56kJ mol A W PV.5 4.5lit atm Work done, W 4.5J 454.5J U q W J 37. A W P V V H U P V 4. A H H H Diamond Graphite Page No.4
5 CPT-_XI (LJ) C N O NO 4. B G.33RT log K C 43. C G GP GR 44. C PV nrt 45. D M U Zx x W H U nrt Q=mst 46. C G H T S 47. B Ice Water is not possible at -5 degree centigrade. So G 48. A U q w 5. D m t t3 W m t3 t 5. B H H H I g I S 53. C H H H Red P Yellow P 54. A A O AO H H P H R H Eq Eq 55. B CH 4 3O CO H O 56. B H S H S H H H 57. A q CP T 58. C G nfe cell 59. A C H H C H R 4 6 H H H P p R q S T Page No.5
6 CPT-_XI (LJ) D NO4 NO Total moles =.85 G.33RT log K p MATHEMATICS 6. C The lines are a( x y ) b(x y 5), a, b x y, x y 5 meet at (-,) 6. A y m m xy m mx 63. B a x x h x x y y b y y 64. A x y y C,3 x y 3 A, B, Lines are x y, x y The angular bisectors x, y Area 65. C n h ab am hlm bl 66. C y x x y A Let the required circle is 68. A 3 x y 6x 4y k and it is passing through (-,4) O 4 Page No.6
7 CPT-_XI (LJ) Tan h ab a b ( a b) ab Squaring a b 69. C) Let the equation of circle, which touches both the axes is (x a) + (y a) = a x + y ax ay + a = a x + y ax ay + a = Since, it is passes through (3, 6), (3) (6) a(6) a a + a = a 8a + 45 = a 5a 3a + 45 = a(a 5) 3(a 5) = (a 3)(a 5) = a = 3, 5 On putting a = 3 in eq. (i), we get x + y (3)y + 3 = x + y 6x 6y + 9 =. 7. (C) The intersection point of the lines x 3y = 5 and 3x 4y = 5 is (, ), which is equal to the centre of the circle. Since, area of circle = r 54 r 7 r 7 r 7 Equation of circle is (x ) + (y + ) = 7 x x + + y + y + = 49 x + y x + y = (A). 4. tan = tan A x y 6x 4y centre ( 3, 7) x y 4x y centre (, 5) Equation of the circle is x 3 x y 7 y 5 x y x y B y = x + 6.(i) x y y 9.(ii) Page No.7
8 CPT-_XI (LJ) 7..7 Putting y = x + 6 in (ii) x 4x 3 x x 4 x x 3 x x 6 Hence equation of circle is x x 3 y 4 y x 4x 3 y 4y x y 4x 4y D Eliminating we get x 3 y 4 x 3 y B 3 x y (x 4 y)( lx my) Where lx my is chord coefficient of x coefficient of l m on comparing Point (,-) 76. C 77. A 5 Apply L S 3 Apply c a m y 78. D Equation of the diameter of given circle having one end at ( 3, ) is given by 5x 7y + 9 =. Solving this equation together with the equation of the circle, We get (, ) as other end point. 79. B Let the equation of the required tangent be x + y = a. Then length of the perpendicular from centre = radius a a 8. C Any line to x + y = 5 and passing through the origin is x + y = Any line to y y 3 and passing through the origin is x + y = combined equation of the pair of lines is (x + y) (x + y) = i.e. x 4xy 4y (c) holds. 8. C Clearly 3 3 a h g h b f 3 g f c 3 3 K Page No.8
9 CPT-_XI (LJ) 7..7 K K 9 K 9 3K C Let m,m be the slopes of the lines h K K m m b 3 3 a mm b 3 Since m m mm K K (C)holds A Tan 3 x y Line is r 3 x Kxy 3y r 3r Any point on line is, where r is distance from (,). Substituting in the curve r r... r... 8 This is a cubic equation in r 4 OA. OB. OC C Since the given lines are coincident h h 6 h 4 4. "h ab" (C) holds 85. (D), 8 such that R.A. of & is AB 86. (D) Page No.9
10 CPT-_XI (LJ) 7..7 Area = sq. units 87. (A) The lines PS and RQ meet at X. The point X lies on the circumference and we know that any diameter subtends a rt. Angle at any point on the circumference S Q X R 9 9 r P PQ tan and tan 9 r PQ. RS tan cot r or d r PQ. RS 88. (B) Let x tangent to the circle, then 3 5 (radius) ,8 Pair of lines ( x )( x 8) 89. (C) x x 6 6 Equation of the circle is x +y =..(i) Let the point be (, h). As it lies on (i) +h = h = 3. There are two points (, 3) and (, -3) on the circle whose abscissa is. Equation of tangent at (, 3) is x+3y= Equation of tangent at (, -3) is x-3y= 9. D For internal point p(, 8) p < p 96 and x intercept = p RS r therefore p p and y intercept 4 p p 4 Page No.
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