NARAYANA I I T / P M T A C A D E M Y. C o m m o n P r a c t i c e T e s t 05 XII-IC SPARK Date:
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1 1. (A). (C). (B). (B) 5. (A) 6. (B) 7. (B) 8. (A) 9. (A) 1. (D) 11. (C) 1. (B) 1. (A) 1. (D) 15. (B) NARAYANA I I T / P M T A C A D E M Y C o m m o n P r a c t i c e T e s t 5 XII-IC SPARK Date: ANSWER PHYSICS CHEMISTRY MATHEMATICS 16. (B) 1. (C) 6. (C) 61. (A) 17. (C). (B) 7. (D) 6. (C) 18. (B). (A) 8. (D) 6. (B) 19. (C). (C) 9. (C) 6. (C). (B) 5. (C) 5. (A) 65. (B) 1. (A) 6. (C) 51. (A) 66. (C). (B) 7. (B) 5. (B) 67. (D). (C) 8. (B) 5. (D) 68. (D). (B) 9. (C) 5. (D) 69. (C) 5. (C). (D) 55. (C) 7. (B) 6. (C) 1. (B) 56. (C) 71. (D) 7. (B). (C) 57. (D) 7. (B) 8. (C). (A) 58. (B) 7. (A) 9. (D). (C) 59. (A) 7. (A). (D) 5. (A) 6. (A) 75. (A) (Hint & Solution) PART A : PHYSICS 76. (C) 77. (A) 78. (C) 79. (B) 8. (D) 81. (B) 8. (C) 8. (C) 8. (C) 85. (C) 86. (C) 87. (D) 88. (B) 89. (A) 9. (D) 1. (A) K I M 1 mr R m. (C) t 5 18cm 175 / 6 o t = 9 rad/s t 9 18 rad N 87. (B) After melting, ice will distribute in whole pan. So, moment of inertia I will increase. Hence, angular speed will decrease (as I = constant).. (B) (1)
2 y x.5 Slope = tan o 7 o r.5sin 5.m L mvr 5. 6kg m / s 5. (A) ma I (in case of square plate) 1 or it is independent of. 6. (B) L 1 K or K (as L = constant) I I As I has doubled, so K will become half. 7. (B) ML L 7 I M ML (A) Decrease in gravitational kinetic energy = increase in rotational kinetic energy 1 1ml mg cos I 9. (A) I 1 1 6g sin L = 6 kg-m 1. (D) L r tan 6 o 1 Area base height 1 L L L ()
3 Mass per unit area or x y dx dy Area of strip M L / M L x dx y dy Mass of strip 11. (C) l l l r I I1 I I I x dx M x dx dm L I di dm X 1 L ML ML sin 5 sin 5 Mr 1 8 ML o o L 8 1. (B) l 5 r l l ml ml I I1 I mr 1 1 m l ()
4 1. (A) x h x r R r R h r Volume of disc dx Mass of disc r dx dm 1 1 di dm r r dx r R r dx x dx h 5 h R h I di h 5 hr 1 m 1 Rh m 1 1 Rh I mr I hr So, Since m and R are same. Therefore I is same. 1. (D) l o l L sec 5 x m Lx l ml o I sin 5 lx l 1 xl 6 ()
5 15. (B) I I I I AB 1 m / a / m / a / m / a / a m / 1 ma (B) Angular impulse (about O) I P R h mr PR R h Now, COM R mr mpr R h P PHinge mcom mr R Solving this equation, we get h 17. (C) mg mg Ti and Tf 18. (B) From energy conservation principle, maximum spring potential energy = maximum decrease in gravitational potential energy (5)
6 1 kx m Mgh Mg x m sin Mg sin xm k 19. (C) m dt dm dx x l T xm dt l l 1 x T m l 1 l. (B) 1 5t d 5 dt At t, 1rad / s and xdx v r ˆj 1j ˆ m / s 5rad / s r OA 1m ˆ a r j r ˆi 1i ˆ 5j ˆm / s 1. (A) I is least and C lies between O and B as density is increasing from A to B. C I A > I B as more mass is concentrated towards B.. (B) R CC 1 where, C = combined centre of mass of ring and particle. From conservation of angular momentum about C. L L i f m.r I mr mr mr Solving this equation we get, (6)
7 R. (C) Angular momentum = Angular impulse IJl Angular impulse or I Jl J m l / ml 1 1 ml J KE= I ml J m. (B) No rotational motion 5. (C) net I l mg mgl m l /.5g l a l.5g coin mg N ma coin N mg ma coin.5g Since, N can't be negative. So, N= and coin can not remain in contact with rod. 6. (C) mg T a.. (i) m l T. l mg. I m l / or 1 mg T ml.. (ii) a l.. (iii) Solving these three equations we get, g a 8 (7)
8 7. (B) F a M FL / 6F ML /1 ML 8. (C) sin sin r b / sin a B L F a (downwards) M sin b is decreasing. Therefore, will decrease. 9. (D) Decrease in gravitational potential energy = increase in gravitational potential energy sin (as and b are constants) Mg. L T Mg M M ac r Mg Mg L g T L. (D) x = tan dx d sec. dt dt or car sec L 1 ML g car sec 1rad / s sec o L L / L r = Mg (8)
9 1. (C). (B) Kc PCl5 Cl PCl. (A). (C) 5. (C) 1 1 K. K1 K 6. (C) 1 K 1 K1 7. (B) NH N H Initial 1 eq: 1 x x x x. K PART B : CHEMISTRY NH N H C 8. (B) Decrease in concentration of products favours the forward rection. 9. (C). (D) PCl5 g PCl g Cl g Initial moles 5 Equi. Moles 5 1= Equi. Mole fraction N g ) ( total no. of moels at equilibrium = 7, including Equ. Partial pressures 7 atm 1 7 atm 1 7 atm Kp (B) n N O g NO g. (C) 1 1 [ degree of dissociation] Total number of moles at equilibrium = 1 + Initial volume = V litres Volume at equilibrium = (1+) V litres As per the given condition 1V 1 V V (9)
10 1. Hence.%. (A) By using K p, first we have to calculate x=.1 mole of NH Formed =.1.6 Total mole at equilibrium NH at equilibrium %.98. (C) At equilibrium G 5. (A) K K K 1 6. (C) H.5.1M H P log 1 H 7. (D) OH 1 OH p H p (D) OH 1 1 poh ph 1 9. (C) ph of KOH=1; ph of NaOH=1 So, poh of NaOH=1 1=; poh of NaOH=1 1 =. NKOH OH 1 ; NNaOH 1 (Because, for strong base, OH =normality=poh) As per the question, 1 x 1 1 x 1 1 H P log (A) S HO HS OH so that 51. (A) H ka Salt p p log Acid 5. (B) In water 7 H ions per liter in 1 ml solution no. of H ions / (D) H p of aq H S is greater than 7. (1)
11 5. ph=; H 1 ph ; H 1 Decreased by 1 times. H ka.c H H P log H P 1.65 V N V N H V V 55. a a b b H H ph log1 ; ph=1 56. (C) ph ; H 1 H kac 1 ka 1 ka 1 6 ka kb 1 6 a OH kbc 1 1 poh= ph= 1 = 1 b (D) N V N V H V V H 1. 1 H P log (B) (A) 6 n 1. 1 K w KaxK b; K K w a Kb (A) Roots are n, ; 6. (C) Real roots does not exist. n n1 PART C : MATHEMATICS b c,. a a (11)
12 6. (B) i 6. (C) c b b c a a 65. (B) LHS always positive RHS always negative LHS RHS. 66. (C) Verify from options. 67. (D) x x x x sin. 1 and sin (D) 1 1 x x A x x B 1 1 x x x x 1 1 x where px x 1 x x x p A.M G.M p 69. (C) The minimum of 1 5 f,f,f,f 7. (B) x 5 x 6 ; x x x ; x 71. (D) x y 9z x 8y 7z 15 1 x 1 y 1 9 z / 1/ 7. (B) 1/ Put x t 7. (A) 1 x x x R x x x x x x (A) y y y y when We have, x1 x1. x1 x1. x 1 y log x log x 5 5 Put log5 x a, then (1)
13 a a a a 1 a 1 or log5 x 1 5 x 5 or 1 x (A) R PQ P Q tan tan a b c 76. (C) sin sin sin f x x sin x sin Given equation is x sin x sin x sin x sin f sin sin sin sin sin f sin sin sin sin sin f sin sin sin sin sin f x has one root between sin and sin. 77. (A) ax bx c n ax 1 bx n cx n 1 n1 n n1 A b c n1 n n1 b c Adding above equations as bs cs 78. (C) n 1 n n x x x w, w 79. (B) x 1 x 1 x x; 1 x 1 x x 16 5x x 5x 16x x 5 8. (D) Eliminate 'z' and convert quadratic in y and use 81. (B) x x x x a; x x x x x x b 1 1 x 1x c Now b c a1 x x x x x x x x 1 x x x x x x 1 x x x x x x x x 8. (C) 1 1 x x 1 (1)
14 Let f x x,g x x f x gx if x 1 x x. 1 x x 1 8. (C) X=111111= f = Sum of digits = 1 8. (C) Cases b b,, (i) a=1, c=1 (ii) a=1, c= b 8 b, (iii) a=, c=1 b 8 b, a 1,c (iv) b 1 b a,c 1 (v) a=, c= b= 1 (vi) a 1,c a,c 1 b= (C) (C) By differentiating we get +6x+1 x which has no real roots given equation has only one real root 1 f f real root lies between (1)
15 1, Only one real root sum of the real roots = the real root. 87. (D) D b a c, D b a c, => a a c c b b the D b b a a c c (B) x1 x... xn Let x 1, x... xn n 1 n A.M. of roots =1 89. (A) x y x y x+y= x+y= x+y=6 x y=87 x y=58 x y=9 91 x x=61 x=1 not an integral 61 1 x x Not an integer not an integer No point with integral coordinates lies on the curve. 9. (D) 9 x x, x 1 9 x x x x x 1 x x x 1 x 9 x x x x 1 > 1 9 x x x x 1 From all case 1 9 x x x x 1 Holds good x R. (15)
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