Jee Main_mock-2 Tnlmj22_key & Solutions_may 2015

TNLMJ22 2015

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JEE MAINS MOCK 2 SOLUTIONS & ANSWER KEY

PART- A PHYSICS

1. d 2. b 3. c 4. a 5. b 6. b 7. a 8. d 9. d 10. b

11. c 12. b 13. c 14. b 15. a 16. a 17. c 18. d 19. b 20. a

21. b 22. a 23. c 24. b 25. c 26. c 27. b 28. b 29. a 30. a

Solutions: 1. Ans: Given circuit reduces to

2. Ans: Circuit can be reduced to

3. Ans: Initially d = 2 – 1.6 = 0.4 mm and C = inally C’ =

Therefore

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4. Ans:

5. Ans:

6. Ans:

q =

V = 9 volts

Charge on

7. Ans: V =

8. Ans: E =

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9. Ans: is in parallel with series combination of

10. Ans:

11. Ans: C =

12. Ans: In parallel connection

13. Ans: 14. Ans: For potential to be made zero after connection the charge on both capacitors must be equal. That is

15. Ans: Since flux entering and flux leaving are the same the net flux is zero.

16. Ans: Electric Flux =

=

17. Ans: Net charge enclosed is zero since it is a dipole. Thus flux is zero. 18. Ans: The charge enclosed is q. 19. Ans: It is the inverse of absolute permittivity since q = 1 C

20. Ans: From Gauss Law, flux leaving the surface =

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21. Ans: From work – energy theorem, work done =

Ratio of work done in the two cases will be

=

22. Ans: As. K.E is same,

23. Ans: W =

24. Ans: When the block is dropped on the spring decrease in potential energy of the block = mg (h + x).

This is stored in the spring as potential energy U =

25. Ans: Work done on each bullet =

Number of bullets fired per second =

Power = work done per second = 3 26. Ans: P = Fv = constant. Since F = ma and v = at,

P = (ma) (at) = constant or

27. Ans Work done = area between the graph and position axis

10 1 20 1 20 1 10 1 20W J

28. Ans: Work done in stretching the rubber band is

W =

29. Ans: Momentum is maximum when kinetic energy is maximum. In this case, the elastic potential energy is converted into kinetic energy.

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30. Ans: For maximum kinetic energy of the particle, its potential energy is minimum. For minimum

potential energy, we have

ie.

ie.

PART – B

CHEMISTRY KEYS:

31 – b 32 – d 33 – a 34 – a 35 – b 36 – b 37 – a 38 – b 39 – d 40 – c

41 – b 42 – d 43 – b 44 – a 45 –b 46 – d 47 – b 48 – c 49 – b 50 – b

51 – c 52 – a 53 – b 54 – d 55 – c 56 – d 57 – b 58 – c 59 – d 60 – c

SOLUTIONS:

31. For conversion of atoms to molecule, heat is liberated so low temp and high pressure is used 32. Energy unutilized = 1560 – 780 = 780 KJ For evoparation of 18g(1mole) energy required is 44KJ so for 780KJ, water evoparated is 18/ 44x 780 = 319g

33. 0

11.70.03

390

c

34. EMF is positive for option a hence such cell works

35. Bufer capacity / 0.02 / 2

0.10.1H

noof molesof acid added litre

P

36. CONCEPTUAL QUESTION

37. 0 0 0

2 3( ) [4( ) 6( ) 4( )] [4( 104) 6( 237) 4( 108)] 574P RS S S NO H O NH

38. Keq = 2c (or)

5

4 2 1 2

0

1.8 10; 18 10 4.243 10 ( ) 16.57

390.7 0.01

c ceq m m

m

m

K xx x or cm

c

39. PNH3 = PH2S = 0.56atm ; KP = PH2S X PNH3 = 0.56 X 0.56 = 0.3136 40. PH = ½ [PKw + PKa – PKb] = ½ [14 + 4.74 – 9.30]= 9.44 / 2 = 4.72

41. One with highest R.P is strongest oxidizing agent 42. conceptual question

43.

2[ ] 0.1 20 /100 0.02 2 10

2 0.3010 1.6990; 14 1.6990 12.3010OH H

OH C x x

P P

44. Increase in number molecules = 0.5, Increase in pressure = 20 so for 1 molecule = 40

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40

220

dp

dt

45. Same electricity eq.wt = 4.2: 5.4 : 19.2 ; valency = at.wt/ eq.wt 46. conceptual question 47. Daniel cell has EMF = 1.1V ;

0 2 96500 1.1 212300 / 212.3 /G nFE x x J mole KJ mole 48. mathematical concept 49. w = - 2.303n RT log v2/ v1 = - 2.303 x 1x 8.314 x 10-3 x 300 log 10= - 5.74 50. acidic nature of oxide oxidation state

51. A) 2 1 0 0

12 ; 2 ( )for Fe Fe e G nFE F E

B) 2 3 1 0 0

2; 1 ( )for Fe Fe e G nFE F E

C) 3 1 0 0

33 ; 3 ( )for Fe Fe e G nFE F E

Substract A from C

0 0 0 0

3 1 3 1[3 2 ]G G F E E

Compare this equation to B;

0 0 0

2 3 1

0 0 0 0

2 3 1 2 3 1[3 2 ] 3 2

G G G

E F F E E so E E E

52.In I3- there is dative bond between I-1 and I2 so I-1 is electron donar

53.During electrolysisof, H+ and Cl-1 are discharged and NaOH is formed, hence PH increases

54. 2 2

2 2

2

[ ] [0.04]160

[ ] [ ] [0.01] [0.1]

NOBrQ whichis also K value

NO Br

55.Compare 1 &2 [A] is doubled but [B] is constant , rate is constant so order w.r.t to [A] is zero Compare 1 &3 [A] is tripled but [B] is doubled, rate has increased by 4 times order w.r.t to [B] is 2 So total order is 2

56. 2 2

1 3. . [ 3( )]

2 2

1 346 [ 712 416 3( )]; 352

2 2

fH B E of react B E of prod N H N H

x x N H N H

57. when Ea has low value, the reaction is fast reaction 58. Copper is in II group and Zn is in IV group Ksp of copper is less hence it will get precipitated first 59. conceptual question 60. PH = 9, POH = 5 ;log [salt]/[base]= 0.3; [salt]/ [base] = 2 ; base : salt = 1 : 2

PART – C MATHS

61.b 62. b 63. a 64. a 65. a 66. d 67. c 68. b 69. b 70. b

71. a 72. b 73. b 74. c 75. c 76. c 77. a 78. d 79. b 80. d

81. c 82. d 83. c 84. a 85. b 86. d 87. b 88. b 89. c 90. b

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1.

2. x

x

3.

Taking modulus on both side, we have

4. Consider G.P as a, ar, .

Given a = ar +

r =

5. Given f(x) = x sin = 0 , x = 0

is continuous at x = 0, x f(x) = x sin (1/x) is continuous

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6.

=

7. X = Y = log x x

8. f (x) = 10 cos x + (13 + 2x) sin x f’(x) = 10 sin x + 13 cos x + 2 x cos x + 2 sin x f ”(x) = 10 cos x - 13 sin x – 2x sin x + 2 cos x + 2 cos x = (10 cox x + (13 + 2x) sin x) + 4 cos x f “(x) = - f(x) + 4 cos x

9. f(x) =

= 3 + 4

10. f(x) =

f’(x) =

11.

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12. First 50 even natural number are 2, 4,6…. 100

Variance of first ‘n’ natural number =

13. x = a (cos )

= a

14.

=

=

15. given

(1) – (2)

16. f(x) = = f(g/x)

h(x) =

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Taking log on bothsides Log (h(x)) =

17.

Z =

Im (z) = 0 .

18.

=

=

= 19. Common herms between the series 3 + 7 + 11 + 15 + ……… and 1 + 6 + 11 + 16 + …. are 11, 31, 51, …..

sum of 20 terms =

= 4020

20.

= 5

21. f(x) =

f’(x) = 1 -

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f ”(x) =

f “ (a) =

f “ (a) =

22. y = a cos (log x) + b sin (logx)

y’ =

= y

23. Given

S. D =

24. Given G =

M =

25. 1 + i = r (cos )

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26.

27. sin y = x sin ( + y)

=

28.

29.

at x = 1

f(x) is continuous or R – { 1}

30. l = = 3

M =

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Jee Main_mock-2 Tnlmj22_key & Solutions_may 2015

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