Jr.IIT-JEE Mains Model Paper Test Date:15/08/2016 Key ...altitudeclasses.org/images/test_sol/02-JR.MAINS-TEST-15-08-2016.pdf · # 2-1-520 to 526 Vijaya Sri Sai Celestia II Floor,

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Jr.IIT-JEE Mains Model Paper

Test Date:15/08/2016

Key & Solutions:

Key

Mathematics

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

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

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

Physics

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

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

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

Chemistry

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

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

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


Answers & Solutions

1.

2.


3.

4.

5.


6.

7.


8.

9.


10.

11.

12.


13.

14.

15.

16.


17.

18.


19.

20.

21.


22.

23.


24.

25

26.

27.


28.

29.

30.


PHYSICS

31. (b) 26

dsv tdt

Impulse = P = m(vf – vi) = 2(6 × 1 – 0) = 12 N-s.

32. (a) After collision 2

1

2

ev u

and 1

1

2

ev u

Here, u = initial speed of ball 1 v2 = 2v1 when 1

3e .

33. (c) Change in linear comentum .p F t

Here, F

is weight, i.e., mg

1 10 1 10 / .p mg t kg m s

34. (a) At maximum extension velocity of both the blocks will be same. Let v be the common velocity of the

blocks (towards right). From conservation of linear momentum.

(3 + 6) v = 6 × 2 – 3 × 1 = 9 or v = 1 m/s

Now let x be the maximum extension in the spring.

From conservation of mechanical energy: 2 2 221 1 1 13 1 6 2 200 9 1

3 2 2 2x

or 23 24 200 9x or 18

200x or x = 0.3 m

or x = 30 cm.

35. (c) Using impulse = change in linear momentum

We have,, .f i

F t m v v

or 2 3 2 1 2fi j k v i j

or 6 3 6fv i j k

2 2 26 3 6 9 / .

fv m s

36. (b) Force on the bullet becomes zero at 0 = 600 – (2 × 105) t or t = 0.003 s

Impulse 0.003

0F dt 0.003

5

0600 2 10 0.9t dt N s .

37. (a) From conservation of mechanical energy

2 21 1

2 2rk x v

Here, = reduced mass of the blocks 2 2

2 3

m mm

m m

and vr = relative velocity of the two.

Substituting in Eq. (1), we gt 2 22

3rkx mv

3

2r

kv x

m

38. (c)


1

sin2 2

r

r

30º

From conservation of linear momentum mu = 2mv cos 30º or 3

uv . . . . .(i)

Now relaive velcoity of separation

erelative velcoity of approach

in common normal direction.

Hence, / 3 2

cos30º 33 / 2

v ueu u

.

39. (d)

Let v be the velocity of ball after collision, collision elastic

e = 1 or relative velocity of separation = relative velocity of approach

v – 1 = 4 + 1 or v = 6 m/s (away from the wall)

40. (c) In perfectly inelastic collision between two particles, linear momentum is conserved. Let be the

angle between the velocities of the two particles before collision. Then

2 2 2

1 2 1 22 cosp p p p p

or 2

2 22 2 cos

2

vm mv mv mv mv

or 1 = 1 + 1 + 2 cos

or 1

cos2

or = 120º.

41. (b) Let u be the velocity of ball before collision.

Speed of ball after collision will become

2 25

.82 2 2

u uv u

Fraction of KE lost in collision

2 22

2

1 1

5 32 2 1 1 .1 8 8

2

mu mvv

umu

42. (b)

Let v1 and v2 be the velocities of the two masses after collision in the same direction. Then

m1v = m1v1 + m2v2 . . . . . .(1)

and v2 – v1= = ev . . . . . .(2)

Solving Eq.s (1) and (2) we get, 1 2

1

1 2

v m emv

m m

For v1 to be positive m1 > em2 or 1

2

me

m .


43. (a) Let speed of block is v. Then from conservation of linear momentum in horizontal direction velocity

of cylinder will be 2v in opposite direction .2

Mas m

Now from conservation of mechanical energy, we have 221 12

2 2mgh Mv m v

Here, h = R – r = 1.0 m

Substituting the values, we get (1)(10)(1) 2 21 12 1 4

2 2v v or

23 10v

10

/ .3

v m s

44. (c) Let vr be the velocity of bullet with respect to gun and v the velocity of gun. Then

From the two figures it is clear that > 45º.

45. (c) 2

2av

F ava

Here, 3 3 6 310 / , 20 10 /kg m av m s and

22

2104 4

da

Substituting the values 2

3 6

3

4

10 20 105.1 10 .

104

F N

46. (b) Block just reaches the top of the wedge, it implies that velocity of block with respect to wedge at the

top of the wedge is zero. Let v be the horizontal velocity of both at this instant.

Then from conservation of linear momentum, we have (2m + m)v = mu

or 3

uv

Now from conservation of mechanical energy, we get 2 21 13

2 2mu m v mgh or

22 3 2

9

uu gh

or 22

2 33u gh or u gh .

47. (b)

48. (b) The rope tension is the same both on the left and right hand side at every

instant, and consequently momentum of both sides are equal

Mv M m v m vr v or 2

r

mv v

M

Momentum of the centre of mass is P = P1 + P2 or 2 comMv Mv Mv

2

com r

mv v v

M .


49. (d) Horizontal and vertical components of initial velocity are:

20 2 cos45º 20 /xu m s and 20 2 sin 45º 20 /yu m s

After 1s horizontal one part comes to rest. Hence, from conservation of linear momentum vertical

component of second part will become 20 /yv m s . Therefore, maximum height attained by the

second part will be H = h1 + h2

Here h1 = height attained in 1s = (20)(1) 2110 1 15

2m

and h2 = height attained after 1s 2220

202 2 10

yvm

g

20 15 35H m

50. (a)

When the string jerks tight both particles being to move with velocity components v in the direction

AB. Using conservation of momentum in the direction AB

mu cos 30º = mv + mv or 3

4

uv

Hence, the velocity of ball A just after the jerk is 3

4

u

51. (a) Relative velocity of separation = relative velocity of approach = v (as e = 1)

Time of next collision 2 r

v

.

52. (a) Inc case of perfectly inelastic collision the velocity of combined mass will be 2

v

21

22 4

vKE m

210.2 0.2

2 4

v

or 2 2 /v m s

In case of perfectly elastic collision

First particle stops and the second acquires the same velocity v 21

2KE mv

or 210.2 0.1

2v or v = 2 m/s

Therefore, minimum value of v is 2 m/s in case of perfectly elastic collision and maximum value of v

is 2 2 /m s in case of perfectly inelastic collision.

53. (c) After collision balls exchange their velocities


i.e., 2Av gh and 2 4 2 2Bv g h gh

Height attained by A will be 2

2

AA

vh h

g

But path of B will be first straight line and then parabola as shown in figure.

After calculation we can shown that 13

4Bh h

4

13

A

B

h

h .

54. (b) j component, i.e., component of velocity parallel to wall remains uncharged while i component will

become 12 .

2i or i

Therefore, velocity vector of the sphere after it hits the walls is 2i j .

55. (a) Retardation due to friction a = g = (0.25) = (10) = 2.5 m/s2.

Collision is elastic, ie, after collision first block comes to rest and the second block acquires the

velocity of first block. Or we can understand it in this manner that second block is permanently at rest

while only the first block moves. Distance travelled by it will be

22 55

2 2 2.5

vs m

a

Final separation will be (5 – 2) = 3 m.

56. (a) Let the masses are unequal

From conservation of linear momentum;

or 1 1 2 2 1 2m v v m v v or 1 2m m

i..e, mass of both the particle should be same

Secondary, relative velcoity of separation

erelative velcoity of approach

1 2

1 2

1v v

v v

i.e., e = 1 or collision is elastic.

57. (a) In an elastic oblique collision both particles move at right angles after collision if mass of both the

particle are same provided one particle is at rest before collision.

58. (b) Let mass of gun is M and that of shell is m. The two cases are shown in figure as below:

Here, v1 and v2 are the recoil speeds of the gun in two cases. Using conservation of linear momentum

in horizontal direction in two cases:

1 1m v v Mv

1

mvv

M m

. . . . . (1)

2 2cos30ºm v v Mv . . . . . (2)

2

3

2

mvv

M m

From Eqs.(1) and (2), we get 1

2

2

3

v

v .

59. (b) 3

2 104 0.0032

2.5r

dmF v N

dt


60. (a) Let P1 and P2 be the momenta of A and B after collision.

Then, applying impulse = change in linear momentum for the two particles, we get

For B : J = P1 . . . . . (1)

For A : J = P – P2 or P2 = P – J . . . . . (2)

Coefficient of restitution 1 2 1 21.

P P P P J J P J Je

P P P P

CHEMISTRY

61. (d) Due to H-bonding in NH3.

62. (b) .

PV PV Rn k

RT kN T N

.PV

n N No of moleculeskT

63. (a) 2

'

O MP P mole fraction of 2O

M

P

P

mole fraction 2

132

3

16 32

w

Ow w

.

64. (c) Greater the value of ‘a’ i.e., intermolecular forces of attraction for gas, the easily the gas will be

liquid.

65. (a) 1 1

2 2

8:AV

u TRTu

M u T

1

2

2

0.3 300 1200, 0.3 0.6

1200 300or u ms

u

.

66. (b) Density of a gas, 1

;PM

e e P and eRT T

.

67. (a) 21

3PV mnu

For one mole, 2 21

3PV Mu P u

68. (b) Use 0

1,

vr P T

t m

2 2 2

2 2 2

32, 20sec

5 2

H O H

O H O

r M r x tt

r M r x .

69. (c) 3

rms

RTU

M

Thus, 2

3 5050

2rms H

RU at K

. . . . . (1)

2

3 800800

32rms O

RU at K

. . . . . (2)

So,

2

2

50 321

2 800

rms H

rms O

U

U


70. (c) According to Grahan’s law of diffusion 1

r PM

at constant temperature.

1

2A A B A B

B B A B A

r P M P M

r P M P M

.

71. (b) 1 1PV

Z ornRT

Thus, 22.4mV litre

72. (c) 1 22 2

3 3

2 28rms rms

RT RTu H and u N

2 27rms rmsu H u N

2 27rms rmsu H u N

or 1 23 3

72 28

RT RT

2 2

1 22 12

2 4N H

T Tor T T or T T

73. (c) 3 3

2rms

RTu and KE RT

M

2

rms

Eu

M

74. (c) For positive deviations Z > 1, i.e., the condition when repulsive forces predominates. Also, at high P,

V is small and b cannot be ignored, but eh factor 2

a

V can be neglected in comparison to P.

Thus, 1Pb

ZRT

75. (b) 3 400

40rms

Ru X

2 60

MPY

Ru

M

4M .

76. (a) 3 8 1C H n mol

4 2CH n mol

Total mole = (n1 + n2)

mole fraction of 13 8

1 2

nC H

n n

n

P RTV

1 2320

760

n nRT

V

3 8 23C H CO ,

4 2CH CO

CO2 formed = (3n1 + n2)


1 23448

760

n n RT

V

1 2

1 2

3 4481.4

320

n n

n n

1 20.25n n

mole fraction of 13 8

1 2

0.2n

C Hn n

Thus, (a)

77. (c) By Avogadro’s law, equal volumes of gases contain equal number of moles.

3 2 24 5 4 6NH g O g NO g H O g

3 24 6mLNH mLH O

3 250 75mLNH mLH O

2 25 6mLO mLH O

2 260 72mLO mLH O

Stoichiometrically, O2 is consumed completely

Thus, 72 mL H2O(g) hence (c).

78. (d)

1 1

2 1

0 01 1

0 02 1 se

x x

y y

A B C D

set Ix x

t IIy y

2

21

2 11c

x yK

y yx

since x = 0.5

2

3y

Fraction of A converted into products 2 / 3 1

0.332 3

Hence, (d)

79. (a) 5 3 2

20.5 2 mol

PCl PCl Cl

Q (reaction quotient)

3 2

5

2 2

2 2 40.5

2

PCl Cl

PCl

Thus, Q = KC, reaction mixture is in equilibrium

1

5

0.50.25

2PCl mol L

Hence (a)

80. (d)

1 2

2 2

2 3

P P

NO O NO OK amd K

NO O NO

1 2

2 3

2 2

1 1

P P

NO O NO

K K NO O NO O

3

2

OK

O O

83

50 35

13.3 10

6 10 5 10K


81. (d) 2 4 22N O NO

1 92 2 46

2 30.61

0.5 .

82. (d) At the initial state if only reactants are used Q = 0, but from initial stage to attain equilibrium is

continuously increase with time till it because equal to K.

83. (d) KP is a characteristic constant for a given reaction and changes only with temperature.

84. (d) For the reaction ; 2 4 22

g gN O NO

2 24..... 1c

aK

a a V

Where, a is initial mole on N2O4 in V liter vessel and is degree of dissociation of N2O4.

If volume is reduced to V/2, the initial concentration of N2O4 becomes 2

.

2

a a

V V

An increase in concentration of reactants leads to forward reaction, i.e., the decomposition of N2O4 to

have constant value of Kp.

85. (c)

2

3

1 3

2 2

NHK

N H . . . . . . . .(i)

2

2

2 2

NOK

N O . . . . . . . .(ii)

2

3 2 1/ 2

2 2

H OK

H O . . . . . . . .(iii)

2 3

2

2 5/2

3 2

NO H OK

NH O

By inspiration of (i), (ii) and (iii)

2 3 332 2 22 3

2 3 3/2

1 2 2 3 2 2

NO N H H OK K

K N O NH H O

2 3

2

2 5/2

3 2

NO H OK

NH O

86. (b) 2 2 3

1;

2SO O SO H ve ; Follow Le Chatelier principle.

87. (b) 2 4 21 0

1 0.20 2 0.20

2N O NO

Mole of 2 4 1N O mole

Dissociated 2 4

1 200.2

100N O mole

2 4NP O at 300 K = 1 atm,

2 4NP O at 600 K = 2 atm

Total mole present at eq. = 1 – 0.20 + 0.40 = 1.20

1 1

2 2

P n

P n (at constant T, V)

2

2 1

1.2P P1 = 2.4 atm


88. (b) 1 1 0

1 1 2

2

x x x

A B C

2

2

481

1

x

x

or

2

91

x

x

1

11x

Thus, 10 10 2

11 11 11

2A B C 10 10 2 22

11 11 11 11Total mole

Mole fraction of

2

11122 11

11

C .

89. (a) The formation of SO3 is exothermic and thus at high temperature backward reaction will proceed to

give lower percentage yield.

90. (a) 2/2 /4

2

2 2P P PP

NOCl NO Cl

r

5P .

2 2 4 4

P P P P

Documents

Jr.IIT-JEE Mains Model Paper Test Date:15/08/2016 Key ...altitudeclasses.org/images/test_sol/02-JR.MAINS-TEST-15-08-2016.pdf · # 2-1-520 to 526 Vijaya Sri Sai Celestia II Floor,