Aiats Jee Main 2015 Test-3 2

Test - 3 (Paper-I) (Code-A) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2015

1/9

ANSWERS

1. (3)

2. (2)

3. (1)

4. (1)

5. (3)

6. (1)

7. (1)

8. (3)

9. (3)

10. (4)

11. (4)

12. (1)

13. (4)

14. (4)

15. (1)

16. (3)

17. (1)

18. (4)

19. (1)

20. (1)

21. (3)

22. (3)

23. (4)

24. (2)

25. (3)

26. (1)

27. (1)

28. (1)

29. (3)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (1)

33. (2)

34. (2)

35. (3)

36. (4)

37. (4)

38. (1)

39. (2)

40. (1)

41. (2)

42. (3)

43. (2)

44. (2)

45. (1)

46. (2)

47. (1)

48. (1)

49. (1)

50. (2)

51. (2)

52. (2)

53. (3)

54. (1)

55. (2)

56. (3)

57. (1)

58. (1)

59. (4)

60. (1)

61. (1)

62. (1)

63. (2)

64. (3)

65. (3)

66. (2)

67. (4)

68. (1)

69. (1)

70. (3)

71. (4)

72. (4)

73. (3)

74. (2)

75. (4)

76. (3)

77. (3)

78. (1)

79. (1)

80. (1)

81. (3)

82. (1)

83. (4)

84. (1)

85. (3)

86. (1)

87. (3)

88. (4)

89. (2)

90. (1)

TEST - 3 (Paper-I)

All India Aakash Test Series for JEE (Main)-2015 Test - 3 (Paper-I) (Code-A) (Answers & Hints)

2/9

1. Answer (3)

2 21 21 2 1 2

FL FL FL FLL

YA Yr rY A A Y r r

2. Answer (2)

Bulk modulus P PV

KV V

V

...(i)

and V = T = 3VT or 13

V

V T ...(ii)

From equations (i) and (ii),

3 3

P PK or T

T K

3. Answer (1)

1

2

0.01 1

0.02 2

P

P

1 1

2

2

4

1or or 2

4 2

T

r r

T r

r

3

1 1

2 2

8V r

V r

4. Answer (1)

Speed of air above the tank will increase, so

pressure will decrease. Hence, v will decrease.

5. Answer (3)

Surface tension + Upthrust = Weight

3 32 423 3

w srT r g r g

Substituting the values we get

1.2 mmr 6. Answer (1)

Angle between initial velocity vector and final

velocity vector is 120.

2 2 2 . cos120 3v v v v v v

Now P dm

F vt dt

3 3dv v av vdt

23 Av

7. Answer (1)

0

1vr

8. Answer (3)

9. Answer (3)

10. Answer (4)

Weight of body displaced by liquid = Upthrust

60 g + v 0.6 103 g = v 1000g

60 + 600 v = 1000 v

60 = 400 v, 360 3

m400 20

v

11. Answer (4)

Stress

StrainO

P

Q

Beyond point P, it will follow dotted line PQ. It

means deformation O, Q will remain permanently.

Hence, final length of the wire will contract but final

length will be greater than original length.

12. Answer (1)

In equilibrium,

Pressure of h cm of oil + Pressure of (20 h) cm

of Hg

= Pressure of 20 cm of CCl4

h 0.9 g + (20 h) 13.6 g = 20 1.6 g

12.7 h = 240

h = 18.9 cm

PART - A (PHYSICS)


3/9

13. Answer (4)

According to Poiseuille's formula

4 2

2Volume, Now

8 Time

pr t r lv r v

vl t

4 2

288

pr pr

vlr vl

3 6

2

2 12

8 0.0015 8 10 5 1015 N/m

4 10

n vlp

r

14. Answer (4)

v = terminal velocity

v0 r2

New terminal velocity

v' = 4v0

3 34 48

3 3 r R

R = 2r

15. Answer (1)

Equation of continuity

Rate = 3 3

p Q

A Ar V A V

VQ = 5 m/s

16. Answer (3)

Thrust force (F) = F1 F

2

2 2

1 2aV aV

= a (2gh1) a (2gh

2)

= 2ag (h1

h2) = 2 agh

17. Answer (1)

Surface area decreases. On the other side small

water drops from shower are cool.

18. Answer (4)

If H is the height of liquid surface then for same

range h2 = H h

1 and for maximum range

1 2

2 2

h hHh

19. Answer (1)

2TP gh

r

23 22

2 7 1010 10 2 10

0.05 10

= 200 + 280 = 480 N/m2

20. Answer (1)

2 21 , orh M r h M r hr

2 1or M r M r

r

21. Answer (3)

Gravitational potential at the surface of earth GmR

Gravitational potential at = 0Potential difference 0

Gm

R

GmVR

Work done W = 'GMm

w m VR

Work done per kg GM gRR

22. Answer (3)

Gravitational field dvdr

Hence, dv kdr r

0 0

v r

v r

kdv dr

r

0

0

logr

v v kr

23. Answer (4)

Gravitational force on each due to other three

particles provides the necessary centripetal force.

45

M

M

MM

R

RR

R

v

F

F'

F

2 2 2

2 2

2cos45

22

GM GM Mv

RRR

2 2 1

4

GMv

R


4/9

PART - B (CHEMISTRY)

24. Answer (2)

A

V

B

V'

30

OR h

O

sin30 'mv R mV R h '( ) '

2 4 4

V V VR R h V

h = R

25. Answer (3)

1 2 E E E

E = Net field at the centre of hole due to entire mass

E = Field due to remaining mass

2E

= Field due to mass in hole = 0

1 3 GME E r

RWhere

2

R

r

22

GMER

26. Answer (1)

27. Answer (1)

28. Answer (1)

29. Answer (3)

30. Answer (3)

31. Answer (1)

2

4MnO 5e Mn

2 2

3 4SO SO 2e

2 2+ 2

4 3 4So, 2MnO +5SO 2Mn +5SO

32. Answer (1)

7 10

4 2HIO ICl I HI

33. Answer (2)

+1 + 0 +(1)3 + x = 0

x = +2

34. Answer (2)

3 3(HPO ) 3( 11 6) 0 ; 5x x

35. Answer (3)

36. Answer (4)

2R

2R

2

2

RPF

44R

Percent vacant space = 100 25

37. Answer (4)

Na2O has antifluorite structure.

38. Answer (1)

C2H

5OH is non-electrolyte.

Tf is lowest.

39. Answer (2)

XA = 0.3 ; X

B = 0.7

YA = 0.4 ; Y

B = 0.60

A B

A B A

P P 20.60

P P P 3 o

B B BP P X or o

A A AP P X

B B

A A

P P2 0.7

3 P 0.3 P

A A AP P X

B A

A B

P P6 2 7or

21 7 2P P

40. Answer (1)

The higher the value of Kb, more be T

b, therefore,

it can be measured more accurately.

41. Answer (2)

Hmix

< 0, exothermic process in negative deviation

because force of attraction between A B > A A

or B B.


5/9

42. Answer (3)

2 22HI H + I

2 0 0

1 1 122

3 3 3

p 2

1 1

13 3K

164

3

43. Answer (2)

232CN + AgNO Ag CN

Initial conc. 0.1 0.03 0

Final conc. 0.1 0.06 0 0.03

2Ag CN Ag 2CN

0.03 0 0.04

0.03a a 0.04 a

219

C

0.04K 4 10

0.03

a

[Since KC is too small in its value hence neglect 'a'

w.r.t. 0.03 and 0.04]

a = 7.5 1018 M

44. Answer (2)

ng

= 0

45. Answer (1)

G RT 2.303 log K 4.606 103 = 2 500 2.303 log K

46. Answer (2)

47. Answer (1)

4 4NH OH NH OH

0.02 x x x

0.02 y y 0.01

5

0.011.8 10

0.02

y

y

y = 3.6 105

48. Answer (1)

49. Answer (1)

3

cell 6

10E 0.059 log 0.177

10

50. Answer (2)

4 2 2 2 42CuSO + 2H O 2Cu + O +2H SODue to formation of H

2SO

4, pH decreases.

51. Answer (2)

b1pOH pK log C2

1 4.73 log 0.12

15.73 2.865

2

pH = 14 2.865 = 11.13

52. Answer (2)

Rate of disappearance of B is 4 times that of A.

Hence, at any point of time [A] will be greater than

[B].

53. Answer (3)

54. Answer (1)

d G= S

dT

cell

d nF E= S

dT

celld E S

=dT nF

55. Answer (2)

1

25 100010 1.86

62 W

1

1.86 25 1000W 75

62 10

Ice separated = 100 75 = 25 g

56. Answer (3)

57. Answer (1)

58. Answer (1)

59. Answer (4)

60. Answer (1)


6/9

PART - C (MATHEMATICS)

61. Answer (1)

2 3 3sec0 sec sec sec sec

7 7 7 7

2sec sec

7 7

A

A = sec0 = 1

B = (cot 1 + cot 179) + (cot 2 + cot 78) + .......

(cot 89 + cot 91)

B = [cot 1 cot 1] + [cot 2 cot 2] + ......

(cot 89 cot 89)

B = 0

A2 + B2 = 162. Answer (1)

Expression 2 2 2 2

2 2

sin cos sin cos 1

sin cos 23 3

1 1 1 3

11 2 3

63. Answer (2)

cos3 (2cos 1)cot3 cot

sin3 (2cos 1) 12L

2sin3 (cos2 1)tan3 tan

2cos3 (cos2 1) 12M

cot tan 2cot 2 312 12 6

L M

64. Answer (3)

Expression 2 2

1

4 tan cot 9

1 1max

4.2 9 17

65. Answer (3)

2 2 2

2

1 tan 36 cos 36 sin 36 cos 72

2 1 cos 722cos 36

sin18 5 1

5 2 5 41 sin18 3 5

a + b = 9

66. Answer (2)

2 2cosec ( ) sin (2 ) 1x y x y

2 2cot ( ) sin (2 ) 0x y x y cot(x + y) = 0 and sin(2x y) = 0

2x y

6x

2x y = 03

y

sinx + cos3y = 1 11

2 2

67. Answer (4)

3 tanA + cotA = 5 cosecA

3 sin2A + cos2A = 5 cos A 2 cos2A + 5 cosA 3 = 0

1cos2

A 3

A

So, 2 sec A + 4 sin2 A

= 3

2.2 4 74

68. Answer (1)

sin3 sin23 2cos13 sin10

sin16 sin4 2sin10 cos6

cos13

cos6

As,19

19

13 6

cos( 6 )

cos6

cos6

1cos6

69. Answer (1)

sin sin sin( )

sin cos sinn

n nf

n n

= tann f

4() = tan4


7/9

4

1cos4

tan32 8

sin4

f

11

22 1

1

2

70. Answer (3)

a : b : c = 4 : 5 : 6

Let a = 4k, b = 5k, c = 6 k

2 2 2 25 3616 9 3

cos2 256 26 4

b c aA

bc

2 2 2 16 3625 27 9

cos2 246 246 16

a c bB

ac

2 2 2 16 2536 5 1

cos2 245 245 8

a b cC

ab

3 9 1cos : cos : cos : :

4 16 8A B C

= 12 : 9 : 2

3x 4y = 36 36 = 071. Answer (4)

22sin cos 1 2sin 4sin 1 4cos

2sin cos sin 4 cos sin cos sin 0 tan 1

4n

7 3 5, , ,

4 4 4 4

Number of solution is 4.

72. Answer (4)

Put 2 cos 2

in 3sin cos sin

y x y

x

33 sin , 2cos

sin

yy x

2 24 4x y

73. Answer (3)

A

B C49

72

2 3

49 72

sin sin3 sin2

c

c = 98 cos

Now, 249 72

sin sin 34sin

2 275 121sin cos

196 196

11cos14

So, 11

98 cos 98 7714

c

74. Answer (2)

A

C B43

2

x x

D

h

In ACD,3

sin sin sin

x h

C ...(i)

in ABD,4

sin sin sin2

x h

B ...(ii)

3 4 2cos

sin sin2 3

75. Answer (4)

Let 15 1215


8/9

Now, sin2 + sin4 + sin7 sin= 2sin3[cos4 + cos]= 2sin36[cos48 + cos12]= 2sin36 2cos30 cos18= 3 sin 54 sin18

= 5 1 5 1 5 15

3 34 4 2 4

a + b = 19

76. Answer (3)

Expression

3

3 3

2

2

64sin cos64sin cos

sin1

cos

3 38 2sin cos 8 sin2

Max. value = 8

77. Answer (3)

13 sin2 cos2 3

f

So, min1

( )5

f78. Answer (1)

2 1 12cos cos

2 2y y 2,

3 3

y

and 2 1

2sin2

x

1 5sin ,2 6 6

x x

,6 3

and 5

,6 3

does not satisfy equation (i),

so answers are 2

,6 3

and 5 2

,6 3

79. Answer (1)

sin sin

b c

B C

sin( )

sin

sin sin

b b A Bc CB B

sin cos cos sin

3sin

A B A B

B

= 3 [sinA.cotB + cosA]

4 5 3 1 3 9 14

3 3 15 12 5 3 5 5 5

80. Answer (1)

sin cos

sin sin2 2

sin sincos sin

2 2

A B A B

A B

A B A B A B

97 2 1

9 7 16 8

a b

a b

81. Answer (3)

sec cosec 1 1

tan cot sin cosE

1 1 2 2 2 2

3 1 3 1sin cos

12 12

22 2 2 2

2

82. Answer (1)

2sin 1 sin 1 0x x 1

sin , sin 12

x x

3, 2 ,6 6 2

x

So, Sum = 9

2

83. Answer (4)

2 2 2 2( ) cos2 2cos 1 sin2 2cos 1p x x x x x 22cos 1 2cos 1x x

= (2cosx + 1)

84. Answer (1)

2sin4 cos3 cos 2cos3 cos3 cosx x x x x x 2sin4 2cos3 cos3 cos 0x x x x


9/9

sin4 cos3 2cos2 cos 0x x x x If cos 2x = 0

4

x

If cos x = 0

2

x

If sin4x = cos3x

Then14

x

85. Answer (3)

2cos2 cos 2x xcos 2x = 0 or cos 2x = 1

3 5 7, , , , 0, , 2

4 4 4 4

x

Sum = 4 + 3 = 786. Answer (1)

87. Answer (3)

88. Answer (4)

89. Answer (2)

90. Answer (1)

Aiats Jee Main 2015 Test-3 2

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