aiats jee main 2015 test-4

Test - 4 (Paper-I) (Code-A) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2015

1/10

1. (2)

2. (4)

3. (3)

4. (4)

5. (3)

6. (2)

7. (2)

8. (1)

9. (2)

10. (4)

11. (3)

12. (2)

13. (2)

14. (2)

15. (2)

16. (1)

17. (1)

18. (2)

19. (4)

20. (2)

21. (4)

22. (1)

23. (2)

24. (2)

25. (4)

26. (1)

27. (1)

28. (2)

29. (4)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (2)

33. (2)

34. (1)

35. (3)

36. (3)

37. (1)

38. (4)

39. (2)

40. (3)

41. (1)

42. (4)

43. (3)

44. (2)

45. (3)

46. (2)

47. (1)

48. (1)

49. (3)

50. (4)

51. (3)

52. (4)

53. (2)

54. (3)

55. (2)

56. (2)

57. (1)

58. (1)

59. (4)

60. (2)

61. (2)

62. (4)

63. (3)

64. (1)

65. (1)

66. (3)

67. (2)

68. (2)

69. (3)

70. (2)

71. (3)

72. (1)

73. (1)

74. (1)

75. (3)

76. (4)

77. (4)

78. (3)

79. (4)

80. (2)

81. (2)

82. (4)

83. (4)

84. (3)

85. (2)

86. (3)

87. (1)

88. (3)

89. (3)

90. (4)

ANSWERS

TEST - 4 (Paper-I) - Code-A

All India Aakash Test Series for JEE (Main)-2015

Test Date : 11-01-2015

All India Aakash Test Series for JEE (Main)-2015 Test - 4 (Paper-I) (Code-A) (Answers & Hints)

2/10

1. Answer (2)

2. Answer (4)

WABC

= 0 0 0 0 0 0 0

1(2 )(3 ) (3 )

2P P V V P V V

= 300 J

UABC

= QABC

– WAB

= 300 J = UAC

WAC

= P0(3V

0 – V

0) = 200 J

QAC

= WAC

+ UAC

= 500 J

3. Answer (3)

Q = dU + W

nCdT = nCvdT + W

( )v

W n C C dT ∫ ∫

0

0

( )

T

v

T

W C C dT

∫ for n = 1

=

0

0

T

v

T

aC dT

T

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 0

1ln

1a RT

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

4. Answer (4)

Work done from B to C = nR(TC – T

B)

= 6R(2200 – 800)

= 6R × 1400

Work done from D to A = nR(TA – T

D)

= 6R(600 – 1200)

= –6R × 600

While WAB

= WCD

= 0

W = 6R × 1400 – 6R × 600

= 6R × 800

= 40,000 J

5. Answer (3)

At t = 0, x1 = 0

At t = 6

, 2sin

6 2

Ax A

⎛ ⎞ ⎜ ⎟⎝ ⎠

<V> = 32

6

A

x A

t

⎛ ⎞

⎜ ⎟⎝ ⎠

6. Answer (2)

Q = dU + W

For isobaric process W = nRT

and dU = nCVT

32

vCU f

W R

dU = 3W = 75 J

Q = 75 + 25 = 100 J

7. Answer (2)

2P0

P0

P 2T0

4T0

CB

A

V

T0

V0

2V0

Total heat absorbed by 1 mole of gas

Q = Cv(2T

0 – T

0) + C

P(4T

0 – 2T

0) = 0

19

2RT

Total change in temperature form state A to C is

T = 3T0

Molar heat capacity = 19

6

QR

T

8. Answer (1)

x = A cos t

So A – = A cos ... (1)

cosA

A

and A – ( + ) = A cos 2 ... (2)

( )

cos 2A

A

2

2

3A

9. Answer (2)

x1 = –A cos t and x

2 = A sin t

B O A

PART - A (PHYSICS)


3/10

A sin t = –A cos t

sin t + cos t = 0

1 1

sin cos 02 2

t t

cos 04

t⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot4 2

3

8

Tt

10. Answer (4)

If mass falls down by x,

so spring extends by 4x.

(4 )4

Tk x

m

4

T

4

T

2

T

2

T

T

T

T = (16k)x

1 16

2

kf

m

2 kf

m

11. Answer (3)

At t = 0, y = 2

10( )

5f x

x

Velocity of pulse v = 2 m/s along positive x-axis.

Hence wave function y(x, t) = f(x – vt)

2

10( , )

5 ( 2 )y x t

x t

12. Answer (2)

13. Answer (2)

0.66 mv

f

The resonance length are

l1 = 0.165 m

4

2

30.495 m

4l

3

50.825 m

4l

4

71.155 m

4l

As 4 is greater than 1 m, so allowed resonance

are only 3.

14. Answer (2)

0 0176 165

22

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

0 = 22 m/s

15. Answer (2)

l y

A Bx

2y

l

Suppose 2y is the distance between two rows of the

buildings. The distance travelled by car in 1 second,

530 1

18x

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 150

m8

The distance travelled by sound in 1 second

2l = 330 × 1 = 330 m

l = 165 m

From the figure

2

2165 m

2

xy l

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

16. Answer (1)

17. Answer (1)

m H

H m

V

V

... (1)

where 0 0

0

H H

m

H

V V

V V

0

0

0

.

4

H

m H

H H

V V

V V

1

4

H

m

11200 600 m/s

2

H

m H

m

V V


4/10

18. Answer (2)

1 129.610

2Lf

... (i)

and 1 160

2Lf

... (ii)

f = 100 Hz

19. Answer (4)

The frequency of sound heard by motorist from

reflection

1.

m

b

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠... (i)

The frequency of sound heard by motorist directly

2

m

b

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠... (ii)

Beat frequency = 1 2 2 2

2 ( ).

( )

m

b

b v vf f f

v v

20. Answer (2)

i f

i f

v v

T T i f

i f

h h

T T

f

f i

i

Th h

T

⎛ ⎞ ⎜ ⎟⎝ ⎠

3734.00 5.09 cm

293f

h⎛ ⎞ ⎜ ⎟⎝ ⎠

21. Answer (4)

4 4

A A B BT T

1/ 4

. 1934 K

A

B A

B

T T⎛ ⎞ ⎜ ⎟⎝ ⎠

A A B BT T

3

BA

But 6

10B A

61.5 10 1.5 m

B

22. Answer (1)

P0

l0

P0

l0

h0

l0 = 45 cm. If the mercury displaces by x, then

l0P0 = (l

0 + x)P

1

l0P0 = (l

0 – x)P

2

and P2 = P

1 + h

0

0 0 0 0

0

0

l P l Ph

l x l x

x = 4.5 cm

23. Answer (2)

I =

2

3

ma and

2

al

22 2

3

I aT

mgl g

24. Answer (2)

The process 3 1 is an isochoric process. So

w = 0

0 0

1 3

0 0

23( )

2v

P M P MU nC T T

⎡ ⎤ ⎢ ⎥ ⎣ ⎦ =

0

0

3

2

P M

P

0

0

3

2

P MQ

25. Answer (4)

By the law of calorimetry

Q1 + Q

2 = Q

3 + Q

4 + Q

5

M3Cw(T2 – T) + M

2Cb(T2 – T)

= M1Ci(0 – T

1) + M

1Lv + M

1Cw (T – 0)

1 1 1 3 2 2 2

3 2 1

( )i v w b

w b w

M C T M L M C T M C TT

M C M C M C

= 10.9°C

26. Answer (1)

Statement-1 is true, statement-2 is true and

statement-2 is correct explanation for statement-1.

27. Answer (1)



28. Answer (2)

Statement-1 is True, Statement-2 is True; Statement-

2 is NOT a correct explanation for Statement-1.

29. Answer (4)

Statement-1 is False, Statement-2 is True.

30. Answer (4)



5/10

PART - B (CHEMISTRY)

31. Answer (4)

4 3 2 2NH NO N O H O

4 2 2 2NH NO N 2H O

4 2 2 7 2 2 3 2(NH ) Cr O N Cr O 4H O

4 2 3 3 2 2(NH ) CO 2NH H O CO

4 4 3 2 2 7 22Mg(NH )PO 2NH Mg P O H O

32. Answer (2)

33. Answer (2)

NO + NO + (NH ) SO + H O 2NH NO + H SO2 4 2 4 2 4 2 2 4

2N + 4H O2 2

Electric2 2 2 Spark

N C H 2HCN

(CN– is a pseudohalide)

34. Answer (1)

1Coagulation Value

Coagulation power

35. Answer (3)

Fe3+ ions coagulate negatively charged blood sol

and seals the cut.

36. Answer (3)

The sols obtained in two cases will be oppositely

charged, so coagulate each other.

37. Answer (1)

34 1000 200N

17 1120 112

Now, volume strength (V) = N × 5.6

= 200

5.6 10 V112

38. Answer (4)

[H+] = 10–3 M

[Ca2+] = 5 × 10–4 M

Weight of CaCO3 = 5 × 10–4 × 1 × 100

= 5 × 10–2 g

So, weight of CaCO3 in 106 ml hard water

=

26

3

5 1010

10

= 50 ppm

39. Answer (2)

RCOOR' + NaOH RCOONa + R'—OH

Rate = k[RCOOR']'[NaOH]'

Hence, molecularity = 2, order = 2

40. Answer (3)

CH4 = non-polar and low molecular weight

SiH4 = non-polar and high molecular weight

NH3 = polar and capable of intermolecular H-bonding

PH3 = polar but incapable of intermolecular H-bonding

41. Answer (1)

Be2C + 4H

2O 2Be(OH)

2 + CH

4

42. Answer (4)

D2O(80) has lower dielectric constant than H

2O(82).

It is due to the reason that ionic salts like NaCl

dissolve more in H2O than in D

2O.

43. Answer (3)

15Se + SeCl4 + 4AlCl

3 2Se

8[AlCl

4]2

Se

ClCl

Cl

Cl

Al

Cl

ClCl

44. Answer (2)

C

OH

C

H

H

C

OH

OOP O4 10

–2H O2O C C C O

(Carbon suboxide)

45. Answer (3)

(BN)x has structure similar to graphite (layer

structure)

46. Answer (2)

Li and H are less dense than kerosene. Other IA

elements are more dense than kerosene.

47. Answer (1)

Due to relative higher I.E. and melting point, as

compared to other alkali metals, lithium, reacts very

slowly with water.


6/10

PART - C (MATHEMATICS)

48. Answer (1)

Triethyl borate; (C2H5)3BO

3 and Tetraethyl silicate;

(C2H5)4SiO

4 are formed due to their diagonal

relationship.

49. Answer (3)

H OO

H OO

O K– +

K+

O–

(structure of KHCO )3

O

H C

ONa

O

H

O

C

ONa

O

H

O

C

O

ONa

Structure of NaHCO3

50. Answer (4)

Minimum weight required to be maintained by 70 kg

man is

= 200 70

140 mg100

2.303 400t log

K 140

51. Answer (3)

B F

F

F

+ B Cl

Cl

Cl

B

F

F

F

Cl

B

Cl

Cl

B Cl + F

F

F

Cl

Cl

B

52. Answer (4)

2 2 2 2pungent smelling gas

CaOCl H O Ca(OH) Cl

2 2(White turbidity)

H S Cl 2HCl S

4 2 4 2 2 4 32FeSO H SO Cl Fe (SO ) 2HCl

53. Answer (2)

If step 2 were RDS

2

3 c

2

[Cl]R k[Cl][CHCl ],K

[Cl ]

1/2 1/2

c 2 3R k (K ) [Cl ] CHCl

1/2

3 2R k '[CHCl ][Cl ]

Step 2 is RDS

54. Answer (3)

Fact.

55. Answer (2)

A(g) 2B(g) 3C(l) 4D(s) Inert gas

t 0 P 0 0 0 i

t 10min (P x) 2x 20 0 i

t 0 2P 20 0 i

P + i = 210

P – x + 2x + 20 + i = 330

Also, 2P + 20 + i = 430

On solving, P = 200, x = 100

Hence, 10 minute must be t1/2

.

56. Answer (2)

Fact.

57. Answer (1)

3

1H has high

n

p ratio.

So, 1 1 0

0 1 1n H

58. Answer (1)

1 1 0

0 1 1n H

59. Answer (4)

Due to high e

p ratio, H– is less stable than He.

60. Answer (2)

Due to availability of vacant d-orbitals on Si, silanes

are more reactive than methane.

61. Answer (2)

S1 x2 + y2 + x + y = 0

1 1,

2 2 g f and c = 0

S2 x2 + y2 + x – y = 0

1 1' , '

2 2

g f and c' = 0

Now

1 1 1 12 ' 2 ' 2 . 2. . 0 '

2 2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

gg ff c c


7/10

Therefore, S = 0 and S' = 0 intersect orthogonally so

that 2

2 tan ( 1)2

y x⎛ ⎞ ⎜ ⎟

⎝ ⎠

1x

62. Answer (4)

The equation of the normal with slope m is

y = mx – 10m – 5m3

The combined equation of the pair of lines

2

3(20 ) 0

10 5

⎛ ⎞ ⎜ ⎟⎝ ⎠

mx yy x

m m

Subtends a right angle at the vertex then coefficient

of x2 + coefficient of y2 = 0

(10m + 5m3)y2 – 20x(mx – y) = 0

3

201 0

10 5

m

m m

m2 + 2 – 4 = 0

2 m

63. Answer (3)

2 23 4 1 4 3 2

5 51

10 2

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x y x y

a2 = 10, b2 = 2

Radius of director circle of axillary circle of

ellipse = 20

64. Answer (1)

Vertex A(a, b) will be mirror image of ex-centre

about x + y – 2 = 0

2 4 2(2 4 2)

41 1 2

a b

a – 2 = 4

a = 6, b + 4 = 4

A(a, b) = A(6, 0)

b = 0

65. Answer (1)

Truth value of (p q) (q p) is always true.

66. Answer (3)

2

2 2 1 3 4 7( 2) ( 3)

4 5

⎛ ⎞ ⎜ ⎟⎝ ⎠

x yx y

Focus (2, –3), directrix 3x – 4y + 7 = 0 and

eccentricity = 1

2

Length of perpendicular from focus to directrix is

3 2 4 ( 3) 75

5

5 aae

e

2 52

aa

10

3a

So length of major axis is 20

3

67. Answer (2)

The line y = ax intersects the lines y = 4 and y = 8

at points 4, 4

⎛ ⎞⎜ ⎟⎝ ⎠

Aa

and 8,8

⎛ ⎞⎜ ⎟⎝ ⎠

Ba

Now AB < 5

2 2

2

165 4

a

21

16 9a

2

169

a

2 160

9 a

i.e. 4

3a and

4

3 a

68. Answer (2)

On solving 1

2

2.4sin

2 1 16

⎛ ⎞ ⎜ ⎟⎝ ⎠

x

x

= 1

2 tan 42

x

1 1

2

2(4 )2 tan 4 sin

1 (4 )

xx

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

∵ 1 1

2

22 tan sin 1

1

yy y

y

As 4x 1

1 1, ,

4 4x x

⎡ ⎞ ⎟⎢⎣ ⎠


8/10

69. Answer (3)

The given lines are

x2 – (y – 1)2 = 0

(x + y – 1)(x – y + 1) = 0

Therefore, the given lines are x – y + 1 = 0 and

x + y – 1 = 0

Whose angle bisectors are

1 1

2 2

x y x y

That is y – 1 = 0 and x = 0, thus the vertices of the

triangle are (0, 1) (0, 5) (4, 1)

14 4

2 = 8 sq. units

70. Answer (2)

The ends of a latus rectum are

2

' ,

bL ae

a

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

,

⎛ ⎞ ⎜ ⎟⎝ ⎠

bL ae

a

Since LL' subtends right angle at the centre then

2 2

4

2 2 2. 1

( )

b b

ba a

ae ae a a e

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

2 2

2

( 1)1

e

e

(e2 – 1)2 = +e2

e4 – 3e2 + 1 = 0

2 3 5

2

e

5 1

2

e

71. Answer (3)

(1)

2 2 2 21 1 1

14

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

x yt t

a b t t

(2)2 2 2 2( ) ( ) 4 t t t t

x y e e e e

(4) x2 – 6 = 2 cos t

y2 + 2 = 2(1 + cos t)

x2 – 6 = y2

x2 – y2 = 6

72. Answer (1)

2 2

2 2

1 1tan

1 1

x x

x x

By componendo and dividendo

2

2

2 1 cos sin

cos sin2 1

x

x

22

2

1 cos sin

cos sin1

⎛ ⎞ ⎜ ⎟ ⎝ ⎠x

x

= 1 sin 2

1 sin 2

⎛ ⎞⎜ ⎟ ⎝ ⎠

2sin 2x

73. Answer (1)

2 3 2 2

1 11

1 (1 )

x

x x x x

So,

2 31 1

2 3

2 2

1

1 1tan tan tan11 1

(1 )

x

x x x

x x

x x

⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠

=

4 2

1

2 4

1tan

( 1)

x x

x x x

= 1 1

tan ⎛ ⎞ ⎜ ⎟⎝ ⎠x

= 3

4

x = 1

74. Answer (1)

By hypothesis, the given lines are parallel

2 2

2

2

( 1)( 1)

1

tt t

t

t = –1


9/10

75. Answer (3)

Centre of the circle is (2, 4) and the radius is 5. The

line meets the circle in two distinct points this

implies that

2 2

| 3(2) 4(4) |5

3 4

k

|10 + k| < 25

–25 < 10 + k < 25

–35 < k < 15

76. Answer (4)

Let 2

y xX

2

y xY

2 2

116 4

X Y

a2 = 165

4e

b2 = 4 X = ±ae and Y = 0

b2 = a2(e2 – 1)5

422

x y

241

16e 0,

2

y xx y

25

4e 2 10x y

2 2 10x

10x

77. Answer (4)

From geometry OA1OA

2 = OB

1.OB

2

then 1 2

1 2

. 1

OB OB

OA OA

m1m2 = 1

78. Answer (3)

sin–1 sin 8 = sin–1 sin(3 – 8) = 3 – 8 = –t

tan–1tan 8 = tan–1tan(8 – 3) = 8 – 3 = t

Then f(t) = 3 1

2t t

3 1( )

2f t t t

f(t) + f(–t) = 1

Then f(sin–1 sin 8) = 1 – a

79. Answer (4)

Let x2 + y2 + 2gx + 2fy + c = 0 ... (i)

y2 = 4ax ... (ii)

Solving (i) and (ii), we get

22 2

22 . 2 0

4 4

y yy g fy c

a a

⎛ ⎞ ⎜ ⎟

⎝ ⎠

4

2

2

21 2 0

4(4 )

y gy fy c

aa

⎛ ⎞ ⎜ ⎟⎝ ⎠

y1 + y

2 + y

3 + y

4 = 0

80. Answer (2)

Putting x = at2 in the equation of the ellipse, we get

2 4 2

2 21

a t y

a b , y2 = b2(1 – t4)

= b2(1 – t2) (1 + t2)

This will give real value of y if

1 – t2 0

|t| 1

81. Answer (2)

Let m1 and m

2 be the slope of lines of first pair

m1 and

2

1

m

be the slope of lines of second

pair.

Comparing pair of lines

(y – m1x) (y – m

2x) with y2 + 2hxy – 9x2 = 0

and pair of lines

(y – m1x)

2

1y x

m

⎛ ⎞⎜ ⎟⎝ ⎠

with ay2 + 10xy + x2 = 0

We get m1 = –1 and m

2 = 9

a = 9

h = –4


10/10

82. Answer (4)

2(3 1) 2x x

|3x + 1| < 2 – x

–(2 – x) < 3x + 1 < 2 – x

–2 + x < 3x + 1 3x + 1 < 2 – x

–3 < 2x 4x < 1

3

2x

1

4x

83. Answer (4)

Equation of tangent is y = 2 2

2 4x a b

This is normal to the circle x2 + y2 + 4x + 1 = 0

This tangent passes through (–2, 0)

2 20 4 4a b

4a2 + b2 = 16

Using AM GM, we get

2 2

2 244

2

a ba b

4ab

84. Answer (3)

Vertices of triangle are (0, 0), 1

2, 2

m

⎛ ⎞⎜ ⎟⎝ ⎠

, 2

2,2

m

⎛ ⎞⎜ ⎟⎝ ⎠

Area = 1 2

1 2

2m m

m m

1 2

( 3 2)

1m m

1 23 1m m

2

1 2 1 2 1 2( ) ( ) 4m m m m m m

= 2( 3 2) 4( 3 1)

= 11 4 3 4 3 11

� � �

Area = 2 11 3 1

11( 3 1)3 1 3 1

= 33 11

85. Answer (2)

1

1

4( ) ( cot ( )

4 cot ( )f x x

x

=

1

1

4 cot ( )4 4 2 2

4 cot ( )

x

x

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠

Equality when 1

cot ( )4

x

1x

86. Answer (3)

Statement-2 is not always true for two parallel line.

87. Answer (1)

Using diametrical form of circle with A(x1, y

1) and

B(x2, y

2) as diameter.

88. Answer (3)

Length of semi-latus rectum

= 1 2

1 2

2l l

l l

= Harmonic mean of l1 and l

2

89. Answer (3)

First law is De-Morgan's law

Statement II neither tautology nor contradiction.

90. Answer (4)

cos–1 x1 + cos–1 x

2 + .......... + cos–1 x

n = n

cos–1 x1 = cos–1 x

2 = ...... = cos–1 x

n =

x1 = x

2 = x

3 .... = x

n = –1

Then sin–1 x1 + sin–1 x

2 ..... + sin–1 x

n =

2

n

Statement-1 is not correct and statement-2 is correct

from definition.

Test - 4 (Paper-I) (Code-B) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2015

1/10

1. (4)

2. (2)

3. (2)

4. (1)

5. (4)

6. (2)

7. (4)

8. (2)

9. (1)

10. (1)

11. (2)

12. (2)

13. (2)

14. (2)

15. (3)

16. (4)

17. (2)

18. (1)

19. (2)

20. (2)

21. (3)

22. (4)

23. (3)

24. (4)

25. (2)

26. (4)

27. (4)

28. (2)

29. (1)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

31. (2)

32. (3)

33. (2)

34. (4)

35. (3)

36. (4)

37. (3)

38. (1)

39. (1)

40. (2)

41. (3)

42. (2)

43. (3)

44. (4)

45. (1)

46. (3)

47. (2)

48. (4)

49. (1)

50. (3)

51. (3)

52. (1)

53. (2)

54. (2)

55. (4)

56. (2)

57. (4)

58. (1)

59. (1)

60. (2)

61. (2)

62. (3)

63. (4)

64. (4)

65. (2)

66. (2)

67. (4)

68. (3)

69. (4)

70. (4)

71. (3)

72. (1)

73. (1)

74. (1)

75. (3)

76. (2)

77. (3)

78. (2)

79. (2)

80. (3)

81. (1)

82. (1)

83. (3)

84. (4)

85. (2)

86. (4)

87. (3)

88. (3)

89. (1)

90. (3)

ANSWERS

TEST - 4 (Paper-I) - Code-B

All India Aakash Test Series for JEE (Main)-2015

Test Date : 11-01-2015

All India Aakash Test Series for JEE (Main)-2015 Test - 4 (Paper-I) (Code-B) (Answers & Hints)

2/10

1. Answer (4)

By the law of calorimetry

Q1 + Q

2 = Q

3 + Q

4 + Q

5

M3Cw(T2 – T) + M

2Cb(T2 – T)

= M1Ci(0 – T

1) + M

1Lv + M

1Cw (T – 0)

1 1 1 3 2 2 2

3 2 1

( )i v w b

w b w

M C T M L M C T M C TT

M C M C M C

= 10.9°C

2. Answer (2)

The process 3 1 is an isochoric process. So

w = 0

0 0

1 3

0 0

23( )

2v

P M P MU nC T T

⎡ ⎤ ⎢ ⎥ ⎣ ⎦ =

0

0

3

2

P M

P

0

0

3

2

P MQ

3. Answer (2)

I =

2

3

ma and

2

al

22 2

3

I aT

mgl g

4. Answer (1)

P0

l0

P0

l0

h0

l0 = 45 cm. If the mercury displaces by x, then

l0P0 = (l

0 + x)P

1

l0P0 = (l

0 – x)P

2

and P2 = P

1 + h

0

0 0 0 0

0

0

l P l Ph

l x l x

x = 4.5 cm

5. Answer (4)

4 4

A A B BT T

1/ 4

. 1934 K

A

B A

B

T T⎛ ⎞ ⎜ ⎟⎝ ⎠

A A B BT T

3

BA

But 6

10B A

61.5 10 1.5 m

B

6. Answer (2)

i f

i f

v v

T T i f

i f

h h

T T

f

f i

i

Th h

T

⎛ ⎞ ⎜ ⎟⎝ ⎠

3734.00 5.09 cm

293f

h⎛ ⎞ ⎜ ⎟⎝ ⎠

7. Answer (4)

The frequency of sound heard by motorist from

reflection

1.

m

b

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠... (i)

The frequency of sound heard by motorist directly

2

m

b

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠... (ii)

Beat frequency = 1 2 2 2

2 ( ).

( )

m

b

b v vf f f

v v

8. Answer (2)

1 129.610

2Lf

... (i)

and 1 160

2Lf

... (ii)

f = 100 Hz

9. Answer (1)

m H

H m

V

V

... (1)

where 0 0

0

H H

m

H

V V

V V

0

0

0

.

4

H

m H

H H

V V

V V

PART - A (PHYSICS)


3/10

1

4

H

m

11200 600 m/s

2

H

m H

m

V V

10. Answer (1)

11. Answer (2)

l y

A Bx

2y

l

Suppose 2y is the distance between two rows of the

buildings. The distance travelled by car in 1 second,

530 1

18x

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 150

m8

The distance travelled by sound in 1 second

2l = 330 × 1 = 330 m

l = 165 m

From the figure

2

2165 m

2

xy l

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

12. Answer (2)

0 0176 165

22

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

0 = 22 m/s

13. Answer (2)

0.66 mv

f

The resonance length are

l1 = 0.165 m

4

2

30.495 m

4l

3

50.825 m

4l

4

71.155 m

4l

As 4 is greater than 1 m, so allowed resonance

are only 3.

14. Answer (2)

15. Answer (3)

At t = 0, y = 2

10( )

5f x

x

Velocity of pulse v = 2 m/s along positive x-axis.

Hence wave function y(x, t) = f(x – vt)

2

10( , )

5 ( 2 )y x t

x t

16. Answer (4)

If mass falls down by x,

so spring extends by 4x.

(4 )4

Tk x

m

4

T

4

T

2

T

2

T

T

T

T = (16k)x

1 16

2

kf

m

2 kf

m

17. Answer (2)

x1 = –A cos t and x

2 = A sin t

B O A

A sin t = –A cos t

sin t + cos t = 0

1 1

sin cos 02 2

t t

cos 04

t⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot4 2

3

8

Tt

18. Answer (1)

x = A cos t

So A – = A cos ... (1)

cosA

A


4/10

and A – ( + ) = A cos 2 ... (2)

( )

cos 2A

A

2

2

3A

19. Answer (2)

2P0

P0

P 2T0

4T0

CB

A

V

T0

V0

2V0

Total heat absorbed by 1 mole of gas

Q = Cv(2T

0 – T

0) + C

P(4T

0 – 2T

0) = 0

19

2RT

Total change in temperature form state A to C is

T = 3T0

Molar heat capacity = 19

6

QR

T

20. Answer (2)

Q = dU + W

For isobaric process W = nRT

and dU = nCVT

32

vCU f

W R

dU = 3W = 75 J

Q = 75 + 25 = 100 J

21. Answer (3)

At t = 0, x1 = 0

At t = 6

, 2sin

6 2

Ax A

⎛ ⎞ ⎜ ⎟⎝ ⎠

<V> = 32

6

A

x A

t

⎛ ⎞

⎜ ⎟⎝ ⎠

22. Answer (4)

Work done from B to C = nR(TC – T

B)

= 6R(2200 – 800)

= 6R × 1400

Work done from D to A = nR(TA – T

D)

= 6R(600 – 1200)

= –6R × 600

While WAB

= WCD

= 0

W = 6R × 1400 – 6R × 600

= 6R × 800

= 40,000 J

23. Answer (3)

Q = dU + W

nCdT = nCvdT + W

( )v

W n C C dT ∫ ∫

0

0

( )

T

v

T

W C C dT

∫ for n = 1

=

0

0

T

v

T

aC dT

T

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 0

1ln

1a RT

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

24. Answer (4)

WABC

= 0 0 0 0 0 0 0

1(2 )(3 ) (3 )

2P P V V P V V

= 300 J

UABC

= QABC

– WAB

= 300 J = UAC

WAC

= P0(3V

0 – V

0) = 200 J

QAC

= WAC

+ UAC

= 500 J

25. Answer (2)

26. Answer (4)


27. Answer (4)


28. Answer (2)

Statement-1 is True, Statement-2 is True; Statement-

2 is NOT a correct explanation for Statement-1.

29. Answer (1)



30. Answer (1)




5/10

PART - B (CHEMISTRY)

31. Answer (2)

A(g) 2B(g) 3C(l) 4D(s) Inert gas

t 0 P 0 0 0 i

t 10min (P x) 2x 20 0 i

t 0 2P 20 0 i

P + i = 210

P – x + 2x + 20 + i = 330

Also, 2P + 20 + i = 430

On solving, P = 200, x = 100

Hence, 10 minute must be t1/2

.

32. Answer (3)

Fact.

33. Answer (2)

If step 2 were RDS

2

3 c

2

[Cl]R k[Cl][CHCl ],K

[Cl ]

1/2 1/2

c 2 3R k (K ) [Cl ] CHCl

1/2

3 2R k '[CHCl ][Cl ]

Step 2 is RDS

34. Answer (4)

2 2 2 2pungent smelling gas

CaOCl H O Ca(OH) Cl

2 2(White turbidity)

H S Cl 2HCl S

4 2 4 2 2 4 32FeSO H SO Cl Fe (SO ) 2HCl

35. Answer (3)

B F

F

F

+ B Cl

Cl

Cl

B

F

F

F

Cl

B

Cl

Cl

B Cl + F

F

F

Cl

Cl

B

36. Answer (4)

Minimum weight required to be maintained by 70 kg

man is

= 200 70

140 mg100

2.303 400t log

K 140

37. Answer (3)

H OO

H OO

O K– +

K+

O–

(structure of KHCO )3

O

H C

ONa

O

H

O

C

ONa

O

H

O

C

O

ONa

Structure of NaHCO3

38. Answer (1)

Triethyl borate; (C2H5)3BO

3 and Tetraethyl silicate;

(C2H5)4SiO

4 are formed due to their diagonal

relationship.

39. Answer (1)

Due to relative higher I.E. and melting point, as

compared to other alkali metals, lithium, reacts very

slowly with water.

40. Answer (2)

Li and H are less dense than kerosene. Other IA

elements are more dense than kerosene.

41. Answer (3)

(BN)x has structure similar to graphite (layer

structure)

42. Answer (2)

C

OH

C

H

H

C

OH

OOP O4 10

–2H O2O C C C O

(Carbon suboxide)

43. Answer (3)

15Se + SeCl4 + 4AlCl

3 2Se

8[AlCl

4]2

Se

ClCl

Cl

Cl

Al

Cl

ClCl

44. Answer (4)

D2O(80) has lower dielectric constant than H

2O(82).

It is due to the reason that ionic salts like NaCl

dissolve more in H2O than in D

2O.

45. Answer (1)

Be2C + 4H

2O 2Be(OH)

2 + CH

4


6/10

PART - C (MATHEMATICS)

46. Answer (3)

CH4 = non-polar and low molecular weight

SiH4 = non-polar and high molecular weight

NH3 = polar and capable of intermolecular H-bonding

PH3 = polar but incapable of intermolecular H-bonding

47. Answer (2)

RCOOR' + NaOH RCOONa + R'—OH

Rate = k[RCOOR']'[NaOH]'

Hence, molecularity = 2, order = 2

48. Answer (4)

[H+] = 10–3 M

[Ca2+] = 5 × 10–4 M

Weight of CaCO3 = 5 × 10–4 × 1 × 100

= 5 × 10–2 g

So, weight of CaCO3 in 106 ml hard water

=

26

3

5 1010

10

= 50 ppm

49. Answer (1)

34 1000 200N

17 1120 112

Now, volume strength (V) = N × 5.6

= 200

5.6 10 V112

50. Answer (3)

The sols obtained in two cases will be oppositely

charged, so coagulate each other.

51. Answer (3)

Fe3+ ions coagulate negatively charged blood sol

and seals the cut.

52. Answer (1)

1Coagulation Value

Coagulation power

53. Answer (2)

NO + NO + (NH ) SO + H O 2NH NO + H SO2 4 2 4 2 4 2 2 4

2N + 4H O2 2

Electric2 2 2 Spark

N C H 2HCN

(CN– is a pseudohalide)

54. Answer (2)

55. Answer (4)

4 3 2 2NH NO N O H O

4 2 2 2NH NO N 2H O

4 2 2 7 2 2 3 2(NH ) Cr O N Cr O 4H O

4 2 3 3 2 2(NH ) CO 2NH H O CO

4 4 3 2 2 7 22Mg(NH )PO 2NH Mg P O H O

56. Answer (2)

Due to availability of vacant d-orbitals on Si, silanes

are more reactive than methane.

57. Answer (4)

Due to high e

p ratio, H– is less stable than He.

58. Answer (1)

1 1 0

0 1 1n H

59. Answer (1)

3

1H has high

n

p ratio.

So, 1 1 0

0 1 1n H

60. Answer (2)

Fact.

61. Answer (2)

1

1

4( ) ( cot ( )

4 cot ( )f x x

x

=

1

1

4 cot ( )4 4 2 2

4 cot ( )

x

x

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠

Equality when 1

cot ( )4

x

1x


7/10

62. Answer (3)

Vertices of triangle are (0, 0), 1

2,2

m

⎛ ⎞⎜ ⎟⎝ ⎠

, 2

2,2

m

⎛ ⎞⎜ ⎟⎝ ⎠

Area = 1 2

1 2

2m m

m m

1 2

( 3 2)

1m m

1 23 1m m

2

1 2 1 2 1 2( ) ( ) 4m m m m m m

= 2( 3 2) 4( 3 1)

= 11 4 3 4 3 11

Area = 2 11 3 1

11( 3 1)3 1 3 1

= 33 11

63. Answer (4)

Equation of tangent is y = 2 2

2 4x a b

This is normal to the circle x2 + y2 + 4x + 1 = 0

This tangent passes through (–2, 0)

2 20 4 4a b

4a2 + b2 = 16

Using AM GM, we get

2 2

2 244

2

a ba b

4ab

64. Answer (4)

2(3 1) 2x x

|3x + 1| < 2 – x

–(2 – x) < 3x + 1 < 2 – x

–2 + x < 3x + 1 3x + 1 < 2 – x

–3 < 2x 4x < 1

3

2x

1

4x

65. Answer (2)

Let m1 and m

2 be the slope of lines of first pair

m1 and

2

1

m

be the slope of lines of second

pair.

Comparing pair of lines

(y – m1x) (y – m

2x) with y2 + 2hxy – 9x2 = 0

and pair of lines

(y – m1x)

2

1y x

m

⎛ ⎞⎜ ⎟⎝ ⎠

with ay2 + 10xy + x2 = 0

We get m1 = –1 and m

2 = 9

a = 9

h = –4

66. Answer (2)

Putting x = at2 in the equation of the ellipse, we get

2 4 2

2 21

a t y

a b , y2 = b2(1 – t4)

= b2(1 – t2) (1 + t2)

This will give real value of y if

1 – t2 0

|t| 1

67. Answer (4)

Let x2 + y2 + 2gx + 2fy + c = 0 ... (i)

y2 = 4ax ... (ii)

Solving (i) and (ii), we get

22 2

22 . 2 0

4 4

y yy g fy c

a a

⎛ ⎞ ⎜ ⎟

⎝ ⎠

4

2

2

21 2 0

4(4 )

y gy fy c

aa

⎛ ⎞ ⎜ ⎟⎝ ⎠

y1 + y

2 + y

3 + y

4 = 0

68. Answer (3)

sin–1 sin 8 = sin–1 sin(3 – 8) = 3 – 8 = –t

tan–1tan 8 = tan–1tan(8 – 3) = 8 – 3 = t

Then f(t) = 3 1

2t t

3 1( )

2f t t t

f(t) + f(–t) = 1

Then f(sin–1 sin 8) = 1 – a


8/10

69. Answer (4)

From geometry OA1OA

2 = OB

1.OB

2

then 1 2

1 2

. 1

OB OB

OA OA

m1m2 = 1

70. Answer (4)

Let 2

y xX

2

y xY

2 2

116 4

X Y

a2 = 165

4e

b2 = 4 X = ±ae and Y = 0

b2 = a2(e2 – 1)5

422

x y

241

16e 0,

2

y xx y

25

4e 2 10x y

2 2 10x

10x

71. Answer (3)

Centre of the circle is (2, 4) and the radius is 5. The

line meets the circle in two distinct points this

implies that

2 2

| 3(2) 4(4) |5

3 4

k

|10 + k| < 25

–25 < 10 + k < 25

–35 < k < 15

72. Answer (1)

By hypothesis, the given lines are parallel

2 2

2

2

( 1)( 1)

1

tt t

t

t = –1

73. Answer (1)

2 3 2 2

1 11

1 (1 )

x

x x x x

So,

2 31 1

2 3

2 2

1

1 1tan tan tan11 1

(1 )

x

x x x

x x

x x

⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠

=

4 2

1

2 4

1tan

( 1)

x x

x x x

= 1 1

tan ⎛ ⎞ ⎜ ⎟⎝ ⎠x

= 3

4

x = 1

74. Answer (1)

2 2

2 2

1 1tan

1 1

x x

x x

By componendo and dividendo

2

2

2 1 cos sin

cos sin2 1

x

x

22

2

1 cos sin

cos sin1

⎛ ⎞ ⎜ ⎟ ⎝ ⎠x

x

= 1 sin 2

1 sin 2

⎛ ⎞⎜ ⎟ ⎝ ⎠

2sin 2x

75. Answer (3)

(1)

2 2 2 21 1 1

14

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

x yt t

a b t t

(2)2 2 2 2( ) ( ) 4 t t t t

x y e e e e

(4) x2 – 6 = 2 cos t

y2 + 2 = 2(1 + cos t)

x2 – 6 = y2

x2 – y2 = 6


9/10

76. Answer (2)

The ends of a latus rectum are

2

' ,

bL ae

a

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

,

⎛ ⎞ ⎜ ⎟⎝ ⎠

bL ae

a

Since LL' subtends right angle at the centre then

2 2

4

2 2 2. 1

( )

b b

ba a

ae ae a a e

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

2 2

2

( 1)1

e

e

(e2 – 1)2 = +e2

e4 – 3e2 + 1 = 0

2 3 5

2

e

5 1

2

e

77. Answer (3)

The given lines are

x2 – (y – 1)2 = 0

(x + y – 1)(x – y + 1) = 0

Therefore, the given lines are x – y + 1 = 0 and

x + y – 1 = 0

Whose angle bisectors are

1 1

2 2

x y x y

That is y – 1 = 0 and x = 0, thus the vertices of the

triangle are (0, 1) (0, 5) (4, 1)

14 4

2 = 8 sq. units

78. Answer (2)

On solving 1

2

2.4sin

2 1 16

⎛ ⎞ ⎜ ⎟⎝ ⎠

x

x

= 1

2 tan 42

x

1 1

2

2(4 )2 tan 4 sin

1 (4 )

xx

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

∵ 1 1

2

22 tan sin 1

1

yy y

y

As 4x 1

1 1, ,

4 4x x

⎡ ⎞ ⎟⎢⎣ ⎠

79. Answer (2)

The line y = ax intersects the lines y = 4 and y = 8

at points 4, 4

⎛ ⎞⎜ ⎟⎝ ⎠

Aa

and 8,8

⎛ ⎞⎜ ⎟⎝ ⎠

Ba

Now AB < 5

2 2

2

165 4

a

21

16 9a

2

169

a

2 160

9 a

i.e. 4

3a and

4

3 a

80. Answer (3)

2

2 2 1 3 4 7( 2) ( 3)

4 5

⎛ ⎞ ⎜ ⎟⎝ ⎠

x yx y

Focus (2, –3), directrix 3x – 4y + 7 = 0 and

eccentricity = 1

2

Length of perpendicular from focus to directrix is

3 2 4 ( 3) 75

5

5 aae

e

2 52

aa

10

3a

So length of major axis is 20

3

81. Answer (1)

Truth value of (p q) (q p) is always true.

82. Answer (1)

Vertex A(a, b) will be mirror image of ex-centre

about x + y – 2 = 0


10/10

2 4 2(2 4 2)

41 1 2

a b

a – 2 = 4

a = 6, b + 4 = 4

A(a, b) = A(6, 0)

b = 0

83. Answer (3)

2 23 4 1 4 3 2

5 51

10 2

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x y x y

a2 = 10, b2 = 2

Radius of director circle of axillary circle of

ellipse = 20

84. Answer (4)

The equation of the normal with slope m is

y = mx – 10m – 5m3

The combined equation of the pair of lines

2

3(20 ) 0

10 5

⎛ ⎞ ⎜ ⎟⎝ ⎠

mx yy x

m m

Subtends a right angle at the vertex then coefficient

of x2 + coefficient of y2 = 0

(10m + 5m3)y2 – 20x(mx – y) = 0

3

201 0

10 5

m

m m

m2 + 2 – 4 = 0

2 m

85. Answer (2)

S1 x2 + y2 + x + y = 0

1 1,

2 2 g f and c = 0

S2 x2 + y2 + x – y = 0

� � �

1 1' , '

2 2

g f and c' = 0

Now

1 1 1 12 ' 2 ' 2 . 2. . 0 '

2 2 2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

gg ff c c

Therefore, S = 0 and S' = 0 intersect orthogonally so

that 2

2 tan ( 1)2

y x⎛ ⎞ ⎜ ⎟

⎝ ⎠

1x

86. Answer (4)

cos–1 x1 + cos–1 x

2 + .......... + cos–1 x

n = n

cos–1 x1 = cos–1 x

2 = ...... = cos–1 x

n =

x1 = x

2 = x

3 .... = x

n = –1

Then sin–1 x1 + sin–1 x

2 ..... + sin–1 x

n =

2

n

Statement-1 is not correct and statement-2 is correct

from definition.

87. Answer (3)

First law is De-Morgan's law

Statement II neither tautology nor contradiction.

88. Answer (3)

Length of semi-latus rectum

= 1 2

1 2

2l l

l l

= Harmonic mean of l1 and l

2

89. Answer (1)

Using diametrical form of circle with A(x1, y

1) and

B(x2, y

2) as diameter.

90. Answer (3)

Statement-2 is not always true for two parallel line.

aiats jee main 2015 test-4

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