Solutions AIATS JEE(Main)-2016 Test-11 (Code-A & B)

8/18/2019 Solutions AIATS JEE(Main)-2016 Test-11 (Code-A & B)

1/21

Test - 11 (Code-A) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2016

1/11

1. (2)

2. (2)

3. (4)

4. (4)

5. (3)

6. (3)

7. (3)

8. (4)

9. (3)

10. (2)

11. (3)

12. (2)

13. (2)

14. (4)

15. (3)

16. (2)

17. (3)

18. (2)

19. (2)

20. (1)

21. (1)

22. (3)

23. (1)

24. (1)

25. (2)

26. (1)

27. (3)

28. (3)

29. (3)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (4)

33. (3)

34. (3)

35. (1)

36. (4)

37. (3)

38. (3)

39. (2)

40. (2)

41. (4)

42. (4)

43. (4)

44. (2)

45. (4)

46. (3)

47. (1)

48. (1)

49. (4)

50. (3)

51. (3)

52. (2)

53. (4)

54. (4)

55. (4)

56. (3)

57. (3)

58. (1)

59. (2)

60. (1)

61. (3)

62. (3)

63. (1)

64. (3)

65. (4)

66. (2)

67. (2)

68. (4)

69. (3)

70. (4)

71. (1)

72. (3)

73. (4)

74. (4)

75. (3)

76. (1)

77. (4)

78. (3)

79. (4)

80. (1)

81. (1)

82. (1)

83. (2)

84. (3)

85. (2)

86. (1)

87. (2)

88. (4)

89. (1)

90. (1)

Test Date : 30/03/2016

ANSWERS

TEST - 11 (Code-A)

All India Aakash Test Series for JEE (Main)-2016

click here for code-B solution


2/21

All India Aakash Test Series for JEE (Main)-2016 Test - 11 (Code-A) (Answers & Hints)

2/11

1. Answer (2)

30°

v 1

v 12

v 12

–v 2

1

2

tan30 v

v

1

2

1

3

v

v

2. Answer (2)

sin2

F

sin2

F

/2 /2

cos

2F

cos

2F

2TR

F F

2 2 sin

2TR F

2 2 2TR F

F = 2TR

FL FR Y

A L A R

22

(2 ).FR R TR

R TR AY AY AY

3. Answer (4)

4. Answer (4)

L

R is time constant. RC is also time constant.

So,/L R

RC is dimensionless.

PART - A (PHYSICS)

5. Answer (3)

v x = by

x dv dy

bdt dt

ax = bv

0

So, 201

( )2

x bv t and y = v 0t

2

0

0

1( )

2

y x bv

v

2

02

by x

v

6. Answer (3)

7. Answer (3)

22

4

Fx

I Mr Mx

For maximum

1 0d

dx

2

2 22 0

4

r x x

2

r

x

8. Answer (4)

B . lvt = constant

C

Blvt

1Bt

9. Answer (3)

2

20

2 2 3/ 2 .

2 ( ) A

i R r

R x

Ad A

edt

=2 2 2 5/20 3 ( ) (2 )

2 2

i R R x x


3/21


3/11

For max value of e A, 0 Ade

dx

2 2 5/2

0( )

d x

dx R x

2 2 5/2 2 2 3/25

( ) ( ) .2 02

x R x R x x

2

R x

10. Answer (2)

11. Answer (3)

(real)(app.)

1

y

y

v

v

v x

(app.) = v x (real)

(app) 4 4 3

tan tan 1(app) 3 3 4

y

x

v

v

12. Answer (2)

steel

2brass

3

3 3. .

2 2 2

s s s b b

b s s

b b

mg ls

l A Y l A Y a

mg l l A Y b c lb A Y

13. Answer (2)

∫ ∫ 3/2 2

0 3/ 2cos cos

3 3S dt dt

=

3 3

2

14. Answer (4)

100 100 2. 100dg dl dT

g l T

=

0.1 2 0.1

100 1.2%

50 2.00

15. Answer (3)

+100 –q

–100 +q

–50 +q

+50 –q

6 6

100 500

5 10 20 10

q q

q = 90 × 10–6 C

Final charge on 5 F top plate is 10 C.

16. Answer (2)

N 1

1 = N

2

2

1 2

2 1

3

2

N

N

17. Answer (3)

I 1

I 2

2 23 1 2I I I

So, 3 1 2( )I I I

18. Answer (2)

DY X

C

10

10

So, R = 5 19. Answer (2)

20. Answer (1)

Electric field 0v dv

E dr r

Centripetal force 2

mv Ee

r

2

0.

v mv e

r r

0v ev m

Now mvr =2

nh

2

nhr

mv

r n


4/21


4/11

21. Answer (1)

22. Answer (3)

In circuit A, both the diodes are forward biased, so

effective resistance is 2 and current 8

4 A4

I

In circuit B, diodes is reverse biased, so effective

resistance is 4 and current 8

2 A4

I

23. Answer (1)

16 mV 8 mV

max16

8mV2

E

min8

4 mV2

E

max min

max min

8 4 1

8 4 3

E E m

E E

24. Answer (1)

0 0

1 1

r r r r

ev

2

2 2.25

r

r

e

v

25. Answer (2)

B = 0(H + I )

0 0

B H I H H

I = (r – 1

)H = (0 – 1

)ni = (1000 – 1) × 500 × 0.5

= 2.5 × 105 A/m

Magnetic moment = I × V = 2.5 × 105 × 10–4

= 2.5 A-m2

26. Answer (1)

27. Answer (3)

28. Answer (3)

29. Answer (3)

30. Answer (1)

PART - B (CHEMISTRY)

31. Answer (4)

At 27°C, 1 × V = nHe × R × 300

nHe

=V

300R

At 127°C, 2 × V = (nHe

+ nP) × R × 400

He PV

n n200R

nP =

V

600R

At 327°C, V V

P V R 6003 00R 30 0R

P = 4 atm

32. Answer (4)

C(1 ) C CHA H A

2

a

CK (1 )

aK (1 )C

a

1[H ] K

a

1log(H ) log K log

a

1pH pK log

apK pH1

10

a

pK pH

1

1 10


5/21


5/11

33. Answer (3)

15Te + TeCl4 + 4AlCl

3 2Te

8[AlCl

4]2

Te

Cl

Cl

Cl

Cl

Al

Cl

Cl

Cl

34. Answer (3)

35. Answer (1)

Boron and silicon show diagonal relationship, hence

similar chemical behaviour.

36. Answer (4)

1 a

xpH pK log

y

2 a

ypH pK log

x

x

3.17y

37. Answer (3)

1 0

1 2

X 2Y

1 0 0

1

Z P Q

1

21

p 2

4 PK

(1 )

2

22

p 2

PK

(1 )

1

2

p 1

p 2

K 4P 1

K P 3

1

2

P 1

P 12

38. Answer (3)

In metal carbonyls with metal as anion, better back

donation of electron is observed from metal to

carbonyl, followed by decrease in metal-carbon bond

length.

39. Answer (2)

partial2 2 2 2conversion

3Cu S O Cu O SO

2

2 2 2Cu S 2Cu O 6Cu SO

(Cu2S is taken in 2 : 1 ratio)

(Two parts in first step and one part in second step)

40. Answer (2)

3 3KI AgNO KNO AgI

Initial: 3 m.moles 2 m.moles 0 0

Final : 1 0 2 2

KI in solution = 1 M40

KNO3 in solution =

2M

40

f

1 2T 1.86 2 1.86 2 0.28

40 40

41. Answer (4)

3 2 2 23CH OH( ) O (g) CO (g) 3H O( )2

G° = –394.4 + 2 × (–237.2) – (–166.2)

= –702.6 kJ mol–1

Efficiency of fuel cell =G

100 97%H

42. Answer (4)

43. Answer (4)

44. Answer (2)

Adsorbed moles of

3

2

0. 03 2.46 10H

0.0821 300

= 3 × 10–6

Number of adsorbed molecules of

H2 = 3 × 10–6 × 6 × 10–23

= 18 × 1017

Total number of surface sites available

= 6 × 1 015 × 1000 = 6 × 1018

Number of surface sites that is occupied by adsorption

of H2 gas =

18 17106 10 6 10

100


6/21


6/11

Number of surface sites occupied by one molecule

of H2

=

17

17

18 10

6 10

= 3

45. Answer (4)

46. Answer (3)

2 12 2

2 2 2 6 2 6 6 0Fe : 1 , 2 , 2 , 3 , 3 , 3 , 4s s p s p d s

47. Answer (1)

Anthracene undergoes addition reactions at 9 and 10

positions because resonance energy per mole of ring

increases.

48. Answer (1)

49. Answer (4)

Q is CH C CH3 3— —

O

, it gives iodoform test.

50. Answer (3)

51. Answer (3)

52. Answer (2)

(i) O

(ii)

3

H O

(iii)2 2

CH COOH2—

+ CO2

(i) O

(ii)

3

H O

(iii)2 2

C O

CH3

+ CO2

(i) O(ii)

3

H O

(iii)2 2

COOH

O

CH3

(i) O

(ii)

3

H O

(iii)2 2

O

O

O

53. Answer (4)

alc.KOHH

H

Ph

HBr

HHO–

H

Ph

3-phenylcyclopentene

54. Answer (4)

o, o'-benzene losses planarity due to steric reason

(bulky group at ortho position) so, the compound does

not have centre of symmetry.

55. Answer (4)

H + Br

NBS/h

Br +

Br

56. Answer (3)

N O (l)2 4

[NO ] [NO ]+ –3

NOCl + NaNO3

⇒

NaCl

57. Answer (3)

O

NH2

(Methyl ion)glycinate

H C CH3

— — —C O

58. Answer (1)

59. Answer (2)

60. Answer (1)


7/21


7/11

PART - C (MATHEMATICS)

61. Answer (3)

D

A

B

P (4, 5)

4

O(2, 1)

Centre (2, 1)

2 2(4 2) (5 1)OP

4 16 20 2 5

4sin

2 5

OD

AO

8

5OD

64

165

AD

4

5

8

5 AB

62. Answer (3)

20

3 20 313 3

1 11 1 x C x

x x

2 20

20 3 20 32 203 3

1 1.....C x C x

x x

Therefore,

2 20

3 3 2 3 3 3 20

3 3 3

1 1 11. .( ) .( ) ....( ) . . ..... x x x x

x x x

All distinct whose number is 1 + 20 + 20 = 41

63. Answer (1)

After solving determinate there are six degree of

polynomial, sum of all roots = 0.

i.e. coefficient of x 5 = 0

Then sum of other five roots = –3

64. Answer (3)

After taking common 1, , 2

we get,

2 4 3 4 3 2(4 3)(3 3 2 ) 1 0 n n n

2 4 3 2(3 3 2 ) 1 0 n n n

Only if, n = 1, 2, 4, 5, 7, 8

Number of values satisfying are 6.

65. Answer (4)Given relation

(2a + 3b)2 – (5c ) = 0

(2a + 3b + 5c )(2a + 3b – 5c ) = 0

2a + 3b + 5c = 0 2 3

05 5

a b c

2a + 3b – 5c = 0 2 3

05 5

a b c

2 3 2 3, ,

5 5 5 5

16 36 1 5252

25 25 5 5D

66. Answer (2)

Given focus S (1, 0)

M (–1, 2t )

where P (t 2, 2t )

Since PSM is equilateral

we have SM = SP

4 + 4t 2

= (t 2

+ 1)2

t 2 + 1 = 4 or 3t

Now, P (3, 2 3 )

Area of PSM =3(16)

4

= 4 3


8/21


8/11

67. Answer (2)

1 1

t a

a

eJ dt

t a

∫

Let,

t = x + a – 1

t = a – 1 x = 0

t = a x = 0

Therefore,

( 1)1

0 2

∫ x a

eJ dx

x

=1

1

0 2

x

a ee dx

x

∫

Put, z = 1 – x

1

0( 1)

(1 2)

∫

z

a eJ e dz

= –e–a(I) = –Ie–a

68. Answer (4)

Let the chosen integers be x 1 and x

2. Let there be

a integer before x i , b integer between x

1 and x

2

and c integer after x 2

a + b + c = 98

where a 0, b 10, c 0

Now, if we consider the choice where difference is at

least 11 then the number of solution is 87 + 3 – 1C 3 – 1

= 89C 2

The number of ways in which b is less than 10 is100C

2 – 90C

2.

69. Answer (3)

We have, f '(C ) = 12C 3 – 8C , now since the function

satisfies L.M.V.T.

3 (1) ( 1)12 8 01 ( 1)

f f C C

C (3C2 – 2) = 0

20,

3C

So,2

3C is one of the values.

70. Answer (4)

2x y y dy e x edx

2

y x x e dy e e dx

3

3 y x

x e e C

From (1, 1)

1 1

3 3e e C C

⇒

31

3 3

y x x e e

71. Answer (1)

(cos7 cos8 ) sin5

sin 5 sin10

x x x dx

x x

∫

15 5 52 sin sin 2 sin .cos

2 2 2 215 5

2 sin .cos2 2

x x x x dx

x x

∫

52 sin .sin (cos 2 cos3 )

2 2

x x dx x x dx ∫ ∫

=1 1

sin 2 sin 32 3

x x C

72. Answer (3)

The probability that P 2 is champion is

1

64 P

2 plays

against 6 other players which can be selected in

63C 6 ways. If P 2 losses to P 1 then P 1 is included in6 players in 62C

5 ways.

Hence probability =

625

636

1.64

C

C

=1

672

73. Answer (4)

0 |a + b + c + d |2 = |a|2 + |b|2 + |c |2 + |d |2 + 2[a.b]

= 4 2( . . . . . . )a b b c c a a d c d b d

( . ) 2a b

2 2 2 2 2| | | | | | | | | |a b b c c d d a c a

2 2 2 2 2| | 3 | | | | | | | | 2( . )b d a b c d a b

12 2 ·a b

16


9/21


9/11

74. Answer (4)

6 sin2 x – 2 cos4 x = cos2 x (2 cos2 x – 1)

4 cos4 x + 5 cos2 x – 6 = 0

cos2 x =3

4

cos 2 x =1

2

75. Answer (3)

2 32

... 1 ....3 9

x x x xy y

also,2 3

1 .... 13 9

x x x

–1 1 + y + y 2 + ..... 1

|y | < 1 | x | < 3

3 1

3 1

x

x y

Only if possible value (3)

76. Answer (1)

As Q lies on the given.

ˆ ˆ ˆ ˆ ˆ ˆ( 2 ) ( 3 5 )OQ i j k i j k

PQ

is parallel to the plane

x – 4y + 3z = 1

ˆ ˆ ˆ( 4 3 ) 0PQ i j k

ˆ ˆ ˆ( ).( 4 3 ) 0OQ OP i j k

(1 + 4 + 6) + (–3 –4 + 15) – (3 – 8 + 18) = 011 + 8 – 13 = 0

1

4

77. Answer (4)

Median of a, 2a, ....50a is1

2(25a + 26a) = (25.5)a

M.D. of given numbers from median

=

50

1

1| (25.5) |

50K

Ka a

∑

=

50

1

| | | || 25.5) | (1 3 ... 49)

50 50K

a aK

∑

=| | 25

(1 49)50 2

a

25

| | 502

a

|a| = 4

78. Answer (3)

O P

Q

R

250 m

50 m

PQ = 50

QR = 250

50tan

x

,300

tan2 x

2

2 tan 300

1 tan x

2

502

300

501

x

x

x

50 325 6

2 x

79. Answer (4)

2

1

2

log ( 4 3) 0 x x

0 < x 2 + 4 x 3 < 1

x (– –3) (–1, )

and 2 6, 2 6 x

2 6, 3 1, 2 6 x

80. Answer (1)

y = ( x + 1)2 – 1, x –1

( x + 1)2 = y + 1

1 1 as 1 x y x

1 1 x y

1( ) 1 1f x x

( x + 1)2 – 1 = –1 + 1 x

31 0 or ( 1) 1

2 x x

x = –1 or x = 0


10/21


10/11

81. Answer (1)

( ) p q p q

F TTT FF F

Contradiction

Thus ( p q) p q can never take value F .

82. Answer (1)

r 1 = 2r

2 = 3r

3.

2 3 1

s a s b s c K

S – a = K

S – b = 2K

S – c = 3K

3S – (a + b + c ) = 6K

S = 6K

a = 5K , b = 4K , c = 3K

191

60

a b c

b c a

83. Answer (2)

2

2

4 3

7 10

x x y

x x

2

2 219 14 19 0( 7 10)

dy x x dx x x

84. Answer (3)

f ( x ) = 2 sin2 x + sin 2 x – 1

[ 2, 2] but sinx + cosx –1

sin2 x – cos2 x – 1

Range: 2, 2 { 1}

85. Answer (2)

Put m = n2 + cos n

So, that m , as n

2

2

2

1 1lim 1 lim 1

cos

n n

n n m m

n n mn n

= e

86. Answer (1)

f ( x ) = ( x – )( x – )

+ = –a

= –b

f (n).f (n + 1) = (n – )(n – )(n + 1 – )(n + 1 – )

= (n(n + 1) + na + b – )[(n(n + 1) + an + b – )]

Put m = n(n + 1) + an + b

then, m is an integer and

f (n) f (n + 1) = (m – ).(m – ) = f (m)

87. Answer (2)

88. Answer (4)

2 2

2 2

cos1, sin 1

x y x y

a ba b

cos sin 13 3

x y

Hence, the intercepts of the tangent on the axes are

3 3 sec and cosec

( ) 3 3 sec cosecf

2 3

2 2

3 3 sin cos'( )

cos .sinf

therefore f '() = 0

1tan

3

6

89. Answer (1)

1, 1,

2....

n – 1 are nth root of unity

1 2 11 1

1

lim lim ( )( )...( )1

n

nx x

x

x x x x

n = (1 – 1)(1 –

2)....(1 –

n – 1)

Statement-2

2 4 2( 1)1 1 .... 1

i i n i

n n nn e e e


11/21


11/11

22 sin sin cos .2 sin .n i

n n n n

2 2 ( 1) ( 1)sin cos .... sin cos

n ni i

n n n n

1 2 ( 1)| | 2 sin .sin ....sin

n n

n

n n n

Statement-1 and Statement-2 are correct and

Statement-2 is correct explanation of Statement-1.

90. Answer (1)

Let1 0

2 2 A

| A – xI | = 0 f ( x ) = x 2 + 3 x + 2 = 0

Then, A2 + 3 A + 2I = 0

A3 + 3 A2 + 2 A + I – I = 0

( A + I )3 + (–I )3 = A

X + Y = A + I – I = A =1 0

2 2


12/21

Test - 11 (Code-B) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2016

1/10

1. (2)

2. (3)

3. (1)

4. (1)

5. (1)

6. (1)

7. (2)

8. (2)

9. (3)

10. (2)

11. (3)

12. (4)

13. (2)

14. (2)

15. (3)

16. (2)

17. (3)

18. (4)

19. (3)

20. (3)

21. (3)

22. (4)

23. (4)

24. (2)

25. (2)

26. (1)

27. (3)

28. (3)

29. (3)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

61. (2)

62. (3)

63. (2)

64. (1)

65. (1)

66. (1)

67. (4)68. (3)

69. (4)

70. (1)

71. (3)

72. (4)

73. (4)

74. (3)

75. (1)

76. (4)

77. (3)

78. (4)

79. (2)

80. (2)

81. (4)

82. (3)

83. (1)

84. (3)

85. (3)

86. (1)

87. (1)

88. (4)

89. (2)

90. (1)

31. (4)

32. (4)

33. (4)

34. (2)

35. (3)

36. (3)

37. (4)

38. (1)

39. (1)

40. (3)

41. (4)

42. (2)

43. (4)

44. (4)

45. (4)

46. (2)

47. (2)

48. (3)

49. (3)

50. (4)

51. (1)

52. (3)

53. (3)

54. (4)

55. (4)

56. (1)

57. (2)

58. (1)

59. (3)

60. (3)

Test Date : 30/03/2016

ANSWERS

TEST - 11 (Code-B)

All India Aakash Test Series for JEE (Main)-2016

click here for code-A solutions


13/21

All India Aakash Test Series for JEE (Main)-2016 Test - 11 (Code-B) (Answers & Hints)

2/10

1. Answer (2)

B = 0(H + I )

0 0

B H I H H

I = (r – 1

)H = (0 – 1

)ni = (1000 – 1) × 500 × 0.5

= 2.5 × 105 A/m

Magnetic moment = I × V = 2.5 × 105 × 10–4

= 2.5 A-m2

2. Answer (3)

In circuit A, both the diodes are forward biased, so

effective resistance is 2 and current 8

4 A4

I

In circuit B, diodes is reverse biased, so effective

resistance is 4 and current 8

2 A4

I

3. Answer (1)

0 0

1 1

r r r r

ev

2

2 2.25

r

r

e

v

4. Answer (1)

16 mV 8 mV

max

168mV

2E

min8

4 mV2

E

max min

max min

8 4 1

8 4 3

E E m

E E

5. Answer (1)

6. Answer (1)

Electric field 0v dv

E dr r

PART - A (PHYSICS)

Centripetal force 2

mv Ee

r

2

0.

v mv e

r r

0v e

v

m

Now mvr =2

nh

2

nhr

mv

r n

7. Answer (2)

8. Answer (2)

DY X

C

10

10

So, R = 5

9. Answer (3)

I 1

I 2

2 23 1 2I I I

So, 3 1 2( )I I I

10. Answer (2)

N 1

1 = N

2

2

1 2

2 1

3

2

N

N

11. Answer (3)

+100 –q

–100 +q

–50 +q

+50 –q


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3/10

6 6

100 500

5 10 20 10

q q

q = 90 × 10–6 C

Final charge on 5 F top plate is 10 C.

12. Answer (4)

100 100 2. 100

dg dl dT

g l T

=

0.1 2 0.1

100 1.2%50 2.00

13. Answer (2)

∫ ∫

3/2 2

0 3/ 2cos cos

3 3S dt dt

=

3 3

2

14. Answer (2)

steel

2brass

3

3 3. .

2 2 2

s s s b b

b s s

b b

mg ls

l A Y l A Y a

mg l l A Y b c lb A Y

15. Answer (3)

(real)(app.)

1

y

y

v

v

v x

(app.) = v x (real)

(app) 4 4 3

tan tan 1(app) 3 3 4

y

x

v

v

16. Answer (2)

17. Answer (3)

220

2 2 3/2 .

2 ( ) A

i R r

R x

Ad A

edt

=2 2 2 5/20 3 ( ) (2 )

2 2

i R R x x

For max value of e A, 0 Ade

dx

2 2 5/2

0( )

d x

dx R x

2 2 5/2 2 2 3/25

( ) ( ) .2 02

x R x R x x

2

R x

18. Answer (4)

B . lvt = constant

C

Blvt

1Bt

19. Answer (3)

22

4

Fx

I Mr Mx

For maximum

1

0d

dx

2

2 22 0

4

r x x

2

r x

20. Answer (3)

21. Answer (3)

v x = by

x dv dy

bdt dt

ax = bv

0

So, 201

( )2

x bv t and y = v 0t

2

0

0

1( )

2

y x bv

v

2

02

by x

v

22. Answer (4)

L

R is time constant. RC is also time constant.

So,/L R

RC is dimensionless.


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4/10

23. Answer (4)

24. Answer (2)

sin2

F

sin2

F

/2 /2

cos

2F

cos

2F

2TR

F F

2 2 sin

2TR F

2 2 2TR F

F = 2TR

FL FR Y

A L A R

22

(2 ).FR R TR

R TR AY AY AY

25. Answer (2)

30°

v 1

v 12

v 12

–v 2

1

2

tan30 v

v

1

2

1

3

v

v

26. Answer (1)

27. Answer (3)

28. Answer (3)

29. Answer (3)

30. Answer (1)

PART - B (CHEMISTRY)

31. Answer (4)

H + Br

NBS/h

Br +

Br

32. Answer (4)

o, o'-benzene losses planarity due to steric reason

(bulky group at ortho position) so, the compound does

not have centre of symmetry.

33. Answer (4)

alc.KOHH

H

Ph

HBr

HHO–

H

Ph

3-phenylcyclopentene

34. Answer (2)

(i) O

(ii)

3

H O

(iii)2 2

CH COOH2—

+ CO2

(i) O

(ii)

3

H O

(iii)2 2

C O

CH3

+ CO2

(i) O

(ii)

3

H O

(iii)2 2

COOH

O

CH3

(i) O

(ii)

3

H O

(iii)2 2

O

O

O


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5/10

35. Answer (3)

36. Answer (3)

37. Answer (4)

Q isCH C CH

3 3

— —

O

, it gives iodoform test.

38. Answer (1)

39. Answer (1)

Anthracene undergoes addition reactions at 9 and 10

positions because resonance energy per mole of ring

increases.

40. Answer (3)

2 12 2

2 2 2 6 2 6 6 0Fe : 1 , 2 , 2 , 3 , 3 , 3 , 4s s p s p d s

41. Answer (4)

42. Answer (2)

Adsorbed moles of

3

2

0.03 2.46 10H

0.0821 300

= 3 × 10–6

Number of adsorbed molecules of

H2 = 3 × 10–6 × 6 × 10–23

= 18 × 1017

Total number of surface sites available

= 6 × 1015 × 1000 = 6 × 1018

Number of surface sites that is occupied by adsorption

of H2

gas =18 1710

6 10 6 10

100

Number of surface sites occupied by one molecule

of H2

=

17

17

18 10

6 10

= 3

43. Answer (4)

44. Answer (4)

45. Answer (4)

3 2 2 2

3CH OH( ) O (g) CO (g) 3H O( )

2

G° = –394.4 + 2 × (–237.2) – (–166.2)

= –702.6 kJ mol–1

Efficiency of fuel cell =G

100 97%H

46. Answer (2)

3 3KI AgNO KNO AgI

Initial: 3 m.moles 2 m.moles 0 0

Final : 1 0 2 2

KI in solution =1M

40

KNO3 in solution =

2M

40

f

1 2T 1.86 2 1.86 2 0.28

40 40

47. Answer (2)

partial2 2 2 2conversion

3Cu S O Cu O SO

2

2 2 2Cu S 2Cu O 6Cu SO

(Cu2S is taken in 2 : 1 ratio)

(Two parts in first step and one part in second step)

48. Answer (3)

In metal carbonyls with metal as anion, better back

donation of electron is observed from metal to

carbonyl, followed by decrease in metal-carbon bond

length.


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49. Answer (3)

1 0

1 2

X 2Y

1 0 0

1

Z P Q

1

21

p 2

4 PK

(1 )

2

22

p 2

PK

(1 )

1

2

p 1

p 2

K 4P 1

K P 3

1

2

P 1

P 12

50. Answer (4)

1 a

xpH pK log

y

2 a

y

pH pK log x

x

3.17y

51. Answer (1)

Boron and silicon show diagonal relationship, hence

similar chemical behaviour.

52. Answer (3)

53. Answer (3)

15Te + TeCl4 + 4AlCl

3 2Te

8[AlCl

4]2

Te

Cl

Cl

Cl

Cl

Al

Cl

Cl

Cl

54. Answer (4)

C(1 ) C CHA H A

2

a

CK (1 )

aK (1 )C

a

1[H ] K

a

1log(H ) log K log

a

1pH pK log

apK pH1

10

a

pK pH

1

1 10

55. Answer (4)

At 27°C, 1 × V = nHe

× R × 300

nHe

=V

300R

At 127°C, 2 × V = (nHe

+ nP) × R × 400

He P Vn n200R

nP =

V

600R

At 327°C, V V

P V R 60030 0R 3 00R

P = 4 atm

56. Answer (1)

57. Answer (2)

58. Answer (1)

59. Answer (3)

O

NH2

(Methyl ion)glycinate

H C CH3

— — —C O

60. Answer (3)

N O (l)2 4

[NO ] [NO ]+ –3

NOCl + NaNO3

⇒

NaCl


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PART - C (MATHEMATICS)

61. Answer (2)

Put m = n2 + cos n

So, that m , as n

2

2

2

1 1lim 1 lim 1

cos

n n

n n m m

n n mn n

= e

62. Answer (3)

f ( x ) = 2 sin2 x + sin 2 x – 1

[ 2, 2] but sinx + cosx –1

sin2 x – cos2 x – 1

Range: 2, 2 { 1}

63. Answer (2)

2

2

4 3

7 10

x x y

x x

2

2 2

19 14 190

( 7 10)

dy x x

dx x x

64. Answer (1)

r 1 = 2r

2 = 3r

3.

2 3 1

s a s b s c K

S – a = K

S – b = 2K

S – c = 3K

3S – (a + b + c ) = 6K

S = 6K

a = 5K , b = 4K , c = 3K

191

60

a b c

b c a

65. Answer (1)

( ) p q p q

F TTT FF F

Contradiction

Thus ( p q) p q can never take value F .

66. Answer (1)

y = ( x + 1)2 – 1, x –1

( x + 1)2 = y + 1

1 1 as 1 x y x

1 1 x y

1( ) 1 1f x x

( x + 1)2 – 1 = –1 + 1 x

31 0 or ( 1) 1

2 x x

x = –1 or x = 0

67. Answer (4)

2

1

2

log ( 4 3) 0 x x

0 < x 2 + 4 x 3 < 1

x (– –3) (–1, )

and 2 6, 2 6 x

2 6, 3 1, 2 6 x

68. Answer (3)

O P

Q

R

250 m

50 m

PQ = 50

QR = 250

50tan

x

,300

tan2 x

2

2 tan 300

1 tan x

2

502

300

501

x

x

x

50 325 6

2 x


19/21


8/10

69. Answer (4)

Median of a, 2a, ....50a is1

2(25a + 26a) = (25.5)a

M.D. of given numbers from median

=

50

1

1| (25.5) |

50K

Ka a

∑

=50

1

| | | || 25.5) | (1 3 ... 49)

50 50K

a aK

∑

=| | 25

(1 49)50 2

a

25

| | 502

a

|a| = 4

70. Answer (1)

As Q lies on the given.

ˆ ˆ ˆ ˆ ˆ ˆ( 2 ) ( 3 5 )OQ i j k i j k

PQ

is parallel to the plane

x – 4y + 3z = 1

ˆ ˆ ˆ( 4 3 ) 0PQ i j k

ˆ ˆ ˆ( ).( 4 3 ) 0OQ OP i j k

(1 + 4 + 6) + (–3 –4 + 15) – (3 – 8 + 18) = 0

11 + 8 – 13 = 0

1

4

71. Answer (3)2 3

2... 1 ....

3 9

x x x xy y

also,2 3

1 .... 13 9

x x x

–1 1 + y + y 2 + ..... 1

|y | < 1 | x | < 3

3 1

3 1

x

x y

Only if possible value (3)

72. Answer (4)

6 sin2 x – 2 cos4 x = cos2 x (2 cos2 x – 1)

4 cos4 x + 5 cos2 x – 6 = 0

cos2 x =3

4

cos 2 x =1

2

73. Answer (4)

0 |a + b + c + d |2 = |a|2 + |b|2 + |c |2 + |d |2 + 2[a.b]

= 4 2( . . . . . . )a b b c c a a d c d b d

( . ) 2a b

2 2 2 2 2| | | | | | | | | |a b b c c d d a c a

2 2 2 2 2| | 3 | | | | | | | | 2( . )b d a b c d a b

12 2 ·a b

16

74. Answer (3)

The probability that P 2 is champion is

1

64 P

2 plays

against 6 other players which can be selected in63C

6 ways. If P

2 losses to P

1 then P

1 is included in

6 players in 62C 5 ways.

Hence probability =62

5

636

1.64

C

C

=1

672

75. Answer (1)

(cos7 cos8 ) sin5

sin 5 sin10

x x x dx

x x

∫

15 5 52 sin sin 2 sin .cos

2 2 2 2

15 52 sin .cos

2 2

x x x x dx

x x

∫

52 sin .sin (cos 2 cos3 )

2 2

x x dx x x dx ∫ ∫

=1 1

sin 2 sin 32 3

x x C

76. Answer (4)

2x y y dy e x edx

2

y x x e dy e e dx

3

3 y x x e e C

From (1, 1)

1 1

3 3e e C C

⇒

31

3 3

y x x e e


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9/10

77. Answer (3)

We have, f '(C ) = 12C 3 – 8C , now since the function

satisfies L.M.V.T.

3 (1) ( 1)12 8 01 ( 1)

f f C C

C (3C2 – 2) = 0

20, 3C

So,2

3C is one of the values.

78. Answer (4)

Let the chosen integers be x 1 and x

2. Let there be

a integer before x i , b integer between x

1 and x

2

and c integer after x 2

a + b + c = 98

where a 0, b 10, c 0

Now, if we consider the choice where difference is at

least 11 then the number of solution is 87 + 3 – 1C 3 – 1 = 89C 2

The number of ways in which b is less than 10 is100C

2 – 90C

2.

79. Answer (2)

1 1

t a

a

eJ dt

t a

∫ Let,

t = x + a – 1

t = a – 1 x = 0

t = a x = 0

Therefore,

( 1)1

0 2

∫ x a

eJ dx

x

=1

1

0 2

x

a ee dx

x

∫

Put, z = 1 – x

1

0( 1)

(1 2)

∫

z

a eJ e dz

= –e

–a

(I) = –Ie

–a

80. Answer (2)

Given focus S (1, 0)

M (–1, 2t )

where P (t 2, 2t )

Since PSM is equilateral

we have SM = SP

4 + 4t 2 = (t 2 + 1)2

t 2 + 1 = 4 or 3t

Now, P (3, 2 3 )

Area of PSM =3(16)

4

= 4 3

81. Answer (4)

Given relation

(2a + 3b)2 – (5c ) = 0

(2a + 3b + 5c )(2a + 3b – 5c ) = 0

2a + 3b + 5c = 0 2 3

05 5

a b c

2a + 3b – 5c = 0 2 3

05 5

a b c

2 3 2 3

, ,5 5 5 5

16 36 1 5252

25 25 5 5D

82. Answer (3)

After taking common 1, , 2

we get,

2 4 3 4 3 2(4 3)(3 3 2 ) 1 0 n n n

2 4 3 2(3 3 2 ) 1 0 n n n

Only if, n = 1, 2, 4, 5, 7, 8

Number of values satisfying are 6.

83. Answer (1)

After solving determinate there are six degree of

polynomial, sum of all roots = 0.

i.e. coefficient of x 5 = 0

Then sum of other five roots = –3

84. Answer (3)

20

3 20 313 3

1 11 1 x C x

x x

2 2020 3 20 3

2 203 3

1 1.....C x C x

x x

Therefore,

2 20

3 3 2 3 3 3 20

3 3 3

1 1 11. .( ) .( ) ....( ) . . ..... x x x x

x x x

All distinct whose number is 1 + 20 + 20 = 41


21/21


85. Answer (3)

D

A

B

P (4, 5)

4

O(2, 1)

Centre (2, 1)

2 2(4 2) (5 1)OP

4 16 20 2 5

4sin

2 5

OD

AO

8

5

OD

64

165

AD

4

5

8

5 AB

86. Answer (1)

Let1 0

2 2 A

| A – xI | = 0 f ( x ) = x 2 + 3 x + 2 = 0

Then, A2 + 3 A + 2I = 0

A3 + 3 A2 + 2 A + I – I = 0

( A + I )3 + (–I )3 = A

X + Y = A + I – I = A =1 0

2 2

87. Answer (1)

1, 1,

2....

n – 1 are nth root of unity

1 2 11 1

1lim lim ( )( )...( )

1

n

nx x

x x x x

x

n = (1 – 1)(1 –

2)....(1 –

n – 1)

Statement-2

2 4 2( 1)1 1 .... 1

i i n i

n n nn e e e

22 sin sin cos .2 sin .n i

n n n n

2 2 ( 1) ( 1)sin cos .... sin cos

n ni i

n n n n

1 2 ( 1)| | 2 sin .sin ....sinn n

n

n n n

Statement-1 and Statement-2 are correct and

Statement-2 is correct explanation of Statement-1.

88. Answer (4)

2 2

2 2

cos1, sin 1

x y x y

a ba b

cos sin 13 3

x y

Hence, the intercepts of the tangent on the axes are

3 3 sec and cosec

( ) 3 3 sec cosecf

2 3

2 2

3 3 sin cos'( )

cos .sinf

therefore f '() = 0

1tan

3

6

89. Answer (2)

90. Answer (1)

f ( x ) = ( x – )( x – )

+ = –a

= –b

f (n).f (n + 1) = (n – )(n – )(n + 1 – )(n + 1 – )

= (n(n + 1) + na + b – )[(n(n + 1) + an + b – )]

Put m = n(n + 1) + an + b

then, m is an integer and

f (n) f (n + 1) = (m – ).(m – ) = f (m)

Solutions AIATS JEE(Main)-2016 Test-11 (Code-A & B)

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