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