All India Aakash Test Series for JEE (Advanced)-2021
Test Date : 04/10/2020
ANSWERS
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TEST -3A (Paper-2) - Code-B
PHYSICS CHEMISTRY MATHEMATICS
1. (A, D)
2. (B, C)
3. (B, C)
4. (A, C)
5. (A, D)
6. (C, D)
7. (A, B, D)
8. (A, D)
9. (A, D)
10. (A, D)
11. (A, D)
12. (D)
13. (A)
14. (D)
15. (A)
16. A (T)
B (P, Q, R)
C (P, Q, R, S, T)
D (P, Q, R, S, T)
17. A (R)
B (R)
C (T)
D (Q)
18. (02)
19. (08)
20. (90)
21. (A, B, C, D)
22. (C, D)
23. (A, B)
24. (B, C)
25. (A, D)
26. (D)
27. (A, B, C, D)
28. (B, C)
29. (A, C)
30. (A, C, D)
31. (A)
32. (A, C, D)
33. (B, C, D)
34. (A, C)
35. (B, C, D)
36. A (P, R)
B (P, Q, R)
C (P, S, T)
D (P, Q, R)
37. A (Q, R, S)
B (P)
C (Q)
D (Q, R, T)
38. (12)
39. (18)
40. (03)
41. (A, B, C)
42. (A, B, C, D)
43. (A, B, C)
44. (C, D)
45. (A, B, C)
46. (C)
47. (A)
48. (A, B)
49. (B, C)
50. (A, B, C, D)
51. (C)
52. (B)
53. (D)
54. (B)
55. (A)
56. A (S, T)
B (Q)
C (P)
D (R)
57. A (Q)
B (P)
C (R)
D (S)
58. (32)
59. (17)
60. (02)
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HINTS & SOLUTIONS
PART - I (PHYSICS)
1. Answer (A, D)
Hint :
1 1 1
v u f
Solution :
Concave mirror cannot form real image between
pole and focus and image of a virtual object is
always real.
2. Answer (B, C)
Hint :
f v
mf
Solution :
f v
mf
1v
mf
Slope = 1 b c
ff c b
Also at m = 0, v = a
0 = f – a
f = a
3. Answer (B, C)
Hint :
1 2
1 1 1( 1)
f R R
Solution :
In air
1 1 1
12f R R
2
3( 1)
Rf
4. Answer (A, C)
Hint :
sin2
sin2
mA
A
Solution :
In case of minimum deviation
sin2
, 60 , 2
sin2
mA
AA
for minimum deviation,
1 2 302
Ar r
30º m
sin i = sinr1
i = 45°
5. Answer (A, D)
Hint :
2 1 2 1
v u R
Solution :
Silvered lens will now behave like a concave
mirror. To coincide the image on the object itself,
the ray should fall normally on silver portion.
3 1 1
2 2v u R
3 1 1
64 64u
u = –16 cm
6. Answer (C, D)
Hint :
For plano-convex lens 1 ( 1)
f R
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Solution :
For plano-convex lens 11
f R
Let F be the focal length of silvered plano-
convex lens.
2 1 2 11 1
F R R
Given 11 1
20f R
1 2 1
20 10F
F = 10 cm
It acts as concave mirror.
Using mirror equation
1 1 1
v u F
7. Answer (A, B, D)
Hint :
296
4.84
x
x
Solution :
D = 96 cm, 2
1
4.84I
I
u + v = 96
11
5
v
u 1 2 12.2O I I I
5
9611
vv 1
1 10 5
2.2 22 11
I
O
v = 66 and u = 30 1
11
5
O
I
Displacement of lens x = v – u = 36 cm
8. Answer (A, D)
Hint :
qR
Solution :
2
B r
qR R
= 2 – 1 = Br2
q B
q r2
1
qR
9. Answer (A, D)
Hint :
vBl
iR
Solution :
In position 1
= Bv × 2 × 10–2
= 1 × 5 × 10–2
× 2 × 10–2
3100.1 mA
10i
In position 2
= Bv × 4 × 10–2
= 2 × 10–3
32 100.2 mA
10l
10. Answer (A, D)
Hint :
| |
d
emfdt
Solution :
= t
Only haft part will be involved in inducing emf so
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2
2
aA
= BAcos
2
cos2
aB t
2
sin ,2
d B at i
dt R
Where = 0° i = 0
When 2
,2 2
B ai
R
11. Answer (A, D)
Hint :
dB
E dl Adt
Solution :
dB
E dl Adt
2
0 2 322
R B tE R
03
2
B RtE
F = QE
12. Answer (D)
Hint :
FR = mgR
Solution :
As, torque acting due to electric field is equal to
frictional force
FR = mgR
0
03
2 2
QB RQB Rt mgR
mg
1
3t
13. Answer (A)
Hint :
2
0net
3
2
QB R tmgR
Solution :
2
0net
3
2
QB R tmgR
20 3
2
QB R tdI mgR
dt
I = mR2
On solving 0
3
QB
m
14. Answer (D)
Hint :
1 21 2
sec sec;
a i b rt t
v v
Solution :
11 1
sec
AO a it
v v
22 2
sec
OB b rt
v v
t = t1 + t2
15. Answer (A)
Hint :
To optimize time 0dt
dr
Solution :
To optimize time 0dt
dr
1 2
sec sec0
d a i b r
dr v v
1 2
sec tan sec tan0
a i i di b r r
v dr v
2
2
cos
cos
di b i
dr a r
16. Answer A(T); B(P, Q, R); C(P, Q, R, S, T);
D(P, Q, R, S, T)
Hint :
For refraction 2 1 2 1
v u R
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Solution :
Solve by using formula of reflection and
refraction.
17. Answer A(R); B(R); C(T); D(Q)
Hint :
| |
d
dt
Solution :
B = 2 – 100t for (0 t 0.02)
= B × 40 × 10–4
4 440 10 40 10 100
dB
dt
440 100 100.08
5I
R
Amp.
Upto t = 0.01 sec
440 10 10.8 mC
5
Q
R
Upto t = 0.02 sec
440 10 2
1.6 mC5
Q
R
18. Answer (02)
Hint :
Z2 = 2R
2
Solution :
2Z R
cosR
Z
Power factor = 1
2
19. Answer (08)
Hint :
A Bq di
V V L iRc dt
Solution :
q = t2 – 4 at t = 3 s q is positive
2 2dq di
i tdt dt
At t = 3 s, q = 5 C
i = 6 A
A Bq di
V V L iRc dt
VA – VB = 8 V
20. Answer (90)
Hint :
1 ( 1)2
f R
Solution :
Apply refraction formula for upper and lower
portion
V1 = 60 cm
V2 = –30 cm
|V1| + |V2| = 90 cm
PART - II (CHEMISTRY)
21. Answer (A, B, C, D)
Hint :
KClO3 + I2 KIO3 + Cl2
Solution :
SF4 – Gas
SeF4 – Liquid
TeF4 – Solid
22. Answer (C, D)
Hint :
CO2, N2O5, SO3 are the most acidic oxide gases
of C, N and S respectively.
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Solution :
P4 + SOCl2 PCl3 + SO2 + S2Cl2
Cu + HNO3 (conc.) Cu(NO3)2 + NO2 + H2O
C + HNO3 (conc.) CO2 + NO2 + H2O
SO2 + O22 5V O SO3
23. Answer (A, B)
Hint :
The correct order of temperature is
T3> T2 > T4 > T1
Solution :
At higher temperature C is a good reducing
agent but at lower temperature CO is a good
option.
24. Answer (B, C)
Hint :
Overall carbon is used to reduce Al2O3
Al2O3 + C Al + CO2
Solution :
To produce 1 kg of Al, around 0.5 kg of carbon is
burnt. Both CO and CO2 produced at anode.
25. Answer (A, D)
Hint :
2
34
H S
24
Cu(CN) No ppt. of Cu
Cd(CN) Cds
Solution :
Cu (CN)2 yellow ppt.
Cd (CN)2 white ppt.
Cu (NH3)42+
, Ni(NH3)62+
both are blue.
26. Answer (D)
Hint :
–Cl and –OMe (at meta position) are
deactivating in nature towards EAS reaction.
Solution :
The ring substituted with –Cl and
–OMe does not form a stable arenium ion and
hence EAS reaction would occur on the other
ring.
27. Answer (A, B, C, D)
Hint :
SO2 bleaches through reduction and Cl2 through
oxidation.
Solution :
Ethanol is less polar as compared to water that’s
why I2 is more soluble in it.
As I– forms complex with I2 (i.e. I3
–) which makes
I2 soluble in water. That’s why continuous
consumption of I– causes decrease in solubility
of I2.
28. Answer (B, C)
Hint :
Fact
Solution :
Gd+3
= 4f7 , 5d
0,6s
0
Lu+3
= 4f15
,5d0,6s
0
29. Answer (A, C)
Hint :
Fe3+
(aq) is yellow in colour
Solution :
Fe3+
= 3d5 configuration maximum paramagnetic
30. Answer (A, C, D)
Hint :
2Br /h Na
4 3 3 3EtherCH CH Br CH CH
Solution :
CH – CH33Br /h2 CH – CH – Br23 CH – CH3
CH C–
CH – C –CH CH3 2 3
OH O
+
3
C C
H
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31. Answer (A)
32. Answer (A, C, D)
33. Answer (B, C, D)
Hint for Q. Nos. 31 to 33
x is NH3
y is [Cu(NH3)4]2+
z is Cu2 [Fe(CN)6]
z is Cu3 [Fe(CN)6]2
Solution for Q. Nos. 31 to 33
34. Answer (A, C)
35. Answer (B, C, D)
Hint for Q. Nos. 34 and 35
The complex is cis-platin
[Pt (NH3)2 Cl2]
Solution for Q. Nos. 34 and 35
Overall it is neutral
There is no chelation in the complex
36. Answer A(P, R); B(P, Q, R); C(P, S, T); D(P, Q,
R)
Hint :
Solution :
37. Answer A(Q, R, S); B(P); C(Q); D(Q, R, T)
Hint :
Solution :
In presence of acid, amines get protonated and
that’s why nucleophilic attack takes place
majorly from oxygen centre.
38. Answer (12)
Hint :
K4 [Fe(CN)6] – d2sp
3, diamagnetic
K3 [Cr(CN)6] – d2sp
3, paramagnetic
Solution :
K3 [Co(CN)6] – d2sp
3, diamagnetic
K2 [Ni(CN)4] – dsp2, diamagnetic
[Co(NH3)6]3+
– d2sp
3, diamagnetic
[Pt(NH3)4]2+
– dsp2, diamagnetic
p = 1, q = 5, r = 4, s = 0, t = 2
39. Answer (18)
Hint :
6CaO + P4O10 2 Ca3 (PO4)2
Solution :
P4 + O2 P4O10
372 g
= 3 mol 3 mol
3 moles of P4O10 are produced required 18 mol
of CaO.
3NH2 23 4Cu [Cu(NH ) ]
PCl + H O5 2 H PO + HCl3 4
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40. Answer (03)
Hint :
It is less stable than iso-propyl carbocation.
Solution :
Carbocation B, C and J are satisfying the asked
conditions.
PART - III (MATHEMATICS)
41. Answer (A, B, C)
Hint :
Divide the integral at each integer value of x.
Solution :
20
0[ ] { }I x x dx
1 2 3 20
0 1 2 190 { } 1 { } 2 { } ... 19 { }x dx x dx x dx x dx
1 19 20 1
(1 2 ...19)2 2 2
= 95
42. Answer (A, B, C, D)
Hint :
Differentiate both sides w.r.t. x.
Solution :
sin( )
sin( )
x ay
x b
2
sin( )cos( ) sin( )cos( )
sin ( )
dy x b x a x a x b
dx x b
2
sin( )0
sin ( )
b a
x b
(b – a) = n or (b – a) = (2n + 1) or b – a = 2n
So 0dy
dx so ( ) 0f x
Therefore f(x) will be constant
43. Answer (A, B, C)
Hint :
02
x
then
sin
( )x
f xx
is decreasing ( ) 0f x
Solution :
sinx < x < tanx, if
0,
2x
then
/2 /2 /2
0 0 0
sin(sin ) sin sin(tan )
sin tan
x x x
x x x
1 2 3I I I
44. Answer (C, D)
Hint :
Put 1
xt
then u = v
Solution :
4 20
7 1
dxu
x x
Put 1
xt
then
0
2
4 2
1
1 71
u dtt
t t
2
2 40 1 7
t dt
t t
u = v
and
3
u v
45. Answer (A, B, C)
Hint :
Differentiate both sides w.r.t. x, then solve
different equation.
Solution :
f(x) = 20
sin ( )(2sin sin )x
x f t t t dt
2( ) cos ( )(2sin sin )f x x f x x x
2( )(sin 2sin 1) cosf x x x x
2
cos( )
(sin 1)
xf x
x
2
cos( )
(sin 1)
xf x dx
x
O
CH2+
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1
( )(sin 1)
f x cx
f(0) = 0 c = –1
1
( ) 1(sin 1)
f xx
sin sin
( )(sin 1) (1 sin )
x xf x
x x
So ( /6) 1f
If g(x) = 0 0
sin( )
1 sin
x x tf t dt dt
t
sin
( ) 01 sin
xg x
x
is increasing in (0, )
f(0) = 0
46. Answer (C)
Hint :
( )
0
( ) ( ) ( ( ))
f td
g u du f t g f tdt
Solution :
20 1
( ) ( ( ) ) x t
g x t f u du dt
21
( ) ( ) ,x
g x x f u du so (1) 0g
21
( ) ( ) ( )x
g x x f x f u du
(1) 1 (1) 0g f
So (1) (1) (1) 0 3 g g f
47. Answer (A)
Hint :
If 2t then 41 1
171 t
Solution :
If 42
1( )
1
x
f x dtt
So 4
1( )
1
f x
x
In [2, 3] applying LMVT
(3) (2)
( ),3 2
f ff c
c (2, 3)
(3) 0
( )1
ff c
4
1(3) ( )
1f f c
c
Now, 2 < c < 3
16 < c4 < 81
17 < 1 + c4 < 82
4
1 1 1
82 1 17c
so
1(3)
17f
48. Answer (A, B)
Hint :
0 0( ) ( )a af x dx f a x dx
Solution :
Area = 1 1
0 0tan tan(1 )x dx x dx
tan1
1
0tan1 tan y dy
49. Answer (B, C)
Hint :
Two curves are orthogonal then 1 2 1m m
Solution :
Let y = ax + b = f(x)
So 1( )x b
f xa
So f(x) = ax + b and f–1
(–x) = x b
a
So they are orthogonal
Similarly f(–x) = –ax + b and 1( )x b
f xa
f(– x) and f–1
(x) are orthogonal
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50. Answer (A, B, C, D)
Hint :
0( ) ( ( ) ( ))a a
af x dx f x f x dx
Solution :
4 7( cos sin ) 3/2a
x
ae x x x dx
4 7cos sin 3/2a a
x
a ae x x x dx
0 3/2a
x
ae
ea – e
–a > 3/2 e
a = t
2 1
3/2t
t
2t2 – 3t – 2 > 0
2t2 – 4t + t – 2 > 0
2t(t – 2) + 1(t – 2) > 0
(t – 2)(2t + 1) > 0 t > 2 or t < –1/2
But t > 0 t > 2 ea > 2
So a = 1, 2, 3, 4
Hints for Q. Nos. 51 to 53
2 2( ) ( )f x f y
x yx y
3( )f x x kx
2 2( ) ( ) ( ) ( )( ) ( )
f x y f x yx y x y
x y x y
2( )f x x kx
3( )f x x xk ,
Now f(1) = 1, so 1 = 1 + k k = 0
f(x) = x3
51. Answer (C)
Here x = 0 is point of inflection.
52. Answer (B)
Area of region = 1 2 3
0( )x x dx
13 4
03 4
x x =
1 1
3 4 =
1
12 sq. units
53. Answer (D)
2
1( )f x dx =
2 3
1x dx =
24
14
x =
15
4
54. Answer (B)
Hint :
2
2
1 1(ln ) ( ) (ln ) 2
e en nnI x d x x xdx
Solution :
2
2
1(ln ) ( )
en
nI x d x
2
2
1
1(2 ln )
2
en
nx dx
2
2 2
1
1(ln )
2
en
nx dx
2
1
1(ln ) ,
2
en
n nI t dt x
2 = t
2
1
1(ln ) 1
2
en
n nI t dt
22
1
1 1
1 1(ln ) (ln )
2
ee
n n
n nI t t n t t dt
t
2
2 1
1
12 (ln )
2
en n
n nI e n t dt
2 11
1[2 2 ]
2
n n
n nnI e nI
2 12n
n
n II e
2In + n In – 1 = 2 e2
So answer is (B)
55. Answer (A)
Hint :
2
21 1 1
(ln ) ( ) ln 2e e
I x d x x xdx
xx
y
y
O
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Solution
2
2
11
(ln )e
I x dx
2
2 2
1
1(ln )
2
e
x dx
2
1
1(ln )
2
e
t dt
2
1
1ln
2
et t t
2 2 21
[ ln (ln1 1)]2
e e e
2 21
[2 1]2
e e
2 1
2
e
So answer is (A)
56. Answer A(S, T); B(Q); C(P); D(R)
Hint :
Put sinx
x = sin2
1 cosx
sin2dx d
Solution :
(A)
2sin cos
(1 sin ) cos
dI
22 (sec tan sec 1)d
2(tan sec ) c
112 sin1
xx c
x
112 cos
1
xx c
x
(B) 2sin cos
(1 sin )cos
dI
22 (1 sec tan sec )d
2( tan sec ) c
1 12 sin1
xx c
x
(C) Let x t
2dx
dtx
1 (1 )
dxI
x x x
22
(1 ) 1
dt
t t
Let sint cosdt d
2
cos 1 sin2 2
(1 sin ) cos cos
dd
2(tan sec ) c
12
1
xc
x
(D) Similarly :
2( 1)
1
xI c
x
57. Answer A(Q); B(P); C(R); D(S)
Hint :
Use the method of integration by substitution, by
partial fraction and properties of ITF.
Solution :
(A)
2
2
sec 1(sec tan )
2(sec tan )
x dxx x c
x x
(B)
cos sin 2
log(sin 1)(sin 2) sin 1
x xdx c
x x x
(C)
1 1
2
2sin 2 tan
1
xdx xdx
x
= 2x tan–1
x – log(1 + x2) + c
58. Answer (32)
Hint :
1
2 2 21 2
0
( ( )) ( )I I x f x x f x dx
21 2 42
0
( )2 4
x xx f x dx
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Solution :
21 16 22
1 2
0 0
( )4 2
x xI I dx x f x dx
21 22
0
1( )
28 2
xx f x dx
(I1 – I2) is maximum if
21 22
0
( ) 82
xx f x dx
2
( )2
xf x
28
(8) 322
f
59. Answer (17)
Hint :
Draw the graph of 3
2( )
xef x
x
Solution :
Let 3
2( )
xef x
x
2 3 3
4
3 2( )
x xx e e xf x
x
3
3
(3 2) xx e
x
29
4
ek
k > 16.625
k = 17
60. Answer (02)
Hint :
If f(x) is decreasing function then
if x1 < x2 then f(x1) > f(x2)
Solution :
f(x) = 30 – 2x – x3
f (x) = –2 – 3x2< 0
So f(x) is decreasing
f(f(f(x)) > f(f(–x))
f(f(x) < f(–x)
f(x) > – x
30 – 2x – x3> –x
x3 + x – 30 < 0
(x – 3)(x2 + 3x + 10) < 0
x < 3
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