(answers) all india aakash test series for jee (main)-2020 ......test - 2 (code-a) (hints &...

Test - 2 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020

Test Date : 11/08/2019

ANSWERS

1/12

TEST - 2 - Code-A

PHYSICS CHEMISTRY MATHEMATICS

1. (4)

2. (3)

3. (3)

4. (3)

5. (3)

6. (3)

7. (3)

8. (2)

9. (2)

10. (2)

11. (4)

12. (1)

13. (3)

14. (2)

15. (4)

16. (1)

17. (2)

18. (4)

19. (4)

20. (3)

21. (3)

22. (3)

23. (3)

24. (2)

25. (4)

26. (3)

27. (2)

28. (1)

29. (1)

30. (2)

31. (2)

32. (1)

33. (3)

34. (4)

35. (3)

36. (2)

37. (4)

38. (4)

39. (3)

40. (2)

41. (1)

42. (4)

43. (2)

44. (1)

45. (4)

46. (3)

47. (4)

48. (1)

49. (1)

50. (4)

51. (2)

52. (1)

53. (3)

54. (3)

55. (2)

56. (1)

57. (3)

58. (1)

59. (4)

60. (2)

61. (2)

62. (4)

63. (2)

64. (2)

65. (4)

66. (1)

67. (4)

68. (2)

69. (3)

70. (4)

71. (2)

72. (3)

73. (2)

74. (3)

75. (4)

76. (4)

77. (2)

78. (2)

79. (2)

80. (1)

81. (2)

82. (1)

83. (1)

84. (3)

85. (1)

86. (4)

87. (1)

88. (3)

89. (1)

90. (1)

All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)

2/12

1. Answer (4)

Hint : 01 2(cos cos )

4

= +

IB

d

Sol. : 2 3

d =

= 30°

03 (cos30 cos30 )

42 3

IB

= +

( )0 2 33 3

4

I =

09

2

I=

2. Answer (3)

Hint : 2= I

TMB

Sol. : 0

603 s

20T = =

Now, 01

0 2 02

BBT

T B B= =

0 3s

2 2

TT = =

3. Answer (3)

Hint : Use Ampere’s circulation law

Sol. : 0 inB dl i = 1forB r r R

30 forB r R=

4. Answer (3)

Hint: Identify the equipotential points and

redraw the circuit.

Sol. : It can be reduced as

RAB = 7

12R by simple calculation

5. Answer (3)

Hint : , 2 ( )mv

r mv m KEqB

= =

Sol.: 2 ( )m KEmv

rqB qB

= =

2 ( . )

,ee

m K Er

exB =

2 ( . )p

p

m K Er

eB=

2 ( )2 ( )

(2 )

pm KEm KEr

e B eB

= =

Clearly, rp > re and rp = r

6. Answer (3)

Hint :

0 (4 2)

42

=

s

IB

l

Sol. :

0 01 2 22 4

242

s

I IB

a a

= =

0

2c

IB

R

=

2 2 2

1s

c

B R

B a

=

PART - A (PHYSICS)

Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/12

Now, 4a = 2R 2

Ra

=

2

4 2 2 8 2S

C

B R

B R

= =

7. Answer (3)

Hint : AV = constant

Sol.: AV = constant

Since A2 < A1

V2 > V1

8. Answer (2)

Hint : cot2 = cot21 + cot22

Sol. : cot2 = cot21 + cot22

1 10

33 3

= + =

10

cot3

=

3

sin13

=

1 3sin

13

− =

9. Answer (2)

Hint : Use concept of effective length

Sol. :

3

0 12

2

R

x R

IdF I dx

x=

=

0 1 2 ln(3)2

I IF

=

10. Answer (2)

Hint and sol. :

VPQ = IPQ RPQ

Apply Kirchhoff law to get current in wire

PQ

|VQ – VP| = I(1)

I = 0.13 A from Q to P

11. Answer (4)

Hint : Baxis 0 0equatorial3 3

2,

4 4

= =

m mB

d d

Sol. : 01 3

2 ˆ4

mB i

a

=

( )02 3

ˆ4

mB i

a

= −

01 2 3

ˆ( )4

p

mB B B i

a

= + =

12. Answer (1)

Hint : At t = 0 all capacitors will act like a

conducting wire

Sol. : total

eq

VI

R=

2

V

R

=

1

3 62

VVI

RR

= =

13. Answer (3)

Hint : V = IR

Sol. : If S1 is closed

3

1

3

3 1 4

E EV

= =

+

Similarly 2

6(6)

7 7

E EV

= =

and 3

2

3

EV =

14. Answer (2)

Hint : Let resistance be X, then 2

12

= ++

XX

X

Sol. :

We have

2

12

XX

X= +

+

2X =


4/12

15. Answer (4)

Hint : Use Kirchhoff law

Sol. : eq

15 304

15 30R

= +

+ = 14

30

2 A14 1

I = =+

16. Answer (1)

Hint : Use condition of balanced Wheatstone

bridge

Sol. : 3 2

PS SQ=

2

100 cm5

SQ = = 40 cm

17. Answer (2)

Hint : Deflection is directly proportional to current

Sol. : 3

max

510 A

4950 50i −= =

+

max imax

For 20 divisions,

3

max

2 210 A

3 3i i − = =

Let R be resistance in series

Then 35 210

50 3R

−= +

R = 7450

18. Answer (4)

Hint : Both fields are perpendicular to each

other

Sol. : 0 1 0 21 2,

2 2

I IB B

d d

= =

B1 ⊥ B2

2 2

net 1 2B B B = +

19. Answer (4)

Hint : Integrate magnetic moment due to an

elementary ring.

Sol. : N

N dxr

=

2

2( )r

x r

d N I x=

=

2

2

r

x r

Nd dx I x

r=

=

20. Answer (3)

Hint : B does not change K.E of particle

Sol. : The charge particle moves on a

cycloidal path when B E⊥

21. Answer (3)


Sol. : 0 in( )B dl i =

= µo (+3 – 6) = –3µo

22. Answer (3)

Hint : q = qo (1 – e–t/)

= ReqC

Sol. : max

2

3q VC=

/22(1– )

3

VCq e−=

23. Answer (3)

Hint : V = IR

Sol. : 12 12 3

| |2 8

− = −C DV V = 1.5 V

24. Answer (2)

Hint : Redraw the circuit

Sol. :

eq

3

RR =

= 1


5/12

25. Answer (4)

Hint : Electromagnet requires temporary

magnet

Sol. : Low retentivity and low coercivity

facilitates temporary magnetic property

in core material.

26. Answer (3)

Hint : At null point potential become same

Sol. : RAB = 10 Vend = 10 Volt

25 cm4

L= =

27. Answer (2)

Hint : Use U = ,i B− then dU

Fdx

−=

Sol. : 2

0 0

2 2 3/22( )

I RB

R x

=

+

µ = r2i

2

20 0

2 2 3/22( )

I RU r i

R x

=

+

2 2

0 0

2 2 5/2

3

2( )d

I R r iddU

R ddx

=

+

28. Answer (1)

Hint : Time spent =

Sol. : qB

m =

0

0

12sin

2

mV

d qB

mVr

qB

= = =

6

=

6

6

mt

qB qB

m

= =

29. Answer (1)

Hint : Force = iB

Sol. : 01

2

IF i a

a

=

02

2 (2 )

IF i a

a

=

0net 1 2

11

2 2

= − = −

IiF F F 0

4

=

Ii

30. Answer (2)

Hint : Magnetic field is uniform

Sol. :

By symmetry,

2 2B d B B = =

0 in( )B d I =

02 ( 1)B i =

0

2

iB

=

31. Answer (2)

Hint: By using gas law and kinetics law, it can

be assumed that at 2 5 2

0N O OV V

2 5 2

2 5 2 2

N O O0

N O O Ott

V VKt ln ln

V V V

= =

−

Sol. : 50

K 10 In50 10

=−

K = 0.02 min–1

Verifying the value of K

50

K 18 In50 15

=−

K = 0.02 min–1

PART - B (CHEMISTRY)


6/12

32. Answer (1)

Hint : a2

1 1 2

Ek 1 1log

k 2.303 R T T

= −

3

a

4

E2 10 1 1log

2.303 R 400 42010

−

−

= −

aE 201.3

2.303 R 400 420=

Ea = 2.303 × 10920 × R

Sol. : a1

1

Elog k log A

RT 2.303= −

4 2.303 10920 Rlog10 log A

2.303 R 400

− = −

log A = 23.3

A 1023

33. Answer (3)

Hint : Chemisorption has high Ea that’s why

rate of chemisorption is enchanced by

increase in temp.

Sol. : But after a certain temperature, rate of

adsorption may decrease as the formed

bonds may dissociate at higher

temperature.

34. Answer (4)

Hint : B 1

C 2

n 2K

n 3K=

Sol. : –2

B 1–2

C 2

P 2K 2 3 101

P 3K 3 2 10

= = =

PB = PC

PB = 2 atm

35. Answer (3)

Hint : x = 1 ; y = 0, z = 2

Sol. : In Graph I , As rate increases linearly

with concentration reaction is of first

order

In Graph II, As concentration decreases

linearly, reaction is of zero order.

In Graph III, Product of half life and initial

concentration is constant, hence

reaction is of second order.

36. Answer (2)

Hint : r = K [R]x [H+]2

Sol. : r = K [R]x . [H+]2

2

A A2

BB

Hr4

r H

+

+

= =

a(HA)

a(HB)

K4

K=

6

a(HB)K 2.5 10−=

37. Answer (4)

Hint : Gases with greater critical temperature

are adsorbed to a greater extent.

Sol. :

SO CH H2 4 2

1 32

P P P

i.e P P P

38. Answer (4)

Hint : At high temperature physisorption may

change into chemisorptions.

Sol. : Physisorption is multilayered whereas

chemisorption is monolayered.

39. Answer (3)

Hint : Lead chamber process employs NO as

a catalyst

Sol. : ( ) ( ) ( ) ( )NO g

2 2 32SO g O g 2SO g+ ⎯⎯⎯⎯→

40. Answer (2)

Hint : Emulsions can be diluted with any

amount of dispersion medium

Sol. : The dispersed liquid on mixing with an

emulsion forms a separate layer

41. Answer (1)

Hint : Preferential adsorption of common ion

present in excess.


7/12

Sol. : I. Fe2O3.xH2O/Fe3+

II. Fe2O3.xH2O/OH–

42. Answer (4)

Hint : (1), (2) and (3) represent adsorption

Sol. : Water vapour are absorbed on

anhydrous calcium chloride.

43. Answer (2)

Hint : Leaching of silver

( )–– –

2 2 24Ag 8CN 2H O O 4 Ag CN 4OH + + + → +

Sol. : Coordination number of Ag in [Ag(CN)2]–

is 2

44. Answer (1)

Hint : It induces the precipitation of hydrated

Al2O3

Sol. : 2Na[Al(OH)4] + CO2 → 2NaHCO3

+ Al2O3xH2O

45. Answer (4)

Hint : ( )

( )( )

2 3 2 2 3 2S x

Fe O .xH O Fe O xH O g

⎯⎯→ +

2 2(y)

ZnS O ZnO SO (g)

+ ⎯⎯→ +

Sol. : Oxidation number of O in H2O is –2 and

of S in SO2 is 4.

Difference = 6

46. Answer (3)

Hint : Composition of the slag formed is

FeSiO3.

Sol. : % of oxygen in FeSiO3

48

100132

= 36%

47. Answer (4)

Hint : Al from Al2O3 is not extracted by carbon

based reduction method.

Sol. : Oxides of Fe, Sn, Zn, Pb are reduced by

carbon

48. Answer (1)

Hint : That characteristic type of copper is

called as blister copper due to the

evolution of SO2

Sol. : Ag and Au impurities setteles as anode

mud.

49. Answer (1)

Hint : 4 2 2 5P SO Cl give PCl+

Sol. : 5 3 2PCl PCl Cl

⎯⎯→ +

4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl+ ⎯⎯→ + +

50. Answer (4)

Hint: NH4NO3 liberates NH3 gas with KOH

and KNO3 on heating dissociates to give

O2 which supports combustion.

Sol. :

4 3 3 2 3NH NO KOH NH H O KNO+ ⎯⎯→ + +

3 3 2 2 2KNO 4Zn 7KOH NH 4K ZnO 2H O+ + ⎯⎯→ + +

3 2 22KNO 2KNO O

⎯⎯→ +

51. Answer (2)

Hint : Red coloured solution show the

presence of Br2.

Sol. : Cl2 being stronger oxidising agent can

oxidise Br– to Br2.

52. Answer (1)

Hint : HI and HBr are strong reducing hydra

acids and hence they reduce H2SO4.

Sol. : HCI is quite stable and hence is oxidised

by strong oxidising agent like KMnO4.

HF is not a reducing agent as in F– ion, the

electron which is to be removed during

oxidation is very close to the nucleus.

53. Answer (3)

Hint : 3O2 → 2O3

( )H 298K 142 kJ / mol

= + process is

endothermic.

Sol. : O3 decompose at high temperature


8/12

54. Answer (3)

Hint : ( ) ( )

6 2 3BA

XeF 3H O XeO 6HF+ ⎯⎯→ +

Sol. : A is pyramidal

55. Answer (2)

Hint : SiO2 is polymeric solid

Sol. : SeO2, TeF4 are solid

56. Answer (1)

Hint : Greater the number of unpaired electrons

greater will be the paramagnetic character

Sol. : Mn2+ in MnSO4 .4H2O is 3d5

Cu2+ in CuSO4.5H2O is 3d9

Fe2+ in FeSO4.6H2O is 3d6

Ni2+ in NiSO4 .6H2O is 3d8

57. Answer (3)

Hint : 2MnO2 + 4KOH + O2 → 2K2MnO4 +

2H2O

Sol. : K2MnO4 is dark green in colour.

58. Answer (1)

Hint : Products are I2 and IO3–

Sol. :

( )

– 24 2 2

in acidicconditions

10 2MnO 16H 2Mn I 8H O+ ++ + ⎯⎯→ + +

( )

– – – –4 2 2 3

in neutralconditions

I 2MnO H O 2MnO 2OH IO+ + ⎯⎯→ + +

59. Answer (4)

Hint : Colour of d-block cations is due to d – d

transitions.

Sol. : Cu2+ (CuCl2) and VOCl2 (V+4) both show

blue colour in aq. medium.

60. Answer (2)

Hint : Lu+2 is 4f14 5d1

Sol. : Ce+2 4f2

Gd+2 4f7 5d1

Lu+2 4f14 5d1

Eu+2 4f7 5d0 6s0

61. Answer (2)

Hint : Use G.P. to get sum then divide by

51/n

Sol. : The given value

=

( )

1

1

1

2 12 2

5 1 55

lim

1 5 1 5

nn

n

nn n

nn

n

n n

+

→

− −

− −

= ( )

111

211

5 1 5 5lim

5 15 111

nn

n

nn

nn

+

→

− +

− −

= –4(log5e)2 + 5log5e

62. Answer (4)

Hint : Factorize then cancel the common term

Sol. :

1

2

4cos ( 1)

(tan 1)(tan tan 1)lim

(tan 1)(tan tan 1)x

x x x

x x x−→ −

+ − +

+ + −

2

4

( 1) ( 1) 13

( 1) 1 1

− − − += = −

− − −

63. Answer (2)

Hint : Properties of G.I.F. and modulus function

Sol. :

2

2( ) cos(| |) | | 3 | 2 |

2 1

xf x x x x x

x

= + + + −

+

Only non differentiable at x = 2

PART - C (MATHEMATICS)


9/12

64. Answer (2)

Hint : Sum of infinite series then differentiation

Sol. : y = sinx y+

y2 = sinx + y

2 cosdy dy

y xdx dx

= +

cos

2 1

dy x

dx y=

−

2

cos

2 1 2sin

dy x

dx y x=

− −

65. Answer (4)

Hint : Algebra of limit then solve simultaneous

eq.

Sol. : p(x) = f(x) + g(x)

and q(x) = f(x) – g(x)

( ) ( )

lim2→

+

x a

p x q x and

( ) ( )lim

2→

−

x a

p x q x exist

( ) ( )

lim 42→

+=

x a

p x q x and

( ) ( )lim 1

2→

−=

x a

p x q x

( ) 4

lim 4( ) 1x a

f x

g x→= =

66. Answer (1)

Hint : is repeated root then f() = f () = 0

Sol. : A six degree polynomial satisfied by

exactly 5 real numbers then exactly one

of the f(0), f (–1), f (1),f (2) or f (–2) is

zero

67. Answer (4)

Hint :

dydy dt

dxdx

dt

= , then t = 1 to get coordinate

Sol. : 224

3 38

dy tt

dx t= = =

t = 1

Required coordinate = (7, 7)

68. Answer (2)

Hint : Differentiation of composite function

then interval

Sol. : Here g(x) = f(cos2x + 4cosx + 7)

On differentiating both sides w.r.t. x we

get :

g (x) = –2sinx(2 + cosx)f (cos2x + 4cosx + 7)

Here g(x) > 0 if sinx < 0

x(, 2)

69. Answer (3)

Hint : f (x) = 0 sin2x = 1

4. Then f(x) has

maxima

Sol. : f(x) = sin2x + cos4x

f (x) = 2cos2x – 4sin4x

= 2cos2x(1 – 4 sin2x)

Here f (x) = 0 if cos2x = 0 or sin2x = 1

4

f(x) has maximum value when sin2x = 1

4

f(x)maximum =

21 1

1 24 4

+ −

=

9

8

70. Answer (4)

Hint : Find the range by maximum and

minimum value

Sol. : We can see that f(x) cannot lie between

25 1and

2 2− −

71. Answer (2)

Hint : f(x) is discontinuous at x = 10

Sol. : The given function is only discontinuous

at x = 10.

10 10

lim ( ) lim ( )x x

f x f x− +→ →


10/12

72. Answer (3)

Hint :

( )

1 1

2 2

1

2x 2xcos sin

21 x 1 x

2tan x2

− −

−

= −

+ +

= − −

Sol. : f(x) = 1

2

2cos

1

xx

x

− −

+

= 1

2

2sin

2 1

xx

x

− − +

+

= 12tan2

x x−− + −

{ x = tan1}

= 12tan2

x x−+ −

f (x) = 2

2 2

2 11

1 1

x

x x

−− =

+ +

and g(x) = 1tan

42

xx

− −

= 2tan–1x +1 – x, as x = tan1

g (x) = 2

2 2

2 11

1 1

x

x x

−− =

+ +

2

2

2

2

1

( ( )) 1 1( ( )) 1

1

x

d f x xd g x x

x

−

+= = −−

+

73. Answer (2)

Hint : 3

2 , 154

dA daa a

dt dt= =

and 0.1=da

dt cm/s

Sol. : 0.1da

dt= cm/sec and

23

4A a=

3 3

0.1 152 2

dA daa

dt dt= = cm2/sec

= 3 3

4

74. Answer (3)

Hint : f(g(x)) = x if f and g are inverse of each

other

f (g(x)) = g (f(x)) = 1

Sol. : g(f(x)) = x

1

( ( ))( )

g f xf x

=

1

( (1))(1)

g ff

=

1 1

(3)6 4 1 11

g = =+ +

75. Answer (4)

Hint : 0 0

(0) lim ( ) lim ( )− +→ →

= =x x

f f x f x

Sol. :

0

lim ( )x

f x→

= tan

0

ln(sec tan )lim

tan

x x

x

e e x x x

x x→

− + + −

−

= tan 2

20

sec sec 1lim

sec 1

x x

x

e x e x

x→

− + −

−

=

( )

tan 2 tan

20

(sec 1) (sec 1)lim

sec 1

x x x

x

e x x e e

x→

− + − + −

−

= tan

20

11 lim

2 tan

x x

x

e e

x→

−+ +

= 3 3

02 2

+ =

76. Answer (4)

Hint : 1 form

Sol. :

1

11

lim2 1

21

lim lim2

n

n n

en

n

n n

ee e

→

−

→ →

+= =

77. Answer (2)

Hint : 2

2

2

2 4( ) ln 4

2 4

+ − = − − − −

xf x x

x

Let x = 2sin


11/12

( ) 2ln cot 2cos2

= −

f x

24( ) ,

− − =

xf x

x

Sol. : 2

2

2

2 4( ) ln 4

2 4

xf x x

x

+ − = − − − −

Let x = 2sin then

24

( )− −

=x

f xx

and x(–2, 2) – {0}

f(x) is decreasing in (0, 1)

78. Answer (2)

Hint : f(a – h) < f(a) > f(a + h) then get b

Sol. : 2

2

2 | 5 6 |, 2( )

1, 2

x x xf x

b x

− − + =

+ =

b2 + 1 > 2 for local maxima

|b| 1

79. Answer (2)

Hint : Find domain

Sol. : –1 x 1 or x(–, –1] [1, )

x = ± 1

4

64

= −

80. Answer (1)

Hint : C1 → C1 + C2

Sol. : C1 → C1 + C2

{ } 1 1 { } 1

– {– } –1 –1 {– } –1 0

0 1 1 0 1

x x x

x x x x

x

= = =

81. Answer (2)

Hint : Coefficient of x in (x) = (0)

Sol. : Coefficient of x in (x) is (0)

2 1

1

21 2

2

1sin –1

1 sin cos –11

( ) cos 0 cos – sin 0 01

tan sec 11

tan 11

xx x x

x x x xx

x x

xx

−

−

−

−

= − + +−

+

0 1 11 0 1

(0) 1 1 0 0 0 (1 1) 22 2 2

1 0 10 1 1

−−

= − + = + + − − = −

82. Answer (1)

Hint : Multiplication of two determinants

Sol. :

2 2

2 2

1 1

( ) 1 1 0

1 1 1 0 0 0

x x x x

f x x x x x= =

( )2 1 0f − =

83. Answer (1)

Hint : Expansion

Sol. :

1 1 1 1 2 31 1

2 2 2 1 2 32 2

1 2 31 0 0

A B C a a aB C

A B C b b bB C

c c c

=

1 1 23

2 21 2 3

0 0

0 0B C

aB C

a a a

= =

1 1

32 2

B Ca

B C =

84. Answer (3)

Hint : = 0

Sol. : For non-trivial solution

1 2

2 1 3 0,

3 1

t t t

t t t

t t t

+ +

+ + + =

+ +

R2→R2 – R1, R3 → R3 –R1


12/12

1 2

1 1 3 0

1 1 1

t t t+ +

− =

−

(t + 1)(–1–3)–(t + 2)(1 + 3) + t(1 – 1) = 0

–4t – 4 – 4t –8 = 0

–8t = 12 3

2t = −

85. Answer (1)

Hint : Property of diagonal matrix

Sol. :

88

0

88

0 1 2 880

88

0

tan( 1) 0 0

... 0 cot( 1) 0

0 0 1

r

r

r

r

A A A A r

=

=

=

+

= +

88

0

tan( 1)r

r=

+ = tan1°tan2°..tan45°…tan88°.

tan89° = 1

88

0

cot( 1)r

r=

+ = cot1°cot2°…cot45°…cot88°

.cot89°= 1

88

0

1 1r =

=

86. Answer (4)

Hint : |A – I| = 0

Sol. :

1 2 3

| | 0, 2 3 1 0

3 1 2

A I

−

− = − =

−

3 – 6 2 – 3 + 18 = 0

A3 – 6A2 – 3A – 18I = 0

A2 – 6A – 18A–1 = 3I

87. Answer (1)

Hint : Property of adj. and inverse

Sol. : (adjA–1)–1(adjA)–1 = ((adjA)(adjA–1))–1

= (adj(A–1A))–1 = (adjI)–1 = (I)–1 = I

88. Answer (3)

Hint : Multiplication of matrices

Sol. :

2

0 0 1 0 0 1 1 0 0

0 1 0 0 1 0 0 1 0

1 0 0 1 0 0 0 0 1

A I

= − − = =

A is involutory matrix

|A| = 1 non-singular matrix

89. Answer (1)

Hint : Properties of adj. and inverse

Sol. :

|(2adj3A–1)–1| = |2adj(3A–1)|–1 = 2–3|adj(3A–1)|–1

= 2–3|32adj(A–1)|–1 = 2–3(32)–3|adj(A–1)|–1

= 18–3|(adjA)–1|–1 = 18–3|adjA| = 18–3|A|2

= 1 1

27 27 218 18 18 4

=

90. Answer (1)

Hint : Find

Sol. :

1 1 1

2 1 3 1( 3 3) 1(6 3) 1(2 1)

1 1 3

= − = − − − − + +

= –6 – 3 + 3 = – 6 0

System has a unique solution

Test - 2 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020

Test Date : 11/08/2019

ANSWERS

1/12

TEST - 2 - Code-B

PHYSICS CHEMISTRY MATHEMATICS

1. (2)

2. (1)

3. (1)

4. (2)

5. (3)

6. (4)

7. (2)

8. (3)

9. (3)

10. (3)

11. (3)

12. (4)

13. (4)

14. (2)

15. (1)

16. (4)

17. (2)

18. (3)

19. (1)

20. (4)

21. (2)

22. (2)

23. (2)

24. (3)

25. (3)

26. (3)

27. (3)

28. (3)

29. (3)

30. (4)

31. (2)

32. (4)

33. (1)

34. (3)

35. (1)

36. (2)

37. (3)

38. (3)

39. (1)

40. (2)

41. (4)

42. (1)

43. (1)

44. (4)

45. (3)

46. (4)

47. (1)

48. (2)

49. (4)

50. (1)

51. (2)

52. (3)

53. (4)

54. (4)

55. (2)

56. (3)

57. (4)

58. (3)

59. (1)

60. (2)

61. (1)

62. (1)

63. (3)

64. (1)

65. (4)

66. (1)

67. (3)

68. (1)

69. (1)

70. (2)

71. (1)

72. (2)

73. (2)

74. (2)

75. (4)

76. (4)

77. (3)

78. (2)

79. (3)

80. (2)

81. (4)

82. (3)

83. (2)

84. (4)

85. (1)

86. (4)

87. (2)

88. (2)

89. (4)

90. (2)

All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)

2/12

1. Answer (2)

Hint : Magnetic field is uniform

Sol. :

By symmetry,

2 2 = = B d B B

0 in( ) = B d I

02 ( 1)B i =

0

2

iB

=

2. Answer (1)

Hint : Force = iB

Sol. : 01

2

IF i a

a

=

02

2 (2 )

IF i a

a

=

0net 1 2

11

2 2

= − = −

IiF F F 0

4

=

Ii

3. Answer (1)

Hint : Time spent =

Sol. : qB

m =

0

0

12sin

2

mV

d qB

mVr

qB

= = =

6

=

6

6

mt

qB qB

m

= =

4. Answer (2)

Hint : Use U = ,i B− then dU

Fdx

−=

Sol. : 2

0 0

2 2 3/22( )

I RB

R x

=

+

µ = r2i

2

20 0

2 2 3/22( )

I RU r i

R x

=

+

2 2

0 0

2 2 5/2

3

2( )d

I R r iddU

R ddx

=

+

5. Answer (3)

Hint : At null point potential become same

Sol. : RAB = 10 Vend = 10 Volt

25 cm4

L= =

6. Answer (4)

Hint : Electromagnet requires temporary

magnet

Sol. : Low retentivity and low coercivity

facilitates temporary magnetic property

in core material.

7. Answer (2)

Hint : Redraw the circuit

Sol. :

eq

3

RR =

= 1

8. Answer (3)

Hint : V = IR

Sol. : 12 12 3

| |2 8

− = −C DV V = 1.5 V

9. Answer (3)

Hint : q = qo (1 – e–t/)

= ReqC

PART - A (PHYSICS)

Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/12

Sol. : max

2

3q VC=

/22(1– )

3

VCq e−=

10. Answer (3)


Sol. : 0 in( )B dl i =

= µo (+3 – 6) = –3µo

11. Answer (3)

Hint : B does not change K.E of particle

Sol. : The charge particle moves on a

cycloidal path when B E⊥

12. Answer (4)

Hint : Integrate magnetic moment due to an

elementary ring.

Sol. : N

N dxr

=

2

2( )r

x r

d N I x=

=

2

2

r

x r

Nd dx I x

r=

=

13. Answer (4)

Hint : Both fields are perpendicular to each

other

Sol. : 0 1 0 21 2,

2 2

I IB B

d d

= =

B1 ⊥ B2

2 2

net 1 2B B B = +

14. Answer (2)

Hint : Deflection is directly proportional to current

Sol. : 3

max

510 A

4950 50i −= =

+

max imax

For 20 divisions,

3

max

2 210 A

3 3i i − = =

Let R be resistance in series

Then 35 210

50 3R

−= +

R = 7450

15. Answer (1)

Hint : Use condition of balanced Wheatstone

bridge

Sol. : 3 2

PS SQ=

2

100 cm5

SQ = = 40 cm

16. Answer (4)

Hint : Use Kirchhoff law

Sol. : eq

15 304

15 30R

= +

+ = 14

30

2 A14 1

I = =+

17. Answer (2)

Hint : Let resistance be X, then 2

12

= ++

XX

X

Sol. :

We have

2

12

XX

X= +

+

2X =

18. Answer (3)

Hint : V = IR

Sol. : If S1 is closed

3

1

3

3 1 4

E EV

= =

+


4/12

Similarly 2

6(6)

7 7

E EV

= =

and 3

2

3

EV =

19. Answer (1)

Hint : At t = 0 all capacitors will act like a

conducting wire

Sol. : total

eq

VI

R=

2

V

R

=

1

3 62

VVI

RR

= =

20. Answer (4)

Hint : Baxis 0 0equatorial3 3

2,

4 4

= =

m mB

d d

Sol. : 01 3

2 ˆ4

mB i

a

=

( )02 3

ˆ4

mB i

a

= −

01 2 3

ˆ( )4

p

mB B B i

a

= + =

21. Answer (2)

Hint and Sol. :

VPQ = IPQ RPQ

Apply Kirchhoff law to get current in wire PQ

|VQ – VP| = I(1)

I = 0.13 A from Q to P

22. Answer (2)

Hint : Use concept of effective length

Sol. :

3

0 12

2

R

x R

IdF I dx

x=

=

0 1 2 ln(3)2

I IF

=

23. Answer (2)

Hint : cot2 = cot21 + cot22

Sol. : cot2 = cot21 + cot22

1 10

33 3

= + =

10

cot3

=

3

sin13

=

1 3sin

13

− =

24. Answer (3)

Hint : AV = constant

Sol.: AV = constant

Since A2 < A1

V2 > V1

25. Answer (3)

Hint :

0 (4 2)

42

=

s

IB

l

Sol. :

0 01 2 22 4

242

s

I IB

a a

= =

0

2c

IB

R

=

2 2 2

1s

c

B R

B a

=

Now, 4a = 2R 2

Ra

=

2

4 2 2 8 2S

C

B R

B R

= =

26. Answer (3)

Hint : , 2 ( )mv

r mv m KEqB

= =

Sol.: 2 ( )m KEmv

rqB qB

= =


5/12

2 ( . )

,ee

m K Er

exB =

2 ( . )p

p

m K Er

eB=

2 ( )2 ( )

(2 )

= =

pm KEm KEr

e B eB

Clearly, rp > re and rp = r

27. Answer (3)

Hint: Identify the equipotential points and

redraw the circuit.

Sol. : It can be reduced as

RAB = 7

12R by simple calculation

28. Answer (3)


Sol. : 0 inB dl i = 1forB r r R

30 forB r R=

29. Answer (3)

Hint : 2= I

TMB

Sol. : 0

603 s

20T = =

Now, 01

0 2 02

BBT

T B B= =

0 3s

2 2

TT = =

30. Answer (4)

Hint : 0

1 2(cos cos )4

= +

IB

d

Sol. : 2 3

d =

= 30°

03 (cos30 cos30 )

42 3

IB

= +

( )0 2 33 3

4

I =

09

2

I=

31. Answer (2)

Hint : Lu+2 is 4f14 5d1

Sol. : Ce+2 4f2

Gd+2 4f7 5d1

Lu+2 4f14 5d1

Eu+2 4f7 5d0 6s0

32. Answer (4)

Hint : Colour of d-block cations is due to d – d

transitions.

Sol. : Cu2+ (CuCl2) and VOCl2 (V+4) both show

blue colour in aq. medium.

33. Answer (1)

Hint : Products are I2 and IO3–

Sol. :

( )

– 24 2 2

in acidicconditions

10 2MnO 16H 2Mn I 8H O+ ++ + ⎯⎯→ + +

( )

– – – –4 2 2 3

in neutralconditions

I 2MnO H O 2MnO 2OH IO+ + ⎯⎯→ + +

PART - B (CHEMISTRY)


6/12

34. Answer (3)

Hint : 2MnO2 + 4KOH + O2 → 2K2MnO4 +

2H2O

Sol. : K2MnO4 is dark green in colour.

35. Answer (1)

Hint : Greater the number of unpaired electrons

greater will be the paramagnetic character

Sol. : Mn2+ in MnSO4 .4H2O is 3d5

Cu2+ in CuSO4.5H2O is 3d9

Fe2+ in FeSO4.6H2O is 3d6

Ni2+ in NiSO4 .6H2O is 3d8

36. Answer (2)

Hint : SiO2 is polymeric solid

Sol. : SeO2, TeF4 are solid

37. Answer (3)

Hint : ( ) ( )

6 2 3BA

XeF 3H O XeO 6HF+ ⎯⎯→ +

Sol. : A is pyramidal

38. Answer (3)

Hint : 3O2 → 2O3

( )H 298K 142 kJ / mol

= + process is

endothermic.

Sol. : O3 decompose at high temperature

39. Answer (1)

Hint : HI and HBr are strong reducing hydra

acids and hence they reduce H2SO4.

Sol. : HCI is quite stable and hence is oxidised

by strong oxidising agent like KMnO4.

HF is not a reducing agent as in F– ion, the

electron which is to be removed during

oxidation is very close to the nucleus.

40. Answer (2)

Hint : Red coloured solution show the

presence of Br2.

Sol. : Cl2 being stronger oxidising agent can

oxidise Br– to Br2.

41. Answer (4)

Hint: NH4NO3 liberates NH3 gas with KOH

and KNO3 on heating dissociates to give

O2 which supports combustion.

Sol. :

4 3 3 2 3NH NO KOH NH H O KNO+ ⎯⎯→ + +

3 3 2 2 2KNO 4Zn 7KOH NH 4K ZnO 2H O+ + ⎯⎯→ + +

3 2 22KNO 2KNO O

⎯⎯→ +

42. Answer (1)

Hint : 4 2 2 5P SO Cl give PCl+

Sol. : 5 3 2PCl PCl Cl

⎯⎯→ +

4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl+ ⎯⎯→ + +

43. Answer (1)

Hint : That characteristic type of copper is

called as blister copper due to the

evolution of SO2

Sol. : Ag and Au impurities setteles as anode

mud.

44. Answer (4)

Hint : Al from Al2O3 is not extracted by carbon

based reduction method.

Sol. : Oxides of Fe, Sn, Zn, Pb are reduced by

carbon

45. Answer (3)

Hint : Composition of the slag formed is

FeSiO3.

Sol. : % of oxygen in FeSiO3

48

100132

= 36%

46. Answer (4)

Hint : ( )

( )( )

2 3 2 2 3 2S x

Fe O .xH O Fe O xH O g

⎯⎯→ +

2 2

(y)

ZnS O ZnO SO (g)

+ ⎯⎯→ +


7/12

Sol. : Oxidation number of O in H2O is –2 and

of S in SO2 is 4.

Difference = 6

47. Answer (1)

Hint : It induces the precipitation of hydrated

Al2O3

Sol. : 2Na[Al(OH)4] + CO2 → 2NaHCO3

+ Al2O3xH2O

48. Answer (2)

Hint : Leaching of silver

( )–– –

2 2 24Ag 8CN 2H O O 4 Ag CN 4OH + + + → +

Sol. : Coordination number of Ag in [Ag(CN)2]–

is 2

49. Answer (4)

Hint : (1), (2) and (3) represent adsorption

Sol. : Water vapour are absorbed on

anhydrous calcium chloride.

50. Answer (1)

Hint : Preferential adsorption of common ion

present in excess.

Sol. : I. Fe2O3.xH2O/Fe3+

II. Fe2O3.xH2O/OH–

51. Answer (2)

Hint : Emulsions can be diluted with any

amount of dispersion medium

Sol. : The dispersed liquid on mixing with an

emulsion forms a separate layer

52. Answer (3)

Hint : Lead chamber process employs NO as

a catalyst

Sol. : ( ) ( ) ( ) ( )NO g

2 2 32SO g O g 2SO g+ ⎯⎯⎯⎯→

53. Answer (4)

Hint : At high temperature physisorption may

change into chemisorptions.

Sol. : Physisorption is multilayered whereas

chemisorption is monolayered.

54. Answer (4)

Hint : Gases with greater critical temperature

are adsorbed to a greater extent.

Sol. :

SO CH H2 4 2

1 32

P P P

i.e P P P

55. Answer (2)

Hint : r = K [R]x [H+]2

Sol. : r = K [R]x . [H+]2

2

A A2

BB

Hr4

r H

+

+

= =

a(HA)

a(HB)

K4

K=

6a(HB)K 2.5 10−=

56. Answer (3)

Hint : x = 1 ; y = 0, z = 2

Sol. : In Graph I , As rate increases linearly

with concentration reaction is of first

order

In Graph II, As concentration decreases

linearly, reaction is of zero order.

In Graph III, Product of half life and initial

concentration is constant, hence

reaction is of second order.

57. Answer (4)

Hint : B 1

C 2

n 2K

n 3K=

Sol. : –2

B 1–2

C 2

P 2K 2 3 101

P 3K 3 2 10

= = =

PB = PC

PB = 2 atm

58. Answer (3)

Hint : Chemisorption has high Ea that’s why

rate of chemisorption is enchanced by

increase in temp.


8/12

Sol. : But after a certain temperature, rate of

adsorption may decrease as the formed

bonds may dissociate at higher

temperature.

59. Answer (1)

Hint : a2

1 1 2

Ek 1 1log

k 2.303 R T T

= −

3

a

4

E2 10 1 1log

2.303 R 400 42010

−

−

= −

aE 201.3

2.303 R 400 420=

Ea = 2.303 × 10920 × R

Sol. : a1

1

Elog k log A

RT 2.303= −

4 2.303 10920 Rlog10 log A

2.303 R 400

− = −

log A = 23.3

A 1023

60. Answer (2)

Hint: By using gas law and kinetics law, it can

be assumed that at 2 5 2

0N O OV V

2 5 2

2 5 2 2

N O O0

N O O Ott

V VKt ln ln

V V V

= =

−

Sol. : 50

K 10 In50 10

=−

K = 0.02 min–1

Verifying the value of K

50

K 18 In50 15

=−

K = 0.02 min–1

61. Answer (1)

Hint : Find

Sol. :

1 1 1

2 1 3 1( 3 3) 1(6 3) 1(2 1)

1 1 3

= − = − − − − + +

= –6 – 3 + 3 = – 6 0

System has a unique solution

62. Answer (1)

Hint : Properties of adj. and inverse

Sol. :

|(2adj3A–1)–1| = |2adj(3A–1)|–1 = 2–3|adj(3A–1)|–1

= 2–3|32adj(A–1)|–1 = 2–3(32)–3|adj(A–1)|–1

= 18–3|(adjA)–1|–1 = 18–3|adjA| = 18–3|A|2

= 1 1

27 27 218 18 18 4

=

63. Answer (3)

Hint : Multiplication of matrices

Sol. :

2

0 0 1 0 0 1 1 0 0

0 1 0 0 1 0 0 1 0

1 0 0 1 0 0 0 0 1

A I

= − − = =

A is involutory matrix

|A| = 1 non-singular matrix

64. Answer (1)

Hint : Property of adj. and inverse

Sol. : (adjA–1)–1(adjA)–1 = ((adjA)(adjA–1))–1

= (adj(A–1A))–1 = (adjI)–1 = (I)–1 = I

65. Answer (4)

Hint : |A – I| = 0

Sol. :

1 2 3

| | 0, 2 3 1 0

3 1 2

A I

−

− = − =

−

3 – 6 2 – 3 + 18 = 0

A3 – 6A2 – 3A – 18I = 0

A2 – 6A – 18A–1 = 3I

PART - C (MATHEMATICS)


9/12

66. Answer (1)

Hint : Property of diagonal matrix

Sol. :

88

0

88

0 1 2 880

88

0

tan( 1) 0 0

... 0 cot( 1) 0

0 0 1

r

r

r

r

A A A A r

=

=

=

+

= +

88

0

tan( 1)r

r=

+ = tan1°tan2°..tan45°…tan88°

tan89° = 1

88

0

cot( 1)r

r=

+ = cot1°cot2°…cot45°…cot88°

. cot89°= 1

88

0

1 1r =

=

67. Answer (3)

Hint : = 0

Sol. : For non-trivial solution

1 2

2 1 3 0,

3 1

t t t

t t t

t t t

+ +

+ + + =

+ +

R2→R2 – R1, R3 → R3 –R1

1 2

1 1 3 0

1 1 1

t t t+ +

− =

−

(t + 1)(–1–3)–(t + 2)(1 + 3) + t(1 – 1) = 0

–4t – 4 – 4t –8 = 0

–8t = 12 3

2t = −

68. Answer (1)

Hint : Expansion

Sol. :

1 1 1 1 2 31 1

2 2 2 1 2 32 2

1 2 31 0 0

A B C a a aB C

A B C b b bB C

c c c

=

1 1 23

2 21 2 3

0 0

0 0B C

aB C

a a a

= =

1 1

32 2

B Ca

B C =

69. Answer (1)

Hint : Multiplication of two determinants

Sol. :

2 2

2 2

1 1

( ) 1 1 0

1 1 1 0 0 0

x x x x

f x x x x x= =

( )2 1 0f − =

70. Answer (2)

Hint : Coefficient of x in (x) = (0)

Sol. : Coefficient of x in (x) is (0)

2 1

1

21 2

2

1sin –1

1 sin cos –11

( ) cos 0 cos – sin 0 01

tan sec 11

tan 11

xx x x

x x x xx

x x

xx

−

−

−

−

= − + +−

+

0 1 11 0 1

(0) 1 1 0 0 0 (1 1) 22 2 2

1 0 10 1 1

−−

= − + = + + − − = −

71. Answer (1)

Hint : C1 → C1 + C2

Sol. : C1 → C1 + C2

{ } 1 1 { } 1

– {– } –1 –1 {– } –1 0

0 1 1 0 1

x x x

x x x x

x

= = =

72. Answer (2)

Hint : Find domain

Sol. : –1 x 1 or x(–, –1] [1, )

x = ± 1

4

64

= −


10/12

73. Answer (2)

Hint : f(a – h) < f(a) > f(a + h) then get b

Sol. : 2

2

2 | 5 6 |, 2( )

1, 2

x x xf x

b x

− − + =

+ =

b2 + 1 > 2 for local maxima

|b| 1

74. Answer (2)

Hint : 2

2

2

2 4( ) ln 4

2 4

+ − = − − − −

xf x x

x

Let x = 2sin

( ) 2ln cot 2cos2

= −

f x

24( ) ,

− − =

xf x

x

Sol. : 2

2

2

2 4( ) ln 4

2 4

xf x x

x

+ − = − − − −

Let x = 2sin then

24( ) ,

− − =

xf x

x and x(–2, 2) – {0}

f(x) is decreasing in (0, 1)

75. Answer (4)

Hint : 1 form

Sol. :

1

11

lim2 1

21

lim lim2

n

n n

en

n

n n

ee e

→

−

→ →

+= =

76. Answer (4)

Hint : 0 0

(0) lim ( ) lim ( )− +→ →

= =x x

f f x f x

Sol. :

0

lim ( )x

f x→

= tan

0

ln(sec tan )lim

tan

x x

x

e e x x x

x x→

− + + −

−

= tan 2

20

sec sec 1lim

sec 1

x x

x

e x e x

x→

− + −

−

= ( )

tan 2 tan

20

(sec 1) (sec 1)lim

sec 1

x x x

x

e x x e e

x→

− + − + −

−

= tan

20

11 lim

2 tan

x x

x

e e

x→

−+ +

= 3 3

02 2

+ =

77. Answer (3)

Hint : f(g(x)) = x if f and g are inverse of each

other

f (g(x)) = g (f(x)) = 1

Sol. : g(f(x)) = x

1

( ( ))( )

g f xf x

=

1

( (1))(1)

g ff

=

1 1

(3)6 4 1 11

g = =+ +

78. Answer (2)

Hint : 3

2 , 154

dA daa a

dt dt= =

and 0.1=da

dt cm/s

Sol. : 0.1da

dt= cm/sec and

23

4A a=

3 3

0.1 152 2

dA daa

dt dt= = cm2/sec

= 3 3

4


11/12

79. Answer (3)

Hint :

( )

1 1

2 2

1

2x 2xcos sin

21 x 1 x

2tan x2

− −

−

= −

+ +

= − −

Sol. : f(x) = 1

2

2cos

1

xx

x

− −

+

= 1

2

2sin

2 1

xx

x

− − +

+

= 12tan2

x x−− + −

{ x = tan1}

= 12tan2

x x−+ −

f (x) = 2

2 2

2 11

1 1

x

x x

−− =

+ +

and g(x) = 1tan

42

xx

− −

= 2tan–1x +1 – x, as x = tan1

g (x) = 2

2 2

2 11

1 1

x

x x

−− =

+ +

2

2

2

2

1

( ( )) 1 1( ( )) 1

1

x

d f x xd g x x

x

−

+= = −−

+

80. Answer (2)

Hint : f(x) is discontinuous at x = 10

Sol. : The given function is only discontinuous

at x = 10.

10 10

lim ( ) lim ( )x x

f x f x− +→ →

81. Answer (4)

Hint : Find the range by maximum and

minimum value

Sol. : We can see that f(x) cannot lie between

25 1and

2 2− −

82. Answer (3)

Hint : f (x) = 0 sin2x = 1

4. Then f(x) has

maxima

Sol. : f(x) = sin2x + cos4x

f (x) = 2cos2x – 4sin4x

= 2cos2x(1 – 4 sin2x)

Here f (x) = 0 if cos2x = 0 or sin2x = 1

4

f(x) has maximum value when sin2x = 1

4

f(x)maximum = 2

1 11 2

4 4

+ −

=

9

8

83. Answer (2)

Hint : Differentiation of composite function

then interval

Sol. : Here g(x) = f(cos2x + 4cosx + 7)

On differentiating both sides w.r.t. x we

get :

g (x) = –2sinx(2 + cosx)f (cos2x + 4cosx + 7)

Here g(x) > 0 if sinx < 0

x(, 2)

84. Answer (4)

Hint :

dydy dt

dxdx

dt

= , then t = 1 to get coordinate

Sol. : 224

3 38

dy tt

dx t= = =

t = 1

Required coordinate = (7, 7)

85. Answer (1)

Hint : is repeated root then f() = f () = 0


12/12

Sol. : A six degree polynomial satisfied by

exactly 5 real numbers then exactly one

of the f(0), f (–1), f (1),f (2) or f (–2) is

zero

86. Answer (4)

Hint : Algebra of limit then solve simultaneous

eq.

Sol. : p(x) = f(x) + g(x)

and q(x) = f(x) – g(x)

( ) ( )

lim2→

+

x a

p x q x and

( ) ( )lim

2→

−

x a

p x q x exist

( ) ( )

lim 42→

+=

x a

p x q x and

( ) ( )lim 1

2→

−=

x a

p x q x

( ) 4

lim 4( ) 1x a

f x

g x→= =

87. Answer (2)

Hint : Sum of infinite series then differentiation

Sol. : y = sinx y+

y2 = sinx + y

2 cosdy dy

y xdx dx

= +

cos

2 1

dy x

dx y=

−

2

cos

2 1 2sin

dy x

dx y x=

− −

88. Answer (2)

Hint : Properties of G.I.F. and modulus function

Sol. :

2

2( ) cos(| |) | | 3 | 2 |

2 1

xf x x x x x

x

= + + + −

+

Only non differentiable at x = 2

89. Answer (4)

Hint : Factorize then cancel the common term

Sol. : 1

2

4cos ( 1)

(tan 1)(tan tan 1)lim

(tan 1)(tan tan 1)x

x x x

x x x−→ −

+ − +

+ + −

2

4

( 1) ( 1) 13

( 1) 1 1

− − − += = −

− − −

90. Answer (2)

Hint : Use G.P. to get sum then divide by

51/n

Sol. : The given value

=

( )

1

1

1

2 12 2

5 1 55

lim

1 5 1 5

nn

n

nn n

nn

n

n n

+

→

− −

− −

= ( )

111

211

5 1 5 5lim

5 15 111

nn

n

nn

nn

+

→

− +

− −

= –4(log5e)2 + 5log5e

(answers) all india aakash test series for jee (main)-2020 ......test - 2 (code-a) (hints &...

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