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Test - 2 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2022

All India Aakash Test Series for JEE (Main)-2022

Test Date : 29/11/2020

ANSWERS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

TEST - 2 (Code-A)

1/11

PHYSICS CHEMISTRY MATHEMATICS

1. (3)

2. (1)

3. (3)

4. (1)

5. (3)

6. (4)

7. (2)

8. (1)

9. (2)

10. (4)

11. (3)

12. (4)

13. (1)

14. (2)

15. (4)

16. (2)

17. (2)

18. (2)

19. (2)

20. (2)

21. (07.00)

22. (02.00)

23. (04.00)

24. (75.00)

25. (02.00)

26. (2)

27. (3)

28. (4)

29. (1)

30. (2)

31. (1)

32. (2)

33. (1)

34. (1)

35. (4)

36. (1)

37. (2)

38. (1)

39. (1)

40. (1)

41. (1)

42. (2)

43. (4)

44. (4)

45. (4)

46. (39.00)

47. (04.00)

48. (05.00)

49. (75.00)

50. (12.50)

51. (2)

52. (2)

53. (3)

54. (4)

55. (4)

56. (2)

57. (1)

58. (2)

59. (3)

60. (1)

61. (4)

62. (3)

63. (1)

64. (3)

65. (3)

66. (2)

67. (2)

68. (1)

69. (3)

70. (4)

71. (13.00)

72. (05.00)

73. (02.00)

74. (21.00)

75. (30.00)

All India Aakash Test Series for JEE (Main)-2022 Test - 2 (Code-A) (Hints & Solutions)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 2/11

PART – A : PHYSICS

1. Answer (3)

Hint : R = 2hV

g

Sol.: R = 20 × 2 = 40 m

2. Answer (1)

Hint : sin60H

l=

Sol.: Let l be the side of the triangle then

sin60H

l=

and 2 2sin

42

u

g

=

2 2sin 10 3u l = …(1)

Range 2 sin2u

lg

= =

2 sin2 10u l = …(2)

From (1) and (2) ( )–1tan 2 3 =

3. Answer (3)

Hint : 2

2 2tan –

2 cos

gxy x

u=

Sol.: The equation to the trajectory is

2

2 2tan –

2 cos=

gxy x

u

2

2tan –

2=

gxx

u (1 + tan2)

(6, 8) should satisfy this

180tan2 – 6u2tan + (180 – 8u 2) = 0

For real , 0

4. Answer (1)

Hint : WP W PV = V –V

Sol.: –WP W PV V V=

It will be at 45°, when W PV V=

5. Answer (3)

Hint : –AB A BV V V=

Sol.: 2 2

AB A BV V V= + = 5 m/s

6. Answer (4)

Hint : 2 2dx dy

ydt dt

= +

Sol.: x = y2 + 2y + 2

2 2dx dy

ydt dt

= +

and

22 2

2 22 2

d x d y dyy

dtdt dt

= +

given 2

22 m/s 0

dy d y

dt dt= =

( )2

2 2

22 2 8 m/s

d x

dt= =

7. Answer (2)

Hint : S = 0.5t2 + 2t comparing with

21

2= +S at u gives at = 1 m/s2 and u = 2 m/s

Sol.: S = 0.5t2 + 2t comparing with 21

2= +S at u

gives at = 1 m/s2 and u = 2 m/s

also

22

2 2 2

t c t

va a a a

R

= + = +

8. Answer (1)

Hint : at = aR

Sol.: at = R

and ( )22

2 2

n

V O R ta R t

R R

+ = = =

according to question, 2 2R R t =

1

t =

9. Answer (2)

Hint : Friction will be limiting.

Sol.: For equation of upper ball

Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2022


2 cos303

mgN mg N = =

For equation of lower left ball

N = Nsin60° + mg

N = cos60° = f N

min

1

3 3 =

10. Answer (4)

Hint : When AB = 2m, max

sf f=

Sol.:

Let be the linear mass density of rope

For equilibrium Fcos37° = 40

and Fsin37° = 20

2

3 =

11. Answer (3)

Hint : tan = a

g

Sol.: 2

sinmv

TR

=

Tcos = mg

2

tan 1v

gR = =

12. Answer (4)

Hint : ap = 3a1 + 2a2

Sol.: 3

3

3m

f fa

m m= = →

2

2

2m

f fa

m m= =

and 1 2

53 2

fa a a

m= + =

13. Answer (1)

Hint : 2

rel rel rel

1

2S u t a t= +

Sol.: 1 2

1 2

– 3

5

m ma g g

m m

= = +

2

rel

612 m/s

5a g= =

Srel = 6

2

rel rel rel

1

2S u t a t= +

t = 1 s

14. Answer (2)

Hint : Force down the incline = macos +

mgsin

Sol.: Force down the incline = macos + mgsin

arel = gsin + acos

15. Answer (4)

Hint : Draw FBD and make equations.

Sol.: Let a be the acceleration of 4 kg mass

then 2a is the upward acceleration of 1 kg

mass. Drawing F.B.D. and writing equations

– 1 1 2

and 4 – 2 4 4

T g a ga

g T a

= =

=



16. Answer (2)

Hint : Acceleration if M and m are same

Sol.:

Let a be the according of the wedge

For equation of block

mgsin = macos

a = gtan

F = (M + m)gtan

17. Answer (2)

Hint : Find Fmin as function of

Sol.:

From F.B.D.

N = mg – Fsin

and Fcos = N = (mg – fsin)

cos sin

=

+

mgF

For fmin, cos + sin should be maximum

and it is equal to 21+

min 21

mgf

=

+

18. Answer (2)

Hint : tan = a

g

Sol.:

Ncos = mg

Nsin = m2r = m2Rsin

2

cosg

R =

19. Answer (2)

Hint : x = ( )3 – 1R

Sol.: Elongation = ( )2 cos30 – 3 – 1R R R x = =

Net force towards centre = kxcos30°+ mgcos30°

20. Answer (2)

Hint : ( )ˆ– ,y x RF mg j F ma= =

Sol.:

(T1 – T 2)cos = mg

(T 1 + T 2)sin = m2r

8

. tan2

g hm

h=

1 2

1 2

4–

T T

T T

+=

21. Answer (07.00)

Hint : Let PQ be along i

Sol.: Let PQ be along i

ˆ10AV i=

ˆ ˆ6cos 6sin= +B

A

V i j 15

cos24

=

ˆ–12=C

B

V i

Together ( ) ( )ˆ ˆ–2 6cos 6sin== + + CV i j

VC = 5 m/s



22. Answer (02.00)

Hint :

Sol.:

( )

1

25 3cos

3sinx

= +

For min

3, 0 cos –

5

dxx

d= =

and min

2km

3x =

23. Answer (04.00)

Hint : N sin30 = T

Sol.:

T = Nsin and N = wcos

sin cosT w=

24. Answer (75.00)

Hint : S = 21

2at

Sol.:

arough = g(sin – kcos)

asmooth = gsin

and 21

2S at=

k = 0.75

25. Answer (02.00)

Hint : If 2

,mv

Mgr

friction acts outward

Sol.:

If 2

,mv

mgr

friction acts outward

2

max

–mv

T mgr

=

and

2

min

mvT mg

r+ =

and T = Mg

PART – B : CHEMISTRY

26. Answer (2)

Hint :

Total no. of bonds between

two atoms in all the structureBond order

Total no. of resonating structure=

Sol. :

27. Answer (3)

Hint : LiCl is more covalent than NaCl.

Sol. : According to Fajan’s rule

28. Answer (4)

Hint : 2 3

3 3 3 3

sp sp

H N + B F H N — BF+

⎯⎯→

Sol. : NH3 and NH4+ both are sp3 hybridised

29. Answer (1)

Hint : –5XeF is planar.

Sol. : –6PCl

– number of atoms one plane = 5

XeF6 number of atoms in one plane = 5

Hyb. sp3d3 bp = 5, lp = 2

Number of atoms in one plane = 6

Number of atoms in one plane = 5

30. Answer (2)

Hint : Draw the structure and find the

hybridizations.



Sol. : NF3, BF3, PF5 [SiF6]–2

sp3, sp2 sp3d sp3d2

31. Answer (1)

Hint : All diatomic molecule containing even

number of total electron till 18 is diamagnetic

except 10 and 16.

Sol. : C2+ → C2

B.O = 1.5 B.O = 2

Paramagnetic Diamagnetic

NO+ → NO

B.O = 3 B.O = 2.5

Diamagnetic Paramagnetic

O2 → O2+

B.O = 2 B.O = 2.5

Paramagnetic Paramagnetic

N2 → N2+

B.O = 3 B.O = 2.5

Diamagnetic Paramagnetic

32. Answer (2)

Hint : has see-saw shape.

Sol. :

According to Bent’s rule more E. N. atom

occupy axial position in trigonal bipyramidal

geometry.

33. Answer (1)

Hint : 2– *2 2P

– *2P

2 2P

HOMO

O

NO

C

Sol. : N2+ M.O. electronic configuration of N2

+ is

1s2, 1s2, 2s2, 2s2, 2px2, 2py

2, 2pz1

H. O. M. O. of N2+ is 2pz

34. Answer (1)

Hint : dxy, dx2-y2 and dz2 are involved.

Sol. : In sp3d3

s + px + py + pz + dx2

– y2 + dxy + dz

2 orbitals take

part.

35. Answer (4)

Hint : (dipole moment) = either of the charge ×

distance between the opposite charge.

Sol. : = q × d, d = 10–10 m = 10–8 cm

1.2 × 10–18 = q × 10–8 then

q = 1.2 × 10–10 esu

% charge = –10

–10

q 1.2 10100 100 25%

e 4.8 10

=

36. Answer (1)

Hint : STP = 273 K and 1 atm, pressure

Sol. : Moles of C2H6 and C2H4 are a and b

respectively.

a + b = 28

22.4 = 1.25 moles

( ) ( )2 6 2 2(g) 2g gC H +3.5O 2CO +3H O(l)→

a 3.5a

( ) ( )2 4 2 2(g) 2g gC H +3O 2CO +2H O(l)→

b 3b

moles of O2 3.5a+3b=4 0.5

a+b=1.25 0.75

a

b

=

=

37. Answer (2)

Hint : After opening of valve of container, mixture

become uniform.

Sol. : 2H

20n 10

2= =

2CO

88n 2

44= =

Hence, composition of H2 = 10

10010 2

+

= 83.33%

38. Answer (1)

Hint : Average molar mass of gaseous mixture

(M) is = 2 2

2 2

O N

O N

32n +28n

n n+

Sol. : Moles of N2 and O2 in mixture are 2Nn

and 2On respectively then



According to ideal gas equation PV = w

RTm

or m = w RT

×V P

m = 1.3×0.0821× 273

1= 29.14 …(2)

by equation (1) and (2)

29.14 = 2 2

2 2

O N

O N

32n +28n

n +n

on solving the relation, 2

2

2 2

O

O

O N

nx = =0.28

n +n

Partial pressure of O2 = 2O totalx P

= 0.28 × 1

{ at NTP, P = 1 atm}

= 0.28 atm

39. Answer (1)

Hint : At low pressure 2

aP + P

v

Sol. : ( )2

aP+ v–b =RT

v

(I) At low pressure, then PV = RT + Pb

(II) At high pressure, then PV = a

RT–v

(III) at STP, for real gas ( )2

aP+ v–b =RT

v

(IV) Condition of ideal behaviour a → 0, b → 0,

then PV = RT

40. Answer (1)

Hint : Slope of line = P

RT = tan

Sol. : n = P

VRT

Temperature = P 16.42

Rtanθ 0.0821tan=

For A, T = 200

K3

For B, T = 200 3K

41. Answer (1)

Hint : Urms = 3RT

M

Sol. : Ump = 2RT

M = 400,

Urms = 1.5 400 m/sec

42. Answer (2)

Hint : (76 × 13.6)Hg = (5.44 × h)

Sol. : Pgas (A) = Patm + Ph 38

1190

+ = 1.2 atm

3A(g) A3(g)

t = 0 1.2 atm –

t = t 1.2 – 0.36 ( )1

0.36 0.12 atm3

=

Ptotal = 1.2 – 0.36 + 0.12 = 0.96 atm

Pressure difference in column 1 – 0.96

= 0.04 atm

The difference in height of the liquid level in two

columns = 0.04 × 190 = 7.6 cm

43. Answer (4)

Hint : It is the highest temperature at which liquid

and vapour can co-exist.

Sol. : At critical temperature densities of liquid

phase and gas phase become same.

44. Answer (4)

Hint : Rate of ( )n

r ×M

;

Sol. : Rate of ( )n

r ×M

; …(1)

r = moles effusing out n

time taken t

=

r × n …(2)

by equation (1) and (2)

n

nM

4 2He CH SOn : n : n

1 2 3

: :4 16 64

4 : 4 : 3

45. Answer (4)

Hint : Evaporation can occur at all temperature.



Sol. : Molecules having higher energy escape

out during evaporation, from the surface. The

remaining molecules are left associated with

lower energy and thus, temperature falls down.

46. Answer (39.00)

Hint : 2NO+ linear

Sol. : NO3– sp2 BA = 120°

NO2+ sp BA = 180°

XeF4 sp 3d2 BA = 90°

47. Answer (04.00)

Hint : Put the value of n from A to E and find the

desired thing.

Sol. : (A) (B) (C) (D) (E)

Comp. XeF4 XeF5+ XeF5

– XeF6 XeF8–2

Bond pair 4 5 5 6 8 = 28

Lone pair 2 1 2 1 1 = 7

P 28

=q 7

= 04

48. Answer (05.00)

Hint : B.O = b aN – N

2

Sol. :

49. Answer (75.00)

Hint : b

aT =

Rb

Sol. : TB = a 3.6

75 KRb 0.08 0.6

= =

50. Answer (12.50)

Hint : b = 3

A

44 r N

3

Sol. : b = –4 3 234

4 10 1000 4 r 6 103

=

r = 5 × 10–9 cm

Distance of closest approach = 2r

= 2 × 5 × 10–9 cm

= 10–8 cm = 10–10 m

y = 10

1.25y = 12.50

PART – C : MATHEMATICS

51. Answer (2)

Hint :

( ) ( )3 3sin – cos sin – cos 1 sin cos = +

Sol.: 6 – 2

sin – cos2

= …(i)

Now ( )2

sin – cos = 2

6 – 2 6 2 – 2 12

2 4

+=

sin.cos = 3 – 1

2 …(ii)

sin3 – cos3 = 6 – 2 3 – 1

12 2

+

= ( )2 3 – 1 3 1 2

2 2 2

+=

(sin3 – cos3)2 = 2

4

( )2

3 34 sin – cos 2 =

52. Answer (2)

Hint : sin8

tan8 =cos8

xx

x then use cross

multiplication.

Sol. : sin8x(cosx + sinx) = cos8x(cosx – sinx)

sin8x.cosx + cos8x.sinx = cos8x.cosx – sin8x

sinx

sin9x = cos9x

tan9x = 1

x = 1

,9 4

n n I

+

x = 5 9 13 17 21

, , , , ,36 36 36 36 36 36

, …..,

69

36

Total number of x = 18



53. Answer (3)

Hint : ( )33 3a b a b+ = + –3ab(a + b)

Sol. : Let 11 – x = a, 13 – x = b

Given equation becomes

3(11 – x) (13 – x) (24 – 2x) = 0

Roots , , are 11, 12 and 13.

+ + = 36

54. Answer (4)

Hint : sin5f(x) has maximum value 1.

Sol. : Then the required maximum value

= 3 + 13 = 16

55. Answer (4)

Hint : i 4n = 1, and i 4n + 1 = i, if iN

Sol. : ( ) ( )

21 3 5 .... 2 1 1n ni i + + + + + + =

= 4 1mi + , if n is even

= i

56. Answer (2)

Hint : if z is purely real then – 0z z =

Sol. : Let z = 3 2 sin

1 – 2 sin

i

i

+

z is purely real – 0z z =

sin = 0

= n, nI

( ) ( )3cos2 4sin2 3cos 2 4sin 2n n + = +

= 3

57. Answer (1)

Hint : z is a complex number then z z+ and

.z z are purely real number.

Solution : –z z is purely imaginary number

but ( )2020

–z z is purely real number.

sum of given number is purely real.

58. Answer (2)

Hint : Take the intersection of solution set of

two inequations.

Sol. : Here 5x + 2 < 3x + 8

2x < 6

x < 3 …(i)

and 2

– 4 0– 1

x

x

+

2 – 4 4

0– 1

x x

x

+ +

6 – 3

0– 1

x

x

( ) ( )– ,1 2, …(ii)

From (i) and (ii)

x(–, 1) (2, 3)

59. Answer (3)

Hint : Both side of inequation a fixed number

can be added or subtracted.

Sol. : 3 3x – 18 18 21 3x 36

7 x 12 8 x + 1 13

60. Answer (1)

Hint : Properties of Linear Inequality.

Sol. : 2 < x – 2 < 4

4 < x < 6

x = 5 as only integral value

61. Answer (4)

Hint : |x|2 = 2x and |x| 0

Sol. x2 – 5|x| + 6 = 0

(|x| – 3) (|x| – 2) = 0

x = 3, – 3, 2 and – 2

Product of roots = 36

62. Answer (3)



Hint : sinx [–1, 0] for – , 02

x

Sol. : –1 sinx 0

1 2 + sinx 2

63. Answer (1)

Hint : Let be a common root then

2 + a + 10 = 0 …(i)

and 2 + b – 10 = 0 …(ii)

Sol. : From (i) and (ii) : (a – b) + 20 = 0

= 20

––a b

From equation (i),

( )2

400 20– 10 0

––+ =

a

a ba b

400 – 20 (a – b).a + 10 (a – b)2 = 0

40 – 2a2 + 2ab + a2 + b2 – 2ab = 0

a 2 – b 2 = 40

64. Answer (3)

Hint : sin + sin3 = 2sin2.cos

Sol. : sin + sin3 + sin = sin

sin2(2cos + 1) = sin …(i)

and cos + cos3 + cos2 = cos

cos2(2cos + 1) = cos ….(ii)

From equation (i) and (ii), we get tan2 = tan

65. Answer (3)

Hint : loga x

a = x and lognm + lognp = ( )logn mp

Sol. :

( ) ( ) ( )2 2 2

log cos20 log cos40 log cos80 + +

= ( )2

log cos20 .cos40 .cos80

= 2

1 1log – cos20 cos40 .cos20

4 2

+

= 2

1 1log – cos20 cos60 cos20

4 4

+ +

= 22

1 –3log log 2 – 6

18

2

= =

and 3log 2

3 = 32log 23 = 4

Required value = ( )2

–6 4+ = 40

66. Answer (2)

Hint : If A + B = 45°

1 – tanA – tanB = tanA .tanB

Sol. : If A + B = 45°

cotA cotB – cotB – cotA = 1

(cotA – 1) (cotB –1) = 2

The given expression = 25 = 32

67. Answer (2)

Hint : arg(x + iy) = –1tany

x

Sol. : z = 13 – 5 13 – 5 4 9

4 – 9 4 – 9 4 9

i i i

i i i

+=

+

= 97 97

197

ii

+= +

arg(z) = tan–1(1) = 4

68. Answer (1)

Hint : (cos ± i sin)n = cosn ± i sin n

Sol. : x2 – 2xcos + 1 = 0 then roots are

x1 or x2 = cos + i sin or cos – i sin

Equation with roots 51x and 5

2x is

x2 – ( ) ( )55 5

1 2 1 2. 0x x x x x+ + =

x2 – 2 cos5 x + 1 = 0

P = cos5 = cos5 ( )5 5 2 5 40 2 4– tan tanC C C +

= 5 3 2 4cos – 10cos sin 5sin .cos +

= 5 316cos – 20cos 5cos +

69. Answer (3)

Hint : Both roots are negative.

So D 0, 2b > 0

Sol. : 2b > 0 and 9b – 14 > 0



0D 4b2 – 4 (9b – 14) 0

b2 – 9b + 14 0

14

, 29

b

[7, )

70. Answer (4)

Hints : cos = cos = 2n ± , nI.

Sol. : cot cot 24

+ + =

coscos 4

sinsin

4

+

+

+

= 2

sin 24

+

= 2sin .sin

4

+

1

sin 2 cos 24 4 2

+ + + =

1

cos22

=

2 = 23

n

= 6

n

71. Answer (13.00)

Hint : cosx = 0 x = ( )2 1

,2

nn I

+

Sol.: 7

cos3 sin 2 – – 26

x x

+ =

7

1 cos3 1 sin 2 – 06

x x

+ + + =

(1 + cos3x) + 2

1– cos 2 – 03

x

=

2 232cos 2sin – 0

2 3

xx

+ =

3

cos 02

x= and sin – 0

3x

=

3 3

, , ...2 2 2

x = and – 0, , 2

3x

= ,…

General solutions is x = 23

k

+

= ( )6 1

,3

kk I

+

Total number of possible solutions are 13.

72. Answer (05.00)

Hint : x + iy = a + ib x = a & y = b

Sol. : x – 2 = 2 and y – 1 = 4

y = 5

73. Answer (02.00)

Hint : ei = cos + i sin

Sol. : z = iee

= ( )cos sinie

+

= ( )cos sinie e

= ( ) ( )( )cos cos sin sin sine i +

cosz e =

( )cos

maxmaxz e =

= 2e =

74. Answer (21.00)

Hint : (a + c) (b + c) = ab + (a + b)c + c2

Put a + b = – 2020 and ab = 1.

Sol. : – 2020a b+ = , c + d = 2018

ab = cd = 1

(a + c) (b + c) (a – d) (b – d)

= ( )( ) ( )( )2 2–ab a b c c ab a b d d+ + + + +

= ( ) ( )2 21– 2020 1 2020c c d d+ + +

= ( ) ( )2018 – 2020 2018 2020c c d d+

= (–2c). (4038)d

= –8076

75. Answer (30.00)

Hint : cos2 = 1 – 2 sin2

Sol. : sinA + sinB = 1 sinA = 1 – sinB

cosA + cosB = 0 cosA = – cosB

Thus cos2A + sin2A = 1

(–cosB)2 + (1 – sinB)2 = 1

cos2B + 1 – 2sinB + sin2B = 1

sinB = 1

2

Thus sinA = 1

2

36cos2A + 24cosB = 36(1 – 2sin2A) +

24(1 – 2sin2B) = 30

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