All India Aakash Test Series for JEE (Advanced)-2021 TEST -3A … · 2020. 4. 10. · Aakash...

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) (Answers) All India Aakash Test Series for JEE (Advanced)-2021

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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)


All India Aakash Test Series for JEE (Advanced)-2021 Test -3A (Paper-2) (Code-B) (Hints & Solutions)

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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


Test -3A (Paper-2) (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2021

3/12

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



4/12

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



5/12

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.



6/12

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.


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



7/12

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



8/12

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


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


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


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+



9/12

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



10/12


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



11/12

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



12/12

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

All India Aakash Test Series for JEE (Advanced)-2021 TEST -3A … · 2020. 4. 10. · Aakash...

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Transcript of All India Aakash Test Series for JEE (Advanced)-2021 TEST -3A … · 2020. 4. 10. · Aakash...