Test - 2A (Paper - 2) (Code-G) (Answers) All India Aakash ... · = 5lnT 4 10 T[ ]600 6002 [ ] 300...

Test - 2A (Paper - 2) (Code-G) (Answers) All India Aakash Test Series for JEE (Advanced)-2020

PHYSICS

1. (A, B, C)

2. (A, B)

3. (C, D)

4. (A, B)

5. (A, B, C)

6. (B)

7. (D)

8. (A)

9. (A)

10. (C)

11. (C)

12. (C)

13. (C)

14. (B)

15. (D)

16. A → (P, T)

B → (P, T)

C → (Q, S)

D → (R)

17. (A) → (Q, T)

(B) → (P)

(C) → (R)

(D) → (S)

18. (03)

19. (08)

20. (04)

CHEMISTRY

21. (A, B, C, D)

22. (B)

23. (A, B, D)

24. (A, C, D)

25. (B, C, D)

26. (A)

27. (A)

28. (C)

29. (B)

30. (B)

31. (C)

32. (C)

33. (A)

34. (B)

35. (B)

36. A → (P, R)

B → (Q, R)

C → (Q, S, T)

D → (Q, R)

37. A → (R, T)

B → (P, Q)

C → (S)

D → (P, Q)

38. (09)

39. (01)

40. (09)

MATHEMATICS

41. (A, B, C, D)

42. (A, C)

43. (A, D)

44. (A, C)

45. (A, B, D)

46. (B)

47. (C)

48. (D)

49. (B)

50. (B)

51. (A)

52. (A)

53. (B)

54. (A)

55. (C)

56. A → (P)

B → (P, Q)

C → (P, Q, R, S, T)

D → (P, Q, R)

57. A → (P, R, T)

B → (Q, S)

C → (Q)

D → (P)

58. (09)

59. (07)

60. (05)

Test Date : 13/01/2019

ANSWERS

TEST - 2A (Paper-2) - Code-G

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

1/10

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

2/10

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, B, C)

Hint: Apply conservation of momentum and conservation

of kinetic energy.

Solution: mu = mV1 + mV

2

u = V2 – V

1

2. Answer (A, B)

Hint: 0 (Initially)P =�

Solution:

2 m

v v

m m

vCM

= 0a

CM= 0

v0

v0

2 2 2

0 0

1 1 12 (2 ) 22 2 2mv m v mv+ = ×

2 2 2

0 0 02v v v v v+ = =

3. Answer (C, D)

Hint: Apply principle of superposition.

Solution: 1 20

E E E= +� � �

10E =

2 2

82

GME

R

=

2 2 2

( )

2(8)

4

G M GME

R R

= =

3 8

2

2

GM

GMV

RR

= − − −

3 1

32 4 2 2

GM GM GM

R R R

= − + = −

5

4

GM

R= −

4. Answer (A, B)

Hint: f = μmgSolution: f = (μ)(ρ)a3g

Longitudinal stress3

2 2(2)

a g ag

a

μρ μρ = =

5. Answer (A, B, C)

Hint: ˆ ˆU U

F i jx y

∂ ∂= − −∂ ∂

�

Solution: ˆ ˆU U

F i jx y

∂ ∂= − −∂ ∂

�

6. Answer (B)

7. Answer (D)

8. Answer (A)

Hint: Draw the FBD in frame of A

Solution: V = 2g�

for first case

T – mg = 2

mv

�

T

mgT = 3mg

for second case

T + mg = ma1

T + ma1 – mg =

2mv

�

T

ma1

mg

9. Answer (A)

10. Answer (C)

11. Answer (C)

Hint: Total linear momentum and angular momentum

of system remain conserved.

Solution: V = 0

2

V

2 2

0

22

4 12 16

m mmV

= + ω

� � �

05

12 4

Vω =�

ω = 03

ˆ

5

Vk−

�

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

3/10

12. Answer (C)

Hint:

2

sin

1CM

ga

I

MR

θ= +

�

Solution: 2

2

sin 33 m/s

2 51

CM

CM

g ga

I

MR

θ = = = +

VCM

= 3t

2

21 3KE (2 )2

tMR

R

=

KE = 9Mt2

13. Answer (C)

Hint: P P

Iτ = α

Solution: p PIτ = α

P

2

2 2(2 ) ( 2)2 2

R MRMg MR M R

= + + α

27

2MgR MR= α

2

7

g

R

α =

2

7

ga R

= α =

14. Answer (B)

Hint: Torque about hinge will be zero.

Solution:Torque due to gravity and buoyancy force will

balance each other.

Fb

M

Mg

O

3 5 5 3( )

5 8 16 5

L L LMg A g

= ρ

⇒ M = 25

128ALρ

15. Answer (D)

Hint: mgh = 2 2 21 1

2 2mv mR+ ω

Solution:

mgh = 2 2 21 1

2 2mv mR+ ω

v = gh

16. Answer A(P, T); B(P, T); C(Q, S); D(R)

Hint: v0 =

J

m and JH = I

CMω

Solution: ω = CM

JH

I

If Rω > v0

⇒ frictional force will act rightward.17. Answer A(Q, T); B(P); C(R); D(S)

Hint: , ( )x ydp dp

a g adx dy

= −ρ = −ρ +

Solution:

ay

ax

B

C

A

(PB – P

A) = ρ(g + a

y)H

y + ρa

xx

18. Answer (03)

Hint: 2mv

T mg− =l

Solution: When string is vertical

2

3− = =mvT mg T mgl

19. Answer (08)

Hint: Surface tension will balance liquid weight

Solution:

2Th

d g=

ρ20. Answer (04)

Hint: Calculate time for upward and downward journey

Solution: a1

= 3

10

g

t1 =

6 102 s

3g

× =

Total time = 4 s


4/10

PART - II (CHEMISTRY)

21. Answer (A, B, C, D)

Hint : H2O + H

+ → H

3O

+

Solution : (H2O)

3 + H

+ → H

7O

3+

(H2O)

2 + H

+ → H

5O

2+

(H2O)

2 – H

+ → H

3O

2–1

(H2O)

3 – H

+ → H

5O

3–1

Hence all can exist.

22. Answer (B)

Hint : ΔH°(neutralisation for S·A and S·B)

= kJ

57.3mole

− .

Solution : For maximum rise in temperature there

should be maximum neutralisation of ⊕H and

–OH.

23. Answer (A, B, D)

Hint : Simultaneous equilibrium

KP = P

(product) / P

(reactant)

Solution : Let PA and P

C and P

B are the partial pressure

at point of equilibrium.

X(s) A(g) 2B(g)+�

PA

2PC + 2P

A

Y(s) 2B(g) C(g)+�

(2PC + 2P

A ) P

C

KP1

= (PB)2 (P

A)

= [2(PA + P

C)]2 P

A…(i)

KP2

= (PC) [(P

A + P

C)2]2 …(ii)

1

2

3

P A

3

P C

K P 9 104

K P 2.25 10

−

−

×= = =×

A

C

P4

P= …(iii)

Now from (i),

2

3 A

A A

P9 10 (4) (P ) P

4

− × = +

= 2

A

A

5P(4) (P )

4

2

3A A25 P ·P

9 104

−= ×

⇒ PA = 0.112 atm

⇒ AC

PP

4

=

= 0.028 atm

[ ]B C AP 2 P +P=

= 2 (0.112 + 0.028)

= 0.28 atm

24. Answer (A, C, D)

Hint : Oxidation number from structure

HO — S — OH

S(–2)

+6

O

Solution :

HO — S — (S) — S — OH

O

O O

O

n

(+5)(+5)0

← Polythionic acid

For all value of n, sum of O.N. of all sulphur = 10

H S O HO — S — S — OH2 2 5

→(+5) (+3)

O O

O

Average oxidation state = +4

In H S O HO — S — S — OH2 2 6

(+5)

O O

O O

(+5)

The O·N. is in between – 2 and +6 hence it behaves

like both oxidising as well as reducing agent.


5/10

25. Answer (B, C, D)

Hint : Vaporisation(boiling point)

HS

T

ΔΔ =

Solution : Vap(BP)

H 30 1000S

T 300

Δ × Δ = =

= 100

When P increase, B·P increase

ΔH = ΔU + Δn(g)

RT

30 = ΔU + (1) (8.3) (300) × 10–3

⇒ ΔU = 30 – 8.3 × 0.3

= 30 – 2.49

= 30 – 2.5

ΔU = 27.5 K/mol

And ΔG at normal boiling point of liquid is zero, sinceliquid and vapor are in equilibrium.

26. Answer (A)

Hint : 2

1

V4

V=

2 2

V

1 1

T VS nC ln nRln

T V

Δ = +

Solution :

33

2 2 2 2 2

3

1 1 1 1 1

T P V KV V64

T PV KV V

= = = =

∴ 2 21 1

T VS 10 ln 8.314 ln

T V

Δ = × +

= 10 × ln 64 + 8.314 ln 4

= 38.314 ln 4

53 J/K�

27. Answer (A)

Hint :dq dU

dsT T

= = [∵ dw = 0]

Solution : V,m

C dTds

T=

and 2mv,m

dUC 5 4 10 T

dT

−= = + ×

∴600

v, m

300

C dTdS

T=

= 600

2

300

54 10 dT

T

− + ×

= [ ] [ ]600 6002300 300

5 ln T 4 10 T−+ ×

= 5 ln 2 + 4 × 10–2 (300)

= 5 ln2 + 12

= 15.465 J/K

28. Answer (C)

Hint :Rev

sys

QS

TΔ =

Solution : Qsys

= Qsurr

= 0

For reversible adiabatic process.

29. Answer (B)

Hint : I is calcium cyanamide

Solution : CaC2 + N

2 → CaCN

2 + C

30. Answer (B)

Hint : CaC2 + H

2O → C

2H

2 + Ca(OH)

2

Solution : 2 3 2

(X) (Y)

Ca N Ca NΔ+ ⎯⎯→

2H O

3 2 2 3W(Z)(Y)

Ca N Ca(OH) NH⎯⎯⎯→ +

2 2E(Z)

Ca(OH) CaO H OΔ⎯⎯→ +

Coke

22000 CE F G

CaO CaC CO°⎯⎯⎯⎯→ +

2 2 2F

CaC N CaCN C+ → +

2H O

2 2 2 2(P)

CaC C H Ca(OH)⎯⎯⎯→ +

31. Answer (C)

Hint : Z = Ca(OH)2

G = CO

Solution : Z is basic

whereas G is neutral.


6/10

32. Answer (C)

Hint : Concept of gram equivalent.

Solution :Ba (MnO )

4 2⎯→ Mn+2

5 × 2 = 10

n factor of Ba(MnO4)2 = 10

K [Fe(CN) ] Fe+3

, NO–1

, +4

CO4 6 3 3

2–⎯→ +2 +2

–3

+5

2 × 6 = 12

8 × 6 = 48

n factor = (48 + 12 + 1)

= 61

⇒ Number of eq. of Ba(MnO4)2 = Number of eq. of

K4[Fe(CN)

6]

⇒ (x) (10) = (10) (61)

⇒ x = 61

33. Answer (A)

Hint : % labeling of oleum.

Solution : meq of base = meq of acid

g2 54 0.2 2 1000

98× × = × ×

g = 1.0584

1 g oleum → 1.0584 g H2SO

4

100 g oleum → 105.84 g H2SO

4

∴ 5.84 g of H2O is added,

3 2SO H O

5.84n n

18= =

∴ Mass of SO3

= 5.84

8018

×

= 25.955

= 26%

34. Answer (B)

Hint : CO2 is produced in water gas shift reaction.

Solution : Coal gasification reaction is

1270 K

2 2C(s) H O (g) CO (g) H (g)+ ⎯⎯⎯⎯→ +

35. Answer (B)

Hint : Cyclic silicates have general formula [ ]2n3 nSiO−

Solution : X = [Si6O

18]12–

Y = [Si8O

24]16–

36. Answer A(P, R); B(Q, R); C(Q, S, T); D(Q, R)

Hint : H

RT

CK e

− Δ

∝

Solution : T increase Kc increase → endothermic

T increase Kc decrease → exothermic

Δn(g) = 0 → No effect of pressure

Addition of inert gas at constant pressure shift

equilibrium towards more number of moles.

37. Answer (A → R, T; B → P, Q; C → S; D → P, Q)

Hint : Concept of acid base theory.

Solution : Arrhenius concept → H+/OH–

Bronsted concept → H+ donor/H+ acceptor

Lewis concept → Lone pair acceptor/donor

38. Answer (09)

Hint :2 2

mass of pure H O% purity 100

Total mass of sample

= ×

Solution : Let mass of H2O

2 in 0.2 gram of sample = (x)

Then, number of eq. of H2O

2 = Number of eq. of KMnO

4

⇒x 0.316

(5)17 158

=

⇒ 5 17 0.316x 0.17 gram158

× ×= =


7/10

⇒ % age purity = 0.17 100

0.20

×

= 85%

39. Answer (01)

Hint : x = 12, y = 20

Solution :5x 60

13y 60

= =

40. Answer (09)

Hint : x = 2, y = 2, z = 5

Solution :

HO

OH

B

B

O O

O O

B

B

OH

OH

O

5 BOB linkage

∴ 5 moles of H2O is required


Hint : Write nth term

Solution :

Tn = (2n + 1)3 – 8n3 = 12n2 + 6n + 1

Sn =

12 ( 1)(2 1) 6 ( 1)

6 2

+ + ++ +n n n n n n

= 4n3 + 9n2 + 6n = n(4n2 + 9n + 6)

42. Answer (A, C)

Hint : AM ≥ GMSolution : AM ≥ GM

4 4 4 4 4 4 2 2 2

6 6 6 6 6 6 3 3 3 2 2

11

+ + + + + + + + + +

a a a a a a b b b c c

≥

1

6 3 2 114 2

6 3 2

a b c

⇒11 7 6 3 2

9

1 2

11 3

≥

a b c

⇒ a6b3c2 ≤ 9

7 11

3

2 (11)

When 4 2

6 3 2= =a b c

⇒3

4= = = λca b

⇒4

4 2 13

λ + λ + λ =

⇒3

22λ =

a = b = 3

22,

2

11=c

43. Answer (A, D)

Hint : f(n) = 2n + 1 – 1

Solution :

f(n) = 1 2

0

( ... )n

r r r n

r r r r

r

C C C C+ +

=+ + + +

= 1 1

0

Coeff. of in (1 ) (1 )+ +

=

+ − + n

r n r

r

x x x

= 1

1

0

++

=n

n

r

r

C

f(n) = 2n + 1 – 1

44. Answer (A, C)

Hint :

9

2

( 1)=

−n

n n

Solution : Since x ≥ 1, then y ≥ 2If y = n then n take the values from 1 to n – 1 and

z can take the value from 0 to n – 1 (i.e. n values),

thus for each value of y(2 < y < 9), x and z take n

(n – 1) values.

9

2

9(10)(19) 9(10)( 1) 240

6 2=− = − =

n

n n

PART - III (MATHEMATICS)


8/10


Hint : SP + S′P = 2a

Solution :

S(3, 1), S′(–1, –2)

SP + S′P = 2a

⇒ a = 5

2ae = 16 9 5+ =

⇒ e = 1

2

Equation of major axis is (y – 1) = 3

4(x – 3)

⇒ 3x – 4y – 5 = 0

46. Answer (B)

Hint : Assume B = (h, 7h + 7) and C = (k, – 8 – 8k)

Solution :

BM : 7x – y + 7 = 0

CN : 8x + y + 8 = 0

Let B : (h, 7h + 7)

A (1, –1)

MN

B CC : (k, – 8 – 8k)

Midpoint of A and B lies on CN.

⇒1 7 6,

2 2

h h+ +

satisfies 8x + y + 8 = 0

⇒ h = –2

B (–2, –7)

midpoint of AC lies on line BM

⇒1 8 8 1,

2 2

k k+ − − −

lies on 7x – y + 7 = 0

⇒ k = –2 ⇒ C (–2, 8)

47. Answer (C)

Hint : 2

2,3

α ∈ −

, [ ]3, 2β ∈ −

Solution :

22,

3

α ∈ −

and [ ]3, 2β ∈ −A (1, –1)

C (–2, 8)B (–2, –7)

m = 3, n = 6

|m + n| = 9

48. Answer (D)

Hint : Image of 1,

tt

is 1

4 ,tt

− −

Solution :

Let point be 1

, tt

Its image in x = –2 is 1

4 ,tt

− −

⇒ Eliminating t, (x + 4)y = –1⇒ xy + 4y + 1 = 0

49. Answer (B)

Hint : p is y2 = 2x

Solution :

(x – 4)2 = 4y

⇒ y2 = 2x is p

Equation of normal is y = 3

2− + + ttx t at (at2, 2at)

It passes through 1

2,2

−

⇒ t3 – 2t + 1 = 0⇒ (t – 1) (t2 + t –1) = 0

t3 = 1, t

1 + t

2 = –1, t

1t2 = –1

Centroid is

2 2

1 2 1 21 1 2, , 0

6 3 3

+ + + + =

t t t t

50. Answer (B)

Hint : Normal on x2 = 4ay is y = mx + 2a + 2

a

m

Solution :

Tangent on y2 = 2x is

1

2= +y mx

m

Normal on x2 = 4ay is y = mx + 2a + 2

a

m

⇒2

12

2+ =aa

mm

⇒ 22 02

mam a− + =

D > 0

⇒ 21

8 04

− ≥a

⇒1

| |4 2


9/10

51. Answer (A)

Hint : Write family of circles

Solution :

Family of circles is

(–y2 + x2 + b2) + λ(y2 – 2x) = 0

⇒ x2 + y2 (λ – 1) – 2λx + b2 = 0

⇒ λ = 2 for circle

x2 + y2 – 4x + b2 = 0

Radius > 0

4 – b2 > 0

b ∈ (–2, 2)

Also parabola and hyperbola should intersect

⇒ 2x – x2 – b2 = 0

⇒ x2 –2x + b2 = 0

D > 0

4 – 4b2 > 0

b ∈ (–1, 1)

52. Answer (A)

Hint : Assume P(a secθ, b tanθ), P′(a secθ, –b tanθ)

Solution :

P(a secθ, b tanθ), P′ (a sec θ, –b tanθ), O(0, 0)Let G(h, k)

⇒ 3h = 2a secθ 3k = 0 ⇒ y = 0

Locus of G is y = 0

53. Answer (B)

Hint : AM ≥ GM

Solution :

α + β + γ + δ = –a

αβγ + βγδ + αγδ + αβδ = –c

αβγδ = 4

AM ≥ GM1

4( )4

α + β + γ + δ ≥ αβγδ

⇒1

444

− ≥a ⇒ a2 ≥ 32

54. Answer (A)

Hint : Line should be tangent to the circle.

Solution : Line is tangent to the circle, the

12

2

− λ =

⇒ λ = –1

55. Answer (C)

Hint : Differentiation of (1 + x)n

Solution :

(1 + x)n = C0 + C

1 + C

2x2 + C

3x3 + .... + C

nxn

⇒ n(1 + x)n–1 = C1 + 2.C

2x + 3.C

3x2 +...+ n.C

nxn–1

⇒ nx(1 + x)n–1 = C1x + 2.C

2x2 + 3C

3x3 +...+ nC

nxn

⇒ nx(n – 1) (1 + x)n–2 + n(1 + x)n–1

= 12C1 + 22C

2x + 32C

3x2 + ..... + n2 C

nxn–1

Put x = 1

⇒2

1=n

n

r

r

r C = n(n + 1)2n–2

56. Answer A(P); B(P, Q); C(P, Q, R, S, T); D(P, Q, R)

Hint : Multiconcept

Solution : 5A, H, I, K, 2S, T

(A) n1 =

5

1

6 5 6 55 6 2 3 5

5 5

× ×× = = × = × ×C

(B) n2 = (4A + 2S) + (4A + 2 diff.)

= C5

2

6 6 16 5 10

4 2 4 2

+ = × +

= 3 × 5 × 21 = 32 × 5 × 7

(C) n3 = (2A + 2S + 2 diff.) + (2A + 4 diff.)

= 4 5

2 4

6 6.2 2 2

+C C

= 6 5 4 3

6 5 6 5 4 32

× × ×× + × × × ×

6 26 5 4 3 (3 5) 2 3 5= × × × + = × ×

(D) n4 = 6 = 6 × 5 × 4 × 3 × 2 = 24 × 32 × 5


10/10

��

57. Answer A(P, R, T); B(Q, S); C(Q); D(P)

Hint : Multiconcept

Solution : Let P(cosθ, sinθ), ‘θ’ is parametric angleof point P

T : x cosθ + y sinθ – 1 = 0, N : x sinθ – y cosθ = 0

(A) Area = 1 1 1 2

2 sin cos 3× × =

θ θ

⇒ 3 2 5

sin2 , , , ,.....2 6 3 3 6

π π π πθ = ± θ =

(B) Area = 1 1 1

2 sin cos× ×

θ θ is minimum, when

3 5sin2 1, , , ,.....

4 4 4

π π πθ = ± θ =

(C) mT = –1 ⇒ –cotθ = –1 ⇒ cotθ = 1

⇒ 5

, ,.....4 4

π πθ =

(D) mN =

1

3 ⇒

1tan

3θ = ⇒

7, ,.....

6 6

π πθ =

58. Answer (09)

Hint : a = –6 + 4 cosθ, b = 3 + 4 sinθSolution :

Equation is (x + 6)2 + (y –3)2 = 42

a = –6 + 4 cosθ, b = 3 + 4 sinθ

a + b = –3 + 4 (cosθ + sinθ)

–3 – 4 2 ≤ (a + b) ≤ –3 + 4 2

|a + b| = 0, 1, 2, .....8

59. Answer (07)

Hint : Coefficient of x17 in (1 + x)20(x + 1)20.

Solution :

Let nCm

= coeff. of x17 in (1 + x)20(x + 1)20

= 40C17

= 40C23

⇒ maximum value of n + m is 63

60. Answer (05)

Hint : Equation of chord of contact is 10 x – 6y + 10 = 0

Solution :

Chord of contact of tangents drawn from P(1, 6) to the

parabola is

6y = 10 (x +1) ⇒ 10x – 6y + 10 = 0

Length of chord of contact = 2

2

4(1 )( )+ −a m a mc

m

= 4 25 5 5

9 5 1 525 9 3 3

× + − ×

= 8

345

Length of perpendicular from P(1, 6) to chord of

contact is 10 36 10 16

136 136

− += =p

1 16 8 3234

2 5 5136Δ = × × =

Test - 2A (Paper - 2) (Code-H) (Answers) All India Aakash Test Series for JEE (Advanced)-2020

PHYSICS

1. (A, B, C)

2. (A, B)

3. (C, D)

4. (A, B)

5. (A, B, C)

6. (B)

7. (D)

8. (A)

9. (A)

10. (C)

11. (C)

12. (D)

13. (B)

14. (C)

15. (C)

16. (A) → (Q, T)

(B) → (P)

(C) → (R)

(D) → (S)

17. A → (P, T)

B → (P, T)

C → (Q, S)

D → (R)

18. (04)

19. (08)

20. (03)

CHEMISTRY

21. (B, C, D)

22. (A, C, D)

23. (A, B, D)

24. (B)

25. (A, B, C, D)

26. (A)

27. (A)

28. (C)

29. (B)

30. (B)

31. (C)

32. (B)

33. (B)

34. (A)

35. (C)

36. A → (R, T)

B → (P, Q)

C → (S)

D → (P, Q)

37. A → (P, R)

B → (Q, R)

C → (Q, S, T)

D → (Q, R)

38. (09)

39. (01)

40. (09)

MATHEMATICS

41. (A, B, D)

42. (A, C)

43. (A, D)

44. (A, C)

45. (A, B, C, D)

46. (B)

47. (C)

48. (D)

49. (B)

50. (B)

51. (A)

52. (C)

53. (A)

54. (B)

55. (A)

56. A → (P, R, T)

B → (Q, S)

C → (Q)

D → (P)

57. A → (P)

B → (P, Q)

C → (P, Q, R, S, T)

D → (P, Q, R)

58. (05)

59. (07)

60. (09)

Test Date : 13/01/2019

ANSWERS

TEST - 2A (Paper-2) - Code-H

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

1/10

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

2/10

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, B, C)

Hint: ˆ ˆU U

F i jx y

∂ ∂= − −∂ ∂

�

Solution: ˆ ˆU U

F i jx y

∂ ∂= − −∂ ∂

�

2. Answer (A, B)

Hint: f = μmg

Solution: f = (μ)(ρ)a3g

Longitudinal stress3

2 2(2)

a g ag

a

μρ μρ = =

3. Answer (C, D)

Hint: Apply principle of superposition.

Solution: 1 20

E E E= +� � �

10E =

2 2

82

GME

R

=

2 2 2

( )

2(8)

4

G M GME

R R

= =

3 8

2

2

GM

GMV

RR

= − − −

3 1

32 4 2 2

GM GM GM

R R R

= − + = −

5

4

GM

R= −

4. Answer (A, B)

Hint: 0 (Initially)P =�

Solution:

2 m

v v

m m

vCM

= 0a

CM= 0

v0

v0

2 2 2

0 0

1 1 12 (2 ) 22 2 2mv m v mv+ = ×

2 2 2

0 0 02v v v v v+ = =

5. Answer (A, B, C)

Hint: Apply conservation of momentum and conservation

of kinetic energy.

Solution: mu = mV1 + mV

2

u = V2 – V

1

6. Answer (B)

7. Answer (D)

8. Answer (A)

Hint: Draw the FBD in frame of A

Solution: V = 2g�

for first case

T – mg = 2

mv

�

T

mg

T = 3mg

for second case

T + mg = ma1

T + ma1 – mg =

2mv

�

T

ma1

mg

9. Answer (A)

10. Answer (C)

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

3/10

11. Answer (C)

Hint: Total linear momentum and angular momentum

of system remain conserved.

Solution: V = 0

2

V

2 2

0

22

4 12 16

m mmV

= + ω

� � �

05

12 4

Vω =�

ω = 03

ˆ

5

Vk−

�

12. Answer (D)

Hint: mgh = 2 2 21 1

2 2mv mR+ ω

Solution:

mgh = 2 2 21 1

2 2mv mR+ ω

v = gh

13. Answer (B)

Hint: Torque about hinge will be zero.

Solution:Torque due to gravity and buoyancy force will

balance each other.

Fb

M

Mg

O

3 5 5 3( )

5 8 16 5

L L LMg A g

= ρ

⇒ M = 25

128ALρ

14. Answer (C)

Hint: P P

Iτ = α

Solution: p PIτ = α

P

2

2 2(2 ) ( 2)2 2

R MRMg MR M R

= + + α

27

2MgR MR= α

2

7

g

R

α =

2

7

ga R

= α =

15. Answer (C)

Hint:

2

sin

1CM

ga

I

MR

θ= +

�

Solution: 2

2

sin 33 m/s

2 51

CM

CM

g ga

I

MR

θ = = = +

VCM

= 3t

2

21 3KE (2 )2

tMR

R

=

KE = 9Mt2

16. Answer A(Q, T); B(P); C(R); D(S)

Hint: , ( )x ydp dp

a g adx dy

= −ρ = −ρ +

Solution:

ay

ax

B

C

A

(PB – P

A) = ρ(g + a

y)H

y + ρa

xx

17. Answer A(P, T); B(P, T); C(Q, S); D(R)

Hint: v0 =

J

m and JH = I

CMω

Solution: ω = CM

JH

I

If Rω > v0

⇒ frictional force will act rightward.18. Answer (04)

Hint: Calculate time for upward and downward journey

Solution: a1

= 3

10

g

t1 =

6 102 s

3g

× =

Total time = 4 s


4/10

PART - II (CHEMISTRY)

21. Answer (B, C, D)

Hint : Vaporisation(boiling point)

HS

T

ΔΔ =

Solution : Vap(BP)

H 30 1000S

T 300

Δ × Δ = =

= 100

When P increase, B·P increase

ΔH = ΔU + Δn(g)

RT

30 = ΔU + (1) (8.3) (300) × 10–3

⇒ ΔU = 30 – 8.3 × 0.3

= 30 – 2.49

= 30 – 2.5

ΔU = 27.5 K/mol

And ΔG at normal boiling point of liquid is zero, sinceliquid and vapor are in equilibrium.

22. Answer (A, C, D)

Hint : Oxidation number from structure

HO — S — OH

S(–2)

+6

O

Solution :

HO — S — (S) — S — OH

O

O O

O

n

(+5)(+5)0

← Polythionic acid

For all value of n, sum of O.N. of all sulphur = 10

H S O HO — S — S — OH2 2 5

→(+5) (+3)

O O

O

Average oxidation state = +4

In H S O HO — S — S — OH2 2 6

(+5)

O O

O O

(+5)

The O·N. is in between – 2 and +6 hence it behaves

like both oxidising as well as reducing agent.


Hint : Simultaneous equilibrium

KP = P

(product) / P

(reactant)

Solution : Let PA and P

C and P

B are the partial pressure

at point of equilibrium.

X(s) A(g) 2B(g)+�

PA

2PC + 2P

A

Y(s) 2B(g) C(g)+�

(2PC + 2P

A ) P

C

KP1

= (PB)2 (P

A)

= [2(PA + P

C)]2 P

A…(i)

KP2

= (PC) [(P

A + P

C)2]2 …(ii)

1

2

3

P A

3

P C

K P 9 104

K P 2.25 10

−

−

×= = =×

A

C

P4

P= …(iii)

Now from (i),

2

3 A

A A

P9 10 (4) (P ) P

4

− × = +

19. Answer (08)

Hint: Surface tension will balance liquid weight

Solution:

2Th

d g=

ρ

20. Answer (03)

Hint: 2mv

T mg− =l

Solution: When string is vertical

2

3− = =mvT mg T mgl


5/10

= 2

A

A

5P(4) (P )

4

2

3A A25 P ·P

9 104

−= ×

⇒ PA = 0.112 atm

⇒ AC

PP

4

=

= 0.028 atm

[ ]B C AP 2 P +P== 2 (0.112 + 0.028)

= 0.28 atm

24. Answer (B)

Hint : ΔH°(neutralisation for S·A and S·B)

= kJ

57.3mole

− .

Solution : For maximum rise in temperature there

should be maximum neutralisation of ⊕H and

–OH.


Hint : H2O + H

+ → H

3O

+

Solution : (H2O)

3 + H

+ → H

7O

3+

(H2O)

2 + H

+ → H

5O

2+

(H2O)

2 – H

+ → H

3O

2–1

(H2O)

3 – H

+ → H

5O

3–1

Hence all can exist.

26. Answer (A)

Hint : 2

1

V4

V=

2 2

V

1 1

T VS nC ln nRln

T V

Δ = +

Solution :

33

2 2 2 2 2

3

1 1 1 1 1

T P V KV V64

T PV KV V

= = = =

∴ 2 21 1

T VS 10 ln 8.314 ln

T V

Δ = × +

= 10 × ln 64 + 8.314 ln 4

= 38.314 ln 4

53 J/K�

27. Answer (A)

Hint :dq dU

dsT T

= = [∵ dw = 0]

Solution : V,m

C dTds

T=

and 2mv,m

dUC 5 4 10 T

dT

−= = + ×

∴600

v, m

300

C dTdS

T=

= 600

2

300

54 10 dT

T

− + ×

= [ ] [ ]600 6002300 300

5 ln T 4 10 T−+ ×

= 5 ln 2 + 4 × 10–2 (300)

= 5 ln2 + 12

= 15.465 J/K

28. Answer (C)

Hint :Rev

sys

QS

TΔ =

Solution : Qsys

= Qsurr

= 0

For reversible adiabatic process.

29. Answer (B)

Hint : I is calcium cyanamide

Solution : CaC2 + N

2 → CaCN

2 + C

30. Answer (B)

Hint : CaC2 + H

2O → C

2H

2 + Ca(OH)

2

Solution : 2 3 2

(X) (Y)

Ca N Ca NΔ+ ⎯⎯→

2H O

3 2 2 3W(Z)(Y)

Ca N Ca(OH) NH⎯⎯⎯→ +

2 2E(Z)

Ca(OH) CaO H OΔ⎯⎯→ +

Coke

22000 CE F G

CaO CaC CO°⎯⎯⎯⎯→ +

2 2 2F

CaC N CaCN C+ → +

2H O

2 2 2 2(P)

CaC C H Ca(OH)⎯⎯⎯→ +


6/10

31. Answer (C)

Hint : Z = Ca(OH)2

G = CO

Solution : Z is basic

whereas G is neutral.

32. Answer (B)

Hint : Cyclic silicates have general formula [ ]2n3 nSiO−

Solution : X = [Si6O

18]12–

Y = [Si8O

24]16–

33. Answer (B)

Hint : CO2 is produced in water gas shift reaction.

Solution : Coal gasification reaction is

1270 K

2 2C(s) H O (g) CO (g) H (g)+ ⎯⎯⎯⎯→ +

34. Answer (A)

Hint : % labeling of oleum.

Solution : meq of base = meq of acid

g2 54 0.2 2 1000

98× × = × ×

g = 1.0584

1 g oleum → 1.0584 g H2SO

4

100 g oleum → 105.84 g H2SO

4

∴ 5.84 g of H2O is added,

3 2SO H O

5.84n n

18= =

∴ Mass of SO3

= 5.84

8018

×

= 25.955

= 26%

35. Answer (C)

Hint : Concept of gram equivalent.

Solution :Ba (MnO )

4 2⎯→ Mn+2

5 × 2 = 10

n factor of Ba(MnO4)2 = 10

K [Fe(CN) ] Fe+3

, NO–1

, +4

CO4 6 3 3

2–⎯→ +2 +2

–3

+5

2 × 6 = 12

8 × 6 = 48

n factor = (48 + 12 + 1)

= 61

⇒ Number of eq. of Ba(MnO4)2 = Number of eq. of

K4[Fe(CN)

6]

⇒ (x) (10) = (10) (61)

⇒ x = 61

36. Answer A(R, T); B(P, Q); C(S); D(P, Q)

Hint : Concept of acid base theory.

Solution : Arrhenius concept → H+/OH–

Bronsted concept → H+ donor/H+ acceptor

Lewis concept → Lone pair acceptor/donor

37. Answer A→(P, R); B→(Q, R); C→(Q, S, T); D→(Q, R)

Hint : H

RT

CK e

− Δ

∝

Solution : T increase Kc increase → endothermic

T increase Kc decrease → exothermic

Δn(g) = 0 → No effect of pressure

Addition of inert gas at constant pressure shift

equilibrium towards more number of moles.

38. Answer (09)

Hint : x = 2, y = 2, z = 5

Solution :

HO

OH

B

B

O O

O O

B

B

OH

OH

O

5 BOB linkage

∴ 5 moles of H2O is required


7/10

39. Answer (01)

Hint : x = 12, y = 20

Solution :5x 60

13y 60

= =

40. Answer (09)

Hint :2 2

mass of pure H O% purity 100

Total mass of sample

= ×

Solution : Let mass of H2O

2 in 0.2 gram of sample = (x)


Hint : SP + S′P = 2a

Solution :

S(3, 1), S′(–1, –2)

SP + S′P = 2a

⇒ a = 5

2ae = 16 9 5+ =

⇒ e = 1

2

Equation of major axis is (y – 1) = 3

4(x – 3)

⇒ 3x – 4y – 5 = 0

42. Answer (A, C)

Hint :

9

2

( 1)=

−n

n n

Solution : Since x ≥ 1, then y ≥ 2

If y = n then n take the values from 1 to n – 1 and

z can take the value from 0 to n – 1 (i.e. n values),

thus for each value of y(2 < y < 9), x and z take n

(n – 1) values.

9

2

9(10)(19) 9(10)( 1) 240

6 2=− = − =

n

n n

43. Answer (A, D)

Hint : f(n) = 2n + 1 – 1

Solution :

f(n) = 1 2

0

( ... )n

r r r n

r r r r

r

C C C C+ +

=+ + + +

= 1 1

0

Coeff. of in (1 ) (1 )+ +

=

+ − + n

r n r

r

x x x

= 1

1

0

++

=n

n

r

r

C

f(n) = 2n + 1 – 1

44. Answer (A, C)

Hint : AM ≥ GM

Solution : AM ≥ GM

4 4 4 4 4 4 2 2 2

6 6 6 6 6 6 3 3 3 2 2

11

+ + + + + + + + + +

a a a a a a b b b c c

≥

1

6 3 2 114 2

6 3 2

a b c

⇒11 7 6 3 2

9

1 2

11 3

≥

a b c

⇒ a6b3c2 ≤ 9

7 11

3

2 (11)

When 4 2

6 3 2= =a b c

⇒3

4= = = λca b

PART - III (MATHEMATICS)

Then, number of eq. of H2O

2 = Number of eq. of KMnO

4

⇒x 0.316

(5)17 158

=

⇒ 5 17 0.316x 0.17 gram158

× ×= =

⇒ % age purity = 0.17 100

0.20

×

= 85%


8/10

⇒4

4 2 13

λ + λ + λ =

⇒3

22λ =

a = b = 3

22,

2

11=c


Hint : Write nth term

Solution :

Tn = (2n + 1)3 – 8n3 = 12n2 + 6n + 1

Sn =

12 ( 1)(2 1) 6 ( 1)

6 2

+ + ++ +n n n n n n

= 4n3 + 9n2 + 6n = n(4n2 + 9n + 6)

46. Answer (B)

Hint : Assume B = (h, 7h + 7) and C = (k, – 8 – 8k)

Solution :

BM : 7x – y + 7 = 0

CN : 8x + y + 8 = 0

Let B : (h, 7h + 7)

A (1, –1)

MN

B CC : (k, – 8 – 8k)

Midpoint of A and B lies on CN.

⇒1 7 6,

2 2

h h+ +

satisfies 8x + y + 8 = 0

⇒ h = –2

B (–2, –7)

midpoint of AC lies on line BM

⇒1 8 8 1,

2 2

k k+ − − −

lies on 7x – y + 7 = 0

⇒ k = –2 ⇒ C (–2, 8)

47. Answer (C)

Hint : 2

2,3

α ∈ −

, [ ]3, 2β ∈ −

Solution :

22,

3

α ∈ −

and [ ]3, 2β ∈ −A (1, –1)

C (–2, 8)B (–2, –7)

m = 3, n = 6

|m + n| = 9

48. Answer (D)

Hint : Image of 1,

tt

is 1

4 ,tt

− −

Solution :

Let point be 1

, tt

Its image in x = –2 is 1

4 ,tt

− −

⇒ Eliminating t, (x + 4)y = –1⇒ xy + 4y + 1 = 0

49. Answer (B)

Hint : p is y2 = 2x

Solution :

(x – 4)2 = 4y

⇒ y2 = 2x is p

Equation of normal is y = 3

2− + + ttx t at (at2, 2at)

It passes through 1

2,2

−

⇒ t3 – 2t + 1 = 0⇒ (t – 1) (t2 + t –1) = 0

t3 = 1, t

1 + t

2 = –1, t

1t2 = –1

Centroid is

2 2

1 2 1 21 1 2, , 0

6 3 3

+ + + + =

t t t t

50. Answer (B)

Hint : Normal on x2 = 4ay is y = mx + 2a + 2

a

m

Solution :

Tangent on y2 = 2x is

1

2= +y mx

m

Normal on x2 = 4ay is y = mx + 2a + 2

a

m

⇒2

12

2+ =aa

mm

⇒ 22 02

mam a− + =

D > 0

⇒ 21

8 04

− ≥a

⇒1

| |4 2


9/10

51. Answer (A)

Hint : Write family of circles

Solution :

Family of circles is

(–y2 + x2 + b2) + λ(y2 – 2x) = 0

⇒ x2 + y2 (λ – 1) – 2λx + b2 = 0

⇒ λ = 2 for circle

x2 + y2 – 4x + b2 = 0

Radius > 0

4 – b2 > 0

b ∈ (–2, 2)

Also parabola and hyperbola should intersect

⇒ 2x – x2 – b2 = 0

⇒ x2 –2x + b2 = 0

D > 0

4 – 4b2 > 0

b ∈ (–1, 1)

52. Answer (C)

Hint : Differentiation of (1 + x)n

Solution :

(1 + x)n = C0 + C

1 + C

2x2 + C

3x3 + .... + C

nxn

⇒ n(1 + x)n–1 = C1 + 2.C

2x + 3.C

3x2 +...+ n.C

nxn–1

⇒ nx(1 + x)n–1 = C1x + 2.C

2x2 + 3C

3x3 +...+ nC

nxn

⇒ nx(n – 1) (1 + x)n–2 + n(1 + x)n–1

= 12C1 + 22C

2x + 32C

3x2 + ..... + n2 C

nxn–1

Put x = 1

⇒2

1=n

n

r

r

r C = n(n + 1)2n–2

53. Answer (A)

Hint : Line should be tangent to the circle.

Solution : Line is tangent to the circle, the

12

2

− λ =

⇒ λ = –1

54. Answer (B)

Hint : AM ≥ GM

Solution :

α + β + γ + δ = –a

αβγ + βγδ + αγδ + αβδ = –c

αβγδ = 4

AM ≥ GM

1

4( )4

α + β + γ + δ ≥ αβγδ

⇒1

444

− ≥a ⇒ a2 ≥ 32

55. Answer (A)

Hint : Assume P(a secθ, b tanθ), P′(a secθ, –b tanθ)

Solution :

P(a secθ, b tanθ), P′ (a sec θ, –b tanθ), O(0, 0)

Let G(h, k)

⇒ 3h = 2a secθ 3k = 0 ⇒ y = 0

Locus of G is y = 0

56. Answer A(P, R, T); B(Q, S); C(Q); D(P)

Hint : Multiconcept

Solution : Let P(cosθ, sinθ), ‘θ’ is parametric angleof point P

T : x cosθ + y sinθ – 1 = 0, N : x sinθ – y cosθ = 0

(A) Area = 1 1 1 2

2 sin cos 3× × =

θ θ

⇒ 3 2 5

sin2 , , , ,.....2 6 3 3 6

π π π πθ = ± θ =

(B) Area = 1 1 1

2 sin cos× ×

θ θ is minimum, when

3 5sin2 1, , , ,.....

4 4 4

π π πθ = ± θ =

(C) mT = –1 ⇒ –cotθ = –1 ⇒ cotθ = 1

⇒ 5

, ,.....4 4

π πθ =

(D) mN =

1

3 ⇒

1tan

3θ = ⇒

7, ,.....

6 6

π πθ =


10/10

��

57. Answer A(P); B(P, Q); C(P, Q, R, S, T); D(P, Q, R)

Hint : Multiconcept

Solution : 5A, H, I, K, 2S, T

(A) n1 =

5

1

6 5 6 55 6 2 3 5

5 5

× ×× = = × = × ×C

(B) n2 = (4A + 2S) + (4A + 2 diff.)

= C5

2

6 6 16 5 10

4 2 4 2

+ = × +

= 3 × 5 × 21 = 32 × 5 × 7

(C) n3 = (2A + 2S + 2 diff.) + (2A + 4 diff.)

= 4 5

2 4

6 6.2 2 2

+C C

= 6 5 4 3

6 5 6 5 4 32

× × ×× + × × × ×

6 26 5 4 3 (3 5) 2 3 5= × × × + = × ×

(D) n4 = 6 = 6 × 5 × 4 × 3 × 2 = 24 × 32 × 5

58. Answer (05)

Hint : Equation of chord of contact is 10 x – 6y + 10 = 0

Solution :

Chord of contact of tangents drawn from P(1, 6) to the

parabola is

6y = 10 (x +1) ⇒ 10x – 6y + 10 = 0

Length of chord of contact = 2

2

4(1 )( )+ −a m a mc

m

= 4 25 5 5

9 5 1 525 9 3 3

× + − ×

= 8

345

Length of perpendicular from P(1, 6) to chord of

contact is 10 36 10 16

136 136

− += =p

1 16 8 3234

2 5 5136Δ = × × =

59. Answer (07)

Hint : Coefficient of x17 in (1 + x)20(x + 1)20.

Solution :

Let nCm

= coeff. of x17 in (1 + x)20(x + 1)20

= 40C17

= 40C23

⇒ maximum value of n + m is 63

60. Answer (09)

Hint : a = –6 + 4 cosθ, b = 3 + 4 sinθSolution :

Equation is (x + 6)2 + (y –3)2 = 42

a = –6 + 4 cosθ, b = 3 + 4 sinθ

a + b = –3 + 4 (cosθ + sinθ)

–3 – 4 2 ≤ (a + b) ≤ –3 + 4 2

|a + b| = 0, 1, 2, .....8

AIATS-2020 for JEE(Advanced)_Test-2A (CODE-G)_Paper-2_13-01-2019_SolAIATS-2020 for JEE(Advanced)_Test-2A (CODE-H)_Paper-2_13-01-2019_Sol

Test - 2A (Paper - 2) (Code-G) (Answers) All India Aakash ... · = 5lnT 4 10 T[ ]600 6002 [ ] 300...

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