Test - 2A (Paper-1) (Code C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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PHYSICS CHEMISTRY MATHEMATICS

Test Date : 14/01/2018

ANSWERS

TEST - 2A (Paper-1) - Code-C

All India Aakash Test Series for JEE (Advanced)-2019

1. (D)

2. (B)

3. (B)

4. (B)

5. (C)

6. (C)

7. (A, B)

8. (A, B)

9. (A, C)

10. (C)

11. (B, C)

12. (A, D)

13. (B, C)

14. (B, C)

15. (B, C)

16. A → (S)

B → (Q, S)

C → (P, T)

D → (R, T)

17. A → (P, R, S, T)

B → (S, T)

C → (S, T)

D → (Q, S, T)

18. (02)

19. (01)

20. (08)

21. (A)

22. (B)

23. (A)

24. (D)

25. (A)

26. (A)

27. (A)

28. (B)

29. (C)

30. (B)

31. (B, C)

32. (B)

33. ( D)

34. (A)

35. (C)

36. A → (P, Q, R)

B → (S, T)

C → (R)

D → (P, S)

37. A → (Q, R, T)

B → (S)

C → (P, Q, R, S, T)

D → (Q, R, T)

38. (03)

39. (06)

40. (04)

41. (B)

42. (C)

43. (D)

44. (B)

45. (D)

46. (D)

47. (A, B, D)

48. (A, B, C, D)

49. (A, D)

50. (A, C)

51. (B)

52. (A, B, D)

53. (A, B)

54. (A, C)

55. (A, B)

56. A → (P, Q, R, T)

B → (Q)

C → (S, T)

D → (P, Q, T)

57. A → (P, Q, R, S, T)

B → (R, S)

C → (Q, S)

D → (P, Q, R, S, T)

58. (07)

59. (04)

60. (08)

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code C) (Answers & Hints)

2/6

1. Answer (D)

Using principle of conservation of mechanical energy

2 22

2 2

1–4 – 2 4

22

4– – 2

2 2 2

GM GMMv

a a

GM GM

R R

× × = ×

× ××

1 11 – 2

2 2 2

GM GMv

R a

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2. Answer (B)

2

0

4 1

3 2

Rmg I= ω

π I0 = MI about point of contact at

instant of minimum potential energy

2 22

0 cm cm

4 4– ; –

3 2 3

R mR RI I m R I m

⎡ ⎤ ⎛ ⎞= + = ⎜ ⎟⎢ ⎥π π⎣ ⎦ ⎝ ⎠

2 22 22

0

4 4 8– –

2 3 3 3

mR R R RI m mR m m

⎡ ⎤ ⎛ ⎞= + + ⎜ ⎟⎢ ⎥π π π⎣ ⎦ ⎝ ⎠

2

0

3 8–

2 3I mR

⎡ ⎤= ⎢ ⎥π⎣ ⎦

16

(9 –16)

g

Rω =

π

3. Answer (B)

3 sin mg θ3 cos mg θf

COM

N

ΣFx = ma

3mg sinθ – f = 3ma

N – 3mg cosθ = 0

Στ = fR = IαFor no slipping, a = Rα

2

min

3

2

sin 3 cos

sin 3 cos

1tan

3

1tan

3

f NI mR

f mg f mg

mg mg

≤ μ=

= θ ≤ μ θθ ≤ μ θ

μ ≥ θ

μ = θ

PART - I (PHYSICS)

ANSWERS & HINTS

4. Answer (B)

ρ(a – x)g + x4ρg = x2ρg + 3ρ(a – x)g

a – x + 4x = 2x + 3a – 3x

4x = 2a

2

ax =

5. Answer (C)

N

mgmg cosθ mg sinθ

μN

a3

θ

21sin

2 3

aI mgω = θ

24

9

maI = ⇒

2 3 sin

2

g

a

θω =

3 cos

4

gd

d a

θω ω = α =θ

Now along the rod

μN – mg sinθ = 2

3

am⎛ ⎞ω⎜ ⎟⎝ ⎠

⇒3 sin

sin3 2

gmaN mg

a

θμ = × + θ

f = 3

sin2

N mgμ = θ ...(i)

⊥ to rod mg cos θ – N = 3 cos

3 3 4

ga am m

a

θ× α = ×

3cos

4N mg= θ ...(ii)

sin3. 4 2tan

2.3 cos

mg

mg

θμ = × = θθ

6. Answer (C)

m

v

v

ω

2

2 2 21 1 1 2 2

2 2 2 3

vmgh mv mv mr

r

⎡ ⎤ ⎛ ⎞= + + × ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

3

7

ghv =

7. Answer (A, B)

Test - 2A (Paper-1) (Code C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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8. Answer (A, B)

Weight of astronaut = force due to gravitation by earth

e

3

eGM gR

vr

= =

e

3

egR GM

r= , r = 3R

e

∴ 2 9(3 )

e

e

GM m mgW

R= =

9. Answer (A, C)

10. Answer (C)

Using Bernoulli's theorem,

XCBA N O

x

v0

2 21 1

2 2N A

N A

P v P v⎛ ⎞ ⎛ ⎞+ ρ = + ρ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

22

0 0–

2 2

vP P v

ρ ρ= +

=

23

2

0 0 3

11– 1

2

RP v

x

⎡ ⎤⎛ ⎞⎢ ⎥+ ρ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

11. Answer (B, C)

0dP

dx=

2 330

3 4

3–2 1 0

2

v RR

x x

⎡ ⎤⎛ ⎞ρ ⎡ ⎤−+ =⎢ ⎥⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦

0dP

dx= at x = –R and x → ± ∞

12. Answer (A, D)

13. Answer (B, C)

14. Answer (B, C)

15. Answer (B, C)

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

( )2

2Ah

d x avdt v

dtdt

ρ ×ρ =

⇒ Acceleration = ( )2 constanta

gA

× =

Time taken to empty tank, t0 =

2A H

a g

∴ Vfinal

= 2

2A H a

ga g A

×

= 2 2gH

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

18. Answer (02)

∵ ( )

muv

M m=

+∴ v 2 = 2gL (1 – cosα)

⇒2 2

2

22 2sin

2( )

m vgL

M m

α⎛ ⎞= × ⎜ ⎟⎝ ⎠+

∴ n = 2

19. Answer (01)

T = Mgsin30° = 2

Mg

and T × R = MR2 × a

R

⎛ ⎞⎜ ⎟⎝ ⎠

, where a = acceleration of

block m

⇒ 2

2

Mg aR MR

R× = ×

⇒2

ga =

And,

( – )1

–2 2

T m g a M

Mg g mm g

=⇒ =

⎛ ⎞= ⎜ ⎟⎝ ⎠

20. Answer (08)

Volume of cone 21

3V R h= π

Volume of protruding part

2

out

1.

3 2 2

R hV

⎛ ⎞= π⎜ ⎟⎝ ⎠

out

8

VV =

Volume of cone in water in

7

8

VV =

Buoyancy force = ρgVin – force missing due to outside

portion

=

27

–8 4

V Rg ghρ π ρ

= 7 3

–8 4

VVg gρ ρ

= 8

Vgρ

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code C) (Answers & Hints)

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PART - II (CHEMISTRY)

21. Answer (A)

ΔG° = –RT lnK

For product to predominate

K > 1 so

lnK = – G

RT

Δ °,

– G

RTK e

Δ °

=

ΔG° < 0 K > 1

22. Answer (B)

23. Answer (A)

KSP

of AgCl ⇒ [Ag+] [Cl–] = 10–9

[Ag+] =

–9 –9

–8

–

10 1010 M

0.1[Cl ]⇒ ⇒

KSP

of Ag2CrO

4 ⇒ [Ag+]2 [CrO

2–

4 ] = 10–11

[Ag+] = 10–5 M

So AgCl need 10–8 mol to start precipitation but Ag2CrO

4

need 10–5 mol to start precipitation.

24. Answer (D)

ΔG° = ΔH° – TΔS° = –RT ln K

ln K = ( )S – H

R RT

Δ ° Δ °+

Slope of lnK versus 1

T is positive if ΔH° = –ve

Intercept is +ve it means ΔS° is positive.

25. Answer (A)

In process A to B, T ⇒ constant, so it is isothermal

process, Entropy increases so ΔS = +ve

ΔS = Q

veT

= +

Q = +ve, W = –ve in isothermal process so work is

done by the gas.

In B to C, entropy ⇒ constant; ΔS = 0 so process may

be adiabatic and temperature decrease it means it is

adiabatic expansion

ΔU = nCvΔT = – P

ext dV for adiabatic.

In C to D temperature is constant so ΔH = 0

In proces D to A. ΔS = 0, it means process may be

adiabatic and temperature increase it is compression.

Hence work is done on the gas.

26. Answer (A)

27. Answer (A)

( )

( ) ( ) ( )

2 6 11 2 2 3

Y

3 2 2 4 7

P ZX

Ca B O .5H O 2Na CO

2CaCO NaBO Na B O

+ ⎯⎯→

↓ + +

( ) ( )2 2 2 3 2 4 7

P Z

NaBO CO Na CO Na B O+ ⎯⎯→ +

28. Answer (B)

29. Answer (C)

30. Answer (B)

31. Answer (B, C)

6.8% H2O

2 means 6.8 g in 100 mL, so 68 g H

2O

2 is

present in 1 litre.

H2O

2 → H

2O +

1

2O

2

Volume strength = 22.4 V

Molarity = 2

1 = 2 M

32. Answer (B)

( )2 2 2excess

H O 2KI 2KOH I+ ⎯⎯→ +

Mole of I2 =

0.508

254 = 2 × 10–3

So mole of H2O

2 = 2 × 10–3

M =

3

3

2 100 25

8 10.

−

−× =×

33. Answer (D)

34. Answer (A)

CaCl2.xH

2O

Δ⎯⎯→ CaCl2 + xH

2O

Mole of CaCl2.xH

2O =

35 10

2

−×= 2.5 × 10–3

Mole of H2O = 2.5 × 10–3x

x = 4

35. Answer (C)

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

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

38. Answer (03)

meq

of metal = meq

of Sn+2

40 × 0.5 × n = 10 × 1 × 2

∴ n = 1

or 4 n 1 3

M M+ = +⎯⎯⎯→

∴ New state = 03

39. Answer (06)

A → B2H

6

40. Answer (04)

SiO2 + NaOH → P + H

2O

P ⇒ Na2SiO

3

Test - 2A (Paper-1) (Code C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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PART - III (MATHEMATICS)

41. Answer (B)

(x – 1)2 = –1 ⇒ x = 1 ± i

α = 1 + i , β = 1 – i

∴ αn + βn = 1

22 cos4

n

n+ π

42. Answer (C)

(i) 0,D ≥ (ii) complement set of set in which both

roots are less than or equal to zero.

From (i) → 4λ ≤ or 10.λ ≥

From (ii) → 15

4λ < or 5λ >

On combining we have 15, [10, )4

⎛ ⎞−∞ ∪ ∞⎜ ⎟⎝ ⎠

43. Answer (D)

Coeff. of xn in

2

11! 2! !

⎛ ⎞+ + + − − − +⎜ ⎟⎜ ⎟

⎝ ⎠

nx x x

n

= ( )1 1 1 1

.........0! ! 1!( 1)! 2! 2 ! !0!

+ + + +− −n n n n

= 1

!n (nC

0 + nC

1 + nC

2 + .....+nC

n )

= 2

!

n

n

44. Answer (B)

2b = a + a2 and a4 = ab

On solving 1 1,

2 8a b

−= = −

45. Answer (D)

y

(0,3)

(2 + 4 )i

(4,0)x

(2 – )i

O

|z – 4|2 + |z – 3i|2 = 25

⇒ x2 + y2 – 8x – 6y = 0

|z – 1| = |z – 3|

⇒ x = 2

⇒ y = –1 and 4

⇒ y1 + y

2 = 3

46. Answer (D)

Equation of the circle is |z + 2 – 3i| = 3

Let W = z + 3 + 2i = z + 2 – 3i + 1 + 5i

|W – 1 – 5i| = |z + 2 – 3i| = 3

47. Answer (A, B, D)

(1 – x)24 = 24C0 – 24C

1x1 + 24C

2 x2 + ..... + (–1)r.24C

r.xr + .....

+ 24C24

.x24

(1 – x)–1 = 1 + x + x2 + x3 + x4 + ..... + x16 + ...... ∞Multiplying both and compare the coefficient of x16 on

both sides

24C0 – 24C

1 + 24C

2 – 24C

3 + ...... + 24C

16 = 23C

16.

⇒ nCk = 23C

16 = 23C

7(∵ k < 10)

⇒ n = 23 and k = 7

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

(1 – x)16 = 16C0 – 16C

1x + 16C

2x2

...... + 16C

16x16

(1 – x)–1 = 1 + x + x2

+ .....+ x16 + ...... ∞

Multiplying both and compare the coefficient of x6 on

both sides we get

16C0 – 16C

1 + 16C

2 ......... = 15C

6

= 15!

6!9! = 5 × 7 × 11 × 13

Which is divisible by 5, 7, 11 and 13

49. Answer (A, D)

(1 – x)20 = 20C0 – 20C

1x + 20C

2x2 + ..... + 20C

20x20

(1 – x)–1 = 1 + x + x2 + ......... ∞Multiplying both sides and compare coefficient of x13

(–1)13 19C13

= – 20C7 + 20C

8 – 20C

9 + ........ + 20C

20

⇒ 20C7 – 20C

8 + 20C

9 – ...... – 20C

20 = 19C

13

n = 19 and m can be 6 or 13.

50. Answer (A, C)

51. Answer (B)

52. Answer (A, B, D)

(0,5)4

3

Ox

y

( )( )5 4 0z z z i+ − ≤ ≤If arg(z) is the least positive then |z| = 3.

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code C) (Answers & Hints)

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53. Answer (A, B)

Coefficient of x4 is 4! ( )3

25

1 1 24542

xx x

⎛ ⎞+ + + =⎜ ⎟

⎝ ⎠

54. Answer (A, C)

3 7

2 2

4!756

2!C C× × =

55. Answer (A, B)

8C4 × 4! = 1680

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

(A) ( )12 11

2 2

1log 1 2 2 log 2 11

2

⎛ ⎞+ − = =⎜ ⎟⎝ ⎠

(B)

10 10 101010 10

0 0 0

1 11

1 11 1k k k

k k k

kC C C

k k= = =

⎛ ⎞ = −⎜ ⎟⎝ + ⎠ +∑ ∑ ∑

= ( )

1010 11 10 11

1

0

1 12 2 2 1

11 11k

k

C +=

⎛ ⎞− = − −⎜ ⎟

⎝ ⎠∑

=

10 1111.2 2 1

11

− +

⇒ a = 9, b = 1 and c = 11

(C) ( )( )( )

( )( )1 1

1 3 2

1 2 2 1 2r r

r r

r r r r r

∞ ∞

= =

+ −++ + + +∑ ∑

= ( ) ( )( )1 1

1 1 3 1 1

1 2 2 1 1 2r r

r r r r r r

∞ ∞

= =

⎡ ⎤⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎢ ⎥⎜ ⎟⎝ + + ⎠ + + +⎝ ⎠⎣ ⎦

∑ ∑

= 1 3 1 1 3 5

0 92 2 2 2 4 4

⎛ ⎞ ⎛ ⎞− + = + = ⇒ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

a b

(D)1 5

1 1 1 12 6 3 8 24

x x xx

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

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

2 21

1 1 1 1 1

2 1 1n nR R n n n n+

⎡ ⎤− = −⎢ ⎥⎣ − − + − ⎦

⇒ Rn = 2(n2 – n – 1)

Now, Sn = 4[(n + 2)2 – (n + 2) – 1] – 3[2(n2 – n – 1)]

= –2n2 + 18n + 10

⇒ Tn =

1

2[(n2 – n – 1) – ((n – 1)2 – (n – 1) – 1)] = (n – 1)

(A) Rn > 0, 2(n2 – n + 1) > 0 for n = 2,3,4,5,6.

(B) Sn will maximum at

9N.

2n = ∉ but due to

symmetry of parabola Sn will be maximum at

n = 4, 5.

(C) ||n – 1| – 3| = 1 ⇒ n = 3, 5

(D) n – 1 < 2(n2 – n + 1) + 2 ⇒ (n – 1)(2n – 1) > 0

n < 1

2 or n > 1 ⇒ n = 2, 3, 4, 5, 6

58. Answer (07)

1 + 2 + 3 + ..... + k = n

⇒( )12

k kn

+=

Now, ( ) ( )21

100 52

k kk

+ + = +

⇒ (k + 25)(k – 6) = 0 ⇒ k = 6

∴ n = 21 ⇒ 73

n =

59. Answer (04)

Required number of ways

= Coefficient of x20 in (x3 + x4 + x5 + x6 + x7)5

= Coefficient of x5 in (1 – x5)5(1 – x)–5 = 121

60. Answer (08)

D < 0, a = 1 > 0

16 – 4(λ – 3) < 0

⇒ 4 – λ + 3 < 0 ⇒ λ > 7

∴ The least integral value = 8

� � �

Test - 2A (Paper-1) (Code D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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PHYSICS CHEMISTRY MATHEMATICS

Test Date : 14/01/2018

ANSWERS

TEST - 2A (Paper-1) - Code-D

All India Aakash Test Series for JEE (Advanced)-2019

1. (C)

2. (C)

3. (B)

4. (B)

5. (B)

6. (D)

7. (A, C)

8. (A, B)

9. (A, B)

10. (A, D)

11. (B, C)

12. (C)

13. (B, C)

14. (B, C)

15. (B, C)

18. A → (P, R, S, T)

B → (S, T)

C → (S, T)

D → (Q, S, T)

17. A → (S)

B → (Q, S)

C → (P, T)

D → (R, T)

18. (08)

19. (01)

20. (02)

21. (A)

22. (A)

23. (D)

24. (A)

25. (B)

26. (A)

27. (C)

28. (B)

29. (A)

30. (B)

31. (B, C)

32. (B)

33. (C)

34. (A)

35. ( D)

36. A → (Q, R, T)

B → (S)

C → (P, Q, R, S, T)

D → (Q, R, T)

37. A → (P, Q, R)

B → (S, T)

C → (R)

D → (P, S)

38. (04)

39. (06)

40. (03)

41. (D)

42. (D)

43. (B)

44. (D)

45. (C)

46. (B)

47. (A, D)

48. (A, B, C, D)

49. (A, B, D)

50. (A, B, D)

51. (B)

52. (A, C)

53. (A, B)

54. (A, C)

55. (A, B)

56. A → (P, Q, R, S, T)

B → (R, S)

C → (Q, S)

D → (P, Q, R, S, T)

57. A → (P, Q, R, T)

B → (Q)

C → (S, T)

D → (P, Q, T)

58. (08)

59. (04)

60. (07)

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code D) (Answers & Hints)

2/6

1. Answer (C)

m

v

v

ω

2

2 2 21 1 1 2 2

2 2 2 3

vmgh mv mv mr

r

⎡ ⎤ ⎛ ⎞= + + × ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

3

7

ghv =

2. Answer (C)

N

mgmg cosθ mg sinθ

μN

a3

θ

21sin

2 3

aI mgω = θ

24

9

maI = ⇒

2 3 sin

2

g

a

θω =

3 cos

4

gd

d a

θω ω = α =θ

Now along the rod

μN – mg sinθ = 2

3

am⎛ ⎞ω⎜ ⎟⎝ ⎠

⇒3 sin

sin3 2

gmaN mg

a

θμ = × + θ

f = 3

sin2

N mgμ = θ ...(i)

⊥ to rod mg cos θ – N = 3 cos

3 3 4

ga am m

a

θ× α = ×

3cos

4N mg= θ ...(ii)

sin3. 4 2tan

2.3 cos

mg

mg

θμ = × = θθ

3. Answer (B)

ρ(a – x)g + x4ρg = x2ρg + 3ρ(a – x)g

a – x + 4x = 2x + 3a – 3x

4x = 2a

2

ax =

PART - I (PHYSICS)

ANSWERS & HINTS

4. Answer (B)

3 sin mg θ3 cos mg θf

COM

N

ΣFx = ma

3mg sinθ – f = 3ma

N – 3mg cosθ = 0

Στ = fR = IαFor no slipping, a = Rα

2

min

3

2

sin 3 cos

sin 3 cos

1tan

3

1tan

3

f NI mR

f mg f mg

mg mg

≤ μ=

= θ ≤ μ θθ ≤ μ θ

μ ≥ θ

μ = θ

5. Answer (B)

2

0

4 1

3 2

Rmg I= ω

π I0 = MI about point of contact at

instant of minimum potential energy

2 22

0 cm cm

4 4– ; –

3 2 3

R mR RI I m R I m

⎡ ⎤ ⎛ ⎞= + = ⎜ ⎟⎢ ⎥π π⎣ ⎦ ⎝ ⎠

2 22 22

0

4 4 8– –

2 3 3 3

mR R R RI m mR m m

⎡ ⎤ ⎛ ⎞= + + ⎜ ⎟⎢ ⎥π π π⎣ ⎦ ⎝ ⎠

2

0

3 8–

2 3I mR

⎡ ⎤= ⎢ ⎥π⎣ ⎦

16

(9 –16)

g

Rω =

π

6. Answer (D)

Using principle of conservation of mechanical energy

2 22

2 2

1–4 – 2 4

22

4– – 2

2 2 2

GM GMMv

a a

GM GM

R R

× × = ×

× ××

Test - 2A (Paper-1) (Code D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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1 11 – 2

2 2 2

GM GMv

R a

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

7. Answer (A, C)

8. Answer (A, B)

Weight of astronaut = force due to gravitation by earth

e

3

eGM gR

vr

= =

e

3

egR GM

r= , r = 3R

e

∴ 2 9(3 )

e

e

GM m mgW

R= =

9. Answer (A, B)

10. Answer (A, D)

11. Answer (B, C)

0dP

dx=

2 330

3 4

3–2 1 0

2

v RR

x x

⎡ ⎤⎛ ⎞ρ ⎡ ⎤−+ =⎢ ⎥⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦

0dP

dx= at x = –R and x → ± ∞

12. Answer (C)

Using Bernoulli's theorem,

XCBA N O

x

v0

2 21 1

2 2N A

N A

P v P v⎛ ⎞ ⎛ ⎞+ ρ = + ρ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

22

0 0–

2 2

vP P v

ρ ρ= +

=

23

2

0 0 3

11– 1

2

RP v

x

⎡ ⎤⎛ ⎞⎢ ⎥+ ρ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

13. Answer (B, C)

14. Answer (B, C)

15. Answer (B, C)

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

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

( )2

2Ah

d x avdt v

dtdt

ρ ×ρ =

⇒ Acceleration = ( )2 constanta

gA

× =

Time taken to empty tank, t0 =

2A H

a g

∴ Vfinal

= 2

2A H a

ga g A

×

= 2 2gH

18. Answer (08)

Volume of cone 21

3V R h= π

Volume of protruding part

2

out

1.

3 2 2

R hV

⎛ ⎞= π⎜ ⎟⎝ ⎠

out

8

VV =

Volume of cone in water in

7

8

VV =

Buoyancy force = ρgVin – force missing due to outside

portion

=

27

–8 4

V Rg ghρ π ρ

= 7 3

–8 4

VVg gρ ρ

= 8

Vgρ

19. Answer (01)

T = Mgsin30° = 2

Mg

and T × R = MR2 × a

R

⎛ ⎞⎜ ⎟⎝ ⎠

, where a = acceleration of

block m

⇒ 2

2

Mg aR MR

R× = ×

⇒2

ga =

And,

( – )1

–2 2

T m g a M

Mg g mm g

=⇒ =

⎛ ⎞= ⎜ ⎟⎝ ⎠

20. Answer (02)

∵ ( )

muv

M m=

+

∴ v 2 = 2gL (1 – cosα)

⇒2 2

2

22 2sin

2( )

m vgL

M m

α⎛ ⎞= × ⎜ ⎟⎝ ⎠+

∴ n = 2

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code D) (Answers & Hints)

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PART - II (CHEMISTRY)

21. Answer (A)

22. Answer (A)

In process A to B, T ⇒ constant, so it is isothermal

process, Entropy increases so ΔS = +ve

ΔS = Q

veT

= +

Q = +ve, W = –ve in isothermal process so work is

done by the gas.

In B to C, entropy ⇒ constant; ΔS = 0 so process may

be adiabatic and temperature decrease it means it is

adiabatic expansion

ΔU = nCvΔT = – P

ext dV for adiabatic.

In C to D temperature is constant so ΔH = 0

In proces D to A. ΔS = 0, it means process may be

adiabatic and temperature increase it is compression.

Hence work is done on the gas.

23. Answer (D)

ΔG° = ΔH° – TΔS° = –RT ln K

ln K = ( )S – H

R RT

Δ ° Δ °+

Slope of lnK versus 1

T is positive if ΔH° = –ve

Intercept is +ve it means ΔS° is positive.

24. Answer (A)

KSP

of AgCl ⇒ [Ag+] [Cl–] = 10–9

[Ag+] =

–9 –9

–8

–

10 1010 M

0.1[Cl ]⇒ ⇒

KSP

of Ag2CrO

4 ⇒ [Ag+]2 [CrO

2–

4 ] = 10–11

[Ag+] = 10–5 M

So AgCl need 10–8 mol to start precipitation but Ag2CrO

4

need 10–5 mol to start precipitation.

25. Answer (B)

26. Answer (A)

ΔG° = –RT lnK

For product to predominate

K > 1 so

lnK = – G

RT

Δ °,

– G

RTK e

Δ °

=

ΔG° < 0 K > 1

27. Answer (C)

28. Answer (B)

29. Answer (A)

( )

( ) ( ) ( )

2 6 11 2 2 3

Y

3 2 2 4 7

P ZX

Ca B O .5H O 2Na CO

2CaCO NaBO Na B O

+ ⎯⎯→

↓ + +

( ) ( )2 2 2 3 2 4 7

P Z

NaBO CO Na CO Na B O+ ⎯⎯→ +

30. Answer (B)

( )2 2 2excess

H O 2KI 2KOH I+ ⎯⎯→ +

Mole of I2 =

0.508

254 = 2 × 10–3

So mole of H2O

2 = 2 × 10–3

M =

3

3

2 100 25

8 10.

−

−× =×

31. Answer (B, C)

6.8% H2O

2 means 6.8 g in 100 mL, so 68 g H

2O

2 is

present in 1 litre.

H2O

2 → H

2O +

1

2O

2

Volume strength = 22.4 V

Molarity = 2

1 = 2 M

32. Answer (B)

33. Answer (C)

34. Answer (A)

CaCl2.xH

2O

Δ⎯⎯→ CaCl2 + xH

2O

Mole of CaCl2.xH

2O =

35 10

2

−×= 2.5 × 10–3

Mole of H2O = 2.5 × 10–3x

x = 4

35. Answer (D)

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

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

38. Answer (04)

SiO2 + NaOH → P + H

2O

P ⇒ Na2SiO

3

39. Answer (06)

A → B2H

6

40. Answer (03)

meq

of metal = meq

of Sn+2

40 × 0.5 × n = 10 × 1 × 2

∴ n = 1

or 4 n 1 3

M M+ = +⎯⎯⎯→

∴ New state = 03

Test - 2A (Paper-1) (Code D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2019

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PART - III (MATHEMATICS)

41. Answer (D)

Equation of the circle is |z + 2 – 3i| = 3

Let W = z + 3 + 2i = z + 2 – 3i + 1 + 5i

|W – 1 – 5i| = |z + 2 – 3i| = 3

42. Answer (D)

y

(0,3)

(2 + 4 )i

(4,0)x

(2 – )i

O

|z – 4|2 + |z – 3i|2 = 25

⇒ x2 + y2 – 8x – 6y = 0

|z – 1| = |z – 3|

⇒ x = 2

⇒ y = –1 and 4

⇒ y1 + y

2 = 3

43. Answer (B)

2b = a + a2 and a4 = ab

On solving 1 1,

2 8a b

−= = −

44. Answer (D)

Coeff. of xn in

2

11! 2! !

⎛ ⎞+ + + − − − +⎜ ⎟⎜ ⎟

⎝ ⎠

nx x x

n

= ( )1 1 1 1

.........0! ! 1!( 1)! 2! 2 ! !0!

+ + + +− −n n n n

= 1

!n (nC

0 + nC

1 + nC

2 + .....+nC

n )

= 2

!

n

n

45. Answer (C)

(i) 0,D ≥ (ii) complement set of set in which both

roots are less than or equal to zero.

From (i) → 4λ ≤ or 10.λ ≥

From (ii) → 15

4λ < or 5λ >

On combining we have 15, [10, )4

⎛ ⎞−∞ ∪ ∞⎜ ⎟⎝ ⎠

46. Answer (B)

(x – 1)2 = –1 ⇒ x = 1 ± i

α = 1 + i , β = 1 – i

∴ αn + βn = 1

22 cos4

n

n+ π

47. Answer (A, D)

(1 – x)20 = 20C0 – 20C

1x + 20C

2x2 + ..... + 20C

20x20

(1 – x)–1 = 1 + x + x2 + ......... ∞Multiplying both sides and compare coefficient of x13

(–1)13 19C13

= – 20C7 + 20C

8 – 20C

9 + ........ + 20C

20

⇒ 20C7 – 20C

8 + 20C

9 – ...... – 20C

20 = 19C

13

n = 19 and m can be 6 or 13.

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

(1 – x)16 = 16C0 – 16C

1x + 16C

2x2

...... + 16C

16x16

(1 – x)–1 = 1 + x + x2

+ .....+ x16 + ...... ∞

Multiplying both and compare the coefficient of x6 on

both sides we get

16C0 – 16C

1 + 16C

2 ......... = 15C

6

= 15!

6!9! = 5 × 7 × 11 × 13

Which is divisible by 5, 7, 11 and 13

49. Answer (A, B, D)

(1 – x)24 = 24C0 – 24C

1x1 + 24C

2 x2 + ..... + (–1)r.24C

r.xr + .....

+ 24C24

.x24

(1 – x)–1 = 1 + x + x2 + x3 + x4 + ..... + x16 + ...... ∞Multiplying both and compare the coefficient of x16 on

both sides

24C0 – 24C

1 + 24C

2 – 24C

3 + ...... + 24C

16 = 23C

16.

⇒ nCk = 23C

16 = 23C

7(∵ k < 10)

⇒ n = 23 and k = 7

50. Answer (A, B, D)

(0,5)4

3

Ox

y

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper-1) (Code D) (Answers & Hints)

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( )( )5 4 0z z z i+ − ≤ ≤If arg(z) is the least positive then |z| = 3.

51. Answer (B)

52. Answer (A, C)

53. Answer (A, B)

8C4 × 4! = 1680

54. Answer (A, C)

3 7

2 2

4!756

2!C C× × =

55. Answer (A, B)

Coefficient of x4 is 4! ( )3

25

1 1 24542

xx x

⎛ ⎞+ + + =⎜ ⎟

⎝ ⎠

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

2 21

1 1 1 1 1

2 1 1n nR R n n n n+

⎡ ⎤− = −⎢ ⎥⎣ − − + − ⎦

⇒ Rn = 2(n2 – n – 1)

Now, Sn = 4[(n + 2)2 – (n + 2) – 1] – 3[2(n2 – n – 1)]

= –2n2 + 18n + 10

⇒ Tn =

1

2[(n2 – n – 1) – ((n – 1)2 – (n – 1) – 1)] = (n – 1)

(A) Rn > 0, 2(n2 – n + 1) > 0 for n = 2,3,4,5,6.

(B) Sn will maximum at

9N.

2n = ∉ but due to

symmetry of parabola Sn will be maximum at

n = 4, 5.

(C) ||n – 1| – 3| = 1 ⇒ n = 3, 5

(D) n – 1 < 2(n2 – n + 1) + 2 ⇒ (n – 1)(2n – 1) > 0

n < 1

2 or n > 1 ⇒ n = 2, 3, 4, 5, 6

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

(A) ( )12 11

2 2

1log 1 2 2 log 2 11

2

⎛ ⎞+ − = =⎜ ⎟⎝ ⎠

(B)

10 10 101010 10

0 0 0

1 11

1 11 1k k k

k k k

kC C C

k k= = =

⎛ ⎞ = −⎜ ⎟⎝ + ⎠ +∑ ∑ ∑

= ( )

1010 11 10 11

1

0

1 12 2 2 1

11 11k

k

C +=

⎛ ⎞− = − −⎜ ⎟

⎝ ⎠∑

=

10 1111.2 2 1

11

− +

⇒ a = 9, b = 1 and c = 11

(C) ( )( )( )

( )( )1 1

1 3 2

1 2 2 1 2r r

r r

r r r r r

∞ ∞

= =

+ −++ + + +∑ ∑

= ( ) ( )( )1 1

1 1 3 1 1

1 2 2 1 1 2r r

r r r r r r

∞ ∞

= =

⎡ ⎤⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎢ ⎥⎜ ⎟⎝ + + ⎠ + + +⎝ ⎠⎣ ⎦

∑ ∑

= 1 3 1 1 3 5

0 92 2 2 2 4 4

⎛ ⎞ ⎛ ⎞− + = + = ⇒ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

a b

(D)1 5

1 1 1 12 6 3 8 24

x x xx

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

58. Answer (08)

D < 0, a = 1 > 0

16 – 4(λ – 3) < 0

⇒ 4 – λ + 3 < 0 ⇒ λ > 7

∴ The least integral value = 8

59. Answer (04)

Required number of ways

= Coefficient of x20 in (x3 + x4 + x5 + x6 + x7)5

= Coefficient of x5 in (1 – x5)5(1 – x)–5 = 121

60. Answer (07)

1 + 2 + 3 + ..... + k = n

⇒( )12

k kn

+=

Now, ( ) ( )21

100 52

k kk

+ + = +

⇒ (k + 25)(k – 6) = 0 ⇒ k = 6

∴ n = 21 ⇒ 73

n =

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