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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)
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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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