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All India Aakash Test Series for JEE (Advanced)-2020
Test Date : 16/06/2019
ANSWERS
1/11
TEST - 1A (Paper-2) - Code-C
Test - 1A (Paper-2) (Code-C) (Answers) All India Aakash Test Series for JEE (Advanced)-2020
PHYSICS CHEMISTRY MATHEMATICS
1. (B, D)
2. (B, C)
3. (A, C)
4. (B, C, D)
5. (B, C, D)
6. (B, C)
7. (25)
8. (24)
9. (36)
10. (32)
11. (10)
12. (18)
13. (13)
14. (30)
15. (D)
16. (B)
17. (C)
18. (D)
19. (A, B, C, D)
20. (B, C, D)
21. (C)
22. (A, B, C, D)
23. (A, B)
24. (A, C, D)
25. (20)
26. (70)
27. (30)
28. (45)
29. (06)
30. (28)
31. (80)
32. (30)
33. (A)
34. (B)
35. (A)
36. (C)
37. (A, B, C)
38. (C, D)
39. (B, C)
40. (A, B, C)
41. (A, C)
42. (A, B, D)
43. (32)
44. (27)
45. (80)
46. (06)
47. (15)
48. (32)
49. (69)
50. (49)
51. (C)
52. (B)
53. (A)
54. (D)
2/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-C) (Hints & Solutions)
PART - I (PHYSICS)
1. Answer (B, D)
Hint : 1 2=Kq q
Ur
Solution :
( )
2 2
sys0 0
4 24 4 2
q qU
a a
= +
WA = –(VA – V) × q
2. Answer (B, C)
Hint : Net field is vertically up.
Solution :
2
04
QE
a=
( )net 2
0
23 cos 3
34
QE E
a = =
2
0
6
4
Q
a=
And, 0
34
QV
a=
3. Answer (A, C)
Hint : Charge = Ceq × V
Solution :
1
2 412 16 C
2 4
= =
+ Q
2
6 312 24 C
6 3
= =
+ Q
1612 4 V
2AV = − =
24
12 8 V6B
V = − =
4 8 4 V − = − = −A B
V V
4. Answer (B, C, D)
Hint : Use the expression of field and potential.
Solution :
( )2 2
33
2= −
KQV R r
R for r < R
3
2
=
C
KQV
R 50% more than VS
and, =KQ
Vr
for r > R
and, 3
=KQ
E rR
for r < R
5. Answer (B, C, D)
Hint : Use Kirchhoff's laws.
Solution :
2 82 F
3 3= + =
ABC
eq
81
83 F8 11
13
= =
+
C
bat
811 8 C
11 = = Q
∴ VAB = 11 – 8 × 1 = 3 V
23 2 V,
3 = =
ACV VCD = 1 V
6. Answer (B, C)
Hint : ,= = x x y yE dE E dE
Solution :
( )0 0
200 0
coscos 2
44x
RdE
RR
=
= − =
Ey = 0
( )20
00
cos0
4C
RdV
R
= =
7. Answer (25)
Hint : 0
2E
r
=
Solution :
10
052
= −
B
A
V
V
dv drr
3/11
Test - 1A (Paper-2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
( ) ( )
0
ln 22
− =
A BV V
= 2 × 10–6 × 2 × 9 × 109 × 0.69
≃25 × 103 V
8. Answer (24)
Hint : in
0
=Q
E ds
Solution :
∵ V = –4ar2 + 3b
8−
= =dv
E ardr
Now, ϕ = EA
⇒ dϕ = dEA + EdA
∵ dE = 8adr, dA = 8𝜋rdr
22
0
48 4 8 8
r dradr r ar rdr
= +
( ) 2
02
32 64
4
+ =
a r dr
r dr
⇒ ρ = 24a0
∴ n = 24
9. Answer (36)
Hint : All charges must experience zero forces.
Solution :
Q should be –ve
Force on ‘q’
2
2 2 2 2
0 0
4 4
4 4
q Qq q Q
l x l x = =
Force on ‘Q’
( ) ( )2 2 2 2
00
4 4 1
4– 4 –
qQ qQ
x xl x l x = =
2 –3
lx l x x = =
2
2
4 4
99
q l qQ Q
l
= =
36 CQ =
10. Answer (32)
Hint : Use concept of electrostatic pressure.
Solution :
Charge gets uniformly distributed over the sphere
24 =
Q
R
Electrostatic pressure, 2
02
P
=
( )2
2
0
Force2
=
R
22
20
1
24
=
QR
R
2
2
032
=
Q
R
∴ n = 32
11. Answer (10)
Hint : Charge flows from one capacitor to the other.
Solution :
0 01 2
0 0
,A A
C Cd vt d vt
= =
+ −
q1 + q2 = 20 C = Qtotal
( )11 total
1 2
= +
Cq Q
C C
( )0total
02
d vtQ
d
−=
( )1total
02
−= =
dq vi Q
dt d
6
–2
10 20 10
2 10
− =
= 10 mA
12. Answer (18)
Hint : Use Kirchhoff's laws.
Solution :
21
1 2
12AB
CV V
C C=
+
12 3
2
ABV
⇒ VAB 18 kV …(i)
and, 12
1 2
8AB
CV V
C C=
+
4/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-C) (Hints & Solutions)
8 3
1
ABV
⇒ VAB 24 kV …(ii)
From (i) and (ii)
(VAB) = 18 kV
13. Answer (13)
Hint : Use Kirchhoff's laws.
Solution :
( )
30 5 10 213 V
2 3 5ABV
− = =
+ +
14. Answer (30)
Hint : E is maximum at 2
Rx =
Solution :
max 3 2
2 022
0
2 32
4 9
42
RQ
QE
RR
R
= =
+
∴ max 2
0
2 3
4 9
q Qa
R m
=
6 6 9
2 3
10 12 10 2 3 9 10
4 10 9 10 10
− −
− −
=
= 30 m/s2
15. Answer (D)
Hint : Use Gauss's theorem.
Solution :
10 0
1
6 4 24
= =
Q Q
20 0
1– 3
3 24
Q Q =
0 0
1 7 7
3 8 24
Q Q = =
For tetrahedron,
3
0 0
1
4 4
Q Q = =
For stripe
4
0 0
90
360 4
Q Q = =
16. Answer (B)
Hint : Electric field is uniform inside cavity.
Solution :
E r for charged solid sphere
E = constant inside cavity
E = 0 inside shells
V = constant inside shells
17. Answer (C)
Hint : Equivalent capacitance increases.
Solution :
So, Q1 increases ⇒ F1 increases ⇒ V1 increases
V2 decreases as V1 + V2 = Vbattery
⇒ Q2 decreases
⇒ F2 decreases
Similarly, for 3 and 4
18. Answer (D)
Hint : Use combination methods.
Solution :
( )0 0
1 2
2A
K A AC
d d
+ = =
( )0 0
2 4
1 3B
K A AC
K d d
= =
+
( )1 2 0 03
2
+ = =
C
K K A AC
d d
( )( )( )1 2 1 2 0 0
1 2
2 19
1 1 12D
K K K K A AC
K K d d
+ + = =
+ +
PART - II (CHEMISTRY)
19. Answer (A, B, C, D)
Hint : In fuel cell, NaOH is used as electrolyte.
Solution:
Reaction
Cathode : 2H2 + 4OH– ⎯→ 4H2O
Anode : O2 + 2H2O ⎯→ 4OH–
Catalyst increase the rate of electrode reaction by providing surface to the electrode.
5/11
Test - 1A (Paper-2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
20. Answer (B, C, D)
Hint : For spontaneous cell
Ecell must be greater than zero.
Solution:
For cell – I
1 2
2 20.1 0.01P P
Pt, H | H || H | H+ +
( )
( )
2
2cell 2
1
0.1 P0.059E log
2 0.01 P= −
2
2cell
1
10 P0.059E log
2 P= −
2
22 1
1 2
10 P P1 so, 10
P P
P1 < 102 P2
So cell I is non-spontaneous.
Cells II, III and IV are spontaneous.
21. Answer (C)
Hint : = iCRT
wh
a
KK
K=
( )
2
a
CK
C 1
=
−
Solution:
= iCRT
3.444 = i × 0.1 × 0.082 × 300
i = 1.4
= i – 1 = 0.4
2 3
HX H O H O X+ −+ +
( )
( )22
a
C 0.1 0.4K 0.0266
1 0.6
= = =
−
14
13h
10K 3.75 10
0.0266
−−= =
22. Answer (A, B, C, D)
Hint : In a closed system overall mole fraction is
remain constant but nv and n may vary.
Solution:
From the graph o oP Qp p
So P is more volatile than Q, so boiling point of Q is more than boiling point of P.
v v
P P Pv v
Q Q Q
x P ·x
x P ·x
=
P
Q
P1
P
v v
P Pv v
Q Q
x x
x x
23. Answer (A, B)
Hint : The operation which divide the crystal in exactly two equal part have electrically neutral crystal.
Solution:
One body diagonal plane contain 4 corner, 2 face centre, one body centre particle and 2 edge centre.
24. Answer (A, C, D)
Hint : Diamond ABCABC
Graphite ABAB or ABCABC
Solution:
Structure of diamond : carbon form C·C·P lattice while other carbon is present at 50% of tetrahedral void.
So total carbon atom in diamond
4 + 4 8.
25. Answer (20)
Hint :
Solution:
S A A B BP y p y p = +
A Bp 100, p 200 = =
A
100 0.5 1y
100 0.5 200 0.5 3
= =
+
B
2y
3=
A
1100
13y1 2 5
100 2003 3
= =
+
6/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-C) (Hints & Solutions)
B
4y
5 =
S
1 4P 100 200
5 5= +
180 mm Hg
26. Answer (70)
Hint : PS = 100 – 50 XB
If xB = 1, PS P°B = 50 mm Hg.
XB = 0, PS P°A = 100 mm Hg.
Solution:
( )
A AA
B A B A
P Xy
P P P X
=
+ −
A
A
100 x4
7 50 50x=
+
200 + 200 xA = 700xA
500 xA = 200
xA = 2/5
27. Answer (30)
Hint :
500 mL of 6% (w/V) urea 30 g urea
500 mL of 18% (w/V) glucose 90 glucose
Mole of urea = 0.5
Mole of glucose = 0.5
Solution:
Total mole = 1
Toatal volume = 2
= Ctotal R T
= 0.5 × R x 300 150 R
150 = 5y
y = 30
28. Answer (45)
Hint : Cathode reaction :
Ag+ + 1e– ⎯→ Ag
Solution:
Total charge
1 1
Q 5 4 5 4 5 42 2
= + +
40 C
96000 C produce = 108 g
40 C produce = 108 40
96000
= 0.045
Mass (in mg) = 45 mg
29. Answer (06)
Hint : ( )Hx
400 =
C = [H+] = 1 × = 0.15
= 0.15
Solution:
m
m 0.15
=
m = 0.15 × 400 = 60
m
1000 K
C
=
K = 6 × 10–2
K GA
=
6 × 10–2 = G × 10–1
G = 0.6 ohm–1
30. Answer (28)
Hint : x = 7
Solution:
cell 2
HE E 0.06 log
10
+
−= −
pH = 7, x = 7, so 4x = 28
31. Answer (80)
Hint : n
MV
=
Solution:
For initial solution
soluten1
0.5=
nsolute = 0.5 mol
Mole of solute in 5 mL = 0.5
100 = 5 × 10–3
Let molar mass of P is x so moles of P is added
0.4
x
3
3
0.45 10
x1010
−
− +
=
2 3 0.410 5 10
x
− −= +
x = 80 g
7/11
Test - 1A (Paper-2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
32. Answer (30)
Hint : 1 1
2 2
w E
w E=
Solution:
Anode 2
1Cl Cl 1e
2
− −⎯⎯→ +
Moles of Cl2 produced = 177.5
2.571
=
5 mol of change is given so anode
A+x + xe ⎯→ A(s)
x mol e– produce ⎯→ 30 g
5 m produce ⎯→ 30 5
30x
=
x = 5
33. Answer (A)
Hint : Active electrode participate only in oxidation reaction.
Solution:
List-I (Electrolysis) List-II (Product)
P. Aq AgNO3 using Cathode → Ag
inert electrode Anode → O2
Q. Aq AgNO3 using Cathode → Ag
Ag electrode Anode → Ag+
R. Aq CuSO4 using Cathode → Cu
inert electrode Anode → O2
S. Aq CuSO3 using Cathode → Cu
inert electrode Anode → Cu2+
34. Answer (B)
Hint : Primary cell Secondary cell
Leclanche cell Lead storage battery
Mercury cell Ni-Cd battery
Solution:
In Leclanche cell, at cathode Mn reduced +4 to +3
Mercury cell : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
35. Answer (A)
Hint : = i CRT
Tb = i m Kb
Tf = i m Kf
Solution:
10% (w/v) 100 g of urea in 1 L
So,
100
60m1
= 1.666 M
= i × 1.666 × R × 300 = 500 R
5.85% (w/v) 58.5 g NaCl in 1000 mL
So, M = 1 molar, m = 1 m
i = 2
= i × 2 × R × 300 600 R
36. Answer (C)
Hint : Fe3O4 - Inverse spinel
Solution:
BaTiO3 – Perovskite structure
MgAl2O4 – Spinel structure
CaF2 - Ca2+ in CCP, F– are located at tetrahedral voids.
PART - III (MATHEMATICS)
37. Answer (A, B, C)
Hint :
h(x) = f(f(x))
Solution :
( )( )g f x x= …(i)
and h(g(g(x))) = x
Replace x by f(x) : h(g(g(f(x)))) = f(x)
h(g(x)) = f(x)
again by replacing x by f(x); we get:
h(x) = f(f(x))
∴ h(1) = f(f(1)) = f(9)
= 777
and h(0) = f(f(0)) = f(3) = 27 + 15 + 3
= 45
∵ f(x) is increasing function hence only one root.
∴ f(f(x)) = h(x) = 0 has only one real root.
38. Answer (C, D)
Hint :
∴ Domain of f(x) = [α, β] = [1, 3]
8/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-C) (Hints & Solutions)
Solution :
( ) 1 1 2sec cosec 9 ln− −= + + − +f x x x x x
∴ Domain of f(x) = [α, β] = [1, 3]
∴ Sum of all integral values in domain of f(x) is 6
And integral values in range of g(x) are 0,1, 2, 3
∴ Sum of integral values of f(x) = 3 + 4 = 7
Now cosec–1(cosec27) + sec–1(sec1)
= 9 – 27 + 1 = 9 – 26
and sec–1(sec10) + cos–1(cos(15))
= 4π – 10 + 15 – 4π = 5
39. Answer (B, C)
Hint :
1, if
0, if
−+ − =
x R Ix x
x I
Solution :
1, if
0, if
−+ − =
x R Ix x
x I
∴ Domain of f(x) is R.
Range of f(x) is 1
0,2
The graph of f(x) is
∴ The number of real solutions of is
10 and number of solution of 5f(x) = x is 5.
40. Answer (A, B, C)
Hint :
(f–1og–1) (17) = 17
Solution :
(f–1og–1) (17) = 17
∴ cosec–1(cosec17) = 5π – 17
and g(4) = 8
Also period of ( ) sin cos16 16
= +
x xh x is 8
But fundamental period of P(x) is also 8
Also f(x) and g(x) intersect each other only once.
41. Answer (A, C)
Hint :
( ) 1 ;nf x x n N =
Solution :
3 3 3 33 .... 4
14 16 641
4
+ + + + = =
−
( )( )
11
= +
f xf x
fx
( ) ( )1 1
1 1
− − + =
f x f f x fx x
( )( ) 11 1 1
− − =
f x f
x
( ) ,1 nf x x n N =
( ) ( ) 44 257 1= = +f f x x
( ) 4 45 (2) 5 2 609f f − = − =
and range of f(x) is [1, ∞)
and f(x) is many-one, into function.
42. Answer (A, B, D)
Hint :
∵ f(g(x)) = g(f(x)) = x
Solution :
Let f(x) = a1x + b1 and g(x) = c1x + d1
∵ f(g(x)) = g(f(x)) = x
( )10
xf x =
9/11
Test - 1A (Paper-2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
∴ a1c1 = 1 and a1d1 + b1 = 0 = b1c1 + d1
Given that f(0) = 7 and g(3) = –2
∴ f(x) = 2x + 7 and ( )7
2
xg x
−=
∴ f(5) = 17, g(135) = 64 and period of sin(g(x)) is 4π
43. Answer (32)
Hint :
2213 sin –
2 16 4
− = +
y x
Solution :
Let ( ) ( )3 3
1 1sin cos− −= +y x x
( )2
1 1 1 1sin cos 3sin sin2 2
x x x x− − − − = + − −
2213 sin –
2 16 4
− = +
y x
3
32y m
=
3
32m
=
44. Answer (27)
Hint :
∵ a2 + b2 = c2 and |x| ≤ 1
Solution :
∵ a2 + b2 = c2 and |x| ≤ 1
Also,
2 2 2 2
2
2 21+ =
a x b xx
c c
2 2 2 2 2 21 1
2 2sin sin− −
− − = +
ax c b x bx c a xx
c cc c
2 2 2 2 2 2
2 2
ax bxx c b x c a x
c c = − + −
2 2 2 2 2 2 20 or = = − + −x c a c b x b c a x
4 2 2 2 2 2 2 22 = + − +c a c b c a b x
4 2 2 2 2 2 2 2 2 42 − − +ab c c a x c b x a b x
( )2 4 2 2 2 2 2 2 4abx c c a b x a b x = − + +
∴ x = 1, –1
∴ nn = 33 = 27
45. Answer (80)
Hint :
∴ x ∈ (42, 43)
Solution :
Here x > 0 and – (log4x)2 + 5(log4x) – 6 > 0
∴ (log4x – 3) (2 – log4x) > 0
∴ x ∈ (42, 43)
∴ a = 16, b = 64
∴ a + b = 80
46. Answer (06)
Hint :
( )1tan3 9 0 and ! 1 ! 0x x x−
− − −
Solution :
( )1tan3 9 0 and ! 1 ! 0x x x−
− − −
∴ x2 – x – 56 < 0
(x – 8) (x + 7) < 0
∴ x ∈ (–7, 8) ∵ 𝑥 > 1
⇒ x = 2, 3, 4, 5, 6, 7
47. Answer (15)
Hint :
fn(x) ( )( )sin cot cot 1= − +x x n x
10/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-C) (Hints & Solutions)
Solution :
( )( )
2
2 21
sin
2 1cos cos
2 2=
= +
−
n
nn
xf x
x n x
fn(x) ( )( )sin cot cot 1= − +x x n x
( )
( )
sin
sin 1
nx
n x=
+
∴ gn (x) = f1(x) . f2(x). f3(x)….fn(x)
( )
( )( )
sin sin2 sin.....
sin2 sin3 sin 1
x x nx
x x n x=
+
( )
sin
sin 1
x
n x=
+
( )
( )
( )sin
sin
n
n
f x nx
g x x =
( )
( )
( )2019
2019
sin 2019
sin
f x x
g x x =
Whose period = T =
∴ [5T] = [5] = 15
48. Answer (32)
Hint :
( )2
1
2
11 3
4tan1
1 3 14
−
− −
=
+ −
xf x
x
Solution :
( )2
1
2
1 4 2 3tan
3 12 2
− − − =
− +
x xf x
x x
( )2
1
2
11 3
4tan1
1 3 14
−
− −
=
+ −
xf x
x
( )1 1
2
1tan 1 tan 3
4x
− −
= − −
3 1
12 3 44 2f
+ = − = −
and 1
4 32 2f
= −
5 53 1 1sec sec 32
34 2 2 2f f
+ + = =
49. Answer (69)
Hint :
f(23) = 0
Solution :
α = 23
β = 23 – α, γ = 23 + α
So, α + β + γ = 69
50. Answer (49)
Hint :
1 1 11 3 2tan tan tan
2 4 2 3x
− − − + = −
Solution :
1 1 11 3 2tan tan tan
2 4 2 3x
− − − + = −
( )1 1 3tan 2 tan , 0
2
xx− −
=
4
3x =
Now ( )1 1 4sin 2tan sin 2tan
3x− −
− = −
1
8
3sin tan16
19
−
= − + −
–1 24sin tan
7
=
–1 24sin sin
25
=
24
25=
∴ m + n = 24 + 25 = 49
11/11
Test - 1A (Paper-2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
51. Answer (C)
Hint :
Draw graph.
Solution :
From graph of each of the given in the given
domain we can find that
( )2
1
xf x
x=
+is one-one into function
( )1
f x xx
= − is onto but many one function
( )1
f x xx
= + is many one into function
and f(x) = 5x – sinx is bijective function
52. Answer (B)
Hint :
By use of graph of different function and their
respective point of intersection we get different
number of real solution.
Solution :
By use of graph of different function and their
respective point of intersection we get different
number of real solution.
53. Answer (A)
Hint :
Then period of f(x) = LCM of (18, 12, 3)
= 36
Solution :
( )
sin cos tan9 6 3
x x xf x
= + +
Then period of f(x) = LCM of (18, 12, 3)
= 36
and f(x) = cos(sinx) + G(cosx)
Then fundamental period of f(x) is 2
For function g(x) = sin(|sinx| – |cosx|)
The period of |sinx| – (cosx) is π
Hence period of g(x) is π
and the period of 4
1 cos22 sin 1
1 sin
xx
x
−+ +
+ is
equal to
54. Answer (D)
Hint :
( ) ( )( )
1 11 1sin tan
41
1
f x x x
xx
− −= + +
+ + +
Solution :
( ) ( )( )
1 11 1sin tan
41
1
f x x x
xx
− −= + +
+ + +
( )3
1,1 ,14
x f x
− −
∴ Absolute difference between maximum and
minimum value = 3 7
14 4
− − =
and g(x) = sin–1([3x] –2) + cos–1([3x] – 1)
( )1 3 2 1and 1 3 1 1x x− − − −
∴ x ∈ [0, 1)
∴ Range of g(x) = {0}
For ( )2 4 35 − + −= x xh x
∴ Maximum and minimum values are 1 and 5
P(x) = cot–1(sgn(x)) + sin–1([x])
then maximum value of P(x) = P(1) = 3
4
and minimum value of P(x) = P(–1) = 4
∴ Their difference = 2
All India Aakash Test Series for JEE (Advanced)-2020
Test Date : 16/06/2019
ANSWERS
1/11
TEST - 1A (Paper-2) - Code-D
Test - 1A (Paper-2) (Code-D) (Answers) All India Aakash Test Series for JEE (Advanced)-2020
PHYSICS CHEMISTRY MATHEMATICS
1. (B, C)
2. (B, C, D)
3. (B, C, D)
4. (A, C)
5. (B, C)
6. (B, D)
7. (30)
8. (13)
9. (18)
10. (10)
11. (32)
12. (36)
13. (24)
14. (25)
15. (D)
16. (C)
17. (B)
18. (D)
19. (A, C, D)
20. (A, B)
21. (A, B, C, D)
22. (C)
23. (B, C, D)
24. (A, B, C, D)
25. (30)
26. (80)
27. (28)
28. (06)
29. (45)
30. (30)
31. (70)
32. (20)
33. (C)
34. (A)
35. (B)
36. (A)
37. (A, B, D)
38. (A, C)
39. (A, B, C)
40. (B, C)
41. (C, D)
42. (A, B, C)
43. (49)
44. (69)
45. (32)
46. (15)
47. (06)
48. (80)
49. (27)
50. (32)
51. (D)
52. (A)
53. (B)
54. (C)
2/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-D) (Hints & Solutions)
PART - I (PHYSICS)
1. Answer (B, C)
Hint : ,= = x x y yE dE E dE
Solution :
( )0 0
200 0
coscos 2
44x
RdE
RR
=
= − =
Ey = 0
( )2
0
00
cos0
4C
RdV
R
= =
2. Answer (B, C, D)
Hint : Use Kirchhoff's laws.
Solution :
2 82 F
3 3= + =
ABC
eq
81
83 F8 11
13
= =
+
C
bat
811 8 C
11 = = Q
∴ VAB = 11 – 8 × 1 = 3 V
23 2 V,
3 = =
ACV VCD = 1 V
3. Answer (B, C, D)
Hint : Use the expression of field and potential.
Solution :
( )2 2
33
2= −
KQV R r
R for r < R
3
2
=
C
KQV
R 50% more than VS
and, =KQ
Vr
for r > R
and, 3
=KQ
E rR
for r < R
4. Answer (A, C)
Hint : Charge = Ceq × V
Solution :
1
2 412 16 C
2 4
= =
+ Q
2
6 312 24 C
6 3
= =
+ Q
1612 4 V
2AV = − =
24
12 8 V6B
V = − =
4 8 4 V − = − = −A B
V V
5. Answer (B, C)
Hint : Net field is vertically up.
Solution :
2
04
QE
a=
( )net 2
0
23 cos 3
34
QE E
a = =
2
0
6
4
Q
a=
And, 0
34
QV
a=
6. Answer (B, D)
Hint : 1 2=Kq q
Ur
Solution :
( )
2 2
sys0 0
4 24 4 2
q qU
a a
= +
WA = –(VA – V) × q
7. Answer (30)
Hint : E is maximum at
2
Rx =
3/11
Test - 1A (Paper-2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution :
max 3 2
2 022
0
2 32
4 9
42
RQ
QE
RR
R
= =
+
∴ max 2
0
2 3
4 9
q Qa
R m
=
6 6 9
2 3
10 12 10 2 3 9 10
4 10 9 10 10
− −
− −
=
= 30 m/s2
8. Answer (13)
Hint : Use Kirchhoff's laws.
Solution :
( )
30 5 10 213 V
2 3 5ABV
− = =
+ +
9. Answer (18)
Hint : Use Kirchhoff's laws.
Solution :
21
1 2
12AB
CV V
C C=
+
12 3
2
ABV
⇒ VAB 18 kV …(i)
and, 12
1 2
8AB
CV V
C C=
+
8 3
1
ABV
⇒ VAB 24 kV …(ii)
From (i) and (ii)
(VAB) = 18 kV
10. Answer (10)
Hint : Charge flows from one capacitor to the other.
Solution :
0 01 2
0 0
,A A
C Cd vt d vt
= =
+ −
q1 + q2 = 20 C = Qtotal
( )11 total
1 2
= +
Cq Q
C C
( )0total
02
d vtQ
d
−=
( )1total
02
−= =
dq vi Q
dt d
6
–2
10 20 10
2 10
− =
= 10 mA
11. Answer (32)
Hint : Use concept of electrostatic pressure.
Solution :
Charge gets uniformly distributed over the sphere
24 =
Q
R
Electrostatic pressure, 2
02
P
=
( )2
2
0
Force2
=
R
22
20
1
24
=
QR
R
2
2
032
=
Q
R
∴ n = 32
12. Answer (36)
Hint : All charges must experience zero forces.
Solution :
Q should be –ve
Force on ‘q’
2
2 2 2 2
0 0
4 4
4 4
q Qq q Q
l x l x = =
Force on ‘Q’
( ) ( )2 2 2 2
00
4 4 1
4– 4 –
qQ qQ
x xl x l x = =
2 –3
lx l x x = =
2
2
4 4
99
q l qQ Q
l
= =
36 CQ =
13. Answer (24)
Hint : in
0
=Q
E ds
4/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-D) (Hints & Solutions)
Solution :
∵ V = –4ar2 + 3b
8−
= =dv
E ardr
Now, ϕ = EA
⇒ dϕ = dEA + EdA
∵ dE = 8adr, dA = 8𝜋rdr
22
0
48 4 8 8
r dradr r ar rdr
= +
( ) 2
02
32 64
4
+ =
a r dr
r dr
⇒ ρ = 24a0
∴ n = 24
14. Answer (25)
Hint : 0
2E
r
=
Solution :
10
052
= −
B
A
V
V
dv drr
( ) ( )
0
ln 22
− =
A BV V
= 2 × 10–6 × 2 × 9 × 109 × 0.69
≃25 × 103 V
15. Answer (D)
Hint : Use combination methods.
Solution :
( )0 0
1 2
2A
K A AC
d d
+ = =
( )0 0
2 4
1 3B
K A AC
K d d
= =
+
( )1 2 0 03
2
+ = =
C
K K A AC
d d
( )
( )( )1 2 1 2 0 0
1 2
2 19
1 1 12D
K K K K A AC
K K d d
+ + = =
+ +
16. Answer (C)
Hint : Equivalent capacitance increases.
Solution :
So, Q1 increases ⇒ F1 increases ⇒ V1 increases
V2 decreases as V1 + V2 = Vbattery
⇒ Q2 decreases
⇒ F2 decreases
Similarly, for 3 and 4
17. Answer (B)
Hint : Electric field is uniform inside cavity.
Solution :
E r for charged solid sphere
E = constant inside cavity
E = 0 inside shells
V = constant inside shells
18. Answer (D)
Hint : Use Gauss's theorem.
Solution :
10 0
1
6 4 24
= =
Q Q
20 0
1– 3
3 24
Q Q =
0 0
1 7 7
3 8 24
Q Q = =
For tetrahedron,
3
0 0
1
4 4
Q Q = =
For stripe
4
0 0
90
360 4
Q Q = =
PART - II (CHEMISTRY)
19. Answer (A, C, D)
Hint : Diamond ABCABC
Graphite ABAB or ABCABC
5/11
Test - 1A (Paper-2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution:
Structure of diamond : carbon form C·C·P lattice while other carbon is present at 50% of tetrahedral void.
So total carbon atom in diamond
4 + 4 8.
20. Answer (A, B)
Hint : The operation which divide the crystal in exactly two equal part have electrically neutral crystal.
Solution:
One body diagonal plane contain 4 corner, 2 face centre, one body centre particle and 2 edge centre.
21. Answer (A, B, C, D)
Hint : In a closed system overall mole fraction is
remain constant but nv and n may vary.
Solution:
From the graph o oP Qp p
So P is more volatile than Q, so boiling point of Q is more than boiling point of P.
v v
P P Pv v
Q Q Q
x P ·x
x P ·x
=
P
Q
P1
P
v v
P Pv v
Q Q
x x
x x
22. Answer (C)
Hint : = iCRT
wh
a
KK
K=
( )
2
a
CK
C 1
=
−
Solution:
= iCRT
3.444 = i × 0.1 × 0.082 × 300
i = 1.4
= i – 1 = 0.4
2 3
HX H O H O X+ −+ +
( )
( )22
a
C 0.1 0.4K 0.0266
1 0.6
= = =
−
14
13h
10K 3.75 10
0.0266
−−= =
23. Answer (B, C, D)
Hint : For spontaneous cell
Ecell must be greater than zero.
Solution:
For cell – I
1 2
2 20.1 0.01P P
Pt, H | H || H | H+ +
( )
( )
2
2cell 2
1
0.1 P0.059E log
2 0.01 P= −
2
2cell
1
10 P0.059E log
2 P= −
2
22 1
1 2
10 P P1 so, 10
P P
P1 < 102 P2
So cell I is non-spontaneous.
Cells II, III and IV are spontaneous.
24. Answer (A, B, C, D)
Hint : In fuel cell, NaOH is used as electrolyte.
Solution:
Reaction
Cathode : 2H2 + 4OH– ⎯→ 4H2O
Anode : O2 + 2H2O ⎯→ 4OH–
Catalyst increase the rate of electrode reaction by providing surface to the electrode.
25. Answer (30)
Hint : 1 1
2 2
w E
w E=
Solution:
Anode 2
1Cl Cl 1e
2
− −⎯⎯→ +
Moles of Cl2 produced = 177.5
2.571
=
5 mol of change is given so anode
A+x + xe ⎯→ A(s)
x mol e– produce ⎯→ 30 g
5 m produce ⎯→ 30 5
30x
=
x = 5
26. Answer (80)
Hint : n
MV
=
6/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-D) (Hints & Solutions)
Solution:
For initial solution
soluten1
0.5=
nsolute = 0.5 mol
Mole of solute in 5 mL = 0.5
100 = 5 × 10–3
Let molar mass of P is x so moles of P is added
0.4
x
3
3
0.45 10
x1010
−
− +
=
2 3 0.410 5 10
x
− −= +
x = 80 g
27. Answer (28)
Hint : x = 7
Solution:
cell 2
HE E 0.06 log
10
+
−= −
pH = 7, x = 7, so 4x = 28
28. Answer (06)
Hint : ( )Hx
400 =
C = [H+] = 1 × = 0.15
= 0.15
Solution:
m
m 0.15
=
m = 0.15 × 400 = 60
m
1000 K
C
=
K = 6 × 10–2
K GA
=
6 × 10–2 = G × 10–1
G = 0.6 ohm–1
29. Answer (45)
Hint : Cathode reaction :
Ag+ + 1e– ⎯→ Ag
Solution:
Total charge
1 1
Q 5 4 5 4 5 42 2
= + +
40 C
96000 C produce = 108 g
40 C produce = 108 40
96000
= 0.045
Mass (in mg) = 45 mg
30. Answer (30)
Hint :
500 mL of 6% (w/V) urea 30 g urea
500 mL of 18% (w/V) glucose 90 glucose
Mole of urea = 0.5
Mole of glucose = 0.5
Solution:
Total mole = 1
Toatal volume = 2
= Ctotal R T
= 0.5 × R x 300 150 R
150 = 5y
y = 30
31. Answer (70)
Hint : PS = 100 – 50 XB
If XB = 1, PS P°B = 50 mm Hg.
XB = 0, PS P°A = 100 mm Hg.
Solution:
( )
A AA
B A B A
P Xy
P P P X
=
+ −
A
A
100 x4
7 50 50x=
+
200 + 200 xA = 700xA
500 xA = 200
xA = 2/5
32. Answer (20)
Hint :
7/11
Test - 1A (Paper-2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution:
S A A B BP y p y p = +
A Bp 100, p 200 = =
A
100 0.5 1y
100 0.5 200 0.5 3
= =
+
B
2y
3=
A
1100
13y1 2 5
100 2003 3
= =
+
B
4y
5 =
S
1 4P 100 200
5 5= +
180 mm Hg
33. Answer (C)
Hint : Fe3O4 - Inverse spinel
Solution:
BaTiO3 – Perovskite structure
MgAl2O4 – Spinel structure
CaF2 - Ca2+ in CCP, F– are located at tetrahedral voids.
34. Answer (A)
Hint : = i CRT
Tb = i m Kb
Tf = i m Kf
Solution:
10% (w/v) 100 g of urea in 1 L
So,
100
60m1
= 1.666 M
= i × 1.666 × R × 300 = 500 R
5.85% (w/v) 58.5 g NaCl in 1000 mL
So, M = 1 molar, m = 1 m
i = 2
= i × 2 × R × 300 600 R
35. Answer (B)
Hint : Primary cell Secondary cell
Leclanche cell Lead storage battery
Mercury cell Ni-Cd battery
Solution:
In Leclanche cell, at cathode Mn reduced +4 to +3
Mercury cell : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
36. Answer (A)
Hint : Active electrode participate only in oxidation reaction.
Solution:
List-I (Electrolysis) List-II (Product)
P. Aq AgNO3 using Cathode → Ag
inert electrode Anode → O2
Q. Aq AgNO3 using Cathode → Ag
Ag electrode Anode → Ag+
R. Aq CuSO4 using Cathode → Cu
inert electrode Anode → O2
S. Aq CuSO3 using Cathode → Cu
inert electrode Anode → Cu2+
PART - III (MATHEMATICS)
37. Answer (A, B, D)
Hint :
∵ f(g(x)) = g(f(x)) = x
Solution :
Let f(x) = a1x + b1 and g(x) = c1x + d1
∵ f(g(x)) = g(f(x)) = x
∴ a1c1 = 1 and a1d1 + b1 = 0 = b1c1 + d1
Given that f(0) = 7 and g(3) = –2
∴ f(x) = 2x + 7 and ( )7
2
xg x
−=
∴ f(5) = 17, g(135) = 64 and period of sin(g(x)) is 4π
38. Answer (A, C)
Hint :
( ) 1 ;nf x x n N =
Solution :
3 3 3 33 .... 4
14 16 641
4
+ + + + = =
−
8/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-D) (Hints & Solutions)
( )( )
11
= +
f xf x
fx
( ) ( )1 1
1 1
− − + =
f x f f x fx x
( )( ) 11 1 1
− − =
f x f
x
( ) ,1 nf x x n N =
( ) ( ) 44 257 1= = +f f x x
( ) 4 45 (2) 5 2 609f f − = − =
and range of f(x) is [1, ∞)
and f(x) is many-one, into function.
39. Answer (A, B, C)
Hint :
(f–1og–1) (17) = 17
Solution :
(f–1og–1) (17) = 17
∴ cosec–1(cosec17) = 5π – 17
and g(4) = 8
Also period of ( ) sin cos16 16
= +
x xh x is 8
But fundamental period of P(x) is also 8
Also f(x) and g(x) intersect each other only once.
40. Answer (B, C)
Hint :
1, if
0, if
−+ − =
x R Ix x
x I
Solution :
1, if
0, if
−+ − =
x R Ix x
x I
∴ Domain of f(x) is R.
Range of f(x) is 1
0,2
The graph of f(x) is
∴ The number of real solutions of is
10 and number of solution of 5f(x) = x is 5.
41. Answer (C, D)
Hint :
∴ Domain of f(x) = [α, β] = [1, 3]
Solution :
( ) 1 1 2sec cosec 9 ln− −= + + − +f x x x x x
∴ Domain of f(x) = [α, β] = [1, 3]
∴ Sum of all integral values in domain of f(x) is 6
And integral values in range of g(x) are 0,1, 2, 3
∴ Sum of integral values of f(x) = 3 + 4 = 7
Now cosec–1(cosec27) + sec–1(sec1)
= 9 – 27 + 1 = 9 – 26
and sec–1(sec10) + cos–1(cos(15))
= 4π – 10 + 15 – 4π = 5
42. Answer (A, B, C)
Hint :
h(x) = f(f(x))
Solution :
( )( )g f x x= …(i)
and h(g(g(x))) = x
Replace x by f(x) : h(g(g(f(x)))) = f(x)
h(g(x)) = f(x)
again by replacing x by f(x); we get:
h(x) = f(f(x))
( )10
xf x =
9/11
Test - 1A (Paper-2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
∴ h(1) = f(f(1)) = f(9)
= 777
and h(0) = f(f(0)) = f(3) = 27 + 15 + 3
= 45
∵ f(x) is increasing function hence only one root.
∴ f(f(x)) = h(x) = 0 has only one real root.
43. Answer (49)
Hint :
1 1 11 3 2tan tan tan
2 4 2 3x
− − − + = −
Solution :
1 1 11 3 2tan tan tan
2 4 2 3x
− − − + = −
( )1 1 3tan 2 tan , 0
2
xx− −
=
4
3x =
Now ( )1 1 4sin 2tan sin 2tan
3x− −
− = −
1
8
3sin tan16
19
−
= − + −
–1 24sin tan
7
=
–1 24sin sin
25
=
24
25=
∴ m + n = 24 + 25 = 49
44. Answer (69)
Hint :
f(23) = 0
Solution :
α = 23
β = 23 – α, γ = 23 + α
So, α + β + γ = 69
45. Answer (32)
Hint :
( )2
1
2
11 3
4tan1
1 3 14
−
− −
=
+ −
xf x
x
Solution :
( )2
1
2
1 4 2 3tan
3 12 2
− − − =
− +
x xf x
x x
( )2
1
2
11 3
4tan1
1 3 14
−
− −
=
+ −
xf x
x
( )1 1
2
1tan 1 tan 3
4x
− −
= − −
3 1
12 3 44 2f
+ = − = −
and 1
4 32 2f
= −
5 53 1 1sec sec 32
34 2 2 2f f
+ + = =
46. Answer (15)
Hint :
fn(x) ( )( )sin cot cot 1= − +x x n x
Solution :
( )( )
2
2 21
sin
2 1cos cos
2 2=
= +
−
n
nn
xf x
x n x
fn(x) ( )( )sin cot cot 1= − +x x n x
( )
( )
sin
sin 1
nx
n x=
+
∴ gn (x) = f1(x) . f2(x). f3(x)….fn(x)
( )
( )( )
sin sin2 sin.....
sin2 sin3 sin 1
x x nx
x x n x=
+
10/11
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper-2) (Code-D) (Hints & Solutions)
( )
sin
sin 1
x
n x=
+
( )
( )
( )sin
sin
n
n
f x nx
g x x =
( )
( )
( )2019
2019
sin 2019
sin
f x x
g x x =
Whose period = T =
∴ [5T] = [5] = 15
47. Answer (06)
Hint :
( )1tan3 9 0 and ! 1 ! 0x x x−
− − −
Solution :
( )1tan3 9 0 and ! 1 ! 0x x x−
− − −
∴ x2 – x – 56 < 0
(x – 8) (x + 7) < 0
∴ x ∈ (–7, 8) ∵ 𝑥 > 1
⇒ x = 2, 3, 4, 5, 6, 7
48. Answer (80)
Hint :
∴ x ∈ (42, 43)
Solution :
Here x > 0 and – (log4x)2 + 5(log4x) – 6 > 0
∴ (log4x – 3) (2 – log4x) > 0
∴ x ∈ (42, 43)
∴ a = 16, b = 64
∴ a + b = 80
49. Answer (27)
Hint :
∵ a2 + b2 = c2 and |x| ≤ 1
Solution :
∵ a2 + b2 = c2 and |x| ≤ 1
Also,
2 2 2 2
2
2 21+ =
a x b xx
c c
2 2 2 2 2 21 1
2 2sin sin− −
− − = +
ax c b x bx c a xx
c cc c
2 2 2 2 2 2
2 2
ax bxx c b x c a x
c c = − + −
2 2 2 2 2 2 20 or = = − + −x c a c b x b c a x
4 2 2 2 2 2 2 22 = + − +c a c b c a b x
4 2 2 2 2 2 2 2 2 42 − − +ab c c a x c b x a b x
( )2 4 2 2 2 2 2 2 4abx c c a b x a b x = − + +
∴ x = 1, –1
∴ nn = 33 = 27
50. Answer (32)
Hint :
2213 sin –
2 16 4
− = +
y x
Solution :
Let ( ) ( )3 3
1 1sin cos− −= +y x x
( )2
1 1 1 1sin cos 3sin sin2 2
x x x x− − − − = + − −
2213 sin –
2 16 4
− = +
y x
3
32y m
=
3
32m
=
51. Answer (D)
Hint :
( ) ( )( )
1 11 1sin tan
41
1
f x x x
xx
− −= + +
+ + +
11/11
Test - 1A (Paper-2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution :
( ) ( )( )
1 11 1sin tan
41
1
f x x x
xx
− −= + +
+ + +
( )3
1,1 ,14
x f x
− −
∴ Absolute difference between maximum and
minimum value = 3 7
14 4
− − =
and g(x) = sin–1([3x] –2) + cos–1([3x] – 1)
( )1 3 2 1and 1 3 1 1x x− − − −
∴ x ∈ [0, 1)
∴ Range of g(x) = {0}
For ( )2 4 35 − + −= x xh x
∴ Maximum and minimum values are 1 and 5
P(x) = cot–1(sgn(x)) + sin–1([x])
then maximum value of P(x) = P(1) = 3
4
and minimum value of P(x) = P(–1) = 4
∴ Their difference = 2
52. Answer (A)
Hint :
Then period of f(x) = LCM of (18, 12, 3)
= 36
Solution :
( )
sin cos tan9 6 3
x x xf x
= + +
Then period of f(x) = LCM of (18, 12, 3)
= 36
and f(x) = cos(sinx) + G(cosx)
Then fundamental period of f(x) is 2
For function g(x) = sin(|sinx| – |cosx|)
The period of |sinx| – (cosx) is π
Hence period of g(x) is π
and the period of 4
1 cos22 sin 1
1 sin
xx
x
−+ +
+ is
equal to
53. Answer (B)
Hint :
By use of graph of different function and their
respective point of intersection we get different
number of real solution.
Solution :
By use of graph of different function and their
respective point of intersection we get different
number of real solution.
54. Answer (C)
Hint :
Draw graph.
Solution :
From graph of each of the given in the given
domain we can find that
( )2
1
xf x
x=
+is one-one into function
( )1
f x xx
= − is onto but many one function
( )1
f x xx
= + is many one into function
and f(x) = 5x – sinx is bijective function
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