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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)


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


8 3

1

ABV

⇒ VAB 24 kV …(ii)

From (i) and (ii)

(VAB) = 18 kV

13. Answer (13)


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


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

−−= =


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


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


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


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


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.


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


∴ 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


Solution :

( )( )

2

2 21

sin

2 1cos cos

2 2=

= +

−

n

nn

xf 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 :


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


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 :




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

= + +


= 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)


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)


Solution :

( )

30 5 10 213 V

2 3 5ABV

− = =

+ +

9. Answer (18)


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


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


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.


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.


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


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


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


Q. Aq AgNO3 using Cathode → Ag

Ag electrode Anode → Ag+

R. Aq CuSO4 using Cathode → Cu


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


( )( )

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.


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


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


∴ 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 :


2 4 2 3x

− − − + = −

Solution :


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 :


Solution :

( )( )

2

2 21

sin

2 1cos cos

2 2=

= +

−

n

nn

xf 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


( )

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


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 :


= 36

Solution :

( )

sin cos tan9 6 3

x x xf x

= + +


= 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 :




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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