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

Test Date : 19/01/2020

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MOCK TEST - 1 (Paper-1) - Code-A

Mock Test - 1 (Paper-1) (Code-A)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (D)

2. (C)

3. (D)

4. (B)

5. (A)

6. (A, C)

7. (A, B, C)

8. (A, C, D)

9. (B, C)

10. (A, D)

11. (A, D)

12. (A, C)

13. (A, C)

14. (A, C)

15. (A, C)

16. A → (P)

B → (Q, S)

C → (R, T)

D → (Q, S)

17. A → (P, S)

B → (Q, T)

C → (P, S)

D → (Q, R)

18. (27)

19. (11)

20. (36)

21. (D)

22. (A)

23. (C)

24. (C)

25. (A)

26. (A, C, D)

27. (A, B, D)

28. (A, B, C)

29. (A, B, C, D)

30. (A, B, C, D)

31. (B)

32. (A)

33. (A, B, C)

34. (B)

35. (A, C, D)

36. A → (Q, T)

B → (P, R)

C → (P, R, S)

D → (P, Q)

37. A → (P, T)

B → (P, Q)

C → (S, T)

D → (S, T)

38. (06)

39. (12)

40. (09)

41. (A)

42. (A)

43. (C)

44. (D)

45. (B)

46. (C, D)

47. (B, D)

48. (B, C, D)

49. (B, C)

50. (A, D)

51. (B)

52. (A, B, C)

53. (A, C, D)

54. (D)

55. (B, C)

56. A → (P, Q, R)

B → (P, Q, R, S)

C → (P, Q, R, S)

D → (P)

57. A → (P, Q, R)

B → (T)

C → (P, Q, R, S, T)

D → (R)

58. (05)

59. (04)

60. (49)


All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)

2/10

PART - I (PHYSICS)

1. Answer (D)

Hint : aAB = aCD = g/3

Solution :

aAB = aCD = g/3

2

rel

1

2a t l=

rel

2

3

ga =

2

3

gt l=

3l

tg

=

2. Answer (C)

Hint : Linear momentum and kinetic energy of

system remains conserved.

Solution :

pB cos = p

pB sin = pA

2 2 2

2 2 2 2

B Ap p p

m m m+ =

2 2 2 2

2sec tan

2 1

p pp

+ =

2 2

23tan

2 2

p p =

1

tan3

=

2

sec3

=

3. Answer (D)

Hint : T2 R3

Solution :

( )

2 312 3

2 4

T R

T R=

2

1

8T

T= T2 = 16 hr

( )2 2 1 1

rel2 1

2 24 –– 16 2

– 3

R RR R

R R R

= =

rel 2 3 6

R

R

= =

rad/hr

4. Answer (B)

Hint : 2 2 sm

Tk

= =

Solution :

( ) ( )sin , sin2

AX A t A t= =

6

t

=

24 6 2

T Tt

= +

4 6

T Tt = +

5

s6

t =

5. Answer (A)

Hint : P1V1 = P2V2

Solution :

0 1

99

100

lP l P

=

1 0

100

99P P=

0 0

100 2–

99

TP P

r=

0 2

99

P T

r=

0

198

P rT =

6. Answer (A, C)

Hint : –KA dT

QL dx

=

Solution :

A BA B

A B

K A K AT T

l l =

A BA B

A B

K KT T

l l =


Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/10

7. Answer (A, B, C)

Hint : Based on question of relative velocity of

approach and separation.

Solution :

Separation is maximum or minimum when

( )

( )1 20 – 0

d rr v v

dt= =

For 1 2

v v= separation will not change.

8. Answer (A, C, D)

Hint : T

v =

Solution :

= 1 m

= 16 Hz.

2

4 3 8sin x

=

2

3x

=

6

x

=

0

3 1– , 2 16

2 6 3X t

= =

0

1s

96t =

9. Answer (B, C)

Hint : Apply work energy theorem.

Solution :

2

2 0 00 0

1–

2 9 3

X XK X mg X

= +

0

3

KX

mg =

Also, 0 –3

KXma mg= a = 0

10. Answer (A, D)

Hint : Fl

lAY

=

Solution :

Stress in both the rod is same.

Stress

StrainY=

Strain is more in aluminium rod.

11. Answer (A, D)

12. Answer (A, C)

13. Answer (A, C)

Hint for Q. no. 11 to 13

Velocity of point of contact is zero and

mechanical energy is conserved.

Solution for Q. no. 11 to 13

21

2 Pmg R I=

2 21 142

2 5

mm R mgR

+ =

2 212

5

RmgR

=

5

12

g

R =

(mg + m2 R) 224

5R mR=

55

8512

24 288

gg

g

R R

+

= =

14. Answer (A, C)

15. Answer (A, C)

Hint for Q. no. 14 and 15

dutotal + dW = 0

Solution for Q. no. 14 and 15

3 5

– 2 200 1 200 02 2

W R+ + =

W = 1100R

1

5

2

RC =

5

1 2 02

RC + =

5

–4

RC =



4/10

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

Hint : Pressure depends on height of liquid

column.

Solution :

Pressure depends on height of liquid column.

More pressure means more normal force by

bottom on liquid.

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

Hint : –p

dyv v

dx=

Solution :

–p

yv v

x

=

–y

p Bx

=

18. Answer (27)

Hint : 1 1

2 2

x

v= +

Solution :

( )

2

1v

x=

+

( ) ( )2

2 –2

1 1

dva v

dx x x= =

+ +

24– m/s

27a =

19. Answer (11)

Hint : It is based on principle of super position.

Solution :

8–3

2

P

GMGM

VRR

= +

1

– 1–12

GM

R

=

3 8–2

2

O

GMGM

VRR

= +

11

15

P

O

V

V=

20. Answer (36)

Hint : Finally pressure will be same.

Solution :

0 0 0 0 0 0 0

0 0 0 0

3

3 3 3

P l P l P l Pl

T T T T+ + =

05

3

PP =

0 00

5

3 3

P lx P =

0

5

lx =

Displacement = 0 0 02

–3 5 15

l l l=

= 12 cm

PART - II (CHEMISTRY)

21. Answer (D)

Hint : Q = oxygen

Solution : Group number = 2 (atomic number)

It means element may be O

22. Answer (A)

Hint : Resonance energy (per ring) of benzene >

naphthalene.

Solution : Resonance energy of naphthalene is

more than benzene but resonance energy per

ring of benzene is more than naphthalene.

23. Answer (C)

Hint : E+ at 2nd position of pyrrole and 3rd position

of indole.

Solution :

24. Answer (C)

Hint : It is an electrophilic aromatic substitution.

Solution :



5/10

25. Answer (A)

Hint : It is the preparation of borax from the ore

2CaO 3B2O3.

Solution :

26. Answer (A, C, D)

Hint : For irreversible process,

system surroundingS S 0 +

Solution :

qsystem = PV

= 1 × 10 = 1013 J

Ssurrounding = 1013

300− = –3.377 JK–1

Ssystem = 1 × 8.314 × ln2 = 5.76 JK–1

27. Answer (A, B, D)

Hint : In N3H and COS no p–d bonding takes

place.

Solution :

28. Answer (A, B, C)

Hint : 2 2 3Cl 3F 2ClF+ ⎯⎯→

Solution : If no limiting reagent in the reaction,

then final compound is ClF3(g). If ClF3 is formed,

then mole of gas become half, it means pressure

decreases to half.

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

Hint : On increasing P, equilibrium shift in that

direction which contain less number of gaseous

mole.

Solution : If O2 is added, then reaction-II and III

move in forward direction.


Hint : Buffer = wA + salt of wA with strong base

= wB + salt of wB with strong acid

Solution :

4 2 4 3 2 4(NH ) SO 2NaOH 2NH Na SO+ ⎯⎯→ +

If mol of 4 2 4(NH ) SO mol of NaOH

4 2 4 4(NH ) SO NH OH Basic buffer+ ⎯⎯→

31. Answer (B)

Hint : M is 80

32. Answer (A)

Hint : mol of Na2CO3 = mol of NaHCO3

= mol of NaOH


Hint : x = 20, x1 = 20

z = 20, z1 = 10

y = 20, y1 = 30

Solution for Q.Nos. 31 to 33 :

x + y + z = 60

x1 + y1 + z1 = 60

From experiment-I,

y + z = 40

x + y = 40

So, x = 20, y = 20, z = 20

From experiment-II,

x1 + z1 = 40

It means y1 = 20

From experiment-III,

x + 2y + z = M 80

From experiment-IV,

2x1 + y1 + z1 = 90

20 + 2x1 + z1 = 90

2x1 + z1 = 70

x1 = 30

z1 = 10

34. Answer (B)

Hint : M is Li.



6/10


Hint : Z

1n

= so velocity of H-atom in ground state

and X1 state of M+q is same.

Solution for Q.Nos. 34 & 35 :

X2 = 1 and n2 – – 1 = 0.

So, n2 = 2

If energy of X1 = ground state energy of H-atom,

it means 1

Z1

n= .

From the velocity of e– in X2,

6 6 Z

3.3 10 2.2 102

=

Z = 3

It means M+q = Li2+

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

Hint : Hybridisation : B2H6 → sp3, Si(OH)4 → sp3,

H3BO3 → sp2

Solution :

3 4 2 6 3BF LiAlH B H LiF AlF+ ⎯⎯→ + +

Heat

2 6 3 3 3 6 2B H NH B N H 12H+ ⎯⎯⎯→ +

4 2 4SiCl H O Si(OH) HCl+ ⎯⎯→ +

2 2 7 2 3 3Na B O HCl H O NaCl H BO+ + ⎯⎯→ +

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

Hint : Solubility of carbonate of 2nd group

decreases top to bottom while for 1st group, it

increases top to bottom.

Solution : Solubility =

BeCO3 > MgCO3 > CaCO3 > SrCO3

Li2CO3 < Na2CO3 < K2CO3 < Cs2CO3

LiF < NaF < KF < CsF

MgF2 < CaF2 < SrF2 < BaF2 < BeF2

38. Answer (06)

Hint : -Cyano, malonic acid, -keto acid and

--unsaturated salt or acid show very high rate

of reaction.

Solution :

39. Answer (12)

Hint : X = 7, Y = 3, Z = 2

Solution :

40. Answer (09)

Hint : Write all the possible arrangement of

p2 configuration and then choose the

configuration which have total spin zero.

Solution :

There are nine possible arrangement with total

spin zero.

PART - III (MATHEMATICS)

41. Answer (A)

Hint : Modulus.

Solution :

1

1+

=

nz

z



7/10

|z + 1| = |z|

z lies on right bisector of the line segment

joining the points (–1, 0) and (0, 0).

42. Answer (A)

Hint : Point of intersection of tangents on

parabola.

Solution :

Locus of P.O.I of tangents to the parabola

y2 = 4ax include angle is

(y2 – 4ax) = tan2 (x + a)2

3

=

y2 = 4ax + 3(x + a)2

43. Answer (C)

Hint : Point of intersection of tangent on ellipse.

Solution :

cos sin 112 8

x y + = …(i)

sin cos 112 8

x y− + = …(ii)

By (i)2 + (ii)2 2 2

212 8

x y+ =

44. Answer (D)

Hint : Companion of C.O.C

Solution :

Equation of C.O.C T = 0

xh – yk = 9 …(i)

x = 9 …(ii)

Comparing 1

1 0 1

h k= =

h = 1; k = 0

Now equation of pair of tangents SS1 = T2

(x2 – y2 – 9) (–8) = (x – 9)2

–8x2 + 8y2 + 72 = x2 – 18x + 81

9x2 – 8y2 – 18x + 9 = 0

45. Answer (B)

Hint : ab + bc + ca – abc

= 1 + (1 – a) (1 – b) (1 – c)

Solution :

s = 1

So, s – a, s – b, s – c are positive numbers.

ab + bc + ca – abc

= 1 + (1 – a) (1 – b) (1 – c)

( ) ( ) ( )– – –

3

s a s b s c+ +

( )( )( )( )1

3– – –s a s b s c

( )( )( )1

1– 1– 1–27

a b c

Also, (1 – a) (1 – b) (1 – c) > 0

So, 28

– 1,27

ab bc ca abc

+ +

46. Answer (C, D)

Hint : The greatest number selected should be a

multiple of perfect square.

Solution :

Let the numbers be a, ar, ar2.

All the three numbers are natural numbers.

So, the largest number out of these three should

be a multiple of perfect square.

The greatest number may be

4, 8, 12, 20, 24, 28, 40, 44 (multiple of 22)

or 9, 18, 27, 45 (multiple of 32)

or 16, 32, 48 (multiple of 42)

or 25, 50, 36, 49 (multiple of 52, 62 and 72)

If the largest number is a multiple of k2, then

common ratio may be 1 2 3 1

, , , ...,k

k k k k

−

So, n = 8 × 1 + 4 × 2 + 3 × 3 + 2 × 4 + 1 × 5 + 1 ×

6

= 8 + 8 + 9 + 8 + 5 + 6 = 44

47. Answer ( B, D)

Hint : Find length of perpendicular from origin on

AB.

Solution :

Let equation of circle C : x2 + y2 = r2 (b < r < a)

Equation of common tangent, y = mx + c



8/10

where 2 2 2 21c r m a m b= + = +

2 2

2 2

r bm

a r

−=

−

Equation of AB, y = mx – aem

Length of perpendicular from origin to AB

21

aem

m=

+

2 2

2 2

2 2

r bae r b

a b

−= = −

−

So, 2 2 22 ( ) 2AB r r b b= − − =

Area of triangle OAB = 2 2b r b−

48. Answer (B, C, D)

Hint : Position of two circles.

Solution :

C1(0, 0); r1 = 2 ; ( )2 2

3 3, 3 ; 4C r =

1 2

27 9 6C C = + =

C1C2 = r1 + r2

Equation transverse common tangent

6 3 6 24x y+ =

3 4x y+ =

3 1

22 2

x y+ =

3 1

cos ; sin2 2

= =

6

=

49. Answer (B, C)

Hint : 1 1

11

nIn n

= + −+

.

Solution :

100

1

1

( 1)rr

r

I−

=

−

= I1 – I2 + I3 – I4 + ... + I99 – I100

2 2 2 2 2 2

1 1 1 1 1 11 1 1 – ...

1 2 2 3 3 4

= + + − + + + + +

2 2

1 1

1 101= −

2

11

101= −

Also, 2 2

1 11

( 1)nI

n n= + +

+

2 1

( 1)

n n

n n

+ +=

+

1 1

1 –1n n

= ++

100

1 2 3 1001

...r

r

I I I I I=

= + + +

1 1 1 1 1 1

1 – 1 – ... 1 –1 2 2 3 100 101

= + + + + + +

1

101–101

=

50. Answer (A, D)

Hint : Let the slope of OP is m, then slope of OQ

will be 1

–m

. Now, find OP and OQ in terms of

m.

Solution :

Let equation of OP : y = mx. On solving with

hyperbola,

2 2 2

2 2– 1

x m x

a b=

2 2

2

2 2 2–

a bx

b a m=

2 2 2OP x m x= +

( )2 2 2

2

2 2 2

1

–

a b mOP

b a m

+=

If we put 1

–m

in place of m, we will get

( )2 2 2

2

2 2 2

1

–

a b mOQ

b m a

+=

Now, ( )( )

( )

2 2 2

2 2 2 2 2

1 1 – 1

1

b a m

OP OQ a b m

++ =

+



9/10

2 2

1 1–

a b=

( )( )

( )

2 2 2

2 2 2 2 2

1 1 1––

1

a b m

OP OQ a b m

+=

+

2

2 2 2

1 1 1–

1

m

a b m

= +

+

2

2

1–0, 1

1

m

m

+

So, 2 2 2 2

max

1 1 1–

a

OP OQ a b= +

51. Answer (B)

Hint : Remainder Theorem.

Solution :

100

118 18 205

− = =

option (B) is correct


Hint : Divisibility concept.

Solution :

In option (D) 1

a bnk

d

−= Also a – b = nk2

2 21

1

nk knk d

d k= = need not N – {1}



Solution :

5x – 2 = 7k

7 2

5

kx

+=

Now verify each and every option.

54. Answer (D)

Hint : Geometrical Probability.

Solution :

D 0 P2 – (P + 2) 0 P2 – P – 2 0

4 2

6 3P = =

55. Answer (B, C)


Solution :

T.C = × 102 = 100

F.C = 100 – 25 = 75

3

4m =

option (B, C) are correct.

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

D(P)

Hint : Use –1

–1

n n

r r

nC C

r= and

0

2n

n n

rr

C=

=

Solution :

(A) ( ) ( )0

0 1 2 ... 1 2 3n n

rC n C+ + + + + + +

( )2

.... ) 2 3 .... ....nn C n+ + + + + + +

( ) ( )

0

1 – 1–

2 2

nn

rr

n n r rC

=

+=

( ) ( )

–21 – 12 – 2

2 2

n nn n n n+=

( )–32 3 5nn n= +

(B) ( ) ( )2 –2

–22 2

– – 1n n

n n

r rr r

N r r C n n C= =

= =

( ) –2– 1 2nn n=

(C) (C0 + C1) + (C0 + C2) + …. + (C0 + Cn)

+ (C1 + C2) + (C1 + C3) + …. + (C1 + Cn)

+ (C2 + C3) + (C2 + C4) +.…+ (C2 + Cn)

……………………………………………..

……………………………………………..

+ (Cn –1 + Cn)

= n(C0 + C1 + C2 + … + Cn) = n2n

(D) [nnC0 + 6(1 + 2 + 3 + …. + n)] +

[(n – 1) nC1 + 6(2 + 3 + 4 + ….+ n)] +

[(n – 2) nC2 + 6(3 + 4 + 5 + ….+ n] + ….

= n2n–1 + 6[12 + 22 + 32 + …. + n2]

= n2n–1 + n(n + 1) (2n + 1)

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

Hint : Property of modulus of complex number.



10/10

Solution :

(A) |iz + 3| = |z – 3i| |z – 2i| + 4

Also |z – 3i| 0

A → P, Q, R

(B) ( )( )( )

4

1 2 3

24

nn n n nC

− − −=

greatest term = middle term

n = 8

B → T

(C) Minimum value occurs at z = 1 + i

minimum value = 5 + 5 + 2 = 12

C → P, Q, R, S, T

(D) For (1 + i)n to be real sin 04

n =

n = 4, 8, 12

3

12P =

1

4P =

1

4P

=

D → R

58. Answer (05)

Hint : Assume the centre of circle on the normal of parabola at point P.

Solution :

Let normal at one end of latus rectum (1, 2) cuts the parabola at point P, then coordinates of P will be

(aT2, 2aT), where 2

– – –3T tt

= =

P(9, –6)

Equation of common normal at P,

y = 3x – (6 + 27) 3x – 33

Consider the centre of required circle C(a, 3a – 33)

CP = CS (S being the focus)

(a – 9)2 + (3a – 27)2 = (a – 1)2 + (3a – 33)2

10(a2 – 18a + 81) = 10a2 – 200a + 1090

20a = 280

a = 14

So, 10 – 9 5 10r a= =

59. Answer (04)

Hint : General Term of H.P

Solution :

1 1 3

– , – , – ,....are in A.P.10 5 10

–10n

nT =

General term of H.P. is 10

–n

10

–n

will be an integer, if n = 1, 2, 5, 10

60. Answer (49)

Hint : General Term.

Solution :

1 2

–1

2 – 12

3 2 2 ....to terms 2 12 – 1

2 2 – 1 2 1

n

n

n n n n

nt

++ + +

= = =+

–1

12

2n n

t = +

11–

22 2

11–

2

n

nS n n

= + =

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

Test Date : 19/01/2020

ANSWERS


MOCK TEST - 1 (Paper-1) - Code-B

Mock Test - 1 (Paper-1) (Code-B)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (A)

2. (B)

3. (D)

4. (C)

5. (D)

6. (A, D)

7. (B, C)

8. (A, C, D)

9. (A, B, C)

10. (A, C)

11. (A, D)

12. (A, C)

13. (A, C)

14. (A, C)

15. (A, C)

16. A → (P, S)

B → (Q, T)

C → (P, S)

D → (Q, R)

17. A → (P)

B → (Q, S)

C → (R, T)

D → (Q, S)

18. (36)

19. (11)

20. (27)

21. (A)

22. (C)

23. (C)

24. (A)

25. (D)

26. (A, B, C, D)

27. (A, B, C, D)

28. (A, B, C)

29. (A, B, D)

30. (A, C, D)

31. (B)

32. (A)

33. (A, B, C)

34. (B)

35. (A, C, D)

36. A → (P, T)

B → (P, Q)

C → (S, T)

D → (S, T)

37. A → (Q, T)

B → (P, R)

C → (P, R, S)

D → (P, Q)

38. (09)

39. (12)

40. (06)

41. (B)

42. (D)

43. (C)

44. (A)

45. (A)

46. (A, D)

47. (B, C)

48. (B, C, D)

49. (B, D)

50. (C, D)

51. (B)

52. (A, B, C)

53. (A, C, D)

54. (D)

55. (B, C)

56. A → (P, Q, R)

B → (T)

C → (P, Q, R, S, T)

D → (R)

57. A → (P, Q, R)

B → (P, Q, R, S)

C → (P, Q, R, S)

D → (P)

58. (49)

59. (04)

60. (05)


All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)

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PART - I (PHYSICS)

1. Answer (A)

Hint : P1V1 = P2V2

Solution :

0 1

99

100

lP l P

=

1 0

100

99P P=

0 0

100 2–

99

TP P

r=

0 2

99

P T

r=

0

198

P rT =

2. Answer (B)

Hint : 2 2 sm

Tk

= =

Solution :

( ) ( )sin , sin2

AX A t A t= =

6

t

=

24 6 2

T Tt

= +

4 6

T Tt = +

5

s6

t =

3. Answer (D)

Hint : T2 R3

Solution :

( )

2 312 3

2 4

T R

T R=

2

1

8T

T= T2 = 16 hr

( )2 2 1 1

rel2 1

2 24 –– 16 2

– 3

R RR R

R R R

= =

rel 2 3 6

R

R

= =

rad/hr

4. Answer (C)

Hint : Linear momentum and kinetic energy of

system remains conserved.

Solution :

pB cos = p

pB sin = pA

2 2 2

2 2 2 2

B Ap p p

m m m+ =

2 2 2 2

2sec tan

2 1

p pp

+ =

2 2

23tan

2 2

p p =

1

tan3

=

2

sec3

=

5. Answer (D)

Hint : aAB = aCD = g/3

Solution :

aAB = aCD = g/3

2

rel

1

2a t l=

rel

2

3

ga =

2

3

gt l=

3l

tg

=

6. Answer (A, D)

Hint : Fl

lAY

=

Solution :

Stress in both the rod is same.

Stress

StrainY=

Strain is more in aluminium rod.


Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/10

7. Answer (B, C)

Hint : Apply work energy theorem.

Solution :

2

2 0 00 0

1–

2 9 3

X XK X mg X

= +

0

3

KX

mg =

Also, 0 –3

KXma mg= a = 0

8. Answer (A, C, D)

Hint : T

v =

Solution :

= 1 m

= 16 Hz.

2

4 3 8sin x

=

2

3x

=

6

x

=

0

3 1– , 2 16

2 6 3X t

= =

0

1s

96t =

9. Answer (A, B, C)

Hint : Based on question of relative velocity of

approach and separation.

Solution :

Separation is maximum or minimum when

( )

( )1 20 – 0

d rr v v

dt= =

For 1 2

v v= separation will not change.

10. Answer (A, C)

Hint : –KA dT

QL dx

=

Solution :

A BA B

A B

K A K AT T

l l =

A BA B

A B

K KT T

l l =

11. Answer (A, D)

12. Answer (A, C)

13. Answer (A, C)

Hint for Q. no. 11 to 13

Velocity of point of contact is zero and

mechanical energy is conserved.

Solution for Q. no. 11 to 13

21

2 Pmg R I=

2 21 142

2 5

mm R mgR

+ =

2 212

5

RmgR

=

5

12

g

R =

(mg + m2 R) 224

5R mR=

55

8512

24 288

gg

g

R R

+

= =

14. Answer (A, C)

15. Answer (A, C)

Hint for Q. no. 14 and 15

dutotal + dW = 0

Solution for Q. no. 14 and 15

3 5

– 2 200 1 200 02 2

W R+ + =

W = 1100R

1

5

2

RC =

5

1 2 02

RC + =

5

–4

RC =



4/10

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

Hint : –p

dyv v

dx=

Solution :

–p

yv v

x

=

–y

p Bx

=

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

Hint : Pressure depends on height of liquid

column.

Solution :

Pressure depends on height of liquid column.

More pressure means more normal force by

bottom on liquid.

18. Answer (36)

Hint : Finally pressure will be same.

Solution :

0 0 0 0 0 0 0

0 0 0 0

3

3 3 3

P l P l P l Pl

T T T T+ + =

05

3

PP =

0 00

5

3 3

P lx P =

0

5

lx =

Displacement = 0 0 02

–3 5 15

l l l=

= 12 cm

19. Answer (11)

Hint : It is based on principle of super position.

Solution :

8–3

2

P

GMGM

VRR

= +

1

– 1–12

GM

R

=

3 8–2

2

O

GMGM

VRR

= +

11

15

P

O

V

V=

20. Answer (27)

Hint : 1 1

2 2

x

v= +

Solution :

( )

2

1v

x=

+

( ) ( )2

2 –2

1 1

dva v

dx x x= =

+ +

24– m/s

27a =

PART - II (CHEMISTRY)

21. Answer (A)

Hint : It is the preparation of borax from the ore

2CaO 3B2O3.

Solution :

22. Answer (C)

Hint : It is an electrophilic aromatic substitution.

Solution :

23. Answer (C)

Hint : E+ at 2nd position of pyrrole and 3rd position

of indole.

Solution :



5/10

24. Answer (A)

Hint : Resonance energy (per ring) of benzene >

naphthalene.

Solution : Resonance energy of naphthalene is

more than benzene but resonance energy per

ring of benzene is more than naphthalene.

25. Answer (D)

Hint : Q = oxygen

Solution : Group number = 2 (atomic number)

It means element may be O


Hint : Buffer = wA + salt of wA with strong base

= wB + salt of wB with strong acid

Solution :

4 2 4 3 2 4(NH ) SO 2NaOH 2NH Na SO+ ⎯⎯→ +

If mol of 4 2 4(NH ) SO mol of NaOH

4 2 4 4(NH ) SO NH OH Basic buffer+ ⎯⎯→


Hint : On increasing P, equilibrium shift in that

direction which contain less number of gaseous

mole.

Solution : If O2 is added, then reaction-II and III

move in forward direction.


Hint : 2 2 3Cl 3F 2ClF+ ⎯⎯→

Solution : If no limiting reagent in the reaction,

then final compound is ClF3(g). If ClF3 is formed,

then mole of gas become half, it means pressure

decreases to half.

29. Answer (A, B, D)

Hint : In N3H and COS no p–d bonding takes

place.

Solution :


Hint : For irreversible process,

system surroundingS S 0 +

Solution :

qsystem = PV

= 1 × 10 = 1013 J

Ssurrounding = 1013

300− = –3.377 JK–1

Ssystem = 1 × 8.314 × ln2 = 5.76 JK–1

31. Answer (B)

Hint : M is 80

32. Answer (A)

Hint : mol of Na2CO3 = mol of NaHCO3

= mol of NaOH


Hint : x = 20, x1 = 20

z = 20, z1 = 10

y = 20, y1 = 30

Solution for Q.Nos. 31 to 33 :

x + y + z = 60

x1 + y1 + z1 = 60

From experiment-I,

y + z = 40

x + y = 40

So, x = 20, y = 20, z = 20

From experiment-II,

x1 + z1 = 40

It means y1 = 20

From experiment-III,

x + 2y + z = M 80

From experiment-IV,

2x1 + y1 + z1 = 90

20 + 2x1 + z1 = 90

2x1 + z1 = 70

x1 = 30

z1 = 10

34. Answer (B)

Hint : M is Li.



6/10


Hint : Z

1n

= so velocity of H-atom in ground state

and X1 state of M+q is same.

Solution for Q.Nos. 34 & 35 :

X2 = 1 and n2 – – 1 = 0.

So, n2 = 2

If energy of X1 = ground state energy of H-atom,

it means 1

Z1

n= .

From the velocity of e– in X2,

6 6 Z

3.3 10 2.2 102

=

Z = 3

It means M+q = Li2+

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

Hint : Solubility of carbonate of 2nd group

decreases top to bottom while for 1st group, it

increases top to bottom.

Solution : Solubility =

BeCO3 > MgCO3 > CaCO3 > SrCO3

Li2CO3 < Na2CO3 < K2CO3 < Cs2CO3

LiF < NaF < KF < CsF

MgF2 < CaF2 < SrF2 < BaF2 < BeF2

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

Hint : Hybridisation : B2H6 → sp3, Si(OH)4 → sp3,

H3BO3 → sp2

Solution :

3 4 2 6 3BF LiAlH B H LiF AlF+ ⎯⎯→ + +

Heat

2 6 3 3 3 6 2B H NH B N H 12H+ ⎯⎯⎯→ +

4 2 4SiCl H O Si(OH) HCl+ ⎯⎯→ +

2 2 7 2 3 3Na B O HCl H O NaCl H BO+ + ⎯⎯→ +

38. Answer (09)

Hint : Write all the possible arrangement of

p2 configuration and then choose the

configuration which have total spin zero.

Solution :

There are nine possible arrangement with total

spin zero.

39. Answer (12)

Hint : X = 7, Y = 3, Z = 2

Solution :

40. Answer (06)

Hint : -Cyano, malonic acid, -keto acid and

--unsaturated salt or acid show very high rate

of reaction.

Solution :

PART - III (MATHEMATICS)

41. Answer (B)

Hint : ab + bc + ca – abc

= 1 + (1 – a) (1 – b) (1 – c)



7/10

Solution :

s = 1

So, s – a, s – b, s – c are positive numbers.

ab + bc + ca – abc

= 1 + (1 – a) (1 – b) (1 – c)

( ) ( ) ( )– – –

3

s a s b s c+ +

( )( )( )( )1

3– – –s a s b s c

( )( )( )1

1– 1– 1–27

a b c

Also, (1 – a) (1 – b) (1 – c) > 0

So, 28

– 1,27

ab bc ca abc

+ +

42. Answer (D)

Hint : Companion of C.O.C

Solution :

Equation of C.O.C T = 0

xh – yk = 9 …(i)

x = 9 …(ii)

Comparing 1

1 0 1

h k= =

h = 1; k = 0

Now equation of pair of tangents SS1 = T2

(x2 – y2 – 9) (–8) = (x – 9)2

–8x2 + 8y2 + 72 = x2 – 18x + 81

9x2 – 8y2 – 18x + 9 = 0

43. Answer (C)

Hint : Point of intersection of tangent on ellipse.

Solution :

cos sin 112 8

x y + = …(i)

sin cos 112 8

x y− + = …(ii)

By (i)2 + (ii)2 2 2

212 8

x y+ =

44. Answer (A)

Hint : Point of intersection of tangents on

parabola.

Solution :

Locus of P.O.I of tangents to the parabola

y2 = 4ax include angle is

(y2 – 4ax) = tan2 (x + a)2

3

=

y2 = 4ax + 3(x + a)2

45. Answer (A)

Hint : Modulus.

Solution :

1

1+

=

nz

z

|z + 1| = |z|

z lies on right bisector of the line segment

joining the points (–1, 0) and (0, 0).

46. Answer (A, D)

Hint : Let the slope of OP is m, then slope of OQ

will be 1

–m

. Now, find OP and OQ in terms of

m.

Solution :

Let equation of OP : y = mx. On solving with

hyperbola,

2 2 2

2 2– 1

x m x

a b=

2 2

2

2 2 2–

a bx

b a m=

2 2 2OP x m x= +

( )2 2 2

2

2 2 2

1

–

a b mOP

b a m

+=

If we put 1

–m

in place of m, we will get

( )2 2 2

2

2 2 2

1

–

a b mOQ

b m a

+=

Now, ( )( )

( )

2 2 2

2 2 2 2 2

1 1 – 1

1

b a m

OP OQ a b m

++ =

+

2 2

1 1–

a b=



8/10

( )( )

( )

2 2 2

2 2 2 2 2

1 1 1––

1

a b m

OP OQ a b m

+=

+

2

2 2 2

1 1 1–

1

m

a b m

= +

+

2

2

1–0, 1

1

m

m

+

So, 2 2 2 2

max

1 1 1–

a

OP OQ a b= +

47. Answer (B, C)

Hint : 1 1

11

nIn n

= + −+

.

Solution :

100

1

1

( 1)rr

r

I−

=

−

= I1 – I2 + I3 – I4 + ... + I99 – I100

2 2 2 2 2 2

1 1 1 1 1 11 1 1 – ...

1 2 2 3 3 4

= + + − + + + + +

2 2

1 1

1 101= −

2

11

101= −

Also, 2 2

1 11

( 1)nI

n n= + +

+

2 1

( 1)

n n

n n

+ +=

+

1 1

1 –1n n

= ++

100

1 2 3 1001

...r

r

I I I I I=

= + + +

1 1 1 1 1 1

1 – 1 – ... 1 –1 2 2 3 100 101

= + + + + + +

1

101–101

=

48. Answer (B, C, D)

Hint : Position of two circles.

Solution :

C1(0, 0); r1 = 2 ; ( )2 2

3 3, 3 ; 4C r =

1 2

27 9 6C C = + =

C1C2 = r1 + r2

Equation transverse common tangent

6 3 6 24x y+ =

3 4x y+ =

3 1

22 2

x y+ =

3 1

cos ; sin2 2

= =

6

=

49. Answer ( B, D)

Hint : Find length of perpendicular from origin on

AB.

Solution :

Let equation of circle C : x2 + y2 = r2 (b < r < a)

Equation of common tangent, y = mx + c

where 2 2 2 21c r m a m b= + = +

2 2

2 2

r bm

a r

−=

−

Equation of AB, y = mx – aem

Length of perpendicular from origin to AB

21

aem

m=

+

2 2

2 2

2 2

r bae r b

a b

−= = −

−

So, 2 2 22 ( ) 2AB r r b b= − − =

Area of triangle OAB = 2 2b r b−

50. Answer (C, D)

Hint : The greatest number selected should be a

multiple of perfect square.

Solution :

Let the numbers be a, ar, ar2.

All the three numbers are natural numbers.



9/10

So, the largest number out of these three should

be a multiple of perfect square.

The greatest number may be

4, 8, 12, 20, 24, 28, 40, 44 (multiple of 22)

or 9, 18, 27, 45 (multiple of 32)

or 16, 32, 48 (multiple of 42)

or 25, 50, 36, 49 (multiple of 52, 62 and 72)

If the largest number is a multiple of k2, then

common ratio may be 1 2 3 1

, , , ...,k

k k k k

−

So, n = 8 × 1 + 4 × 2 + 3 × 3 + 2 × 4 + 1 × 5 + 1 ×

6

= 8 + 8 + 9 + 8 + 5 + 6 = 44

51. Answer (B)


Solution :

100

118 18 205

− = =

option (B) is correct


Hint : Divisibility concept.

Solution :

In option (D) 1

a bnk

d

−= Also a – b = nk2

2 21

1

nk knk d

d k= = need not N – {1}



Solution :

5x – 2 = 7k

7 2

5

kx

+=

Now verify each and every option.

54. Answer (D)


Solution :

D 0 P2 – (P + 2) 0 P2 – P – 2 0

4 2

6 3P = =

55. Answer (B, C)


Solution :

T.C = × 102 = 100

F.C = 100 – 25 = 75

3

4m =

option (B, C) are correct.

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

Hint : Property of modulus of complex number.

Solution :

(A) |iz + 3| = |z – 3i| |z – 2i| + 4

Also |z – 3i| 0

A → P, Q, R

(B) ( )( )( )

4

1 2 3

24

nn n n nC

− − −=

greatest term = middle term

n = 8

B → T

(C) Minimum value occurs at z = 1 + i

minimum value = 5 + 5 + 2 = 12

C → P, Q, R, S, T

(D) For (1 + i)n to be real sin 04

n =

n = 4, 8, 12

3

12P =

1

4P =

1

4P

=

D → R

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

D(P)

Hint : Use –1

–1

n n

r r

nC C

r= and

0

2n

n n

rr

C=

=

Solution :

(A) ( ) ( )0

0 1 2 ... 1 2 3n n

rC n C+ + + + + + +

( )2

.... ) 2 3 .... ....nn C n+ + + + + + +

( ) ( )

0

1 – 1–

2 2

nn

rr

n n r rC

=

+=



10/10

( ) ( )

–21 – 12 – 2

2 2

n nn n n n+=

( )–32 3 5nn n= +

(B) ( ) ( )2 –2

–22 2

– – 1n n

n n

r rr r

N r r C n n C= =

= =

( ) –2– 1 2nn n=

(C) (C0 + C1) + (C0 + C2) + …. + (C0 + Cn)

+ (C1 + C2) + (C1 + C3) + …. + (C1 + Cn)

+ (C2 + C3) + (C2 + C4) +.…+ (C2 + Cn)

……………………………………………..

……………………………………………..

+ (Cn –1 + Cn)

= n(C0 + C1 + C2 + … + Cn) = n2n

(D) [nnC0 + 6(1 + 2 + 3 + …. + n)] +

[(n – 1) nC1 + 6(2 + 3 + 4 + ….+ n)] +

[(n – 2) nC2 + 6(3 + 4 + 5 + ….+ n] + ….

= n2n–1 + 6[12 + 22 + 32 + …. + n2]

= n2n–1 + n(n + 1) (2n + 1)

58. Answer (49)

Hint : General Term.

Solution :

1 2

–1

2 – 12

3 2 2 ....to terms 2 12 – 1

2 2 – 1 2 1

n

n

n n n n

nt

++ + +

= = =+

–1

12

2n n

t = +

11–

22 2

11–

2

n

nS n n

= + =

59. Answer (04)

Hint : General Term of H.P

Solution :

1 1 3

– , – , – ,....are in A.P.10 5 10

–10n

nT =

General term of H.P. is 10

–n

10

–n

will be an integer, if n = 1, 2, 5, 10

60. Answer (05)

Hint : Assume the centre of circle on the normal of parabola at point P.

Solution :

Let normal at one end of latus rectum (1, 2) cuts the parabola at point P, then coordinates of P will be

(aT2, 2aT), where 2

– – –3T tt

= =

P(9, –6)

Equation of common normal at P,

y = 3x – (6 + 27) 3x – 33

Consider the centre of required circle C(a, 3a – 33)

CP = CS (S being the focus)

(a – 9)2 + (3a – 27)2 = (a – 1)2 + (3a – 33)2

10(a2 – 18a + 81) = 10a2 – 200a + 1090

20a = 280

a = 14

So, 10 – 9 5 10r a= =

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