test - 6 (code-c) (answers) all india aakash test …...all india aakash test series for jee...

Test - 6 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/14

1. (4)

2. (3)

3. (4)

4. (2)

5. (4)

6. (1)

7. (3)

8. (3)

9. (1)

10. (1)

11. (2)

12. (3)

13. (4)

14. (2)

15. (1)

16. (1)

17. (4)

18. (4)

19. (3)

20. (1)

21. (2)

22. (1)

23. (2)

24. (1)

25. (4)

26. (2)

27. (1)

28. (2)

29. (4)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (1)

33. (1)

34. (2)

35. (3)

36. (2)

37. (3)

38. (4)

39. (2)

40. (2)

41. (3)

42. (3)

43. (3)

44. (1)

45. (4)

46. (2)

47. (3)

48. (4)

49. (3)

50. (1)

51. (2)

52. (3)

53. (3)

54. (3)

55. (2)

56. (2)

57. (2)

58. (3)

59. (4)

60. (3)

61. (4)

62. (3)

63. (3)

64. (2)

65. (3)

66. (3)

67. (2)

68. (4)

69. (2)

70. (4)

71. (3)

72. (1)

73. (3)

74. (1)

75. (3)

76. (1)

77. (1)

78. (3)

79. (2)

80. (1)

81. (2)

82. (4)

83. (2)

84. (2)

85. (2)

86. (3)

87. (3)

88. (1)

89. (4)

90. (3)

Test Date : 17/02/2019

ANSWERS

TEST - 6 - Code-C

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

All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)

2/14

1. Answer (4)

Hint : x = A sin(t + )

Sol. : x = A sin(t + )

20 sin 2A

T

0 sin4

A

4

cos( )dx

A tdt

2 24 cos 4

16 16 4A

32cos sin

2 4 4A A

32 2mA

2. Answer (3)

Hint : 12

kf

m

Sol. : 12

kf

m

12 1kg

kf

eq12 8

kf

eq

1 12 4 2

k k kk k

k k

1Hz

4 f

3. Answer (4)

Hint : KeA = K1A1 = K2A2

Sol. : Let the amplitude of oscillation is A

3 63

3 6 e p

K KK A A K A

K K

23

pA A

23

pv v

PART - A (PHYSICS)

4. Answer (2)

Hint : dyv

dt

Sol. : y = A sint

v = Acost

2 2 21KE cos

2 mA t

2 21 1 cos22 2

tmA

2 T

T

5. Answer (4)

Hint : Use projection of particle on circle and findphase.

Sol. :A

Al22

/6

1

x1 = A cos(t)

2 sin6

x A t

x1 = x2

cos( ) sin6

t t

22 6

t

32

6 t

43 t

T

12T

t

6. Answer (1)

Hint : The phase difference must be 2 when theymeet again.

Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/14

Sol. : 6( 1)

5 T

nT n

5n = 6n – 6

n = 6

Time period = 6T

Method 2

2 25 2

6t

T T

1 51

6t

T T

t = 6T

7. Answer (3)

Hint : 2

k

Sol. :4

0.5 sin 2011

y t x

2 411

11cm 5.5 cm

2

12011 55 cm s

4

v

k

8. Answer (3)

Hint : Draw standing wave in string.

Sol. :

9. Answer (1)

Hint : Use dimensional analysis

Sol. :1

2

T

fL

2

1 12

4

T Tf

L LDD

1f

LD

10. Answer (1)

Hint :2

5

Sol. :2

5

= 10 cm

5 cm2

11. Answer (2)

Hint : The frequency will be maximum whenapproaching the sound tangentially.

Sol. :4 m

5 m1

37°

4 m37°

3 m

3 m

7410

180 t

t = 0.13 s

12. Answer (3)

Hint : vp = A

Sol. : vp (maximum) = 2n x0

22wn

v n

2nx0 = (n)2

022

x

= x0

13. Answer (4)

Hint : v2 = 2(A2 – x2)

Sol. : v2 = 2(4a2 – a2)

22 23

34

aa

12

at t = 0, x = a

1sin

2

6


4/14

x = 1

2 sin2 6

a t

= 1 1

2 sin cos cos sin2 6 2 6

a t t

= 3 1

2 sin cos2 2 2 2

t ta

= 3 sin cos2 2t t

a

14. Answer (2)

Hint : f0 = f1 – f2Sol. : 2f1 = 410

14102

f

24542

f

454 410 44Beat 7 7 Hz

2 2 2 22

15. Answer (1)

Hint : Wavelength does not change after reflectionfrom the wall.

Sol. : = 0 – V T

= 0 1s

VV

= 3320 20

1 104 320

= 330010

4

= 75 × 10–3 m

= 75 mm

16. Answer (1)

Hint :

12

Tf [Second harmonic of wire]

Sol. :1 2

12V TL L

320 1 1002 0.8 0.5

1100

1

0.5 100m

m = 5 g

17. Answer (4)

Hint : 2 1 cos(2 )sin

2

Sol. :2

1 cos2 22

a xy nt

41 cos 4

2

a xnt

18. Answer (4)

Hint :60°

x

x

Fnet = 2Kx + 2Kx cos60°

Sol. :

KK

2K

60° x

x

x = x cos60°

–ma = 2Kx + 2K x cos60°

–ma = 2Kx + 2Kx cos260°

=

21

2 22

Kx Kx

–ma = 2

24Kx

Kx

–ma = 22

KxKx

–ma = 5

2Kx

52

Ka x

m

22

5 m

TK

19. Answer (3)

Hint : Use formula B


5/14

Sol. :3

5

100 10

5 10

PB

VV

B = 20 × 108

8

3

20 10

10

B

v

v � 1400 ms–1

20. Answer (1)

Hint : a = –2A

Sol. : = –2x1 = –2x2

( + ) = –2(x1 + x2)

u2 = 2A2 – 2x12

v2 = 2A2 – 2x22

v2 – u2 = 2(x1 + x2)(x1 – x2)

= –( + )(x1 – x2)

2 2

1 2

u v

x x

21. Answer (2)

Hint : Total energy remain conserved.

Sol. :2

2 21 1 12 2 4 2

AkA k mV

22 21 1 1

' 42 2 4 2

AkA k mV

2 22 21 1 1 1

' 42 2 4 2 2 4

A AkA k kA k

A2 = 2

234

AA

A2 = 2134

A

A = 132

A

22. Answer (1)

Hint : amax = 2A

Sol. :2

5AA

= 5 rad s–1

5 sin6 A

5 = 12

A A = 10

a = (5)2 × 10 = 250 m/s2

23. Answer (2)

Hint : 2 2net 1 2 1 22 cosA A A A A

Sol. :

A2

A1

1 sin4

x a t

22

2 sin63

ax a t

22

sin63

x a t

1

2

32

AA

12

24. Answer (1)

Hint : 2max 1 2( )I I I

Sol. :2

max 1 2

min 1 2

I A A

I A A

1 2

1 2

31

A A

A A

3A1 – 3A2 = A1 + A2

2A1 = 4A2

1

2

2A

A

25. Answer (4)

Hint : 2 2v A x

Sol. : 2 213 3 A

2 212 4 A

(13)2 – (12)2 = 2(42 – 32)

25 = 2 (16 – 9)

2 125 5rad s

7 7


6/14

26. Answer (2)

Hint : TE = KE + PE

Sol. :2

21 1 3KE

2 2 4

kA k A

21 7KE

2 16

kA

780 KE

16

KE = 35 J

27. Answer (1)

Hint : Amplitude of oscillation = 2L

Sol. :

2L

m1

m2

2KL – KL = (m1 + m2)g

KL = (m1 + m2)g

1 2( )

m mK g

L

Amplitude of oscillation

2

2

m g LK

1 22

( )

2

m mm g g

2m2 = m1 + m2

m2 = m1

2

1

1m

m

28. Answer (2)

Hint : 2

T TV

r

Sol. :1

2T

fLr

11 1 1

2 1 1 1

12 92

rf L T

f L r T

1

2

3f

f

2450

Hz 150 Hz3

f

29. Answer (4)

Hint : 1 2

2m

Tk k

Sol. : 1 2 mt

k

2 21

1

4m

tk

21 2

1

4 mk

t

22 2

2

4 mk

t

1 2

2 m

Tk k

22 21 2

21 1

4

mT

mt t

2 2 21 2

1 1 1 T t t

2 2 21 2

T t t

30. Answer (1)

Hint : I1 = 0.8I0

Sol. : I = (0.8I0)2 = 0.64I0

I = 36%


7/14

PART - B (CHEMISTRY)

31. Answer (3)

Hint : Check all possible positions available andcheck stereochemistry for each isomer.

Sol. : Here are all the possible isomers.

Br

,

Br

,

CH Br2

Br

,

CH —Br2

Br

CH—Br

CH Br2

,

Br

,

CH Br2

Br

,

CH — Br2

Br

Br

CH — Br2

BrCH2

32. Answer (1)

Hint : — NH — C — CH 3

O

is ortho/para director.

Sol. : NHCOCH3 NHCOCH3

NO2

(Major product)

HNO /H SO3 2 4

33. Answer (1)

Hint : Alkyne having acidic H.

Sol. : White precipitates of HC C–Ag.

34. Answer (2)

Hint : Benzene is reactive toward ozonolysis butdoes not react with Br2.

Sol. : Benzene generally does not give additionreaction but it reacts with H2/Pt and showsaddition reaction.

35. Answer (3)

Hint : With alc. KOH both cis and trans butene isformed.

Sol. :

OH

OH

P4

CH3

OH

OH

CH3

,36. Answer (2)

Hint : FeCl3 is a Lewis acid.

Sol. : Cl — Cl + FeCl3L.B L.A

37. Answer (3)

Hint : Dehydrohalogenation

Sol. :

(ii) K/dry ether

CH Cl3

(i) NaNH2

Cl

Br

Br

Br

CH3

38. Answer (4)

Hint : Different carbides gives differenthydrocarbons upon hydrolysis.

Sol. : 2H O4 3 4Al C CH

CaC2 and Na2C2 on hydrolysis producesethyne.

Mg2C3 on hydrolysis gives propyne.

39. Answer (2)

Hint : 3AlCl6 6 2 6 5

(gas)BenzeneC H Cl C H Cl HCl

Sol. : The difference in the molar mass of productsis 76 g/mol.

40. Answer (2)

Hint : Alkenes on reaction with dil. H2SO4 formcarbocation, which has tendency ofrearrangement.

Sol. :(+)

(+)

dil. H SO2 4

OH H O2 :

Methylshift

41. Answer (3)

Hint : Alkynes undergo 2 times of addition reactionwith HCl.

Sol. : CH C CH CH — C — CH3 3 3

HCl

—Cl

—

Cl


8/14

42. Answer (3)

Hint : Alcohols with conc. H2SO4 undergodehydration.

Sol. :

(+)

H+

OHH+

43. Answer (3)

Hint : Decarboxylation of carboxylates reduces1 C atom from the parent chain.

Sol. : NaOH CaO3 2 2

3 2 3

CH CH CH COOH

CH CH CH

44. Answer (1)

Hint : Decarboxylation

Sol. : Trimethyl cyclohexane is formed when (R)undergo sodalime decarboxylation.

45. Answer (4)

Hint :H+

+ CH – CH = CH3 2

CH – CH – CH3 3

Sol. : + CH — CH = CH3 2

CH — CH — CH3 3

(A)

H+

(P)

Br /h2

Br

CH — C — CH3 3

——

(Q)

Alc. KOH

(R)dil. H SO2 4

(S)

CH — C = CH3 2 CH — C — CH3 3

OH

46. Answer (2)

Hint : 1,2-dihalo molecules give alkene on reactionwith zinc.

Sol. : A : CH3 – CH = CH2

B : 3 3CH – CH – CH|

Cl

47. Answer (3)

Hint : Tert. carbocation is most stable.

Sol. : 3° alcohols produces alkenes fastest.

48. Answer (4)

Hint : Ozonolysis reaction

Sol. :

CH2

O3

Zn/H O2

O

+ HCOH

49. Answer (3)

Hint : Warm KMnO4 in acidic medium oxidise thecompound.

Sol. : A CH — CH — CH — CH3 3—

OH

—

OHB CH — C — OH3 —

O

—

50. Answer (1)

Hint : CH — C — CH3 3

O

(P)

CH — C CH3 dil. H SO /HgSO2 4 4

— —

Sol. : Product P is a ketone and degree ofunsaturation of P is one.

51. Answer (2)

Hint: Both naphthalene and phenanthrene arearomatic rings and have e– participating inresonance and after alkylation e– remainsame.

Sol. : Naphthalene

Phenanthrene

52. Answer (3)

Hint : Correct formula of the given molecule accordingto the question is (CH3)3C — C(CH3)3.

Sol. : CH — C — C — CH3 3

CH3 CH3

CH3 CH3


9/14

53. Answer (3)

Hint : To produce formaldehyde molecule must have= CH2 group.

Sol. : , ,

54. Answer (3)

Hint : F– being a poor leaving group, cannot leaveeasily.

Br– is a good leaving group, so can produceSaytzeff product.

Sol. :

CH — CH — CH — CH3 2 2—

F

CH — CH — CH == CH3 2 2

CH — CH — CH — CH3 2 2

—H

NH2—

F

More stable anion

CH — CH == CH — CH3 3

—H

—

Br

B

CH — CH — CH — CH3 3

any strong base

More substitutedalkene is formed

which is more stable

55. Answer (2)

Hint : With Lindlar's catalyst syn addition occurs.

Sol. :

CH — C C — CH3 3Lindlar's catalyst

C == CH

CH3CH3

Hcis-2-butene

56. Answer (2)

Hint : HCl does not show peroxide effect.

Sol. : Br

HBr

HBr

HCl

Peroxide

Peroxide

CH — CH == CH2 2

Br

Cl

57. Answer (2)

Hint : Cl2/H2O gives Markovnikov addition of OHand Cl+.

Sol. : CH — CH = CH CH — CH — CH3 2 3 2

Cl /H O2 2

OH Cl(P)

58. Answer (3)

Hint : NBS produce Br.

which majorly attacksallylic and benzylic position.

Sol. :

CH3

NBS

CH — Br2 CH2

+ Br—

CH3 CH3 CH3

++

Br

Br

+

Br

59. Answer (4)

Hint : Grignard reagents first undergoes acid basereaction and produces a salt.

Sol. :

Br /h2

0.5 mol 0.5 mol

Mg/dry ether

Br

0.5 mol

Mg Br

(P)( )R

S — CH — C C — H2

—H

two acidic

(A)

2 moles of P are required for 1 mole of A

S — CH — C C2 0.25 mol

(Salt)(Q)

+

60. Answer (3)

Hint : Hydroboration reaction.

Sol. :

CH — CH == CH3 2

B H /THF2 6

D O , OD2 2– —

H

—

OD(A)

CH — CH — CH3 2

CH — CH == CH3 2

B H /THF2 6

H O /OH2 2

–

D(B)

CH — CH — CH — OH3 2


10/14

PART - C (MATHEMATICS)

61. Answer (4)

Hint : 5cos(2 ) cos2

Sol. : As cos(25) = cos 3264

= cos2

= 0

cos cos2 cos22 ... cos29 = 0

62. Answer (3)

Hint : cosine law

Sol. : Triangle DEF is called the pedal triangle ofthe triangle ABC.

Circumradius of pedal triangle = 12

(Circumradius of triangle ABC) = 2R

cosC = 2 2 2

2 a b c

ab

= 25(3 1 2 3) 50 100

25 2( 3 1)8

4

= 50( 3 1) 1

50 2( 3 1) 2

C = 45°, similarly B = 30°, A = 105°

5 512sin 222

cR

C

5 5 2

2 42 2 R

R

63. Answer (3)

Hint : Formula

Sol. : Apply direct formula

64. Answer (2)

Hint : Counting

Sol. : 12 11 11( )

52 51 221P E

65. Answer (3)

Hint : sin + sin( + ) + sin( + 2) + ... upton terms

Sol :13

1

(sin cos )

r

r r

13sin

2[sin7 cos7 ] 0

sin2

13

sin 02

or tan7 = 1 and sin 02

2 4 6, ,

13 13 13

5 9 13, , ,

28 28 28 28

66. Answer (3)

Hint : Sum of digits should be divisible by 3.

Sol. : If the sum of digits of a number is divisible by3, then the number is divisible by 3.

The number should be formed by using

1, 2, 3, 6, 9 or 2, 3, 4, 6, 9

6

5

2 5 2 1( )

6 35P E

C

67. Answer (2)

Hint : cosine law

Sol. :2 2 2

cos2

b c aA

bc

b2 – 2bc cosA + c2 – a2 = 0

2 4 2 8 0b b

2( 2 2) 0 2 2 b b

68. Answer (4)

Hint : Combinations

Sol. :18 17

2 235

2

9 17 17 8 17( )

35 17 35

C CP E

C


11/14

69. Answer (2)

Hint : Square both sides

Sol. : 1 + sin + cos = sincos

sin + cos = sincos– 1

Squaring on both sides,

1 + 2sincos

= 1 + sin2cos2 – 2sincos

212sin2 sin 2

4

sin2(sin2 – 8) = 0

3sin2 0 , ,

2 2

Only and 32

satisfy the original equation.

3 5Sum

2 2

70. Answer (4)

Hint : Adding a number in every observation doesnot change .

Sol. :2( )ix x

N

11 2 3 ... 10 11

10 2 x

25 6 7 ... 14 95 19 11

410 10 2 2

x

1 = 2

71. Answer (3)

Hint : cosine law

Sol. :2 2 26 (3 5) 3

cos2 6 3 5

36 45 9 2

36 5 5

1

tan 2 12

1 1tan

2 3

tan tan8 tan tan

6

8

6

72. Answer (1)

Hint : Venn diagram

Sol. : The probability of solving the question by

three students A, B and C are 2 1

,3 3

and 16

respectively.

2 1 1( ) , ( ) , ( )

3 3 6 P A P B P C

Probability that the question is solved byexactly two students is

P(E) = ( ) ( ) P A B C P A B C

( ) P A B C

= ( ) ( ) ( ) ( ) ( ) ( ) P A P B P C P A P B P C

( ) ( ) ( )P A P B P C

= 1 1 1 2 2 1 2 1 5 53 3 6 3 3 6 3 3 6 18

73. Answer (3)

Hint : Solve equation

Sol. :sin1 sin 22

sin cos2

sin2(1 sin ) 2

2sin cos cos2 2 2

22(1 sin ) 2sin2

2(1 + sin) = 1 – cos

2sin = –(1 + cos)

24 sin cos 2 cos2 2 2

2 sin cos2 2

tan 12

74. Answer (1)

Hint :2

2 2( )ixx

N


12/14

Sol. : x4 + x5 = 30 – (2 + 4 + 9) = 15 ...(i)

2 22 4 54 16 81

36 9.25

x x

x42 + x5

2 = 125 ...(ii)

From (i) and (ii) gives,

x4 = 5, x5 = 10

x4 : x5 = 1 : 2

75. Answer (3)

Hint : Venn diagram

Sol. : E

55

25

H

Percentage of population which readexactly one newspaper is

( ) ( )n E H n E H

= n(E) – n(E H) + n(H) – n(E H)

= 10 + 30 – 2 x 5 = 30%

76. Answer (1)

Hint : sec2 = 1 + tan2

Sol. : R = sec2Asec2B + tan2Atan2B +

2 secAsecBtanAtanB – sec2Atan2B –

sec2Btan2A – 2 secAsecBtanA tanB

= sec2A(sec2B – tan2B) +

tan2A(tan2B – sec2B)

= sec2A – tan2A = 1

77. Answer (1)

Hint : P(E) = 1 – P(X = 1) – P(X = 0)

Sol. : S = {HHH, HHT, HTH, THH, HTT, THT, TTH}

E = {HTT, THT, TTH}

( ) 3 4( ) 1 1

( ) 7 7n E

P En S

78. Answer (3)

Hint :( )( )

sin2

A s b s cbc

( )cos

2

A s s abc

Sol. : 22cos cos 4sin2 2 2

B C B C A

22sin cos 4sin2 2 2A B C A

cos 2sin2 2

B C A

cos cos sin sin 2sin2 2 2 2 2

B C B C A

( ) ( )s s b s s cac ab

( )( ) ( )( )s a s c s a s bac ab

( )( )

2s b s c

bc

2 2b c

b c aa

79. Answer (2)

Hint : 2 2 21( ) ix x

n

Sol. : 22

1122

xx

N N

2 21 1 1 22x x x x

21 100 80 20 x

2 20 1001

10 100

= 1

80. Answer (1)

Hint : sin2x = 1, cos2x = 0

Sol. : sin2x – cos2x = 1

sin2x – 1 + sin2x = 1

sin2x = 1 and cos2x = 0

2 2

1

(sin cos )

n

r r

r

x x

= (sin2x + sin4x + ... + sin2nx)

+ (cos2x + cos4x + ... + cos2nx)

= n + 0 = n


13/14

81. Answer (2)

Hint : Circular permutations

Sol. :4

45 4( )

1

nn CP E

n

5( 4)( 5)( 6)( 7)( 1)( 2)( 3) 4

n n n n n

n n n n

( 5)( 6)( 7)( )

( 1)( 2)( 3)n n n

P En n n

82. Answer (4)

Hint : Perpendiculartan

Base

Sol. :

10 m x

H

45° 60°

tan45 , tan6010

H Hx x

10 ,3

HH x x

103

HH

11 10

3

H

10 3 10 3( 3 1)

23 1

H

5(3 3)mH

83. Answer (2)

Hint : P(E) = 1 – P(A B)

Sol. : P(E) =

5 54

2 21

62 2

= 60 60 24

1180

= 96

1180

= 84 7

180 15

84. Answer (2)

Hint : Check domain

Sol. : tan3x = tanx

3 (2 1) , (2 1)2 2

x n x n

3x = n + x

3 5 7 11, , , , ,

2 2 6 6 6 6 x

2 n

x

3

0, , , , 22 2 x

x = 0, , 2

Sum = 3

85. Answer (2)

Hint : D = 0

Sol. : For exactly one solution,

D = 0 a2 = 4b

4 4( )

11 11 121P E

86. Answer (3)

Hint : sine law

Sol. : 2b2 = a2 + c2

2sin2B = sin2A + sin2C

2[1 – cos2B] = [1 – cos2A] + [1 – cos2C]

2cos2B = cos2A + cos2C

cos2B – cos2A = cos2C – cos2B

2sin(A + B) sin(A – B)

= 2sin(B + C) sin (B – C)

sinC [sinA cosB – cosAsinB]

= sinA[sinB cosC – cosBsinC]

cotB – cotA = cotC – cotB

2cotB = cotA + cotC

cotA, cotB, cotC are in A.P.

tanA, tanB, tanC are in H.P.


14/14

��

87. Answer (3)

Hint : Venn diagram formula

Sol. :1 1 13 6 6

A

B

16

C

1 1 12 6 3

P(A B C)

= P(A) + P(B) + P(C) – P(A B) – P(B C)

– P(C A) + P(A B C)

1 1 1 3 2

( )3 2 6 6 3

P A B C

2 1 1 1 13 2 3 2 6

8 6 4 6 2 112 6

1 1 1( )

6 3 2P E

88. Answer (1)

Hint : 2sin cos 2 cos 12 2

Sol. : ( ) sin 2 cos4 8

f x x x

sin 2 cos8 2 8

x x

cos2 cos8 8

x x

22cos cos 18 8

x x

21 9

2 cos8 4 8

x

min9

( ( ))8

f x

89. Answer (4)

Hint : P(E1) + P(E2) + P(E3) = 1

and 0 P(E1), P(E2), P(E3) 1

Sol. :

–1 143

16

23

56 1 2

,6 3

P

1 2 3 1 62 6 6 P P P

3 3 2 3 1 61

6 P P P

10 1

2P 2 3

0 16

P 1 60 1

6P

0 P – 1 – 2 0 3P – 2 – 6 0 6P + 1 6

1 P – 1 2 3P – 4 –1 6P 5

2 43 3

P 1 56 6

P

90. Answer (3)

Hint : Factorization

Sol. : (sin2x – y2) – (sinx – y) = 0

(sinx – y)(sinx + y – 1) = 0

y = sinx OR sinx = 1 – y

–1 sinx 1 –1 sinx 1

y = –1, 0, 1 –1 1 – y 1

–2 –y 0

0 y 2

y = 0, 1, 2

y = {–1, 0, 1, 2}

Test - 6 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/14

1. (1)

2. (4)

3. (2)

4. (1)

5. (2)

6. (4)

7. (1)

8. (2)

9. (1)

10. (2)

11. (1)

12. (3)

13. (4)

14. (4)

15. (1)

16. (1)

17. (2)

18. (4)

19. (3)

20. (2)

21. (1)

22. (1)

23. (3)

24. (3)

25. (1)

26. (4)

27. (2)

28. (4)

29. (3)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (4)

33. (3)

34. (2)

35. (2)

36. (2)

37. (3)

38. (3)

39. (3)

40. (2)

41. (1)

42. (3)

43. (4)

44. (3)

45. (2)

46. (4)

47. (1)

48. (3)

49. (3)

50. (3)

51. (2)

52. (2)

53. (4)

54. (3)

55. (2)

56. (3)

57. (2)

58. (1)

59. (1)

60. (3)

61. (3)

62. (4)

63. (1)

64. (3)

65. (3)

66. (2)

67. (2)

68. (2)

69. (4)

70. (2)

71. (1)

72. (2)

73. (3)

74. (1)

75. (1)

76. (3)

77. (1)

78. (3)

79. (1)

80. (3)

81. (4)

82. (2)

83. (4)

84. (2)

85. (3)

86. (3)

87. (2)

88. (3)

89. (3)

90. (4)

Test Date : 17/02/2019

ANSWERS

TEST - 6 - Code-D

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

All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)

2/14

1. Answer (1)

Hint : I1 = 0.8I0Sol. : I = (0.8I0)

2 = 0.64I0I = 36%

2. Answer (4)

Hint : 1 2

2m

Tk k

Sol. : 1 2 mt

k

2 21

1

4m

tk

21 2

1

4 mk

t

22 2

2

4 mk

t

1 2

2 m

Tk k

22 21 2

21 1

4

mT

mt t

2 2 21 2

1 1 1 T t t

2 2 21 2

T t t

3. Answer (2)

Hint : 2

T TV

r

Sol. :1

2T

fLr

11 1 1

2 1 1 1

12 92

rf L T

f L r T

1

2

3f

f

2450

Hz 150 Hz3

f

PART - A (PHYSICS)

4. Answer (1)

Hint : Amplitude of oscillation = 2L

Sol. :

2L

m1

m2

2KL – KL = (m1 + m2)g

KL = (m1 + m2)g

1 2( )

m mK g

L

Amplitude of oscillation

2

2

m g LK

1 22

( )

2

m mm g g

2m2 = m1 + m2

m2 = m1

2

1

1m

m

5. Answer (2)

Hint : TE = KE + PE

Sol. :2

21 1 3KE

2 2 4

kA k A

21 7KE

2 16

kA

780 KE

16

KE = 35 J

6. Answer (4)

Hint : 2 2v A x

Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/14

Sol. : 2 213 3 A

2 212 4 A

(13)2 – (12)2 = 2(42 – 32)

25 = 2 (16 – 9)

2 125 5rad s

7 7

7. Answer (1)

Hint : 2max 1 2( )I I I

Sol. :2

max 1 2

min 1 2

I A A

I A A

1 2

1 2

31

A A

A A

3A1 – 3A2 = A1 + A2

2A1 = 4A2

1

2

2A

A

8. Answer (2)

Hint : 2 2net 1 2 1 22 cosA A A A A

Sol. :

A2

A1

1 sin4

x a t

22

2 sin63

ax a t

22

sin63

x a t

1

2

32

AA

12

9. Answer (1)

Hint : amax = 2A

Sol. :2

5AA

= 5 rad s–1

5 sin6 A

5 = 12

A A = 10

a = (5)2 × 10 = 250 m/s2

10. Answer (2)

Hint : Total energy remain conserved.

Sol. :2

2 21 1 12 2 4 2

AkA k mV

22 21 1 1

' 42 2 4 2

AkA k mV

2 22 21 1 1 1

' 42 2 4 2 2 4

A AkA k kA k

A2 = 2

234

AA

A2 = 2134

A

A = 132

A

11. Answer (1)

Hint : a = –2A

Sol. : = –2x1 = –2x2

( + ) = –2(x1 + x2)

u2 = 2A2 – 2x12

v2 = 2A2 – 2x22

v2 – u2 = 2(x1 + x2)(x1 – x2)

= –( + )(x1 – x2)

2 2

1 2

u v

x x

12. Answer (3)

Hint : Use formula B

Sol. :3

5

100 10

5 10

PB

VV

B = 20 × 108

8

3

20 10

10

B

v

v � 1400 ms–1


4/14

13. Answer (4)

Hint :60°

x

x

Fnet = 2Kx + 2Kx cos60°

Sol. :

KK

2K

60° x

x

x = x cos60°

–ma = 2Kx + 2K x cos60°

–ma = 2Kx + 2Kx cos260°

=

21

2 22

Kx Kx

–ma = 2

24Kx

Kx

–ma = 22

KxKx

–ma = 5

2Kx

52

Ka x

m

22

5 m

TK

14. Answer (4)

Hint : 2 1 cos(2 )sin

2

Sol. :2

1 cos2 22

a xy nt

41 cos 4

2

a xnt

15. Answer (1)

Hint :

12

Tf [Second harmonic of wire]

Sol. :1 2

12V TL L

320 1 1002 0.8 0.5

1100

1

0.5 100m

m = 5 g

16. Answer (1)

Hint : Wavelength does not change after reflectionfrom the wall.

Sol. : = 0 – V T

= 0 1s

VV

= 3320 20

1 104 320

= 330010

4

= 75 × 10–3 m

= 75 mm

17. Answer (2)

Hint : f0 = f1 – f2Sol. : 2f1 = 410

14102

f

24542

f

454 410 44Beat 7 7 Hz

2 2 2 22

18. Answer (4)

Hint : v2 = 2(A2 – x2)

Sol. : v2 = 2(4a2 – a2)

22 23

34

aa

12

at t = 0, x = a

1sin

2

6


5/14

x = 1

2 sin2 6

a t

= 1 1

2 sin cos cos sin2 6 2 6

a t t

= 3 1

2 sin cos2 2 2 2

t ta

= 3 sin cos2 2t t

a

19. Answer (3)

Hint : vp = A

Sol. : vp (maximum) = 2n x0

22wn

v n

2nx0 = (n)2

022

x

= x0

20. Answer (2)

Hint : The frequency will be maximum whenapproaching the sound tangentially.

Sol. :4 m

5 m1

37°

4 m37°

3 m

3 m

7410

180 t

t = 0.13 s

21. Answer (1)

Hint :2

5

Sol. :2

5

= 10 cm

5 cm2

22. Answer (1)

Hint : Use dimensional analysis

Sol. :1

2

T

fL

2

1 12

4

T Tf

L LDD

1f

LD

23. Answer (3)

Hint : Draw standing wave in string.

Sol. :

24. Answer (3)

Hint : 2

k

Sol. :4

0.5 sin 2011

y t x

2 411

11cm 5.5 cm

2

12011 55 cm s

4

v

k

25. Answer (1)

Hint : The phase difference must be 2 when theymeet again.

Sol. : 6( 1)

5 T

nT n

5n = 6n – 6

n = 6

Time period = 6T

Method 2

2 25 2

6t

T T

1 51

6t

T T

t = 6T


6/14

26. Answer (4)

Hint : Use projection of particle on circle and findphase.

Sol. :A

Al22

/6

1

x1 = A cos(t)

2 sin6

x A t

x1 = x2

cos( ) sin6

t t

22 6

t

32

6 t

43 t

T

12T

t

27. Answer (2)

Hint : dyv

dt

Sol. : y = A sint

v = Acost

2 2 21KE cos

2 mA t

2 21 1 cos22 2

tmA

2 T

T

28. Answer (4)

Hint : KeA = K1A1 = K2A2

Sol. : Let the amplitude of oscillation is A

3 63

3 6 e p

K KK A A K A

K K

23

pA A

23

pv v

29. Answer (3)

Hint : 12

kf

m

Sol. : 12

kf

m

12 1kg

kf

eq12 8

kf

eq

1 12 4 2

k k kk k

k k

1Hz

4 f

30. Answer (4)

Hint : x = A sin(t + )

Sol. : x = A sin(t + )

20 sin 2A

T

0 sin4

A

4

cos( )dx

A tdt

2 24 cos 4

16 16 4A

32cos sin

2 4 4A A

32 2mA


7/14

PART - B (CHEMISTRY)

31. Answer (3)

Hint : Hydroboration reaction.

Sol. :

CH — CH == CH3 2

B H /THF2 6

D O , OD2 2– —

H

—

OD(A)

CH — CH — CH3 2

CH — CH == CH3 2

B H /THF2 6

H O /OH2 2

–

D(B)

CH — CH — CH — OH3 2

32. Answer (4)

Hint : Grignard reagents first undergoes acid basereaction and produces a salt.

Sol. :

Br /h2

0.5 mol 0.5 mol

Mg/dry ether

Br

0.5 mol

Mg Br

(P)( )R

S — CH — C C — H2

—H

two acidic

(A)

2 moles of P are required for 1 mole of A

S — CH — C C2 0.25 mol

(Salt)(Q)

+

33. Answer (3)

Hint : NBS produce Br.

which majorly attacksallylic and benzylic position.

Sol. :

CH3

NBS

CH — Br2 CH2

+ Br—

CH3 CH3 CH3

++

Br

Br

+

Br

34. Answer (2)

Hint : Cl2/H2O gives Markovnikov addition of OHand Cl+.

Sol. : CH — CH = CH CH — CH — CH3 2 3 2

Cl /H O2 2

OH Cl(P)

35. Answer (2)

Hint : HCl does not show peroxide effect.

Sol. : Br

HBr

HBr

HCl

Peroxide

Peroxide

CH — CH == CH2 2

Br

Cl

36. Answer (2)

Hint : With Lindlar's catalyst syn addition occurs.

Sol. :

CH — C C — CH3 3Lindlar's catalyst

C == CH

CH3CH3

Hcis-2-butene

37. Answer (3)

Hint : F– being a poor leaving group, cannot leaveeasily.

Br– is a good leaving group, so can produceSaytzeff product.

Sol. :

CH — CH — CH — CH3 2 2—

F

CH — CH — CH == CH3 2 2

CH — CH — CH — CH3 2 2

—H

NH2—

F

More stable anion

CH — CH == CH — CH3 3

—H

—

Br

B

CH — CH — CH — CH3 3

any strong base

More substitutedalkene is formed

which is more stable

38. Answer (3)

Hint : To produce formaldehyde molecule must have= CH2 group.

Sol. : , ,


8/14

39. Answer (3)

Hint : Correct formula of the given molecule accordingto the question is (CH3)3C — C(CH3)3.

Sol. : CH — C — C — CH3 3

CH3 CH3

CH3 CH3

40. Answer (2)

Hint: Both naphthalene and phenanthrene arearomatic rings and have e– participating inresonance and after alkylation e– remainsame.

Sol. : Naphthalene

Phenanthrene

41. Answer (1)

Hint : CH — C — CH3 3

O

(P)

CH — C CH3 dil. H SO /HgSO2 4 4

— —

Sol. : Product P is a ketone and degree ofunsaturation of P is one.

42. Answer (3)

Hint : Warm KMnO4 in acidic medium oxidise thecompound.

Sol. : A CH — CH — CH — CH3 3—

OH

—

OHB CH — C — OH3 —

O

—

43. Answer (4)

Hint : Ozonolysis reaction

Sol. :

CH2

O3

Zn/H O2

O

+ HCOH

44. Answer (3)

Hint : Tert. carbocation is most stable.

Sol. : 3° alcohols produces alkenes fastest.

45. Answer (2)

Hint : 1,2-dihalo molecules give alkene on reactionwith zinc.

Sol. : A : CH3 – CH = CH2

B : 3 3CH – CH – CH|

Cl

46. Answer (4)

Hint :H+

+ CH – CH = CH3 2

CH – CH – CH3 3

Sol. : + CH — CH = CH3 2

CH — CH — CH3 3

(A)

H+

(P)

Br /h2

Br

CH — C — CH3 3

——

(Q)

Alc. KOH

(R)dil. H SO2 4

(S)

CH — C = CH3 2 CH — C — CH3 3

OH

47. Answer (1)

Hint : Decarboxylation

Sol. : Trimethyl cyclohexane is formed when (R)undergo sodalime decarboxylation.

48. Answer (3)

Hint : Decarboxylation of carboxylates reduces1 C atom from the parent chain.

Sol. : NaOH CaO3 2 2

3 2 3

CH CH CH COOH

CH CH CH

49. Answer (3)

Hint : Alcohols with conc. H2SO4 undergodehydration.


9/14

Sol. :

(+)

H+

OHH+

50. Answer (3)

Hint : Alkynes undergo 2 times of addition reactionwith HCl.

Sol. : CH C CH CH — C — CH3 3 3

HCl

—Cl

—

Cl

51. Answer (2)

Hint : Alkenes on reaction with dil. H2SO4 formcarbocation, which has tendency ofrearrangement.

Sol. :(+)

(+)

dil. H SO2 4

OH H O2 :

Methylshift

52. Answer (2)

Hint : 3AlCl6 6 2 6 5

(gas)BenzeneC H Cl C H Cl HCl

Sol. : The difference in the molar mass of productsis 76 g/mol.

53. Answer (4)

Hint : Different carbides gives differenthydrocarbons upon hydrolysis.

Sol. : 2H O4 3 4Al C CH

CaC2 and Na2C2 on hydrolysis producesethyne.

Mg2C3 on hydrolysis gives propyne.

54. Answer (3)

Hint : Dehydrohalogenation

Sol. :

(ii) K/dry ether

CH Cl3

(i) NaNH2

Cl

Br

Br

Br

CH3

55. Answer (2)

Hint : FeCl3 is a Lewis acid.

Sol. : Cl — Cl + FeCl3L.B L.A

56. Answer (3)

Hint : With alc. KOH both cis and trans butene isformed.

Sol. :

OH

OH

P4

CH3

OH

OH

CH3

,

57. Answer (2)

Hint : Benzene is reactive toward ozonolysis butdoes not react with Br2.

Sol. : Benzene generally does not give additionreaction but it reacts with H2/Pt and showsaddition reaction.

58. Answer (1)

Hint : Alkyne having acidic H.

Sol. : White precipitates of HC C–Ag.

59. Answer (1)

Hint : — NH — C — CH3

O

is ortho/para director.

Sol. : NHCOCH3 NHCOCH3

NO2

(Major product)

HNO /H SO3 2 4

60. Answer (3)

Hint : Check all possible positions available andcheck stereochemistry for each isomer.

Sol. : Here are all the possible isomers.

Br

,

Br

,

CH Br2

Br

,

CH —Br2

Br

CH—Br

CH Br2

,

Br

,

CH Br2

Br

,

CH — Br2

Br

Br

CH — Br2

BrCH2


10/14

PART - C (MATHEMATICS)

61. Answer (3)

Hint : Factorization

Sol. : (sin2x – y2) – (sinx – y) = 0

(sinx – y)(sinx + y – 1) = 0

y = sinx OR sinx = 1 – y

–1 sinx 1 –1 sinx 1

y = –1, 0, 1 –1 1 – y 1

–2 –y 0

0 y 2

y = 0, 1, 2

y = {–1, 0, 1, 2}

62. Answer (4)

Hint : P(E1) + P(E2) + P(E3) = 1

and 0 P(E1), P(E2), P(E3) 1

Sol. :

–1 143

16

23

56 1 2

,6 3

P

1 2 3 1 62 6 6 P P P

3 3 2 3 1 61

6 P P P

10 1

2P 2 3

0 16

P 1 60 1

6P

0 P – 1 – 2 0 3P – 2 – 6 0 6P + 1 6

1 P – 1 2 3P – 4 –1 6P 5

2 43 3

P 1 56 6

P

63. Answer (1)

Hint : 2sin cos 2 cos 12 2

Sol. : ( ) sin 2 cos4 8

f x x x

sin 2 cos8 2 8

x x

cos2 cos8 8

x x

22cos cos 18 8

x x

21 9

2 cos8 4 8

x

min9

( ( ))8

f x

64. Answer (3)

Hint : Venn diagram formula

Sol. :1 1 13 6 6

A

B

16

C

1 1 12 6 3

P(A B C)

= P(A) + P(B) + P(C) – P(A B) – P(B C)

– P(C A) + P(A B C)

1 1 1 3 2

( )3 2 6 6 3

P A B C

2 1 1 1 13 2 3 2 6

8 6 4 6 2 112 6

1 1 1( )

6 3 2P E

65. Answer (3)

Hint : sine law

Sol. : 2b2 = a2 + c2

2sin2B = sin2A + sin2C

2[1 – cos2B] = [1 – cos2A] + [1 – cos2C]

2cos2B = cos2A + cos2C

cos2B – cos2A = cos2C – cos2B


11/14

2sin(A + B) sin(A – B)

= 2sin(B + C) sin (B – C)

sinC [sinA cosB – cosAsinB]

= sinA[sinB cosC – cosBsinC]

cotB – cotA = cotC – cotB

2cotB = cotA + cotC

cotA, cotB, cotC are in A.P.

tanA, tanB, tanC are in H.P.

66. Answer (2)

Hint : D = 0

Sol. : For exactly one solution,

D = 0 a2 = 4b

4 4( )

11 11 121P E

67. Answer (2)

Hint : Check domain

Sol. : tan3x = tanx

3 (2 1) , (2 1)2 2

x n x n

3x = n + x

3 5 7 11, , , , ,

2 2 6 6 6 6 x

2 n

x

3

0, , , , 22 2 x

x = 0, , 2

Sum = 3

68. Answer (2)

Hint : P(E) = 1 – P(A B)

Sol. : P(E) =

5 54

2 21

62 2

= 60 60 24

1180

= 96

1180

= 84 7

180 15

69. Answer (4)

Hint : Perpendiculartan

Base

Sol. :

10 m x

H

45° 60°

tan45 , tan6010

H Hx x

10 ,3

HH x x

103

HH

11 10

3

H

10 3 10 3( 3 1)

23 1

H

5(3 3)mH

70. Answer (2)

Hint : Circular permutations

Sol. :4

45 4( )

1

nn CP E

n

5( 4)( 5)( 6)( 7)( 1)( 2)( 3) 4

n n n n n

n n n n

( 5)( 6)( 7)( )

( 1)( 2)( 3)n n n

P En n n

71. Answer (1)

Hint : sin2x = 1, cos2x = 0

Sol. : sin2x – cos2x = 1

sin2x – 1 + sin2x = 1

sin2x = 1 and cos2x = 0

2 2

1

(sin cos )

n

r r

r

x x

= (sin2x + sin4x + ... + sin2nx)

+ (cos2x + cos4x + ... + cos2nx)

= n + 0 = n


12/14

72. Answer (2)

Hint : 2 2 21( ) ix x

n

Sol. : 22

1122

xx

N N

2 21 1 1 22x x x x

21 100 80 20 x

2 20 1001

10 100

= 1

73. Answer (3)

Hint :( )( )

sin2

A s b s cbc

( )cos

2

A s s abc

Sol. : 22cos cos 4sin2 2 2

B C B C A

22sin cos 4sin2 2 2A B C A

cos 2sin2 2

B C A

cos cos sin sin 2sin2 2 2 2 2

B C B C A

( ) ( )s s b s s cac ab

( )( ) ( )( )s a s c s a s bac ab

( )( )

2s b s c

bc

2 2b c

b c aa

74. Answer (1)

Hint : P(E) = 1 – P(X = 1) – P(X = 0)

Sol. : S = {HHH, HHT, HTH, THH, HTT, THT, TTH}

E = {HTT, THT, TTH}

( ) 3 4( ) 1 1

( ) 7 7n E

P En S

75. Answer (1)

Hint : sec2 = 1 + tan2

Sol. : R = sec2Asec2B + tan2Atan2B +

2 secAsecBtanAtanB – sec2Atan2B –

sec2Btan2A – 2 secAsecBtanA tanB

= sec2A(sec2B – tan2B) +

tan2A(tan2B – sec2B)

= sec2A – tan2A = 1

76. Answer (3)

Hint : Venn diagram

Sol. : E

55

25

H

Percentage of population which readexactly one newspaper is

( ) ( )n E H n E H

= n(E) – n(E H) + n(H) – n(E H)

= 10 + 30 – 2 x 5 = 30%

77. Answer (1)

Hint :2

2 2( )ixx

N

Sol. : x4 + x5 = 30 – (2 + 4 + 9) = 15 ...(i)

2 22 4 54 16 81

36 9.25

x x

x42 + x5

2 = 125 ...(ii)

From (i) and (ii) gives,

x4 = 5, x5 = 10

x4 : x5 = 1 : 2

78. Answer (3)

Hint : Solve equation

Sol. :sin1 sin 22

sin cos2

sin2(1 sin ) 2

2sin cos cos2 2 2


13/14

22(1 sin ) 2sin2

2(1 + sin) = 1 – cos

2sin = –(1 + cos)

24 sin cos 2 cos2 2 2

2 sin cos2 2

tan 12

79. Answer (1)

Hint : Venn diagram

Sol. : The probability of solving the question by

three students A, B and C are 2 1

,3 3

and 16

respectively.

2 1 1( ) , ( ) , ( )

3 3 6 P A P B P C

Probability that the question is solved byexactly two students is

P(E) = ( ) ( ) P A B C P A B C

( ) P A B C

= ( ) ( ) ( ) ( ) ( ) ( ) P A P B P C P A P B P C

( ) ( ) ( )P A P B P C

= 1 1 1 2 2 1 2 1 5 53 3 6 3 3 6 3 3 6 18

80. Answer (3)

Hint : cosine law

Sol. :2 2 26 (3 5) 3

cos2 6 3 5

36 45 9 2

36 5 5

1

tan 2 12

1 1tan

2 3

tan tan8 tan tan

6

8

6

81. Answer (4)

Hint : Adding a number in every observation doesnot change .

Sol. :2( )ix x

N

11 2 3 ... 10 11

10 2 x

25 6 7 ... 14 95 19 11

410 10 2 2

x

1 = 2

82. Answer (2)

Hint : Square both sides

Sol. : 1 + sin + cos = sincos

sin + cos = sincos– 1

Squaring on both sides,

1 + 2sincos

= 1 + sin2cos2 – 2sincos

212sin2 sin 2

4

sin2(sin2 – 8) = 0

3sin2 0 , ,

2 2

Only and 32

satisfy the original equation.

3 5Sum

2 2

83. Answer (4)

Hint : Combinations

Sol. :18 17

2 235

2

9 17 17 8 17( )

35 17 35

C CP E

C

84. Answer (2)

Hint : cosine law

Sol. :2 2 2

cos2

b c aA

bc

b2 – 2bc cosA + c2 – a2 = 0

2 4 2 8 0b b

2( 2 2) 0 2 2 b b


14/14

��

85. Answer (3)

Hint : Sum of digits should be divisible by 3.

Sol. : If the sum of digits of a number is divisible by3, then the number is divisible by 3.

The number should be formed by using

1, 2, 3, 6, 9 or 2, 3, 4, 6, 9

6

5

2 5 2 1( )

6 35P E

C

86. Answer (3)

Hint : sin + sin( + ) + sin( + 2) + ... upton terms

Sol :13

1

(sin cos )

r

r r

13sin

2[sin7 cos7 ] 0

sin2

13

sin 02

or tan7 = 1 and sin 02

2 4 6, ,

13 13 13

5 9 13, , ,

28 28 28 28

87. Answer (2)

Hint : Counting

Sol. : 12 11 11( )

52 51 221P E

88. Answer (3)

Hint : Formula

Sol. : Apply direct formula

89. Answer (3)

Hint : cosine law

Sol. : Triangle DEF is called the pedal triangle ofthe triangle ABC.

Circumradius of pedal triangle = 12

(Circumradius of triangle ABC) = 2R

cosC = 2 2 2

2 a b c

ab

= 25(3 1 2 3) 50 100

25 2( 3 1)8

4

= 50( 3 1) 1

50 2( 3 1) 2

C = 45°, similarly B = 30°, A = 105°

5 512sin 222

cR

C

5 5 2

2 42 2 R

R

90. Answer (4)

Hint : 5cos(2 ) cos2

Sol. : As cos(25) = cos 3264

= cos2

= 0

cos cos2 cos22 ... cos29 = 0

test - 6 (code-c) (answers) all india aakash test …...all india aakash test series for jee...

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