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Test - 6 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2021

All India Aakash Test Series for JEE (Main)-2021

Test Date : 16/02/2020

ANSWERS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

TEST-6 - Code-C

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (3)

2. (1)

3. (2)

4. (3)

5. (2)

6. (3)

7. (4)

8. (4)

9. (4)

10. (3)

11. (2)

12. (1)

13. (4)

14. (2)

15. (2)

16. (1)

17. (3)

18. (2)

19. (4)

20. (2)

21. (10)

22. (66)

23. (25)

24. (30)

25. (25)

26. (2)

27. (1)

28. (4)

29. (3)

30. (3)

31. (1)

32. (2)

33. (2)

34. (1)

35. (4)

36. (1)

37. (2)

38. (3)

39. (1)

40. (2)

41. (4)

42. (2)

43. (3)

44. (4)

45. (2)

46. (01)

47. (05)

48. (28)

49. (10)

50. (06)

51. (4)

52. (2)

53. (2)

54. (3)

55. (2)

56. (3)

57. (2)

58. (4)

59. (2)

60. (4)

61. (3)

62. (3)

63. (4)

64. (4)

65. (1)

66. (2)

67. (2)

68. (4)

69. (3)

70. (2)

71. (06)

72. (03)

73. (02)

74. (04)

75. (00)

All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)


2/10

HINTS & SOLUTIONS

PART - A (PHYSICS)

1. Answer (3)

Hint: 8

22

= =

Sol.: red 22 2 s

8

MT

k= = =

cm

4 4 16m/s , V 4 m/s

3 3 3ACV = → = + =

4 8

4 m/s3 3

PxV = − = →

KE in C frame

2 21 16 8

3 62 3 3

= +

1 256 6 64

2 3 3 3

= +

128 64

643

+= =

sinA t =

2 s6 12

t t

= =

2164 8 4 m

2A A= =

2. Answer (1)

Hint: 2

3 2

2

42

3

d xR R xg

dt = −

Sol.: 2

2

6 605

4 4

d x xg

R Rdt

− = = =

3. Answer (2)

Hint: 0 0.1mF

Ak

= =

Sol.: x = 0.1 – 0.1 cos(t – 0.2)

= 0.1[1 – cos5(t – 0.2)]

4. Answer (3)

Hint: 3 31 1 2 2

2

k xk x k x= =

Sol.: 3 3 3 31 23 3

1 22 4 4

k x k xx xx x x

k k

+= + = + +

2 3 3 1 1 23

1 2

( 4 )

4

k k k k k kx x

k k

+ +=

1 23

2 3 3 1 1 2

4

4

k k xx

k k k k k k=

+ +

2

1 2 3

22 3 3 1 1 2

4

( 4 )

k k k xd xM

k k k k k kdt= −

+ +

1 2 3

2 3 3 1 1 2

4

( 4 )

k k k

M k k k k k k =

+ +

5. Answer (2)

Hint: 2

cos3

dx t= −

Sol.: 3

cos2 2 3

t t

= − = +

13

t

=

12M

T tk

= +

2

3

M M

k k

= +

5

3

M

k

=

6. Answer (3)

Hint: y(x, t) = 5 sin(kx – t)

Sol.: y(x, t = 0) = 5 mm sinkx

y(x, t) = y(x – vt, t = 0) = 5 mm sink(x – vt)

= 5 mm sin(kx – t)

5 mm cos( )y

kx tt

= − −

At 3

0, cos2

t kx= =

3 ˆ5 mm 100 250 3 mm/s

2PV j= − = −

Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021


3/10

7. Answer (4)

Hint: 2

3 2 3

GMrm d r GMrma

R dt R= − = −

Sol.: 3

3, 2

GM RT

GMR = =

3

4 2

= =

T Rt

GM

8. Answer (4)

Hint:

210

2 20 0

btt

mA A e A e

−

− −

= =

Sol.: 35 10 31

ln10 5 1010

te t−− − = =

t = 200 ln10 s = 461 s

9. Answer (4)

Hint: 3

0cm 2

0

2 2

33

L

L

x kx dxL L

xL

kx dx

= = =

Sol.: 4

2

04

= =L

kLI kx dx x

4

2

3 22 2

4 334 2

2 2

kL LT

gkL gL

= =

10. Answer (3)

Hint: 2sindU

F a ax a xdx

= − = − = −

Sol.: 2 2

2

2

d x a am a x

mdt m= − = =

2

T ma

=

11. Answer (2)

Hint: 0

2( , 0) siny x t a x

= =

Sol.: = 4b

12. Answer (1)

Hint: 202 sin

2

dT rd r

=

Sol.: 2 20T r =

rel 0

Tv r= =

v = 20r

13. Answer (4)

Hint: v xg=

Sol.: 1

22

dv ga xg g

dt x= = =

14. Answer (2)

Hint: 2 2

cos ( )P

y vv A vt x

dt

= = −

Sol.: , max wave

2,Pv Av v v

= =

2

2A A

= =

15. Answer (2)

Hint: 330 330

4000 4250330 330

v

v

+ =

−

Sol.: 16(330 + v) = 17(330 – v)

33v = 330 v = 10 m/s

16. Answer (1)

Hint: 110 10 1 10 0

0

20 10log 2 log logI

I II

= = −

Sol.: 210 10 2 10 0

0

30 10log 3 log logI

I II

= = −

210 2 1

1

1 log 10= =I

I II

17. Answer (3)

Hint: 1 0 2 0,4 4

u uT T = + = −

Sol.:

00

01

02 00

21 214 5 2020[19 ] 19

4 5

u

uu

u

+

= = =

−

05uT

=



4/10

18. Answer (2)

Hint: nf0 = 225, (n + 1)f0 = 375

Sol.: f0 = 150 Hz

n is not an integer

(2n – 1)f0 = 225, (2n + 1)f0 = 375

0 0

225 3751 1

f f+ = −

00

1502 75 Hzf

f= =

n = 2

75 1.1m4 300

v vL

L= = =

19. Answer (4)

Hint: 24

PI

r=

Sol.: 20

02 224 2

pP VPI p

PVr r

= = =

P0 = BkS0

00 2 22

p V VPS

Bk V r

= =

2

1

2

P

V r=

51 20m 1.4 10 m

6000 1.3 340 2 3.14

−= =

20. Answer (2)

Hint: 2 280 18 6400 324+ = +

6724 82 m= =

Sol.: 2 m 4 m2

x

= = =

340

85 Hz4

f = =

21. Answer (10)

Hint: 1 2 20 0

1 1,

2

D D at t

v f v f v= = + −

Sol.: 02 1 2 2

0 0 0

21

2 2

f v aaT t t

f f v f v

− = − = − =

2 2

0 0

0 0

2 101

2 10 5

f v f vf

T f v a f v a = = =

− −

22. Answer (66)

Hint: 100

2 5 cm/s 10 m/sr = =

Sol.: max min

330 3301088

320 340

− = −

f f

= 2

1088 3332 34

= 66 Hz

23. Answer (25)

Hint: 340 3

3 85 Hz 255 Hz4

f

= = =

Sol.: 1 260.1 0.4

2550.8 m

=

260.1 0.4

255 255 0.8 0.8m

=

3260.1 0.410 2.5 g

255 255 0.8 0.8m

= =

10m = 25

24. Answer (30)

Hint: cos 602 4

= =

Sol.: 3 3 300 md

d DD

= = =

30 m10

d=

25. Answer (25)

Hint: 2

50 cm25

= =

Sol.: min 25 cm2

L

= =

PART - B (CHEMISTRY)

26. Answer (2)

Hint : B.P. surface area

Sol. : Boiling point for the isomers same

number of C-atoms follows the order of

normal > iso > neo.



5/10

27. Answer (1)

Hint : In propagation step R is formed

Sol. : R X R X+ ⎯⎯→ − is a termination step

28. Answer (4)

Hint : This is the competition between the

reactivity and stability Cl and Br

Sol. : Br is more stable, hence less reactive

majorly attacks at 3° carbon position

that why S is formed in very less amount.

29. Answer (3)

Hint :

Sol. : Molar mass of CH2O is in lesser but

carbon is in higher oxidation state as

compared to X.

30. Answer (3)

Hint : To show geometrical isomerism C = C

bond must have different groups.

Sol. :

No geometrical isomerism

Can show geometrical isomerism

31. Answer (1)

Hint : -anti elimination of fluoroalkanes give

kinetically controlled product in which

transition state of intermediate is more

stable (i.e. carbanion)

Sol. :

Only two hyperconjugable H’s

32. Answer (2)

Hint : NBS cause substitution of Br at allylic

position

Sol. : P2 is

33. Answer (2)

Hint : Lindlar’s catalyst form cis product

Sol. :

34. Answer (1)

Hint : In presence of Pd and BaSO4, partial

addition takes place

Sol. :

35. Answer (4)

Hint & Sol. :

Benzene do not give test for unsaturation

36. Answer (1)

Hint : General mechanism of ozonolysis is

Sol. :



6/10

37. Answer (2)

Hint : Cl2 in water exist as

HOCl (HO– + Cl+)

Sol. :

38. Answer (3)

Hint : AlCl3 is a Lewis acid

Sol. :

39. Answer (1)

Hint :

Sol. : x is an aromatic compound and Cl is ortho-para directing

40. Answer (2)

Hint :

Sol.

41. Answer (4)

Hint : –OCH3 is ortho-para directing

Sol. :

42. Answer (2)

Hint : Dehydrohalogenation is elimination type

of reaction.

Sol. :

43. Answer (3)

Hint : 4KMnO cause the addition of –OH to

both the double bonded (atoms)

Sol. :

44. Answer (4)

Hint : Free radical substitution reaction takes

place

Sol. :

45. Answer (2)

Hint : HBr/ROOR cause anti-Markovnikoff

addition

Sol. :



7/10

46. Answer (01)

Hint : The hydro carbon with molar weight

72 g/mol and 12 primary hydrogens is

C5H12

Sol. : Only one isomer is possible.

47. Answer (05)

Hint :

⎯⎯⎯⎯→ +alc.KOH

4 7 4 6C H Br C H HBr

Sol. : Possible isomers with molar formula C4H6

48. Answer (28)

Hint : In Kolbe’s electrolysis method, reaction

goes via free radical intermediate.

Sol. :

49. Answer (10)

Hint : Naphthalene has 10 e’s

Sol. :

50. Answer (06)

Hint : The site which is less hindered and more

electron dense will participate most in

EAS.

Sol. :

is electron withdrawing group

and –NH – is electron donating in

nature.

So EAS will occur according to –NH–

group.

PART - C (MATHEMATICS)

51. Answer (4)

Hint: Range of both sides.

Sol.: L.H.S. 5 and R.H.S. 5

equation holds when

2 29sec 4cosec 5y y+ =

2 2 1tan cos(2 )

3 5y y= =

52. Answer (2)

Hint: P(A) + P(B) = P(A B) + P(A B)

Sol.: P(A B) = 0.2, P(A B) = 0.6

P(A) + P(B) = 0.8

( ) ( ) 2 0.8 1.2+ = − =P A P B

53. Answer (2)

Hint: Modulus Property.

Sol.: For maximum value cos2x = 0

max value ( 3 1)= −

54. Answer (3)

Hint: Probability few cases

total case=

Sol.: Few cases = {(1, 1), (1, 2), (2, 1), (1, 4),

(4, 1), (2, 3), (3, 2), (3, 4), (4, 3), (1, 6),

(6, 1), (2, 5), (5, 2), (6, 5), (5, 6)}

15 5

Pr36 12

= =

55. Answer (2)

Hint: Replace r → n – r.

Sol.: Let 1000

1

sin( )cos(1001 )r

S rx r x=

= − …(1)

Replace r → 1001 – r

1000

1

sin(1001 ) cos( )r

S r x rx=

= − …(2)

By (1) and (2)

2 1000 sin(1001 )S x=

500 sin(1001 )S x=



8/10

56. Answer (3)

Hint: Simplify.

Sol.: 2 2 4 2 2

2 2 2

4sin cos 4sin 4sin cos 1

94 4sin 4sin cos

x x x x x

x x x

+ −=

− −

4

4

sin 1 1tan

9 6cos 3

= = =

xx x n

x

57. Answer (2)

Hint: 21 cos 2cos2

xx+ =

21 cos 2sin2

xx− =

Sol.: cos sin

1 cos 1 cos 2 2

1 cos 1 coscos sin

2 2

++ + −

=+ − −

−

x x

x x

x xx x

( )tan cot2 4 4 2

x x = − − = +

58. Answer (4)

Hint: tan tanA A = .

Sol.: tanA tanC = 2 and tanB tanC = 18

Let tanC = x

2 18

tan and tanA Bx x

= =

tanA + tanB + tanC = tanA tanB × tanC

x = 4 = tanC

2

2

1 tan 1 16 15cos2

1 17 171 tan

CC

C

− − −= = =

++

59. Answer (2)

Hint: 10cot 45 10 =

Sol.: 10cot 45 10 =

60. Answer (4)

Hint: cos3 = 4cos3 – 3cos

Sol.: 10 10

3

0 0

1cos 3cos cos( )

3 4 3r r

r rr

= =

= +

10 11cos sin

3 16 6

4 4sin

6

= +

1

8

−=

61. Answer (3)

Hint: 5 1

cos364

+ =

Sol.: ( ) 1 cos 1 sin4 4

F kk k

= + +

1 sin 1 cos4 4k k

− −

2 21 cos 1 sin4 4k k

= − −

21

sin4 2k

=

1 3 5

(5) (1 cos36 )8 32

F−

= − =

62. Answer (3)

Hint: A.M. G.M.

Sol.:

1 1 2 2( )

cos cos6 6

f xa b ab

x x

+

− +

2( ) 0,

31 cos(2 )

4 2

f x xx

+

max

4( )

3 =f x

63. Answer (4)

Hint: Number of relation = 2mn,

Number of function = mn

Sol.:

44

12

3 3Pr

82

= =



9/10

64. Answer (4)

Hint: 4 sin sin sin2 2

A Cr R B=

Sol.: 8 8 4 sin sin sin2 2 2

A B CR r R

= =

1

2sin sin sin2 2 2 16

=A B C

1

sin 1 sin2 2 16

C C − =

1

sin2 4

C=

65. Answer (1)

Hint: 2 2 2

cos2

b c aA

bc

+ −=

Sol.: 2 2 2 2 cos

3AC OA OC AOOC

= + −

2 2d r=

66. Answer (2)

Hint: Mean.

Sol.: 1 2 1

( 2 )2 1 2

nx a a nd a nd

n

+ = + + = + +

Mean deviation from mean is

2

0

1( ) ( )

2 1

n

r

a rd a ndn =

+ − ++

2

0

1

2 1

n

r

r n dn =

= −+

( 1)

2 1

n nd

n

+=

+

67. Answer (2)

Hint: 2var( ) var( )ax a x=

Sol.: 2var( ) var( )ax b a x+ =

2 2

2

2 2var var( )

ax b a ax

c c c

+ = =

Standard deviation 2

2

2

a a

cc= =

68. Answer (4)

Hint: Probability by classical theory.

Sol.: T.C = 25

Set of F.C = {2, 3, 5, 7, 11, 13, 17, 19, 23}

9

25P =

69. Answer (3)

Hint: Probability P and C.

Sol.: T.C = 30C3

F.C = apart from 9 : 25 one number is to

be chosen from 10, 11, …… 24 which are

selected in 15 ways

30

3

15 15 6 3

30 29 28 812

= = =

P

C

70. Answer (2)

Hint: Required probability 1

2=

Sol.: Required probability 12 1

22

−

= =n

n

71. Answer (06)

Hint: A.M H.M

Sol.: a + b + c = 2s

Given expression is a b c

s a s b s c+ +

− − −

3s s s

s a s b s c= − + + +

− − −

A.M H.M

3

3

+ +− − −

− − −+ +

s s s

s a s b s cs a s b s c

s s s

9s s s

s a s b s c+ +

− − −

Minimum value of given expression is 6



10/10

72. Answer (03)

Hint: Distance of circumcenter from sides.

Sol.: Distance of circumcenter from side

AC = R cosB

Distance of orthocenter from side AC =

2R cosA cosC

2cos cos cos( )A C A C= − +

tan tan 3A C =

73. Answer (02)

Hint: Cosine rule in cyclic quadrilateral.

Sol.:

In PQR PR2 = 4 + 25 – 20cos60° = 19

In PRS PR2 = 9 + x2 + 3x = 19

(x + 5) (x – 2) = 0 x = 2

74. Answer (04)

Hint: Cosine rule.

Sol.: 2 2

1 22 2cos( )a b+ = + +

minimum value = 0; maximum value = 4

75. Answer (00)

Hint: Compound angle formulas.

Sol.: sin( )cos( ) sin cos

sin( )cos( ) sin cos

+ − − +=

− + + −

A B C A B C C B

A B C A B C B C

Apply componendo and dividendo

( )sin 2( ) sin( ) sin(2 )sin( ) 0B C B C A B C − + + − =

sin( ) 2cos( )sin( ) sin2 0B C B C B C A − − + + =

sin2 sin2 sin2 0 ( )A B C B C + + =

Test - 6 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2021

All India Aakash Test Series for JEE (Main)-2021

Test Date : 16/02/2020

ANSWERS


TEST-6 - Code-D

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (2)

2. (4)

3. (2)

4. (3)

5. (1)

6. (2)

7. (2)

8. (4)

9. (1)

10. (2)

11. (3)

12. (4)

13. (4)

14. (4)

15. (3)

16. (2)

17. (3)

18. (2)

19. (1)

20. (3)

21. (25)

22. (30)

23. (25)

24. (66)

25. (10)

26. (2)

27. (4)

28. (3)

29. (2)

30. (4)

31. (2)

32. (1)

33. (3)

34. (2)

35. (1)

36. (4)

37. (1)

38. (2)

39. (2)

40. (1)

41. (3)

42. (3)

43. (4)

44. (1)

45. (2)

46. (06)

47. (10)

48. (28)

49. (05)

50. (01)

51. (2)

52. (3)

53. (4)

54. (2)

55. (2)

56. (1)

57. (4)

58. (4)

59. (3)

60. (3)

61. (4)

62. (2)

63. (4)

64. (2)

65. (3)

66. (2)

67. (3)

68. (2)

69. (2)

70. (4)

71. (00)

72. (04)

73. (02)

74. (03)

75. (06)

All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)


2/10

HINTS & SOLUTIONS

PART - A (PHYSICS)

1. Answer (2)

Hint: 2 280 18 6400 324+ = +

6724 82 m= =

Sol.: 2 m 4 m2

x

= = =

340

85 Hz4

f = =

2. Answer (4)

Hint: 24

PI

r=

Sol.: 20

02 224 2

pP VPI p

PVr r

= = =

P0 = BkS0

00 2 22

p V VPS

Bk V r

= =

2

1

2

P

V r=

51 20m 1.4 10 m

6000 1.3 340 2 3.14

−= =

3. Answer (2)

Hint: nf0 = 225, (n + 1)f0 = 375

Sol.: f0 = 150 Hz

n is not an integer

(2n – 1)f0 = 225, (2n + 1)f0 = 375

0 0

225 3751 1

f f+ = −

00

1502 75 Hzf

f= =

n = 2

75 1.1m4 300

v vL

L= = =

4. Answer (3)

Hint: 1 0 2 0,4 4

u uT T = + = −

Sol.:

00

01

02 00

21 214 5 2020[19 ] 19

4 5

u

uu

u

+

= = =

−

05uT

=

5. Answer (1)

Hint: 110 10 1 10 0

0

20 10log 2 log logI

I II

= = −

Sol.: 210 10 2 10 0

0

30 10log 3 log logI

I II

= = −

210 2 1

1

1 log 10= =I

I II

6. Answer (2)

Hint: 330 330

4000 4250330 330

v

v

+ =

−

Sol.: 16(330 + v) = 17(330 – v)

33v = 330 v = 10 m/s

7. Answer (2)

Hint: 2 2

cos ( )P

y vv A vt x

dt

= = −

Sol.: , max wave

2,Pv Av v v

= =

2

2A A

= =

8. Answer (4)

Hint: v xg=

Sol.: 1

22

dv ga xg g

dt x= = =

9. Answer (1)

Hint: 202 sin

2

dT rd r

=

Sol.: 2 2

0T r =

rel 0

Tv r= =

v = 20r

Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021


3/10

10. Answer (2)

Hint: 0

2( , 0) siny x t a x

= =

Sol.: = 4b

11. Answer (3)

Hint: 2sindU

F a ax a xdx

= − = − = −

Sol.: 2 2

2

2

d x a am a x

mdt m= − = =

2

T ma

=

12. Answer (4)

Hint: 3

0cm 2

0

2 2

33

L

L

x kx dxL L

xL

kx dx

= = =

Sol.: 4

2

04

= =L

kLI kx dx x

4

2

3 22 2

4 334 2

2 2

kL LT

gkL gL

= =

13. Answer (4)

Hint:

210

2 20 0

btt

mA A e A e

−

− −

= =

Sol.: 35 10 31

ln10 5 1010

te t−− − = =

t = 200 ln10 s = 461 s

14. Answer (4)

Hint: 2

3 2 3

GMrm d r GMrma

R dt R= − = −

Sol.: 3

3, 2

GM RT

GMR = =

3

4 2

= =

T Rt

GM

15. Answer (3)

Hint: y(x, t) = 5 sin(kx – t)

Sol.: y(x, t = 0) = 5 mm sinkx

y(x, t) = y(x – vt, t = 0) = 5 mm sink(x – vt)

= 5 mm sin(kx – t)

5 mm cos( )y

kx tt

= − −

At 3

0, cos2

t kx= =

3 ˆ5 mm 100 250 3 mm/s

2PV j= − = −

16. Answer (2)

Hint: 2

cos3

dx t= −

Sol.: 3

cos2 2 3

t t

= − = +

13

t

=

12M

T tk

= +

2

3

M M

k k

= +

5

3

M

k

=

17. Answer (3)

Hint: 3 31 1 2 2

2

k xk x k x= =

Sol.: 3 3 3 31 23 3

1 22 4 4

k x k xx xx x x

k k

+= + = + +

2 3 3 1 1 23

1 2

( 4 )

4

k k k k k kx x

k k

+ +=

1 23

2 3 3 1 1 2

4

4

k k xx

k k k k k k=

+ +

2

1 2 3

22 3 3 1 1 2

4

( 4 )

k k k xd xM

k k k k k kdt= −

+ +

1 2 3

2 3 3 1 1 2

4

( 4 )

k k k

M k k k k k k =

+ +



4/10

18. Answer (2)

Hint: 0 0.1mF

Ak

= =

Sol.: x = 0.1 – 0.1 cos(t – 0.2)

= 0.1[1 – cos5(t – 0.2)]

19. Answer (1)

Hint: 2

3 2

2

42

3

d xR R xg

dt = −

Sol.: 2

2

6 605

4 4

d x xg

R Rdt

− = = =

20. Answer (3)

Hint: 8

22

= =

Sol.: red 22 2 s

8

MT

k= = =

cm

4 4 16m/s , V 4 m/s

3 3 3ACV = → = + =

4 8

4 m/s3 3

PxV = − = →

KE in C frame

2 21 16 8

3 62 3 3

= +

1 256 6 64

2 3 3 3

= +

128 64

643

+= =

sinA t =

2 s6 12

t t

= =

2164 8 4 m

2A A= =

21. Answer (25)

Hint: 2

50 cm25

= =

Sol.: min 25 cm2

L

= =

22. Answer (30)

Hint: cos 602 4

= =

Sol.: 3 3 300 md

d DD

= = =

30 m10

d=

23. Answer (25)

Hint: 340 3

3 85 Hz 255 Hz4

f

= = =

Sol.: 1 260.1 0.4

2550.8 m

=

260.1 0.4

255 255 0.8 0.8m

=

3260.1 0.410 2.5 g

255 255 0.8 0.8m

= =

10m = 25

24. Answer (66)

Hint: 100

2 5 cm/s 10 m/sr = =

Sol.: max min

330 3301088

320 340

− = −

f f

= 2

1088 3332 34

= 66 Hz

25. Answer (10)

Hint: 1 2 20 0

1 1,

2

D D at t

v f v f v= = + −

Sol.: 02 1 2 2

0 0 0

21

2 2

f v aaT t t

f f v f v

− = − = − =

2 2

0 0

0 0

2 101

2 10 5

f v f vf

T f v a f v a = = =

− −



5/10

PART - B (CHEMISTRY)

26. Answer (2)

Hint : HBr/ROOR cause anti-Markovnikoff

addition

Sol. :

27. Answer (4)

Hint : Free radical substitution reaction takes

place

Sol. :

28. Answer (3)

Hint : 4KMnO cause the addition of –OH to

both the double bonded (atoms)

Sol. :

29. Answer (2)

Hint : Dehydrohalogenation is elimination type

of reaction.

Sol. :

30. Answer (4)

Hint : –OCH3 is ortho-para directing

Sol. :

31. Answer (2)

Hint :

Sol.

32. Answer (1)

Hint :

Sol. : x is an aromatic compound and Cl is

ortho-para directing

33. Answer (3)

Hint : AlCl3 is a Lewis acid

Sol. :



6/10

34. Answer (2)

Hint : Cl2 in water exist as

HOCl (HO– + Cl+)

Sol. :

35. Answer (1)

Hint : General mechanism of ozonolysis is

Sol. :

36. Answer (4)

Hint & Sol. :

Benzene do not give test for unsaturation

37. Answer (1)

Hint : In presence of Pd and BaSO4, partial

addition takes place

Sol. :

38. Answer (2)

Hint : Lindlar’s catalyst form cis product

Sol. :

39. Answer (2)

Hint : NBS cause substitution of Br at allylic

position

Sol. : P2 is

40. Answer (1)

Hint : -anti elimination of fluoroalkanes give

kinetically controlled product in which

transition state of intermediate is more

stable (i.e. carbanion)

Sol. :

Only two hyperconjugable H’s

41. Answer (3)

Hint : To show geometrical isomerism C = C

bond must have different groups.

Sol. :

No geometrical isomerism

Can show geometrical isomerism

42. Answer (3)

Hint :

Sol. : Molar mass of CH2O is in lesser but

carbon is in higher oxidation state as

compared to X.



7/10

43. Answer (4)

Hint : This is the competition between the

reactivity and stability Cl and Br

Sol. : Br is more stable, hence less reactive

majorly attacks at 3° carbon position

that why S is formed in very less amount.

44. Answer (1)

Hint : In propagation step R is formed

Sol. : R X R X+ ⎯⎯→ − is a termination step

45. Answer (2)

Hint : B.P. surface area

Sol. : Boiling point for the isomers same

number of C-atoms follows the order of

normal > iso > neo.

46. Answer (06)

Hint : The site which is less hindered and more

electron dense will participate most in

EAS.

Sol. :

is electron withdrawing group

and –NH – is electron donating in

nature.

So EAS will occur according to –NH–

group.

47. Answer (10)

Hint : Naphthalene has 10 e’s

Sol. :

48. Answer (28)

Hint : In Kolbe’s electrolysis method, reaction

goes via free radical intermediate.

Sol. :

49. Answer (05)

Hint :

⎯⎯⎯⎯→ +alc.KOH

4 7 4 6C H Br C H HBr

Sol. : Possible isomers with molar formula C4H6

50. Answer (01)

Hint : The hydro carbon with molar weight

72 g/mol and 12 primary hydrogens is

C5H12

Sol. : Only one isomer is possible.

PART - C (MATHEMATICS)

51. Answer (2)

Hint: Required probability 1

2=

Sol.: Required probability 12 1

22

−

= =n

n

52. Answer (3)

Hint: Probability P and C.

Sol.: T.C = 30C3

F.C = apart from 9 : 25 one number is to

be chosen from 10, 11, …… 24 which are

selected in 15 ways

30

3

15 15 6 3

30 29 28 812

= = =

P

C

53. Answer (4)

Hint: Probability by classical theory.

Sol.: T.C = 25

Set of F.C = {2, 3, 5, 7, 11, 13, 17, 19, 23}

9

25P =



8/10

54. Answer (2)

Hint: 2var( ) var( )ax a x=

Sol.: 2var( ) var( )ax b a x+ =

2 2

2

2 2var var( )

ax b a ax

c c c

+ = =

Standard deviation 2

2

2

a a

cc= =

55. Answer (2)

Hint: Mean.

Sol.: 1 2 1

( 2 )2 1 2

nx a a nd a nd

n

+ = + + = + +

Mean deviation from mean is

2

0

1( ) ( )

2 1

n

r

a rd a ndn =

+ − ++

2

0

1

2 1

n

r

r n dn =

= −+

( 1)

2 1

n nd

n

+=

+

56. Answer (1)

Hint: 2 2 2

cos2

b c aA

bc

+ −=

Sol.: 2 2 2 2 cos

3AC OA OC AOOC

= + −

2 2d r=

57. Answer (4)

Hint: 4 sin sin sin2 2

A Cr R B=

Sol.: 8 8 4 sin sin sin2 2 2

A B CR r R

= =

1

2sin sin sin2 2 2 16

=A B C

1

sin 1 sin2 2 16

C C − =

1

sin2 4

C=

58. Answer (4)

Hint: Number of relation = 2mn,

Number of function = mn

Sol.:

44

12

3 3Pr

82

= =

59. Answer (3)

Hint: A.M. G.M.

Sol.:

1 1 2 2( )

cos cos6 6

f xa b ab

x x

+

− +

2( ) 0,

31 cos(2 )

4 2

f x xx

+

max

4( )

3 =f x

60. Answer (3)

Hint: 5 1

cos364

+ =

Sol.: ( ) 1 cos 1 sin4 4

F kk k

= + +

1 sin 1 cos4 4k k

− −

2 21 cos 1 sin4 4k k

= − −

21

sin4 2k

=

1 3 5

(5) (1 cos36 )8 32

F−

= − =

61. Answer (4)

Hint: cos3 = 4cos3 – 3cos

Sol.: 10 10

3

0 0

1cos 3cos cos( )

3 4 3r r

r rr

= =

= +

10 11cos sin

3 16 6

4 4sin

6

= +

1

8

−=



9/10

62. Answer (2)

Hint: 10cot 45 10 =

Sol.: 10cot 45 10 =

63. Answer (4)

Hint: tan tanA A = .

Sol.: tanA tanC = 2 and tanB tanC = 18

Let tanC = x

2 18

tan and tanA Bx x

= =

tanA + tanB + tanC = tanA tanB × tanC

x = 4 = tanC

2

2

1 tan 1 16 15cos2

1 17 171 tan

CC

C

− − −= = =

++

64. Answer (2)

Hint: 21 cos 2cos2

xx+ =

21 cos 2sin2

xx− =

Sol.: cos sin

1 cos 1 cos 2 2

1 cos 1 coscos sin

2 2

++ + −

=+ − −

−

x x

x x

x xx x

( )tan cot2 4 4 2

x x = − − = +

65. Answer (3)

Hint: Simplify.

Sol.: 2 2 4 2 2

2 2 2

4sin cos 4sin 4sin cos 1

94 4sin 4sin cos

x x x x x

x x x

+ −=

− −

4

4

sin 1 1tan

9 6cos 3

= = =

xx x n

x

66. Answer (2)

Hint: Replace r → n – r.

Sol.: Let 1000

1

sin( )cos(1001 )r

S rx r x=

= − …(1)

Replace r → 1001 – r

1000

1

sin(1001 ) cos( )r

S r x rx=

= − …(2)

By (1) and (2)

2 1000 sin(1001 )S x=

500 sin(1001 )S x=

67. Answer (3)

Hint: Probability few cases

total case=

Sol.: Few cases = {(1, 1), (1, 2), (2, 1), (1, 4),

(4, 1), (2, 3), (3, 2), (3, 4), (4, 3), (1, 6),

(6, 1), (2, 5), (5, 2), (6, 5), (5, 6)}

15 5

Pr36 12

= =

68. Answer (2)

Hint: Modulus Property.

Sol.: For maximum value cos2x = 0

max value ( 3 1)= −

69. Answer (2)

Hint: P(A) + P(B) = P(A B) + P(A B)

Sol.: P(A B) = 0.2, P(A B) = 0.6

P(A) + P(B) = 0.8

( ) ( ) 2 0.8 1.2+ = − =P A P B

70. Answer (4)

Hint: Range of both sides.

Sol.: L.H.S. 5 and R.H.S. 5

equation holds when

2 29sec 4cosec 5y y+ =

2 2 1tan cos(2 )

3 5y y= =



10/10

71. Answer (00)

Hint: Compound angle formulas.

Sol.: sin( )cos( ) sin cos

sin( )cos( ) sin cos

+ − − +=

− + + −

A B C A B C C B

A B C A B C B C

Apply componendo and dividendo

( )sin 2( ) sin( ) sin(2 )sin( ) 0B C B C A B C − + + − =

sin( ) 2cos( )sin( ) sin2 0B C B C B C A − − + + =

sin2 sin2 sin2 0 ( )A B C B C + + =

72. Answer (04)

Hint: Cosine rule.

Sol.: 2 2

1 22 2cos( )a b+ = + +

minimum value = 0; maximum value = 4

73. Answer (02)

Hint: Cosine rule in cyclic quadrilateral.

Sol.:

In PQR PR2 = 4 + 25 – 20cos60° = 19

In PRS PR2 = 9 + x2 + 3x = 19

(x + 5) (x – 2) = 0 x = 2

74. Answer (03)

Hint: Distance of circumcenter from sides.

Sol.: Distance of circumcenter from side

AC = R cosB

Distance of orthocenter from side AC =

2R cosA cosC

2cos cos cos( )A C A C= − +

tan tan 3A C =

75. Answer (06)

Hint: A.M H.M

Sol.: a + b + c = 2s

Given expression is a b c

s a s b s c+ +

− − −

3s s s

s a s b s c= − + + +

− − −

A.M H.M

3

3

+ +− − −

− − −+ +

s s s

s a s b s cs a s b s c

s s s

9s s s

s a s b s c+ +

− − −

Minimum value of given expression is 6

Test - 6 (Code-C) (Answers) All India Aakash Test Series ... › Aakash › s3fs-public ›...

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Transcript of Test - 6 (Code-C) (Answers) All India Aakash Test Series ... › Aakash › s3fs-public ›...