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Test - 4 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 17/11/2019

ANSWERS

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

TEST - 4 - Code-A

1/11

PHYSICS CHEMISTRY MATHEMATICS

1. (4)

2. (1)

3. (2)

4. (2)

5. (1)

6. (4)

7. (2)

8. (1)

9. (3)

10. (1)

11. (3)

12. (1)

13. (2)

14. (3)

15. (4)

16. (1)

17. (3)

18. (1)

19. (2)

20. (4)

21. (16)

22. (10)

23. (14)

24. (07)

25. (88)

26. (3)

27. (3)

28. (1)

29. (3)

30. (1)

31. (2)

32. (3)

33. (2)

34. (3)

35. (2)

36. (3)

37. (3)

38. (1)

39. (2)

40. (3)

41. (4)

42. (1)

43. (2)

44. (2)

45. (1)

46. (02)

47. (09)

48. (02)

49. (12)

50. (20)

51. (4)

52. (2)

53. (1)

54. (2)

55. (4)

56. (4)

57. (2)

58. (1)

59. (2)

60. (3)

61. (1)

62. (4)

63. (2)

64. (4)

65. (1)

66. (3)

67. (1)

68. (4)

69. (4)

70. (2)

71. (24)

72. (01)

73. (00)

74. (22)

75. (04)

All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)


2/11

1. Answer (4)

Hint : A = λN

Sol. : 3λ1NA = λ2NB

1 2

3 2 1T T×

=

12 6

TT =

2. Answer (1)

Hint : hv = φ + eV0

Sol. : φ = 5.2 eV

E = 6.2 eV

V0 = 1 V

01

4ne

r=

πε

9 19

29 10 1.6 10 1

2 10n −

−

× × × ×=

×

10 210 2 109 1.6

n−× ×

=×

7 71.38 10 1.4 10= × ≈ ×

3. Answer (2)

Hint : 2

21 602

k emvd

=

Sol. : 2

2 120( ) ke mmvd

=

2120dhmke

λ =

4. Answer (2)

Hint : Fringes formed will be symmetric on screen

Sol. : Number of Fringes are 5d = λ .

5. Answer (1)

Hint : 2

1En

∝

Sol. : 2

1nE

n∝

1 2 21 1

( 1)n nE E kn n−

− = − −

31hc

n=

λ

3nλ ∝

6. Answer (4)

Hint : ( ) ( )Y A B A B= + ⋅ +

Sol. : ( ) ( ) 1Y A B A B= + + + =

7. Answer (2)

Hint : 24 cos2oI I φ

=

Sol. : 3π

φ =

1 6Dyd

λ=

2 6Dyd

λ= −

8. Answer (1)

Hint : Effective phase difference must be 2nπ

Sol. : 2 sec2

t r x λµ − =

2 tan sinx t r= θ

2 2 sin sincos cos 2

t t rr r

µ θ λ− =

22 sincos 2

tr

θ λµ − =

µ

2 2

2 2

2 [ sin ]2sin

tµ µ − θ λ⇒ =

µ µ − θ

2 24 sint λ

⇒ =µ − θ

PART - A (PHYSICS)

Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020


3/11

9. Answer (3)

Hint : For polarizing angle reflected and refracted rays are mutually perpendicular.

Sol. : sincos

p

p

θ= µ

θ

tan pθ = µ

1sin cθ =µ

tan sin 1p cθ θ =

10. Answer (1)

Hint : 00 00.7 tN N e−λ=

Sol. : 00.7 te−λ=

03(1 )tx e− λ= −

31 (0.7)x = −

1 0.49 0.7x = − ×

65.7%x =

11. Answer (3)

Hint : 2

2hc P

m= φ +

λ

Sol. : 2

21

35 2

hc hc hm

= +λ λ λ

2

1

25 2hc h

m=

λ λ

221

54

hhmc

λλ =

21

54

hmc

λλ =

11 52

hmc

λλ =

12. Answer (1)

Hint : 24PId

=π

Sol. : 24PId

=π

2Pressure4

I PL d c

= =π

13. Answer (2)

Hint : 22 21 2

1 1 1Rzn n

= − λ

Sol. : 22

1 1 1 164 16 3

RzRz

= − ⇒ λ = λ

2

least

1 1116

Rz = − λ

2

least

1 1516

Rz=

λ

least 216

15Rzλ =

14. Answer (3)

Hint : 1/2ln2t =λ

Sol. : 0tN N e−λ=

ln22

0N N eλ

−λ=

0

2N

N =

00

0

2 12Decayed2

NN

N

−−

= =

15. Answer (4)

Hint : duFdr

= −

Sol. : 2 2

3mv ke

r r=

2

2K.E.2ke

r=

2

22keU

r= −

Total energy = zero



4/11

16. Answer (1)

Hint : hp

λ =

Sol. : 1 2

fh hp = −λ λ

1 2

h h h= −

λ λ λ

1 2

2 1

λ λλ =

λ − λ

17. Answer (3)

Hint : E = (B.E)l – (B.E)R

Sol. : E = E1 – 2E2

18. Answer (1)

Hint : Apply KVL and KCL

Sol. : 210 4 1.2 0I− − =

2 5 mAI =

14 0I− =

1 4 mAI =

2 1 1mA= − =zI I I

19. Answer (2)

Hint and Sol. : max min

max min

13

v vm

v v−

= =+

20. Answer (4)

Hint : Reading = MSR + n × LC

Sol. : 0.5LC 0.005 mm100

= =

Reading 3 40 0.005= + ×

= 3.200 mm

21. Answer (16)

Hint : C

B

II

β =

Sol. : 0C C

b b i

I R VI R V

=

250b

iC

RV

R×

=×

Vi = 16 mV

22. Answer (10)

Hint : It will vary from fc – fm to fc + fm

Sol. : 2 mf f∆ = 10 kHz=

23. Answer (14)

Hint : For intensity to be 1 th4

223

n πφ = π ±

Sol. : 5 cos 53λ λ θ = λ −

145 cos3

λλ θ =

1 14cos15

− θ =

24. Answer (07)

Hint : Path difference will be 3λ corresponding to that point.

Sol. : 4 cos 3λ θ = λ

1 3cos4

− θ =

tan yD

θ =

73

y=

25. Answer (88)

Hint : Analyse the circuit for half of the cycle

Sol. : 2

01 1 42 2 2 11

VPR

= × +

2019

88VR

=



5/11

26. Answer (3)

Hint : A strong acid can produce weak acid.

Sol. : Acidic strength of carbolic acid (Phenol) is less than H2CO3 so, it not produce CO2.

27. Answer (3)

Hint : Hemiacetal can reduce Tollen’s reagent

Sol. : If A give positive iodoform test so it must contain

group and A does not reduce

Tollen’s reagent it means A not contain

or hemiacetal

28. Answer (1)

Hint :

Sol. :

29. Answer (3)

Hint : Carbylamine test is given by primary amines.

Sol. : Carbylamine test is not given by secondary amines.

30. Answer (1)

Hint : Gabriel phthalimide synthesis Sol. :

31. Answer (2)

Hint : A and D → α

Sol. : CH2OH at C–5 and OH at C–1 decide the α or β nomenclature.

CH2OH and OH same side → β-form

CH2OH and OH opposite side → α-form

32. Answer (3)

Hint : β–Keto acid show decarboxylation on heating.

Sol. :

33. Answer (2)

Hint : B not cleaved with HI

Sol. :

34. Answer (3)

Hint : 4-methylpent-3-en-2-one

Sol. :

35. Answer (2)

Hint : p-chlorophenol is most acidic

Sol. :

36. Answer (3)

Hint : Reaction is intermolecular, it is a Cannizzaro reaction.

PART - B (CHEMISTRY)



6/11

Sol . :

37. Answer (3)

Hint :

Sol. :

38. Answer (1)

Hint :

Sol. :

39. Answer (2)

Hint : B.P of isomeric amine 1°> 2° > 3°

Sol. : A → 1° amine

B → 2° amine

C → 3° amine

Due to less extent of H–Bonding in 3° amine,

B.P. of 3° amine is minimum.

40. Answer (3)

Hint : (iv), (v), (vi) are not Sandmeyer’s reaction

Sol. :

41. Answer (4)

Hint : As positive charge on carbonyl carbon decrease, rate of reaction decrease.

Sol. : sp2 ‘N’ in 3 membered ring and bridge head is not possible.

42. Answer (1)

Hint : When NH2 and OH are at axial position then form compound P.

Sol. :

43. Answer (2)

Hint : HNO3Glucose Saccharic acid→

Sol. : Aspartic acid



7/11

44. Answer (2)

Hint : Melamine and formaldehyde

Sol. :

45. Answer (1)

Hint : Teflon

Sol. :

46. Answer (02)

Hint : PCC oxidises alcohol to carbonyl group.

Sol. : MnO2 oxidises allylic alcohol.

47. Answer (09)

Hint : CH3MgBr react with acidic hydrogen

Sol. :

48. Answer (02)

Hint : Glyptal is used in manufacture of paints and lacquers.

Sol. : Polyester is a condensation polymer. Two statements are incorrect (ii) & (vii)

49. Answer (12)

Hint : β - D ribose has 4 chiral centre.

Sol. :

xy = 4 × 3 = 12

50. Answer (20)

Hint :

Sol. : Chloramphenicol has two –OH groups, two chiral carbon and zero –COOH.

5(2 + 0 + 2) = 20

51. Answer (4)

Hint : Find the d.r. of normal to the plane

Sol. : 1 2 1

1 2 3 02 1 5

x y z− + −=

−

13( 1) ( 2) 5( 1) 0x y z⇒ − + + − − =

13 5 6x y z⇒ + − =

52. Answer (2)

Hint : ( 2 ) 0a b c+ ⋅ =

Sol. : 3a b=

ˆ ˆ ˆ ˆ ˆ ˆ2 3 3 3 3p i j k i q j k⇒ + + = − +

23,3

p q⇒ = = −

ˆ ˆ ˆ( 2 ) ( 2) (2 2 ) (3 2)a b p i q j k+ = + + − + +

2ˆ ˆ ˆ5 53

i j k= + +

( 2 ) 0a b c+ ⋅ =

PART - C (MATHEMATICS)



8/11

139

r −⇒ =

2 13( , , ) 3, ,3 9− − =

p q r

53. Answer (1)

Hint : Use section formula

Sol. : A = (2, 4, 5), B = (–2, 0, 1)

2 2 4 5, ,1 1 1

P − λ + λ + λ + λ + λ +

33 2 2⇒ λ + = λ +

31⇒ λ =

⇒ 31 : 1

54. Answer (2)

Hint : Find the probability of complementary event

Sol. : Required probability

2 2 3 3 313 3 4 4 4

= − × × × =

55. Answer (4)

Hint : d.r. of line segment is same as that of normal to the plane

Sol. : Equation of plane

( 5)2 ( 0)6 ( 2)10 0x x y− + − + − =

3 5 0x y z⇒ + + =

56. Answer (4)

Hint : Proceed with parametric form of point

Sol. : Point Q on the line = (–5k – 9, k + 2, 2k + 5)

Q lies on plane 5 9 2 2 5 2k k k⇒ − − + + + + =

2k⇒ = −

(1, 0,1)Q∴ =

Distance, 2 2 25 1 2 30PQ = + + =

57. Answer (2)

Hint : variance = 22

i ix xN N

∑ ∑ −

Sol. : 5

120 5 100i

ix

== × =∑

6

1100 100 0i

ix

=⇒ = − =∑

52

21 (20) 165

ii

x= − =∑

5

2

12080i

ix

=⇒ =∑

6

2 2

12080 ( 100) 12080i

ix

=∴ = + − =∑

Variance

26 62

1 16 6

i ii i

x x= =

= −

∑ ∑

12080 60406 3

= =

58. Answer (1)

Hint : Make cases


5 5 5 5

2 2 2 35 7

2 3

( )C C C CC C

+ +=

×

10 10 10 635 7

+ += =

59. Answer (2)

Hint : | || | cosa b a b⋅ = θ

Sol. : | | , | |a a b b= =

ab(1 + cosθ) = 12

ab(1 – cosθ) = 6

1 cos 21 cos

+ θ⇒ =

− θ

1cos and 93

ab⇒ θ = =

| | | sin |a b ab∴ × = θ

∴ option (2) can be a b×



9/11

60. Answer (3)

Hint : Use the concept of family of planes

Sol. : Required plane is

( 3) (3 2 5) 0x y z x y z+ + − + λ − + + =

Parallel to xz plane 0(1 3 ) 1(1 2 ) 0(1 ) 0⇒ + λ + − λ + + λ =

12

⇒ λ =

∴ Plane:

1( 3) (3 2 5) 02

x y z x y z+ + − + − + + =

5 3 1 0x z⇒ + − =

61. Answer (1)

Hint : Find the d.r. of line.

Sol. : Any point on line is (–λ + 2, –3λ – 1, 2λ + 3) which also passes through (1, 1, –2)

∴ d.r. of line = (–λ + 1, –3λ – 2, 2λ + 5)

This line is parallel to the given plane.

⇒ (–λ + 1)(–2) + (–3λ – 2)(1) + (2λ + 5)2 = 0

⇒ 3λ + 6 = 0 ⇒ λ = –2

∴ d.r. = (3, 4, 1)

∴ Equation of line: 1 1 23 4 1

x y z− − += =

62. Answer (4)

Hint : Find the d.r. of lines

Sol. : First line is 10 32 1

x zy− −= =

− and second

line is 5 23 7

x y z− −= =

− as

2 ( 3) 1 7 ( 1) 1 0× − + × + − × =

⇒ L1 is perpendicular to L2

63. Answer (2) Hint : Break 5 into 3 non-zero parts

Sol. : 5 3 5 31 2( ) 3 (2 2)n E C C= − − −

5( ) 3n S =

( ) 50( ) 81

n EPn S

= =

64. Answer (4)

Hint : Use total probability theorem


( ) ( ) = ⋅ + ⋅ W WP W P P B PW B

3 4 2 3 18 35 6 5 6 30 5

= × + × = =

65. Answer (1)

Hint : Find the angle between ( )OA OB×

and

( )AB AC×

.

Sol. : 1n OA OB= ×

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 3 51 3 2

i j ki j k= = − − +

2n AB AC= ×

ˆ ˆ ˆˆ ˆ ˆ1 2 1 2 2 6

3 0 1

i j ki j k= − = − +

−

1 2

1 2cos

| || |n nn n

⋅θ =

2 6 30

1 9 25 4 4 36− + +

=+ + + +

34 17

35 44 385= =

66. Answer (3)

Hint : Use triangle rule of vector addition

Sol. :



10/11

AB BM AM+ =

…(1)

AC CM AM+ =

…(2)

(1) + (2)

⇒ 2AB BC AM+ =

67. Answer (1)

Hint : , ,a b c

do not make a triangle.

Sol. : 1 | | 7a b c≤ + + ≤

68. Answer (4)

Hint : P1 + λP2 = 0

Sol. : Intersection point of

1 2 31 2 3

x y z− − −= = and xy plane

is P(λ + 1, 2λ + 2, 3λ + 3) where 3λ + 3 = 0

⇒ P(0, 0, 0)

Plane passing through intersection of

P1 = 0 and P2 = 0 is

P1 + tP2 = 0

⇒ (1 + t)x + (1 – t)y + (1 + t)z + t – 1 = 0

Passes through P(0, 0, 0) ⇒ 1t =

⇒ required plane is 2x + 2z = 0

0x z⇒ + =

69. Answer (4)

Hint : 1 2n b b= ×

Sol. : ˆ ˆ ˆ

ˆ ˆ1 4 5 40 81 4 5

i j kn i k= = −

−

70. Answer (2)

Hint : Use Bayes theorem

Sol. : Let Bi be the number of black balls transferred (i = 0, 1, 2, 3). B is the event of drawing a black ball. Therefore,

0 15 30( ) , ( ) ,

70 70P B P B= =

2 330 5( ) , ( ) ,70 70

P B P B= =

Also 0 1

10, ,4

B BP PB B

= =

2 3

2 3, ,4 4

B BP PB B

= =

∴ By Bayes theorem

3

5 370 4

5 30 1 30 2 5 3070 70 4 70 4 70 4

BP

B

× =

× + × + × + ×

17

=

71. Answer (24)

Hint : Find P(x = 1) + P(x = 2)

Sol. : ( 2) ( 1) ( 2)P X P X P X≤ = = + =

12 40 12

1 1 252 52

2 2

C C CC C×

= +

6 91 751 26 17

×= =

×

72. Answer (01)

Hint : 22

2 i ix xN N

∑ ∑ σ = −

Sol. : 2(2 1) 21ix n+ =∑

4 4 29A n B n⇒ + + =

7A B n⇒ + =

Where 2

1 1,

n n

i ii i

A x B x= =

= =∑ ∑

2(3 1) 34ix n− =∑

9 6 34A n B n⇒ + − =

3 2 11A B n⇒ − =

5 , 2A n B n∴ = =

2

2 21 1i ix x

n n σ = − ∑ ∑

2

2 5 4 1A Bn n

= − = − =



11/11

73. Answer (00)

Hint : 0 (acute), 0 (obtuse)a b a b⋅ > ⋅ <

Sol. : x2 – x – x > 0 …(1)

–x < 0 …(2)

2

cos1 1

x

x

−θ =

+ + …(3)

( 2) 0, 0 2x x x x− > > ⇒ > …(a)

52 6π π

< θ <

50 cos cos6π

> θ >

2

3 02 2

x

x

− −⇒ < <

+

2

32 2

x

x

− −⇒ <

+

2

3 6 62 2

x xx

⇒ > ⇒ − < <+

…(b)

and2

0 {0}2

x x Rx

−< ⇒ ∈ −

+ …(c)

From (a), (b) and (c), 2 6x< <

74. Answer (22)

Hint : Make cases

Sol. : 5 6 5 3 6 3

1 1 1 1 1 114

2( ) C C C C C C

P EC

⋅ + ⋅ + ⋅=

(30 15 18) 63 913 14 13 7 13

2

+ += = =

⋅ ⋅

75. Answer (04)

Hint : 22 1 ( )ix xn

σ = −∑

Sol. : 1 21

... nx x xxn

+ + +=

2 2 2

2 1 1 1 2 1( ) ( ) ... ( )nx x x x x xn

− + − + + −σ =

1 22 1

... 4ny y yx xn

+ + += =

2 2 2

2 2 1 2 2 2( ) ( ) ... ( )( ) − + − + + −′σ = nx y x y x yn

2 2 2

1 1 1 2 1( ) ( ) ... ( )16 nx x x x x xn

− + − + + −=

2 216 4′ ′σ = σ ⇒ σ = σ

Test - 4 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 17/11/2019

ANSWERS


TEST - 4 - Code-B

1/11

PHYSICS CHEMISTRY MATHEMATICS

1. (4)

2. (2)

3. (1)

4. (3)

5. (1)

6. (4)

7. (3)

8. (2)

9. (1)

10. (3)

11. (1)

12. (3)

13. (1)

14. (2)

15. (4)

16. (1)

17. (2)

18. (2)

19. (1)

20. (4)

21. (88)

22. (07)

23. (14)

24. (10)

25. (16)

26. (1)

27. (2)

28. (2)

29. (1)

30. (4)

31. (3)

32. (2)

33. (1)

34. (3)

35. (3)

36. (2)

37. (3)

38. (2)

39. (3)

40. (2)

41. (1)

42. (3)

43. (1)

44. (3)

45. (3)

46. (20)

47. (12)

48. (02)

49. (09)

50. (02)

51. (2)

52. (4)

53. (4)

54. (1)

55. (3)

56. (1)

57. (4)

58. (2)

59. (4)

60. (1)

61. (3)

62. (2)

63. (1)

64. (2)

65. (4)

66. (4)

67. (2)

68. (1)

69. (2)

70. (4)

71. (04)

72. (22)

73. (00)

74. (01)

75. (24)

All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)


2/11

1. Answer (4)

Hint : Reading = MSR + n × LC

Sol. : 0.5LC 0.005 mm100

= =

Reading 3 40 0.005= + ×

= 3.200 mm

2. Answer (2)

Hint and Sol. : max min

max min

13

v vm

v v−

= =+

3. Answer (1)

Hint : Apply KVL and KCL

Sol. : 210 4 1.2 0I− − =

2 5 mAI =

14 0I− =

1 4 mAI =

2 1 1mA= − =zI I I

4. Answer (3)

Hint : E = (B.E)l – (B.E)R

Sol. : E = E1 – 2E2

5. Answer (1)

Hint : hp

λ =

Sol. : 1 2

fh hp = −λ λ

1 2

h h h= −

λ λ λ

1 2

2 1

λ λλ =

λ − λ

6. Answer (4)

Hint : duFdr

= −

Sol. : 2 2

3mv ke

r r=

2

2K.E.2ke

r=

2

22keU

r= −

Total energy = zero

7. Answer (3)

Hint : 1/2ln2t =λ

Sol. : 0tN N e−λ=

ln22

0N N eλ

−λ=

0

2N

N =

00

0

2 12Decayed2

NN

N

−−

= =

8. Answer (2)

Hint : 22 21 2

1 1 1Rzn n

= − λ

Sol. : 22

1 1 1 164 16 3

RzRz

= − ⇒ λ = λ

2

least

1 1116

Rz = − λ

2

least

1 1516

Rz=

λ

least 216

15Rzλ =

9. Answer (1)

Hint : 24PId

=π

Sol. : 24PId

=π

2Pressure4

I PL d c

= =π

PART - A (PHYSICS)

Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020


3/11

10. Answer (3)

Hint : 2

2hc P

m= φ +

λ

Sol. : 2

21

35 2

hc hc hm

= +λ λ λ

2

1

25 2hc h

m=

λ λ

221

54

hhmc

λλ =

21

54

hmc

λλ =

11 52

hmc

λλ =

11. Answer (1)

Hint : 00 00.7 tN N e−λ=

Sol. : 00.7 te−λ=

03(1 )tx e− λ= −

31 (0.7)x = −

1 0.49 0.7x = − ×

65.7%x =

12. Answer (3)

Hint : For polarizing angle reflected and refracted rays are mutually perpendicular.

Sol. : sincos

p

p

θ= µ

θ

tan pθ = µ

1sin cθ =µ

tan sin 1p cθ θ =

13. Answer (1)

Hint : Effective phase difference must be 2nπ

Sol. : 2 sec2

t r x λµ − =

2 tan sinx t r= θ

2 2 sin sincos cos 2

t t rr r

µ θ λ− =

22 sincos 2

tr

θ λµ − =

µ

2 2

2 2

2 [ sin ]2sin

tµ µ − θ λ⇒ =

µ µ − θ

2 24 sint λ

⇒ =µ − θ

14. Answer (2)

Hint : 24 cos2oI I φ

=

Sol. : 3π

φ =

1 6Dyd

λ=

2 6Dyd

λ= −

15. Answer (4)

Hint : ( ) ( )Y A B A B= + ⋅ +

Sol. : ( ) ( ) 1Y A B A B= + + + =

16. Answer (1)

Hint : 2

1En

∝

Sol. : 2

1nE

n∝

1 2 21 1

( 1)n nE E kn n−

− = − −

31hc

n=

λ

3nλ ∝



4/11

17. Answer (2)

Hint : Fringes formed will be symmetric on screen

Sol. : Number of Fringes are 5d = λ .

18. Answer (2)

Hint : 2

21 602

k emvd

=

Sol. : 2

2 120( ) ke mmvd

=

2120dhmke

λ =

19. Answer (1)

Hint : hv = φ + eV0

Sol. : φ = 5.2 eV

E = 6.2 eV

V0 = 1 V

01

4ne

r=

πε

9 19

29 10 1.6 10 1

2 10n −

−

× × × ×=

×

10 210 2 109 1.6

n−× ×

=×

7 71.38 10 1.4 10= × ≈ ×

20. Answer (4)

Hint : A = λN

Sol. : 3λ1NA = λ2NB

1 2

3 2 1T T×

=

12 6

TT =

21. Answer (88)

Hint : Analyse the circuit for half of the cycle

Sol. : 2

01 1 42 2 2 11

VPR

= × +

2019

88VR

=

22. Answer (07)

Hint : Path difference will be 3λ corresponding to that point.

Sol. : 4 cos 3λ θ = λ

1 3cos4

− θ =

tan yD

θ =

73

y=

23. Answer (14)

Hint : For intensity to be 1 th4

223

n πφ = π ±

Sol. : 5 cos 53λ λ θ = λ −

145 cos3

λλ θ =

1 14cos15

− θ =

24. Answer (10)

Hint : It will vary from fc – fm to fc + fm

Sol. : 2 mf f∆ = 10 kHz=

25. Answer (16)

Hint : C

B

II

β =

Sol. : 0C C

b b i

I R VI R V

=

250b

iC

RV

R×

=×

Vi = 16 mV



5/11

26. Answer (1)

Hint : Teflon

Sol. :

27. Answer (2)

Hint : Melamine and formaldehyde

Sol. :

28. Answer (2)

Hint : HNO3Glucose Saccharic acid→

Sol. : Aspartic acid

29. Answer (1)

Hint : When NH2 and OH are at axial position then form compound P.

Sol. :

30. Answer (4)

Hint : As positive charge on carbonyl carbon decrease, rate of reaction decrease.

Sol. : sp2 ‘N’ in 3 membered ring and bridge head is not possible.

31. Answer (3)

Hint : (iv), (v), (vi) are not Sandmeyer’s reaction

Sol. :

32. Answer (2)

Hint : B.P of isomeric amine 1°> 2° > 3°

Sol. : A → 1° amine

B → 2° amine

C → 3° amine

Due to less extent of H–Bonding in 3° amine,

B.P. of 3° amine is minimum.

33. Answer (1)

Hint :

Sol. :

PART - B (CHEMISTRY)



6/11

34. Answer (3)

Hint :

Sol. :

35. Answer (3)

Hint : Reaction is intermolecular, it is a Cannizzaro reaction.

Sol . :

36. Answer (2)

Hint : p-chlorophenol is most acidic

Sol. :

37. Answer (3)

Hint : 4-methylpent-3-en-2-one

Sol. :

38. Answer (2)

Hint : B not cleaved with HI

Sol. :

39. Answer (3)

Hint : β–Keto acid show decarboxylation on heating.

Sol. :

40. Answer (2)

Hint : A and D → α

Sol. : CH2OH at C–5 and OH at C–1 decide the α or β nomenclature.

CH2OH and OH same side → β-form

CH2OH and OH opposite side → α-form

41. Answer (1)

Hint : Gabriel phthalimide synthesis

Sol. :

42. Answer (3)

Hint : Carbylamine test is given by primary amines.

Sol. : Carbylamine test is not given by secondary amines.

43. Answer (1)

Hint :



7/11

Sol. :

44. Answer (3)

Hint : Hemiacetal can reduce Tollen’s reagent

Sol. : If A give positive iodoform test so it must contain

group and A does not reduce

Tollen’s reagent it means A not contain

or hemiacetal

45. Answer (3)

Hint : A strong acid can produce weak acid.

Sol. : Acidic strength of carbolic acid (Phenol) is less than H2CO3 so, it not produce CO2.

46. Answer (20)

Hint :

Sol. : Chloramphenicol has two –OH groups, two chiral carbon and zero –COOH.

5(2 + 0 + 2) = 20

47. Answer (12)

Hint : β - D ribose has 4 chiral centre. Sol. :

xy = 4 × 3 = 12

48. Answer (02) Hint : Glyptal is used in manufacture of paints and lacquers. Sol. : Polyester is a condensation polymer. Two statements are incorrect (ii) & (vii)

49. Answer (09)

Hint : CH3MgBr react with acidic hydrogen

Sol. :

50. Answer (02)

Hint : PCC oxidises alcohol to carbonyl group.

Sol. : MnO2 oxidises allylic alcohol.

51. Answer (2)

Hint : Use Bayes theorem

Sol. : Let Bi be the number of black balls transferred (i = 0, 1, 2, 3). B is the event of drawing a black ball. Therefore,

0 15 30( ) , ( ) ,

70 70P B P B= =

2 330 5( ) , ( ) ,70 70

P B P B= =

Also 0 1

10, ,4

B BP PB B

= =

2 3

2 3, ,4 4

B BP PB B

= =

∴ By Bayes theorem

PART - C (MATHEMATICS)



8/11

3

5 370 4

5 30 1 30 2 5 3070 70 4 70 4 70 4

BP

B

× =

× + × + × + ×

17

=

52. Answer (4)

Hint : 1 2n b b= ×

Sol. : ˆ ˆ ˆ

ˆ ˆ1 4 5 40 81 4 5

i j kn i k= = −

−

53. Answer (4)

Hint : P1 + λP2 = 0

Sol. : Intersection point of

1 2 31 2 3

x y z− − −= = and xy plane

is P(λ + 1, 2λ + 2, 3λ + 3) where 3λ + 3 = 0

⇒ P(0, 0, 0)

Plane passing through intersection of

P1 = 0 and P2 = 0 is

P1 + tP2 = 0

⇒ (1 + t)x + (1 – t)y + (1 + t)z + t – 1 = 0

Passes through P(0, 0, 0) ⇒ 1t =

⇒ required plane is 2x + 2z = 0

0x z⇒ + =

54. Answer (1)

Hint : , ,a b c

do not make a triangle.

Sol. : 1 | | 7a b c≤ + + ≤

55. Answer (3)

Hint : Use triangle rule of vector addition

Sol. :

AB BM AM+ =

…(1)

AC CM AM+ =

…(2)

(1) + (2)

⇒ 2AB BC AM+ =

56. Answer (1)

Hint : Find the angle between ( )OA OB×

and

( )AB AC×

.

Sol. : 1n OA OB= ×

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 3 51 3 2

i j ki j k= = − − +

2n AB AC= ×

ˆ ˆ ˆ

ˆ ˆ ˆ1 2 1 2 2 63 0 1

i j ki j k= − = − +

−

1 2

1 2cos

| || |n nn n

⋅θ =

2 6 301 9 25 4 4 36

− + +=

+ + + +

34 1735 44 385

= =

57. Answer (4)

Hint : Use total probability theorem


( ) ( ) = ⋅ + ⋅ W WP W P P B PW B

3 4 2 3 18 35 6 5 6 30 5

= × + × = =



9/11

58. Answer (2) Hint : Break 5 into 3 non-zero parts

Sol. : 5 3 5 31 2( ) 3 (2 2)n E C C= − − −

5( ) 3n S =

( ) 50( ) 81

n EPn S

= =

59. Answer (4)

Hint : Find the d.r. of lines

Sol. : First line is 10 32 1

x zy− −= =

− and second

line is 5 23 7

x y z− −= =

− as

2 ( 3) 1 7 ( 1) 1 0× − + × + − × =

⇒ L1 is perpendicular to L2

60. Answer (1)

Hint : Find the d.r. of line.

Sol. : Any point on line is (–λ + 2, –3λ – 1, 2λ + 3) which also passes through (1, 1, –2)

∴ d.r. of line = (–λ + 1, –3λ – 2, 2λ + 5)

This line is parallel to the given plane.

⇒ (–λ + 1)(–2) + (–3λ – 2)(1) + (2λ + 5)2 = 0

⇒ 3λ + 6 = 0 ⇒ λ = –2

∴ d.r. = (3, 4, 1)

∴ Equation of line: 1 1 23 4 1

x y z− − += =

61. Answer (3)

Hint : Use the concept of family of planes

Sol. : Required plane is

( 3) (3 2 5) 0x y z x y z+ + − + λ − + + =

Parallel to xz plane 0(1 3 ) 1(1 2 ) 0(1 ) 0⇒ + λ + − λ + + λ =

12

⇒ λ =

∴ Plane:

1( 3) (3 2 5) 02

x y z x y z+ + − + − + + =

5 3 1 0x z⇒ + − =

62. Answer (2)

Hint : | || | cosa b a b⋅ = θ

Sol. : | | , | |a a b b= =

ab(1 + cosθ) = 12

ab(1 – cosθ) = 6

1 cos 21 cos

+ θ⇒ =

− θ

1cos and 93

ab⇒ θ = =

| | | sin |a b ab∴ × = θ

∴ option (2) can be a b×

63. Answer (1)

Hint : Make cases


5 5 5 5

2 2 2 35 7

2 3

( )C C C CC C

+ +=

×

10 10 10 635 7

+ += =

64. Answer (2)

Hint : variance = 22

i ix xN N

∑ ∑ −

Sol. : 5

120 5 100i

ix

== × =∑

6

1100 100 0i

ix

=⇒ = − =∑

52

21 (20) 165

ii

x= − =∑

5

2

12080i

ix

=⇒ =∑

6

2 2

12080 ( 100) 12080i

ix

=∴ = + − =∑



10/11

Variance

26 62

1 16 6

i ii i

x x= =

= −

∑ ∑

12080 60406 3

= =

65. Answer (4)

Hint : Proceed with parametric form of point

Sol. : Point Q on the line = (–5k – 9, k + 2, 2k + 5)

Q lies on plane 5 9 2 2 5 2k k k⇒ − − + + + + =

2k⇒ = −

(1, 0,1)Q∴ =

Distance, 2 2 25 1 2 30PQ = + + =

66. Answer (4)

Hint : d.r. of line segment is same as that of normal to the plane

Sol. : Equation of plane

( 5)2 ( 0)6 ( 2)10 0x x y− + − + − =

3 5 0x y z⇒ + + =

67. Answer (2)

Hint : Find the probability of complementary event


2 2 3 3 313 3 4 4 4

= − × × × =

68. Answer (1)

Hint : Use section formula

Sol. : A = (2, 4, 5), B = (–2, 0, 1)

2 2 4 5, ,1 1 1

P − λ + λ + λ + λ + λ +

33 2 2⇒ λ + = λ +

31⇒ λ =

⇒ 31 : 1

69. Answer (2)

Hint : ( 2 ) 0a b c+ ⋅ =

Sol. : 3a b=

ˆ ˆ ˆ ˆ ˆ ˆ2 3 3 3 3p i j k i q j k⇒ + + = − +

23,3

p q⇒ = = −

ˆ ˆ ˆ( 2 ) ( 2) (2 2 ) (3 2)a b p i q j k+ = + + − + +

2ˆ ˆ ˆ5 53

i j k= + +

( 2 ) 0a b c+ ⋅ =

139

r −⇒ =

2 13( , , ) 3, ,3 9

p q r − − =

70. Answer (4)

Hint : Find the d.r. of normal to the plane

Sol. : 1 2 1

1 2 3 02 1 5

x y z− + −=

−

13( 1) ( 2) 5( 1) 0x y z⇒ − + + − − =

13 5 6x y z⇒ + − =

71. Answer (04)

Hint : 22 1 ( )ix xn

σ = −∑

Sol. : 1 21

... nx x xxn

+ + +=

2 2 2

2 1 1 1 2 1( ) ( ) ... ( )nx x x x x xn

− + − + + −σ =

1 22 1

... 4ny y yx xn

+ + += =

2 2 2

2 2 1 2 2 2( ) ( ) ... ( )( ) − + − + + −′σ = nx y x y x yn

2 2 2

1 1 1 2 1( ) ( ) ... ( )16 nx x x x x xn

− + − + + −=

2 216 4′ ′σ = σ ⇒ σ = σ



11/11

72. Answer (22)

Hint : Make cases

Sol. : 5 6 5 3 6 3

1 1 1 1 1 114

2( ) C C C C C C

P EC

⋅ + ⋅ + ⋅=

(30 15 18) 63 913 14 13 7 13

2

+ += = =

⋅ ⋅

73. Answer (00)

Hint : 0 (acute), 0 (obtuse)a b a b⋅ > ⋅ <

Sol. : x2 – x – x > 0 …(1)

–x < 0 …(2)

2

cos1 1

x

x

−θ =

+ + …(3)

( 2) 0, 0 2x x x x− > > ⇒ > …(a)

52 6π π

< θ <

50 cos cos6π

> θ >

2

3 02 2

x

x

− −⇒ < <

+

2

32 2

x

x

− −⇒ <

+

2

3 6 62 2

x xx

⇒ > ⇒ − < <+

…(b)

and2

0 {0}2

x x Rx

−< ⇒ ∈ −

+ …(c)

From (a), (b) and (c), 2 6x< <

74. Answer (01)

Hint : 22

2 i ix xN N

∑ ∑ σ = −

Sol. : 2(2 1) 21ix n+ =∑

4 4 29A n B n⇒ + + =

7A B n⇒ + =

Where 2

1 1,

n n

i ii i

A x B x= =

= =∑ ∑

2(3 1) 34ix n− =∑

9 6 34A n B n⇒ + − =

3 2 11A B n⇒ − =

5 , 2A n B n∴ = =

2

2 21 1i ix x

n n σ = − ∑ ∑

2

2 5 4 1A Bn n

= − = − =

75. Answer (24)

Hint : Find P(x = 1) + P(x = 2)

Sol. : ( 2) ( 1) ( 2)P X P X P X≤ = = + =

12 40 12

1 1 252 52

2 2

C C CC C×

= +

6 91 751 26 17

×= =

×

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