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Test - 3 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/16

1. (2)

2. (4)

3. (2)

4. (2)

5. (4)

6. (3)

7. (3)

8. (2)

9. (3)

10. (3)

11. (4)

12. (3)

13. (2)

14. (3)

15. (4)

16. (3)

17. (4)

18. (4)

19. (3)

20. (2)

21. (3)

22. (1)

23. (2)

24. (3)

25. (2)

26. (2)

27. (4)

28. (3)

29. (3)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (2)

32. (1)

33. (1)

34. (2)

35. (3)

36. (2)

37. (3)

38. (4)

39. (4)

40. (2)

41. (4)

42. (1)

43. (1)

44. (4)

45. (2)

46. (3)

47. (4)

48. (2)

49. (2)

50. (1)

51. (3)

52. (2)

53. (4)

54. (4)

55. (4)

56. (1)

57. (3)

58. (4)

59. (3)

60. (2)

61. (2)

62. (3)

63. (1)

64. (2)

65. (4)

66. (2)

67. (2)

68. (3)

69. (4)

70. (1)

71. (3)

72. (3)

73. (4)

74. (2)

75. (2)

76. (3)

77. (4)

78. (2)

79. (1)

80. (4)

81. (2)

82. (2)

83. (3)

84. (2)

85. (1)

86. (2)

87. (4)

88. (3)

89. (4)

90. (1)

Test Date : 18/11/2018

ANSWERS

TEST - 3 - Code-A

All India Aakash Test Series for JEE (Main)-2019

All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)

2/16

1. Answer (2)

Hint : Pressure due to radiation = I

C where I is

intensity

Sol. :0 0

2 sin,

sin

av

I A AF S

c

P =

22 sin av

F I

S c

2. Answer (4)

Hint : KE can vary from 0 to Kmax

.

Sol. : KE of ejected photoelectrons varies from 0 to

Kmax

and its variation is non-linear

3. Answer (2)

Hint : Cut-off wavelength is maximum wavelength

which can produce photoelectrons

Sol. :hc

= 0

0

hceV

...(i)

hcn

=

2

0

0

hcn eV

...(ii)

0 =

1n

n

⎛ ⎞⎜ ⎟⎝ ⎠

4. Answer (2)

Hint : Use equation of photoelectric effect

Sol. :1240

500 = +

1

2mv

1

2 ...(i)

1240

620 = +

1

2mv

2

2 ...(ii)

= 1.6 eV

5. Answer (4)

Hint : Revolving electron can be equivalent to a

current carrying loop.

Sol. : 2

c

vB

r

r n2

1v

n

PART - A (PHYSICS)

6. Answer (3)

Hint : For inelastic collision to occur, loss of energy

should be equal to energy difference of two

Bohr's orbits.

Sol. : (Kloss

)max

= 2

0

1 4

2 5

m mv

m

= 4( )

5i

K

For inelastic collision,

4( )

5i

K 10.2 4

Ki 51 eV

7. Answer (3)


Sol. :

1

hc

= + K

2

hc

= + 3 K

2 >

1

3

8. Answer (2)

Hint : Use equation 2 2

1 2

1 1 1R

n n

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Sol. : Series is Paschen

2 2

min

1 1 1

3R⎛ ⎞ ⎜ ⎟ ⎝ ⎠

min

= 820 nm.

9. Answer (3)

Hint : In -decay mass decreases by 4 a.m.u and

Z by 2.

Sol. : No. of -particles = 238 206

4

= 8

No. of -particles = 8 2 = 16

Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/16

10. Answer (3)

Hint : Q-value of a reaction.

Sol. : 64Cu decay gives +ve Q-value

11. Answer (4)

Hint : Higher the half life, slower will be decay rate.

Sol. : Higher the half life, slower will be decay rate

12. Answer (3)

Hint : Angular momentum is integral multiple of 2

h

and

2

2Z

En

Sol. : � = 2

hn

�

Be > �

Li

and E

2

2

Z

n |E

Li| > |E

Be|

13. Answer (2)

Hint : Q-value = Kinetic energy produced

Sol. : KE =

2 2

2

1

2 2

p h

m m

⎛ ⎞ ⎜ ⎟

⎝ ⎠

and Q = (KE )4

A

A

=

2

2

232

228 2

h

m

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

=

2

2

29

57

h

m

14. Answer (3)

Hint : Modulation index = m

c

A

A

Sol. :

c m

c m

A A

A A =

x

y

c

m

A

A =

x y

x y

m

c

A

A =

x y

x y

15. Answer (4)

Hint : Power gain = 2 ⎛ ⎞ ⎜ ⎟

⎝ ⎠

L

B

R

R

Sol. : Power gain = 2 ⎛ ⎞ ⎜ ⎟

⎝ ⎠

L

B

R

R

200 = 2 2

101

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Vin =

4

20 = 0.2 V

16. Answer (3)

Hint : NAND gates can be used to form OR gate

Sol. : Given logic gate represents OR gate

17. Answer (4)

Hint : Use 1 1 1 u v f

Sol. : v1 =

15 1030 cm

(15 10)

v2 =

25 10 50cm

(25 10) 3

� = 50 40

30 cm3 3

18. Answer (4)

Hint : Refraction through curved surface

2 1 2 1 v u R

Sol. :1.2 (1.5 1.8) (1.2 1.5)

20 ( 20)f

f =

19. Answer (3)

Hint : Use image formation concept by ray diagram

Sol. :1 1 1

22.5 22.5 10x x

x = 7.5 cm


4/16

20. Answer (2)

Hint : Wavelength in a medium = 0

µ

Sol. : 1 sin 60° = μ sin(45°)

μ = 3

2

= 0 450 2

367 nmµ 3

21. Answer (3)

Hint : For maximum magnification, the image should

be formed at least distance of clear vision.

Sol. : v = –25 cm

1 1 1

( 25) 10u

u = 50

cm7

22. Answer (1)

Hint : Interference can be analysed by vector algebra.

Sol. : By vector representation:

2E0

3E0

E0

Er =

02 E

as I (Er)2

23. Answer (2)

Hint : I = Imax

2

cos2

⎛ ⎞⎜ ⎟⎝ ⎠

and 2

( )

x

Sol. : I1 =

0

0

4

4

II

I0 =

2

0(4 )cos

2I

⎛ ⎞⎜ ⎟⎝ ⎠

= 2

3

2

3

⎛ ⎞⎜ ⎟⎝ ⎠

= 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

y d

D

y = 1

0.50 mm3

⎛ ⎞ ⎜ ⎟⎝ ⎠

D

d

24. Answer (3)

Hint : Number of minima = No. of maxima

Sol. :

No. of minima = 5 4 = 20

25. Answer (2)

Hint : Length of tube = f0 + f

e

Sol. : � = f0 + f

e, m =

0

e

f

f

10 = 20

2 cme

e

ff

⇒

� = 20 + 2 = 22 cm.

26. Answer (2)

Hint : Consider three lenses are in contact

Sol. : R = 20 cm

2

1 2 3(1.3 1)

20 100f

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 1 350 cm

20 100e

e

ff ⇒

27. Answer (4)

Hint : If two mirrors are at 90°, then ray reflects back

antiparallel.

Sol. : Ray becomes anti-parallel if and only if = 90°.

28. Answer (3)

Hint : Use

sin2µ

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

mA

A and i + i = + A

Sol. :

sin2µ

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

mA

A

μ = 2cos2

A⎛ ⎞⎜ ⎟⎝ ⎠

cos2

A⎛ ⎞⎜ ⎟⎝ ⎠

= µ

1 µ 22 ⇒

and, + A = i + i < 180° (


5/16

PART - B (CHEMISTRY)

A < 90°

μ > 2 cos90

2

⎛ ⎞⎜ ⎟⎝ ⎠

μ > 2

29. Answer (3)

Hint : Use fundamental concept to calculate zero error.

Sol. : 1 VSD = 9

mm10

Zero error = 6 (1 mm) – 6 9

mm10

⎛ ⎞⎜ ⎟⎝ ⎠

= +0.6 mm

31. Answer (2)

Hint : A =

CH3

C =

H C5 2

C

H

C C

CH3

H

Sol. :

CH3 CH3

H / t2P

(A) (B)

(C) (D)

H C5 2

C

H

C C

CH3

H

H / t2 P

CH —(CH ) —CH3 2 4 3

32. Answer (1)

Hint : (CH3)3CBr is 3° while other are 1° or 2° halide

Sol. : 3° halide show SN1 while other show S

N2

reaction.

33. Answer (1)

Hint : Ease of SN2 decrease as hindrance

increases.

Sol. : Correct order of ease of SN2 is IV > II > I > III

34. Answer (2)

Hint : When —NH2 is present on equatorial then

ring contract to five membered.

Sol. :

OH

NH2

HNO2

CHO

OH

NH2

NaNO2

C — H

O

HCl

35. Answer (3)

Hint : Same M.F but different functional group

Sol. : Aldehydes and ketones are functional isomers

36. Answer (2)

Hint : OH– is a good nucleophile whereas H2O is a

poor nucleophile

Sol. :

+Cl

H O2

S TypeN1

OH

SN2

OH

(A)

(B)

–

::

OH

(A) and (B) are positional isomers.

30. Answer (3)

Hint : Meter Bridge works on Wheatstone

balancing

Sol. :40 60

P Q

and

50

50

40 10 (60 10)

⎛ ⎞⎜ ⎟⎝ ⎠

Q

P Q

P = 50 50

,3 2

Q


6/16

37. Answer (3)

Hint : Boiling point of isomeric alcohol varies as

1° > 2° > 3°.

Sol. : Alcohols have higher BP as compared to

ethers because of H bonding.

38. Answer (4)

Hint : Fact Based.

Sol. : Antibiotics have either cidal effect or a static

effect on microbes on this basis drugs are

divided into Bactericidal or Bacteriostatic.

39. Answer (4)

Hint : End Product of reaction (4) is phenol

Sol. :

OH

NO2

Fe

HCl

OH

NH2

OH

N Cl2

OH

H

NaNO2

HCl

H PO3 2

(Phenol)

+

40. Answer (2)

Hint : The reaction is SN1 type

Sol. :

CH — CH — CH — CH3 2 3

CH 3

CH — CH — CH — CH3 3

CH 3

Cl

Cl2

h

OH (alc)

PhCO H3

CH 3

CH 3

C CH — CH3

CH 3

CH 3

C CH

3

H O

H/ H O2

::

18

(S Type)N1

CH — C — CH — CH3 3

CH3

OHOH18

41. Answer (4)

Hint : Aldehydes are more reactive than ketones as

ketones are hindered.

Sol. : The attack happens by lone pair over sulphur.

The rate of addition is much greater for

aldehydes and can be used for purification/

separation bisulphite addition can be

reversed by acidic as well as basic

hydrolysis.

42. Answer (1)

Hint : Aldol condensation

Sol. :

C

43. Answer (1)

Hint : P Sucrose

R and S Glucose and fructose

Sol. : Sucrose is formed by formation of glycosidic

linkage between C1 of -D-glucose and C

2

and -D-fructose


7/16

44. Answer (4)

Hint : Carbene is neutral

Sol. : Amine on reaction with HNO2 form diazonium

salt

45. Answer (2)

Hint : hydroxy acid with H+/ gives , unsaturated acid

Sol. :

CH — CH — CH —CH — COOH3 2 2

CH — CH — CH = CH — COOH3 2

OH H +,

46. Answer (3)

Hint : Nitrene rearrangement

Sol. :

CH — CH3

CH2

HF, 0°C

CH3

CH

CH3

KMnO , H , 4

+

COOH

PCl5

C — Cl

O

N 3 OH , ,

(Curtius rearrangement)

NH2

47. Answer (4)

Hint : H of esters are acidic

Sol. :

CH — C — OEt3

O

CH — C — CH — C — OEt3 2

O O

EtONa

H(Claisen condensation)

CH – I3

EtO Na

CH — C — CH — C — OEt3 2

O O

CH — C — CH — C — OEt3

O O

—

CH3

CH — C — CH — CH3 2 3

O

H /+

48. Answer (2)

Hint : Ammoniolysis of amine

Sol. :

+

49. Answer (2)

Hint : The NaOBr formed will oxidise the Benzylic

Alcohol also

Sol. :

C

Br2

O

CH 2 NH

2

OH

NaOH

OHC NH2

50. Answer (1)

Hint : —CH3

C— C—Cl

——

CH3

CH3

— —

O

AlCl3 —CH3

C— C (+)

——

CH3

CH3

— —

O

—CH3

C (+)

——

CH3

CH3


8/16

Sol. :

+ —CH3

C— C— Cl

——

CH3

CH3

— —

O

AlCl3

—

C—

——

CH3

CH3

CH3

51. Answer (3)

Hint : Rate of reaction will depend upon electrophilic

nature of the electrophile

Sol. : NO2 is EWG whereas OCH

3 is EDG so III is

the most reactive and I is the least

52. Answer (2)

Hint : Cu is used in Gatterman’s reaction

Sol. :

Cu

HCl

N2

Cl

53. Answer (4)

Hint : Novolac is used to make paints

Sol. : Phenol + formaldehyde forms Novolac in

presence of acid or base catalyst which

undergoes cross linking on further heating

with HCHO to form Bakelite.

54. Answer (4)

Hint : Neoprene is a polymer of chloroprene

Sol. : Terylene is polyester, Buna–S is a polymer of

butadiene and styrene.

55. Answer (4)

Hint : Phenol > alcohol > alkyne

Sol. : Phenoxide is resonance stabilised.

56. Answer (1)

Hint : a

1pK

acidic strength

Sol. :

OH

> CH OH3 > H O2 >CH CH CH OH3 2 2

Acidic strength

57. Answer (3)

Hint : On the basis of product stability

Sol. : III leads to formation of benzene.

58. Answer (4)

Hint : POCl3 remove H

2O by -elimination

Sol. : In -elimination H and OH must be present

on axial.

59. Answer (3)

Hint : DNA has double strand helix whereas RNA

has single strand.

Sol. :

Base + sugar Nucleoside

Nucleoside + phosphoric acid Nucleotide

RNA does not contain thymine.

60. Answer (2)

Hint : Methyl ketone on reaction with NaOH/Br2 give

haloform reaction while ester give carboxylic

acid.

Sol. :


9/16

PART - C (MATHEMATICS)

61. Answer (2)

Hint : Use condition for coplanarities.

Sol. : Since vectors are coplanar,

So, 0 1 1 0

4

a a c

c c b

4ab = c2

c2 – 4ab = 0 ...(i)

And

Discriminant of given equation is

= c2 – 4ab = 0, [from (i)]

Roots are real and equal

62. Answer (3)

Hint : If a�

and

�

b are non-collinear vectors such that

0 � �

xa yb x = 0, y = 0

Sol. : Here, (1 – tan)

�

a – 1 2 sin 0b � �

and a�

and b�

are non-collinear

1 = tan and 1 2 sin 0

4

or 5

4

and

5

4

or 7

4

Most general value of 5

24

n , n I

63. Answer (1)

Hint : Use concept of coplanarity of vectors

Sol. : Since given vectors are coplanar

So, 61�

p + 40�

q + 21�

r = 0�

�

p , �

q , �

r are coplanar.

�

q × �

r and �

r × �

p are collinear

(�

q × �

r ) × {(�

r × �

p ) = 0�

64. Answer (2)

Hint : Use formula for 'dot' and cross product of two

vectors

Sol. : Let be the angle between b�

and c�

,

Then,

5b c � �

b c� �

sin = 5

1sin =

2

3

cos =2

Now, given,

3( ) 0 � �

� � �

a b a c ( 3 ) 0 � �

� �

a b c

3 ||�

� �

b c a

3 �

� �

b c a

2 2

9 – 6 ·b c b c� � � �

= 2

2a�

25 + 36 – 6 cosb c � �

= 2·4

2361– 6 5.2· 4

2

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

2 61– 30 3

4

65. Answer (4)

Hint : Use formula for cross product of three vectors

Sol. : From given relation

2 · – ·a c b a b c

� � � � � �

3b c � �

1

·2

a c � �

and 3

· –2

a b � �

3

cos –2

a b � �

5

cos cos – cos6 6

⎛ ⎞ ⎜ ⎟⎝ ⎠

angle between a�

and b�

is 5

6


10/16

66. Answer (2)

Hint :

Volume of parallelepiped

· · ·

· · ·

· ·

·

��

��

��

��

� � � � � �

a a a b a c

b a b b b c

c a c cc b

Sol. : Volume of parallelepiped

· · ·

· · ·

· ·

·

a a a b a c

V b a b b b c

c a c cc b

��

��

��

��

� � � � � �

4 3 4288 cu units

3 9 6

12 2 cu units4 6 16

67. Answer (2)

Hint : Mode = 3 Median – 2 Mean

Sol. : We have mode = 3 Median – 2 Mean

Mode – Median = 2(Median – Mean)

(Median – Mean) = 1Mode – Median

2

136 18

2

68. Answer (3)

Hint :sum of variables

Mean n

n

If n is odd, then median item

th1

2

n ⎛ ⎞ ⎜ ⎟⎝ ⎠

term

Sol. : Given,

3 3 3 31 2 3 ...

1900n

n

22

1

4·

n n

n

21 ·1900

4

n n

7600 = n(n + 1)2

19 × (20)2 = n(n + 1)2

n = 19

Median of 1, 2, 3 ..., 18, 19 is

th19 1

2

⎛ ⎞⎜ ⎟⎝ ⎠

item = 10

69. Answer (4)

Hint : 9!

2! 3!n S

7 6

3 3

6!– 5!

2! n E C C

Sol. : 9! 9 8 7 6 5 4 3!

3! 2! 3! 2!

9 8 7 6 5 2 9 8 7 6 10

n S

To find n(E)

Case I.

position for T is denoted by ×

× A × E × N × I × O × N ×

No two T's come together but N may or may not

Total words fomed 3

6! 7! 6 5 4!7

2! 3! 4! 2C

7 6 5 4 3 2 542 300 12600

2

Case. II

× A × E × NN × I × O ×

No two T's come together but two N come together

['×' is the position for T]

In this case total words formed

6

3

6 5 4 3!= × 5! =

3! 3!

C × 5 × 4 × 3 × 2 × 1

= 20 × 120 = 2400

n(E) = 12600 – 2400 = 10200

Required probability

n E

n S

10200

9 8 7 6 10

85 85

18 14 252

70. Answer (1)

Hint : Use

2

22

1 2

1 2

( )

( ) ( )

RP E P

EEP

R R RP E P P P E

E E

⎛ ⎞ ⎜ ⎟

⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


11/16

Sol. : Let

E1 the event when die A is used

E2 the event when die B is used

R when red face appears

1

3

5

n

RP

E

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

and

2

2

5

n

RP

E

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

,

P(E1) =

1

2, P(E

2) =

1

2

2

1 2

162 5

973 2 1

5 5 2

n

n n

EP

R

⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎡ ⎤⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

2 16

973 2

n

n n

n = 4

71. Answer (3)

Hint : Use Bayes Theorem

Sol. :

3

33

1 2 3

1 2 3

·

· · ·

BP B P

BBP

B B B BP B P P B P P B P

B B B

⎛ ⎞⎜ ⎟

⎛ ⎞ ⎝ ⎠⎜ ⎟⎛ ⎞⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

2

3

2 2 2

2 2 2

1 – 4 7

3 – 4 10

1 – 4 8 – 4 9 – 4 7

3 – 4 10 – 4 10 – 4 10

⎛ ⎞ ⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

a a

a aBP

B a a a a a a

a a a a a a

2

3

22

– 4 7 1 1 1–

3 33 – 4 8 – 2 4

B a aP

B a a a

⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

1 1 1–

3 12 4

72. Answer (3)

Hint : n(S) = 66, n(E) = 6

6 5

6 1( ) = 1 – 1 –

6 6P E

Sol. : n(S) = 66, n(E) = 6C1 × 1 × 1 × 1 × 1 × 1


6

61 –

6

5

5 5 5

1 6 –1 77751 –

6 6 6

73. Answer (4)

Hint :

5 2 4 2 3

5 1 3 2 1

5! 5!5!

2! 2! 2!

n S C C C C C

2 3

2 1

5!

2! 2!n E C C

Sol. : To find n(S)

Letters to given word are

K 2

O 1

L 1

A 2

T 1

Case I.

When all 5 letters are distinct

Required words formed = 5C5 × 5! = 1 × 120 = 120

Case II.

When 2 alike letters and 3 different letters are

taken

Required words formed 2 4

1 3

5!= × ×

2!C C

5 4 3 2!2 4 480

2!

Case III.

When 2 alike, 2 alike and 1 different letter are

taken


2 1

5!= × ×

2! 2!C C

5 4 3 2!1 3 90

2! 2

n(S) = 120 + 480 + 90 = 690

n(E) = total number of words formed with all

the repeated letters used = 90


90 3

690 23

n E

n S


12/16

74. Answer (2)

Hint : Probability

10 11 11 11 11

11

1 1 1 1 1 7

2 2 2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Sol. :

H or T (appearing)

10 times

H H....... H ��

��

10 101 1

1 1 1 1 12 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

T

H or T (appearing)

10 times

H H....... H ��

��

10 111 1 1

1 1 1 12 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

H or T (appearing)

10 times

T H.........HH��

��

10 111 1 1

1 1 1 12 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


10 11 11 11 11 111 1 1 1 1 1

2 2 2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

10 11 11

11

1 1 1 75 2 5

2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

75. Answer (2)

Hint : Required probability

6 5 4

6 9 9 6

n

n n

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

Sol. : Bag 1 Bag 2

6W 4W

nB 5B

16

6P W

n

2

4

9P W

16

nP B

n

2

5

9P B

P(one white, one black)

6 5 4

6 9 9 6

n

n n

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

5030 4

(given)9 6 99

n

n

15 2 25

6 11

n

n

150 + 25n = 165 + 22n

3n = 165 – 150 = 15

n = 5

76. Answer (3)

Hint :

P

A

5

B

O

a�

a bOP

a b

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

� �

��

� �

Sol. :

A

B C

Dˆ ˆ–i k

ˆ ˆ( )i j

ˆ ˆ( )k iˆ ˆ( )j kˆ ˆ( – )i j

ˆ ˆ ˆ ˆ ˆ ˆ ˆ– – 2 – –

2 2 2

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

�� i k i j i j kBD

1 64 1 1 3

2 2BD ��

Unit vector along BD

ˆ ˆ ˆ2 – –

6 i j k

77. Answer (4)

Hint : Volume of tetrahedron

= 1

3(Area of base) × height


13/16

Sol. : Volume of tetrahedron, is

D

h

Q

A

B

C

1

3V

⎛ ⎞ ⎜ ⎟⎝ ⎠

(Area of base) (Height)

1120 20

3h

120 3

20h

h = 18 units

78. Answer (2)

Hint : 2a + 3b + 10c = 0

2a – 3b – 7 = 0

where a, b, c are the d.r' s of the line of

intersection of planes

Sol. : Planes are

2x + 3y + 10z = 8 ... (i)

2x – 3y – 7z = 2 ... (ii)

Let a, b, c be the d.r's of the line of intersection

of planes (i) and (ii)

2a + 3b + 10c = 0

2a – 3b – 7c = 0

–

–21 30 –14 – 20 –6 – 6

a b c

9 34 –12

a b c i.e., a : b : c = 9 : 34 : –12

79. Answer (1)

Hint :

2 1–

n

a a b

S

b

��

�

Sol. : Here lines are parallel and they are

ˆ ˆ ˆ ˆ ˆ ˆ3 2 2 2r i j k i j k �

and

ˆ ˆ ˆ ˆ ˆ ˆ4 3 2 2r i j k i j k �

Distance between them

2 1–a a b

b

��

�

ˆ ˆ ˆ ˆ ˆ ˆ( – 2 ) (2 2 )

4 1 4

i j k i j k

�

�

2 1 2

1 –1 2 4 – 2 – 3

4 1 4 9

i j k

i j k

� �

� �

16 4 9 29

39

80. Answer (4)

Hint : Use concept of the distance of any point from

line parallel to any plane

Sol. :

P(5, 4, 3)

Q R

2 3 4

3 4 5

x y z

x – y + z =2 3 0

– – –

Coordinates of R are (3 + 2, 4 + 3, 5+ 4)

and are satisfying the equation x – 2y + 3z = 0

i,e, (3 + 2) – 8 – 6 + 15 + 12 = 0

10 = –8

= 4

–5

Coordinates of R are 2 1

– , – , 05 5

⎛ ⎞⎜ ⎟⎝ ⎠

Here equation of line PQ is

– 5 – 4 – 3

3 4 5

x y zsay

Coordinates of Q are (3 + 5, 4 + 4, 5 + 3)

and the point Q lies on the plane


14/16

x – 2y + 3z = 0

(3 + 5) – (8 + 8) + (15+ 9) = 0

10= –6 = –3

5

Coordinates of Q are 16 8

, , 05 5

⎛ ⎞⎜ ⎟⎝ ⎠

Required distance

= 9

5QR

81. Answer (2)

Hint : Take image of given point A(1, 2, 3) due to

plane then find the distance between image

point and the point B.

Sol. :

D

C

AB (5, 8 , 11) (1, 2, 3)

(–1, –2, –3)

plane

x y z + 2 + 3 = 0

Image of A is D (x, y, z) and co-ordinates of D

can be obtained as

–2 1 4 9–1 – 2 – 3

1 2 3 1 4 9

x y z

–1 – 2 – 3

– 2

1 2 3

x y z

x – 1 = – 2, y – 2 = –4, z – 3 = –6

x = –1, y = –2, z = –3

D (–1, –2, – 3)

Minimum length of AC + CB

= 36 100 196 332BD

82. Answer (2)

Hint : Put x + y = v –1dy dx

dx dn

Sol. : From given equation,

cos ,dy

x ydx

Put, x y v

1dy dv

dx dx

From given equation, dv

dx – 1 = cosv

21 cos 2cos

2

dv vv

dx

21sec

2 2∫ ∫

vdv dx

tan2

vx c

tan2

x yx c

⎛ ⎞ ⎜ ⎟⎝ ⎠

83. Answer (3)

Hint : Solve by equation by method of inspection

Sol. : Here,

–

x ydy

dx x

xdy + y dx + xdx = 0

d(xy) + xdx = 0

2

2

xxy c ... (i)

If (i) passes through the point (2, 1), then

42

2c

c = 4

Put c = 4 in equation (i)

2xy + y2 = 2 × 4 = 8

84. Answer (2)

Hint : Differentiate solution and then use method of

separation of variables

Sol. : Differentiating both sides w.r.t.x

x(1 – x) f(x) + 0

1–

x

t f t xdt x f x∫


15/16

2

0

1–

x

t f t dt x f x∫

Again differentiating w.r.t. x we get

(1 – x)f(x) = 2xf(x) + x2f'(x)

2' 1– 3x f x x f x

2

' 1 3–

f x

f x xx

ln f(x) = 1

– – 3lnx c

x

(integrating both sides)

1ln ( ) = – – 3 ln f x x + c

x...(i)

Put x = 1 and f(1) = 1, in (i) we get

0 –1– 0 c 1c

1ln – – 3 ln 1f x x

x

3

1 1 11 – ln 1 –

3

1x x xf x e e

x

1

1 –2

12

8 8

ef e

85. Answer (1)

Hint : Convert equation into linear form

Sol. : From given differential equation, we have

2 3

1,

dy y

dx x x

1–

IF = xe

Solution of equation (i) is

1–

·

xy e =

1–x

e +

1–x

ec

x

Put x = 1 and y (1) = 1, we get c = –1

e

1

11 –

xey

x e

2

11 2 – 3 –

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

ey e

e

86. Answer (2)

Hint : Put x – 2 = X, y – 2 = Y

dY dy

dX dx and put Y = vX

Sol. : From given differential equation

2 2– 2 – 2

– 2 – 2

x ydy

dx x y

...(i)

Let x – 2 = X & y – 2 = Y dY dy

dX dx

From equation (i)

2 2dY X Y

dX XY

...(ii)

(homogeneous differential equation)

Put Y = VX, then dY dV

V XdX dX

Substituting these values in (ii) and then solving

we get,

2

22 In | |

YX c

X

2

2

– 2

– 2

y

x

= 2 ln |x – 2| + c

87. Answer (4)

Hint : Differentiate both sides w.r.t. x and eliminate

arbitrary constants

Sol. :

5

2y cx c , dyc

dx

5

2dy dyy x

dx dx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 5

–

dy dyy x

dx dx

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Degree of differential equation = 5

88. Answer (3)

Hint : Length of subsnormal dy

ydx


16/16

Sol. : Given length of subnormal dy

ydx

= k

dy

y kdx

2

2

ykx c

Which passes through the origin O(0, 0)

0 = 0 + c

c = 0

Curve is 22y kx

89. Answer (4)

Hint : Here 5a + 3b + 3c = 0 and 2a + 3b + 4c = 0

and solve for a, b, c

Sol. : DRs of normal to the plane containing lines

5 2 3

x y z and 2 3 4 x y z

are a, b, c, then

5a + 2b + 3c = 0 and 2a + 3b + 4c = 0 and

so d.r.'s of nomral to the plane are –1, –14, 11

Also plane contains the line 1 2 3

x y z

Required equation of plane will be given by

1 2 3 0

–1 –14 11

x y z

32x – 7y – 6z = 0

90. Answer (1)

Hint : Use condition for coplanarity for two line

Sol. : Given lines are

– 5

0 4 – –1

x y z

and –

0 –2 2 –

x y z are coplanar

5 – 0 0

0 4 – –1 0

0 –2 2 –

3 – 112 + 36 – 30 = 0

sum of all possible values of

1 2 3

– 11

11

1

� � �

Test - 3 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/16

1. (3)

2. (3)

3. (3)

4. (4)

5. (2)

6. (2)

7. (3)

8. (2)

9. (1)

10. (3)

11. (2)

12. (3)

13. (4)

14. (4)

15. (3)

16. (4)

17. (3)

18. (2)

19. (3)

20. (4)

21. (3)

22. (3)

23. (2)

24. (3)

25. (3)

26. (4)

27. (2)

28. (2)

29. (4)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (2)

32. (3)

33. (4)

34. (3)

35. (1)

36. (4)

37. (4)

38. (4)

39. (2)

40. (3)

41. (1)

42. (2)

43. (2)

44. (4)

45. (3)

46. (2)

47. (4)

48. (1)

49. (1)

50. (4)

51. (2)

52. (4)

53. (4)

54. (3)

55. (2)

56. (3)

57. (2)

58. (1)

59. (1)

60. (2)

61. (1)

62. (4)

63. (3)

64. (4)

65. (2)

66. (1)

67. (2)

68. (3)

69. (2)

70. (2)

71. (4)

72. (1)

73. (2)

74. (4)

75. (3)

76. (2)

77. (2)

78. (4)

79. (3)

80. (3)

81. (1)

82. (4)

83. (3)

84. (2)

85. (2)

86. (4)

87. (2)

88. (1)

89. (3)

90. (2)

Test Date : 18/11/2018

ANSWERS

TEST - 3 - Code-B

All India Aakash Test Series for JEE (Main)-2019

All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)

2/16

1. Answer (3)

Hint : Meter Bridge works on Wheatstone

balancing

Sol. :40 60

P Q

and

50

50

40 10 (60 10)

⎛ ⎞⎜ ⎟⎝ ⎠

Q

P Q

P = 50 50

,3 2

Q

2. Answer (3)

Hint : Use fundamental concept to calculate zero error.

Sol. : 1 VSD = 9

mm10

Zero error = 6 (1 mm) – 6 9

mm10

⎛ ⎞⎜ ⎟⎝ ⎠

= +0.6 mm

3. Answer (3)

Hint : Use

sin2µ

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

mA

A and i + i = + A

Sol. :

sin2µ

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

mA

A

μ = 2cos2

A⎛ ⎞⎜ ⎟⎝ ⎠

cos2

A⎛ ⎞⎜ ⎟⎝ ⎠

= µ

1 µ 22 ⇒

and, + A = i + i < 180° (

A < 90°

μ > 2 cos90

2

⎛ ⎞⎜ ⎟⎝ ⎠

μ > 2

PART - A (PHYSICS)

4. Answer (4)

Hint : If two mirrors are at 90°, then ray reflects back

antiparallel.

Sol. : Ray becomes anti-parallel if and only if = 90°.

5. Answer (2)

Hint : Consider three lenses are in contact

Sol. : R = 20 cm

2

1 2 3(1.3 1)

20 100f

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 1 350 cm

20 100e

e

ff ⇒

6. Answer (2)

Hint : Length of tube = f0 + f

e

Sol. : � = f0 + f

e, m =

0

e

f

f

10 = 20

2 cme

e

ff

⇒

� = 20 + 2 = 22 cm.

7. Answer (3)

Hint : Number of minima = No. of maxima

Sol. :

No. of minima = 5 4 = 20

8. Answer (2)

Hint : I = Imax

2

cos2

⎛ ⎞⎜ ⎟⎝ ⎠

and 2

( )

x

Sol. : I1 =

0

0

4

4

II

I0 =

2

0(4 )cos

2I

⎛ ⎞⎜ ⎟⎝ ⎠

= 2

3

2

3

⎛ ⎞⎜ ⎟⎝ ⎠

= 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

y d

D

y = 1

0.50 mm3

⎛ ⎞ ⎜ ⎟⎝ ⎠

D

d

Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/16

9. Answer (1)

Hint : Interference can be analysed by vector algebra.

Sol. : By vector representation:

2E0

3E0

E0

Er =

02 E

as I (Er)2

10. Answer (3)

Hint : For maximum magnification, the image should

be formed at least distance of clear vision.

Sol. : v = –25 cm

1 1 1

( 25) 10u

u = 50

cm7

11. Answer (2)

Hint : Wavelength in a medium = 0

µ

Sol. : 1 sin 60° = μ sin(45°)

μ = 3

2

= 0 450 2

367 nmµ 3

12. Answer (3)

Hint : Use image formation concept by ray diagram

Sol. :1 1 1

22.5 22.5 10x x

x = 7.5 cm

13. Answer (4)

Hint : Refraction through curved surface

2 1 2 1 v u R

Sol. :1.2 (1.5 1.8) (1.2 1.5)

20 ( 20)f

f =

14. Answer (4)

Hint : Use 1 1 1 u v f

Sol. : v1 =

15 1030 cm

(15 10)

v2 =

25 10 50cm

(25 10) 3

� = 50 40

30 cm3 3

15. Answer (3)

Hint : NAND gates can be used to form OR gate

Sol. : Given logic gate represents OR gate

16. Answer (4)

Hint : Power gain = 2 ⎛ ⎞ ⎜ ⎟

⎝ ⎠

L

B

R

R

Sol. : Power gain = 2 ⎛ ⎞ ⎜ ⎟

⎝ ⎠

L

B

R

R

200 = 2 2

101

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Vin =

4

20 = 0.2 V

17. Answer (3)

Hint : Modulation index = m

c

A

A

Sol. :

c m

c m

A A

A A =

x

y

c

m

A

A =

x y

x y

m

c

A

A =

x y

x y

18. Answer (2)

Hint : Q-value = Kinetic energy produced

Sol. : KE =

2 2

2

1

2 2

p h

m m

⎛ ⎞ ⎜ ⎟

⎝ ⎠


4/16

and Q = (KE )4

A

A

=

2

2

232

228 2

h

m

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

=

2

2

29

57

h

m

19. Answer (3)

Hint : Angular momentum is integral multiple of 2

h

and

2

2Z

En

Sol. : � = 2

hn

�

Be > �

Li

and E

2

2

Z

n |E

Li| > |E

Be|

20. Answer (4)

Hint : Higher the half life, slower will be decay rate.

Sol. : Higher the half life, slower will be decay rate

21. Answer (3)

Hint : Q-value of a reaction.

Sol. : 64Cu decay gives +ve Q-value

22. Answer (3)

Hint : In -decay mass decreases by 4 a.m.u and

Z by 2.

Sol. : No. of -particles = 238 206

4

= 8

No. of -particles = 8 2 = 16

23. Answer (2)

Hint : Use equation 2 2

1 2

1 1 1R

n n

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Sol. : Series is Paschen

2 2

min

1 1 1

3R⎛ ⎞ ⎜ ⎟ ⎝ ⎠

min

= 820 nm.

24. Answer (3)


Sol. :

1

hc

= + K

2

hc

= + 3 K

2 >

1

3

25. Answer (3)

Hint : For inelastic collision to occur, loss of energy

should be equal to energy difference of two

Bohr's orbits.

Sol. : (Kloss

)max

= 2

0

1 4

2 5

m mv

m

= 4( )

5i

K

For inelastic collision,

4( )

5i

K 10.2 4

Ki 51 eV

26. Answer (4)

Hint : Revolving electron can be equivalent to a

current carrying loop.

Sol. : 2

c

vB

r

r n2

1vn

27. Answer (2)


Sol. :1240

500 = +

1

2mv

1

2 ...(i)

1240

620 = +

1

2mv

2

2 ...(ii)

= 1.6 eV

28. Answer (2)

Hint : Cut-off wavelength is maximum wavelength

which can produce photoelectrons


5/16

PART - B (CHEMISTRY)

31. Answer (2)

Hint : Methyl ketone on reaction with NaOH/Br2 give

haloform reaction while ester give carboxylic

acid.

Sol. :

32. Answer (3)

Hint : DNA has double strand helix whereas RNA

has single strand.

Sol. :

Base + sugar Nucleoside

Nucleoside + phosphoric acid Nucleotide

RNA does not contain thymine.

33. Answer (4)

Hint : POCl3 remove H

2O by -elimination

Sol. : In -elimination H and OH must be present

on axial.

34. Answer (3)

Hint : On the basis of product stability

Sol. : III leads to formation of benzene.

35. Answer (1)

Hint : a

1pK

acidic strength

Sol. :

OH

> CH OH3 > H O2 >CH CH CH OH3 2 2

Acidic strength

36. Answer (4)

Hint : Phenol > alcohol > alkyne

Sol. : Phenoxide is resonance stabilised.

37. Answer (4)

Hint : Neoprene is a polymer of chloroprene

Sol. : Terylene is polyester, Buna–S is a polymer of

butadiene and styrene.

Sol. :hc

= 0

0

hceV

...(i)

hcn

=

2

0

0

hcn eV

...(ii)

0 =

1n

n

⎛ ⎞⎜ ⎟⎝ ⎠

29. Answer (4)

Hint : KE can vary from 0 to Kmax

.

Sol. : KE of ejected photoelectrons varies from 0 to

Kmax

and its variation is non-linear

30. Answer (2)

Hint : Pressure due to radiation = I

C where I is

intensity

Sol. :0 0

2 sin,

sin

av

I A AF S

c

P =

22 sin av

F I

S c


6/16

38. Answer (4)

Hint : Novolac is used to make paints

Sol. : Phenol + formaldehyde forms Novolac in

presence of acid or base catalyst which

undergoes cross linking on further heating

with HCHO to form Bakelite.

39. Answer (2)

Hint : Cu is used in Gatterman’s reaction

Sol. :

Cu

HCl

N2

Cl

40. Answer (3)

Hint : Rate of reaction will depend upon electrophilic

nature of the electrophile

Sol. : NO2 is EWG whereas OCH

3 is EDG so III is

the most reactive and I is the least

41. Answer (1)

Hint : —CH3

C— C—Cl

——

CH3

CH3

— —

O

AlCl3 —CH3

C— C (+)

——

CH3

CH3

— —

O

—CH3

C (+)

——

CH3

CH3

Sol. :

+ —CH3

C— C— Cl

——

CH3

CH3

— —

O

AlCl3

—

C—

——

CH3

CH3

CH3

42. Answer (2)

Hint : The NaOBr formed will oxidise the Benzylic

Alcohol also

Sol. :

C

Br2

O

CH 2 NH

2

OH

NaOH

OHC NH2

43. Answer (2)

Hint : Ammoniolysis of amine

Sol. :

+

44. Answer (4)

Hint : H of esters are acidic

Sol. :

CH — C — OEt3

O

CH — C — CH — C — OEt3 2

O O

EtONa

H(Claisen condensation)

CH – I3

EtO Na

CH — C — CH — C — OEt3 2

O O

CH — C — CH — C — OEt3

O O

—

CH3

CH — C — CH — CH3 2 3

O

H /+


7/16

45. Answer (3)

Hint : Nitrene rearrangement

Sol. :

CH — CH3

CH2

HF, 0°C

CH3

CH

CH3

KMnO , H , 4

+

COOH

PCl5

C — Cl

O

N 3 OH , ,

(Curtius rearrangement)

NH2

46. Answer (2)

Hint : hydroxy acid with H+/ gives , unsaturated acid

Sol. :

CH — CH — CH —CH — COOH3 2 2

CH — CH — CH = CH — COOH3 2

OH H +,

47. Answer (4)

Hint : Carbene is neutral

Sol. : Amine on reaction with HNO2 form diazonium

salt

48. Answer (1)

Hint : P Sucrose

R and S Glucose and fructose

Sol. : Sucrose is formed by formation of glycosidic

linkage between C1 of -D-glucose and C

2

and -D-fructose

49. Answer (1)

Hint : Aldol condensation

Sol. :

C

50. Answer (4)

Hint : Aldehydes are more reactive than ketones as

ketones are hindered.

Sol. : The attack happens by lone pair over sulphur.

The rate of addition is much greater for

aldehydes and can be used for purification/

separation bisulphite addition can be reversed

by acidic as well as basic hydrolysis.

51. Answer (2)

Hint : The reaction is SN1 type

Sol. :

CH — CH — CH — CH3 2 3

CH 3

CH — CH — CH — CH3 3

CH 3

Cl

Cl2

h

OH (alc)

PhCO H3

CH 3

CH 3

C CH — CH3

CH 3

CH 3

C CH

3

H O

H/ H O2

::

18

(S Type)N1

CH — C — CH — CH3 3

CH3

OHOH18


8/16

52. Answer (4)

Hint : End Product of reaction (4) is phenol

Sol. :

OH

NO2

Fe

HCl

OH

NH2

OH

N Cl2

OH

H

NaNO2

HCl

H PO3 2

(Phenol)

+

53. Answer (4)

Hint : Fact Based.

Sol. : Antibiotics have either cidal effect or a static

effect on microbes on this basis drugs are

divided into Bactericidal or Bacteriostatic.

54. Answer (3)

Hint : Boiling point of isomeric alcohol varies as

1° > 2° > 3°.

Sol. : Alcohols have higher BP as compared to

ethers because of H bonding.

55. Answer (2)

Hint : OH– is a good nucleophile whereas H2O is a

poor nucleophile

Sol. :

+Cl

H O2

S TypeN1

OH

SN2

OH

(A)

(B)

–

::

OH

(A) and (B) are positional isomers.

56. Answer (3)

Hint : Same M.F but different functional group

Sol. : Aldehydes and ketones are functional isomers

57. Answer (2)

Hint : When —NH2 is present on equatorial then

ring contract to five membered.

Sol. :

OH

NH2

HNO2

CHO

OH

NH2

NaNO2

C — H

O

HCl

58. Answer (1)

Hint : Ease of SN2 decrease as hindrance

increases.

Sol. : Correct order of ease of SN2 is IV > II > I > III

59. Answer (1)

Hint : (CH3)3CBr is 3° while other are 1° or 2° halide

Sol. : 3° halide show SN1 while other show S

N2

reaction.

60. Answer (2)

Hint : A =

CH3

C =

H C5 2

C

H

C C

CH3

H

Sol. :

CH3 CH3

H / t2P

(A) (B)

(C) (D)

H C5 2

C

H

C C

CH3

H

H / t2 P

CH —(CH ) —CH3 2 4 3


9/16

PART - C (MATHEMATICS)

61. Answer (1)

Hint : Use condition for coplanarity for two line

Sol. : Given lines are

– 5

0 4 – –1

x y z

and –

0 –2 2 –

x y z are coplanar

5 – 0 0

0 4 – –1 0

0 –2 2 –

3 – 112 + 36 – 30 = 0

sum of all possible values of

1 2 3

– 11

11

1

62. Answer (4)

Hint : Here 5a + 3b + 3c = 0 and 2a + 3b + 4c = 0

and solve for a, b, c

Sol. : DRs of normal to the plane containing lines

5 2 3

x y z and 2 3 4 x y z

are a, b, c, then

5a + 2b + 3c = 0 and 2a + 3b + 4c = 0 and

so d.r.'s of nomral to the plane are –1, –14, 11

Also plane contains the line 1 2 3

x y z

Required equation of plane will be given by

1 2 3 0

–1 –14 11

x y z

32x – 7y – 6z = 0

63. Answer (3)

Hint : Length of subsnormal dy

ydx

Sol. : Given length of subnormal dy

ydx

= k

dy

y kdx

2

2

ykx c

Which passes through the origin O(0, 0)

0 = 0 + c

c = 0

Curve is 22y kx

64. Answer (4)

Hint : Differentiate both sides w.r.t. x and eliminate

arbitrary constants

Sol. :

5

2y cx c , dyc

dx

5

2dy dyy x

dx dx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 5

–

dy dyy x

dx dx

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Degree of differential equation = 5

65. Answer (2)

Hint : Put x – 2 = X, y – 2 = Y

dY dy

dX dx and put Y = vX

Sol. : From given differential equation

2 2– 2 – 2

– 2 – 2

x ydy

dx x y

...(i)

Let x – 2 = X & y – 2 = Y dY dy

dX dx

From equation (i)

2 2dY X Y

dX XY

...(ii)

(homogeneous differential equation)

Put Y = VX, then dY dV

V XdX dX


10/16

Substituting these values in (ii) and then solving

we get,

2

22 In | |

YX c

X

2

2

– 2

– 2

y

x

= 2 ln |x – 2| + c

66. Answer (1)

Hint : Convert equation into linear form

Sol. : From given differential equation, we have

2 3

1,

dy y

dx x x

1–

IF = xe

Solution of equation (i) is

1–

·

xy e =

1–x

e +

1–x

ec

x

Put x = 1 and y (1) = 1, we get c = –1

e

1

11 –

xey

x e

2

11 2 – 3 –

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

ey e

e

67. Answer (2)

Hint : Differentiate solution and then use method of

separation of variables

Sol. : Differentiating both sides w.r.t. x

x(1 – x) f(x) + 0

1–

x

t f t xdt x f x∫

2

0

1–

x

t f t dt x f x∫

Again differentiating w.r.t. x we get

(1 – x)f(x) = 2xf(x) + x2f'(x)

2' 1– 3x f x x f x

2

' 1 3–

f x

f x xx

ln f(x) = 1

– – 3ln x c

x

(integrating both sides)

1ln ( ) = – – 3 ln f x x + c

x...(i)

Put x = 1 and f(1) = 1, in (i) we get

0 –1– 0 c 1c

1ln – – 3 ln 1f x x

x

3

1 1 11 – ln 1 –

3

1x x xf x e e

x

1

1 –2

12

8 8

ef e

68. Answer (3)

Hint : Solve by equation by method of inspection

Sol. : Here,

–

x ydy

dx x

xdy + y dx + xdx = 0

d(xy) + xdx = 0

2

2

xxy c ... (i)

If (i) passes through the point (2, 1), then

42

2c

c = 4

Put c = 4 in equation (i)

2xy + y2 = 2 × 4 = 8

69. Answer (2)

Hint : Put x + y = v –1dy dx

dx dn

Sol. : From given equation,

cos ,dy

x ydx

Put, x y v

1dy dv

dx dx


11/16

From given equation, dv

dx – 1 = cosv

21 cos 2cos

2

dv vv

dx

21sec

2 2∫ ∫

vdv dx

tan2

vx c

tan2

x yx c

⎛ ⎞ ⎜ ⎟⎝ ⎠

70. Answer (2)

Hint : Take image of given point A(1, 2, 3) due to

plane then find the distance between image

point and the point B.

Sol. :

D

C

AB (5, 8 , 11) (1, 2, 3)

(–1, –2, –3)

plane

x y z + 2 + 3 = 0

Image of A is D (x, y, z) and co-ordinates of D

can be obtained as

–2 1 4 9–1 – 2 – 3

1 2 3 1 4 9

x y z

–1 – 2 – 3

– 2

1 2 3

x y z

x – 1 = – 2, y – 2 = –4, z – 3 = –6

x = –1, y = –2, z = –3

D (–1, –2, – 3)

Minimum length of AC + CB

= 36 100 196 332BD

71. Answer (4)

Hint : Use concept of the distance of any point from

line parallel to any plane

Sol. :

P(5, 4, 3)

Q R

2 3 4

3 4 5

x y z

x – y + z =2 3 0

– – –

Coordinates of R are (3 + 2, 4 + 3, 5+ 4)

and are satisfying the equation x – 2y + 3z = 0

i,e, (3 + 2) – 8 – 6 + 15 + 12 = 0

10 = –8

= 4

–5

Coordinates of R are 2 1

– , – , 05 5

⎛ ⎞⎜ ⎟⎝ ⎠

Here equation of line PQ is

– 5 – 4 – 3

3 4 5

x y zsay

Coordinates of Q are (3 + 5, 4 + 4, 5 + 3)

and the point Q lies on the plane

x – 2y + 3z = 0

(3 + 5) – (8 + 8) + (15+ 9) = 0

10= –6 = –3

5

Coordinates of Q are 16 8

, , 05 5

⎛ ⎞⎜ ⎟⎝ ⎠

Required distance = 9

5QR

72. Answer (1)

Hint :

2 1–

n

a a b

S

b

��

�

Sol. : Here lines are parallel and they are

ˆ ˆ ˆ ˆ ˆ ˆ3 2 2 2r i j k i j k �

and

ˆ ˆ ˆ ˆ ˆ ˆ4 3 2 2r i j k i j k �


12/16

Distance between them

2 1–a a b

b

��

�

ˆ ˆ ˆ ˆ ˆ ˆ( – 2 ) (2 2 )

4 1 4

i j k i j k

�

�

2 1 2

1 –1 2 4 – 2 – 3

4 1 4 9

i j k

i j k

� �

� �

16 4 9 29

39

73. Answer (2)

Hint : 2a + 3b + 10c = 0

2a – 3b – 7 = 0

where a, b, c are the d.r' s of the line of

intersection of planes

Sol. : Planes are

2x + 3y + 10z = 8 ... (i)

2x – 3y – 7z = 2 ... (ii)

Let a, b, c be the d.r's of the line of intersection

of planes (i) and (ii)

2a + 3b + 10c = 0

2a – 3b – 7c = 0

–

–21 30 –14 – 20 –6 – 6

a b c

9 34 –12

a b c i.e., a : b : c = 9 : 34 : –12

74. Answer (4)

Hint : Volume of tetrahedron

= 1

3(Area of base) × height

Sol. : Volume of tetrahedron, is

D

h

Q

A

B

C

1

3V

⎛ ⎞ ⎜ ⎟⎝ ⎠

(Area of base) (Height)

1120 20

3h

120 3

20h

h = 18 units

75. Answer (3)

Hint :

P

A

5

B

O

a�

a bOP

a b

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

� �

��

� �

Sol. :

A

B C

Dˆ ˆ–i k

ˆ ˆ( )i j

ˆ ˆ( )k iˆ ˆ( )j kˆ ˆ( – )i j

ˆ ˆ ˆ ˆ ˆ ˆ ˆ– – 2 – –

2 2 2

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

�� i k i j i j kBD

1 64 1 1 3

2 2BD ��

Unit vector along BD

ˆ ˆ ˆ2 – –

6 i j k

76. Answer (2)

Hint : Required probability

6 5 4

6 9 9 6

n

n n

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

Sol. : Bag 1 Bag 2

6W 4W

nB 5B

16

6P W

n

2

4

9P W

16

nP B

n

2

5

9P B


13/16

P(one white, one black)

6 5 4

6 9 9 6

n

n n

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

5030 4

(given)9 6 99

n

n

15 2 25

6 11

n

n

150 + 25n = 165 + 22n

3n = 165 – 150 = 15

n = 5

77. Answer (2)

Hint : Probability

10 11 11 11 11

11

1 1 1 1 1 7

2 2 2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Sol. :

H or T (appearing)

10 times

H H....... H ��

��

10 101 1

1 1 1 1 12 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

T

H or T (appearing)

10 times

H H....... H ��

��

10 111 1 1

1 1 1 12 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

H or T (appearing)

10 times

T H.........HH��

��

10 111 1 1

1 1 1 12 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


10 11 11 11 11 111 1 1 1 1 1

2 2 2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

10 11 11

11

1 1 1 75 2 5

2 2 2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

78. Answer (4)

Hint :

5 2 4 2 3

5 1 3 2 1

5! 5!5!

2! 2! 2!

n S C C C C C

2 3

2 1

5!

2! 2!n E C C

Sol. : To find n(S)

Letters to given word are

K 2

O 1

L 1

A 2

T 1

Case I.

When all 5 letters are distinct

Required words formed = 5C5 × 5! = 1 × 120 = 120

Case II.

When 2 alike letters and 3 different letters are

taken


1 3

5!= × ×

2!C C

5 4 3 2!2 4 480

2!

Case III.

When 2 alike, 2 alike and 1 different letter are

taken


2 1

5!= × ×

2! 2!C C

5 4 3 2!1 3 90

2! 2

n(S) = 120 + 480 + 90 = 690

n(E) = total number of words formed with all

the repeated letters used = 90


90 3

690 23

n E

n S

79. Answer (3)

Hint : n(S) = 66, n(E) = 6

6 5

6 1( ) = 1 – 1 –

6 6P E


14/16

Sol. : n(S) = 66, n(E) = 6C1 × 1 × 1 × 1 × 1 × 1

Required probability 6

61 –

6

5

5 5 5

1 6 –1 77751 –

6 6 6

80. Answer (3)

Hint : Use Bayes Theorem

Sol. :

3

33

1 2 3

1 2 3

·

· · ·

BP B P

BBP

B B B BP B P P B P P B P

B B B

⎛ ⎞⎜ ⎟

⎛ ⎞ ⎝ ⎠⎜ ⎟⎛ ⎞⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

2

3

2 2 2

2 2 2

1 – 4 7

3 – 4 10

1 – 4 8 – 4 9 – 4 7

3 – 4 10 – 4 10 – 4 10

⎛ ⎞ ⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

a a

a aBP

B a a a a a a

a a a a a a

2

3

22

– 4 7 1 1 1–

3 33 – 4 8 – 2 4

B a aP

B a a a

⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

1 1 1–

3 12 4

81. Answer (1)

Hint : Use

2

22

1 2

1 2

( )

( ) ( )

RP E P

EEP

R R RP E P P P E

E E

⎛ ⎞ ⎜ ⎟

⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Sol. : Let

E1 the event when die A is used

E2 the event when die B is used

R when red face appears

1

3

5

n

RP

E

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

and

2

2

5

n

RP

E

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

,

P(E1) =

1

2, P(E

2) =

1

2

2

1 2

162 5

973 2 1

5 5 2

n

n n

EP

R

⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎡ ⎤⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

2 16

973 2

n

n n

n = 4

82. Answer (4)

Hint : 9!

2! 3!n S

7 6

3 3

6!– 5!

2! n E C C

Sol. : 9! 9 8 7 6 5 4 3!

3! 2! 3! 2!

9 8 7 6 5 2 9 8 7 6 10

n S

To find n(E)

Case I.

position for T is denoted by ×

× A × E × N × I × O × N ×

No two T's come together but N may or may not

Total words fomed 3

6! 7! 6 5 4!7

2! 3! 4! 2C

7 6 5 4 3 2 542 300 12600

2

Case. II

× A × E × NN × I × O ×

No two T's come together but two N come together

['×' is the position for T]

In this case total words formed

6

3

6 5 4 3!= × 5! =

3! 3!

C × 5 × 4 × 3 × 2 × 1

= 20 × 120 = 2400

n(E) = 12600 – 2400 = 10200


n E

n S

10200

9 8 7 6 10

85 85

18 14 252

83. Answer (3)

Hint :sum of variables

Mean n

n

If n is odd, then median item

th1

2

n ⎛ ⎞ ⎜ ⎟⎝ ⎠

term


15/16

Sol. : Given,

3 3 3 31 2 3 ...

1900n

n

22

1

4·

n n

n

21 ·1900

4

n n

7600 = n(n + 1)2

19 × (20)2 = n(n + 1)2

n = 19

Median of 1, 2, 3 ..., 18, 19 is

th19 1

2

⎛ ⎞⎜ ⎟⎝ ⎠

item = 10

84. Answer (2)

Hint : Mode = 3 Median – 2 Mean

Sol. : We have mode = 3 Median – 2 Mean

Mode – Median = 2(Median – Mean)

(Median – Mean) = 1Mode – Median

2

136 18

2

85. Answer (2)

Hint :

Volume of parallelepiped

· · ·

· · ·

· ·

·

��

��

��

��

� � � � � �

a a a b a c

b a b b b c

c a c cc b

Sol. : Volume of parallelepiped

· · ·

· · ·

· ·

·

a a a b a c

V b a b b b c

c a c cc b

��

��

��

��

� � � � � �

4 3 4288 cu units

3 9 6

12 2 cu units4 6 16

86. Answer (4)

Hint : Use formula for cross product of three vectors

Sol. : From given relation

2 · – ·a c b a b c

� � � � � �

3b c � �

1

·2

a c � �

and 3

· –2

a b � �

3

cos –2

a b � �

5

cos cos – cos6 6

⎛ ⎞ ⎜ ⎟⎝ ⎠

angle between a�

and b�

is 5

6

87. Answer (2)

Hint : Use formula for 'dot' and cross product of two

vectors

Sol. : Let be the angle between b�

and c�

,

Then,

5b c � �

b c� �

sin = 5

1sin =

2

3

cos =2

Now, given,

3( ) 0 � �

� � �

a b a c ( 3 ) 0 � �

� �

a b c

3 ||�

� �

b c a

3 �

� �

b c a

2 2

9 – 6 ·b c b c� � � �

= 2

2a�

25 + 36 – 6 cosb c � �

= 2·4

2361– 6 5.2· 4

2

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

2 61– 30 3

4

88. Answer (1)

Hint : Use concept of coplanarity of vectors

Sol. : Since given vectors are coplanar

So, 61�

p + 40�

q + 21�

r = 0�

�

p , �

q , �

r are coplanar.


16/16

� � �

�

q × �

r and �

r × �

p are collinear

(�

q × �

r ) × {(�

r × �

p ) = 0�

89. Answer (3)

Hint : If a�

and

�

b are non-collinear vectors such that

0 � �

xa yb x = 0, y = 0

Sol. : Here, (1 – tan)

�

a – 1 2 sin 0b � �

and a

�

and b�

are non-collinear

1 = tan and 1 2 sin 0

4

or 5

4

and

5

4

or 7

4

Most general value of 5

24

n , n I

90. Answer (2)

Hint : Use condition for coplanarities.

Sol. : Since vectors are coplanar,

So, 0 1 1 0

4

a a c

c c b

4ab = c2

c2 – 4ab = 0 ...(i)

And

Discriminant of given equation is

= c2 – 4ab = 0, [from (i)]

Roots are real and equal

test - 3 (code-a) (answers) all india aakash test series ... · all india aakash test series for...

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