mock test - 4 (code-a) (answers) all india aakash test ...€¦ · all india aakash test series for...

Mock Test - 4 (Code-A)(Answers) All India Aakash Test Series for JEE (Main)-2019

1/12

1. (4)

2. (2)

3. (1)

4. (2)

5. (1)

6. (2)

7. (4)

8. (2)

9. (1)

10. (3)

11. (2)

12. (3)

13. (2)

14. (3)

15. (3)

16. (1)

17. (1)

18. (2)

19. (2)

20. (3)

21. (2)

22. (2)

23. (2)

24. (1)

25. (3)

26. (1)

27. (1)

28. (3)

29. (2)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (3)

33. (1)

34. (1)

35. (4)

36. (4)

37. (3)

38. (2)

39. (4)

40. (4)

41. (1)

42. (2)

43. (1)

44. (3)

45. (2)

46. (1)

47. (3)

48. (3)

49. (3)

50. (1)

51. (4)

52. (3)

53. (4)

54. (3)

55. (4)

56. (2)

57. (1)

58. (4)

59. (3)

60. (2)

61. (3)

62. (1)

63. (2)

64. (2)

65. (1)

66. (4)

67. (3)

68. (2)

69. (4)

70. (2)

71. (4)

72. (2)

73. (4)

74. (1)

75. (3)

76. (4)

77. (4)

78. (4)

79. (4)

80. (3)

81. (3)

82. (4)

83. (2)

84. (4)

85. (4)

86. (1)

87. (1)

88. (1)

89. (3)

90. (4)

Test Date : 31/03/2019

ANSWERS

MOCK TEST - 4 Code-A

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

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 4 (Code-A) (Hints & Solutions)

2/12

1. Answer (4)

Hint :2

xv T

x

Sol. : ta =

0v

g

x = (v0 cot)t

a

v0

2. Answer (2)

Hint :

2

c

vR

a

Sol. : vy = 2xv

x

ay = 2v

x2 + 2 × a

x

ax = 0

ay = 200 m/s2

R = 2

0.5 my

v

a

3. Answer (1)

Hint : Write down equation of motion for each block.

Sol. : Fsp

+ 5 g = 5a

Fsp

= 1 g

a2 kg

= g

4. Answer (2)

Hint : a = 0 for velocity to be maximum

Sol. : For vmax

a = 0

g sin – kx g cos = 0

x = tank

5. Answer (1)

Hint : ( ) dvh x v gh

dx

Sol. : gh = ( ) dvh x v

dx

PART - A (PHYSICS)

( )

0( )

∫ ∫

� hdx

gh v dvh x

v = 2 ln⎛ ⎞⎜ ⎟⎝ ⎠

lgh

h

6. Answer (2)

Hint :

2sin2 u

Rg

Sol. : R =

2 sin(2 )40 m

u

g

x = (R + eR + e2R + ...)

= 60m1

⎛ ⎞ ⎜ ⎟⎝ ⎠

R

e

7. Answer (4)

Hint : Acceleration of each will be in opposite

direction.

Sol. : arel

= µ 13

⎛ ⎞⎜ ⎟

⎝ ⎠

mg

m

= 240

m/s9

•

µmg

µmg

. 3 m

8. Answer (2)

Hint : cmF ma∑

Sol. : J = mv0

J

2

�J =

2

12

m �

Mock Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/12

06v �

T =

2

06

2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

�

vm

=

2

09

2�

mv

9. Answer (1)

Hint : Gravitational force is responsible for centripetal

acceleration.

Sol. :2( ) R

2

MRd

R

=

2 ( )

(2 )

GM Rd

R R

2 = 2

2 G

R

10. Answer (3)

Hint : Acceleration need not be in horizontal

direction

Sol. :

geff

Free surface45°

g

a

45°

11. Answer (2)

Hint : 0( ) dT

k T Tdt

Sol. : dTdt

= k(T – T0)

80 0ln( 20)

T tT kt

ln2 = k(10 min)

60ln (20 min)

20k

T

60

ln4 ln20T

T – 20 = 15

T = 35°C

12. Answer (3)

Hint : AB and CD are isobaric process.

Sol. : AB and CD are isobaric process.

Point 'A and C' are at same volume, also

point 'B and D' are at same volume. So dv in

both the process are same.

So, Pdv = NRdT

3Pov = NRdT|

for CD

and Pov = NRdT|

AB

3dT/AB

= dT/CD

3(TB – T

A) = T

D – T

C

13. Answer (2)

Hint : Frequency given are 3rd and 5th harmonic

respectively.

Sol. :4

= 9

= 36 cm

v = v= 360 m/s

14. Answer (3)

Hint : Distance of either of extreme from mean

position is same

Sol. : emax

= 1 2

( )m m g

k (extension)

cmax

= 1 2

( )m m g

k (compression)

m g k, /M. Position

N length

m g k, /m g

2

k

5 m

4 m

1 m

15. Answer (3)

Hint : Find phase angle at the time of maximum

separation.

Sol. : Draw phasor diagram,

A

A

P

Q

60°

60°


4/12

vp

= 10 cos 60°

= 5 m/s

16. Answer (1)

Hint : Apply equation of refraction at surfaces

consecutively

Sol. :1

µ 4

3

v R = 0

2 1

1 µ

( )

v R v =

(1 µ)R

v2 = μ = 3/2

17. Answer (1)

Hint : me = D/f

e

Sol. : me = D/f

e = 5

18. Answer (2)

Hint : 2 2

1 2 1 22 cos A A A A A d

Sol. : I0 = 4I

I = 4 2 .4 cos2

I I I I

19. Answer (2)

Hint :dU

dt has to be maximum

Sol. : U = 21

2Li

dU

dt =

diLi

dt

i = /

0(1 ) tR L

i e

diidt

is maximum, when / 1

2

t RLe

i = 1

12 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

v v

R R

20. Answer (3)

Hint : Direction will not change if ||�

�

v B

Sol. : As velocity is along E�

so speed will keep on

increasing. And, since B�

will not have any

effect on v�

. So, direction will not change.

21. Answer (2)

Hint : ∫Q idt

Sol. : i = /0 (1 ) Rt Lv

eR

Q =

/

0

∫L R

i dt

Q = 0 1

1⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦

v L L

R R R e

wd =

2

0

2

Lv

eR

22. Answer (2)

Hint : For maximum power r = Rext

Sol. : Circuit can be reduced to

4Eeq

R

for maximum power, R = 4

eq

4

E =

10 10

5 20

Eeq

= 10 V

P =

210

4 6.25 W8

⎛ ⎞ ⎜ ⎟⎝ ⎠

23. Answer (2)

Hint :1 2

1 1 1 R R R

Sol. :1

R =

6 6

100 (100)5

R = 500

36

24. Answer (1)

Hint : Potential along axis is constant

Sol. : Since net charge is zero, apply properties of

dipole.

25. Answer (3)

Hint : Q = A + Be–t/e


5/12

Sol. : Initial P.D = 2

Q

C

Charge = /

2

t RCQe

Net charge = /3

2 2

t RCQ Qe

26. Answer (1)

Hint : Based on conservation of mass number and

charge

Sol. : 92 – (8 × 2) + x = 82

x = 6

27. Answer (1)

Hint : N = Noe–t

Sol. : Half life = 1 day

1000 = Aoe–2

500 = Aoe–3

2 = e

Now, o

21000

( )

A

e

Ao = 4000 dps

28. Answer (3)

Hint : hv = + eV0

PART - B (CHEMISTRY)

31. Answer (1)

Hint : Rate of dehydration depends upon stability

of carbocation, Aromatisation stabilises

carbocation.

Sol. :

H OH

H SO2 4

H O—H

H

—

+

does not occur×

+

not formed

(4 e)(antiaromatic)

+ H O2

–

II, III, IV produces stable carbocation

OHH

+

–H O2

+

(Aromatic compound) Highly stable

CH2

H+

–H O2

+

OH CH2

(Highly stable)

H

OH

H+

–H O2

+

(Aromatic) Highly stable

32. Answer (3)

Hint : Friedel craft alkylation followed by benzylic

bromination

Sol. : 0

hCeV

0

2 3

eVhC

22

hC

= 4hC

v = 8 10

153 10 10 110

4 6000 8

= 1.25 × 1014 Hz

29. Answer (2)

Hint : (Signal will have frequency range from c –

m

to c +

m)

Sol. : Signal will have frequency range from c –

m

to c +

m

30. Answer (3)

Hint : In n-type semiconductors, electrons are

majority charge carrier and are being made

by doping with atoms having 5 electrons in

outer most shell.

Sol. : In n-type semiconductors, electrons are

majority charge carrier and are being made

by doping with atoms having 5 electrons in

outer most shell.


6/12

Sol. :

+ OHBF

3

Br

NBS, CCl 4

33. Answer (1)

Hint : Nucleophilic substitution at haloarene and

NO2 group at ortho-para position favours

SNAr

Sol. :

Cl

NO2

NO2

Step 1.OH

–

rate daterming step

HO Cl

..–

–

N+

O

O

NO2

HO Cl

–

O

O

NO2

N+

–

HO Cl

..–

NO2

NO2

Step 2.

HO Cl..–

–

N+

O

O

NO2

Fast

NO2

NO2

OH

34. Answer (1)

Hint : Birch reduction of benzene ring

Sol. : The two carbon atoms that are reduced go

through anionic intermediates. Electron

withdrawing substituent stabilises carbanion;

while electron donating destabilises.

3 3(solvated e)

Na NH NH e Na ��⇀

↽��

•e–

•

..

–

–RO–

H—O—R

•

H H

•e

H H

..–

R—O—H

H H

H H

35. Answer (4)

Hint : Reaction with hot KMnO4 of side chain of

benzene ring

Sol. :

Br

(B) Contain chiral carbon

2Br (1mole)

4

hot

KMnO

COOH

COOH

(A) (No chiral carbon)

36. Answer (4)

Hint : HCOOH is more acidic than

COOH

Sol. :

COOH

OHOH

>

COOH

OH

>

COOH

>

OH


7/12

37. Answer (3)

Hint : Reductive ozonolysis form ketone and

aldehyde of the given compound.

Sol. :

CH3

3

3 2

(i) O

(ii) (CH ) S

C

C

O

O

CH3

H

+

3 2Ag(NH )

Tollen's test

C

C

O

O

CH3

O–

Tollen's test by —CHO group

38. Answer (2)

Hint : Preparation of ketone and Baeyer-Villiger

oxidation

Sol. :

2PhLi

OH

O

C C

O Li– +

O Li– +

Dianion

H O3+

C

OH

OH

–H O2

Ph

O

CR C O O H

O

Baeyer-Villiger oxidation

O

O

C Ph

Migratory aptitude

H > –Ph > 3° alkyl > 2° alkyl > 1° alkyl

> Methyl

39. Answer (4)

Hint : Ozonolysis followed by aldol condensation

Sol. : 3

2

(i) O

(ii) Zn;H O

CHO

CHO

A

OH

(Intra aldol. condensation)

C

H

O

B

40. Answer (4)

Hint : Decarboxylation rate and stability of

carbanion

Sol. :

CO H2

O

×

OH

+ CO2

C-atom at bridge head position cannot be

sp2 hybridised according to Bredt’s rule.

41. Answer (1)

Hint : Elevation in boiling point

Sol. : Tb = K

bm =

bK Y 1000

M 250 gm

=

b4K Y

M

42. Answer (2)

Hint : I2 + 2Na

2S

2O

3 Na

2S

4O

6 + 2NaI

Sol. : Let mmol of CaCO3 x

mmol of CO2 = x, mmol of CO = 2x

mMol of I2 =

2x 400 0.02 8

5 2 2

⇒

x = 10

Mass of CaCO3 = 10 × 10–3 × 100 = 1 g


8/12

43. Answer (1)

Hint : Vapour pressure of liquid solution

Sol. : o o

A A B BP X P X = 550

o o

A B

1 3P P

4 4 = 550 ...(i)

After addition of 1 mol of B

o o

A B

1 4P P

5 5 = 560 ...(ii)

From (i) and (ii)

o

AP = 400 mm Hg

o

BP = 600 mm Hg

44. Answer (3)

Hint : [Fe(CN)6]3–

Sol. : Splitting increases with increase in charge on

the central metal ion and strength of ligand

i.e. [Fe(CN)6]3–

45. Answer (2)

Hint : Extraction of silver

Sol. : Cyanide process or Mc Arthur Forest

Cyanide process

2Ag2S + 8NaCN + O

2 + 2H

2O

4Na[Ag(CN)2] + 4NaOH + 2S

Soluble silver complex is filtered and treated

with zinc dust and silver gets precipitated.

2Na[Ag(CN)2] + Zn

Na2[Zn(CN)

4] + 2Ag

46. Answer (1)

Hint : Reactivity towards O2

Sol. :2 2 2 3 2

4KO 2CO 2K CO 3O

47. Answer (3)

Hint : Solubility of sulphate of group II

Sol. : On moving down water solubility of alkaline

earth metal sulphates decreases.

Oxides and hydroxides of alkaline earth

metals are basic, except Be, which gives

amphoteric. Hence metal is Be.

48. Answer (3)

Hint : Hybridisation of element

Sol. :3 4

BF BF

2 3sp sp

49. Answer (3)

Hint : 7

Sol. :11

Na 1s2 2s2 2p6 3s1

n 1 2 2 3

� 0 0 1 0

n + � 1 2 3 3

50. Answer (1)

Hint : Valence bond theory

Sol. : Magnetic moment = n(n 2) BM

3.83 = n(n 2)

n = 3

M2+ has three unpaired electrons

M = Co [ Co = [Ar] 3d74s2]

Co2+ = [Ar] 3d7

n = 3 (3 unpaired e)–

51. Answer (4)

Hint : Dependence of temperature on rate of

reaction

Sol. : K = a

E

RTAe

lnK = ln aE

ART

So the lnK versus 1

T is linear with negative

slope and positive intercept.

52. Answer (3)

Hint : A = Cu2[Fe(CN)

6]

Sol. : CuSO4 + K

4[Fe(CN)

6]

Cu2[Fe(CN)

6] + K

2SO

4

(chocolate brown)

53. Answer (4)

Hint : Le-chatelier's principle

Sol. : Ammonia react with HCl.


9/12

PART - C (MATHEMATICS)

61. Answer (3)

Hint : z = i

Sol. : z = i

z45 = i

62. Answer (1)

Hint : Equation of line in vector form

Sol. : Equations of lines are

r

�

= 1

2 (2 )b a b c � �

� �

r

�

= 2

2 (3 2 2 )a b a b c � �

� � �

21 = 3

2 + 1, –2 –

1 = –2 – 2

2,

1 = 2

2

1 = 2,

2 = 1

63. Answer (2)

Hint : Formula

Sol. : x – 2 = 3

2

y

= z – 1 = 2( 4)

6

(x, y, z) 10 1 7

, ,3 3 3

⎛ ⎞⎜ ⎟⎝ ⎠

64. Answer (2)

Hint : Range = co-domain

Sol. : –3 3sinx – 4cosx + 2 7

65. Answer (1)

Hint : Make 0

0 form

54. Answer (3)

Hint : Ssys

= nR ln(V2/V

1)

Sol. : W = –nRT lnV2/V

1

= –2R × 300 ln(10) = –1381.55 R

q = –W (for isothermal process)

S = +2 × R ln(10) = 4.06 R

55. Answer (4)

Hint : Preparation of B2H

6

Sol. : Balanced equation

4 2 2 6 22NaBH I B H 2NaI H

56. Answer (2)

Hint : Anode:

2CH3—C—O

–

— —

O

—C—O•

— —

O

+2e–

2CH3

2 6 2C H (g) 2CO (g)

2 2

1H O 1e H (g) OH

2

Sol. : q = i × t = 965 × 10 = 9650C

No. of faraday = 0.1

1 mol e– produce = 2 mol of gases

0.1 mol e– produce = 0.2 mol of gases

V = 0.2 × 22.4 = 4.48 L

57. Answer (1)

Hint : Statement S2 is incorrect

Sol. : Smicellisation

= +ve

SAdsorption

= –ve

HAdsorption

= –ve

58. Answer (4)

Hint : Ni Mond process

Sol. : Ni Mond process

Zr Van - Arkel process

Zn Distillation

59. Answer (3)

Hint : BiH3 > SbH

3 > NH

3 > AsH

3 > PH

3

Sol. : B.P molecular mass

H-bonding

So order is

BiH3 > SbH

3 > NH

3 > AsH

3 > PH

3

60. Answer (2)

Hint : X SO2Cl

2

Sol. : P4 + SO

2Cl

2 PCl

5 + SO

2

X =

— —

—

S

O•

•

•

•

Cl ••

•

•

• •

—

—

O•

•

•

•

—

Cl ••

•

•

•

•


10/12

Sol. : � = 0

logcoslim

tanx

x

x

0

0 form

= 0

66. Answer (4)

Hint : Substitution method

Sol. : 1

2 3 8( 1)( 3 )x x x dx ∫

Let x3 + 3x = t

3(x2 + 1)dx = dt

I =

91 9

838 8

1 8 8[ 3 ]

3 3 9 27

tt dt x x c ∫

67. Answer (3)

Hint : Parametric point on the line

Sol. : Point on the line is (2 – 3, – 1, 3 + 1)

17 = 2 2 2(2 5) ( 2) (3 2)

17 = 42 + 25 – 20 + 2 + 4 – 4 + 92 + 4 – 12

= 2, 4

7

Point is (1, 1, 7)

68. Answer (2)

Hint : Trigonometric substitution

Sol. : 2 2[( 1) 2]

dx

x ∫

Let (x + 1)2 = 2tan2

= 4

2

0

2 2 2cos

4 32 16d

∫

69. Answer (4)

Hint : Counting

Sol. : 24C15C

1 × 7 × 6 = 24 × 31 × 51 × 71

Number of divisors = 40

70. Answer (2)

Hint : Break the function

Sol. : A =

1 0 1 2

2 2

2 1 0 1

x dx xdx xdx x dx

∫ ∫ ∫ ∫

= 17

3 sq. units

71. Answer (4)

Hint : Cramer's rule

Sol. : = 0

a = 2, –1

x

= b(7a + 6)

y

= b(7a2 – 12)

z

= –7b(a + 2)

For no solution a = 2, –1, b 0

72. Answer (2)

Hint : Find f(–x)

Sol. : f(–x) = f(x)

Hence, an even function

73. Answer (4)

Hint : Definition of GIF and modulus

Sol. : f(x) is discontinuous at

x = 1, 2, 3, 4, 5, 6, 7, 8

i.e. 16 points

74. Answer (1)

Hint : Modulus function

Sol. : f(x) = 1

sin cos2 2

tan

sin cos2 2

x x

x x

= tan–1 tan

4 2

x⎛ ⎞⎜ ⎟⎝ ⎠

= 4 2

x

dy

dx=

1

2

1( )

4 2y x

x + 2y = 2

75. Answer (3)

Hint : Maxima/minima


11/12

Sol. : f (x) = e–2x(2x – 2x2 + 4)

f (x) = 0 for x = –1, 2

f (x) = e–2x(4x2 – 8x – 6)

f(–1) = –e2

76. Answer (4)

Hint : Linear differential equation

Sol. : IF =

1

log logdx

x xe x

∫

ylogx = 2

log

(1 )

xdx C

x

∫

= log 1

1 (1 )

xdx

x x x

∫

ylogx = log log( 1)1

xx x C

x

77. Answer (4)

Hint : Counting

Sol. : P(E) =

8 7

8 8

0 1

2 1 2

3 3 3C C

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

2 6 3 5

8 8

2 3

1 2 1 2

3 3 3 3C C

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

=

82

193

⎛ ⎞ ⎜ ⎟⎝ ⎠

78. Answer (4)

Hint : Series

Sol. : Sn

= 1 2

1

cot ( 1)n

r

r r

∑

= tan–1(n + 1) – tan–11 = 1tan

2

n

n

79. Answer (4)

Hint : Coefficients are complex (not all real)

Sol. : Sum of roots = 1 + 3i = 1 – 2i + x2

x2 = 5i

80. Answer (3)

Hint : Binomial theorem

Sol. : Coefficient of x9 in 8 3 4 5 4

2 3

(1 ) (1 )(1 )

(1 )

x x x

x

= (1 – 3x8)(1 + x4)(1 + 4x5)(1 – x2)–3

= (1 + 4x5 + x4 + 4x9 – 3x8)

(1 + 3x2 + 6x4 + 10x6 + 15x8)

= 24 + 4 = 28

81. Answer (3)

Hint : Prime number

Sol. : Order of the matrix is n × 1 or 1 × n

Number of matrices are 2n + 2n = 2n + 1

82. Answer (4)

Hint : Make diagram

Sol. : 5 and 1020

h h

x x

h = 200 m

83. Answer (2)

Hint : Telescopic method

Sol. : Let S = 2 3 4

2 7 12 17...

4 4 4 4

1

4S =

2 3 4

2 7 12...

4 4 4

3

4

S =

2 3 4

2 5 5 5...

4 4 4 4

3

4

S =

11

3

S = 11 20

19 9

84. Answer (4)

Hint : Trigonometric formulae

Sol. :

2 5sin 3 4sin

2 2

x x⎡ ⎤⎢ ⎥⎣ ⎦

= 15

2sin sin2 2

x x⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 5sin 3 4sin

2 2

x x⎡ ⎤⎢ ⎥⎣ ⎦

= 25 5

2sin 3 4sin sin2 2 2

x x x⎡ ⎤ ⎢ ⎥⎣ ⎦

2 5 5 1sin 0, 3 4sin 0, sin

2 2 2 2

x x x

5 5

, , ( 1)2 2 3 2 6

nx x xn n n

⎛ ⎞ ⎜ ⎟⎝ ⎠


12/12

12 2 22 , , ( 1)

5 15 5 15

nn nx n x x

85. Answer (4)

Hint : Parametric equation of a line

Sol. : Mid point of BC is D11

2,2

⎛ ⎞⎜ ⎟⎝ ⎠

A

B C

(2, 1)

(2, 4)

3 3 112,

2 2

⎛ ⎞⎜ ⎟⎝ ⎠

B, C 3 3 11

2 ,2 2

⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠

86. Answer (1)

Hint : C1C

2 = r

1 + r

2

Sol. : (h – 1)2 +

2

223

3k

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

2

( , )h k

21,

3

⎛ ⎞⎜ ⎟⎝ ⎠

9x2 + 9y2 – 18x + 12y – 68 = 0

87. Answer (1)

Hint : Parametric form of tangent

Sol. : 1 1cos , sin

3 2x y

3cosx + 2siny – 1 = 0

4 3cos

3 2sin

tan = 9

8

(x, y) 8 9 8 9

, or ,3 145 2 145 3 145 2 145

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

88. Answer (1)

Hint : Counting

Sol. : 6

8

2 8

6

7C

89. Answer (3)

Hint : Use formulae

Sol. : 45

135

p q

p + q = 20 ...(i)

88 1

5 5 (9 + 4 + 49 + (13 – p)2 + (13 – q)2)

(13 – p)2 + (13 – q)2 = 26 ...(ii)

From (i) and (ii) p = 8, q = 12, or p = 12, q = 8

90. Answer (4)

Hint : Definition

Sol. : ∵ p (q r) is false

case (I)

p is true and q r is false

p is true and at least one out of q and r is false

or

Case (II)

p is false and q r is true

p is false and q and r both are true

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