(Answers) All India Aakash Test Series for JEE (Main)-2020 ......Sep 06, 2019 · U CV Sol. : 12 12...

Test - 1 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 09/06/2019

ANSWERS

1/13

TEST - 1 - Code-A

PHYSICS CHEMISTRY MATHEMATICS

1. (3)

2. (4)

3. (3)

4. (1)

5. (3)

6. (4)

7. (1)

8. (1)

9. (1)

10. (2)

11. (4)

12. (3)

13. (4)

14. (3)

15. (3)

16. (4)

17. (2)

18. (2)

19. (2)

20. (4)

21. (2)

22. (4)

23. (1)

24. (2)

25. (4)

26. (4)

27. (3)

28. (2)

29. (1)

30. (1)

31. (4)

32. (3)

33. (3)

34. (2)

35. (4)

36. (2)

37. (2)

38. (4)

39. (3)

40. (1)

41. (4)

42. (3)

43. (4)

44. (2)

45. (1)

46. (2)

47. (4)

48. (1)

49. (2)

50. (2)

51. (3)

52. (2)

53. (4)

54. (3)

55. (2)

56. (4)

57. (2)

58. (1)

59. (1)

60. (2)

61. (4)

62. (3)

63. (4)

64. (4)

65. (3)

66. (4)

67. (1)

68. (3)

69. (4)

70. (3)

71. (1)

72. (2)

73. (2)

74. (1)

75. (4)

76. (2)

77. (1)

78. (1)

79. (3)

80. (4)

81. (3)

82. (3)

83. (3)

84. (2)

85. (2)

86. (2)

87. (2)

88. (3)

89. (4)

90. (3)

All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)

2/13

1. Answer (3)

Hint : EA = 2

KQ

r due to shell at outside point

Sol. : 2

(3 )A

A

K QE

r=

rA is distance of A from O

⟹ EA = constant

2. Answer (4)

Hint : VP = V1 + V2

Sol. : 0 0

2 (2 )2 2

P

RR

V

= +

0

2 R =

3. Answer (3)

Hint : Parallel arrangement

Sol. : Ceq = ...2 4 8

C C CC + + + + ∞

1

21

12

C C= =

−

4. Answer (1)

Hint : 0Q =

Sol. : 0Q =

⟹ Q + 2Q – Q + Q1 = 0

⟹ Q1 = –2Q

5. Answer (3)

Hint : 2

2

QU

C

=

Sol. : 2

0

(2 )

(2){4 2 }

QU

a=

2 2

0 0

4

16 4

Q QU

a a= =

6. Answer (4)

Hint : 1 2=kq q

Ur

Sol. : 2 2 2

2 ...2 3

kq kq kqU

a a a

= − + − +

2 2 1 1

1 ...2 3

kq

a

= − − + −

Note: 2 3 4

ln(1 ) ...2 3 4

x x xx x+ = − + − +

2 2

ln(2)kq

Ua

= −

2

0

ln(2)2

qU

a = −

7. Answer (1)

Hint : 0 =kQ

VR

Sol. : V at center is 00

3 3

4 2 2

QV

R =

30

0 00

3 1 1 4

2 2 2 4 3

VV V V R

R = − = =

2

06

RV

=

8. Answer (1)

Hint : Vinside = Vsurface

Sol. : Potential at inside point will be same as

potential at the surface of inside sphere.

PART - A (PHYSICS)

Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/13

9. Answer (1)

Hint : 0 0 0QU V Q=

Sol. : 00 0

( )ln2 ln2

4 4

LV

L

= =

0

0

ln2

4

QU

=

10. Answer (2)

Hint : 0

inq =

Sol. : 11

0

q =

22

0

q = −

3 0 =

11. Answer (4)

Hint : 21

2U CV=

Sol. : 1 2

1 2

Q Q

C C=

21 48

(2)2 3

U

=

= 256 µJ

12. Answer (3)

Hint : Electric field near a point charge will

dominated by nearer charge.

Sol. : Electric field at x = ± ∞ will tend to zero.

13. Answer (4)

Hint : Use symmetry

Sol. : Field due to each spherical shell will be

along –y direction.

14. Answer (3)

Hint : Use super position principle

Sol. : sheet holePE E E= −

02 2

PE

=

15. Answer (3)

Hint : Work done in closed loop is zero due to

conservative field.

Sol. : Electric field lines cannot form a closed

loop.

16. Answer (4)

Hint : Use Gauss’ law

Sol. : Use concept of solid angle

Total flux = 0

2

10

Q

05

Q =

17. Answer (2)

Hint : VB = 0 after switch S is closed

Sol. :

VB = 0

2

A

kQ kQV

R R= −

2

kQ

R=

18. Answer (2)

Hint : Use combination of capacitor

Sol. : 2 8

23 3

PQ

CC C C= + =

1 1

8 31

3

ABV V

=

+

1

110 10 V11

= =

10 VABV =


4/13

19. Answer (2)

Hint : 0

=Q

E ds

Sol. : 0

4Q

=

0

1

4

Q =

20. Answer (4)

Hint : 0

=

P

RV

Sol. : 0

P

RV

=

( )2

2Q

R =

2

0

2P

R QV

R

=

2

0

2Q

R

=

21. Answer (2)

Hint : 2ABeqC C=

Sol. : ABeqQ C V=

= (2C) V

= 2 × 1 × 12

= 24 µC

22. Answer (4)

Hint : Potential increases opposite to the

direction of electric field

Sol. : EA > EB

Since field lines are more dense at A and

VA > VB

23. Answer (1)

Hint : Use symmetry

Sol. : 1 1 1 1

...1 3 9C

= + + + ∞

1 3

1 21

3

= =

−

2

F3

C =

4

2 F3

ABC C = =

24. Answer (2)

Hint : Show charge distribution on ring

Sol. :

25. Answer (4)

Hint : 0

inqE ds =

Sol. : Total0

(2 )Q =

0 0

1 2

3 4

Q Q = −

= 0

7

12

Q


5/13

26. Answer (4)

Hint : Potential become same

Sol. : V1 = V2

⟹ Field will be zero in between them

27. Answer (3)

Hint : For maximum field

0dE

dz=

Sol. : Calculate electric field at z

Then 02

d RE z

dz= =

28. Answer (2)

Hint :

=

KQV

R

Sol. : 2

2 2 3B

KQ K Q KQV

a a a= + −

1 1

12 3

KQ

a

= + −

3 6 2

6

KQ

a

+ − =

7

6

KQ

a=

29. Answer (1)

Hint : | |P Q=

Sol. : ( 3)rP Q=

30. Answer (1)

Hint : 03

rE

=

Sol. : 03

rE

=

0

| |6

RE

=

31. Answer (4)

Hint : Body diagonal plane contains 2 Zn2+

ions.

Sol. : Body diagonal plane will consist of four

S2– ions at the corners 1

48

and two

S2– ions at the face centre 1

22

.

32. Answer (3)

Hint : Mole of H2 = 2 mole of e–

Sol. : 2

3

H

1.93 5 60n 3 10

2 96500

− = =

33. Answer (3)

Hint : LHE is anode.

RHE is cathode.

Sol. :

11

22

1 22 1

2M

P

2M

P

2 2M M

P P

A : 2Cl Cl 2e

C : Cl 2e 2Cl

2Cl Cl Cl 2Cl

− −

−−

−−

⎯→ +

+ ⎯→

+ ⎯→ +

34. Answer (2)

Hint : In the titration, AD is precipitated so A+ are replaced by C+.

Sol. : Since, conductance does not change so mobility of C+ is comparable to A+.

35. Answer (4)

Hint : The liquids may be immiscible or showing positive deviation from Raoults law.

Sol. : For solution with positive deviation, the more volatile component may have very low mole fraction in the liquid phase.

36. Answer (2)

Hint : 0

23 3

z(M )d

6 10 a=

PART - B (CHEMISTRY)


6/13

M0 = 58

d = 2.48 gcm–3

Sol. : 3 22

23

4 58a 1.56 10

6 10 2.48

−= =

a = 5.38 × 10–8 cm = 538 pm

a

269 pm2

=

37. Answer (2)

Hint : Mole fraction of solute will be equal in both the beakers after a long time.

Sol. : Initially

Beaker A : n moles solute

4 moles of water

Beaker B : 2 moles of solute

3 moles of water

Finally

There is shifting of 1

2 mole of water to

beaker B.

n 2

1 1n 4 2 3

2 2

=

+ − + +

n = 2

28 g of X contains 2 moles

M = 14 g/mol

38. Answer (4)

Hint : o o ocell cathode AnodeE E E= −

Sol. : For (4), Eo = 2.126 V.

39. Answer (3)

Hint : Electrolyte is paste of KOH and ZnO

Sol. : It does not involve any ion whose conc. can change during its life time

40. Answer (1)

Hint : For positive deviation, Pactual > PRaoult

Sol. : ∆H > 0 and ∆V > 0

41. Answer (4)

Hint : van't Hoff factor = number of ions furnished by 1 mole.

Sol. : Ca3(PO4)2, i = 5

Na4[Fe(CN)6], i = 5

42. Answer (3)

Hint : 24 2

MnO 8H 5e Mn 4H O− + − ++ + ⎯→ +

Sol. : E = 8

0.06 1E log

5 [H ]+ −

= E° – 0.096 pH

= 1.51 – 0.096 (pH)

43. Answer (4)

Hint : meq of NaOH = meq of H2SO4

Sol. : Neutralisation

1.21000 M 20 2

40 =

M = 0.75

Now, wt. of H2SO4 in 1 lit solution = 0.75 × 98 = 73.5 and wt. of 1 lit solution = 1024.5

wt. of solvent = 951

Mass percent of solute

= 73.5

10001024.5

= 7.17%

0.75 1000

m 0.79951

= =

And mole fraction

= 0.75

0.014951

0.7518

=

+

44. Answer (2)

Hint : P = KH · X

Sol. : H

500K

0.01= = 5 × 10+4

= 50 k torr

As temperature increases, Henry’s

constant increases. As solvent solute

interactions become stronger, value of

Henry’s constant decreases.

45. Answer (1)

Hint : When the last trace of liquid disappears,

the vapor contains 6 moles of A and 4

moles of B.


7/13

Sol. : oA

P 30= , oB

P 50=

xA = ?, xB = ?

A

3y

5= ,

B

2y

5=

Now o

A A AA o o

T A A B B

P P xy

P P x P x= =

+

( )

o

A AA o o o

A B A B

P xy

P P x P=

− +

A

A

30x3

5 50 20x=

−

10 – 4xA = 10xA

A

10x

14=

∴ T

10 4 500P 30 50

14 14 14= + =

46. Answer (2)

Hint : meq of acid = meq of base

1.51 1000 25 0.2

M =

M = 300

Sol. : Molality = 2.5 1000 1

300 250 30

=

Tf = iKfm

10.1 i(1.86)

30=

i = 1.61

1 + = 1.61

= 0.61

47. Answer (4)

Hint : Tb (solution) = Tb (solvent) + Kb · m.

Sol. : Solution with the least boiling point will be decided by the value of ‘m’.

48. Answer (1)

Hint : Entropy of solid is lesser than liquid.

Sol. : As Ssolvent < Ssolution

So ( ) ( )fusion fusionsolvent solutionS S

49. Answer (2)

Hint : Reactions take place in basic medium.

Sol. : Cathode: O2(g) + H2O(l) + 4e– ⎯→

4OH–(aq)

Anode: 2H2(g) + 4OH–(aq) ⎯→

4H2O(l) + 4e–

50. Answer (2)

Hint : Charge in faradays passed = g eq of species produced

Sol. : Charge passed =1.93 2500 40

100

= 1930 C

Number of m moles of e– = 20

Now, A : H2O ⎯→ 1

2O2 + 2H+ + 2e–

C : Cu2+ + 2e– ⎯→ Cu

m moles of Cu deposited = 10

m moles of H+ produced = 20

3

220 10 2[H ] 10

3 3

−+ −

= =

pH = 2 + log3 – log2 = 2.18

51. Answer (3)

Hint : At Cathode, reduction takes place

Sol. : At Anode, oxidation takes place

52. Answer (2)

Hint : In conductivity cell, AC is used.

Sol. : DC changes the composition of solution

and KCl cannot be used as Cl– react with

Ag+ to form AgCl.

53. Answer (4)

Hint : o

cell cell

0.06E E logQ

2= −

Sol. :

3 2

2

3 2 2

C : B e B

A : A A 2e

A 2B A 2B

+ − +

+ −

+ + +

+ ⎯→

⎯→ +

+ ⎯→ +

2 2 2 2

3 2

[B ] [A ] xQ

0.1[B ]

+ +

+

= =

2o 0.06 x

E E log2 0.1

= −

2x2.03 2 0.03 log

0.1

= −


8/13

2xlog 1

0.1

= −

x = 0.1

54. Answer (3)

Hint : m

K 1000

C

=

Sol. : K 1000

1000.1

0.82

=

K = 4 × 10–3

55. Answer (2)

Hint : Tetragonal system has all angles equal to 90°.

Sol. : Trigonal system has all angles not equal to 90°.

56. Answer (4)

Hint : Electrical neutrality must be maintained in ionic solids.

Sol. : Presence of F-centres impart colour.

57. Answer (2)

Hint : If edge length is a, then 3 a 4R= .

Sol. : 4R

a3

=

Now 2R + 2x = a

x = a 2R

2

−

=

4R2R

3

2

−

=

2 3R

3

−

58. Answer (1)

Hint : In AB type structure, the cation A+ can occupy either all octahedral voids or half the tetrahedral voids.

Sol. : In AB2 type structure, co-ordination number ratio of A2+ : B– is 2 : 1.

59. Answer (1)

Hint : In HCP, there are 6 atoms per unit cell.

Sol. : 8 tetrahedral voids are completely inside. There are tetrahedral voids on the vertical edges which are shared.

60. Answer (2)

Hint : Paramagnetic substances get magnetised in a magnetic field and lose their magnetism when the field is removed.

Sol. : Ferrimagnetic substances have domains oriented oppositely in unequal numbers.

61. Answer (4)

Hint : Draw graph

Sol. :

n(A ∩ B) = 3

⟹ n(P (A ∩ B)) = 23 = 8

62. Answer (3)

Hint : Venn diagram

Sol. :

63. Answer (4)

Hint : y = x

Sol. : ∵ sin–1x + cos–1y =

2 …(1)

PART - C (MATHEMATICS)


9/13

(x, y) ∈ R

∵ sin–1x + cos–1x =

2 ⟹ (x, x) ∈ R

⟹ R is reflexive

Let sin–1y + cos–1x = k …(2)

From (1) + (2)

− − = + =1 1sin cos

2 2k y x

⟹ (y, x) ∈ R ⟹ R is symmetric

Let (y, z) ∈ R ⟹ − − + =1 1sin cos

2y z …(3)

From (1) + (3),

− − + = 1 1sin cos ( , )

2x z x z R

⟹ R is transitive

⟹ R is equivalence relation

64. Answer (4)

Hint : − −= −1 1cos sin

2x x

Sol. :

− − − − − + − −

−

1 1 1 1

2

sin sin sin sin2 4 2

016

x x x x

− − − − + − −

2 21 1 2 1sin (sin ) sin 0

2 8 2 16x x x

− −

21 2(sin ) 0

16x

− − 1sin

4 4x

1 1

2 2x−

65. Answer (3)

Hint : + 1

2,tt

when t > 0

Sol. :

1 1| | 2, | | 2,

| | | |

1| | | | 2

| || |

x yx y

x yx y

+ +

+

+ + + + + 1 1 1

| | | | | || | 6| | | | | || |

x y x yx y x y

+ + + + + =1 1 1

| | | | | || | 6| | | | | || |

x y x yx y x y

= = = = = =1 1 1

| | | | | || | 1| | | | | || |

x y x yx y x y

⟹ |x| = |y| = 1 ⟹ x = ± 1, y = ± 1

Ordered pairs are (1, 1), (–1, 1), (1, –1), (–1, –1)

66. Answer (4)

Hint : Put 9x = t

Sol. : 34x + 9|x – 1| – 10 ≤ 0

81x + 9|x – 1| – 10 ≤ 0

Put 9x = t

If x – 1 ≤ 0 If x – 1 ≥ 0

+ − 2 910 0t

t + − 2 10 0

9

tt

− + 3 10 9 0t t 1 9 9xx

(t – 1)(t2 + t – 9) ≤ 0 9t

− + − −

21 1

( 1) 9 02 4

t t + 2 829

tt

− + +

+ −

1 37( 1)

2 2

1 370

2 2

t t

t

+ − 2 10 09

tt

Has no solution.

− − −

37 1( 1) 0

2t t

−

37 11

2t


10/13

−

37 11 9

2

x

−

9

37 10 log

2x

67. Answer (1)

Hint : Draw graph

Sol. : 1

11 1

xy

x x= = −

+ +

(y – 1)(x + 1) = –1

[0,1)y

f(x) is one-one and into function

68. Answer (3)

Hint : sinx ∈ [–1, 1]

Sol. : 2sinx + 2y = 1

− sin11 sin 1 2 2

2

xx

− − −sin12 2

2

x

− − − − sin 111 2 1 1 2 2

2

x y

− −( , 1]y

69. Answer (4)

Hint : ln(1 + x) is an integer

Sol. : x + 1 > 0 1x −

ln(1 + x) + [(1+ x)2] – 3 is an integer

⇒ ln(1 + x) is an integer

⇒ [ln(1 + x)] = ln(1 + x)

⇒ [(1 + x)2] = 3

3 ≤ (1 + x)2 < 4

+ 3 |1 | 2x

ln 3 ln(1 x) ln2 +

ln(1 + x) is integer for no value of x

70. Answer (3)

Hint : Break G.I.F

Sol. :

3, [ 1, sin1)

2, [ sin1, 0)( )

0, [0, sin1)

1, [sin1,1]

− − −

− −=

x

xf x

x

x

71. Answer (1)

Hint : |sinx| + |cosx| ∈ [1, 2]

Sol. : + | sin | | cos | [1, 2]x x

+ =[| sin | | cos |] 1x x

⟹ domain of f(x) is ϕ (empty set)

72. Answer (2)

Hint : 0 ≤ {x} < 1

Sol. : ∵ sgn(sin–1x) = {–1, 0, 1} and {2x} ∈ [0, 1)

⟹ sgn(sin–1x) = {2x} = 0 ⟹ x = 0

73. Answer (2)

Hint : − −=−

1 1

2

22tan tan

1

xx

x

Sol. : − − =

1 11 14tan 2 2tan

5 5

− −

= = −

1 1

12

552 tan 2tan1 12

125


11/13

− −

= = −

1 1

52

12012tan tan25 119

1144

− − −

1 114tan cot (239)

5

− − = −

1 1120 1tan tan

119 239

− −

−

= = = +

1 1

120 1

119 239tan tan (1)120 1 4

1119 239

74. Answer (1)

Hint : Find domain

Sol. : − − 1 1

1 2 12 2

x x

Let f(x) = sin–1(2x) – cos–1x + tan–1(2x)

sin–12x is increasing

cos–1x is decreasing

and tan–1(2x) is increasing function

−

= − = − − − =

min

1 2 17( )

2 2 3 4 12f x f

= = − + =

max.

1 5( )

2 2 3 4 12f x f

=( )2

f x does not have any solution

75. Answer (4)

Hint : Draw graph of sin–1(sinx)

Sol. :

76. Answer (2)

Hint : − − −

=

1 1 1tan tan tan1

x yx y

xy

Sol. : tan cot 1 1x x = −

1 1 1 tan cottan (tan ) tan (cot ) tan

1 tan cot

− − − − − =

+

x xx x

x x

21 1tan 1

tan tan ( cot 2 )2 tan

− − −

= = −

xx

x

1 1tan (cot2 ) tan tan 22

− − = − = − −

x x

1tan tan 22

− = −

x

=

2T

77. Answer (1)

Hint : − − 1sin

2 2x

Sol. : − − 12tan

2 2a

− − − 1tan 1 1

4 4a a

78. Answer (1)

Hint : − −+= −

+ +

1 1( 1)tan tan

1 1n

n r nrT

n n

Sol. : − +

=+ + +

1

2 2

( 1)tan

( 1) ( 1)n

n nT

n n r r

−

+=

+ + + +

1 1tan( 1)

11 1

n

nn r nr

n n

1

( 1)

1 1tan( 1)

11 1

n r nr

n nn r nr

n n

−

+ − + +=

+ + + +

1 1( 1)tan tan

1 1

n r nr

n n

− −+= −

+ +


12/13

− − = −

+

1 1tan ( ) tan1

nS n

n

→

= − =lim

2 4 4nS

79. Answer (3)

Hint : Break the function at –1, 0, 1, 2, 3

Sol. : [ ] 0 [0,1) x x

1, 1

2, 1 0

( ) 2, 1 2

1, 2 3

0, 3

− − −

− −

=

x

x

f x x

x

x

80. Answer (4)

Hint : − − + =1 1sin cos

2x x

Sol. : − −+1 2 1 2(sin ) (cos )x x

− − − − = + − −

1 1 2 1 1(sin cos ) 2sin sin2

x x x x

− −= + −

21 2 12(sin ) sin

4x x

− −

= + − + −

2 2 21 2 12 (sin ) sin

4 2 16 8x x

= −

+ −

2212 sin

8 4x

1 13sin sin

2 2 4 4 4

− − − − − x x

− −

2 21 9

0 sin4 16

x

− − +

2 21 2 1 2 10

(sin ) (cos )8 8

x x

81. Answer (3)

Hint : − − − −− =

+

1 1 1tan tan tan1

x yx y

xy

Sol. : − − −

+ =

1 1 11tan tan tan (3)x

y

− − − − − = − = +

1 1 1 11 3tan tan 3 tan tan

1 3

xx

y x

+= = − +

− −

1 3 103

3 3

xy

x x

For positive integer y, x = 1, 2

⟹ Solutions are (1, 2)(2, 7)

82. Answer (3)

Hint : 1 – sin2 = (cos1 – sin1)2

Sol. : 1 sin2 1

tancos2

− −

1 1 tan1tan

1 tan1

− − = −

+

1tan tan 1

4

− = − −

14

= −

83. Answer (3)

Hint : Solve graphically

Sol. : −

− [ ] 1

1 1, 0x

xx

If x > 0 If x < 0

–x ≤ [x] – 1 ≤ x –x ≥ [x] – 1 ≥ x

1 – x ≤ [x] ≤ x + 1 1 – x ≥ [x] ≥ 1+ x

1 – x ≤ x – {x} ≤ x + 1 1 – x ≥ x – {x} ≥ 1 + x

− − − −1 { } 1x x x x − − − −1 { } 1x x x x

− −2 1 { } 1x x

− −2 1 { } 1x x

1x = x

[1, )x


13/13

84. Answer (2)

Hint : f(x + T) = f(x)

Sol. : ( )2

f x f x

+ =

85. Answer (2)

Hint : Find domain

Sol. : –1 ≤ x ≤ 1, –1 ≤ x + 2 ≤ 1

⟹ x = –1

( 1) 0 02 2

f

− = − − + = −

86. Answer (2)

Hint : Onto function = 0

Sol. : Number of onto functions = 0

Number of functions = 43

⟹ Number of functions which are not onto = 64

87. Answer (2)

Hint : Reflexive relations = −2

2n n

Sol. : Number of relations = 2 252 2n =

Number of reflexive relations

= 2 25 5 202 2 2n n− −= =

88. Answer (3)

Hint : Draw graph

Sol. :

• y = sin–1(sinx)

• y = sin–1(cosx) = 1cos (cos )2

x−−

89. Answer (4)

Hint : R–1 is also an equivalence relation

Sol. : The inverse of an equivalence relation is

also an equivalence relation.

90. Answer (3)

Hint : Venn diagram

Sol. :

n(P ∪ C ∪ M) = 640

⟹ n(who did not opt)

= 800 – 640 = 160

Test - 1 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 09/06/2019

ANSWERS

1/13

TEST - 1 - Code-B

PHYSICS CHEMISTRY MATHEMATICS

1. (1)

2. (1)

3. (2)

4. (3)

5. (4)

6. (4)

7. (2)

8. (1)

9. (4)

10. (2)

11. (4)

12. (2)

13. (2)

14. (2)

15. (4)

16. (3)

17. (3)

18. (4)

19. (3)

20. (4)

21. (2)

22. (1)

23. (1)

24. (1)

25. (4)

26. (3)

27. (1)

28. (3)

29. (4)

30. (3)

31. (2)

32. (1)

33. (1)

34. (2)

35. (4)

36. (2)

37. (3)

38. (4)

39. (2)

40. (3)

41. (2)

42. (2)

43. (1)

44. (4)

45. (2)

46. (1)

47. (2)

48. (4)

49. (3)

50. (4)

51. (1)

52. (3)

53. (4)

54. (2)

55. (2)

56. (4)

57. (2)

58. (3)

59. (3)

60. (4)

61. (3)

62. (4)

63. (3)

64. (2)

65. (2)

66. (2)

67. (2)

68. (3)

69. (3)

70. (3)

71. (4)

72. (3)

73. (1)

74. (1)

75. (2)

76. (4)

77. (1)

78. (2)

79. (2)

80. (1)

81. (3)

82. (4)

83. (3)

84. (1)

85. (4)

86. (3)

87. (4)

88. (4)

89. (3)

90. (4)

All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)

2/13

1. Answer (1)

Hint : 03

rE

=

Sol. : 03

rE

=

0

| |6

RE

=

2. Answer (1)

Hint : | |P Q=

Sol. : ( 3)rP Q=

3. Answer (2)

Hint :

=

KQV

R

Sol. : 2

2 2 3B

KQ K Q KQV

a a a= + −

1 1

12 3

KQ

a

= + −

3 6 2

6

KQ

a

+ − =

7

6

KQ

a=

4. Answer (3)

Hint : For maximum field

0dE

dz=

Sol. : Calculate electric field at z

Then 02

d RE z

dz= =

5. Answer (4)

Hint : Potential become same

Sol. : V1 = V2

⟹ Field will be zero in between them

6. Answer (4)

Hint : 0

inqE ds =

Sol. : Total0

(2 )Q =

0 0

1 2

3 4

Q Q = −

= 0

7

12

Q

7. Answer (2)

Hint : Show charge distribution on ring

Sol. :

8. Answer (1)

Hint : Use symmetry

Sol. : 1 1 1 1

...1 3 9C

= + + + ∞

1 3

1 21

3

= =

−

2

F3

C =

4

2 F3

ABC C = =

PART - A (PHYSICS)

Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/13

9. Answer (4)

Hint : Potential increases opposite to the

direction of electric field

Sol. : EA > EB

Since field lines are more dense at A and

VA > VB

10. Answer (2)

Hint : 2ABeqC C=

Sol. : ABeqQ C V=

= (2C) V

= 2 × 1 × 12

= 24 µC

11. Answer (4)

Hint : 0

=

P

RV

Sol. : 0

P

RV

=

( )2

2Q

R =

2

0

2P

R QV

R

=

2

0

2Q

R

=

12. Answer (2)

Hint : 0

=Q

E ds

Sol. : 0

4Q

=

0

1

4

Q =

13. Answer (2)

Hint : Use combination of capacitor

Sol. : 2 8

23 3

PQ

CC C C= + =

1 1

8 31

3

ABV V

=

+

1

110 10 V11

= =

10 VABV =

14. Answer (2)

Hint : VB = 0 after switch S is closed

Sol. :

VB = 0

2

A

kQ kQV

R R= −

2

kQ

R=

15. Answer (4)

Hint : Use Gauss’ law

Sol. : Use concept of solid angle

Total flux = 0

2

10

Q

05

Q =

16. Answer (3)

Hint : Work done in closed loop is zero due to

conservative field.

Sol. : Electric field lines cannot form a closed

loop.


4/13

17. Answer (3)

Hint : Use super position principle

Sol. : sheet holePE E E= −

02 2

PE

=

18. Answer (4)

Hint : Use symmetry

Sol. : Field due to each spherical shell will be

along –y direction.

19. Answer (3)

Hint : Electric field near a point charge will

dominated by nearer charge.

Sol. : Electric field at x = ± ∞ will tend to zero.

20. Answer (4)

Hint : 21

2U CV=

Sol. : 1 2

1 2

Q Q

C C=

21 48

(2)2 3

U

=

= 256 µJ

21. Answer (2)

Hint : 0

inq =

Sol. : 11

0

q =

22

0

q = −

3 0 =

22. Answer (1)

Hint : 0 0 0QU V Q=

Sol. : 00 0

( )ln2 ln2

4 4

LV

L

= =

0

0

ln2

4

QU

=

23. Answer (1)

Hint : Vinside = Vsurface

Sol. : Potential at inside point will be same as

potential at the surface of inside sphere.

24. Answer (1)

Hint : 0 =kQ

VR

Sol. : V at center is 00

3 3

4 2 2

QV

R =

30

0 00

3 1 1 4

2 2 2 4 3

VV V V R

R = − = =

2

06

RV

=

25. Answer (4)

Hint : 1 2=kq q

Ur

Sol. : 2 2 2

2 ...2 3

kq kq kqU

a a a

= − + − +

2 2 1 11 ...

2 3

kq

a

= − − + −

Note: 2 3 4

ln(1 ) ...2 3 4

x x xx x+ = − + − +

2 2ln(2)

kqU

a

= −

2

0

ln(2)2

qU

a = −


5/13

26. Answer (3)

Hint : 2

2

QU

C

=

Sol. : 2

0

(2 )

(2){4 2 }

QU

a=

2 2

0 0

4

16 4

Q QU

a a= =

27. Answer (1)

Hint : 0Q =

Sol. : 0Q =

⟹ Q + 2Q – Q + Q1 = 0

⟹ Q1 = –2Q

28. Answer (3)

Hint : Parallel arrangement

Sol. : Ceq = ...2 4 8

C C CC + + + + ∞

1

21

12

C C= =

−

29. Answer (4)

Hint : VP = V1 + V2

Sol. : 0 0

2 (2 )2 2

P

RR

V

= +

0

2 R =

30. Answer (3)

Hint : EA = 2

KQ

r due to shell at outside point

Sol. : 2

(3 )A

A

K QE

r=

rA is distance of A from O

⟹ EA = constant

31. Answer (2)

Hint : Paramagnetic substances get magnetised in a magnetic field and lose their magnetism when the field is removed.

Sol. : Ferrimagnetic substances have domains oriented oppositely in unequal numbers.

32. Answer (1)

Hint : In HCP, there are 6 atoms per unit cell.

Sol. : 8 tetrahedral voids are completely inside. There are tetrahedral voids on the vertical edges which are shared.

33. Answer (1)

Hint : In AB type structure, the cation A+ can occupy either all octahedral voids or half the tetrahedral voids.

Sol. : In AB2 type structure, co-ordination number ratio of A2+ : B– is 2 : 1.

34. Answer (2)

Hint : If edge length is a, then 3 a 4R= .

Sol. : 4R

a3

=

Now 2R + 2x = a

x = a 2R

2

−

=

4R2R

3

2

−

=

2 3R

3

−

35. Answer (4)

Hint : Electrical neutrality must be maintained in ionic solids.

Sol. : Presence of F-centres impart colour.

36. Answer (2)

Hint : Tetragonal system has all angles equal to 90°.

Sol. : Trigonal system has all angles not equal to 90°.

PART - B (CHEMISTRY)


6/13

37. Answer (3)

Hint : m

K 1000

C

=

Sol. : K 1000

1000.1

0.82

=

K = 4 × 10–3

38. Answer (4)

Hint : o

cell cell

0.06E E logQ

2= −

Sol. :

3 2

2

3 2 2

C : B e B

A : A A 2e

A 2B A 2B

+ − +

+ −

+ + +

+ ⎯→

⎯→ +

+ ⎯→ +

2 2 2 2

3 2

[B ] [A ] xQ

0.1[B ]

+ +

+

= =

2o 0.06 x

E E log2 0.1

= −

2x2.03 2 0.03 log

0.1

= −

2xlog 1

0.1

= −

x = 0.1

39. Answer (2)

Hint : In conductivity cell, AC is used.

Sol. : DC changes the composition of solution

and KCl cannot be used as Cl– react with

Ag+ to form AgCl.

40. Answer (3)

Hint : At Cathode, reduction takes place

Sol. : At Anode, oxidation takes place

41. Answer (2)

Hint : Charge in faradays passed = g eq of species produced

Sol. : Charge passed =1.93 2500 40

100

= 1930 C

Number of m moles of e– = 20

Now, A : H2O ⎯→ 1

2O2 + 2H+ + 2e–

C : Cu2+ + 2e– ⎯→ Cu

m moles of Cu deposited = 10

m moles of H+ produced = 20

3220 10 2

[H ] 103 3

−+ −

= =

pH = 2 + log3 – log2 = 2.18

42. Answer (2)

Hint : Reactions take place in basic medium.

Sol. : Cathode: O2(g) + H2O(l) + 4e– ⎯→

4OH–(aq)

Anode: 2H2(g) + 4OH–(aq) ⎯→

4H2O(l) + 4e–

43. Answer (1)

Hint : Entropy of solid is lesser than liquid.

Sol. : As Ssolvent < Ssolution

So ( ) ( )fusion fusionsolvent solutionS S

44. Answer (4)

Hint : Tb (solution) = Tb (solvent) + Kb · m.

Sol. : Solution with the least boiling point will be decided by the value of ‘m’.

45. Answer (2)

Hint : meq of acid = meq of base

1.51 1000 25 0.2

M =

M = 300

Sol. : Molality = 2.5 1000 1

300 250 30

=

Tf = iKfm

10.1 i(1.86)

30=

i = 1.61

1 + = 1.61

= 0.61

46. Answer (1)

Hint : When the last trace of liquid disappears,

the vapor contains 6 moles of A and 4

moles of B.

Sol. : oA

P 30= , oB

P 50=

xA = ?, xB = ?


7/13

A

3y

5= ,

B

2y

5=

Now o

A A AA o o

T A A B B

P P xy

P P x P x= =

+

( )

o

A AA o o o

A B A B

P xy

P P x P=

− +

A

A

30x3

5 50 20x=

−

10 – 4xA = 10xA

A

10x

14=

∴ T

10 4 500P 30 50

14 14 14= + =

47. Answer (2)

Hint : P = KH · X

Sol. : H

500K

0.01= = 5 × 10+4

= 50 k torr

As temperature increases, Henry’s

constant increases. As solvent solute

interactions become stronger, value of

Henry’s constant decreases.

48. Answer (4)

Hint : meq of NaOH = meq of H2SO4

Sol. : Neutralisation

1.21000 M 20 2

40 =

M = 0.75

Now, wt. of H2SO4 in 1 lit solution = 0.75 × 98 = 73.5 and wt. of 1 lit solution = 1024.5

wt. of solvent = 951

Mass percent of solute

= 73.5

10001024.5

= 7.17%

0.75 1000

m 0.79951

= =

And mole fraction

= 0.75

0.014951

0.7518

=

+

49. Answer (3)

Hint : 24 2

MnO 8H 5e Mn 4H O− + − ++ + ⎯→ +

Sol. : E = 8

0.06 1E log

5 [H ]+ −

= E° – 0.096 pH

= 1.51 – 0.096 (pH)

50. Answer (4)

Hint : van't Hoff factor = number of ions furnished by 1 mole.

Sol. : Ca3(PO4)2, i = 5

Na4[Fe(CN)6], i = 5

51. Answer (1)

Hint : For positive deviation, Pactual > PRaoult

Sol. : ∆H > 0 and ∆V > 0

52. Answer (3)

Hint : Electrolyte is paste of KOH and ZnO

Sol. : It does not involve any ion whose conc. can change during its life time

53. Answer (4)

Hint : o o ocell cathode AnodeE E E= −

Sol. : For (4), Eo = 2.126 V.

54. Answer (2)

Hint : Mole fraction of solute will be equal in both the beakers after a long time.

Sol. : Initially

Beaker A : n moles solute

4 moles of water

Beaker B : 2 moles of solute

3 moles of water

Finally

There is shifting of 1

2 mole of water to

beaker B.

n 2

1 1n 4 2 3

2 2

=

+ − + +


8/13

n = 2

28 g of X contains 2 moles

M = 14 g/mol

55. Answer (2)

Hint : 0

23 3

z(M )d

6 10 a=

M0 = 58

d = 2.48 gcm–3

Sol. : 3 22

23

4 58a 1.56 10

6 10 2.48

−= =

a = 5.38 × 10–8 cm = 538 pm

a

269 pm2

=

56. Answer (4)

Hint : The liquids may be immiscible or showing positive deviation from Raoults law.

Sol. : For solution with positive deviation, the more volatile component may have very low mole fraction in the liquid phase.

57. Answer (2)

Hint : In the titration, AD is precipitated so A+ are replaced by C+.

Sol. : Since, conductance does not change so mobility of C+ is comparable to A+.

58. Answer (3)

Hint : LHE is anode.

RHE is cathode.

Sol. :

11

22

1 22 1

2M

P

2M

P

2 2M M

P P

A : 2Cl Cl 2e

C : Cl 2e 2Cl

2Cl Cl Cl 2Cl

− −

−−

−−

⎯→ +

+ ⎯→

+ ⎯→ +

59. Answer (3)

Hint : Mole of H2 = 2 mole of e–

Sol. : 2

3

H

1.93 5 60n 3 10

2 96500

− = =

60. Answer (4)

Hint : Body diagonal plane contains 2 Zn2+

ions.

Sol. : Body diagonal plane will consist of four

S2– ions at the corners 1

48

and two

S2– ions at the face centre 1

22

.

61. Answer (3)

Hint : Venn diagram

Sol. :

n(P ∪ C ∪ M) = 640

⟹ n(who did not opt)

= 800 – 640 = 160

62. Answer (4)

Hint : R–1 is also an equivalence relation

Sol. : The inverse of an equivalence relation is

also an equivalence relation.

63. Answer (3)

Hint : Draw graph

Sol. :

• y = sin–1(sinx)

• y = sin–1(cosx)

= 1cos (cos )2

x−−

PART - C (MATHEMATICS)


9/13

64. Answer (2)

Hint : Reflexive relations = −2

2n n

Sol. : Number of relations = 2 252 2n =

Number of reflexive relations

= 2 25 5 202 2 2n n− −= =

65. Answer (2)

Hint : Onto function = 0

Sol. : Number of onto functions = 0

Number of functions = 43

⟹ Number of functions which are not onto = 64

66. Answer (2)

Hint : Find domain

Sol. : –1 ≤ x ≤ 1, –1 ≤ x + 2 ≤ 1

⟹ x = –1

( 1) 0 02 2

f

− = − − + = −

67. Answer (2)

Hint : f(x + T) = f(x)

Sol. : ( )2

f x f x

+ =

68. Answer (3)

Hint : Solve graphically

Sol. : −

− [ ] 1

1 1, 0x

xx

If x > 0 If x < 0

–x ≤ [x] – 1 ≤ x –x ≥ [x] – 1 ≥ x

1 – x ≤ [x] ≤ x + 1 1 – x ≥ [x] ≥ 1+ x

1 – x ≤ x – {x} ≤ x + 1 1 – x ≥ x – {x} ≥ 1 + x

− − − −1 { } 1x x x x − − − −1 { } 1x x x x

− −2 1 { } 1x x

− −2 1 { } 1x x

1x = x

[1, )x

69. Answer (3)

Hint : 1 – sin2 = (cos1 – sin1)2

Sol. : 1 sin2 1

tancos2

− −

1 1 tan1tan

1 tan1

− − = −

+

1tan tan 1

4

− = − −

14

= −

70. Answer (3)

Hint : − − − −− =

+

1 1 1tan tan tan1

x yx y

xy

Sol. : − − −

+ =

1 1 11tan tan tan (3)x

y

− − − − − = − = +

1 1 1 11 3tan tan 3 tan tan

1 3

xx

y x

+= = − +

− −

1 3 103

3 3

xy

x x


10/13

For positive integer y, x = 1, 2

⟹ Solutions are (1, 2)(2, 7)

71. Answer (4)

Hint : − − + =1 1sin cos

2x x

Sol. : − −+1 2 1 2(sin ) (cos )x x

− − − − = + − −

1 1 2 1 1(sin cos ) 2sin sin2

x x x x

− −= + −

21 2 12(sin ) sin

4x x

− −

= + − + −

2 2 21 2 12 (sin ) sin

4 2 16 8x x

= −

+ −

2212 sin

8 4x

1 13sin sin

2 2 4 4 4

− − − − − x x

− −

2 21 9

0 sin4 16

x

− − +

2 21 2 1 2 10

(sin ) (cos )8 8

x x

72. Answer (3)

Hint : Break the function at –1, 0, 1, 2, 3

Sol. : [ ] 0 [0,1) x x

1, 1

2, 1 0

( ) 2, 1 2

1, 2 3

0, 3

− − −

− −

=

x

x

f x x

x

x

73. Answer (1)

Hint : − −+= −

+ +

1 1( 1)tan tan

1 1n

n r nrT

n n

Sol. : − +

=+ + +

1

2 2

( 1)tan

( 1) ( 1)n

n nT

n n r r

−

+=

+ + + +

1 1tan( 1)

11 1

n

nn r nr

n n

1

( 1)

1 1tan( 1)

11 1

n r nr

n nn r nr

n n

−

+ − + +=

+ + + +

1 1( 1)tan tan

1 1

n r nr

n n

− −+= −

+ +

− − = −

+

1 1tan ( ) tan1

nS n

n

→

= − =lim

2 4 4nS

74. Answer (1)

Hint : − − 1sin

2 2x

Sol. : − − 12tan

2 2a

− − − 1tan 1 1

4 4a a

75. Answer (2)

Hint : − − −

=

1 1 1tan tan tan1

x yx y

xy

Sol. : tan cot 1 1x x = −

1 1 1 tan cottan (tan ) tan (cot ) tan

1 tan cot

− − − − − =

+

x xx x

x x

21 1tan 1

tan tan ( cot 2 )2 tan

− − −

= = −

xx

x

1 1tan (cot2 ) tan tan 22

− − = − = − −

x x

1tan tan 22

− = −

x

=

2T

76. Answer (4)

Hint : Draw graph of sin–1(sinx)


11/13

Sol. :

77. Answer (1)

Hint : Find domain

Sol. : − − 1 1

1 2 12 2

x x

Let f(x) = sin–1(2x) – cos–1x + tan–1(2x)

sin–12x is increasing

cos–1x is decreasing

and tan–1(2x) is increasing function

−

= − = − − − =

min

1 2 17( )

2 2 3 4 12f x f

= = − + =

max.

1 5( )

2 2 3 4 12f x f

=( )2

f x does not have any solution

78. Answer (2)

Hint : − −=−

1 1

2

22tan tan

1

xx

x

Sol. : − − =

1 11 14tan 2 2tan

5 5

− −

= = −

1 1

12

552 tan 2tan1 12

125

− −

= = −

1 1

52

12012tan tan25 119

1144

− − −

1 114tan cot (239)

5

− − = −

1 1120 1tan tan

119 239

− −

−

= = = +

1 1

120 1

119 239tan tan (1)120 1 4

1119 239

79. Answer (2)

Hint : 0 ≤ {x} < 1

Sol. : ∵ sgn(sin–1x) = {–1, 0, 1} and {2x} ∈ [0, 1)

⟹ sgn(sin–1x) = {2x} = 0 ⟹ x = 0

80. Answer (1)

Hint : |sinx| + |cosx| ∈ [1, 2]

Sol. : + | sin | | cos | [1, 2]x x

+ =[| sin | | cos |] 1x x

⟹ domain of f(x) is ϕ (empty set)

81. Answer (3)

Hint : Break G.I.F

Sol. :

3, [ 1, sin1)

2, [ sin1, 0)( )

0, [0, sin1)

1, [sin1,1]

− − −

− −=

x

xf x

x

x

82. Answer (4)

Hint : ln(1 + x) is an integer

Sol. : x + 1 > 0 1x −

ln(1 + x) + [(1+ x)2] – 3 is an integer

⇒ ln(1 + x) is an integer

⇒ [ln(1 + x)] = ln(1 + x)

⇒ [(1 + x)2] = 3

3 ≤ (1 + x)2 < 4

+ 3 |1 | 2x

ln 3 ln(1 x) ln2 +

ln(1 + x) is integer for no value of x


12/13

83. Answer (3)

Hint : sinx ∈ [–1, 1]

Sol. : 2sinx + 2y = 1

− sin11 sin 1 2 2

2

xx

− − −sin12 2

2

x

− − − − sin 111 2 1 1 2 2

2

x y

− −( , 1]y

84. Answer (1)

Hint : Draw graph

Sol. : 1

11 1

xy

x x= = −

+ +

(y – 1)(x + 1) = –1

[0,1)y

f(x) is one-one and into function

85. Answer (4)

Hint : Put 9x = t

Sol. : 34x + 9|x – 1| – 10 ≤ 0

81x + 9|x – 1| – 10 ≤ 0

Put 9x = t

If x – 1 ≤ 0 If x – 1 ≥ 0

+ − 2 910 0t

t + − 2 10 0

9

tt

− + 3 10 9 0t t 1 9 9xx

(t – 1)(t2 + t – 9) ≤ 0 9t

− + − −

21 1

( 1) 9 02 4

t t + 2 829

tt

− + +

+ −

1 37( 1)

2 2

1 370

2 2

t t

t

+ − 2 10 09

tt

Has no solution.

− − −

37 1( 1) 0

2t t

−

37 11

2t

−

37 11 9

2

x

−

9

37 10 log

2x

86. Answer (3)

Hint : + 1

2,tt

when t > 0

Sol. :

1 1| | 2, | | 2,

| | | |

1| | | | 2

| || |

x yx y

x yx y

+ +

+

+ + + + + 1 1 1

| | | | | || | 6| | | | | || |

x y x yx y x y

+ + + + + =1 1 1

| | | | | || | 6| | | | | || |

x y x yx y x y

= = = = = =1 1 1

| | | | | || | 1| | | | | || |

x y x yx y x y

⟹ |x| = |y| = 1 ⟹ x = ± 1, y = ± 1

Ordered pairs are (1, 1), (–1, 1), (1, –1), (–1, –1)


13/13

87. Answer (4)

Hint : − −= −1 1cos sin

2x x

Sol. :

− − − − − + − −

−

1 1 1 1

2

sin sin sin sin2 4 2

016

x x x x

− − − − + − −

2 21 1 2 1sin (sin ) sin 0

2 8 2 16x x x

− −

21 2(sin ) 0

16x

− − 1sin

4 4x

1 1

2 2x−

88. Answer (4)

Hint : y = x

Sol. : ∵ sin–1x + cos–1y =

2 …(1)

(x, y) ∈ R

∵ sin–1x + cos–1x =

2 ⟹ (x, x) ∈ R

⟹ R is reflexive

Let sin–1y + cos–1x = k …(2)

From (1) + (2)

− − = + =1 1sin cos

2 2k y x

⟹ (y, x) ∈ R ⟹ R is symmetric

Let (y, z) ∈ R ⟹ − − + =1 1sin cos

2y z …(3)

From (1) + (3),

− − + = 1 1sin cos ( , )

2x z x z R

⟹ R is transitive

⟹ R is equivalence relation

89. Answer (3)

Hint : Venn diagram

Sol. :

90. Answer (4)

Hint : Draw graph

Sol. :

n(A ∩ B) = 3

⟹ n(P (A ∩ B)) = 23 = 8

(Answers) All India Aakash Test Series for JEE (Main)-2020 ......Sep 06, 2019 · U CV Sol. : 12 12...

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Transcript of (Answers) All India Aakash Test Series for JEE (Main)-2020 ......Sep 06, 2019 · U CV Sol. : 12 12...