Mock Test - 5 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020 MOCK ... · 2020. 4. 7. · Mock Test - 5 (Code-A) (Answers) All India Aakash Test Series for JEE

Mock Test - 5 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 22/03/2020

ANSWERS

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

MOCK TEST - 5 - Code-A

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (3)

2. (4)

3. (1)

4. (2)

5. (4)

6. (1)

7. (4)

8. (1)

9. (2)

10. (4)

11. (2)

12. (4)

13. (3)

14. (1)

15. (3)

16. (4)

17. (1)

18. (3)

19. (1)

20. (1)

21. (03)

22. (07)

23. (09)

24. (06)

25. (02)

26. (1)

27. (4)

28. (1)

29. (1)

30. (1)

31. (3)

32. (1)

33. (3)

34. (1)

35. (1)

36. (2)

37. (3)

38. (3)

39. (1)

40. (1)

41. (4)

42. (4)

43. (3)

44. (2)

45. (3)

46. (11)

47. (05)

48. (04)

49. (03)

50. (16)

51. (1)

52. (3)

53. (3)

54. (1)

55. (2)

56. (4)

57. (2)

58. (4)

59. (1)

60. (2)

61. (4)

62. (2)

63. (1)

64. (2)

65. (2)

66. (3)

67. (4)

68. (1)

69. (1)

70. (4)

71. (02)

72. (39)

73. (33)

74. (08)

75. (16)

All India Aakash Test Series for JEE (Main)-2020 Mock Test - 5 (Code-A) (Hints & Solutions)


2/10

1. Answer (3)

Hint : c ta a a= +

Sol. : 2

2 3MMg = α

32 4ct

ga α= =

2

212 3 2

M Mgω =

2 32 2c

ga = ω =

5 34

ga∴ =

2. Answer (4)

Hint : drvdt

=

dvadt

=

Sol. : =

drvdt

dvadt

=

cos| || |

a va v

⋅θ =

3. Answer (1)

Hint : Resolve the dipole moment along X-axis and Y-axis respectively.

Sol. : ˆ ˆcos sinp p i p j= θ − θ

3 30 0

cos 2 sinˆ ˆ4 4p pE i j

d d− θ θ

= −πε πε

4. Answer (2)

Hint : Apply condition of equilibrium in respective direction.

Sol. : Tcos30° = mg

Tsin30° = 02

σεQ

5. Answer (4)

Hint : 1 1 813.6 13.61 9 9

E = − = ×

Sol. : 0E eV= φ +

012.1 1.5 eV= +

010.6 V=

6. Answer (1)

Hint : 20

1,2

Fa x atm

= =

Sol. : P F V= ⋅

2P F ax= ⋅

P x∝

7. Answer (4)

Hint : Initially, T1cos30° = mg

Sol. : T1cos30° = mg

T2 = mgcos30°

8. Answer (1)

Hint : Apply circuit diagram and change = final reading – initial reading

Sol. : eqeq

eq 1

20, 50i

RV R

R R= = Ω

+

20 100 10 V200fV ×

= =

9. Answer (2)

Hint : 2E A∝

Sol. : 2i iEK U= =

Finally, total energy is 3 .2E

2

23 ( )2 ( )

E AE A

′=

3 A2

′ =A

PART - A (PHYSICS)

Mock Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020


3/10

10. Answer (4)

Hint : For minimum velocity, velocity of person should be equal to, velocity at the time of boarding the bus.

Sol. : 2v ax=

21 ,2

vs at ta

= =

11. Answer (2)

Hint : Write equation of line as v = a + bT

Sol. : pv = nRT

nRTpa bT

=+

12. Answer (4)

Hint : Path difference increases with θ.

Sol. : max4( ) sin 43

∆ = × × θ = λx d

Number of maxima = (4 × 2 + 1) = 9

13. Answer (3)

Hint : Wavelength is dependent on length only.

Sol. : Wavelength is dependent on length of tube.

14. Answer (1)

Hint : Find current in individual branch using phasor. Add the current based on vector addition.

Sol. :

0 0 sin302

= °ωv v

L R

ωL = 4R

4=

ωRL

15. Answer (3)

Hint : 1 2 1 1 1 1 0 0M L T , L T , M M L T− −= = = v

Sol. : v Mα β γ= P

1 1 2M L TP − −=

16. Answer (4)

Hint : 0 [sin sin ]4

IBd

µ= α + β

π

Sol. : 02 [sin45 sin90]4

IBa

µ= +

π

0

7 7

24 2 2

2(10 ) 10 T2

IBa

− −

µ = π −

= =

17. Answer (1)

Hint : 1 2B B B= +

Sol. : 01 3

24

MB

dµ

=π

02 34

MB

dµ

=π

2 21 2B B B= +

18. Answer (3)

Hint : 4

8π

=

PrQn

Sol. : 4•

1 11 8

π=

P rQ

n

4•

2 32 8

π=

P rQ

n

• •

1 2 1 216Q Q P P= ⇒ =

19. Answer (1)

Hint : Let distance be x from end P.

Sol. : (200 0 ) 80° − °=

kA dmx dt

(200 100) 540( )

−=

−kA dm

x dt

Find x.



4/10

20. Answer (1)

Hint : Silvered lens will act as converging mirror of focal length 20 cm.

Sol. : For image to be formed on object itself, rays should fall normally on mirror.

d – 40 = 20

d = 60

21. Answer (03)

Hint : Redraw it to form the equivalent circuit.

Sol. : All capacitors are connected in parallel.

0eq

3 ACdε

=

22. Answer (07)

Hint : 2 ,iv gR= write velocity as a function

of h.

Sol. : 21 02 ( )

GMmmvR x

− =+

2GMvR x

=+

( )2dx GM

dt R x=

+

3

0 0

2R t

R x dx GM dt+ =∫ ∫

Solve for t.

23. Answer (09)

Hint : Let temperature be T.

2001 1450T

T− = −

Sol. : 2001 1450T

T− = −

T = 300 K

1 213

n n= =

Total work doneHeat given

η =

24. Answer (06)

Hint : 1 1 1, vmv u f u

− = =

Sol. : 1 1 1 , 30 cm15 10

+ = =vv

m = –2

Images formed will be 2 mm above and 2 mm below respective to P axis.

2 2 2 6 mmd = + + =

25. Answer (02)

Hint : At the time of maximum angular displacement, kinetic energy is zero.

Sol. : Initially

2

20

1(1 cos )2 2 3

− θ = ωL MLMg

Finally

2

21

1(1 cos )2 4 2 2 4 3

− θ = ω×

M L M Lg

Compare and get the result.

26. Answer (1)

Hint : ureanM V (L) 2 LV (L)

= ⇒ =

Mass of solution = 2000 × 2 = 4000 g

Sol. : Mass of solvent = 4000 – 120 = 3880 g

urean 2m 0.52 mMass of solvent 3.880

= = =

27. Answer (4) Hint : Number of photoelectron ∝ intensity

Sol. : Energy of an orbital ∝ (Z2)

28. Answer (1)

Hint : 2NO 180+ → °

2NO 120− < °

Sol. : 2 2 2NO NO NO+ −> >

PART - B (CHEMISTRY)



5/10

29. Answer (1)

Hint : M occupies less volume than N. O is more compressible than P.

Sol. : If two gases have same value of ‘b’ but different values of ‘a’, then the gas having a larger value ‘a’ will occupy lesser volume.

If two gases have the same value of ‘a’ but different values of ‘b’, then smaller the value of b, larger will be the compressibility.

30. Answer (1)

Hint : ∆UC→A = 0

Sol. : ∆H ∝ ∆T

31. Answer (3)

Hint : NaH, LiH and KH are ionic hydride.

Sol. : Covalent hydride Ionic Metallic

NH3 NaH CrH

HF LiH YbH2.55

CH4 ZrH1.3

BeH2

32. Answer (1)

Hint : For fluoride ∆fH° becomes less negative

as we move top to bottom.

Sol. : ∆fH° value of chloride, bromide and iodide

become more negative as we move top to

bottom.

33. Answer (3)

Hint : 2 6A B H⇒

3 3 6D : B N H

Sol. : B2H6 contains 3c–2e– as well as 2c–2e–

bonds while D contains all equal bond lengths.

34. Answer (1)

Hint : +I power of –CH3 is less than –CD3 while

+H power of –CH3 is more than –CD3.

Sol. : Correct stability order is

D > E > C > B > A

35. Answer (1)

Hint : In E2 reaction, anti elimination of H and Br takes place.

Sol. :

36. Answer (2)

Hint : Compound contains double bond.

Sol. :

37. Answer (3)

Hint : A and B are identical.

Sol. :

Configuration of chiral centre Molecule

A R

B R

C S

38. Answer (3)

Hint : β-D-ribose and β-D-2-deoxyribose are reducing sugar.

Sol. :

39. Answer (1)

Hint : Only serotonin has OH group.



7/10

50. Answer (16)

Hint : x = +1 y = +3 z = +4

(1 + 3)4 = 16

A = AgCl

B = PCl3

C = SnCl4

Sol. : Reaction I

5 32Ag PCl 2AgCl PCl+ → +

Reaction II

5 4 3Sn 2PCl SnCl 2PCl+ → +

51. Answer (1)

Hint : Derivative of function.

Sol. : 2

2 21 1 ( 1 )( ) 0

1 ( 1)x xf x

xx x x− + −′ = − = <

+ +

1 , 33

x ∀ ∈

min ( 3) ln 33

f f π∴ = = −

max1 1ln ln 3.

6 63 3f f π π = = − = +

min max 6 3 2f f π π π

∴ + = + =

52. Answer (3)

Hint : Graphical approach

Sol. : ( ) cosf x x a′ = +

If 1a < −

0b <

∴ Option (3) is False.

53. Answer (3)

Hint : Take common x2q from Nq.

Sol. : 1 1

2( )

( )

p q

p qpx qx dx

x x

− − −

−−

+∫

Put xp + x–q = t

1I ct

∴ = − +

1

q

p qxI

x += −+

54. Answer (1)

Hint : Integration by parts.

Sol. : 1 1

1 1 1

0 0

(tan ) ( tan )n nnI x x dx x x x dx− − −= =∫ ∫

12 11 1

01 2

2 1 1

0

tantan2 2 2

( 1) tan tan2 2 2

n

n

x xx x x

x x xn x x dx

−− −

− −

= − +

− − − +

∫

11

20

1 ( 1)4 2 2

( 1) ( 1)2 2

n n

nn

nI I

n nx dx I−−

π −= − − +

− −−∫

2( 1) 1 1 1 1( 1)

2 4 2 2 2 2n nn I n I

n −+ π

⇒ = − + − − −

21( 1) ( 1)

2n nn I n In−

π⇒ + + − = −

Now verify each option.

55. Answer (2)

Hint : Substitute t = x + 1

Sol. : Put x + 1 = t ⇒ dx = dt

2 2

242

2 11 142

t t It−

+ +⇒ =

+∫

PART - C (MATHEMATICS)



8/10

2 6 5 4 2

1 2 42

3 7 2 11 14( 2)

x x x t tI I dtt−

+ + + + +∴ + =

+∫

2 2 4 4

1 2 42

( ( 2) 7( 2))( 2)−

+ + ++ =

+∫t t tI I dt

t

22

1 22

22

023

0

( 7)

2 ( 7)

8 1002 7 2 143 3 3

I I t dt

t dt

t t

−

+ = +

= × +

= × + = × + =

∫

∫

56. Answer (4)

Hint : Make [⋅] free by limits

Sol. :

Req. area

( )2 4 2

1 2

2 (3 ) 24xx x dx x dx

= − − + −

∫ ∫

2 42 3

3/2 3/2

1 2

4 433 2 3 12

x xx x x

= − + + −

19 sq. units6

=

57. Answer (2)

Hint : Exponential series

Sol. : lndyxydx dyx e x xy

dx

= ⇒ =

2 2ln (ln )

2 2x x ydx y dy c

x= ⇒ = +∫

2 2(ln )x y c⇒ = +

58. Answer (4)

Hint : By Rolle’s theorem.

Sol. : ( ) ( )f x g x c′ ′= +

( ) (1) (1) 2f x g c g c c′ ′ ′⇒ = + ⇒ = ⇒ =

( ) ( ) 2f x g x x d∴ = + +

(2) 9(2) 4 9 3 4 2f d d d⇒ = + + ⇒ = + + ⇒ =

( ) 9( ) 2 2f x x x⇒ − = +

Now verify each options.

59. Answer (1)

Hint : Real roots condition

Sol. : For two points of non-differentiability

x2 + 2ax + 1 – b = 0 has real and distinct roots

2 1a b∴ > −

60. Answer (2)

Hint : L. ‘Hospital rule.

Sol. :

2

2

1sinln

sinlim 1

1cossin 11 sinsin

lim 1

x

x

ax

a

x

aa xa xa

x

x

→∞

→∞

+

+ − × × + =

−

cota=

61. Answer (4)

Hint : Range method

Sol. : 2 [0, )x x a x R+ + ∈ ∞ ∀ ∈

21 1 [0, )

2 4x a ⇒ + + − ∈ ∞

14

a⇒ =



10/10

68. Answer (1)

Hint : Cross product formula

Sol. : | | 2 4 1 sin 30b c× = = × × θ ⇒ θ = °

2 2 2 24 4b c b c a⇒ + − ⋅ = λ

264 1 4 4cos⇒ + − × θ = λ

2 365 162

⇒ λ = − ×

65 8 3⇒ λ = −

69. Answer (1)

Hint : If then statement

Sol. : ~ ~ :p q→ If I do not take medicine, then

I cannot sleep.

70. Answer (4)

Hint : Mean and variance formula

Sol. : 8 5 10 30a b+ + + + =

7a b⇒ + =

2 2( 6) ( 6) 4 1 16 5 6.8a b− + − + + + = ×

2 2 12 7 34 93a b⇒ + = × + −

2 2 25a b+ =

∴ a = 3 and b = 4 will satisfy

71. Answer (02)

Hint : Plane in intercept form

Sol. : ( , , 0), (0, , );L a b M b c≡ ( , 0, )N a c

Equation of LMNA

0 00

x a y b za b

b c

− −− =

−

( ) ( ) 0bc x a ac y b abz⇒ − + − + =

2x y za b c

⇒ + + =

2k∴ =

72. Answer (39)

Hint : Number of triangle formula

Sol. : Req. ways − + =10 43 3 1 117C C

39k∴ =

73. Answer (33)

Hint : Divisibility by 5.

Sol. : T.C. = 19 and F.C. = 14

14 3319

P m n∴ = ⇒ + =

74. Answer (08)

Hint : Evaluation of determinant

Sol. : tan 1 or tan 2x x= = −

From graph it is deal that equation has 8 distinct

real roots.

75. Answer (16)

Hint : Factorisation

Sol. : 2 2( 1)( 1) 0z z z+ + + =

| | 1z =

4 4 4(| | 1) (| | 1) 2 16z z∴ + + − = =

Mock Test - 5 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 22/03/2020

ANSWERS


MOCK TEST - 5 - Code-B

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (1)

2. (1)

3. (3)

4. (1)

5. (4)

6. (3)

7. (1)

8. (3)

9. (4)

10. (2)

11. (4)

12. (2)

13. (1)

14. (4)

15. (1)

16. (4)

17. (2)

18. (1)

19. (4)

20. (3)

21. (02)

22. (06)

23. (09)

24. (07)

25. (03)

26. (3)

27. (2)

28. (3)

29. (4)

30. (4)

31. (1)

32. (1)

33. (3)

34. (3)

35. (2)

36. (1)

37. (1)

38. (3)

39. (1)

40. (3)

41. (1)

42. (1)

43. (1)

44. (4)

45. (1)

46. (16)

47. (03)

48. (04)

49. (05)

50. (11)

51. (4)

52. (1)

53. (1)

54. (4)

55. (3)

56. (2)

57. (2)

58. (1)

59. (2)

60. (4)

61. (2)

62. (1)

63. (4)

64. (2)

65. (4)

66. (2)

67. (1)

68. (3)

69. (3)

70. (1)

71. (16)

72. (08)

73. (33)

74. (39)

75. (02)

All India Aakash Test Series for JEE (Main)-2020 Mock Test - 5 (Code-B) (Hints & Solutions)


2/10

1. Answer (1)

Hint : Silvered lens will act as converging mirror of focal length 20 cm.

Sol. : For image to be formed on object itself, rays should fall normally on mirror.

d – 40 = 20

d = 60

2. Answer (1)

Hint : Let distance be x from end P.

Sol. : (200 0 ) 80° − °=

kA dmx dt

(200 100) 540( )

−=

−kA dm

x dt

Find x.

3. Answer (3)

Hint : 4

8π

=

PrQn

Sol. : 4•

1 11 8

π=

P rQ

n

4•

2 32 8

π=

P rQ

n

• •

1 2 1 216Q Q P P= ⇒ =

4. Answer (1)

Hint : 1 2B B B= +

Sol. : 01 3

24

MB

dµ

=π

02 34

MB

dµ

=π

2 21 2B B B= +

5. Answer (4)

Hint : 0 [sin sin ]4

IBd

µ= α + β

π

Sol. : 02 [sin45 sin90]4

IBa

µ= +

π

0

7 7

24 2 2

2(10 ) 10 T2

IBa

− −

µ = π −

= =

6. Answer (3)

Hint : 1 2 1 1 1 1 0 0M L T , L T , M M L T− −= = = v

Sol. : v Mα β γ= P

1 1 2M L TP − −=

7. Answer (1)

Hint : Find current in individual branch using phasor. Add the current based on vector addition.

Sol. :

0 0 sin302

= °ωv v

L R

ωL = 4R

4=

ωRL

8. Answer (3)

Hint : Wavelength is dependent on length only.

Sol. : Wavelength is dependent on length of tube.

9. Answer (4)

Hint : Path difference increases with θ.

Sol. : max4( ) sin 43

∆ = × × θ = λx d

Number of maxima = (4 × 2 + 1) = 9

10. Answer (2)

Hint : Write equation of line as v = a + bT

PART - A (PHYSICS)

Mock Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020


3/10

Sol. : pv = nRT

nRTpa bT

=+

11. Answer (4) Hint : For minimum velocity, velocity of person

should be equal to, velocity at the time of boarding the bus.

Sol. : 2v ax=

21 ,2

vs at ta

= =

12. Answer (2)

Hint : 2E A∝

Sol. : 2i iEK U= =

Finally, total energy is 3 .2E

2

23 ( )2 ( )

E AE A

′=

3 A2

′ =A

13. Answer (1)

Hint : Apply circuit diagram and change = final reading – initial reading

Sol. : eqeq

eq 1

20, 50i

RV R

R R= = Ω

+

20 100 10 V200fV ×

= =

14. Answer (4)

Hint : Initially, T1cos30° = mg

Sol. : T1cos30° = mg

T2 = mgcos30°

15. Answer (1)

Hint : 20

1,2

Fa x atm

= =

Sol. : P F V= ⋅

2P F ax= ⋅

P x∝

16. Answer (4)

Hint : 1 1 813.6 13.61 9 9

E = − = ×

Sol. : 0E eV= φ +

012.1 1.5 eV= +

010.6 V=

17. Answer (2)

Hint : Apply condition of equilibrium in respective direction.

Sol. : Tcos30° = mg

Tsin30° = 02

σεQ

18. Answer (1)

Hint : Resolve the dipole moment along X-axis and Y-axis respectively.

Sol. : ˆ ˆcos sinp p i p j= θ − θ

3 30 0

cos 2 sinˆ ˆ4 4p pE i j

d d− θ θ

= −πε πε

19. Answer (4)

Hint : drvdt

=

dvadt

=

Sol. : =

drvdt

dvadt

=

cos| || |

a va v

⋅θ =

20. Answer (3)

Hint : c ta a a= +

Sol. : 2

2 3MMg = α

32 4ct

ga α= =

2

212 3 2

M Mgω =



4/10

2 32 2c

ga = ω =

5 34

ga∴ =

21. Answer (02)

Hint : At the time of maximum angular displacement, kinetic energy is zero.

Sol. : Initially

2

20

1(1 cos )2 2 3

− θ = ωL MLMg

Finally

2

21

1(1 cos )2 4 2 2 4 3

− θ = ω×

M L M Lg

Compare and get the result.

22. Answer (06)

Hint : 1 1 1, vmv u f u

− = =

Sol. : 1 1 1 , 30 cm15 10

+ = =vv

m = –2 Images formed will be 2 mm above and 2 mm

below respective to P axis. 2 2 2 6 mmd = + + =

23. Answer (09)

Hint : Let temperature be T.

2001 1450T

T− = −

Sol. : 2001 1450T

T− = −

T = 300 K

1 213

n n= =

Total work doneHeat given

η =

24. Answer (07)

Hint : 2 ,iv gR= write velocity as a function

of h.

Sol. : 21 02 ( )

GMmmvR x

− =+

2GMvR x

=+

( )2dx GM

dt R x=

+

3

0 0

2R t

R x dx GM dt+ =∫ ∫

Solve for t.

25. Answer (03)

Hint : Redraw it to form the equivalent circuit.

Sol. : All capacitors are connected in parallel.

0eq

3 ACdε

=

26. Answer (3) Hint : This trend is followed by density. Sol. : Density order Sc Ti V Cr Mn Fe Co Cu< < < < < < <

27. Answer (2)

Hint : Strength of field absorption

1∝

λ

Sol. : If 3 2 1λ > λ > λ then strength of ligand

1 2 3L L L> >

28. Answer (3)

Hint : Oxidation number of Au and Zn in P and Q is +1 and +2 respectively.

Sol. : P is [Au(CN)2]–, Q is [Zn(CN)4]2–

29. Answer (4)

Hint : Water move towards anode it means water contains negatively charged ions.

Sol. : Sol must have positive charge.

PART - B (CHEMISTRY)



5/10

30. Answer (4)

Hint : ocellE remain independent on [CI–].

Sol. : Anode : 2Zn Zn 2e+ −→ +

Cathode : AgCl 1e Ag(s) Cl− −+ → +

If [Cl–]↑, Ecell↓ 31. Answer (1)

Hint : Distance between A to A layer is 3 (edge length).

Sol. :

3a 3 100 173 pm= × =

32. Answer (1)

Hint : Only serotonin has OH group.

Sol. : Serotonin

33. Answer (3)

Hint : β-D-ribose and β-D-2-deoxyribose are reducing sugar.

Sol. :

34. Answer (3) Hint : A and B are identical. Sol. :

Configuration of chiral centre Molecule

A R

B R

C S

35. Answer (2) Hint : Compound contains double bond. Sol. :

36. Answer (1) Hint : In E2 reaction, anti elimination of H and Br

takes place. Sol. :

37. Answer (1)

Hint : +I power of –CH3 is less than –CD3 while

+H power of –CH3 is more than –CD3.

Sol. : Correct stability order is

D > E > C > B > A

38. Answer (3)

Hint : 2 6A B H⇒

3 3 6D : B N H

Sol. : B2H6 contains 3c–2e– as well as 2c–2e–

bonds while D contains all equal bond lengths.

39. Answer (1) Hint : For fluoride ∆fH° becomes less negative

as we move top to bottom.

Sol. : ∆fH° value of chloride, bromide and iodide become more negative as we move top to bottom.

40. Answer (3)

Hint : NaH, LiH and KH are ionic hydride.

Sol. : Covalent hydride Ionic Metallic

NH3 NaH CrH

HF LiH YbH2.55

CH4 ZrH1.3

BeH2



6/10

41. Answer (1)

Hint : ∆UC→A = 0

Sol. : ∆H ∝ ∆T 42. Answer (1)

Hint : M occupies less volume than N. O is more compressible than P.

Sol. : If two gases have same value of ‘b’ but different values of ‘a’, then the gas having a larger value ‘a’ will occupy lesser volume.

If two gases have the same value of ‘a’ but different values of ‘b’, then smaller the value of b, larger will be the compressibility.

43. Answer (1)

Hint : 2NO 180+ → °

2NO 120− < °

Sol. : 2 2 2NO NO NO+ −> >

44. Answer (4) Hint : Number of photoelectron ∝ intensity

Sol. : Energy of an orbital ∝ (Z2)

45. Answer (1)

Hint : ureanM V (L) 2 LV (L)

= ⇒ =

Mass of solution = 2000 × 2 = 4000 g

Sol. : Mass of solvent = 4000 – 120 = 3880 g

urean 2m 0.52 mMass of solvent 3.880

= = =

46. Answer (16)

Hint : x = +1 y = +3 z = +4

(1 + 3)4 = 16

A = AgCl

B = PCl3

C = SnCl4

Sol. : Reaction I

5 32Ag PCl 2AgCl PCl+ → +

Reaction II

5 4 3Sn 2PCl SnCl 2PCl+ → +

47. Answer (03)

Hint : 12

0.693 0.693t 25 mink 0.02772

= = =

Sol. :

A → nB

t = 0 ⇒ 700 0

t = t1/2 350 350 n

total pressure ⇒ 350 + 350 n = 1400

n = 3

48. Answer (04)

Hint : Product is

Sol. : The IUPAC name of the product is 3-methylcyclopent-1-ene.

49. Answer (05)

Hint : Eqv. of KMnO4 = Eqv. of (A–1 to A+n)

Sol. : 500 × 2 × 3 = 500 × 1(n + 1)

n 1 6⇒ + =

n = 5

50. Answer (11) Hint : (NH4)2SO4 on reaction with NaOH can

form a buffer solution.

Sol. :

4 2 4 2 4 2 3(NH ) SO 2NaOH Na SO H O 2NH+ → + +

1 mol 1 0 0

0.5 0 1

4b

3

NHpOH pK logNH

0.5 24.74 log 4.741

+ = +

× ⇒ + =



7/10

51. Answer (4)

Hint : Mean and variance formula

Sol. : 8 5 10 30a b+ + + + =

7a b⇒ + =

2 2( 6) ( 6) 4 1 16 5 6.8a b− + − + + + = ×

2 2 12 7 34 93a b⇒ + = × + −

2 2 25a b+ =

∴ a = 3 and b = 4 will satisfy

52. Answer (1)

Hint : If then statement

Sol. : ~ ~ :p q→ If I do not take medicine, then

I cannot sleep. 53. Answer (1)

Hint : Cross product formula

Sol. : | | 2 4 1 sin 30b c× = = × × θ ⇒ θ = °

2 2 2 24 4b c b c a⇒ + − ⋅ = λ

264 1 4 4cos⇒ + − × θ = λ

2 365 162

⇒ λ = − ×

65 8 3⇒ λ = −

54. Answer (4)

Hint : Height and distance

Sol. : tan603

h hrr

° = ⇒ =

2

2

3hA m

∴ = π ×

Area of the hexagon 236

4 3h = × ×

236

4 3h

= × ×

3 32

AA′∴ = ×π

55. Answer (3)

Hint : Equation of line in space

Sol. : Equation of line is

2 1 21 1 13 3 3

x y z r− + −= = =

2 , 1, 23 3 3r r rQ ⇒ + − +

satisfy plane

24 1 2 93 3 3r r r

⇒ + + − + + =

4 4 33r r⇒ = ⇒ =

(3,0,0)Q∴

1 1 1 3PQ∴ = + + =

56. Answer (2) Hint : Internal touching of two circles

Sol. : | 1| 5r − =

1 5 6r r⇒ − = ⇒ =

∴ Option (2) is correct.

57. Answer (2)

Hint : Distance formulas in ellipse

Sol. : = − =

2 22 2 16ae a b

ellipse is touching x-axis ∴ b = 15

2 2 256a b⇒ − =

2 256 2254

a = +

2 64 225a = +

17a =

2 34a∴ =

58. Answer (1)

Hint : Equation of normal

Sol. : Normal at P to the ellipse is

2 2

cos sinax b a b− = −

θ θ

PART - C (MATHEMATICS)



8/10

Normal at Q, y = (tanθ)x

2 2(tan )cos sinax b x a bθ

⇒ − = −θ θ

2 2( ) ( )cos ( )cos .a b x a b x a b⇒ − = − θ ⇒ = + θ

tan ( )sin .y x y a b∴ = θ ⇒ = + θ

2 2 2( )x y a b∴ + = +

59. Answer (2)

Hint : C.O.C in terms of M.P

Sol. : Equation of chord in terms of middle point (h, k)

T = S1

( ) 22 4yk a x h k ah⇒ − + = −

22 2yk ax k ah⇒ = + −

22 2ax k ahy

k k−

⇒ = +

It touches y2 = –4ax

22 2;a b k ahm

k m k− −

∴ = =

2 2

2b k k ah

a k− × −

⇒ =

2 2 22 4k k a h⇒ − = −

2 24 (2 )a h a b k⇒ = +

2

2 42

a xya b

⇒ =+

60. Answer (4)

Hint : Range method

Sol. : 2 [0, )x x a x R+ + ∈ ∞ ∀ ∈

21 1 [0, )

2 4x a ⇒ + + − ∈ ∞

14

a⇒ =

61. Answer (2)

Hint : L. ‘Hospital rule.

Sol. :

2

2

1sinln

sinlim 1

1cossin 11 sinsin

lim 1

x

x

ax

a

x

aa xa xa

x

x

→∞

→∞

+

+ − × × + =

−

cota=

62. Answer (1)

Hint : Real roots condition

Sol. : For two points of non-differentiability

x2 + 2ax + 1 – b = 0 has real and distinct roots

2 1a b∴ > −

63. Answer (4)

Hint : By Rolle’s theorem.

Sol. : ( ) ( )f x g x c′ ′= +

( ) (1) (1) 2f x g c g c c′ ′ ′⇒ = + ⇒ = ⇒ =

( ) ( ) 2f x g x x d∴ = + +

(2) 9(2) 4 9 3 4 2f d d d⇒ = + + ⇒ = + + ⇒ =

( ) 9( ) 2 2f x x x⇒ − = +

Now verify each options.

64. Answer (2)

Hint : Exponential series

Sol. : lndyxydx dyx e x xy

dx

= ⇒ =

2 2ln (ln )

2 2x x ydx y dy c

x= ⇒ = +∫

2 2(ln )x y c⇒ = +

65. Answer (4)

Hint : Make [⋅] free by limits



9/10

Sol. :

Req. area

( )2 4 2

1 2

2 (3 ) 24xx x dx x dx

= − − + −

∫ ∫

2 42 3

3/2 3/2

1 2

4 433 2 3 12

x xx x x

= − + + −

19 sq. units6

=

66. Answer (2)

Hint : Substitute t = x + 1

Sol. : Put x + 1 = t ⇒ dx = dt

2 2

242

2 11 142

t t It−

+ +⇒ =

+∫

2 6 5 4 2

1 2 42

3 7 2 11 14( 2)

x x x t tI I dtt−

+ + + + +∴ + =

+∫

2 2 4 4

1 2 42

( ( 2) 7( 2))( 2)−

+ + ++ =

+∫t t tI I dt

t

22

1 22

22

023

0

( 7)

2 ( 7)

8 1002 7 2 143 3 3

I I t dt

t dt

t t

−

+ = +

= × +

= × + = × + =

∫

∫

67. Answer (1)

Hint : Integration by parts.

Sol. : 1 1

1 1 1

0 0

(tan ) ( tan )n nnI x x dx x x x dx− − −= =∫ ∫

12 11 1

01 2

2 1 1

0

tantan2 2 2

( 1) tan tan2 2 2

n

n

x xx x x

x x xn x x dx

−− −

− −

= − +

− − − +

∫

11

20

1 ( 1)4 2 2

( 1) ( 1)2 2

n n

nn

nI I

n nx dx I−−

π −= − − +

− −−∫

2( 1) 1 1 1 1( 1)

2 4 2 2 2 2n nn I n I

n −+ π

⇒ = − + − − −

21( 1) ( 1)

2n nn I n In−

π⇒ + + − = −

Now verify each option.

68. Answer (3) Hint : Take common x2q from Nq.

Sol. : 1 1

2( )

( )

p q

p qpx qx dx

x x

− − −

−−

+∫

Put xp + x–q = t

1I ct

∴ = − +

1

q

p qxI

x += −+

69. Answer (3) Hint : Graphical approach Sol. : ( ) cosf x x a′ = +

If 1a < −

0b <

∴ Option (3) is False.



10/10

70. Answer (1) Hint : Derivative of function.

Sol. : 2

2 21 1 ( 1 )( ) 0

1 ( 1)x xf x

xx x x− + −′ = − = <

+ +

1 , 33

x ∀ ∈

min ( 3) ln 33

f f π∴ = = −

max1 1ln ln 3.

6 63 3f f π π = = − = +

min max 6 3 2f f π π π

∴ + = + =

71. Answer (16) Hint : Factorisation

Sol. : 2 2( 1)( 1) 0z z z+ + + =

| | 1z =

4 4 4(| | 1) (| | 1) 2 16z z∴ + + − = =

72. Answer (08)

Hint : Evaluation of determinant

Sol. : tan 1 or tan 2x x= = −

From graph it is deal that equation has 8 distinct

real roots.

73. Answer (33)

Hint : Divisibility by 5.

Sol. : T.C. = 19 and F.C. = 14

14 3319

P m n∴ = ⇒ + =

74. Answer (39)

Hint : Number of triangle formula

Sol. : Req. ways − + =10 43 3 1 117C C

39k∴ =

75. Answer (02)

Hint : Plane in intercept form

Sol. : ( , , 0), (0, , );L a b M b c≡ ( , 0, )N a c

Equation of LMNA

0 00

x a y b za b

b c

− −− =

−

( ) ( ) 0bc x a ac y b abz⇒ − + − + =

2x y za b c

⇒ + + =

2k∴ =

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