mock test - 4 (code-e) (answers) all india aakash test series for … · 2020. 3. 2. · mock test...

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

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

Test Date : 01/03/2020

ANSWERS

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

MOCK TEST - 4 - Code-E

1/9

PHYSICS CHEMISTRY MATHEMATICS

1. (4)

2. (3)

3. (1)

4. (4)

5. (4)

6. (3)

7. (1)

8. (4)

9. (1)

10. (2)

11. (4)

12. (2)

13. (1)

14. (3)

15. (1)

16. (4)

17. (3)

18. (4)

19. (1)

20. (4)

21. (05)

22. (04)

23. (32)

24. (04)

25. (09)

26. (1)

27. (3)

28. (1)

29. (2)

30. (3)

31. (4)

32. (1)

33. (1)

34. (4)

35. (2)

36. (1)

37. (3)

38. (2)

39. (4)

40. (2)

41. (3)

42. (4)

43. (2)

44. (3)

45. (2)

46. (20)

47. (80)

48. (85)

49. (24)

50. (60)

51. (3)

52. (1)

53. (1)

54. (3)

55. (2)

56. (2)

57. (4)

58. (4)

59. (4)

60. (4)

61. (2)

62. (3)

63. (4)

64. (1)

65. (4)

66. (3)

67. (1)

68. (1)

69. (3)

70. (4)

71. (06)

72. (00)

73. (02)

74. (04)

75. (02)

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


2/9

1. Answer (4)

Hint : Apply conservation of angular momentum about the point of contact.

Sol. 20

2 53 3

vmv R mvR mR mvRR

= + ⋅ =

0 03 35 5v v

vR

= ⇒ ω =

2. Answer (3) Hint : BAV

= –B AV V

Sol. : BAV

is 16 m/s along positive y-axis. 3. Answer (1) Hint : Use constraint relation

Sol. : 21 2

3cos53

5a

a a= ° =

21

62 2

5a

T g a− = = …(i)

2320 35T a− = …(ii)

⇒ 2100 5

3T a− = …(iii)

2 2100 6 3120 5

3 5 5a a − = + =

22 2

40 31 200 m/s3 5 93

a a= ⇒ =

4. Answer (4) Hint : Apply net gain of thermal energy of

middle plate is zero.

Sol. ( ) ( )4 44 44 3T T T T′ ′− = −

⇒ ( )4 4 42 256 81 337T T T′ = + =

⇒ 14337

2T T ′ =

5. Answer (4) Hint : Heat is extracted in process A → B → C Sol. : AB BC v B A p C BQ Q nC T T nC T T∆ + ∆ = − + −

0 0 0 0 0 0 0 02 6 25 72 2

P V P V P V P Vn R RnnR nR

− − = + ⋅

0 00 0 0 0

335 142 2

P VP V P V= + =

6. Answer (3)

Hint : Use vmu

= and 1 1 1v u f

− =

Sol. : 3 3 34 4

v uvu

= − ⇒ = − =

⇒ u = – 4 m

1 1 7 1 1 23 4 12 2f R

+ = = = ×

⇒ 127

R = m

7. Answer (1) Hint : LC = 1 MSD – 1 VSD Sol. : 1 MSD = 0.05 cm

2.450.05 0.001 cm50

LC = − =

D = 6.20 + 12 × 0.001 = 6.20 + 0.012 = 6.212 cm 8. Answer (4)

Hint : E = n hc Epc pc

= ⇒ =λ

Sol. : 3 9

860 10 500 10

3 10p

− −× × ×=

×

= 2 × 5 × 10–17 kg m/s = 10–16 kg m/s 9. Answer (1) Hint : φ21 = M21 i1

Sol. : ( )

20 0 0

32 2

24 2 8 164

I II RBR RR

µ µ µ⋅ π= = =

π × ×

2 2 0 15 516 2

Ir B r

Rµ

φ = π = π ⋅

⇒ 2

0516 2

rM

Rπµ

=

PART - A (PHYSICS)

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


3/9

10. Answer (2)

Hint : 2

GM mE V Kr

= + −

Sol. : ,2 4f iGMm GMmE E

R R= − = −

×

1 718 8f i

GMm GMmE ER R

− = − + =

11. Answer (4)

Hint : Reff = 2000 Ω

Sol. : 20 50 C 1t

Q e− τ = × µ −

τ = Reff C = 2000 × 50 µs = 0.1 s

0.11000 C 1t

Q e−

= µ −

0.13 4 4ln 0.1 ln4 3 0.1 3

t te t−

= ⇒ = ⇒ =

12. Answer (2)

Hint : Remember 0 1 2 3 4 5 6 7 8 9BBRO YGB V GW Gold

silver

Sol. : R = 34 × 10° ± 5% = (34 ± 5%) Ω

13. Answer (1)

Hint : Q = 20 cos ωt

Sol. : 1 11cos2 3

t t πω ⇒ ω =

⇒ 1 3t π

=ω

81

3 6

1 1 10 1000 s10040 10 25 10LC

−

− −ω = = = =

× × ×

31

3.14 10 s 1.053

t −= × =

14. Answer (3)

Hint : ( )T L x gv −

= =µ µ

Sol. : ( )dx L x gdt

= −

( )2 1

2

0 0

L t

L x dx g dt−− =∫ ∫

( )1 22 02

L

L x g t− − =

⇒ 22L L g t

− − =

12 12

Ltg

= −

15. Answer (1)

Hint : 11 wV f V− ρ = ρ

Sol. : 21 oV f V− ρ = ρ

⇒ ( ) ( )1 21 1w of f− ρ = − ρ

⇒ 1

2

11

o

w

ff

ρ −=

ρ −

16. Answer (4)

Hint : a sinθ = λ

Sol. : 650 m 2.5 msin 0.259

a λ= = µ = µ

θ

17. Answer (3)

Hint: 1240 eV 3.1 eV400

hcE = = =λ

Sol. : 2

maxKE 1.1 eV2pm

= =

⇒ 31 192 2 9 10 1.1 1.6 10p m KE − −= = × × × × ×

= 5.7 × 10–25 kg m/s

18. Answer (4)

Hint : 2nhmvr =

π

Sol. : 2mvevB

r=

⇒ 2

eBr nhvm mr

= =π

⇒ 2

nhveB

=π



4/9

19. Answer (1) Hint : 0

tR R e−λ=

Sol. : 0

tRCQ Q e

−=

1RC

λ =

⇒ 3

36

1 40 10 4 10550 10

RC

−

−×

= = = ×λ ×

= 800 Ω 20. Answer (4) Hint : Check truth table

Sol. : A B X

1 1 1

1 0 1

0 1 1

0 0 0

21. Answer (05)

Hint : 0BM C

T=

Sol. : ⇒ 0.8 0.4 5 K10

TT

= ⇒ =

22. Answer (04)

Hint : Apply conservation of mechanical energy

Sol. : 2

22 122 2

R MRMg × ⋅ = ⋅ ωπ

16 4g gR R

ω = =π π

23. Answer (32)

Hint : 2 22

1 1 22 2

kx mv=

Sol. : 2

2 2kxv

m=

2 2cm

2 2 23 3 3 2mv v x kvm m

= = =

24. Answer (04)

Hint : ( )2 14nf v

l−

=

Sol. : ( ) ( )2 1 340 2 1 100Hz4 0.85nf n−

= = −×

100Hz, 300Hz, 500Hz, 700Hz,f =

25. Answer (09)

Hint : 00

s

v vf f

v v+

=−

Sol. 345 350306340 345

f = × ×

306 35 315 Hz34

×= =

Beat frequency = 315 – 306 = 9 Hz

26. Answer (1) Hint : NaBH4 converts carbonyl compounds to

alcohols. Sol. :

27. Answer (3)

Hint :

4 33 4 4 4Zr (PO ) 3Zr 4PO

3x 4x

+ −+

Sol. : Given 3x = S

⇒ Sx3

=

∴ 34

4S[PO ]3

− =

∴ 3 4spK (3x) (4x)= ⋅

4

3 4S(S)3

= ⋅

7256 S81

=

BrNaBH4

O–

BrO

OHCNOH

CN

PART - B (CHEMISTRY)



5/9

28. Answer (1) Hint : RNA contains uracil and not thymine.

Sol. : RNA is single stranded and adenine is

H-bonded to uracil.

29. Answer (2)

Hint : Lanthanoid contraction.

Sol. : Zr and Hf have almost similar size.

30. Answer (3)

Hint : Claisen rearrangement.

Sol. :

31. Answer (4)

Hint : Fe2+ has d6 configuration.

Sol. : dipy is a strong field ligand and all except

Fe2+ lose electron from eg level upon

oxidation.

32. Answer (1)

Hint : Self reduction of copper.

Sol. : 2 2 2Cu O Cu S Cu SO+ → +

33. Answer (1)

Hint : At t = 1 hr, Nt = N0(e)

Sol. : 2dN 2Ndt

= −

⇒ 2dN 2dtN

= −

⇒ 0

0

N2 1t1

N (e)

N 2[t]2 1

− + = − − +

⇒ 0 0

1 1 2(t 1)N N (e)

− = −

⇒ 0

1 1t 1 12N e

= + −

34. Answer (4)

Hint : Aldehydes, ketones and esters contain

hydrogen.

Sol. : C12O9 is mellitic anhydride.

35. Answer (2)

Hint : i = 1 + (n – 1) α

Sol. : 2CaCli 1 2(0.6) 2.2= + =

xyi 1 1(0.75) 1.75= + =

∴ 4 × 2.2 × 0.01 = 1.75 × C

C 0.05 M

36. Answer (1)

Hint : Fact based.

Sol. : Salvarsan is arsphenamine.

37. Answer (3)

Hint : H4P2O6 contains both phosphorus in +4

state.

Sol. :

O

O

OO

OO

OO

O

HO P P OH

OH OH

O O



6/9

38. Answer (2)

Hint : Cannizzaro reaction.

Sol. :

39. Answer (4)

Hint : Carbocation is formed and can undergo rearrangement.

Sol. : In , the carbocation

formed cannot undergo ring expansion to form 6 membered ring.

40. Answer (2)

Hint : 0

2RTMPSM

=

Sol. : As T increases, MPS increases As M0 increases, MPS decreases

41. Answer (3) Hint : PCl5 solid exists as [PCl4]

+ [PCl6]–

Sol. : PBr5 solid exists as [PBr4]+[Br]–

42. Answer (4) Hint : Fact based. Sol. : Portland cement majorly contains

dicalcium silicate, tricalcium silicate and tricalcium aluminate.

43. Answer (2) Hint : p 0E W KE= +

Sol. : (Ep = W0 + k1) × 2 2Ep = W0 + k2 k2 = 2k1 + W0 i.e. k2 > 2k1

44. Answer (3) Hint : Adiabatic processes are isentropic.

Sol. : ∆S = 0 for path C D→ .

45. Answer (2) Hint : CIP rules for assigning priority. Sol. : Configuration of both chiral centres is R.

46. Answer (20) Hint : Given ligand is EDTA4–.

Sol. : Ethylenediamine tetra acetate (EDTA4–) is a hexadentate ligand and y is 4. 2(x + y) is 20.

47. Answer (80)

Hint : In[In ]pK log pH[HIn]

−

+ =

Sol. : From dissociation of HA,

2 6 2[H ] 10 10+ − −= ×

4[H ] 10+ −=

pH = 4

∴ [In ]4.6 log 4[HIn]

−

+ =

⇒ 0.6In 1 1

[HIn] 410

−

= =

∴ 20% In– and 80% HIn.

48. Answer (85)

Hint : Volume strengthMolarity = 11.2

Sol. : M = 28 2.511.2

=

1 L solution contains 85 g H2O2

∴ (w/v)% = 8.5%

49. Answer (24)

Hint : 32 6[Al(H O) ] + and 4[Al(OH) ]−

Sol. : Al3+ is hexa co-ordinated with water while it is tetra co-ordinated with hydroxide ions.

50. Answer (60)

Hint : O2– at face centre touches Ca2+ at all

corners of the face. ∴ y = 4

Sol. : O2– at face centre touches two Ti4+ at

adjacent body centres. ∴ x = 2

Ph CH CH2

KMnO4

0ºC, OH– Ph CH CH2

OH OH

HIO4

Ph CHO + HCHOOH–

(conc.)PhCH OH + HCOO2

CH3

CH3CH3



7/9

51. Answer (3) Hint : Newton Leibnitz Formula

Sol. : ( ) sin xf xx

′ =

⇒ f(x) increases in

x∈(0, π) and f(x) decreases in x ∈ (π, 2π)

52. Answer (1) Hint : Location of roots

Sol. : 2 lies between interval (e1 > e2)

∴ ( )2 0F <

⇒ 2 – 2 2 0a + <

⇒ 2 2a >

53. Answer (1) Hint : Discriminant = 0

Sol. : P satisfy x2 2

–3y = 1

Let m is slope of common tangent then m2 – 3 = r2(1 + m2)

⇒ m2 = 2

23

1–e

r+

2 4r =

∴ no common tangent exists. 54. Answer (3)

Hint : Assumption of normal

Sol. : A normal at 44 ,A tt

is

( )24– – 4y t x tt

=

⇒ 4 34 – – 4 0t t h tk+ =

( ),p h k satisfy it given

2 2 2 21 2 3 4 1 2 3 4t t t t y y y y+ + + = + + +

⇒ 2

16h k=

⇒ 2 16x y= (Parabola)

55. Answer (2) Hint : Concept of tangency.

Sol. : On solving both the equations

( )2 2 2 2 2 2– 2 – 0b a x a x a b+ =

For least area, circle must touch the ellipse.

⇒ 4a4 + 4a2b2(b2 – a2) = 0

⇒ 1 – b2e2 = 0

⇒ 1be

=

⇒ a2 = ( )

2

2 2

11– 1–

be e e

=

Area = A = 2 21–e e

π = f(e) say

Area is minimum if e4 – e6 is maximum.

Let g(e) = e4 – e6

⇒ g′(e) = 4e3 – 6e5 = 0

⇒ 23

e =

∴ ellipse is 2 22 6 9x y+ =

56. Answer (2) Hint : Touching concept

Sol. : Equation of any circle (x – α) (x – β) + y2

+ λy = 0

⇒ x2 + y2 – (α + β)x + λy + αβ = 0 …(1)

∠ACB is maximum

∴ It touches y-axis at (0, γ).

⇒ 2 0y y+ λ + αβ = have repeated

roots

⇒ 2γ = αβ

PART - C (MATHEMATICS)



8/9

57. Answer (4) Hint : Location of roots Sol. : x4 + 25 = mx2

⇒ 2

2 5xx

+

= m

⇒ 25– – 10x m

x =

For m = 10 equation has four real roots but two are repeated one

∴ m ∈ φ 58. Answer (4) Hint : Geometrical application

Sol. : Req. Area = 1 2 42

× × = 4 sq. units.

59. Answer (4) Hint : Telescopic method. Sol. : The given series is

( )( )

215–1

1

12

1r

r

r

r r=

+

+∑ = 15–1

1

2 – 1– 21

r

r

r rr r=

+

∑

= ( )15–1

1

– 12 – 2

1r r

r

rrr r=

+ ∑

= v15 – v0 = 1515 216

× = 1115 2×

60. Answer (4)

Hint : Choosing the 4 vertex.

Sol. : nC4 = 35

⇒ n(n –1) (n – 2) (n – 3) = 24 × 35

⇒ n = 7

61. Answer (2)

Hint : Numerically greatest term.

Sol. : 4

31T

T≥ and 5

41T

T≤

⇒ ( )11– 3 3 13 16

x≥

and 11– 4 3 14 16

x ≤

⇒ 2 ≤ x ≤ 6421

Required sum = (2 + 3) = 5 62. Answer (3) Hint : Property of determinants. Sol. : Take x5 common from R3

⇒ x5

1010 1012 2020

1010 1 2020

1 n

n

x x x

x n

x x x+

= 0

⇒ n + 1 = 1012 ⇒ n = 1011 63. Answer (4) Hint : Graphical approach. Sol. : The above equation has always two real

roots by observing the graph of the function in Domain.

∴ U ∈φ. 64. Answer (1) Hint : Scalar triple product.

Sol. : ( )( ) ( ) ( )( ).a a b b a b a b+ × + × × ×

= ( )( ) ( )( ).a a b b a b+ × × ×

= ( ). . 1a a a a b+ × =

65. Answer (4) Hint : Assumption of plane.

Sol. : Given line 6 4 3x y z

= =

Let plane is ax + by + cz = 1

⇒ 6 4 3a b c+ + = 0 …(i)

2a – b = 1 …(ii) 3a – 4b + 5c = 1 …(iii)

⇒ a = 2985

; b = –2785

c = –2285

∴ Eqn. of plane is

29 – 27 – 22 85x y z =



9/9

66. Answer (3)Hint : Definition of orthogonal matrices.

Sol. : – –2 2

T I IA A

= I … (i)

and 2 2

T I IA A + +

= I …(ii)

⇒ A + AT = 0 (By subtracting (i) & (ii)

⇒ AT = –A ⇒ 2 3–4

A I=

⇒ |A|2 = –34

n

⇒ is evenn

67. Answer (1)Hint : Total law of probabilitySol. : P(2 white + 1 black)

= P(W1 W2 B3) + P(W1 B2 W3) + P(B1 W2 W3)

= 3 2 3 3 2 1 1 2 14 4 4 4 4 4 4 4 4

× × + × × + × ×

= 1332

68. Answer (1)Hint : Continuity at discrete point.

Sol. : If f(x) is continuous at x = 6π

⇒ cos2x = sinx

⇒ x = 6π is one of the solution

∴ Option 1 is correct. 69. Answer (3)

Hint : ( )~ ~ ~p q p q∨ = ∧ . Sol. : By definition of negation option (3) is

correct

70. Answer (4)Hint : Mean formulae.

Sol. : New mean of the series is X nm

+

⇒X nm

m+

71. Answer (06)Hint : Inverse function concept.

Sol. : Let y = 23 2x x+

⇒ x = – 1 + ( )1

3 21y +

⇒ f –1(y) = ( )1

3 2–1 1y+ +

∴ I = ( ) ( )( )2

–1

0

1f u f u du+ +∫

= ( )( ) ( )2 200

6uf u u+ =

72. Answer (00)Hint : Leibnitz formula.

Sol. : 1 3 1 7 02 2 4 4

f f f f ′ = ′ = ′ = ′ =

Also ( )1f x′ + = ( )– 1–f x′

⇒ g(1) = 0 = 1 32 4

g g ′ = ′

73. Answer (02)

Hint : By using (AOD).

Sol. : Take log both side.

74. Answer (04)

Hint : ∆ = 12

ah

Sol. : ah1 = bh2 = ch3 = 2∆

⇒ 1 2 3

1 1 1h h h

+ + = 2

a b c+ +∆

75. Answer (02)

Hint : Rationalisation Method.

Sol. : ( ) ( ) ( )1 2 .... –x x x x n x+ + +

= ( )1 12 2

n n nn+ +

=

∴ L = 3 12+ = 2

Anil

Cross-Out

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

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

Test Date : 01/03/2020

ANSWERS


MOCK TEST - 4 - Code-F

1/9

PHYSICS CHEMISTRY MATHEMATICS

1. (4)

2. (1)

3. (4)

4. (3)

5. (4)

6. (1)

7. (3)

8. (1)

9. (2)

10. (4)

11. (2)

12. (1)

13. (4)

14. (1)

15. (3)

16. (4)

17. (4)

18. (1)

19. (3)

20. (4)

21. (09)

22. (04)

23. (32)

24. (04)

25. (05)

26. (2)

27. (3)

28. (2)

29. (4)

30. (3)

31. (2)

32. (4)

33. (2)

34. (3)

35. (1)

36. (2)

37. (4)

38. (1)

39. (1)

40. (4)

41. (3)

42. (2)

43. (1)

44. (3)

45. (1)

46. (60)

47. (24)

48. (85)

49. (80)

50. (20)

51. (4)

52. (3)

53. (1)

54. (1)

55. (3)

56. (4)

57. (1)

58. (4)

59. (3)

60. (2)

61. (4)

62. (4)

63. (4)

64. (4)

65. (2)

66. (2)

67. (3)

68. (1)

69. (1)

70. (3)

71. (02)

72. (04)

73. (02)

74. (00)

75. (06)

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


2/9

1. Answer (4) Hint : Check truth table

Sol. : A B X

1 1 1

1 0 1

0 1 1

0 0 0 2. Answer (1) Hint : 0

tR R e−λ=

Sol. : 0

tRCQ Q e

−=

1RC

λ =

⇒ 3

36

1 40 10 4 10550 10

RC

−

−×

= = = ×λ ×

= 800 Ω

3. Answer (4)

Hint : 2nhmvr =

π

Sol. : 2mvevB

r=

⇒ 2

eBr nhvm mr

= =π

⇒ 2

nhveB

=π

4. Answer (3)

Hint: 1240 eV 3.1 eV400

hcE = = =λ

Sol. : 2

maxKE 1.1 eV2pm

= =

⇒ 31 192 2 9 10 1.1 1.6 10p m KE − −= = × × × × ×

= 5.7 × 10–25 kg m/s

5. Answer (4)

Hint : a sinθ = λ

Sol. : 650 m 2.5 msin 0.259

a λ= = µ = µ

θ

6. Answer (1)

Hint : 11 wV f V− ρ = ρ

Sol. : 21 oV f V− ρ = ρ

⇒ ( ) ( )1 21 1w of f− ρ = − ρ

⇒ 1

2

11

o

w

ff

ρ −=

ρ −

7. Answer (3)

Hint : ( )T L x gv −

= =µ µ

Sol. : ( )dx L x gdt

= −

( )2 1

2

0 0

L t

L x dx g dt−− =∫ ∫

( )1 22 02

L

L x g t− − =

⇒ 22L L g t

− − =

12 12

Ltg

= −

8. Answer (1)

Hint : Q = 20 cos ωt

Sol. : 1 11cos2 3

t t πω ⇒ ω =

⇒ 1 3t π

=ω

8

13 6

1 1 10 1000 s10040 10 25 10LC

−

− −ω = = = =

× × ×

31

3.14 10 s 1.053

t −= × =

9. Answer (2)

Hint : Remember 0 1 2 3 4 5 6 7 8 9BBRO YGB V GW Gold

silver

Sol. : R = 34 × 10° ± 5% = (34 ± 5%) Ω

PART - A (PHYSICS)

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


3/9

10. Answer (4)

Hint : Reff = 2000 Ω

Sol. : 20 50 C 1t

Q e− τ = × µ −

τ = Reff C = 2000 × 50 µs = 0.1 s

0.11000 C 1t

Q e−

= µ −

0.13 4 4ln 0.1 ln4 3 0.1 3

t te t−

= ⇒ = ⇒ =

11. Answer (2)

Hint : 2

GM mE V Kr

= + −

Sol. : ,2 4f iGMm GMmE E

R R= − = −

×

1 718 8f i

GMm GMmE ER R

− = − + =

12. Answer (1) Hint : φ21 = M21 i1

Sol. : ( )

20 0 0

32 2

24 2 8 164

I II RBR RR

µ µ µ⋅ π= = =

π × ×

2 2 0 15 516 2

Ir B r

Rµ

φ = π = π ⋅

⇒ 2

0516 2

rM

Rπµ

=

13. Answer (4)

Hint : E = n hc Epc pc

= ⇒ =λ

Sol. : 3 9

860 10 500 10

3 10p

− −× × ×=

×

= 2 × 5 × 10–17 kg m/s

= 10–16 kg m/s

14. Answer (1)

Hint : LC = 1 MSD – 1 VSD

Sol. : 1 MSD = 0.05 cm

2.450.05 0.001 cm50

LC = − =

D = 6.20 + 12 × 0.001 = 6.20 + 0.012

= 6.212 cm

15. Answer (3)

Hint : Use vmu

= and 1 1 1v u f

− =

Sol. : 3 3 34 4

v uvu

= − ⇒ = − =

⇒ u = – 4 m

1 1 7 1 1 23 4 12 2f R

+ = = = ×

⇒ 127

R = m

16. Answer (4) Hint : Heat is extracted in process A → B → C Sol. : AB BC v B A p C BQ Q nC T T nC T T∆ + ∆ = − + −

0 0 0 0 0 0 0 02 6 25 72 2

P V P V P V P Vn R RnnR nR

− − = + ⋅

0 00 0 0 0

335 142 2

P VP V P V= + =

17. Answer (4) Hint : Apply net gain of thermal energy of

middle plate is zero.

Sol. ( ) ( )4 44 44 3T T T T′ ′− = −

⇒ ( )4 4 42 256 81 337T T T′ = + =

⇒ 14337

2T T ′ =

18. Answer (1) Hint : Use constraint relation

Sol. : 21 2

3cos53

5a

a a= ° =

21

62 2

5a

T g a− = = …(i)



4/9

2320 35T a− = …(ii)

⇒ 2100 5

3T a− = …(iii)

2 2100 6 3120 5

3 5 5a a − = + =

22 2

40 31 200 m/s3 5 93

a a= ⇒ =

19. Answer (3) Hint : BAV

= –B AV V

Sol. : BAV

is 16 m/s along positive y-axis. 20. Answer (4)

Hint : Apply conservation of angular momentum about the point of contact.

Sol. 20

2 53 3

vmv R mvR mR mvRR

= + ⋅ =

0 03 35 5v v

vR

= ⇒ ω =

21. Answer (09)

Hint : 00

s

v vf f

v v+

=−

Sol. 345 350306340 345

f = × ×

306 35 315 Hz34

×= =

Beat frequency = 315 – 306 = 9 Hz

22. Answer (04)

Hint : ( )2 14nf v

l−

=

Sol. : ( ) ( )2 1 340 2 1 100Hz4 0.85nf n−

= = −×

100Hz, 300Hz, 500Hz, 700Hz,f =

23. Answer (32)

Hint : 2 22

1 1 22 2

kx mv=

Sol. : 2

2 2kxv

m=

2 2cm

2 2 23 3 3 2mv v x kvm m

= = =

24. Answer (04) Hint : Apply conservation of mechanical energy

Sol. : 2

22 122 2

R MRMg × ⋅ = ⋅ ωπ

16 4g gR R

ω = =π π

25. Answer (05)

Hint : 0BM C

T=

Sol. : ⇒ 0.8 0.4 5 K10

TT

= ⇒ =

26. Answer (2) Hint : CIP rules for assigning priority. Sol. : Configuration of both chiral centres is R.

27. Answer (3) Hint : Adiabatic processes are isentropic.

Sol. : ∆S = 0 for path C D→ .

28. Answer (2)

Hint : p 0E W KE= +

Sol. : (Ep = W0 + k1) × 2 2Ep = W0 + k2 k2 = 2k1 + W0 i.e. k2 > 2k1

29. Answer (4) Hint : Fact based. Sol. : Portland cement majorly contains

dicalcium silicate, tricalcium silicate and

tricalcium aluminate.

30. Answer (3) Hint : PCl5 solid exists as [PCl4]

+ [PCl6]–

Sol. : PBr5 solid exists as [PBr4]+[Br]–

31. Answer (2)

Hint : 0

2RTMPSM

=

Sol. : As T increases, MPS increases As M0 increases, MPS decreases

PART - B (CHEMISTRY)



5/9

32. Answer (4) Hint : Carbocation is formed and can undergo

rearrangement.

Sol. : In , the carbocation

formed cannot undergo ring expansion

to form 6 membered ring.

33. Answer (2) Hint : Cannizzaro reaction. Sol. :

34. Answer (3)

Hint : H4P2O6 contains both phosphorus in +4 state.

Sol. :

35. Answer (1) Hint : Fact based. Sol. : Salvarsan is arsphenamine.

36. Answer (2) Hint : i = 1 + (n – 1) α

Sol. : 2CaCli 1 2(0.6) 2.2= + =

xyi 1 1(0.75) 1.75= + =

∴ 4 × 2.2 × 0.01 = 1.75 × C

C 0.05 M 37. Answer (4)

Hint : Aldehydes, ketones and esters contain hydrogen.

Sol. : C12O9 is mellitic anhydride.

38. Answer (1)

Hint : At t = 1 hr, Nt = N0(e)

Sol. : 2dN 2Ndt

= −

⇒ 2dN 2dtN

= −

⇒ 0

0

N2 1t1

N (e)

N 2[t]2 1

− + = − − +

⇒ 0 0

1 1 2(t 1)N N (e)

− = −

⇒ 0

1 1t 1 12N e

= + −

39. Answer (1)

Hint : Self reduction of copper.

Sol. : 2 2 2Cu O Cu S Cu SO+ → +

40. Answer (4)

Hint : Fe2+ has d6 configuration.

Sol. : dipy is a strong field ligand and all except Fe2+ lose electron from eg level upon oxidation.

41. Answer (3)

Hint : Claisen rearrangement.

Sol. :

CH3

CH3CH3

Ph CH CH2

KMnO4

0ºC, OH– Ph CH CH2

OH OH

HIO4

Ph CHO + HCHOOH–

(conc.)PhCH OH + HCOO2

HO P P OH

OH OH

O O

O

O

OO

OO

OO

O



6/9

42. Answer (2)

Hint : Lanthanoid contraction. Sol. : Zr and Hf have almost similar size.

43. Answer (1) Hint : RNA contains uracil and not thymine. Sol. : RNA is single stranded and adenine is

H-bonded to uracil. 44. Answer (3)

Hint :

4 33 4 4 4Zr (PO ) 3Zr 4PO

3x 4x

+ −+

Sol. : Given 3x = S

⇒ Sx3

=

∴ 34

4S[PO ]3

− =

∴ 3 4spK (3x) (4x)= ⋅

4

3 4S(S)3

= ⋅

7256 S81

=

45. Answer (1)

Hint : NaBH4 converts carbonyl compounds to

alcohols.

Sol. :

46. Answer (60) Hint : O2– at face centre touches Ca2+ at all

corners of the face. ∴ y = 4 Sol. : O2– at face centre touches two Ti4+ at

adjacent body centres. ∴ x = 2 47. Answer (24)

Hint : 32 6[Al(H O) ] + and 4[Al(OH) ]−

Sol. : Al3+ is hexa co-ordinated with water while it is tetra co-ordinated with hydroxide ions.

48. Answer (85)

Hint : Volume strengthMolarity = 11.2

Sol. : M = 28 2.511.2

=

1 L solution contains 85 g H2O2 ∴ (w/v)% = 8.5%

49. Answer (80)

Hint : In[In ]pK log pH[HIn]

−

+ =

Sol. : From dissociation of HA, 2 6 2[H ] 10 10+ − −= ×

4[H ] 10+ −= pH = 4

∴ [In ]4.6 log 4[HIn]

−

+ =

⇒ 0.6In 1 1

[HIn] 410

−

= =

∴ 20% In– and 80% HIn 50. Answer (20)

Hint : Given ligand is EDTA4–. Sol. : Ethylenediamine tetra acetate (EDTA4–)

is a hexadentate ligand and y is 4.

2(x + y) is 20.

51. Answer (4) Hint : Mean formulae.

Sol. : New mean of the series is X nm

+

⇒ X nmm+

52. Answer (3) Hint : ( )~ ~ ~p q p q∨ = ∧ .

Sol. : By definition of negation option (3) is correct

53. Answer (1) Hint : Continuity at discrete point.

Sol. : If f(x) is continuous at x = 6π

⇒ cos2x = sinx

⇒ x = 6π is one of the solution

∴ Option (1) is correct.

BrNaBH4

O–

BrO

OHCNOH

CN

PART - C (MATHEMATICS)



7/9

54. Answer (1) Hint : Total law of probability Sol. : P(2 white + 1 black)

= P(W1 W2 B3) + P(W1 B2 W3)

+ P(B1 W2 W3)

= 3 2 3 3 2 1 1 2 14 4 4 4 4 4 4 4 4

× × + × × + × ×

= 1332

55. Answer (3)

Hint : Definition of orthogonal matrices.

Sol. : – –2 2

T I IA A

= I … (i)

and 2 2

T I IA A + +

= I …(ii)

⇒ A + AT = 0 (By subtracting (i) & (ii)

⇒ AT = –A ⇒ 2 3–4

A I=

⇒ |A|2 = –34

n

⇒ is evenn

56. Answer (4)

Hint : Assumption of plane.

Sol. : Given line 6 4 3x y z

= =

Let plane is ax + by + cz = 1

⇒ 6 4 3a b c+ + = 0 …(i)

2a – b = 1 …(ii)

3a – 4b + 5c = 1 …(iii)

⇒ a = 2985

; b = –2785

c = –2285

∴ Eqn. of plane is

29 – 27 – 22 85x y z =

57. Answer (1) Hint : Scalar triple product.

Sol. : ( )( ) ( ) ( )( ).a a b b a b a b+ × + × × ×

= ( )( ) ( )( ).a a b b a b+ × × ×

= ( ). . 1a a a a b+ × =

58. Answer (4) Hint : Graphical approach. Sol. : The above equation has always two real

roots by observing the graph of the function in Domain.

∴ U ∈φ. 59. Answer (3) Hint : Property of determinants. Sol. : Take x5 common from R3

⇒ x5

1010 1012 2020

1010 1 2020

1 n

n

x x x

x n

x x x+

= 0

⇒ n + 1 = 1012 ⇒ n = 1011

60. Answer (2)

Hint : Numerically greatest term.

Sol. : 4

31T

T≥ and 5

41T

T≤

⇒ ( )11– 3 3 13 16

x≥

and 11– 4 3 14 16

x ≤

⇒ 2 ≤ x ≤ 6421

Required sum = (2 + 3) = 5

61. Answer (4)

Hint : Choosing the 4 vertex.

Sol. : nC4 = 35

⇒ n(n –1) (n – 2) (n – 3) = 24 × 35

⇒ n = 7

62. Answer (4)

Hint : Telescopic method.

Sol. : The given series is



8/9

( )( )

215–1

1

12

1r

r

r

r r=

+

+∑ = 15–1

1

2 – 1– 21

r

r

r rr r=

+

∑

= ( )15–1

1

– 12 – 2

1r r

r

rrr r=

+ ∑

= v15 – v0 = 1515 216

× = 1115 2×

63. Answer (4) Hint : Geometrical application

Sol. : Req. Area = 1 2 42

× × = 4 sq. units.

64. Answer (4)

Hint : Location of roots Sol. : x4 + 25 = mx2

⇒ 2

2 5xx

+

= m

⇒ 25– – 10x m

x =

For m = 10 equation has four real roots but two are repeated one

∴ m ∈ φ

65. Answer (2) Hint : Touching concept

Sol. : Equation of any circle (x – α) (x – β) + y2

+ λy = 0

⇒ x2 + y2 – (α + β)x + λy + αβ = 0 …(1)

∠ACB is maximum

∴ It touches y-axis at (0, γ).

⇒ 2 0y y+ λ + αβ = have repeated

roots

⇒ 2γ = αβ

66. Answer (2) Hint : Concept of tangency.

Sol. : On solving both the equations

( )2 2 2 2 2 2– 2 – 0b a x a x a b+ =

For least area, circle must touch the ellipse.

⇒ 4a4 + 4a2b2(b2 – a2) = 0

⇒ 1 – b2e2 = 0

⇒ 1be

=

⇒ a2 = ( )

2

2 2

11– 1–

be e e

=

Area = A = 2 21–e e

π = f(e) say

Area is minimum if e4 – e6 is maximum.

Let g(e) = e4 – e6

⇒ g′(e) = 4e3 – 6e5 = 0

⇒ 23

e =

∴ ellipse is 2 22 6 9x y+ =

67. Answer (3) Hint : Assumption of normal

Sol. : A normal at 44 ,A tt

is

( )24– – 4y t x tt

=

⇒ 4 34 – – 4 0t t h tk+ =

( ),p h k satisfy it

given

2 2 2 21 2 3 4 1 2 3 4t t t t y y y y+ + + = + + +

⇒ 2

16h k=

⇒ 2 16x y= (Parabola)



9/9

68. Answer (1) Hint : Discriminant = 0

Sol. : P satisfy x2 2

–3y = 1

Let m is slope of common tangent then m2 – 3 = r2(1 + m2)

⇒ m2 = 2

23

1–e

r+

2 4r =

∴ no common tangent exists. 69. Answer (1)

Hint : Location of roots Sol. : 2 lies between interval (e1 > e2)

∴ ( )2 0F <

⇒ 2 – 2 2 0a + <

⇒ 2 2a > 70. Answer (3)

Hint : Newton Leibnitz Formula

Sol. : ( ) sin xf xx

′ =

⇒ f(x) increases in

x∈(0, π) and f(x) decreases in x ∈ (π, 2π) 71. Answer (02)

Hint : Rationalisation Method.

Sol. : ( ) ( ) ( )1 2 .... –x x x x n x+ + +

= ( )1 12 2

n n nn+ +

=

∴ L = 3 12+ = 2

72. Answer (04)

Hint : ∆ = 12

ah

Sol. : ah1 = bh2 = ch3 = 2∆

⇒ 1 2 3

1 1 1h h h

+ + = 2

a b c+ +∆

73. Answer (02)

Hint : By using (AOD).

Sol. : Take log both sides.

74. Answer (00) Hint : Leibnitz formula.

Sol. : 1 3 1 7 02 2 4 4

f f f f ′ = ′ = ′ = ′ =

Also ( )1f x′ + = ( )– 1–f x′

⇒ g(1) = 0 = 1 32 4

g g ′ = ′

75. Answer (06) Hint : Inverse function concept.

Sol. : Let y = 23 2x x+

⇒ x = – 1 + ( )1

3 21y +

⇒ f –1(y) = ( )1

3 2–1 1y+ +

∴ I = ( ) ( )( )2

–1

0

1f u f u du+ +∫

= ( )( ) ( )2 200

6uf u u+ =

mock test - 4 (code-e) (answers) all india aakash test series for … · 2020. 3. 2. · mock test...

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