Test - 5 (Code-C) (Answers) All India Aakash Test Series for JEE … · 2020. 2. 3. · All India...

Test - 5 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2021

All India Aakash Test Series for JEE (Main)-2021

Test Date : 02/02/2020

ANSWERS

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

TEST-5 - Code-C

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (3)

2. (1)

3. (2)

4. (3)

5. (3)

6. (4)

7. (2)

8. (1)

9. (2)

10. (1)

11. (2)

12. (4)

13. (1)

14. (4)

15. (4)

16. (3)

17. (3)

18. (1)

19. (4)

20. (2)

21. (24)

22. (04)

23. (25)

24. (31)

25. (07)

26. (2)

27. (1)

28. (2)

29. (3)

30. (3)

31. (4)

32. (3)

33. (2)

34. (2)

35. (1)

36. (3)

37. (2)

38. (3)

39. (3)

40. (1)

41. (2)

42. (4)

43. (1)

44. (3)

45. (1)

46. (02)

47. (06)

48. (06)

49. (60)

50. (56)

51. (3)

52. (3)

53. (3)

54. (3)

55. (2)

56. (4)

57. (4)

58. (1)

59. (2)

60. (1)

61. (3)

62. (3)

63. (4)

64. (2)

65. (1)

66. (3)

67. (4)

68. (4)

69. (3)

70. (3)

71. (65)

72. (09)

73. (61)

74. (10)

75. (36)

All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-C) (Hints & Solutions)


2/10

HINTS & SOLUTIONS

PART - A (PHYSICS)

1. Answer (3)

Hint: Let R = thermal resistance of each rod

of y

then 2R = thermal resistance of each rod of x.

Sol.:

1 16A BT T T RI = − =

2 14B ET T T RI = − =

1 12

3 3100 60 C

2 5

TT

T

= = =

2 40 = T C

60 60 20 C= − = − = B A B B CT T T T T

TC = 40°C

2. Answer (1)

Hint: 140 z = 100°C

Sol.: 14 7

1 C10 5

z = =

1 1 14200

4200 J kg J kg7

5

ws k

z

− − −= =

= 3000 J kg–1 z–1

3. Answer (2)

Hint: 2

12

mlI =

Sol.: 2

2I l

TI l

= =

4. Answer (3)

Hint: V = VT

Sol.: 1 dV

V dT =

3nRT

PV nRT KV

= =

ln( ) 3 ln ln ln+ = +nR T K V

3 1 dV

T V dT= =

5. Answer (3)

Hint: W = Area under P-V graph

Sol.: In general W may be positive or

negative.

6. Answer (4)

Hint: dT dt

H KA aATdx dx

= − = −

Sol.: 0

0

2 20 0

4 0

16

2

T l

T

T THdx HlTdT

aA aA

− −= − =

2

015

2

TaAH

l=

2 2 2

0 016 15

2 2 2

T T TaA l

l aA

− −=

2

2 2 00

1516

2

TT T= −

2

017

2

T=

017

2T T=

7. Answer (2)

Hint: PV = RT

Sol.: 22 2

Constant ConstantP

PVP V

= =

3

2 2 1 2

R R RC = − =

−

02

RQ C T T = =

8. Answer (1)

Hint: dQ = nCdT

Sol.: dQ = nCvdT + PdV

PdV = nTdT

nRT

dV n TdTV

=

Test - 5 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021


3/10

dV dT

V R

=

ln lnT

k VR

+ =

=

T

RkV e

Constant

−

=

T

RVe

9. Answer (2)

Hint: 4 4[ ]sP A T T= −

Sol.: P = 5 × 10–4 × 6.0 × 10–8 [6004 – 3004]

= 3 × 10–11 × 108 [1296 – 81] = 3 × 10–3 ×

1215 W

= 3.645 W

10. Answer (1)

Hint: dQ = nCdT

Sol.: nCdT = nCvdT + nVdT

PdV = nVdT

nRT

PV nRT PV

= =

dV

nRT n VdTV

=

2

=

dT RdV

T V

1

ln( )

= −

kT RV

R

VkT e−

=

constant =

R

VTe

11. Answer (2)

Hint: U = nCvT

Sol.: 15

2i

RU n T=

1 2 25 3

( ) 2 2 22 2

= − + fR R

U n n T n T

1 2 2( )5 6n n RT n RT= − +

1 12 2

5 5

2 2

= − = + = +

f i

n RT nU U U n RT n RT

12. Answer (4)

Hint: The decrease in the KE due to ordered

motion is equal to increase in internal

energy.

Sol.: 2

2 21 1

2 2 4

= − =

VU nm V nmV

1

v

RnC T n T= =

−

2( 1)

4

− =

mVT

R

13. Answer (1)

Hint: p = nRT

Sol.: 1 1 1 2

2 2 1 2

p n m M

p n M m= =

2 1 2

1 2 1

4 2

3 3

M M

M M

= =

1

2

1

2

M

M=

14. Answer (4)

Hint: TV = constant PV2 = constant

Sol.: 22 2

k kP

v m = =

15. Answer (4)

Hint: SlopeP

V

= −

Sol.:

12 22 1

2

f

f

f f f

+ + = = = +

16. Answer (3)

Hint: dp = –gdh

Sol.: nRT mRT RT

PV MV M

= = =

RdT dT Mg

gdhM dh R

−= − =

dT Mg

dh R=



4/10

17. Answer (3)

Hint: A = A0[1 + T]

Sol.: 3 5[1 ] 1 12 2

= + + = +

A lb T T lb T

5

2 =

18. Answer (1)

Hint: =

FLY

A L

Sol.: L F F

T Y TL A T AY

= = =

19. Answer (4)

Hint: extracted

Wn

Q=

Sol.: 0 02W PV=

0 0 0 033

2 2AB

P V P Vn RQ

nR= =

0 0 0 045

102

BC

P Vn RQ P V

nR= =

0 0

0 0

2 42 100%

23 23

P Vn

P V= =

20. Answer (2)

Hint: 2Q

W =

Sol.: 2 2

1 2 1 2

279 93

24 8

Q T

Q Q T T = = = =

− −

= 11.6

21. Answer (24)

Hint: 21

0.6 1002

m

Sol.: 20.6

1002

ms T m =

0.3 104

24 C125

T

= =

22. Answer (04)

Hint: isothermal isothermalQ W =

Sol.: ln = CBCB

VQ nRT

V

23. Answer (25)

Hint: Maximum temperature is attained in

process AB

Sol.: 0

0

= − +P

P V CV

00 0 00

4 5= − + =P

P V C C PV

0 00

5P

P V PV

= − +

20 00

5P

PV RT V P V RTV

= − + =

0 000

2 55 0

2

P VV P V

V− + = =

2

0 0 0 0max

0

25 251

4 2

= − +

P V P VT

R V

0 025

4

P V

R=

24. Answer (31)

Hint: Q = nCPT

Sol.: 5

1252

RT=

125 2

3 6 K5 25

= =

T

25. Answer (07)

Hint: 5

2

RU n T =

Sol.: 7

2

RQ n T =

W = nRT



5/10

PART - B (CHEMISTRY)

26. Answer (2)

Hint :

possible tautomers

Sol. : A is poly hydroxy aromatic compound.

B is poly keto non-aromatic compound

27. Answer (1)

Hint :

Sol. : Acidic strength order is Hb > Ha > Hc

28. Answer (2)

Hint :

According to lowest sum rule correct path

of numbering is given above

Sol. : Correct IUPAC name is

4-Amino-4-hydroxy oct-2-en-5-yne-1,

8-dioic acid.

29. Answer (3)

Hint : To show geometrical isomerism molecule

must have a group which can be oriented

in different spatial arrangements.

Sol. :

Two geometrical isomers

30. Answer (3)

Hint : Alcohols and ethers are functional

isomers

Sol. :

and CH3CH2 – O – CH3

are the two compounds which have same

molecular formula with different functional

groups.

31. Answer (4)

Hint & Sol. :

1, 2 and 3 are conjugated system.

32. Answer (3)

Hint : Energy of structure I and V and II and IV

are same

Sol. : III is least stable

33. Answer (2)

Hint : Lesser is the double bond character

lower would be the energy required for

rotation.

Sol. : Both ring becomes aromatic after

dissociation of -bond hence have least

double bond order.

34. Answer (2)

Hint : Greater the electron density on nitrogen

greater will be the basic strength



6/10

Sol. :

Lone pair of nitrogen is in conjugation

with benzene rings

35. Answer (1)

Hint & Sol. :

Hyperconjugation participate with in

toluene and increases the resonance

energy.

36. Answer (3)

Hint : Greater the stability of conjugate base

lesser will be strength of conjugate base

Sol. :

Most stable conjugate base due to

identical resonating structure.

37. Answer (2)

Hint : The species which have complete octet

cannot act as an electrophile.

Sol. : NH3 has lone pair of electron.

38. Answer (3)

Hint : Negative charge on more electronegative

atom is more stable.

Sol. :

Negative charge on N is more stable than

at carbon.

39. Answer (3)

Hint :

Sol. : Lesser the charge density on cation

greater will be the stability.

40. Answer (1)

Hint : Dipole moment is a vector quantity and to

calculate the net dipole moment, vector

sum formula can be used.

Sol. : Dipole moment would be maximum when

the angle between the two vectors is

180°.

41. Answer (2)

Hint & Sol. :

2 46Fe CN [Fe(CN) ]

+ − −+ ⎯⎯→

Some of the Fe2+ is converted into Fe3+

when heated with conc. H2SO4

3 4

6 4 6 3Prussian blue

Fe [Fe(CN) ] Fe [Fe(CN) ]+ −+ ⎯⎯→

42. Answer (4)

Hint & Sol. :

S2– + Pb2+ ⎯⎯→ PbS black ppt.

43. Answer (1)

Hint : Equivalent of BaSO4 = Equivalents of S

in organic compound

Sol. : 0.9625 g of BaSO4 contains

= 32 0.9625

g233

of sulphur

% of sulphur = 32 0.9625 100

233 0.314

42%

44. Answer (3)

Hint : In compound P the degree of

unsaturation is 2 where as in Q is 4.

Sol. : Both have 4 primary carbon atoms.

45. Answer (1)

Hint : Consider the involvement of lone pair in

resonance.

Sol. :



7/10

46. Answer (02)

Hint : To show geometrical isomerism along a

C=C, both C-atom should posses

different groups

Sol. :

47. Answer (06)

Hint : Except 3-chlorocyclopropene all have

DBE 3.

Sol. : DBE = 2.

48. Answer (06)

Hint : Check all the possibilities

Sol. :

49. Answer (60)

Hint : Let the given compound is CxHy

x y 2 2 2

4.4g2g

C H O CO H O+ ⎯⎯→ +

Sol. : 4.4 g of CO2 1.2 g of C

Composition of C = 1.2

1002

= 60%

50. Answer (56)

Hint : Equivalents of H2SO4 acid = eq. of NH3 +

eq. of NaOH

Sol. : 100 × 0.25 × 2 = eq. of NH3 + 60 × 0.5

eq. of NH3 = 20 m eq. = 20 m moles

% of N = 314 20 10 100

56%0.5

− =

PART - C (MATHEMATICS)

51. Answer (3)

Hint: Use chain rule.

Sol.: y = f(x3)

3 2( ) 3= dy

f x xdx

6 2 13 6 3 9 3 9== + = =xx x

52. Answer (3)

Hint: Use binomial theorem.

Sol.: 5

01 5

5(1 )5 ... 5 4

+= + + +

CxC C

x x

5

41 2 5

(1 ) 15 5 ... 5

+ = + + + +

C C C

5(1 ) 1+ −

53. Answer (3)

Hint: Make truth table.

Sol.:

54. Answer (3)

Hint: Express equation as factors.

Sol.: P i(2i) = –2

z4 + 5z3 + 18z2 – 4z2 + 5 = (z – z1)

(z – z2) (z – z3) (z – z4)

16 – 40 + 72 – 16 + 5 = (–2 – z1)

(–2 – z2) (–2 – z3) (–2 – z4)

37 = ((PA) (PC))2



8/10

55. Answer (2)

Hint: G.P of infinite series.

Sol.: 1

11 2 1

12 12

=

= = −

rr

56. Answer (4)

Hint: Section formula.

Sol.:

9 10 4

6 3 41 3

+= = − = −

+

4(1) 1( 3)

133 131 1

3

z− + −

−= = =

−−

57. Answer (4)

Hint: Rationalise.

Sol.:

2 2

2 21

(2 3 4) ( 6 2)lim

( 1)( 2 3 4 6 2)→

+ + − + +

− + + + + +x

x x x x

x x x x x

1

( 2)( 1) 1lim

( 1)(3 3) 6x

x x

x→

− −= −

− +

58. Answer (1)

Hint: Locus.

Sol.: |z1| = |z2| = |z1 – z2|

2 21 2 1 2z z z z+ =

1 2

2 1

1z z

z z+ =

59. Answer (2)

Hint: Condition of line intersection.

Sol.: z lies on line or segment joining z1 & z2

and also on line or segment joining 1z

and z2. Hence only 1 possible solution.

60. Answer (1)

Hint: Chain rule.

Sol.: f(x) = f(2 – x)

f (x) = –f (2 – x)

f (x) + f (2 – x) = 0

Put x = 3 f (3) + f (–1) = 0

61. Answer (3)

Hint: Negation of statement.

Sol.: ~( (~ )) ~(~ )→ = q q p q q p

( ~ ) ~q q p q p= =

62. Answer (3)

Hint: Truth table.

Sol.:

63. Answer (4)

Hint: 1 form.

Sol.:

1 1 1 1

2 3 5 7 4lim 2

4→

+ + + −

x x x x

xx

e

( ) ( ) ( ) ( )1 1 1 12 1 3 1 5 1 7 1lim

12

→

− + − + − + −

x x x x

x

xe

1ln(2 3 5 7)

2 210e

=

64. Answer (2)

Hint: Create Indeterminacy.

Sol.: 2 2 ( ) (1 )

lim 31x

x ax b a x b

x→

− + + + −=

−

a = 1 and b – a = 3 b = 2



9/10

65. Answer (1)

Hint: Locus is a straight line.

Sol.: 1

| 1|2021

z z− = −

Locus

11

101120212 2021

x+

= = =

66. Answer (3)


Sol.: Let B divides AC in ratio : 1, then for

y-coordinates

( 2) 1(4)

21

− +=

+

–2 + 4 = 2 + 2

1

4 22

= =

5(1) ( )

61

a+ =

+

5 9 82

aa+ = =

67. Answer (4)

Hint: Locus based.

Sol.: 21

(3 ) (4 )2

+ − −k i i

21

1 22

i − +

5 5

,2 2

k

68. Answer (4)

Hint: 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

cosa a b b c c

a b c a b c

+ + =

+ + + +

Sol.: For perpendicular lines

2( 6) 5( ) 6( 1) 0 − + + − =

3 26 11 6 0 1, 2, 3 − + − = =

69. Answer (3)

Hint: Put z = x + iy

Sol.: 2 2 2 4x iy x y i+ − + = − −

24 and 16 2y x x= + − + = −

2 24 4 16x x x+ + = +

x = 3

14

3 4 Arg( ) tan3

z i z −

= + =

70. Answer (3)

Hint: Check L.H.L. and R.H.L.

Sol.: R.H.L. = e–1 and L.H.L. = 1

Limit does not exist

71. Answer (65)

Hint: | x | < a –a < x < a

Sol.: | | 13 2z −

2 | | 13 2z− −

11 | | 15z

72. Answer (09)

Hint: 0

form0

Sol.:

23

3

2

3

2 3

1 1tan

3 7

1

1lim

7 81

→−

+ + − = =

+ +

x

xxx x

xL

xx x

22

1

L=

73. Answer (61)

Hint: u

v rule of differentiation



10/10

Sol.: 22

( )( 1) (3 1)

xf x

x x= −

+ −

2

2 2

2 (3 2 )( )

( 1) (3 1)

x xf x

x x

− − = −

+ −

2 8 50 72 122

(2)9 25 225 225

f− − −

= − − = =

74. Answer (10)

Hint: Locus based

Sol.: |z – 4| = |z – 8| x = 6

and 3|z – 12| = 5|z – 8i|

9(x2 + y2 – 24x + 144) = 25(x2 + y2 – 16y

+ 64)

y = 8, 17

| z |min = 10

75. Answer (36)

Hint: V = abc

Sol.: a = 3 b = 6 c = 2

V = abc = 3 6 2 = 36

Test - 5 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2021

All India Aakash Test Series for JEE (Main)-2021

Test Date : 02/02/2020

ANSWERS


TEST-5 - Code-D

1/10

PHYSICS CHEMISTRY MATHEMATICS

1. (2)

2. (4)

3. (1)

4. (3)

5. (3)

6. (4)

7. (4)

8. (1)

9. (4)

10. (2)

11. (1)

12. (2)

13. (1)

14. (2)

15. (4)

16. (3)

17. (3)

18. (2)

19. (1)

20. (3)

21. (07)

22. (31)

23. (25)

24. (04)

25. (24)

26. (1)

27. (3)

28. (1)

29. (4)

30. (2)

31. (1)

32. (3)

33. (3)

34. (2)

35. (3)

36. (1)

37. (2)

38. (2)

39. (3)

40. (4)

41. (3)

42. (3)

43. (2)

44. (1)

45. (2)

46. (56)

47. (60)

48. (06)

49. (06)

50. (02)

51. (3)

52. (3)

53. (4)

54. (4)

55. (3)

56. (1)

57. (2)

58. (4)

59. (3)

60. (3)

61. (1)

62. (2)

63. (1)

64. (4)

65. (4)

66. (2)

67. (3)

68. (3)

69. (3)

70. (3)

71. (36)

72. (10)

73. (61)

74. (09)

75. (65)

All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-D) (Hints & Solutions)


2/10

HINTS & SOLUTIONS

PART - A (PHYSICS)

1. Answer (2)

Hint: 2Q

W =

Sol.: 2 2

1 2 1 2

279 93

24 8

Q T

Q Q T T = = = =

− −

= 11.6

2. Answer (4)

Hint: extracted

Wn

Q=

Sol.: 0 02W PV=

0 0 0 033

2 2AB

P V P Vn RQ

nR= =

0 0 0 045

102

BC

P Vn RQ P V

nR= =

0 0

0 0

2 42 100%

23 23

P Vn

P V= =

3. Answer (1)

Hint: =

FLY

A L

Sol.: L F F

T Y TL A T AY

= = =

4. Answer (3)

Hint: A = A0[1 + T]

Sol.: 3 5[1 ] 1 12 2

= + + = +

A lb T T lb T

5

2 =

5. Answer (3)

Hint: dp = –gdh

Sol.: nRT mRT RT

PV MV M

= = =

RdT dT Mg

gdhM dh R

−= − =

dT Mg

dh R=

6. Answer (4)

Hint: SlopeP

V

= −

Sol.:

12 22 1

2

f

f

f f f

+ + = = = +

7. Answer (4)

Hint: TV = constant PV2 = constant

Sol.: 2

2 2

k kP

v m = =

8. Answer (1)

Hint: p = nRT

Sol.: 1 1 1 2

2 2 1 2

p n m M

p n M m= =

2 1 2

1 2 1

4 2

3 3

M M

M M

= =

1

2

1

2

M

M=

9. Answer (4)

Hint: The decrease in the KE due to ordered

motion is equal to increase in internal

energy.

Sol.: 2

2 21 1

2 2 4

= − =

VU nm V nmV

1

v

RnC T n T= =

−

2( 1)

4

− =

mVT

R

Test - 5 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021


3/10

10. Answer (2)

Hint: U = nCvT

Sol.: 15

2i

RU n T=

1 2 25 3

( ) 2 2 22 2

= − + fR R

U n n T n T

1 2 2( )5 6n n RT n RT= − +

1 12 2

5 5

2 2

= − = + = +

f i

n RT nU U U n RT n RT

11. Answer (1)

Hint: dQ = nCdT

Sol.: nCdT = nCvdT + nVdT

PdV = nVdT

nRT

PV nRT PV

= =

dV

nRT n VdTV

=

2

=

dT RdV

T V

1

ln( )

= −

kT RV

R

VkT e−

=

constant =

R

VTe

12. Answer (2)

Hint: 4 4[ ]sP A T T= −

Sol.: P = 5 × 10–4 × 6.0 × 10–8 [6004 – 3004]

= 3 × 10–11 × 108 [1296 – 81] = 3 × 10–3 ×

1215 W

= 3.645 W

13. Answer (1)

Hint: dQ = nCdT

Sol.: dQ = nCvdT + PdV

PdV = nTdT

nRT

dV n TdTV

=

dV dT

V R

=

ln lnT

k VR

+ =

=

T

RkV e

Constant

−

=

T

RVe

14. Answer (2)

Hint: PV = RT

Sol.: 22 2

Constant ConstantP

PVP V

= =

3

2 2 1 2

R R RC = − =

−

02

RQ C T T = =

15. Answer (4)

Hint: dT dt

H KA aATdx dx

= − = −

Sol.: 0

0

2 20 0

4 0

16

2

T l

T

T THdx HlTdT

aA aA

− −= − =

2

015

2

TaAH

l=

2 2 2

0 016 15

2 2 2

T T TaA l

l aA

− −=

2

2 2 00

1516

2

TT T= −

2

017

2

T=

017

2T T=

16. Answer (3)

Hint: W = Area under P-V graph

Sol.: In general W may be positive or

negative.



4/10

17. Answer (3)

Hint: V = VT

Sol.: 1 dV

V dT =

3nRT

PV nRT KV

= =

ln( ) 3 ln ln ln+ = +nR T K V

3 1 dV

T V dT= =

18. Answer (2)

Hint: 2

12

mlI =

Sol.: 2

2I l

TI l

= =

19. Answer (1)

Hint: 140 z = 100°C

Sol.: 14 7

1 C10 5

z = =

1 1 14200

4200 J kg J kg7

5

ws k

z

− − −= =

= 3000 J kg–1 z–1

20. Answer (3)

Hint: Let R = thermal resistance of each rod

of y

then 2R = thermal resistance of each rod of x.

Sol.:

1 16A BT T T RI = − =

2 14B ET T T RI = − =

1 12

3 3100 60 C

2 5

TT

T

= = =

2 40 = T C

60 60 20 C= − = − = B A B B CT T T T T

TC = 40°C

21. Answer (07)

Hint: 5

2

RU n T =

Sol.: 7

2

RQ n T =

W = nRT

22. Answer (31)

Hint: Q = nCPT

Sol.: 5

1252

RT=

125 2

3 6 K5 25

= =

T

23. Answer (25)

Hint: Maximum temperature is attained in

process AB

Sol.: 0

0

= − +P

P V CV

00 0 00

4 5= − + =P

P V C C PV

0 00

5P

P V PV

= − +

20

00

5P

PV RT V P V RTV

= − + =

0 000

2 55 0

2

P VV P V

V− + = =

2

0 0 0 0max

0

25 251

4 2

= − +

P V P VT

R V

0 025

4

P V

R=

24. Answer (04)

Hint: isothermal isothermalQ W =

Sol.: ln = CBCB

VQ nRT

V



5/10

25. Answer (24)

Hint: 21

0.6 1002

m

Sol.: 20.6

1002

ms T m =

0.3 104

24 C125

T

= =

PART - B (CHEMISTRY)

26. Answer (1)

Hint : Consider the involvement of lone pair in

resonance.

Sol. :

27. Answer (3)

Hint : In compound P the degree of

unsaturation is 2 where as in Q is 4.

Sol. : Both have 4 primary carbon atoms.

28. Answer (1)

Hint : Equivalent of BaSO4 = Equivalents of S

in organic compound

Sol. : 0.9625 g of BaSO4 contains

= 32 0.9625

g233

of sulphur

% of sulphur = 32 0.9625 100

233 0.314

42%

29. Answer (4)

Hint & Sol. :

S2– + Pb2+ ⎯⎯→ PbS black ppt.

30. Answer (2)

Hint & Sol. :

2 46Fe CN [Fe(CN) ]

+ − −+ ⎯⎯→

Some of the Fe2+ is converted into Fe3+

when heated with conc. H2SO4

3 4

6 4 6 3Prussian blue

Fe [Fe(CN) ] Fe [Fe(CN) ]+ −+ ⎯⎯→

31. Answer (1)

Hint : Dipole moment is a vector quantity and to

calculate the net dipole moment, vector

sum formula can be used.

Sol. : Dipole moment would be maximum when

the angle between the two vectors is

180°.

32. Answer (3)

Hint :

Sol. : Lesser the charge density on cation

greater will be the stability.

33. Answer (3)

Hint : Negative charge on more electronegative

atom is more stable.

Sol. :

Negative charge on N is more stable than

at carbon.

34. Answer (2)

Hint : The species which have complete octet

cannot act as an electrophile.

Sol. : NH3 has lone pair of electron.

35. Answer (3)

Hint : Greater the stability of conjugate base

lesser will be strength of conjugate base

Sol. :

Most stable conjugate base due to

identical resonating structure.

36. Answer (1)

Hint & Sol. :

Hyperconjugation participate with in

toluene and increases the resonance

energy.



6/10

37. Answer (2)

Hint : Greater the electron density on

nitrogen greater will be the basic

strength

Sol. :

Lone pair of nitrogen is in conjugation

with benzene rings

38. Answer (2)

Hint : Lesser is the double bond character

lower would be the energy required

for rotation.

Sol. : Both ring becomes aromatic after

dissociation of -bond hence have

least double bond order.

39. Answer (3)

Hint : Energy of structure I and V and II and

IV are same

Sol. : III is least stable

40. Answer (4)

Hint & Sol. :

1, 2 and 3 are conjugated system.

41. Answer (3)

Hint : Alcohols and ethers are functional

isomers

Sol. :

and CH3CH2 – O – CH3

are the two compounds which have

same molecular formula with different

functional groups.

42. Answer (3)

Hint : To show geometrical isomerism molecule

must have a group which can be oriented

in different spatial arrangements.

Sol. :

Two geometrical isomers

43. Answer (2)

Hint :

According to lowest sum rule correct path

of numbering is given above

Sol. : Correct IUPAC name is

4-Amino-4-hydroxy oct-2-en-5-yne-1,

8-dioic acid.

44. Answer (1)

Hint :

Sol. : Acidic strength order is Hb > Ha > Hc

45. Answer (2)

Hint :

possible tautomers

Sol. : A is poly hydroxy aromatic compound.

B is poly keto non-aromatic compound



7/10

46. Answer (56)

Hint : Equivalents of H2SO4 acid = eq. of NH3 +

eq. of NaOH

Sol. : 100 × 0.25 × 2 = eq. of NH3 + 60 × 0.5

eq. of NH3 = 20 m eq. = 20 m moles

% of N = 314 20 10 100

56%0.5

− =

47. Answer (60)

Hint : Let the given compound is CxHy

x y 2 2 2

4.4g2g

C H O CO H O+ ⎯⎯→ +

Sol. : 4.4 g of CO2 1.2 g of C

Composition of C = 1.2

1002

= 60%

48. Answer (06)

Hint : Check all the possibilities

Sol. :

49. Answer (06)

Hint : Except 3-chlorocyclopropene all have

DBE 3.

Sol. : DBE = 2.

50. Answer (02)

Hint : To show geometrical isomerism along a

C=C, both C-atom should posses

different groups

Sol. :

PART - C (MATHEMATICS)

51. Answer (3)

Hint: Check L.H.L. and R.H.L.

Sol.: R.H.L. = e–1 and L.H.L. = 1

Limit does not exist

52. Answer (3)

Hint: Put z = x + iy

Sol.: 2 2 2 4x iy x y i+ − + = − −

24 and 16 2y x x= + − + = −

2 24 4 16x x x+ + = +

x = 3

14

3 4 Arg( ) tan3

z i z −

= + =

53. Answer (4)

Hint: 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

cosa a b b c c

a b c a b c

+ + =

+ + + +

Sol.: For perpendicular lines

2( 6) 5( ) 6( 1) 0 − + + − =

3 26 11 6 0 1, 2, 3 − + − = =

54. Answer (4)

Hint: Locus based.

Sol.: 21

(3 ) (4 )2

+ − −k i i

21

1 22

i − +

5 5

,2 2

k



8/10

55. Answer (3)


Sol.: Let B divides AC in ratio : 1, then for

y-coordinates

( 2) 1(4)

21

− +=

+

–2 + 4 = 2 + 2

1

4 22

= =

5(1) ( )

61

a+ =

+

5 9 82

aa+ = =

56. Answer (1)

Hint: Locus is a straight line.

Sol.: 1

| 1|2021

z z− = −

Locus

11

101120212 2021

x+

= = =

57. Answer (2)

Hint: Create Indeterminacy.

Sol.: 2 2 ( ) (1 )

lim 31x

x ax b a x b

x→

− + + + −=

−

a = 1 and b – a = 3 b = 2

58. Answer (4)

Hint: 1 form.

Sol.:

1 1 1 1

2 3 5 7 4lim 2

4→

+ + + −

x x x x

xx

e

( ) ( ) ( ) ( )1 1 1 12 1 3 1 5 1 7 1lim

12

→

− + − + − + −

x x x x

x

xe

1ln(2 3 5 7)

2 210e

=

59. Answer (3)

Hint: Truth table.

Sol.:

60. Answer (3)

Hint: Negation of statement.

Sol.: ~( (~ )) ~(~ )→ = q q p q q p

( ~ ) ~q q p q p= =

61. Answer (1)

Hint: Chain rule.

Sol.: f(x) = f(2 – x)

f (x) = –f (2 – x)

f (x) + f (2 – x) = 0

Put x = 3 f (3) + f (–1) = 0

62. Answer (2)

Hint: Condition of line intersection.

Sol.: z lies on line or segment joining z1 & z2

and also on line or segment joining 1z

and z2. Hence only 1 possible solution.

63. Answer (1)

Hint: Locus.

Sol.: |z1| = |z2| = |z1 – z2|

2 21 2 1 2z z z z+ =

1 2

2 1

1z z

z z+ =



9/10

64. Answer (4)

Hint: Rationalise.

Sol.:

2 2

2 21

(2 3 4) ( 6 2)lim

( 1)( 2 3 4 6 2)→

+ + − + +

− + + + + +x

x x x x

x x x x x

1

( 2)( 1) 1lim

( 1)(3 3) 6x

x x

x→

− −= −

− +

65. Answer (4)


Sol.:

9 10 4

6 3 41 3

+= = − = −

+

4(1) 1( 3)

133 131 1

3

z− + −

−= = =

−−

66. Answer (2)

Hint: G.P of infinite series.

Sol.: 1

11 2 1

12 12

=

= = −

rr

67. Answer (3)

Hint: Express equation as factors.

Sol.: P i(2i) = –2

z4 + 5z3 + 18z2 – 4z2 + 5 = (z – z1)

(z – z2) (z – z3) (z – z4)

16 – 40 + 72 – 16 + 5 = (–2 – z1)

(–2 – z2) (–2 – z3) (–2 – z4)

37 = ((PA) (PC))2

68. Answer (3)

Hint: Make truth table.

Sol.:

69. Answer (3)

Hint: Use binomial theorem.

Sol.: 5

01 5

5(1 )5 ... 5 4

+= + + +

CxC C

x x

5

41 2 5

(1 ) 15 5 ... 5

+ = + + + +

C C C

5(1 ) 1+ −

70. Answer (3)

Hint: Use chain rule.

Sol.: y = f(x3)

3 2( ) 3= dy

f x xdx

6 2 13 6 3 9 3 9== + = =xx x

71. Answer (36)

Hint: V = abc

Sol.: a = 3 b = 6 c = 2

V = abc = 3 6 2 = 36

72. Answer (10)

Hint: Locus based

Sol.: |z – 4| = |z – 8| x = 6

and 3|z – 12| = 5|z – 8i|

9(x2 + y2 – 24x + 144) = 25(x2 + y2 – 16y

+ 64)

y = 8, 17

| z |min = 10



10/10

73. Answer (61)

Hint: u

v rule of differentiation

Sol.: 22

( )( 1) (3 1)

xf x

x x= −

+ −

2

2 2

2 (3 2 )( )

( 1) (3 1)

x xf x

x x

− − = −

+ −

2 8 50 72 122

(2)9 25 225 225

f− − −

= − − = =

74. Answer (09)

Hint: 0

form0

Sol.:

23

3

2

3

2 3

1 1tan

3 7

1

1lim

7 81

→−

+ + − = =

+ +

x

xxx x

xL

xx x

22

1

L=

75. Answer (65)

Hint: | x | < a –a < x < a

Sol.: | | 13 2z −

2 | | 13 2z− −

11 | | 15z

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Test - 5 (Code-C) (Answers) All India Aakash Test Series for JEE … · 2020. 2. 3. · All India...

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