test - 4 (code-c) (answers) all india aakash test series ... · test - 4 (code-c) (hints &...

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

1/12

1. (4)

2. (3)

3. (3)

4. (1)

5. (1)

6. (2)

7. (1)

8. (2)

9. (3)

10. (2)

11. (2)

12. (1)

13. (1)

14. (2)

15. (3)

16. (4)

17. (4)

18. (4)

19. (1)

20. (3)

21. (1)

22. (2)

23. (4)

24. (3)

25. (1)

26. (3)

27. (2)

28. (1)

29. (3)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (3)

33. (2)

34. (4)

35. (2)

36. (4)

37. (4)

38. (2)

39. (2)

40. (3)

41. (1)

42. (4)

43. (3)

44. (4)

45. (3)

46. (3)

47. (3)

48. (2)

49. (3)

50. (1)

51. (2)

52. (3)

53. (2)

54. (4)

55. (4)

56. (4)

57. (1)

58. (4)

59. (1)

60. (4)

61. (1)

62. (2)

63. (2)

64. (4)

65. (3)

66. (4)

67. (2)

68. (1)

69. (4)

70. (1)

71. (3)

72. (2)

73. (4)

74. (1)

75. (3)

76. (3)

77. (4)

78. (1)

79. (1)

80. (3)

81. (3)

82. (4)

83. (3)

84. (2)

85. (2)

86. (3)

87. (1)

88. (3)

89. (2)

90. (2)

Test Date : 06/01/2019

ANSWERS

TEST - 4 - Code-C

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

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

2/12

1. Answer (4)

Hint : ˆ ˆ

V VE i j

x y

�

Sol. :1ˆ ˆ( 3 4 )N kgE i j

�

ˆ ˆ6 8 NF mE i j � �

2. Answer (3)

Hint : For the two moving particles to escape each

other's gravitational field, their mechanical

energy in centre of mass frame should

become zero.

Sol. :0

cm4

vv ,

03

4

PC

vv

0

4

QC

vv

COME

2 2 2

0 031 1 3

3 02 4 2 4

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v v G mm m

d

3. Answer (3)

Hint : Excess pressure, P = 4

r

Sol. :2 1

1 2

4 43 3

⇒ r r

r r

2

2 2

1 1

9⎛ ⎞

⎜ ⎟⎝ ⎠

S r

S r

4. Answer (1)

Hint : In order to escape, U + K = 0

Sol. :21 2

03 2 3

GMm GMmv v

R R ⇒

5. Answer (1)

Hint & Sol. : E = 8r2T

16⎛ ⎞ ⎜ ⎟⎝ ⎠

dE drP T r

dt dt

6. Answer (2)

Hint :2

FL FL

YA L r L

Sol. :

2 2

1 1 2

2 2 1

1.51.5

1.5

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

L L R L R

L L R L R

PART - A (PHYSICS)

7. Answer (1)

Hint : Buoyant force = weight of liquid displaced

Sol. :2 32

3

CF gR R R g

35

3

CF gR

8. Answer (2)

Hint & Sol. : TA = 3mg

3 2,a b

a b

a a b b

mg mgY Y

A A

l l

l l

So, 2

3 3

2 2

a a b b

b b a a

A Y a

A Y b c

l l

l l

9. Answer (3)

Hint : g = g – 2Rcos2

Sol. :2

37

16

25 g g R

10. Answer (2)

Hint :2

PR

where P = Pressure on the concave side –

Pressure on the convex side

Sol. : PB = P

C = P,

2A

P PR

11. Answer (2)

Hint : Av1 = A

2v2

and 2 2

1 1 1 2 2 2

1 1

2 2 P gh v P gh v

Sol. : P0 + g 0.15 +

1

212 = P

0 + 0 +

1

2v2

2

v2 = 2 m/s

2 2

1 1

2

10 1

4 4 2

Avd

v

10

5 2 mm2

d

12. Answer (1)

Hint :

2FL Mg L L Mg L

YA L m L m L

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

3/12

Sol. : Y = 3 3

8 10 7800 5 5

0.625 10 7.8 10 4 4

= 10 118 25

10 2 10 Pa0.625 16

13. Answer (1)

Hint :V p

BV

Sol. :V p gh

V B B

8

2

4

9 100.2 10

10

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

B Vh

g V

= 180 m

14. Answer (2)

Hint : TensionL

F YSL

Sol. : F = YS T

Force = 2F = 2YST

15. Answer (3)

Hint : r = r0 is stable position.

Sol. : For r > r0, as r increase, Uincrease

r < r0, as rincrease, U decrease

16. Answer (4)

Hint : L = L1 + L

2

Sol. : L = 2 3 5 FL FL FL

AY AY AY

17. Answer (4)

Hint : T(x) = 2(2 ) (2 )

2 2

M L x L x

L

Sol. :

x

2L

O

T =

2

2 2(4 )4

ML x

L

L =

2 2

2 2[4 ]4

L

L

Mdl L x dx

LAY

∫ ∫

=

2 25

12

L M

AY

18. Answer (4)

Hint : Conserve mechanical energy

Sol. :2 21

2

e

GMm GMmm v

R r

2 2e

GMv

R

19. Answer (1)

Hint : dE = 2

G Rd

R

Sol. :

dE

x

y

dEy = sin

GM d

L R

E = Ey =

0

2sin

∫

GM GMd

L R LR

2

2 2 ⇒

GM GMR L E

L L L

20. Answer (3)

Hint :

2

2

rdA d

Sol. :

d

2 2

2 2

dA r d r

dt dt

22 dA

L Mr Mdt

21. Answer (1)

Hint : There is a pushing force due to pressure

and pulling force due to surface tension

Sol. : 0| | | |

2

ghF P ha ah aT

22. Answer (2)

Hint : Rate of decrease of water level,

2dx a

gxdt A


4/12

PART - B (CHEMISTRY)

31. Answer (1)

Hint : Hybridisation of C in diamond is sp3.

Sol. : Hybridisation of C in graphite is sp2.

Hybridisation of C in C60

is sp2.

32. Answer (3)

Hint : Standard potential follows the order (M/M+) :

Li > Rb > K > Na

Sol. : Hydration enthalpy Charge on ion

Size of ion

Sol. :

2

02

∫ ∫

H

t

H

dx adt

Ax g =

1 2

22

2

H

H

x at

Ag

t = 2

2

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

A HH

a g

23. Answer (4)

Hint : mg = B + Fviscous

when v = vT

Sol. : V1g = V

2g + kv

T

kvT = V[

1 –

2]g

vT =

1 2[ ]

Vg

k

24. Answer (3)

Hint & Sol. : The acceleration of lift will not change

the fraction of volume submerged.

25. Answer (1)

Hint : F = 6rv

Sol. :

2

1 1

1

[MLT ][ ] = [ML T ]

[L LT ]

26. Answer (3)

Hint : 2F sin 2

d = 2S Rd

Sol. : F = 2SR

Stress = 2F SR

A A

27. Answer (2)

Hint :2 2

3[3 ],

2

GMV R r r R

R

Sol. : Alternate

2

2 3

4 2 R

v R R

2

2

2 3

d x GMx x

dt R

3

3 3

2 2 GM GM

v RR R

28. Answer (1)

Hint : Equation of continuity and Bernoulli's equation

Sol. : v1 =

1

1000100 cm/s

10

dV

dt

A

v2 =

1000250 cm/s

4

2 2

1 2

1 11 2.5

2 2

w wP P

3 2 2

1 2

110 [2.5 1 ]

2 P P

29. Answer (3)

Hint :2

hgr

Sol. : h =

5

2 3 3

2 40 101000 mm

10 10 10 2 10

= 4 mm

30. Answer (2)

Hint :34

63

rv r g

Sol. : =

22

9

r g

v

= 1.85 10–5

Ionisation enthalpy decreases down the group

for alkali metals.

Order of density : Rb > Na > K > Li

33. Answer (2)

Hint : In tetrathionate ion, sulphur atoms are

present in +5 and zero oxidation state and all

the O-atoms are present in –2 each.

Sol. : Tetrathionate ion (S4O6

2–)

–

–

6-Oxygen atoms have same state of –2.


5/12

34. Answer (4)

Hint : Balanced reaction

6Fe2+ + Cr2O7

2– + 14H+

6Fe3+ + 2Cr3+ + 7H2O

Sol. : Balanced reaction involving lowest integral

stoichiometric ratio

6Fe2+ + Cr2O7

2– + 14H+

6Fe3+ + 2Cr3+ + 7H2O

a = x, y = 2b and z = 7b

c = 2z

35. Answer (2)

Hint : Hydrogen exist in three isotopic form H, D

and T.

T is radioactive in nature.

Sol. : Deuterium or heavy hydrogen exists mostly

in the form of HD.

36. Answer (4)

Hint : H2O2 is miscible with H

2O in all proportions.

Sol. : H2O2 has open book like structure

(Non-planar) and it is stored in waxlined glass

vessel in dark.

37. Answer (4)

Hint : In a compound, if the oxidation state of

element lies between maximum and

minimum value, then it can act as oxidising

as well as reducing agent.

Sol. : N2O3

+3

N2H4

–2

FeSO3

+3

H4P2O6

+4

H2S2O7

+6

38. Answer (2)

Hint : In case of metal displacement reaction, a

more reactive metal will replace a less

reactive metal from its salt.

Sol. : Reaction

2 2Mg(s) 2HCl(aq) MgCl (aq) H (g)

is an example of non-metal displacement

reaction.

39. Answer (2)

Hint : Law of equivalence,

Equivalents reactants reacted

= Equivalents of products formed

Sol. : Using law of equivalence,

Equivalent of metal X = Equivalent of metal Y

1

Weight of metal X

E =

2

w

E

Weight of metal X = 1

2

Ew g

E

40. Answer (3)

Hint : For complete neutralisation,

Total equivalents of acid

= Total equivalents of base


Total equivalent of base

= Total equivalent of acid

65x = 0.2 × 20 × 1 + 10 × 0.3 × 3

x = 13 1

0.2M65 5

41. Answer (1)

Hint :

5 0

23KIO I

Sol. : n-factor of KIO3 is +5.

2Na2S2O3 + I

2 Na

2S4O6 + 2NaI

1 mole of I2 react with 2 moles of Na

2S2O3.

42. Answer (4)

Hint : Volume strength of H2O2 solution

= Normality × 5.6

Sol. : Normality of diluted solution = 10 0.2

20

= 1

N10

Normality of original solution

= 1

1010

= 1 N

Volume strength = Normality × 5.6

= 5.6 V


6/12

43. Answer (3)

Hint : In the molecule CuSO45H

2O, only one H

2O

molecule is hydrogen bonded.

Sol. : (1)2

H O H OH ��⇀

↽��

(2)2 2 6 12 6 2

6CO 6H O C H O 6O

44. Answer (4)

Hint : Ionic hydride – Formed by s-block elements

Molecular hydride – Formed by p-block

elements

Metallic hydride – Formed by d and f-block

metals

Sol. : Metallic hydrides are non-stoichiometric and

they conduct heat and electricity.

45. Answer (3)

Hint : CaCl2 and Ca(OCl)

2 are product containing

Cl-atom.

Sol. : 2Ca(OH)2 + 2Cl

2

CaCl2 + Ca(OCl)

2 + 2H

2O

Oxidation state of Cl in product = –1 and +1

46. Answer (3)

Hint : meq of reductant = meq of oxidant

Sol. : 33.33 × M × (x – 3) = 20 × M × (x – 1)

x = 6

47. Answer (3)

Hint : Ca(HCO3)2 when reacts with Ca(OH)

2 forms

ppt. of CaCO3.

Sol. : 103 g of water contain 324 mg Ca(HCO3)2.

106 g 324 g Ca(HCO3)2

Moles of Ca(HCO3)2 = 2

Ca(HCO3)2 + Ca(OH)

2 2CaCO

3 + 2H

2O

Moles of Ca(OH)2 required = 2

Mass of Ca(OH)2 = 148 g in 106 g water

Mass of Ca(OH)2 in 1000 g water = 0.148 g

48. Answer (2)

Hint : Fact based.

Sol. : Except beryllium all other group-2 elements

are known as alkaline earth metals.

49. Answer (3)

Hint : 4ClO3

– Cl– + 3ClO4

–

Sol. :3 4

1 75

4ClO Cl 3ClO

50. Answer (1)

Hint : Dead burnt plaster is CaSO4.

Sol. : Statement-IV is incorrect.

4 2 4 2 4Gypsum POP Dead burnt

plaster

1CaSO 2H O CaSO H O CaSO

2

Four statements are correct.

51. Answer (2)

Hint : Correct order – 1, 3, 4

Incorrect order – 2

Sol. : Solubility of carbonates of group-1 increases

down the group

Li2CO

3 < Na

2CO

3 < K

2CO

3 < Rb

2CO

3

< Cs2CO

3

52. Answer (3)

Hint : Diborane on combustion forms boric oxide.

Sol. : B2H6 + 3O

2 B

2O3 + 3H

2O

53. Answer (2)

Hint : In borax, 2 B-atoms are connected to

4 O-atoms and 2 B-atoms are connected to

3 O-atoms.

Sol. :

54. Answer (4)

Hint :

(three Si—O—Si linkage)

2-units of (CH3)3SiCl and 2-units of

(CH3)2SiCl

2.


7/12

Sol. :


2-units of (CH3)3SiCl and 2-units of

(CH3)2SiCl.

55. Answer (4)

Hint : Fact based.

Sol. : At atmospheric pressure ice crystallises in

the hexagonal form, but at very low

temperatures it condenses to cubic form.

56. Answer (4)

Hint : Diborane has electron deficient bridge bond.

Sol. : B

H

H H

H

B

H

H

2-bridge bond (3c – 2e– bond)

All the B–H bonds are not identical.

Maximum 6-atoms are planar.

57. Answer (1)

Hint : Inert pair effect is observed as we move down

the group 14.

Sol. : GeCl4 is more stable than GeCl

2 and PbCl

2

is more stable than PbCl4.

58. Answer (4)

Hint : Reaction of boron trihalide with H2 in

presence of Ta forms boron.

Sol. :Tantalu

3 2

m2BX 3H 2B 6HX

59. Answer (1)

Hint : In the presence of phenolphthalein indicator,

NaOH will react with HCl and form NaCl and

Na2CO

3 will react with HCl and convert into

NaHCO3.

If methyl orange is used as indicator, then all

the NaHCO3 will convert into NaCl and H

2CO

3.

Sol. : In phenolphthalein

milliequi. of HCl = milliequi. of NaOH

+ 1

2 milliequi. of Na

2CO

3

7 × 0.25 = x + 1

2y

x + 1

2y = 1.75 ...(i)

In methyl orange,

1 × 0.25 = 1

2 milliequi. of Na

2CO

3

0.25 = 1

2y ...(ii)

x = 1.5, y = 0.5

mmole of Na+ = 2

V = 10 mL

N = 0.2

60. Answer (4)

Hint : Producer gas is CO + N2

Sol. : 2C(s) + O2 + 4N2

air

1273 K2CO(g) + 4N (g)2

Producer gas

PART - C (MATHEMATICS)

61. Answer (1)

Hint : Ratio of circumference.

Sol. :

BAx y + – 1 = 0

C(0, 0)

1

1

2

x y2 2 + = 1

Perpendicular (d) = | 1| 1

2 2

1sin 45

2 ⇒

= 90°

Circumference AB subtends right angle at

centre.

Line x + y – 1 = 0 divides circumference

in the ratio 1 : 3.

62. Answer (2)

Hint : Find common normal.


8/12

Sol. : y2 = 4x, point P(t2, 2t),

normal y = –tx + 2t + t3 ...(i)

For minimum distance (i) passes through the

centre (0, 3) of the circle.

3 = 2t + t3

(t – 1)(t2 + t + 3) = 0

t = 1

Point is P(1, 2)

Minimum distance = CP – r = 2 1

63. Answer (2)

Hint : Line with equal intercept has slope ±1.

Sol. :

O

B

A

Equations of line AB is x + y = a.

∵

0 02

2

a

2 2 a

Equation of line is x + y = 2 2

But (2 2, 0) and (0, 2 2)A B

Area of OAB = 1

2 2 2 2 42

64. Answer (4)

Hint : Eliminate t from given equations.

Sol. :1 1,x t y t

t t

x + y = 2t and x – y = 2

t

Eliminating t,

x2 – y2 = –4 (Rectangular hyperbola)

2e

65. Answer (3)

Hint : a, b and c are in G.P. and positive real

numbers b2 = ac.

Sol. :2 ⇒ b ac ac b

ax + by + b = 0

ax + b(1 + y) = 0

(1 ) 0 bx y

a

Fixed point is F(0, –1)

66. Answer (4)

Hint : From director circle perpendicular tangents

are drawn.

Sol. : Ellipse,

2 2

11 1

3 2

x y

Its director circle is 2 2 1 1

3 2x y .

6x2 + 6y2 = 5

∵

1 1,

2 3

⎛ ⎞⎜ ⎟⎝ ⎠

lies on the director circle.

Angle between the pair of tangents is 2

.

67. Answer (2)

Hint : Focus = (1, 0)

Sol. : Equation of line y – 0 = 1(x – 1).

x – y = 1

68. Answer (1)

Hint : 2 2 and 2 3 2ae a

Sol. :5 5,

2 2C

⎛ ⎞ ⎜ ⎟⎝ ⎠

SS = 2ae

12 2

3ae e ⇒

2 2 2 9 1(1 ) 1 4

2 9b a e

⎛ ⎞ ⎜ ⎟⎝ ⎠

b = 2

Length of latus rectum

=

22 2 4 8 2

3 3

2

b

a


9/12

69. Answer (4)

Hint : Given pair of lines forms a parallelogram.

Sol. : x2 – xy – 6y2 = 0

x + 2y = 0 and x – 3y = 0

x2 – xy – 6y2 + 5y – 1 = 0

x + 2y – 1 = 0 and x – 3y + 1 = 0

Formed quadrilateral is a parallelogram.

Area of parallelogram is P1P2 cosec

1 2

1 1, ,

5 10P P

1 1

52 3tan 1

1 1 51

2 3

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 45°, cosec = 2

1 1 12 sq. units

55 10A

70. Answer (1)

Hint : Ends of latus rectum are (±x1, ±x

1)

Sol. :

P x x( , )1 1

y

x45°

O(0, 0)

P(x1, x

1) lies on the hyperbola.

2 2

1 1

2 21

2

x x

aba b

2

1 2 2

1 11

2x

aba b

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2

2

1 2

2 ( )

( )

ab a bx

a b

x1

2 cannot be negative.

There are no such double ordinates.

71. Answer (3)

Hint : Centre of circle (2, –3) lies on given line.

Sol. : ∵ Line is normal to the circle.

Line passes through the centre C(2, –3)

of the circle.

(m2 – 1)2 – 3(m2 + 1) – m(m – 7) = 0

2m2 – 2 – 3m2 – 3 – m2 + 7m = 0

2m2 – 7m + 5 = 0

2m2 – 2m – 5m + 5 = 0

2m(m – 1) – 5(m – 1) = 0

5

1,2

m

1 2

7

2m m

72. Answer (2)

Hint : Let equation of tangent in parametric form.

Sol. : Equation of tangent to

2 2

2 21

x y

a b is

cos sin 1x y

a b

, 0 , 0,cos sin

a bA B⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Let mid-point is M(h, k).

,2 cos 2 sin

a bh k

cos2 + sin2 = 1

2 2

12 2

a b

h k

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Locus of M(h, k) is

2 2

2 24

a b

x y .

73. Answer (4)

Hint : Slope of directrix is 1

5

.

Sol. :

S(0, 2)

V(–1, –3)T


10/12

V(–1, –3), S(0, 2)

V is the mid-point of TS.

T(–2, –8)

Slope of directrix is 1 1

5VS

m

m

.

Equation of directrix is x + 5y + 42 = 0.

74. Answer (1)

Hint : Asymptotes are perpendicular.

Sol. : Rectangular hyperbola has perpendicular

asymptotes.

Hence, eccentricity is 2 .

75. Answer (3)

Hint : At focus right angle formed.

Sol. : S(1, 0)

Triangle PTS is a right-angled triangle at

point S(1, 0).

Orthocentre of triangle PTS is S(1, 0) a fixed point on latus rectum.

76. Answer (3)

Hint : Equation of tangent is 2 2 2y mx a m b .

Sol. : Equations of tangents to

2 2

2 21

x y

a b and

2 2

2 21

x y

b a are

2 2 2

1 1y m x a m b and

2 2 2

2 2y m x b m a respectively for

common tangent m1 = m

2 and

2 2 2 2 2 2

1 2a m b b m a

m1 = m

2 = ±1

Common tangents are

2 21y x a b y x ⇒

Area of quadrilateral is 2.

77. Answer (4)

Hint : PA and PB are parallel to coordinate axes.

Sol. : y

x

y x =

P(2, 2)

P(1, 1)B(2, 1)

A(1, 2)

Coordinates of point P are (1, 1) or (2, 2).

78. Answer (1)

Hint : Locus be concentric circle with radius half of

the given circle.

Sol. :

C(2, 3)

4 ( – 2) + ( – 3) = 16x y2 2

Equation of locus of centre of a circle

which passes through C(2, 3) and touches

the circle (x – 2)2 + (y – 3)2 = 16 internally

is (x – 2)2 + (y – 3)2 = 4.

79. Answer (1)

Hint : Given two sides are perpendicular.

Sol. :

C(0, 0)

B(3, –6)

A(–2, –1)x y – 2 = 0

2 + = 0x y

x y + + 3 = 0

Triangle formed by the lines is right-angled

triangle.

AB will be the diameter of circle passing

through the intersection points of the lines.

Equation of the required circle is

x2 + y2 – x + 7y = 0.

80. Answer (3)

Hint : Draw the graph of ellipse and circle.

Sol. :

Intersection points of circle and ellipse are

27 5,

8 8

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠.

81. Answer (3)

Hint : Direct and transverse tangent intersect at

point joining centre of circles.


11/12

Sol. : C1 (0, 0), r

1 = 1, C

2 (5, 0), r

2 = 2

Points D and T divides C1C2 in the ratio

1 : 2 externally and internally respectively.

5 20

( 5, 0), , 03 3

D T DT⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

82. Answer (4)

Hint : Use concept of orthogonal curves m1m2 = –1.

Sol. : xy = a2

d.w.r.t. x,

0 ⇒ y

xy y yx

...(i)

d.w.r.t. x,

2 2 0 ⇒ xx yy y

y...(ii)

From (i) and (ii),

m1m2 = –1

Hyperbolas are orthogonal to each other

for all real values of a and b.

Infinitely many ordered pairs (a, b).

83. Answer (3)

Hint : Write equation of normal in parametric form.

Sol. : Normal to the parabola y2 = 4x is

y = –tx + 2t + t3 ...(i)

Equation (i) is also tangent to

x2 = 4(y + 2) ...(ii)

Equations (i) and (ii) have exactly one

intersection point

x2 – 4(–tx + 2t + t3) – 8 = 0

x2 + 4tx – 8t – 4t3 – 8 = 0

D = 0 16t2 = –32t – 16t3 – 32

t3 + t2 + 2t + 2 = 0

t = –1

The equation of required line is y = x – 3.

84. Answer (2)

Hint : Let the point be (2 sec, 3 tan), where is

eccentric angle.

Sol. : Equation of tangent to the hyperbola is

sec tan1

2 3

x y...(i)

Comparing equation (i) with 3x – y = 3 3

6

85. Answer (2)

Hint : For two lines a1x + b

1y + c

1 = 0 and

a2x + b

2y + c

2 = 0, if

1 1

2 2

a b

a b

, then only

one point of intersection.

Sol. : If 1

1

1

1

, infinitely many solutions.

L1 = 0 and L

2 = 0 have same intercept on

y-axis and different slopes.

Only one intersection point.

86. Answer (3)

Hint : Distance of line ax + by + c = 0 from point

(x1, y

1) is

1 1

2 2

x by c

a b

.

Sol. : Distance of each line from (0, 0) is 2 units.

87. Answer (1)

Hint :2 2

g c f c

Sol. :2 2

2 2g c f c

2

2 2 20

4g f

⇒ ⇒

When = 0, equation becomes x2 + y2 + 1

= 0 (which is not an equation of circle).

88. Answer (3)

Hint : Write parametric equation of normal.

Sol. : xy = c2 ...(i)

xy + y = 0

yy

x

Let point ,

cP ct

t

⎛ ⎞⎜ ⎟⎝ ⎠

equation of normal to (i)

at P.

at

1( )

P

cy x ct

t y

⎛ ⎞ ⎜ ⎟⎝ ⎠

( )c ct

y x ctct

t


12/12

� � �

2( )c

y t x ctt

3 2cy ct t x

t ...(ii)

If equation (ii) is tangent to

xy = –c2 ...(iii)

then equations (ii) and (iii) will have exactly

one intersection point.

3 2 2cx ct t x c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2 3 20

ct x ct x c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

D = 0

2

3 2 24

cct t c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

3 214t t

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

(1 – t4)2 = 4t4

Put t4 =

4 = (1 – )2 2 + 1 – 2 = 4

2 – 6 + 1 = 0

6 36 43 2 2

2

43 2 2t

There are exactly 4 real values of t.

89. Answer (2)

Hint : Combine equation of lines L1 = 0 and L

2 = 0

is L1 L

2 = 0.

Sol. : (1) L1L2 = 0 L

1 = 0 or L

2 = 0

(2) L1

2 + L2

2 = 0 L1 = 0 and L

2 = 0

Represents intersection point of

L1 = 0 and L

2 = 0

(3) L1

2 = L2

2 L1 = L

2or L

1 + L

2 = 0

(4) (L1 + L

2)2 = (L

1 – L

2)2

4L1L2 = 0 L

1 = 0 or L

2 = 0

90. Answer (2)

Hint : Use equation of chord of given mid-point.

Sol. : Let centre of the circle be C(x1, y

1).

Equation of chord whose mid-point is

C(x1, y

1) is T = S

1

yy1 – 2(x + x

1) = y

1

2 – 4x1

2x – yy1 – 2x

1 + y

1

2 = 0 ...(i)

x – y – 1 = 0 ...(ii)

Equations (i) & (ii) represent equations of

same chord.

2

1 1 122

1 1 1

y x y

y1 = 2 and x

1 = 3 Centre C(3, 2).

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

1/12

1. (2)

2. (3)

3. (1)

4. (2)

5. (3)

6. (1)

7. (3)

8. (4)

9. (2)

10. (1)

11. (3)

12. (1)

13. (4)

14. (4)

15. (4)

16. (3)

17. (2)

18. (1)

19. (1)

20. (2)

21. (2)

22. (3)

23. (2)

24. (1)

25. (2)

26. (1)

27. (1)

28. (3)

29. (3)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (1)

33. (4)

34. (1)

35. (4)

36. (4)

37. (4)

38. (2)

39. (3)

40. (2)

41. (1)

42. (3)

43. (2)

44. (3)

45. (3)

46. (3)

47. (4)

48. (3)

49. (4)

50. (1)

51. (3)

52. (2)

53. (2)

54. (4)

55. (4)

56. (2)

57. (4)

58. (2)

59. (3)

60. (1)

61. (2)

62. (2)

63. (3)

64. (1)

65. (3)

66. (2)

67. (2)

68. (3)

69. (4)

70. (3)

71. (3)

72. (1)

73. (1)

74. (4)

75. (3)

76. (3)

77. (1)

78. (4)

79. (2)

80. (3)

81. (1)

82. (4)

83. (1)

84. (2)

85. (4)

86. (3)

87. (4)

88. (2)

89. (2)

90. (1)

Test Date : 06/01/2019

ANSWERS

TEST - 4 - Code-D

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

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

2/12

1. Answer (2)

Hint :34

63

rv r g

Sol. : =

22

9

r g

v

= 1.85 10–5

2. Answer (3)

Hint :2

hgr

Sol. : h =

5

2 3 3

2 40 101000 mm

10 10 10 2 10

= 4 mm

3. Answer (1)

Hint : Equation of continuity and Bernoulli's equation

Sol. : v1

=

1

1000100 cm/s

10

dV

dt

A

v2

= 1000

250 cm/s4

2 2

1 2

1 11 2.5

2 2

w wP P

3 2 2

1 2

110 [2.5 1 ]

2 P P

4. Answer (2)

Hint :2 2

3[3 ],

2

GMV R r r R

R

Sol. : Alternate

2

2 3

4 2 R

v R R

2

2

2 3

d x GMx x

dt R

3

3 3

2 2 GM GM

v RR R

5. Answer (3)

Hint : 2F sin 2

d = 2S Rd

Sol. : F = 2SR

Stress = 2F SR

A A

PART - A (PHYSICS)

6. Answer (1)

Hint : F = 6rv

Sol. :

2

1 1

1

[MLT ][ ] = [ML T ]

[L LT ]

7. Answer (3)

Hint & Sol. : The acceleration of lift will not change

the fraction of volume submerged.

8. Answer (4)

Hint : mg = B + Fviscous

when v = vT

Sol. : V1

g = V2

g + kvT

kvT = V[

1

– 2

]g

vT =

1 2[ ]

Vg

k

9. Answer (2)

Hint : Rate of decrease of water level,

2dx a

gxdt A

Sol. :

2

02

∫ ∫

H

t

H

dx adt

Ax g =

1 2

22

2

H

H

x at

Ag

t = 2

2

⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦

A HH

a g

10. Answer (1)

Hint : There is a pushing force due to pressure

and pulling force due to surface tension

Sol. : 0| | | |

2

ghF P ha ah aT

11. Answer (3)

Hint :

2

2

rdA d

Sol. :

d

2 2

2 2

dA r d r

dt dt

22 dA

L Mr Mdt

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

3/12

12. Answer (1)

Hint : dE = 2

G Rd

R

Sol. :

dE

x

y

dEy = sin

GM d

L R

E = Ey =

0

2sin

∫

GM GMd

L R LR

2

2 2 ⇒

GM GMR L E

L L L

13. Answer (4)

Hint : Conserve mechanical energy

Sol. :2 21

2

e

GMm GMmm v

R r

2 2e

GMv

R

14. Answer (4)

Hint : T(x) = 2(2 ) (2 )

2 2

M L x L x

L

Sol. :

x

2L

O

T =

2

2 2(4 )4

ML x

L

L =

2 2

2 2[4 ]4

L

L

Mdl L x dx

LAY

∫ ∫

=

2 25

12

L M

AY

15. Answer (4)

Hint : L = L1

+ L2

Sol. : L = 2 3 5 FL FL FL

AY AY AY

16. Answer (3)

Hint : r = r0

is stable position.

Sol. : For r > r0

, as r increase, Uincrease

r < r0

, as rincrease, U decrease

17. Answer (2)

Hint : TensionL

F YSL

Sol. : F = YS T

Force = 2F = 2YST

18. Answer (1)

Hint :V p

BV

Sol. :V p gh

V B B

8

2

4

9 100.2 10

10

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

B Vh

g V

= 180 m

19. Answer (1)

Hint :

2FL Mg L L Mg L

YA L m L m L

Sol. : Y = 3 3

8 10 7800 5 5

0.625 10 7.8 10 4 4

= 10 118 25

10 2 10 Pa0.625 16

20. Answer (2)

Hint : Av1

= A2

v2

and 2 2

1 1 1 2 2 2

1 1

2 2 P gh v P gh v

Sol. : P0

+ g 0.15 + 1

212 = P

0

+ 0 + 1

2v2

2

v2

= 2 m/s

2 2

1 1

2

10 1

4 4 2

Avd

v

10

5 2 mm2

d

21. Answer (2)

Hint :2

PR

where P = Pressure on the concave side –

Pressure on the convex side

Sol. : PB = P

C = P,

2A

P PR


4/12

PART - B (CHEMISTRY)

31. Answer (4)

Hint : Producer gas is CO + N2

Sol. : 2C(s) + O2 + 4N2

air

1273 K2CO(g) + 4N (g)2

Producer gas

32. Answer (1)

Hint : In the presence of phenolphthalein indicator,

NaOH will react with HCl and form NaCl and

Na2

CO3

will react with HCl and convert into

NaHCO3

.

If methyl orange is used as indicator, then all

the NaHCO3

will convert into NaCl and H2

CO3

.

Sol. : In phenolphthalein

milliequi. of HCl = milliequi. of NaOH

+ 1

2 milliequi. of Na

2

CO3

7 × 0.25 = x + 1

2y

x + 1

2y = 1.75 ...(i)

In methyl orange,

1 × 0.25 = 1

2 milliequi. of Na

2

CO3

22. Answer (3)

Hint : g = g – 2Rcos2

Sol. :2

37

16

25 g g R

23. Answer (2)

Hint & Sol. : TA = 3mg

3 2,a b

a b

a a b b

mg mgY Y

A A

l l

l l

So, 2

3 3

2 2

a a b b

b b a a

A Y a

A Y b c

l l

l l

24. Answer (1)

Hint : Buoyant force = weight of liquid displaced

Sol. :2 32

3

CF gR R R g

35

3

CF gR

25. Answer (2)

Hint :2

FL FL

YA L r L

Sol. :

2 2

1 1 2

2 2 1

1.51.5

1.5

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

L L R L R

L L R L R

26. Answer (1)

Hint & Sol. : E = 8r2T

16⎛ ⎞ ⎜ ⎟⎝ ⎠

dE drP T r

dt dt

27. Answer (1)

Hint : In order to escape, U + K = 0

Sol. :21 2

03 2 3

GMm GMmv v

R R ⇒

28. Answer (3)

Hint : Excess pressure, P = 4

r

Sol. :2 1

1 2

4 43 3

⇒ r r

r r

2

2 2

1 1

9⎛ ⎞

⎜ ⎟⎝ ⎠

S r

S r

29. Answer (3)

Hint : For the two moving particles to escape each

other's gravitational field, their mechanical

energy in centre of mass frame should

become zero.

Sol. :0

cm4

vv ,

03

4

PC

vv

0

4

QC

vv

COME

2 2 2

0 031 1 3

3 02 4 2 4

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v v G mm m

d

30. Answer (4)

Hint : ˆ ˆ

V VE i j

x y

�

Sol. :1ˆ ˆ( 3 4 )N kgE i j

�

ˆ ˆ6 8 NF mE i j � �


5/12

0.25 = 1

2y ...(ii)

x = 1.5, y = 0.5

mmole of Na+ = 2

V = 10 mL

N = 0.2

33. Answer (4)

Hint : Reaction of boron trihalide with H2

in

presence of Ta forms boron.

Sol. :Tantalu

3 2

m2BX 3H 2B 6HX

34. Answer (1)

Hint : Inert pair effect is observed as we move down

the group 14.

Sol. : GeCl4

is more stable than GeCl2

and PbCl2

is more stable than PbCl4

.

35. Answer (4)

Hint : Diborane has electron deficient bridge bond.

Sol. : B

H

H H

H

B

H

H

2-bridge bond (3c – 2e– bond)

All the B–H bonds are not identical.

Maximum 6-atoms are planar.

36. Answer (4)

Hint : Fact based.

Sol. : At atmospheric pressure ice crystallises in

the hexagonal form, but at very low

temperatures it condenses to cubic form.

37. Answer (4)

Hint :


2-units of (CH3

)3

SiCl and 2-units of

(CH3

)2

SiCl2

.

Sol. :


2-units of (CH3

)3

SiCl and 2-units of

(CH3

)2

SiCl.

38. Answer (2)

Hint : In borax, 2 B-atoms are connected to

4 O-atoms and 2 B-atoms are connected to

3 O-atoms.

Sol. :

39. Answer (3)

Hint : Diborane on combustion forms boric oxide.

Sol. : B2

H6

+ 3O2

B2

O3

+ 3H2

O

40. Answer (2)

Hint : Correct order – 1, 3, 4

Incorrect order – 2

Sol. : Solubility of carbonates of group-1 increases

down the group

Li2

CO3

< Na2

CO3

< K2

CO3

< Rb2

CO3

< Cs2

CO3

41. Answer (1)

Hint : Dead burnt plaster is CaSO4

.

Sol. : Statement-IV is incorrect.

4 2 4 2 4Gypsum POP Dead burnt

plaster

1CaSO 2H O CaSO H O CaSO

2

Four statements are correct.

42. Answer (3)

Hint : 4ClO3

– Cl– + 3ClO4

–

Sol. :3 4

1 75

4ClO Cl 3ClO

43. Answer (2)

Hint : Fact based.

Sol. : Except beryllium all other group-2 elements

are known as alkaline earth metals.

44. Answer (3)

Hint : Ca(HCO3

)2

when reacts with Ca(OH)2

forms

ppt. of CaCO3

.


6/12

Sol. : 103 g of water contain 324 mg Ca(HCO3

)2

.

106 g 324 g Ca(HCO3

)2

Moles of Ca(HCO3

)2

= 2

Ca(HCO3

)2

+ Ca(OH)2

2CaCO3

+ 2H2

O

Moles of Ca(OH)2

required = 2

Mass of Ca(OH)2

= 148 g in 106 g water

Mass of Ca(OH)2

in 1000 g water = 0.148 g

45. Answer (3)

Hint : meq of reductant = meq of oxidant

Sol. : 33.33 × M × (x – 3) = 20 × M × (x – 1)

x = 6

46. Answer (3)

Hint : CaCl2

and Ca(OCl)2

are product containing

Cl-atom.

Sol. : 2Ca(OH)2

+ 2Cl2

CaCl2

+ Ca(OCl)2

+ 2H2

O

Oxidation state of Cl in product = –1 and +1

47. Answer (4)

Hint : Ionic hydride – Formed by s-block elements

Molecular hydride – Formed by p-block

elements

Metallic hydride – Formed by d and f-block

metals

Sol. : Metallic hydrides are non-stoichiometric and

they conduct heat and electricity.

48. Answer (3)

Hint : In the molecule CuSO4

5H2

O, only one H2

O

molecule is hydrogen bonded.

Sol. : (1)2

H O H OH ��⇀

↽��

(2)2 2 6 12 6 2

6CO 6H O C H O 6O

49. Answer (4)

Hint : Volume strength of H2

O2

solution

= Normality × 5.6

Sol. : Normality of diluted solution = 10 0.2

20

= 1

N10

Normality of original solution

= 1

1010

= 1 N

Volume strength = Normality × 5.6

= 5.6 V

50. Answer (1)

Hint :

5 0

23KIO I

Sol. : n-factor of KIO3

is +5.

2Na2

S2

O3

+ I2

Na2

S4

O6

+ 2NaI

1 mole of I2

react with 2 moles of Na2

S2

O3

.

51. Answer (3)

Hint : For complete neutralisation,

Total equivalents of acid

= Total equivalents of base


Total equivalent of base

= Total equivalent of acid

65x = 0.2 × 20 × 1 + 10 × 0.3 × 3

x = 13 1

0.2M65 5

52. Answer (2)

Hint : Law of equivalence,

Equivalents reactants reacted

= Equivalents of products formed


Equivalent of metal X = Equivalent of metal Y

1

Weight of metal X

E =

2

w

E

Weight of metal X = 1

2

Ew g

E

53. Answer (2)

Hint : In case of metal displacement reaction, a

more reactive metal will replace a less

reactive metal from its salt.

Sol. : Reaction

2 2Mg(s) 2HCl(aq) MgCl (aq) H (g)

is an example of non-metal displacement

reaction.


7/12

54. Answer (4)

Hint : In a compound, if the oxidation state of

element lies between maximum and

minimum value, then it can act as oxidising

as well as reducing agent.

Sol. : N2

O3

+3

N2

H4

–2

FeSO3

+3

H4

P2

O6

+4

H2

S2

O7

+6

55. Answer (4)

Hint : H2

O2

is miscible with H2

O in all proportions.

Sol. : H2

O2

has open book like structure

(Non-planar) and it is stored in waxlined glass

vessel in dark.

56. Answer (2)

Hint : Hydrogen exist in three isotopic form H, D

and T.

T is radioactive in nature.

Sol. : Deuterium or heavy hydrogen exists mostly

in the form of HD.

57. Answer (4)

Hint : Balanced reaction

6Fe2+ + Cr2

O7

2– + 14H+

6Fe3+ + 2Cr3+ + 7H2

O

PART - C (MATHEMATICS)

61. Answer (2)

Hint : Use equation of chord of given mid-point.

Sol. : Let centre of the circle be C(x1

, y1

).

Equation of chord whose mid-point is

C(x1

, y1

) is T = S1

yy1

– 2(x + x1

) = y1

2 – 4x1

2x – yy1

– 2x1

+ y1

2 = 0 ...(i)

x – y – 1 = 0 ...(ii)

Equations (i) & (ii) represent equations of

same chord.

2

1 1 122

1 1 1

y x y

y1

= 2 and x1

= 3 Centre C(3, 2).

62. Answer (2)

Hint : Combine equation of lines L1

= 0 and L2

= 0

is L1

L2

= 0.

Sol. : Balanced reaction involving lowest integral

stoichiometric ratio

6Fe2+ + Cr2

O7

2– + 14H+

6Fe3+ + 2Cr3+ + 7H2

O

a = x, y = 2b and z = 7b

c = 2z

58. Answer (2)

Hint : In tetrathionate ion, sulphur atoms are

present in +5 and zero oxidation state and all

the O-atoms are present in –2 each.

Sol. : Tetrathionate ion (S4

O6

2–)

–

–

6-Oxygen atoms have same state of –2.

59. Answer (3)

Hint : Standard potential follows the order (M/M+) :

Li > Rb > K > Na

Sol. : Hydration enthalpy Charge on ion

Size of ion

Ionisation enthalpy decreases down the group

for alkali metals.

Order of density : Rb > Na > K > Li

60. Answer (1)

Hint : Hybridisation of C in diamond is sp3.

Sol. : Hybridisation of C in graphite is sp2.

Hybridisation of C in C60

is sp2.


8/12

Sol. : (1) L1

L2

= 0 L1

= 0 or L2

= 0

(2) L1

2 + L2

2 = 0 L1

= 0 and L2

= 0

Represents intersection point of

L1

= 0 and L2

= 0

(3) L1

2 = L2

2 L1

= L2

or L1

+ L2

= 0

(4) (L1

+ L2

)2 = (L1

– L2

)2

4L1

L2

= 0 L1

= 0 or L2

= 0

63. Answer (3)

Hint : Write parametric equation of normal.

Sol. : xy = c2 ...(i)

xy + y = 0

yy

x

Let point ,

cP ct

t

⎛ ⎞⎜ ⎟⎝ ⎠

equation of normal to (i)

at P.

at

1( )

P

cy x ct

t y

⎛ ⎞ ⎜ ⎟⎝ ⎠

( )c ct

y x ctct

t

2( )c

y t x ctt

3 2cy ct t x

t ...(ii)

If equation (ii) is tangent to

xy = –c2 ...(iii)

then equations (ii) and (iii) will have exactly

one intersection point.

3 2 2cx ct t x c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2 3 20

ct x ct x c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

D = 0

2

3 2 24

cct t c

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

3 214t t

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

(1 – t4)2 = 4t4

Put t4 =

4 = (1 – )2 2 + 1 – 2 = 4

2 – 6 + 1 = 0

6 36 43 2 2

2

43 2 2t

There are exactly 4 real values of t.

64. Answer (1)

Hint :2 2

g c f c

Sol. :2 2

2 2g c f c

2

2 2 20

4g f

⇒ ⇒

When = 0, equation becomes x2 + y2 + 1

= 0 (which is not an equation of circle).

65. Answer (3)

Hint : Distance of line ax + by + c = 0 from point

(x1

, y1

) is 1 1

2 2

x by c

a b

.

Sol. : Distance of each line from (0, 0) is 2 units.

66. Answer (2)

Hint : For two lines a1

x + b1

y + c1

= 0 and

a2

x + b2

y + c2

= 0, if 1 1

2 2

a b

a b

, then only

one point of intersection.

Sol. : If 1

1

1

1

, infinitely many solutions.

L1

= 0 and L2

= 0 have same intercept on

y-axis and different slopes.

Only one intersection point.

67. Answer (2)

Hint : Let the point be (2 sec, 3 tan), where is

eccentric angle.

Sol. : Equation of tangent to the hyperbola is

sec tan1

2 3

x y...(i)

Comparing equation (i) with 3x – y = 3 3

6


9/12

68. Answer (3)

Hint : Write equation of normal in parametric form.

Sol. : Normal to the parabola y2 = 4x is

y = –tx + 2t + t3 ...(i)

Equation (i) is also tangent to

x2 = 4(y + 2) ...(ii)

Equations (i) and (ii) have exactly one

intersection point

x2 – 4(–tx + 2t + t3) – 8 = 0

x2 + 4tx – 8t – 4t3 – 8 = 0

D = 0 16t2 = –32t – 16t3 – 32

t3 + t2 + 2t + 2 = 0

t = –1

The equation of required line is y = x – 3.

69. Answer (4)

Hint : Use concept of orthogonal curves m1

m2

= –1.

Sol. : xy = a2

d.w.r.t. x,

0 ⇒ y

xy y yx

...(i)

d.w.r.t. x,

2 2 0 ⇒ xx yy y

y...(ii)

From (i) and (ii),

m1

m2

= –1

Hyperbolas are orthogonal to each other

for all real values of a and b.

Infinitely many ordered pairs (a, b).

70. Answer (3)

Hint : Direct and transverse tangent intersect at

point joining centre of circles.

Sol. : C1

(0, 0), r1

= 1, C2

(5, 0), r2

= 2

Points D and T divides C1

C2

in the ratio

1 : 2 externally and internally respectively.

5 20

( 5, 0), , 03 3

D T DT⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

71. Answer (3)

Hint : Draw the graph of ellipse and circle.

Sol. :

Intersection points of circle and ellipse are

27 5,

8 8

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠.

72. Answer (1)

Hint : Given two sides are perpendicular.

Sol. :

C(0, 0)

B(3, –6)

A(–2, –1)x y – 2 = 0

2 + = 0x y

x y + + 3 = 0

Triangle formed by the lines is right-angled

triangle.

AB will be the diameter of circle passing

through the intersection points of the lines.

Equation of the required circle is

x2 + y2 – x + 7y = 0.

73. Answer (1)

Hint : Locus be concentric circle with radius half of

the given circle.

Sol. :

C(2, 3)

4 ( – 2) + ( – 3) = 16x y2 2

Equation of locus of centre of a circle

which passes through C(2, 3) and touches

the circle (x – 2)2 + (y – 3)2 = 16 internally

is (x – 2)2 + (y – 3)2 = 4.


10/12

74. Answer (4)

Hint : PA and PB are parallel to coordinate axes.

Sol. : y

x

y x =

P(2, 2)

P(1, 1)B(2, 1)

A(1, 2)

Coordinates of point P are (1, 1) or (2, 2).

75. Answer (3)

Hint : Equation of tangent is 2 2 2y mx a m b .

Sol. : Equations of tangents to

2 2

2 21

x y

a b and

2 2

2 21

x y

b a are

2 2 2

1 1y m x a m b and

2 2 2

2 2y m x b m a respectively for

common tangent m1

= m2

and

2 2 2 2 2 2

1 2a m b b m a

m1

= m2

= ±1

Common tangents are

2 21y x a b y x ⇒

Area of quadrilateral is 2.

76. Answer (3)

Hint : At focus right angle formed.

Sol. : S(1, 0)

Triangle PTS is a right-angled triangle at

point S(1, 0).

Orthocentre of triangle PTS is S(1, 0) a fixed point on latus rectum.

77. Answer (1)

Hint : Asymptotes are perpendicular.

Sol. : Rectangular hyperbola has perpendicular

asymptotes.

Hence, eccentricity is 2 .

78. Answer (4)

Hint : Slope of directrix is 1

5

.

Sol. :

S(0, 2)

V(–1, –3)T

V(–1, –3), S(0, 2)

V is the mid-point of TS.

T(–2, –8)

Slope of directrix is 1 1

5VS

m

m

.

Equation of directrix is x + 5y + 42 = 0.

79. Answer (2)

Hint : Let equation of tangent in parametric form.

Sol. : Equation of tangent to

2 2

2 21

x y

a b is

cos sin 1x y

a b

, 0 , 0,cos sin

a bA B⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Let mid-point is M(h, k).

,2 cos 2 sin

a bh k

cos2 + sin2 = 1

2 2

12 2

a b

h k

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Locus of M(h, k) is

2 2

2 24

a b

x y .

80. Answer (3)

Hint : Centre of circle (2, –3) lies on given line.

Sol. : ∵ Line is normal to the circle.

Line passes through the centre C(2, –3)

of the circle.

(m2 – 1)2 – 3(m2 + 1) – m(m – 7) = 0

2m2 – 2 – 3m2 – 3 – m2 + 7m = 0

2m2 – 7m + 5 = 0

2m2 – 2m – 5m + 5 = 0

2m(m – 1) – 5(m – 1) = 0

5

1,2

m

1 2

7

2m m


11/12

81. Answer (1)

Hint : Ends of latus rectum are (±x1

, ±x1

)

Sol. :

P x x( , )1 1

y

x45°

O(0, 0)

P(x1

, x1

) lies on the hyperbola.

2 2

1 1

2 21

2

x x

aba b

2

1 2 2

1 11

2x

aba b

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2

2

1 2

2 ( )

( )

ab a bx

a b

x1

2 cannot be negative.

There are no such double ordinates.

82. Answer (4)

Hint : Given pair of lines forms a parallelogram.

Sol. : x2 – xy – 6y2 = 0

x + 2y = 0 and x – 3y = 0

x2 – xy – 6y2 + 5y – 1 = 0

x + 2y – 1 = 0 and x – 3y + 1 = 0

Formed quadrilateral is a parallelogram.

Area of parallelogram is P1

P2

cosec

1 2

1 1, ,

5 10P P

1 1

52 3tan 1

1 1 51

2 3

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= 45°, cosec = 2

1 1 12 sq. units

55 10A

83. Answer (1)

Hint : 2 2 and 2 3 2ae a

Sol. :5 5,

2 2C

⎛ ⎞ ⎜ ⎟⎝ ⎠

SS = 2ae

12 2

3ae e ⇒

2 2 2 9 1(1 ) 1 4

2 9b a e

⎛ ⎞ ⎜ ⎟⎝ ⎠

b = 2

Length of latus rectum

=

22 2 4 8 2

3 3

2

b

a

84. Answer (2)

Hint : Focus = (1, 0)

Sol. : Equation of line y – 0 = 1(x – 1).

x – y = 1

85. Answer (4)

Hint : From director circle perpendicular tangents

are drawn.

Sol. : Ellipse,

2 2

11 1

3 2

x y

Its director circle is 2 2 1 1

3 2x y .

6x2 + 6y2 = 5

∵

1 1,

2 3

⎛ ⎞⎜ ⎟⎝ ⎠

lies on the director circle.

Angle between the pair of tangents is 2

.

86. Answer (3)

Hint : a, b and c are in G.P. and positive real

numbers b2 = ac.

Sol. :2 ⇒ b ac ac b

ax + by + b = 0

ax + b(1 + y) = 0

(1 ) 0 bx y

a

Fixed point is F(0, –1)


12/12

� � �

87. Answer (4)

Hint : Eliminate t from given equations.

Sol. :1 1,x t y t

t t

x + y = 2t and x – y = 2

t

Eliminating t,

x2 – y2 = –4 (Rectangular hyperbola)

2e

88. Answer (2)

Hint : Line with equal intercept has slope ±1.

Sol. :

O

B

A

Equations of line AB is x + y = a.

∵

0 02

2

a

2 2 a

Equation of line is x + y = 2 2

But (2 2, 0) and (0, 2 2)A B

Area of OAB = 1

2 2 2 2 42

89. Answer (2)

Hint : Find common normal.

Sol. : y2 = 4x, point P(t2, 2t),

normal y = –tx + 2t + t3 ...(i)

For minimum distance (i) passes through the

centre (0, 3) of the circle.

3 = 2t + t3

(t – 1)(t2 + t + 3) = 0

t = 1

Point is P(1, 2)

Minimum distance = CP – r = 2 1

90. Answer (1)

Hint : Ratio of circumference.

Sol. :

BAx y + – 1 = 0

C(0, 0)

1

1

2

x y2 2 + = 1

Perpendicular (d) = | 1| 1

2 2

1sin 45

2 ⇒

= 90°

Circumference AB subtends right angle at

centre.

Line x + y – 1 = 0 divides circumference

in the ratio 1 : 3.

test - 4 (code-c) (answers) all india aakash test series ... · test - 4 (code-c) (hints &...

Documents