WBJEE Previous Year - 2012 | 160 Questions MCQ Test WBJEE Sample Papers, Section Wise & Full Mock Tests JEE

QUESTION: 1

In a mercury thermometer the ice point (lower fixed point) is marked as 10^o and the steam point (upper fixed point) is marked as 130^o. At 40^oC temperature, what will this thermometer read?

A.
78^o
B.
66^o
C.
62^o
D.
58^o

Solution:

QUESTION: 2

The magnetic flux linked with a coil satisfies the relation = 4t²+ 6t + 9 Wb, where t is the time in second. The e.m.f. induced in the coil at t = 2 second is

A.
22 V
B.
18 V
C.
16 V
D.
40 V

Solution:

QUESTION: 3

Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as

A.
Relative velocity
B.
Terminal velocity
C.
Critical velocity
D.
Particle velocity

Solution:

QUESTION: 4

If the velocity of light in vacuum is 3 x 10⁸ ms^–1, the time taken (in nanosecond) to travel through a glass of thickness 10 cm and refractive index 1.5 is

A.
0.5
B.
1.0
C.
2.0
D.
3.0

Solution:

QUESTION: 5

A charge +q is placed at the orgin O of X – Y axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is

A.
B.
C.
D.

Solution:

QUESTION: 6

What current will flow through 2 k resistor in the circuit shown in the figure?

A.
3 mA
B.
6 mA
C.
12 mA
D.
36 mA

Solution:

I₁= current through 2 kΩ

QUESTION: 7

In a region, the intensity of an electric field is given by in NC^–1. The electric flux through a surface in the region is

A.
5 Nm²C^–1
B.
10 Nm²C^–1
C.
15 Nm²C^–1
D.
20 Nm²C^–1

Solution:

QUESTION: 8

A train approaching a railway platform with a speed of 20 ms^–1starts blowing the whistle. Speed of sound in air is 340 ms^–1. If the frequency of the emitted sound from the whistle is 640 Hz, the frequency of sound to a person standing on the platform will appear to be

A.
600 Hz
B.
640 Hz
C.
680 Hz
D.
720 Hz

Solution:

QUESTION: 9

A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of 45^o with the magnetic field, the applied force on the wire will be

A.
1.5 N
B.
3 N
C.
3√2N
D.
3/√2 N

Solution:

QUESTION: 10

What is the phase difference between two simple harmonic motions represented by and x₂ = A cos (ωt)?

A.
π/6
B.
π/3
C.
π/2
D.
2π/3

Solution:

QUESTION: 11

Heat is produced at a rate given by H in a resistor when it is connected across a supply of voltage V. If now the resistance of the resistor is doubled and the supply voltage is made V/3 then the rate of production of heat in the resistor will be

A.
H/18
B.
H/9
C.
6H
D.
18H

Solution:

QUESTION: 12

Two elements A and B with atomic numbers Z_A and Z_B are used to produce characteristic x–rays with frequencies A and B respectively. If Z_A : Z_B= 1 : 2, then v_A : v_B will be

A.
1 :√2
B.
1 : 8
C.
4 : 1
D.
1 : 4

Solution:

√v = a(z - b) , Ignoring screening effect (i.e. b=0)

QUESTION: 13

The de Broglie wavelength of an electron moving with a velocity c/2 (c = velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

A.
1 : 4
B.
1 : 2
C.
1 : 1
D.
2 : 1

Solution:

QUESTION: 14

Two infinite parallel metal planes, contain electric charges with charge densities + and – respectively and they are separated by a small distance in air. If the permittivity of air is 0 then the magnitude of the field between the two planes with its direction will be

A.
σ/∈_o towards the positively charged plane
B.
σ/∈_o towards the negatively charged plane
C.
σ/2∈_o towards the positively charged plane
D.
0 and towards any direction

Solution:

QUESTION: 15

A box of mass 2 kg is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and g = 10 ms^–2.This maximum acceleration of the car, for the box to remain stationary, is

A.
8 ms^–2
B.
6 ms^–2
C.
4 ms^–2
D.
2 ms^–2

Solution:

a_max = μg =0.2x 10 = 2 m/s²

QUESTION: 16

The dimension of angular momentum is

A.
M⁰L¹T^–1
B.
M¹L²T^–2
C.
M¹L²T^–1
D.
M²L¹T^–2

Solution:

[L]= [m r v] = [M¹L²T^–1]

QUESTION: 17

have scalar magnitudes of 5, 4 , 3 units respectivley then the angle between A and C is

A.
cos^–1(3/5)
B.
cos^–1(4/5)
C.
π/2
D.
sin^–1 (3/4)

Solution:

QUESTION: 18

A particle is travelling along a straight line OX. The distance x (in meters) of the particle from O at a time t is given by x = 37 + 27 t – t³ where t is time in seconds. The distance of the particle from O when it comes to rest is

A.
81 m
B.
91 m
C.
101 m
D.
111 m

Solution:

t = 3 sec. x = 37 + 27 x 3 – 3³, = 37 + 81 – 27 = 91 m

QUESTION: 19

A particle is projected from the ground with kinetic energy E at an angle of 60⁰ with the horizontal. Its kinetic energy at the highest point of its motion will be

A.
E /√2
B.
E / 2
C.
E / 4
D.
E / 8

Solution:

QUESTION: 20

A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distance will it penetrate into the target before it comes to rest?

A.
30 cm
B.
20 cm
C.
10 cm
D.
5 cm

Solution:

From (1) and (2) s = 10 m

QUESTION: 21

When a spring is stretched by 10 cm, the potential energy stored is E. When the spring is stretched by 10 cm more, the potential energy stored in the spring becomes

A.
2 E
B.
4 E
C.
6 E
D.
10 E

Solution:

QUESTION: 22

Average distance of the earth from the Sun is L_1. If one year of the Earth = D days, one year of another planet whose average distance from the Sun is L₂ will be

A.
B.
C.
D.

Solution:

QUESTION: 23

A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest.
After the collision, A and B move with velocities v₁ ms^–1 and v₂ ms^–1respectively making angles of 30^o and 60^owith respect to the original direction of motion of A. The ratio v₁/v₂ will be

A.
√3/4
B.
4/√3
C.
1/√3
D.
√3

Solution:

Apply conservation of momentum along Normal

QUESTION: 24

The decimal number equivalent ot a binary number 1011001 is

A.
13
B.
17
C.
89
D.
178

Solution:

(1011001)₂= 1 x 2⁶ + 0 x 2⁵+1x 2⁴ + 1x 2³ + 0x 2² + 0x 2¹ + 1x 2⁰ = 64 + 16 + 8 + 1 =89

QUESTION: 25

The frequency of the first overtone of a closed pipe of length l₁is equal to that of the first overtone of an open pipe of length l₂. The ratio of their lenghts (l₁ : l₂) is

A.
2 : 3
B.
4 : 5
C.
3 : 5
D.
3 : 4

Solution:

QUESTION: 26

The I – V characteristics of a metal wire at two different temperatures (T₁ and T₂) are given in the adjoining figure. Here, we can conclude that

A.
T₁ > T₂
B.
T₁ < T₂
C.
T₁ = T₂
D.
T₁= 2T₂

Solution:

1/V = 1/R

slope of curve, R ∞T greater the slope smaller will be temperature.

T₁ < T₂

QUESTION: 27

In a slide calipers, (m + 1) number of vernier divisions is equal to m number of smallest main scale divisions. If d unit is the magnitude of the smallest main scale division, then the magnitude of the vernier constant is

A.
d/(m+1) unit
B.
d/m unit
C.
md/(m+1) unit
D.
(m+1)d/m unit

Solution:

QUESTION: 28

From the top of a tower, 80 m high from the ground, a stone is thrown in the horizontal direction with a velocity of 8 ms^–1. The stone reaches the ground after a time ‘t’ and falls at a distance of ‘d’ from the foot of the tower. Assuming g = 10 ms^–2, the time t and distance d are given respectively by

A.
6 s, 64 m
B.
6 s, 48 m
C.
4 s, 32 m
D.
4 s, 16 m

Solution:

time of flight (t) =

QUESTION: 29

A Wheatstone bridge has the resistances 10 Ω , 10 Ω , 10Ω and 30 Ω in its four arms. What resistance joined in parallel to the 30 resistance will bring it to the balanced condition?

A.
2Ω
B.
5Ω
C.
10Ω
D.
15Ω

Solution:

for a balanced bridge. R = 10 Ω

QUESTION: 30

An electric bulb marked as 50 W–200 V is connected across a 100 V supply. The present power of the bulb is

A.
37.5 W
B.
25 W
C.
12.5 W
D.
10 W

Solution:

QUESTION: 31

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 600 and the work done is W. The torque on the magnetic needle at this poition is

A.
2√3 W
B.
√3 W
C.
√3/2 W
D.
√3/4 W

Solution:

W = MB(1 - cos ) = MB(1 - cos 60^o ) = MB/2

QUESTION: 32

In the adjoining figure the potential difference between X and Y is 60 V. The potential difference between the points M and N will be

A.
10 V
B.
15 V
C.
20 V
D.
30 V

Solution:

QUESTION: 33

A body when fully immersed in a liquid of specific gravity 1.2 weighs 44 gwt. The same body when fully immersed in water weighs 50 gwt. The mass of the body is

A.
36 g
B.
48 g
C.
64 g
D.
80 g

Solution:

mg – σvg = 44, σ = density of liquid,
mg –1.2
ρvg = 44 ..............(1),
ρ = density of water
mg – ρvg = 50.............(2),
(1) – 1.2 x(2)
M = 80 g

QUESTION: 34

When a certain metal surface is illuminated with light of frequency ν , the stopping potential for photoelectric current is ν_o When the same surface is illuminated by light of frequency v/2 the stopping potential is v₀/4 The threshold frequency for photoelectric emission is

A.
v/6
B.
v/3
C.
2v/3
D.
4v/3

Solution:

solving equation (1) and (2) ,v/3

QUESTION: 35

Three blocks of mass 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of 14 N is applied on the 4 kg block, the contact force between the 4 kg block, the contact force between the 4 kg and the 2 kg block will be

A.
2 N
B.
6 N
C.
8 N
D.
14 N

Solution:

QUESTION: 36

Let L be the length and d be the diameter of cross section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

A.
L = 200 cm, d = 0.5 mm
B.
L = 300 cm, d = 1.0 mm
C.
L = 50 cm, d = 0.05 mm
D.
L = 100 cm, d = 0.2 mm

Solution:

QUESTION: 37

An object placed in front of a concave mirror at a distance of x cm from the pole gives a 3 times magnified real image.If it is moved to a distance of (x + 5) cm, the magnification of the image becomes 2. The focal length of the mirror is

A.
15 cm
B.
20 cm
C.
25 cm
D.
30 cm

Solution:

so f = 30 cm

QUESTION: 38

22320 cal of heat is supplied to 100 g of ice at 0^oC If the latent heat of fusion of ice is 80 cal g^–1and latent heat of vaporization of water is 540 cal g^–1, the final amount of water thus obtained and its temperature respectively are

A.
8g, 100^oC
B.
100 g, 90^oC
C.
92g, 100^oC
D.
82g, 100^oC

Solution:

total energy required to melt and boil at 100^oC is

(Q)min = 8000 + 10000 = 18000 cal,

(Qrequired)min < Qgiven for amount of vapour 22320=18000+ m x 540

= m = 8 gm. temperature = 100^oC and water remaining = 92 gm.

QUESTION: 39

A progressive wave moving along x–axis is represented by The wavelength (λ) at which themaximum particle velocity is 3 times the wave velocity is

A.
A / 3
B.
2A / (3π )
C.
(3/4) πA
D.
(2/3) πA

Solution:

QUESTION: 40

Two radioactive substances A and B have decay constant 5λ and λ respectively. At t = 0, they have the same number of nuclei the ratio of number of nuclei of A to that of B will be (1/e)² after a time interval of

A.
1/λ
B.
1/2λ
C.
1/3λ
D.
1/4λ

Solution:

QUESTION: 41

Li occupies higher position in the electrochemical series of metals as compared to Cu since

A.
the standard reduction potential of Li⁺/Li is lower than that of Cu²⁺/Cu
B.
the standard reduction potential of Cu²⁺/Cu is lower than that of Li⁺/Li
C.
the standard oxidation potential of Li/Li⁺ is lower than that of Cu/Cu²⁺
D.
Li is smaller in size as compared to Cu

Solution:

QUESTION: 42

₁₁Na²⁴ is radioactive and it decays to

A.
₉F²⁰ and -particles
B.
₁₃Al²⁴ and positron
C.
₁₁Na²³ and neutron
D.
₁₂Mg²⁴ and -particles

Solution:

QUESTION: 43

The paramagnetic behavior of B₂ is due to the presence of

A.
2 unpaired electrons in π_b MO
B.
2 unpaired electrons in π^* MO
C.
2 unpaired electrons in σ^* MO
D.
2 unpaired electrons in σ_b MO

Solution:

QUESTION: 44

A 100 ml 0.1 (M) solution of ammonium acetate is diluted by adding 100 ml of water. The pH of the resulting solution will be (pK_a of acetic acid is nearly equal to pK_b of NH₄OH)

A.
4.9
B.
5.0
C.
7.0
D.
10.0

Solution:

Hydrolysis of a salt of weak acid and weak base

pH = 7 + ½ (pK_a– pK_b) = 7 [ pK_a = pK_b ]

QUESTION: 45

In 2-butene, which one of the following statements is true ?

A.
C₁-C₂ bond is a sp³- sp³ -bond
B.
C₂-C₃ bond is a sp³ - sp² -bond
C.
C₁-C₂ bond is a sp³ - sp² -bond
D.
C₁-C₂ bond is a sp² - sp² -bond

Solution:

QUESTION: 46

The well known compounds, (+)- lactic acid and (–) lactic acid have the same molecular formula, C₃H₆O₃. The correct relationship between them is

A.
constitutional isomerism
B.
geometrical isomerism
C.
identicalness
D.
optical isomerism

Solution:

They are enantiomers and correct relationship between them is optical isomerism.

QUESTION: 47

The stability of Me₂C=CH₂ is more than that of MeCH₂CH = CH₂ due to

A.
inductive effect of the Me group
B.
resonance effect of the Me group
C.
hyperconjugative effect of the Me group
D.
resonance as well as inductive effect of the Me group

Solution:

Me₂C = CH₂ is more stable than MeCH₂CH = CH₂due to hyperconjugative effect.

QUESTION: 48

Which one of the following characteristics belongs to an electrophile ?

A.
It is any species having electron deficiency, which reacts at an electron rich C-centre
B.
It is any species having electron enrichment that reacts at an electron deficient C-centre
C.
It is cationic in nature
D.
It is anionic in nature

Solution:

QUESTION: 49

Which one of the following methods is used to prepare Me₃COEt with a good yield?

A.
Mixing EtONa with Me₃CCl
B.
Mixing Me₃CONa with EtCl
C.
Heating a mixture of (1:1) EtOH and Me₃COH in presence of conc. H₂SO₄
D.
Treatment of Me₃COH with EtMgl

Solution:

Williamson’s synthesis (SN₂) given by 1° alkyl halide.

QUESTION: 50

58.5 gm of NaCl and 180 gm of glucose were separately dissolved in 1000 ml of water. Identify the correct statement regarding the elevation of boiling point (b.p.) of the resulting solutions.

A.
NaCl solution will show higher elevation of b.p.
B.
Glucose solution will show higher elevation of b.p.
C.
Both the solutions will show equal elevation of b.p.
D.
The b.p. elevation will be shown by neither of the solutions

Solution:

For NaCl (i = 2) for glucose (i = 1), Hence NaCl solution will show higher elevation of bp (Colligative property) as both are one Molar Solution.

QUESTION: 51

Equal weights of CH₄ and H₂ are mixed in an empty container at 25^oC. The fraction of the total pressure exerted by H₂is

A.
1/9
B.
1/2
C.
8/9
D.
16/17

Solution:

Let equal weights be w g.

Partial pressure = mole fraction × Total pressure

P_H₂ = 8/9 x Total pressure

QUESTION: 52

Which of the following will show a negative deviation from Raoult’s law?

A.
Acetone-benzene
B.
Acetone-ethanol
C.
Benzene-methanol
D.
Acetone-chloroform

Solution:

Acetone – chloroform due to formation of intermolecular Hydrogen bonding

QUESTION: 53

In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will

A.
also be doubled
B.
be halved
C.
remain the same
D.
become one-fourth

Solution:

Equilibrium constant does not depend on conc. It is only a function of temperature

QUESTION: 54

Identify the correct statement from the following in a chemical reaction.

A.
The entropy always increases
B.
The change in entropy along with suitable change in enthalpy decides the fate of a reaction
C.
The enthalpy always decreases
D.
Both the enthalpy and the entropy remain constant

Solution:

Spontaneity of a reaction is decided by a combined effect of suitable change in values of enthalpy and entropy. [ ΔG = ΔH – TΔS]

QUESTION: 55

Which one of the following is wrong about molecularity of a reaction?

A.
It may be whole number or fractional
B.
It is calculated from reaction mechanism
C.
It is the number of molecules of the reactants taking part in a single step chemical reaction
D.
It is always equal to the order of elementary reaction

Solution:

Molecularity can never be a fraction.

QUESTION: 56

Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?

A.
Δx. Δp ≥ h/(4π )
B.
Δx. Δx≥ h/(4πm)
C.
ΔEΔt≥ h/(4π )
D.
ΔE. Δx≥ h/(4π)

Solution:

QUESTION: 57

The stable bivalency of Pb and trivalency of Bi is

A.
due to d contraction in Pb and Bi
B.
due to relativistic contraction of the 6s orbitals of Pb and Bi, leading to inert pair effect
C.
due to screening effect
D.
due to attainment of noble liquid configuration

Solution:

Lower oxidation state becomes more stable for group 14 and group 15 elements as we move down the group due to inert pair effect. Hence Pb⁺²and Bi⁺³ are more stable.

QUESTION: 58

The equivalent weight of K₂Cr₂O₇in acidic medium is expressed in terms of its molecular weight (M) as

A.
M/3
B.
M/4
C.
M/6
D.
M/7

Solution:

Number of electrons gained by one ion = 6
Eq.wt = M/6

QUESTION: 59

Which of the following is correct ?

A.
radius of Ca²⁺< Cl^-< S^2-
B.
radius of Cl^- < S^2- < Ca²⁺
C.
radius of S^2-= Cl^- = Ca²⁺
D.
radius of S²- < Cl^- < Ca²⁺

Solution:

Ca⁺², S^–2, Cl^–are isoelectronic (consist of 18e^–) and for isoelectronic species, ionic radii ∞e/z

QUESTION: 60

CO is practically non-polar since

A.
the σ-electron drift from C to O is almost nullified by the -electron drift from O to C
B.
the σ-electron drift from O to C is almost nullified by the σ-electron drift from C to O
C.
the bond moment is low
D.
there is a triple bond between C and O

Solution:

QUESTION: 61

The number of acidic protons in H₃PO₃are

A.
0
B.
1
C.
2
D.
3

Solution:

Number of OH group is 2 hence dibasic in nature.

QUESTION: 62

When H₂O₂ is shaken with an acidified solution of K₂Cr₂O₇ in presence of ether, the ethereal layer turns blue due to the formation of

A.
Cr₂O₃
B.
CrO₄^2-
C.
Cr₂(SO₄)₃
D.
CrO₅

Solution:

QUESTION: 63

The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl₃ are

A.
sp, 0
B.
sp², 0
C.
sp³, 0
D.
dsp², 1

Solution:

Number of lone pairs on P atom = 0

QUESTION: 64

Upon treatment with l₂ and aqueous NaOH, which of the following compounds will form iodoform?

A.
CH₃CH₂CH₂CH₂CHO
B.
CH₃CH₂COCH₂CH₃
C.
CH₃CH₂CH₂CH₂CH₂OH
D.
CH₃CH₂CH₂CH(OH)CH₃

Solution:

Iodoform reaction is given by all alcohols having

QUESTION: 65

Upon treatment with Al(OEt)₃ followed by usual reactions (work up), CH₃CHO will produce

A.
only CH₃COOCH₂CH₃
B.
a mixture of CH₃COOH and EtOH
C.
only CH₃COOH
D.
only EtOH

Solution:

This reaction is known as Tischenko reaction

QUESTION: 66

Friedel-Craft’s reaction using MeCl and anhydrous AlCl₃ will take place most efficiently with

A.
Benzene
B.
Nitrobenzene
C.
Acetophenone
D.
Toluene

Solution:

As – CH₃ group in toluene activates the benzene ring for electrophilic substitution reaction.

QUESTION: 67

Which one of the following properties is exhibited by phenol?

A.
It is soluble in aq. NaOH and evolves CO₂ with aq. NaHCO₃
B.
It is soluble in aq. NaOH and does not evolve CO₂ with aq. NaHCO₃
C.
It is not soluble in aq. NaOH but evolves CO₂ with aq. NaHCO₃
D.
It is insoluble in aq. NaOH and does not evolve CO₂with aq. NaHCO₃

Solution:

Phenol is less acidic than H₂CO_3. So it cannot liberate CO₂ from NaHCO₃ (aq).

QUESTION: 68

The basicity of aniline is weaker in comparison to that of methyl amine due to

A.
hyperconjugative effect of Me-group in MeNH₂
B.
resonance effect of phenyl group in aniline
C.
lower molecular weight of methyl amine as compared to that of aniline
D.
resonance effect of – NH₂group in MeNH₂

Solution:

QUESTION: 69

Under identical conditions, the SN₁ reaction will occur most efficiently with

A.
tert-butyl chloride
B.
1-chlorobutane
C.
2-methyl-1-chloropropane
D.
2-chlorobutane

Solution:

Reactivity order for SN₁ reaction in case of alkyl halide is 3°> 2°> 1°

QUESTION: 70

Identify the method by which Me₃CCO₂H can be prepared

A.
Treating 1 mol of MeCOMe with 2 moles of MeMgl
B.
Treating 1 mol of MeCO₂Me with 3 moles of MeMgl
C.
Treating 1 mol of MeCHO with 3 moles of MeMgl
D.
Treating 1 mol of dry ice with 1 mol of Me₃CMgl

Solution:

QUESTION: 71

20 ml 0.1 (N) acetic acid is mixed with 10 ml 0.1 (N) solution of NaOH. The pH of the resulting solution is (pK_a of acetic acid is 4.74)

A.
3.74
B.
4.74
C.
5.74
D.
6.74

Solution:

CH₃COOH + NaOH ⇒ CH₃COONa + H₂O

QUESTION: 72

In the brown ring complex [Fe(H₂O)₅(NO)]SO₄, nitric oxide behaves as

A.
NO⁺
B.
neutral NO molecule
C.
NO^–
D.
NO^2–

Solution:

‘NO’ ligand when attached to Fe behaves as positively charged ligand.

QUESTION: 73

The most contributing tautomeric enol form of MeCOCH₂CO₂Et is

A.
CH₂= C(OH)CH₂CO₂Et
B.
MeC(OH) = CHCO₂Et
C.
MeCOCH = C(OH)OEt
D.
CH₂ = C(OH)CH = C(OH) OEt

Solution:

QUESTION: 74

By passing excess Cl₂(g) in boiling toluene, which one of the following compounds is exclusively formed?

A.
A
B.
B
C.
C
D.
D

Solution:

QUESTION: 75

An equimolar mixture of toluene and chlorobenzene is treated with a mixture of conc. H₂SO₄ and conc. HNO₃.Indicate the correct statement from the following

A.
p-nitrotoluene is formed in excess
B.
equimolar amounts of p-nitrotoluene and p-nitrochlorobenzene are formed
C.
p-nitrochlorobenzene is formed in excess
D.
m-nitrochlorobenzene is formed in excess

Solution:

– CH₃ is activating group due to +I effect and hyperconjugation.
But for – Cl, – I > + R effect.

∴CH₃ is weakly activating and – Cl is weakly deactivating

QUESTION: 76

Among the following carbocations :

Ph₂C*CH₂Me(I),

PhCH₂CH₂CH*Ph (II),

Ph₂CHCH*Me (III)

and Ph₂C(Me)CH₂⁺(IV),

the order of stability is

A.
IV > II > I > III
B.
I > II > III > IV
C.
II > I > IV > III
D.
I > IV > III > II

Solution:

QUESTION: 77

Which of the following is correct?

A.
Evaporation of water causes an increase in disorder of the system
B.
Melting of ice causes a decrease in randomness of the system
C.
Condensation of steam causes an increase in disorder of the system
D.
There is practically no change in the randomness of the system when water is evaporated

Solution:

Evaparation increases randomness as liquid water is converted to water vapours.

QUESTION: 78

On passing ‘C’ Ampere of current for time ‘t’ sec through 1 litre of 2 (M) CuSO₄ solution (atomic weight of Cu = 63.5), the amount ‘m’ of Cu (in gm) deposited on cathode will be

A.
m = Ct / (63.5 × 96500)
B.
m = Ct/ (31.25 × 96500)
C.
m = (C × 96500) / (31.25 × t)
D.
m = (31.25 × C × t) / 96500

Solution:

1000 ml 2(M) CuSO₄ ≡ 2(M) CuSO₄ solution contains 2 moles Cu⁺²

Cu²+ 2 →Cu
2F 63.5g

q = c × t coulomb

or, 2 × 96500C

QUESTION: 79

If the 1st ionization energy of H atom is 13.6 eV, then the 2nd ionization energy of He atom is

A.
27.2 eV
B.
40.8 eV
C.
54.4 eV
D.
108.8 eV

Solution:

QUESTION: 80

The weight of oxalic acid that will be required to prepare a 1000 ml (N/20) solution is

A.
126/100 gm
B.
63 / 40 gm
C.
63 / 20 gm
D.
126 / 20 gm

Solution:

QUESTION: 81

The maximum value of |z| when the complex number z satisfies the condition |z + Z/2| = 2 is

A.
√3
B.
√3 +√2
C.
√3 +1
D.
√3 -1

Solution:

QUESTION: 82

where x and y are real, then the ordered pair (x, y) is

A.
(–3, 0)
B.
(0,3)
C.
(0,-3)
D.
(1/2,√3/2)

Solution:

QUESTION: 83

A.
|Z| = 1/2
B.
|Z| = 1
C.
|Z| = 2
D.
|Z| = 3

Solution:

QUESTION: 84

There are 100 students in a class. In an examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failed in Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then the number of students failing in all the three subjects

A.
is 12
B.
is 4
C.
is 2
D.
cannot be determined from given information

Solution:

n(MUPUB) = n(M) + n(P) + n(B) -n(M ∩ P) - n(PMB) - n(B ∩ P) + n(M ∩ P ∩ B)

given n(M) = 50 , n (P) = 45, n(B) = 40;

n(M∩ P) + n(P∩ B) + n(B∩ M) - 3(M∩ P ∩ B) = 32

99 = 50 + 45 + 40 – (32 + 3 n (M ∩ P∩ B) )+ n(M ∩ P∩ B) ;

2 n(M ∩ P ∩ B) = 36 – 32;

n(M∩ P ∩ B) = 2

QUESTION: 85

A vehicle registration number consists of 2 letters of English alphabets followed by 4 digits, where the first digit is not zero. Then the total number of vehicles with distinct registration numbers is

A.
26² x 10⁴
B.
²⁶P₂ x ¹⁰P⁴
C.
²⁶P₂ x 9 x 10 P³
D.
26² x 9 x 10³

Solution:

Total No. of ways = 26² x 9 x 10³

QUESTION: 86

The number of words that can be written using all the letters of the word ‘IRRATIONAL’ is

A.
B.
C.
D.

Solution:

IRRATIONAL. There are 2 I, 2 R, 2 A, one T, one O, One N, One L. Total Number of words

QUESTION: 87

Four speakers will address a meeting where speaker Q will always speak after speaker P. Then the number of ways in which the order of speakers can be prepared is

A.
256
B.
128
C.
24
D.
12

Solution:

The number of ways in which speaker P and Q can deliver their speach is ⁴C₂.Total number of ways = ⁴C₂ x 2! = 12

QUESTION: 88

The number of diagonals in a regular polygon of 100 sides is

A.
4950
B.
4850
C.
4750
D.
4650

Solution:

Number of Diagonals in n sided Polygon is equal to ⁿC₂ – n. Total number of Diagonals = ¹⁰⁰C₂ – 100=4850

QUESTION: 89

Let the coefficients of powers of x in the 2nd, 3rd and 4th terms in the expansion of (1+x)ⁿ, where n is a positive integer, be in arithmetic progression. Then the sum of the coefficients of odd powers of x in the expansion is

A.
32
B.
64
C.
128
D.
256

Solution:

Co-efficient of 2nd, 3rd and 4th terms are ⁿC₁, ⁿC₂, ⁿC₃. It is given 2.ⁿC₂= ⁿC₂+ ⁿC₃ = n = 7.

Sum of Co-efficient of odd power of x = 2⁶

QUESTION: 90

Let f(x) = ax² + bx + c, g(x) = px² + qx + r, such that f(1) = g(1), f(2) = g(2) and f(3) – g(3) = 2. Then f(4) – g(4) is

A.
4
B.
5
C.
6
D.
7

Solution:

Let a – p = w, b – q = y and c – r = z

f(1) = g(1) w + y + z = 0;

f(2) = g(2) 4w + 2y + z = 0;

f(3) = g(3) + 2 9w + 2y + z = 2w = 1,

y = –3, z = 2

f(4) – g(4) = 16 w + 4y + z = 6

QUESTION: 91

The sum 1x 1! + 2x 2! + .... + 50x 50! equals

A.
51!
B.
51! – 1
C.
51! + 1
D.
2 x 51!

Solution:

QUESTION: 92

Six numbers are in A.P. such that their sum is 3. The first term is 4 times the third term. Then the fifth term is

A.
–15
B.
–3
C.
9
D.
–4

Solution:

Let α, α - 5d, α - 3d, α - d, + d, + 3d, α + 5d are six numbers in A.P.

QUESTION: 93

The sum of the infinite series

A.
√2
B.
√3
C.
√3/2
D.
√1/3

Solution:

QUESTION: 94

The equations x² +x +a=0 and x² +ax +1=0 have a common real root

A.
for no value of a
B.
for exactly one value of a
C.
for exactly two values of a
D.
for exactly three values of a

Solution:

x² +x +a=0 .....(A) and x² +ax +1=0 ...(B)

A - B

(1–a)x + (a–1) = 0; (1–a) (x–1) = 0 if a = 1, x = 1, so, only sol is a = -2

QUESTION: 95

If 64, 27, 36 are the Pth , Qth and Rth terms of a G.P., then P + 2Q is equal to

A.
R
B.
2R
C.
3 R
D.
4 R

Solution:

t_P = 64,

t_R = 36,

a . r^P–1 = 64.............(1);

t_Q = 27, a . r^Q–1 = 27.............(2)

a . r^R–1 = 36.............(3)

(2)²x (1)/(3)³ ; 2 Q + P = 3R

QUESTION: 96

If (α+√β) and (α-√β) are the roots of the equation x² + px + q = 0 where α,β p and q are real, then the roots of the equation (p²– 4q) (p²x² + 4 px) – 16 q = 0 are

A.
B.
C.
D.

Solution:

2α = -p, α² - β = q,

so the given equation is 4β(4α²x² - 8αx) - 16α² 16+b = 0

QUESTION: 97

The number of solutions of the equation log₂(x² + 2x –1) = 1 is

A.
0
B.
1
C.
2
D.
3

Solution:

x²+ 2x –1 = 2

x² + 2x – 3=0

(x + 3) (x – 1) = 0;

x = 1, –3

QUESTION: 98

The sum of the series

A.
B.
C.
D.

Solution:

QUESTION: 99

A.
e
B.
2e
C.
e/2
D.
3e/2

Solution:

QUESTION: 100

then the value of the determinant of Q is equal to

A.
2
B.
–2
C.
1
D.
0

Solution:

QUESTION: 101

The remainder obtained when 1!+2!+...+95! is divided by 15 is

A.
14
B.
3
C.
1
D.
0

Solution:

Starting from 5! all the other numbers are divisible by 15 since 15 will be a factor So 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 So the remainder on dividing 33 by 15 is 3.

QUESTION: 102

If P,Q,R are angles of triangle PQR, then the value of

A.
-1
B.
0
C.
1/2
D.
1

Solution:

On expanding the determinant,

Δ = – 1 + cos²p + cos²Q + cos²R + 2cosP cosQ cosR

= – 1 cos² P + cos²Q + cos²Q + cos²R + (cos(P+Q) + cos(P – Q)) cosR

now P + Q = π – R

So Δ = – 1+ cos²P + cos²Q + cos2R – cos2R – cos(P – Q) cos(P + Q)

= – 1 + cos²P + cos²Q – cos.²P + sin²Q = – 1 + 1 = 0

QUESTION: 103

The number of real values of for which the system of equations

x + 3y+5z = αx
5x+y+3z =α y
3x+5y+z = αz

A.
1
B.
2
C.
4
D.
6

Solution:

QUESTION: 104

The total number of injections (one-one into mappings) from {a₁, a₂, a₃, a₄} to {b₁,b₂ ,b₃ ,b₄ ,b₅ ,b₆ , b₇ is}

A.
400
B.
420
C.
800
D.
840

Solution:

So total = 7 × 6 × 5 × 4 = 840 one-one into functions

QUESTION: 105

A.
4
B.
8
C.
16
D.
32

Solution:

QUESTION: 106

Two decks of playing cards are well schuffled and 26 cards are randomly distributed to a player. Then the probability that the player gets all distinct cards is

A.
⁵²C₂₆ / ¹⁰⁴C₂₆
B.
2 X ⁵²C₂₆ / ¹⁰⁴C₂₆
C.
2¹³ x ⁵²C₂₆ / ¹⁰⁴C₂₆
D.
2²⁶x ⁵²C₂₆/ ¹⁰⁴C₂₆

Solution:

Total no. of way of distribution = ¹⁰⁴ C₂₆ Favourable no. of ways of ditribution =

(selecting any one decki) × (selecting 26 cards out of it) = 2 x ⁵²C₂₆

QUESTION: 107

An urn contains 8 red and 5 white balls. Three balls are drawn at random. Then the probability that balls of both colours are drawn is

A.
40/143
B.
70/143
C.
3/13
D.
10/13

Solution:

Total no. of selection = ¹³C₃

QUESTION: 108

Two coins are avilable, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probalility 3/4. Given that the outcome is head, the probability that the two-headed coin waschosen is

A.
3/5
B.
2/5
C.
1/5
D.
2/7

Solution:

B is the event of selecting a biased coin
UB is the event of selecting an unbiased coin

QUESTION: 109

Let be the set of real numbers and the funtions f: R → R and g: R → R be defined by f(x)=x² + 2x-3 and g(x) = x+1. Then the value of x for which f(g(x)) = g(f(x)) is

A.
-1
B.
0
C.
1
D.
2

Solution:

fog(x) = (x + 1)²+ 2(x + 1) – 3 = x² + 4x

= x² + 4x

gof(x) = x² + 2x – 3 + 1 = x² + 2x – 2

fog(x) = gof(x)

x² + 4x = x² + 2x – 2 x = – 1

QUESTION: 110

If a,b,c are in arithmetic progression, then the roots of the equation ax² - 2bx + c = 0 are

A.
1 and c/a
B.
-1/a and -c
C.
-1 and -c/a
D.
-2 and -c/2a

Solution:

a – 2b + c = 0

So x = 1 is a root

Now product of roots = c/a

So the other root is = c/a

QUESTION: 111

If sin^-1x + sin^-1y + sin^-1z = 3π/2, then the value of

A.
0
B.
1
C.
2
D.
3

Solution:

sin^-1x + sin^-1y + sin^-1z = π/2

x = y = z = 1

QUESTION: 112

Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR, if r²sin P sin Q = pq, then the triangle is

A.
equilateral
B.
acute angled but not equilateral
C.
obtuse angled
D.
right angled

Solution:

QUESTION: 113

Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR. Then

A.
p²+q²+r²
B.
p²+r²-q²
C.
q²+r²-p²
D.
p²+q²-r²

Solution:

QUESTION: 114

Let P (2, 3) , Q (2,1) be the vertices of the triangle PQR. If the centroid of PQR lies on the line 2x+3y = 1, then the locus of R is

A.
2x + 3y=9
B.
2x - 3y =7
C.
3x + 2y=5
D.
3x - 2y=5

Solution:

Let R(h, k)

QUESTION: 115

A.
does not exist
B.
equals log_e( π²)
C.
equals 1
D.
Lies between 10 and 11

Solution:

QUESTION: 116

If f is a real-valued differentiable function such that f(x) f’(x) < 0 for all real x, then

A.
f(x) must be an increasing function
B.
f(x) must be a decreasing function
C.
|f (x )| must be an increasing function
D.
|f (x)| must be a drecreasing function

Solution:

f(x) f (x) < 0 ....(i)
f(x) < 0 ; f (x) > 0 or f(x) > 0 ; f (x) < 0 ............... (ii)

So possible graphs of f(x) are

|f(x)| is decreasing function in both cases.

QUESTION: 117

Rolle’s theorem is applicable in the interval [-2,2] for the function.

A.
(A) f(x) = x³
B.
f(x) = 4x⁴
C.
f(x) = 2x³+3
D.
f(x) = x

Solution:

For f(x) = 4x⁴ f(–2) = f(2) & f'(0) exist

QUESTION: 118

The solution of

y(0) = 1,
y(1) = 2e^1/5 is

A.
y = e^5x+ e^–5x
B.
y = (1 + x)e^5x
C.
y = (1 + x)e^x/5
D.
y = (1 + x)e^-^x/5

Solution:

Let y = e^mxis a trial solution

dy/dx = m

25m² - 10m - 1 = 0

m = 1/5(equal roots)

The differential solution y = (1 + x)e^x/5 (A & B are two arbitrary constant)

Initially y (0) = 1 ( A + 0) e^o

1 ⇒ A = 1

QUESTION: 119

Let P be the midpoint of a chord joining the vertex of the parabola y² = 8x to another point on it. Then the locus of P is

A.
y² = 2x
B.
y² = 4x
C.
x²/4 + y² = 1
D.
x² + y² /4= 1

Solution:

Let P(h, k) be the midpoint of chord AB through vertex (0, 0) of parabola y² = 8x

Co-ordinate of B= (2h - 0, 2k - 0) = (2h, 2k )
B (2h, 2k) lies on y² = 8x
⇒ 4k² = 8 × 2h
⇒ k² = 4h

QUESTION: 120

The line x = 2y intersects the ellipse x²/4 + y² = 1 at the points P and Q. The equation of the circle with PQ as diameter is

A.
x² + y² = 1/2
B.
x² + y² = 1
C.
x² + y² = 2
D.
x² + y² = 5/2

Solution:

Solving the given equations then common points are

Eqn. of Circle PQ as
diameter

QUESTION: 121

The eccentric angle in the first quadrant of a point on the ellipse x² /10 + y²/8 = 1 at a distance 3 units from the centre of the ellipse is

A.
π/6
B.
π/8
C.
π/3
D.
π/2

Solution:

QUESTION: 122

The transverse axis of a hyperbola is along the x-axis and its length is 2a. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

A.
6x² – y²= 3a²
B.
x² – 3y² = 3a²
C.
x²– 6y² = 3a²
D.
3x² – y²= 3a²

Solution:

QUESTION: 123

A point moves in such a way that the difference of its distance from two points (8, 0) and (–8, 0) always remains 4.Then the locus of the point is

A.
a circle
B.
a parabola
C.
an ellipse
D.
a hyperbola

Solution:

Let P(h, k) be the moving point

Conjugate equation of (i)

QUESTION: 124

The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is

A.
0
B.
2
C.
4
D.
1

Solution:

From given equation x = 5/3+4m, x is an integer

3 + 4m = ± 5
or ± 1 m = – 2
or m = – 1

QUESTION: 125

If a straight line passes through the point (α,β ) and the portion of the line intercepted between the axes is divided equally at that point, then x/α + y/β

A.
0
B.
1
C.
2
D.
4

Solution:

Let, the equation of line be x/a + y/b = 1 , passes through P(α,β )

(α,β ),is the mid point of the portion intercepted between the axes

QUESTION: 126

The equation y² + 4x + 4y + k = 0 represents a parabola whose latus rectum is

A.
1
B.
2
C.
3
D.
4

Solution:

y² + 4x + 4y + k = 0

⇒ (y + 2)² = 4 (–x + 1 – k/4)

Latus rectum = 4

QUESTION: 127

If the circles x² + y² + 2x + 2ky + 6 = 0 and x² + y² + 2ky + k = 0 intersect orthogonally, then k is equal to

A.
2 or -3/2
B.
-2 or -3/2
C.
2 or 3/2
D.
-2 or 3/2

Solution:

2g₁g₂+ 2f₁f₂

= c₁ + c₂
⇒ 2k₂ – k – 6 = 0
⇒ k = 2, –3/2

QUESTION: 128

If four distinct points (2k, 3k), (2, 0), (0, 3) (0, 0) lie on a circle, then

A.
k < 0
B.
o < k < 1
C.
k = 1
D.
k > 1

Solution:

Equation of circle x(x – 2) + y (y – 3) = 0
(2k, 3k) will satisfy
So, K = 1

QUESTION: 129

The line joining A (b cosα , b sinα ) and B (a cosβ , a sinβ ), where a ≠ b , is produced to the point M(x, y) so that AM : MB = b : a.

A.
0
B.
1
C.
–1
D.
a² + b²

Solution:

QUESTION: 130

Let the foci of the ellipse x²/9+ y² = 1 subtend a right angle at a point P. Then the locus of P is

A.
x² + y² = 1
B.
x² + y²= 2
C.
x² + y² = 4
D.
x² + y² = 8

Solution:

P(h, k) be a point on the ellipse,, e = 2√2/3

= h²+ k² = 8

QUESTION: 131

The general solution of the differential equation

A.
log_e|3x + 3y + 2| + 3x + 6y = c
B.
log_e|3x + 3y + 2| – 3x + 6y = c
C.
log_e|3x + 3y + 2| – 3x – 6y = c
D.
log _e|3x + 3y + 2| + 3x – 6y = c

Solution:

Put x + y = θ

QUESTION: 132

The value of the integral

A.
16
B.
8
C.
4
D.
1

Solution:

QUESTION: 133

the value of the integral

A.
1
B.
π/6
C.
π/8
D.
π/4

Solution:

I = π/4

QUESTION: 134

The integrating factor of the differential equation

A.
(log_e x)³
B.
log_e (log_e x)
C.
log_ex
D.
(log_e x)^1/3

Solution:

QUESTION: 135

Number of solutions of the equation tan x + sec x = 2 cos x, x ∈ [0, π] is

A.
0
B.
1
C.
2
D.
3

Solution:

tanx + secx = 2cosx
2sin²x + sinx – 1 = 0
sinx = 1/2

QUESTION: 136

The value of the integral

A.
log_e2
B.
log_e3
C.
1/4 log_e2
D.
1/4 log_e3

Solution:

QUESTION: 137

A.
1
B.
0
C.
–1
D.
–2

Solution:

QUESTION: 138

Maximum value of function f(x) = x/8 + 2/x on the interval [1, 6] is

A.
1
B.
9/8
C.
13/12
D.
17/8

Solution:

QUESTION: 139

A.
1/2
B.
-1/2
C.
1
D.

Solution:

QUESTION: 140

The value of the integral

A.
0
B.
e² - 1
C.
2(e² - 1)
D.
1

Solution:

QUESTION: 141

The points representing the complex number z for which

A.
a circle
B.
a straight line
C.
an ellipse
D.
a parabola

Solution:

QUESTION: 142

Let a, b, c, p,q, r be positive real numbers such that a, b, c are in G.P. and a^p = b^q = c^r. Then

A.
p, q, r are in G.P.
B.
p, q, r are in A.P.
C.
p, q, r are in H.P.
D.
p², q² ,r² are in A.P.

Solution:

b² = ac ; 2logb = loga + log c

a^p= b^q

loga/logb = q/p....(ii)

QUESTION: 143

Let S_k be the sum of an infinite G.P. series whose first term is k and common ratio is k/k+1.Then the value of

A.
log_e4
B.
log_e2 - 1
C.
1 – log_e2
D.
1 – log_e4

Solution:

QUESTION: 144

The quadratic equation 2x² – (a³+ 8a – 1)x + a² – 4a = 0 possesses roots of opposite sign. Then

A.
a ≤ 0
B.
0 < a < 4
C.
4 ≤ a < 8
D.
a ≥ 8

Solution:

QUESTION: 145

If log_e(x²– 16) log_e(4x – 11), then

A.
4 < x ≤ 5
B.
x < – 4 or x > 4
C.
– 1 ≤ x ≤ 5
D.
x < – 1 or x > 5

Solution:

x²– 16 > 0

QUESTION: 146

The coefficient of x10 in the expansion of 1 + (1 + x) + ......... + (1 + x)²⁰ is

A.
¹⁹C₉
B.
²⁰C₁₀
C.
²¹C₁₁
D.
²²C₁₂

Solution:

1 + (1 + x) + ......... + (1 + x)²⁰

QUESTION: 147

The coefficient of x10 in the expansion of 1 + (1 + x) + ......... + (1 + x)²⁰ is

A.
¹⁹C₉
B.
²⁰C₁₀
C.
²¹C₁₁
D.
²²C₁₂

Solution:

1 + (1 + x) + ......... + (1 + x)²⁰

QUESTION: 148

The system of linear equations

λx + y + z = 3
x – y – 2z = 6
– x + y + z = μ has

A.
infinite number of solutions for λ ≠ – 1 and all μ
B.
infinite number of solutions of λ = – 1 and μ = 3
C.
no solution for λ ≠ – 1
D.
unique solution for λ = –1 and μ = 3

Solution:

if λ = – 1
then Δ = 0 so equation has infinite number of solution for λ = – 1 and μ = 3

QUESTION: 149

Let A and B be two events with P(A^C) = 0.3, P(B) = 0.4 and P(A ∩ B^C) = 0.5. Then P(B/A U B^C) is equal to

A.
1/4
B.
1/3
C.
1/2
D.
2/3

Solution:

P(A ∩ B^C) = P(A) - P(A ∩ B) ; P(A ∩ B) = 0.2

QUESTION: 150

Let p, q, r be the altitudes of a triangle with area S and perimeter 2t. Then the value of 1/p + 1/q +1/r is

A.
s/t
B.
t/s
C.
s/2t
D.
2s/t

Solution:

QUESTION: 151

Let C₁ and C₂ denote the centres of the circles x² + y² = 4 and (x – 2)² + y²= 1 respectively and let P and Q be their points of intersection. Then the areas of triangles C₁PQ and C₂PQ are in the ratio

A.
3 : 1
B.
5 : 1
C.
7 : 1
D.
9 : 1

Solution:

Equation of chord S₂ – S₁= 0 ; x₂ + y₂ – 4x + 3 = 0

x = 7/4

C₁T = Perpendicular distance of point (0, 0) from x = 7/4 is 7/4

C₂T = Perpendicular distance of point (0, 0) from x = 7/4 is 1/4

QUESTION: 152

A straight line through the point of intersection of the lines x + 2y = 4 and 2x + y = 4 meets the coordinate axes at A and B. The locus of the midpoint of AB is

A.
3(x + y) = 2xy
B.
2(x + y) = 3xy
C.
2(x + y) = xy
D.
x + y = 3xy

Solution:

Point of intersection of given lines x = 4/3, y = 4/3

QUESTION: 153

Let P and Q be the points on the parabola y² = 4x so that the line segment PQ subtends right angle at the vertex.If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is

A.
1
B.
2
C.
4
D.
6

Solution:

PQ subtants right angle at vertex.

so t₁,t₂ = -4

equation of PQ is

it intersect the x axis so, put y = 0 we get x = 4

QUESTION: 154

The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle is