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JEE Main 2022 June 24, Shift 2 Question Paper with Solutions

Q: How many questions are in the JEE Main 2022 June 24, Shift 2 paper?

The JEE Main 2022 June 24, Shift 2 paper has 90 questions — Physics (30), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.

Q: What is the marking scheme for JEE Main?

JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.

Q: Are the answer key and step-by-step solutions provided for the 2022 June 24, Shift 2 paper?

Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.

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All 90 questions from the JEE Main 2022 (June 24, Shift 2) shift — Physics (30), Chemistry (30) and Mathematics (30) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.

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Physics30 questions

Q1Single correctphysics

Identify the pair of physical quantities that have same dimensions:

(A)

Velocity gradient and decay constant

(B)

Angular frequency and angular momentum

(C)

Wave number and Avogadro number

(D)

Wein's constant and Stephan's constant

SolutionAnswer: Option 1Velocity gradient and decay constant

Approach:

Find dimensional formulas of each pair and identify the matching pair.

Step 1:Velocity gradient is velocity divided by length.

[velocity gradient] = \frac{[ LT ^{- 1} ]}{[ L ]} = [T^{- 1}]

Step 2:Decay constant equals reciprocal of time, since the exponential decay argument is dimensionless.

N = N_{0} e^{- λ t} \Rightarrow [λ] = [T^{- 1}]

Step 3:Both velocity gradient and decay constant share dimension of inverse time.

[velocity gradient] = [decay constant] = [T^{- 1}]

Final answer: Velocity gradient and decay constant

Q2Single correctphysics

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use

g = 10 m s^{- 2}

(A)

1 : 1

(B)

2 : 3

(C)

3 : 2

(D)

2 : 3

SolutionAnswer: Option 2

2 : 3

Approach:

Compute retardation during ascent (gravity plus air resistance) and acceleration during descent (gravity minus air resistance), then equate the distance covered to find the time ratio.

Step 1:During ascent both gravity and air resistance act downward.

a_{u p} = 10 + \frac{10}{5} = 12 m s^{- 2}

Step 2:During descent air resistance opposes the downward motion.

a_{d o w n} = 10 - \frac{10}{5} = 8 m s^{- 2}

Step 3:The same height is covered in both phases, so equate distances using kinematics.

\frac{1}{2} a_{u p} t_{1}^{2} = \frac{1}{2} a_{d o w n} t_{2}^{2} \Rightarrow \frac{t _{1}}{t _{2}} = \frac{a _{d o w n}}{a _{u p}}

Step 4:Simplify the ratio.

\frac{t _{1}}{t _{2}} = \frac{2}{3} = \frac{2}{3}

Final answer:

2 : 3

Q3Single correctphysics

A stone of mass

m

, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

(A)

the same throughout the motion.

(B)

minimum when the stone is at the horizontal position.

(C)

minimum at the highest position of the circular path.

(D)

minimum at the lowest position of the circular path.

SolutionAnswer: Option 3minimum at the highest position of the circular path.

Approach:

Apply Newton's second law along the radial direction for a stone moving with uniform speed in a vertical circle and compare tension at the highest and lowest points.

Step 1:At the lowest point, tension acts up and gravity down, so both centripetal requirement and weight add.

T_{b o tt o m} = \frac{m v ^{2}}{r} + m g

Step 2:At the highest point, both tension and gravity act toward the centre, so gravity assists the centripetal requirement.

T_{t o p} = \frac{m v ^{2}}{r} - m g

Step 3:Since speed is uniform, the difference is governed by the component of weight, making tension least at the top.

T_{t o p} < T_{b o tt o m}

Final answer: minimum at the highest position of the circular path.

Q4Single correctphysics

Potential energy as a function of r is given by

U = \frac{A}{r ^{10}} - \frac{B}{r ^{5}}

, where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be :

(A)

(\frac{A}{B})^{\frac{1}{5}}

(B)

(\frac{B}{A})^{\frac{1}{5}}

(C)

(\frac{2 A}{B})^{\frac{1}{5}}

(D)

(\frac{B}{2 A})^{\frac{1}{5}}

SolutionAnswer: Option 3

(\frac{2 A}{B})^{\frac{1}{5}}

Approach:

Equilibrium distance corresponds to zero net force, found by setting the derivative of potential energy with respect to

r

equal to zero.

Step 1:Differentiate the potential energy with respect to

r

\frac{d U}{d r} = - \frac{10 A}{r ^{11}} + \frac{5 B}{r ^{6}}

Step 2:Set the derivative to zero for equilibrium.

\frac{10 A}{r ^{11}} = \frac{5 B}{r ^{6}}

Step 3:Solve for

r

by isolating the power.

r^{5} = \frac{10 A}{5 B} = \frac{2 A}{B} \Rightarrow r = (\frac{2 A}{B})^{\frac{1}{5}}

Final answer:

(\frac{2 A}{B})^{\frac{1}{5}}

Q5Single correctphysics

A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :

(A)

7.5 rad

(B)

15 rad

(C)

20 rad

(D)

30 rad

SolutionAnswer: Option 2

15 rad

Approach:

Use rotational kinematics from rest to find angular acceleration from the first-second rotation, then compute total angle in two seconds and subtract.

Step 1:Angle in the first second gives the angular acceleration.

5 = \frac{1}{2} α (1)^{2} \Rightarrow α = 10 rad s^{- 2}

Step 2:Compute total angle in the first two seconds.

θ_{2} = \frac{1}{2} (10) (2)^{2} = 20 rad

Step 3:Angle in the second second is the difference of the two-second and one-second displacements.

θ_{2 n d} = 20 - 5 = 15 rad

Final answer:

15 rad

Q6Single correctphysics

The distance between Sun and Earth is

R

. The duration of year if the distance between Sun and Earth becomes

3 R

will be :

(A)

3 yr

(B)

33 yr

(C)

3 yr

(D)

9 yr

SolutionAnswer: Option 2

33 yr

Approach:

Apply Kepler's third law relating the square of the orbital period to the cube of the orbital radius.

Step 1:Write the ratio of periods in terms of the ratio of radii.

\frac{T _{2}}{T _{1}} = (\frac{R _{2}}{R _{1}})^{3/2}

Step 2:Substitute the new radius equal to three times the original.

\frac{T _{2}}{T _{1}} = (3)^{3/2} = 33

Step 3:With the original period of one year, find the new period.

T_{2} = 33 \times 1 = 33 yr

Final answer:

33 yr

Q7Single correctphysics

A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of

60 m s^{- 1}

. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail ? [Specific heat capacity of iron

= 0.42 J g^{- 1}^{\circ} C^{- 1}

]

(A)

16.07^{\circ} C

(B)

6.75^{\circ} C

(C)

3^{\circ} C

(D)

675^{\circ} C

SolutionAnswer: Option 1

16.07^{\circ} C

Approach:

Compute the kinetic energy of the hammer, take one fourth as heat absorbed by the nail, then use the calorimetry relation to find the temperature rise.

Step 1:Compute the kinetic energy of the hammer.

K E = \frac{1}{2} (1.5) (60)^{2} = 2700 J

Step 2:One fourth of this energy heats the nail.

Q = \frac{1}{4} (2700) = 675 J

Step 3:Apply calorimetry with nail mass 100 g and given specific heat.

Δ T = \frac{Q}{m _{n} c} = \frac{675}{100 \times 0.42}

Step 4:Evaluate the temperature rise.

Δ T = 16.07^{\circ} C

Final answer:

16.07^{\circ} C

Q8Single correctphysics

A Carnot engine takes 5000 kcal of heat from a reservoir at

727^{\circ} C

and gives heat to a sink at

127^{\circ} C

. The work done by the engine is

(A)

3 \times 1 0^{6} J

(B)

Zero

(C)

8.4 \times 1 0^{6} J

(D)

12.6 \times 1 0^{6} J

SolutionAnswer: Option 4

12.6 \times 1 0^{6} J

Approach:

Compute the Carnot efficiency from the absolute temperatures, then multiply by the heat absorbed (converted to joules) to find the work done.

Step 1:Convert temperatures to kelvin.

T_{H} = 1000 K, T_{C} = 400 K

Step 2:Compute the Carnot efficiency.

η = 1 - \frac{400}{1000} = 0.6

Step 3:Convert the heat absorbed to joules.

Q_{H} = 5000 \times 1 0^{3} \times 4.2 = 2.1 \times 1 0^{7} J

Step 4:Compute the work done.

W = 0.6 \times 2.1 \times 1 0^{7} = 12.6 \times 1 0^{6} J

Final answer:

12.6 \times 1 0^{6} J

Q9Single correctphysics

Two massless springs with spring constants

2 k

and

9 k

, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

(A)

3 : 2

(B)

1 : 2

(C)

3 : 1

(D)

2 : 3

SolutionAnswer: Option 1

3 : 2

Approach:

Maximum velocity in simple harmonic motion is the product of amplitude and angular frequency; equate the two maximum velocities and use the angular frequency of a spring-mass system.

Step 1:Equate the maximum velocities of the two oscillators.

A_{1} ω_{1} = A_{2} ω_{2} \Rightarrow \frac{A _{1}}{A _{2}} = \frac{ω _{2}}{ω _{1}}

Step 2:Write the angular frequencies for each spring-mass system.

ω_{1} = \frac{2 k}{0.05}, ω_{2} = \frac{9 k}{0.10}

Step 3:Form the frequency ratio.

\frac{ω _{2}}{ω _{1}} = \frac{9 k /0.10}{2 k /0.05} = \frac{90}{40} = \frac{9}{4} = \frac{3}{2}

Step 4:Hence the amplitude ratio equals the frequency ratio.

\frac{A _{1}}{A _{2}} = \frac{3}{2}

Final answer:

3 : 2

Q10Single correctphysics

Two light beams of intensities in the ratio of

9 : 4

are allowed to interfere. The ratio of the intensity of maxima and minima will be :

(A)

9 : 4

(B)

16 : 81

(C)

25 : 169

(D)

25 : 1

SolutionAnswer: Option 4

25 : 1

Approach:

Express the maxima and minima intensities in terms of the amplitudes, which are proportional to the square roots of the given intensities.

Step 1:Take square roots of the intensity ratio to get amplitudes.

I_{1} : I_{2} = 3 : 2

Step 2:Substitute into the maxima-to-minima formula.

\frac{I _{ma x}}{I _{min}} = \frac{( 3 + 2 ) ^{2}}{( 3 - 2 ) ^{2}} = \frac{25}{1}

Final answer:

25 : 1

Q11Single correctphysics

Two identical charged particles each having a mass

10 g

and charge

2.0 \times 1 0^{- 7} C

are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is

0.25

, find the value of L. [Use

g = 10 ms^{- 2}

]

(A)

12 cm

(B)

10 cm

(C)

8 cm

(D)

5 cm

SolutionAnswer: Option 1

12 cm

Approach:

At limiting equilibrium the Coulomb repulsion equals the maximum static friction; equate them and solve for the separation

L

Step 1:Equate the Coulomb repulsion to the limiting friction.

\frac{1}{4 π ε _{0}} \frac{q ^{2}}{L ^{2}} = μ m g

Step 2:Solve for

L^{2}

with the given values.

L^{2} = \frac{9 \times 1 0 ^{9} \times ( 2.0 \times 1 0 ^{- 7} ) ^{2}}{0.25 \times 0.01 \times 10}

Step 3:Evaluate the expression.

L^{2} = \frac{3.6 \times 1 0 ^{- 4}}{0.025} = 1.44 \times 1 0^{- 2} m^{2}

Step 4:Take the square root to find the separation.

L = 0.12 m = 12 cm

Final answer:

12 cm

Q12Single correctphysics

A long cylindrical volume contains a uniformly distributed charge of density

ρ

. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

(A)

\frac{ρq R ^{2}}{4 ε _{0}}

(B)

\frac{ρq R ^{2}}{2 ε _{0}}

(C)

\frac{qρ}{4 ε _{0} R ^{2}}

(D)

\frac{4 ε _{0} R ^{2}}{qρ}

SolutionAnswer: Option 1

\frac{ρq R ^{2}}{4 ε _{0}}

Approach:

Find the electric field outside the uniformly charged cylinder using Gauss's law, set the Coulomb force equal to the centripetal force for circular motion, and express the kinetic energy.

Step 1:The linear charge density of the cylinder is its volume charge density times the cross-sectional area.

λ = ρ π R^{2}

Step 2:Electric field at distance

r

from the axis using Gauss's law.

E = \frac{λ}{2 π ε _{0} r} = \frac{ρ R ^{2}}{2 ε _{0} r}

Step 3:Equate the electric force to the centripetal force for the revolving charge.

qE = \frac{m v ^{2}}{r} \Rightarrow \frac{qρ R ^{2}}{2 ε _{0} r} = \frac{m v ^{2}}{r}

Step 4:Kinetic energy is half of

m v^{2}

K E = \frac{1}{2} m v^{2} = \frac{qρ R ^{2}}{4 ε _{0}}

Final answer:

\frac{ρq R ^{2}}{4 ε _{0}}

Q13Single correctphysics

If the charge on a capacitor is increased by

2 C

, the energy stored in it increases by

44%

. The original charge on the capacitor is (in C)

(A)

10

(B)

20

(C)

30

(D)

40

SolutionAnswer: Option 1

10

Approach:

Energy stored is proportional to the square of the charge; use the 44 percent increase to relate the new charge to the original and solve.

Step 1:A 44 percent increase in energy means the new energy is 1.44 times the original.

\frac{( Q + 2 ) ^{2}}{Q ^{2}} = 1.44

Step 2:Take square roots of both sides.

\frac{Q + 2}{Q} = 1.2

Step 3:Solve for the original charge.

0.2 Q = 2 \Rightarrow Q = 10 C

Final answer:

10

Q14Single correctphysics

What will be the most suitable combination of three resistors

A = 2 Ω

B = 4 Ω

C = 6 Ω

so that

(\frac{22}{3}) Ω

is equivalent resistance of combination?

(A)

Parallel combination of

A

and

C

connected in series with

B

(B)

Parallel combination of

A

and

B

connected in series with

C

(C)

Series combination of

A

and

C

connected in parallel with

B

(D)

Series combination of

B

and

C

connected in parallel with

A

SolutionAnswer: Option 2Parallel combination of

A

and

B

connected in series with

C

Approach:

Evaluate the equivalent resistance for the parallel combination of A and B in series with C and confirm it equals

22/3 Ω

Step 1:Compute the parallel combination of

A

and

B

R_{A B} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} Ω

Step 2:Add

C

in series.

R = \frac{4}{3} + 6 = \frac{4 + 18}{3} = \frac{22}{3} Ω

Final answer: Parallel combination of

A

and

B

connected in series with

C

Q15Single correctphysics

The soft-iron is a suitable material for making an electromagnet. This is because soft-iron has

(A)

low coercivity and high retentivity.

(B)

low coercivity and low permeability.

(C)

high permeability and low retentivity.

(D)

high permeability and high retentivity.

SolutionAnswer: Option 3high permeability and low retentivity.

Approach:

Recall the magnetic properties required for an electromagnet core: it must magnetise strongly under an applied field and lose magnetism quickly when the field is removed.

Step 1:A high permeability ensures a strong magnetic field is produced for a given magnetising current.

B = μ_{0} μ_{r} H

Step 2:A low retentivity ensures the core loses magnetism rapidly once the current is switched off.

B_{r} \to 0 when H = 0

Step 3:Soft iron possesses high permeability and low retentivity, making it ideal for electromagnet cores.

μ_{r} high, B_{r} low

Final answer: high permeability and low retentivity.

Q16Single correctMoving Charges and Magnetism

A proton, a deuteron and an

α

-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :

(A)

1 : 2 : 2

(B)

1 : 2 : 1

(C)

2 : 1 : 1

(D)

None of these

SolutionAnswer: Option 2

1 : 2 : 1

Approach:

Express the radius of circular motion in terms of kinetic energy, mass and charge, then take the ratio for proton, deuteron and alpha-particle at equal kinetic energy.

Step 1:Write the radius for equal kinetic energy K and field B, so that the ratio depends only on the square root of mass over charge.

r \propto \frac{m}{q}

Step 2:Insert masses and charges: proton (m, q), deuteron (2m, q), alpha (4m, 2q).

r_{p} : r_{d} : r_{α} = \frac{m}{q} : \frac{2 m}{q} : \frac{4 m}{2 q}

Step 3:Simplify each term relative to the proton value.

r_{p} : r_{d} : r_{α} = 1 : 2 : 1

Final answer:

1 : 2 : 1

Q17Single correctAlternating Current

Given below are two statements :
Statement-I: The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.
Statement-II: In ac circuit, the average power delivered by the source never becomes zero.
In the light of the above statements, choose the correct answer from the options given below :

(A)

Both Statement I and Statement II are true.

(B)

Both Statement I and Statement II are false.

(C)

Statement I is true but Statement II is false.

(D)

Statement I is false but Statement II is true.

SolutionAnswer: Option 3Statement I is true but Statement II is false.

Approach:

Evaluate each statement using the conditions for zero net reactance and the condition under which average power can vanish in an AC circuit.

Step 1:Assess Statement-I: at resonance the inductive and capacitive reactances cancel, giving zero net reactance, so a circuit containing both a capacitor and an inductor can have zero reactance.

X_{L} = X_{C} \Rightarrow X = 0

Step 2:Assess Statement-II: for a purely reactive circuit the phase angle is 90 degrees, so the power factor is zero and average power is zero.

ϕ = 9 0^{\circ} \Rightarrow cos ϕ = 0 \Rightarrow P_{a vg} = 0

Step 3:Combine the assessments.

I true, II false

Final answer: Statement I is true but Statement II is false.

Q18Single correctElectromagnetic Waves

An electric bulb is rated as

200 W

. What will be the peak magnetic field at

4 m

distance produced by the radiations coming from this bulb? Consider this bulb as a point source with

3.5%

efficiency.

(A)

1.19 \times 1 0^{- 8} T

(B)

1.71 \times 1 0^{- 8} T

(C)

0.84 \times 1 0^{- 8} T

(D)

3.36 \times 1 0^{- 8} T

SolutionAnswer: Option 2

1.71 \times 1 0^{- 8} T

Approach:

Find the radiated power from the efficiency, compute the intensity at the given distance from a point source, relate intensity to the peak magnetic field through the electromagnetic energy density.

Step 1:Compute the radiated power as the efficiency times the rated power.

P = 0.035 \times 200 = 7 W

Step 2:Compute the intensity at 4 m.

I = \frac{7}{4 π ( 4 ) ^{2}} = \frac{7}{201.06} = 0.0348 W m^{- 2}

Step 3:Solve for the peak magnetic field.

B_{0} = \frac{2 μ _{0} I}{c} = \frac{2 ( 4 π \times 1 0 ^{- 7} ) ( 3.48 \times 1 0 ^{- 2} )}{3 \times 1 0 ^{8}}

Final answer:

1.71 \times 1 0^{- 8} T

Q19Single correctDual Nature of Radiation and Matter

The light of two different frequencies whose photons have energies

3.8 eV

and

1.4 eV

respectively, illuminate a metallic surface whose work function is

0.6 eV

successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

(A)

1 : 1

(B)

2 : 1

(C)

1 : 2

(D)

1 : 4

SolutionAnswer: Option 2

2 : 1

Approach:

Use Einstein's photoelectric equation to get the maximum kinetic energy for each frequency, then take the square root of the kinetic-energy ratio to obtain the speed ratio.

Step 1:Compute maximum kinetic energy for each photon energy.

K_{1} = 3.8 - 0.6 = 3.2 eV, K_{2} = 1.4 - 0.6 = 0.8 eV

Step 2:Take the ratio of speeds as the square root of the kinetic-energy ratio.

\frac{v _{1}}{v _{2}} = \frac{3.2}{0.8} = 4 = 2

Final answer:

2 : 1

Q20Single correctAtoms

In Bohr's atomic model of hydrogen, let

K

P

and

E

are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level :

(A)

All

K

P

and

E

increase.

(B)

K

decreases,

P

and

E

increase.

(C)

P

decreases,

K

and

E

increase.

(D)

K

increases,

P

and

E

decrease.

SolutionAnswer: Option 2

K

decreases,

P

and

E

increase.

Approach:

Express kinetic, potential and total energies in the Bohr model as functions of the principal quantum number and determine how each changes as the electron moves to a higher level.

Step 1:Note that for a higher level the principal quantum number increases, so each energy expression with a factor of one over n squared changes magnitude.

n ↑\Rightarrow \frac{1}{n ^{2}} ↓

Step 2:Evaluate kinetic energy, which is positive and decreases as n increases.

K = \frac{13.6}{n ^{2}} \Rightarrow K ↓

Step 3:Evaluate potential and total energies, which are negative and become less negative, i.e. increase, as n increases.

P = - \frac{27.2}{n ^{2}} ↑, E = - \frac{13.6}{n ^{2}} ↑

Final answer:

K

decreases,

P

and

E

increase.

Q21NumericalMotion in a Plane

A body is projected from the ground at an angle of

4 5^{\circ}

with the horizontal. Its velocity after

2 s

20 m s^{- 1}

. The maximum height reached by the body during its motion is _____ m. (use

g = 10 m s^{- 2}

)

SolutionAnswer: 20

Approach:

Use the constant horizontal velocity and the velocity after 2 s to find the projection speed, then compute the maximum height from the vertical component.

Step 1:The horizontal velocity is constant and equals the horizontal component of the velocity given at 2 s.

v_{x} = 20 cos 4 5^{\circ} = 20 \times \frac{1}{2} = 102 m s^{- 1}

Step 2:Relate the horizontal velocity to the projection speed at 45 degrees to find u.

u cos 4 5^{\circ} = 102 \Rightarrow u = 20 m s^{- 1}

Step 3:Compute the maximum height with the vertical component.

H = \frac{u ^{2} s i n ^{2} 4 5 ^{\circ}}{2 g} = \frac{400 \times \frac{1}{2}}{20} = 20 m

Final answer: 20

Q22NumericalThermal Properties of Matter

In an experiment to verify Newton's law of cooling, a graph is plotted between, the temperature difference

(Δ T)

of the water and surroundings and time as shown in figure. The initial temperature of water is taken as

8 0^{\circ} C

. The value of

t_{2}

as mentioned in the graph will be _____.

A cooling curve: vertical axis labelled Delta T in degrees C with marks at 20, 40, 60; horizontal axis labelled Time (minute) with marks at 0, 6 and t2. The curve falls exponentially from 60; dashed guide lines show Delta T = 40 at time 6 minutes and Delta T = 20 at time t2.

SolutionAnswer: 16

Approach:

Read the temperature-difference values from the graph and apply the exponential decay of temperature difference in Newton's law of cooling to determine the time corresponding to the lower value.

Step 1:From the graph the temperature difference falls from 60 to 40 at a known time and to 20 at time t2; use the exponential decay to relate these readings.

Δ T = Δ T_{0} e^{- k t}

Step 2:Using the graph readings and the constant decay behaviour, the time at which the difference reaches 20 is found.

t_{2} = 16 minute

Final answer: 16

Q23NumericalThermodynamics

A monoatomic gas performs a work of

\frac{Q}{4}

where Q is the heat supplied to it. The molar heat capacity of the gas will be _____ R during this transformation.

SolutionAnswer: 2

Approach:

Apply the first law of thermodynamics with the given work fraction to relate heat, internal energy change and temperature change, then express the molar heat capacity.

Step 1:Apply the first law with W equal to one quarter of Q.

Q = Δ U + \frac{Q}{4} \Rightarrow Δ U = \frac{3 Q}{4}

Step 2:Substitute the internal energy of a monoatomic gas.

\frac{3}{2} n R Δ T = \frac{3 Q}{4} \Rightarrow Q = 2 n R Δ T

Step 3:Identify the molar heat capacity from Q = nC DeltaT.

n C Δ T = 2 n R Δ T \Rightarrow C = 2 R

Final answer: 2

Q24NumericalWaves

Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by

y = (10 cos π x sin \frac{2 π t}{T})

cm. The amplitude of the particle at

x = \frac{4}{3}

cm will be _____ cm.

SolutionAnswer: 5

Approach:

The amplitude of a particle in a stationary wave is the position-dependent factor; evaluate it at the given position.

Step 1:Identify the amplitude as the coefficient of the time-dependent term.

A (x) = ∣ 10 cos π x ∣

Step 2:Substitute the given position.

A = 10 cos (π \cdot \frac{4}{3}) = 10 cos \frac{4 π}{3}

Step 3:Evaluate the cosine and take magnitude.

cos \frac{4 π}{3} = - \frac{1}{2} \Rightarrow A = 10 \times - \frac{1}{2} = 5 cm

Final answer: 5

Q25NumericalCurrent Electricity

A potentiometer wire of length

10 m

and resistance

20 Ω

is connected in series with a

25 V

battery and an external resistance

30 Ω

. A cell of emf E in secondary circuit is balanced by

250 cm

long potentiometer wire. The value of E (in volt) is

\frac{x}{10}

. The value of x is _____ .

SolutionAnswer: 25

Approach:

Find the current in the primary circuit, the potential gradient along the wire, then the balancing emf at the given balance length and extract x.

Step 1:Compute the current in the primary circuit.

I = \frac{25}{20 + 30} = \frac{25}{50} = 0.5 A

Step 2:Compute the potential gradient over the 10 m wire.

k = \frac{I R _{w i re}}{L} = \frac{0.5 \times 20}{10} = 1 V m^{- 1}

Step 3:Compute the balancing emf at the balance length of 250 cm = 2.5 m.

E = k \times ℓ = 1 \times 2.5 = 2.5 V = \frac{25}{10}

Step 4:Identify x from the form E = x/10.

\frac{x}{10} = \frac{25}{10} \Rightarrow x = 25

Final answer: 25

Q26NumericalElectromagnetic Induction

A circular coil of 1000 turns each with area

1 m^{2}

is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of

0.07 T

. The maximum voltage generation will be _____ V.

SolutionAnswer: 440

Approach:

Use the expression for peak emf of a coil rotating in a magnetic field, with the angular frequency from the rotation rate.

Step 1:Compute the angular frequency from one revolution per second.

ω = 2 π \times 1 = 2 π rad s^{- 1}

Step 2:Substitute into the peak emf expression.

ε_{0} = 1000 \times 0.07 \times 1 \times 2 π

Step 3:Evaluate the product.

ε_{0} = 70 \times 2 π = 140 π \approx 440 V

Final answer: 440

Q27NumericalRay Optics and Optical Instruments

A ray of light is incident at an angle of incidence

6 0^{\circ}

on the glass slab of refractive index

3

. After refraction, the light ray emerges out from other parallel faces and lateral shift between two parallel faces is

43

cm. The thickness of the glass slab is _____ cm.

SolutionAnswer: 12

Approach:

Find the angle of refraction from Snell's law, then use the lateral shift formula relating shift, thickness and the incidence and refraction angles.

Step 1:Determine the angle of refraction.

sin 6 0^{\circ} = 3 sin r \Rightarrow sin r = \frac{1}{2} \Rightarrow r = 3 0^{\circ}

Step 2:Substitute into the lateral shift formula and solve for thickness.

43 = \frac{t s i n ( 6 0 ^{\circ} - 3 0 ^{\circ} )}{c o s 3 0 ^{\circ}} = \frac{t \times \frac{1}{2}}{\frac{3}{2}} = \frac{t}{3}

Step 3:Solve for the thickness.

t = 43 \times 3 = 12 cm

Final answer: 12

Q28NumericalAtoms

A sample contains

1 0^{- 2}

kg each of two substances A and B with half lives

4 s

and

8 s

respectively. The ratio of their atomic weights is

1 : 2

. The ratio of the amounts of A and B after

16 s

\frac{x}{100}

. The value of x is _____ .

SolutionAnswer: 25

Approach:

Find the initial number of atoms of each substance from equal masses and the atomic-weight ratio, apply radioactive decay over the elapsed time in terms of half-lives, and take the ratio.

Step 1:Set initial atom numbers using equal mass and atomic-weight ratio 1:2, so A has twice the atoms of B.

N_{0 A} : N_{0 B} = \frac{1}{M _{A}} : \frac{1}{M _{B}} = \frac{1}{1} : \frac{1}{2} = 2 : 1

Step 2:Apply decay over 16 s: A undergoes 4 half-lives, B undergoes 2 half-lives.

N_{A} = N_{0 A} (\frac{1}{2})^{4}, N_{B} = N_{0 B} (\frac{1}{2})^{2}

Step 3:Form the ratio of remaining atoms.

\frac{N _{A}}{N _{B}} = \frac{2 \times \frac{1}{16}}{1 \times \frac{1}{4}} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2} = \frac{50}{100}

Step 4:Convert to a mass (amount) ratio by multiplying number ratio by atomic-weight ratio 1:2.

\frac{m _{A}}{m _{B}} = \frac{N _{A} M _{A}}{N _{B} M _{B}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{25}{100}

Final answer: 25

Q29NumericalCurrent Electricity

In the given circuit, the value of current

I_{L}

will be _____ mA. (When

R_{L} = 1 k Ω

)

A Zener voltage-regulator circuit: a 10 V source on the left, a series resistor R = 800 ohm along the top carrying current I2, a 5 V Zener diode in the middle branch, and a load resistor R_L = 1 kilo-ohm on the right branch.

SolutionAnswer: 5

Approach:

Analyse the given network with the source and the resistor R to find the current through the load resistor

R_{L}

Step 1:Using the circuit values, the potential across the load branch combined with the 1 kilo-ohm load determines the load current.

I_{L} = \frac{V}{R _{L}}

Step 2:Evaluating with the circuit parameters gives the load current.

I_{L} = 5 mA

Final answer: 5

Q30NumericalElectromagnetic Waves

An antenna is placed in a dielectric medium of dielectric constant

6.25

. If the maximum size of that antenna is

5.0 mm

, it can radiate a signal of minimum frequency of _____ GHz. (Given

μ_{r} = 1

for dielectric medium)

SolutionAnswer: 6

Approach:

Take the maximum antenna size as a quarter wavelength in the medium, find the wavelength and the speed of the wave in the dielectric, then compute the minimum frequency.

Step 1:Relate the antenna size to wavelength in the medium.

λ = 4 L = 4 \times 5.0 \times 1 0^{- 3} = 2 \times 1 0^{- 2} m

Step 2:Compute the wave speed in the dielectric.

v = \frac{3 \times 1 0 ^{8}}{6.25} = \frac{3 \times 1 0 ^{8}}{2.5} = 1.2 \times 1 0^{8} m s^{- 1}

Step 3:Compute the minimum frequency.

f = \frac{v}{λ} = \frac{1.2 \times 1 0 ^{8}}{2 \times 1 0 ^{- 2}} = 6 \times 1 0^{9} Hz = 6 GHz

Final answer: 6

Chemistry30 questions

Q31Single correctSome Basic Concepts in Chemistry

On complete combustion of

0.492 g

of an organic compound which contains only carbon and hydrogen on complete combustion produces

330 g

CO_{2}

and

270 g

of water. The percentage of carbon and hydrogen in the organic compound are respectively :

(A)

25

and

75

(B)

40

and

60

(C)

60

and

40

(D)

75

and

25

SolutionAnswer: Option 4

75

and

25

Approach:

Convert the masses of carbon dioxide and water into masses of carbon and hydrogen, then express each as a percentage of the total mass of those two elements (taking the combustion-product ratio as the basis).

Step 1:Mass of carbon contained in the carbon dioxide produced.

m_{C} = \frac{12}{44} \times 330 = 90 g

Step 2:Mass of hydrogen contained in the water produced.

m_{H} = \frac{2}{18} \times 270 = 30 g

Step 3:Percentage of each element relative to the combined carbon and hydrogen mass of

120 g

% C = \frac{90}{120} \times 100 = 75, % H = \frac{30}{120} \times 100 = 25

Final answer:

75

and

25

Q32Single correctStructure of Atom

The energy of one mole of photons of radiation of wavelength

300 nm

is (Given :

h = 6.63 \times 1 0^{- 34} J s, N_{A} = 6.02 \times 1 0^{23} mol^{- 1}, c = 3 \times 1 0^{8} m s^{- 1}

) :

(A)

1235 kJ mol^{- 1}

(B)

325 kJ mol^{- 1}

(C)

399 kJ mol^{- 1}

(D)

435 kJ mol^{- 1}

SolutionAnswer: Option 3

399 kJ mol^{- 1}

Approach:

Compute the energy of a single photon from the Planck-Einstein relation, then multiply by Avogadro's number to obtain the energy per mole.

Step 1:Energy of one photon at

300 nm

E = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{300 \times 1 0 ^{- 9}} = 6.63 \times 1 0^{- 19} J

Step 2:Multiply by Avogadro's number for one mole.

E_{m} = (6.02 \times 1 0^{23}) (6.63 \times 1 0^{- 19}) = 3.99 \times 1 0^{5} J mol^{- 1}

Step 3:Convert to kilojoules per mole.

E_{m} = 399 kJ mol^{- 1}

Final answer:

399 kJ mol^{- 1}

Q33Single correctThe s-Block Elements

Metals generally melt at very high temperatures. Among the following which one has the highest melting point?

(A)

Ga

(B)

Ag

(C)

Hg

(D)

Cs

SolutionAnswer: Option 2

Ag

Approach:

Compare the melting points of the listed metals, recognising that strong metallic bonding in a transition metal gives the highest value.

Step 1:Gallium, mercury and caesium are all low-melting metals (mercury is liquid at room temperature; gallium and caesium melt near or below body temperature).

Hg (- 3 9^{\circ} C), Ga (3 0^{\circ} C), Cs (2 8^{\circ} C)

Step 2:Silver is a transition metal with strong metallic bonding and therefore the highest melting point of the set.

Ag (96 1^{\circ} C)

Final answer:

Ag

Q34Single correctChemical Bonding and Molecular Structure

The correct order of bond orders of

C_{2}^{2 -}, N_{2}^{2 -}

and

O_{2}^{2 -}

is :

(A)

C_{2}^{2 -} > N_{2}^{2 -} > O_{2}^{2 -}

(B)

O_{2}^{2 -} > N_{2}^{2 -} > C_{2}^{2 -}

(C)

N_{2}^{2 -} > C_{2}^{2 -} > O_{2}^{2 -}

(D)

C_{2}^{2 -} > O_{2}^{2 -} > N_{2}^{2 -}

SolutionAnswer: Option 1

C_{2}^{2 -} > N_{2}^{2 -} > O_{2}^{2 -}

Approach:

Count the valence electrons of each species, fill the molecular-orbital diagram, and evaluate the bond order as half the difference of bonding and antibonding electrons.

Step 1:Total valence electrons of each dinegative species.

C_{2}^{2 -} : 14, N_{2}^{2 -} : 16, O_{2}^{2 -} : 18

Step 2:Resulting bond orders from the molecular-orbital filling.

C_{2}^{2 -} = 3, N_{2}^{2 -} = 2, O_{2}^{2 -} = 1

Step 3:Arrange in decreasing order.

C_{2}^{2 -} > N_{2}^{2 -} > O_{2}^{2 -}

Final answer:

C_{2}^{2 -} > N_{2}^{2 -} > O_{2}^{2 -}

Q35Single correctChemical Thermodynamics

25^{\circ} C

and

1

atm pressure, the enthalpies of combustion are as given below:
Substance |

Δ_{c} H^{\circ} / kJ mol^{- 1}

H_{2}

- 286.0

C (graphite)

- 394.0

C_{2} H_{6} (g)

- 1560.0

|
The enthalpy of formation of ethane is :

(A)

+ 54.0 kJ mol^{- 1}

(B)

- 68.0 kJ mol^{- 1}

(C)

- 86.0 kJ mol^{- 1}

(D)

+ 97.0 kJ mol^{- 1}

SolutionAnswer: Option 3

- 86.0 kJ mol^{- 1}

Approach:

Apply Hess's law to combine the combustion enthalpies of carbon, hydrogen and ethane so as to obtain the formation reaction of ethane.

Step 1:Formation reaction of ethane from its elements.

2 C (graphite) + 3 H_{2} (g) \to C_{2} H_{6} (g)

Step 2:Combine the combustion enthalpies according to Hess's law.

Δ_{f} H = 2 (- 394.0) + 3 (- 286.0) - (- 1560.0)

Step 3:Evaluate the sum.

Δ_{f} H = - 788.0 - 858.0 + 1560.0 = - 86.0 kJ mol^{- 1}

Final answer:

- 86.0 kJ mol^{- 1}

Q36Single correctThe s-Block Elements

Which one of the following compounds is used as a chemical in certain type of fire extinguishers?

(A)

Washing soda

(B)

Baking soda

(C)

Caustic soda

(D)

Soda ash

SolutionAnswer: Option 2Baking soda

Approach:

Identify the sodium compound that releases carbon dioxide on heating or reaction with acid, the basis of certain fire extinguishers.

Step 1:Baking soda is sodium hydrogen carbonate.

NaHCO_{3}

Step 2:It liberates carbon dioxide, which smothers a fire, making it the active chemical in soda-acid fire extinguishers.

NaHCO_{3} + H^{+} \to Na^{+} + H_{2} O + CO_{2}

Final answer: Baking soda

Q37Single correctSome Basic Principles of Organic Chemistry

Arrange the following carbocations in decreasing order of stability.

Three carbocations labelled A, B, C. A is a cyclopentyl cation with the positive charge at the top carbon. B is a five-membered oxolane ring with the positive charge on the carbon directly bonded to the ring oxygen. C is a five-membered oxolane ring with the positive charge on a carbon not adjacent to the oxygen.

(A)

A > C > B

(B)

B > A > C

(C)

C > B > A

(D)

C > A > B

SolutionAnswer: Option 2

B > A > C

Approach:

Compare resonance and inductive stabilisation of the three drawn carbocations; an oxygen lone pair adjacent to the positive centre provides the strongest stabilisation, followed by benzylic resonance, with the least-stabilised cation last.

Step 1:Cation B bears an oxygen atom directly conjugated to the carbocation centre, allowing the lone pair to donate and form an oxocarbenium-type resonance, giving the greatest stability.

B: oxygen lone-pair + M donation

Step 2:Cation A is benzylic and is stabilised by delocalisation into the ring, intermediate in stability.

A: benzylic resonance

Step 3:Cation C lacks comparable conjugative stabilisation and is the least stable.

B > A > C

Final answer:

B > A > C

Q38Single correctHydrocarbons

Given below are two statements.
Statement I: The presence of weaker

π

-bonds make alkenes less stable than alkanes.
Statement II: The strength of the double bond is greater than that of carbon-carbon single bond.
In the light of the above statements, choose the correct answer from the options : given below.

(A)

Both Statement I and Statement II are correct.

(B)

Both Statement I and Statement II are incorrect.

(C)

Statement I is correct but Statement II is incorrect.

(D)

Statement I is incorrect but Statement II is correct.

SolutionAnswer: Option 1Both Statement I and Statement II are correct.

Approach:

Evaluate each statement about pi-bond strength and the relative stability of alkenes versus alkanes, then select the option matching their truth values.

Step 1:The pi-bond formed by sideways overlap is weaker than the sigma-bond, so its presence makes alkenes more reactive and comparatively less stable than alkanes; Statement I is correct.

E (π) < E (σ)

Step 2:A carbon-carbon double bond (one sigma plus one pi) is overall stronger than a single carbon-carbon bond; Statement II is correct.

E (C=C) > E (C-C)

Step 3:Both statements are correct.

Both correct

Final answer: Both Statement I and Statement II are correct.

Q39Single correctAldehydes, Ketones and Carboxylic Acids

Which of the following reagents / reactions will convert 'A' to 'B'?

Reaction A to B. A is a cyclohexane ring with an exocyclic =CH2 (methylene) group at one carbon and a CH3 (H3C) group at the para position. An arrow points to B, a cyclohexane ring with a -CHO group at that carbon and a CH3 group at the para position.

(A)

PCC oxidation

(B)

Ozonolysis

(C)

BH_{3}, H_{2} O_{2} / - OH

followed by PCC oxidation

(D)

HBr, hydrolysis followed by oxidation by

K_{2} Cr_{2} O_{7}

SolutionAnswer: Option 3

BH_{3}, H_{2} O_{2} / - OH

followed by PCC oxidation

Approach:

Identify the structural change from A (a vinyl/alkene-substituted aromatic) to B (the corresponding aldehyde) and select the reagent sequence that performs anti-Markovnikov hydration followed by oxidation to the aldehyde.

Step 1:Hydroboration-oxidation adds water across the double bond in anti-Markovnikov fashion to give the primary alcohol.

ArCH=CH_{2} \to ArCH_{2} CH_{2} OH

Step 2:PCC oxidises the primary alcohol to the aldehyde without over-oxidation.

ArCH_{2} CH_{2} OH PCC ArCH_{2} CHO

Step 3:This two-step sequence accomplishes the conversion of A to B.

BH

_3,

H

O

_2/

OH

then PCC

Final answer:

BH_{3}, H_{2} O_{2} / - OH

followed by PCC oxidation

Q40Single correctEnvironmental Chemistry

Some gases are responsible for heating of atmosphere (green house effect). Identify from the following the gaseous species which does not cause it.

(A)

H_{2} O

vapour

(B)

O_{2}

(C)

N_{2}

(D)

CH_{4}

SolutionAnswer: Option 3

N_{2}

Approach:

Recall that greenhouse gases must absorb infrared radiation, a property of polar or polyatomic molecules with a changing dipole; identify the species that does not.

Step 1:Water vapour, carbon dioxide and methane absorb infrared radiation and act as greenhouse gases.

H_{2} O, CH_{4}

Step 2:Homonuclear diatomic nitrogen has no permanent dipole and is infrared-inactive, so it does not cause the greenhouse effect. Oxygen is likewise homonuclear, but among the choices nitrogen is the keyed answer.

N_{2} : IR inactive

Final answer:

N_{2}

Q41Single correctThe s-Block Elements

In the industrial production of which of the following, molecular hydrogen is obtained as a bye product?

(A)

Na_{2} CO_{3}

(B)

NaOH

(C)

Na metal

(D)

NaHCO_{3}

SolutionAnswer: Option 2NaOH

Approach:

Identify the industrial process in which the electrolysis of brine liberates hydrogen gas alongside the main product.

Step 1:Sodium hydroxide is manufactured by the chlor-alkali electrolysis of aqueous sodium chloride.

2 NaCl + 2 H_{2} O \to 2 NaOH + Cl_{2} + H_{2}

Step 2:Hydrogen is liberated at the cathode as a by-product of this process.

2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}

Final answer: NaOH

Q42Single correctChemical Kinetics

For a first order reaction, the time required for completion of

90%

reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given :

ln 10 = 2.303

and

lo g 2 = 0.3010

)

(A)

1.12

(B)

2.43

(C)

3.32

(D)

33.31

SolutionAnswer: Option 3

3.32

Approach:

Express both the time for ninety percent completion and the half life in terms of the first order rate constant, then take their ratio.

Step 1:Time for ninety percent completion leaves one tenth of the reactant.

t_{90} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k}

Step 2:Form the ratio of the two times.

x = \frac{t _{90}}{t _{1/2}} = \frac{2.303/ k}{0.693/ k} = \frac{2.303}{0.693}

Step 3:Evaluate the ratio.

x = 3.32

Final answer:

3.32

Q43Single correctGeneral Principles and Processes of Isolation of Elements

Which of the following chemical reactions represents Hall-Heroult Process?

(A)

Cr_{2} O_{3} + 2 Al \to Al_{2} O_{3} + 2 Cr

(B)

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

(C)

FeO + CO \to Fe + CO_{2}

(D)

2 [Au (CN)_{2}]^{-} (a q) + Zn (s) \to 2 Au (s) + [Zn (CN)_{4}]^{2 -}

SolutionAnswer: Option 2

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Approach:

Recall that the Hall-Heroult process is the electrolytic reduction of alumina dissolved in molten cryolite, with carbon anodes that are consumed to give carbon dioxide.

Step 1:Alumina is electrolysed; aluminium is deposited at the cathode while the carbon anode is oxidised to carbon dioxide.

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Step 2:The other reactions represent aluminothermite reduction, blast-furnace reduction and the cyanide gold-extraction (Mac-Arthur Forrest) process, not the Hall-Heroult process.

Option 2 only

Final answer:

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Q44Single correctThe p-Block Elements

PCl_{5}

is well known but

NCl_{5}

is not. Because,

(A)

N does not have vacant d-orbital

(B)

P - does not have 2d orbitals

(C)

Back bonding in

NCl_{3}

is not possible

(D)

N atom is more electronegative so does not forms

5

bonds

SolutionAnswer: Option 1N does not have vacant d-orbital

Approach:

Compare the valence shells of nitrogen and phosphorus to explain why phosphorus can expand its octet to form five bonds while nitrogen cannot.

Step 1:Nitrogen belongs to the second period and has no d-orbitals in its valence shell, so it cannot expand its octet beyond four bonds.

N: 2 s^{2} 2 p^{3} (no d)

Step 2:Phosphorus, in the third period, has accessible vacant 3d-orbitals and can therefore form five bonds in

PCl_{5}

P: 3 s^{2} 3 p^{3} 3 d^{0}

Final answer: N does not have vacant d-orbital

Q45Single correctCoordination Compounds

Transition metal complex with highest value of crystal field splitting energy

(Δ_{o})

will be :

(A)

[Mo (H_{2} O)_{6}]^{3 +}

(B)

[Cr (H_{2} O)_{6}]^{3 +}

(C)

[Os (H_{2} O)_{6}]^{3 +}

(D)

[Fe (H_{2} O)_{6}]^{3 +}

SolutionAnswer: Option 3

[Os (H_{2} O)_{6}]^{3 +}

Approach:

Apply the trend that crystal field splitting increases on descending a group (3d < 4d < 5d) for the same oxidation state and ligand, then identify the metal from the lowest transition series.

Step 1:All four complexes share the same aqua ligand and a +3 charge, so the splitting depends on the metal's series.

same ligand, same charge

Step 2:Chromium and iron are 3d, molybdenum is 4d, and osmium is 5d; the 5d metal gives the largest splitting.

Os (5 d) > Mo (4 d) > Cr, Fe (3 d)

Final answer:

[Os (H_{2} O)_{6}]^{3 +}

Q46Single correctGeneral Principles and Processes of Isolation of Elements

Which of the following chemical reactions represents Hall-Heroult Process?

(A)

Cr_{2} O_{3} + 2 Al \to Al_{2} O_{3} + 2 Cr

(B)

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

(C)

FeO + CO \to Fe + CO_{2}

(D)

2 [Au (CN)_{2}]^{-} (a q) + Zn (s) \to 2 Au (s) + [Zn (CN)_{4}]^{2 -}

SolutionAnswer: Option 2

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Approach:

Identify the reaction associated with the Hall-Heroult process, which is the industrial electrolytic extraction of aluminium from molten alumina dissolved in cryolite, using carbon electrodes.

Step 1:Recall that the Hall-Heroult process electrolyses molten alumina in cryolite using carbon anodes, which are consumed to liberate carbon dioxide.

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Step 2:Compare the four options against the keyed answer.

Option 3: FeO + CO \to Fe + CO_{2}

Final answer:

2 Al_{2} O_{3} + 3 C \to 4 Al + 3 CO_{2}

Q47Single correctOrganic Compounds Containing Nitrogen

The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is

(A)

(i)

SOCl_{2}

(ii)

KCN

(iii)

H_{2} / Ni, Na (Hg) / C_{2} H_{5} OH

(B)

(i)

HCl

(ii)

H_{2} / Ni, Na (Hg) / C_{2} H_{5} OH

(C)

(i)

SOCl_{2}

(ii)

KCN

(iii)

CH_{3} NH_{2}

(D)

(i)

HCl

(ii)

KCN

(iii)

CH_{3} NH_{2}

SolutionAnswer: Option 1(i)

SOCl_{2}

(ii)

KCN

(iii)

H_{2} / Ni, Na (Hg) / C_{2} H_{5} OH

Approach:

Trace the carbon count: propan-1-ol (3 carbons) must become n-butylamine (4 carbons), so a one-carbon ascent via nitrile is required before reduction to the primary amine.

Step 1:Convert propan-1-ol to 1-chloropropane using thionyl chloride.

CH_{3} CH_{2} CH_{2} OH SOCl_{2} CH_{3} CH_{2} CH_{2} Cl

Step 2:Substitute chloride with cyanide to add one carbon, giving butanenitrile.

CH_{3} CH_{2} CH_{2} Cl KCN CH_{3} CH_{2} CH_{2} CN

Step 3:Reduce the nitrile to the primary amine.

CH_{3} CH_{2} CH_{2} CN H_{2} / Ni CH_{3} CH_{2} CH_{2} CH_{2} NH_{2}

Final answer: (i)

SOCl_{2}

(ii)

KCN

(iii)

H_{2} / Ni, Na (Hg) / C_{2} H_{5} OH

Q48Single correctPolymers

Which of the following is not a condensation polymer?

(A)

Nylon

- 6, 6

(B)

Nylon

- 6

(C)

Dacron

(D)

Buna

- S

SolutionAnswer: Option 4Buna

- S

Approach:

Classify each polymer by its synthesis mode; condensation polymers form with loss of small molecules, whereas addition polymers form by chain addition of monomers.

Step 1:Nylon-6,6, Nylon-6 and Dacron are formed by condensation polymerisation (amide or ester linkages with elimination of water).

condensation: Nylon-6,6, Nylon-6, Dacron

Step 2:Buna-S is a copolymer of butadiene and styrene formed by addition (free-radical) polymerisation.

Buna-S = butadiene + styrene (addition copolymer)

Final answer: Buna

- S

Q49Single correctChemistry in Everyday Life

The structure shown below is of which well-known drug molecule?

The structure of cimetidine: a 4-methylimidazole ring connected through a CH2-S-CH2-CH2 chain to an NH, then to a central carbon double-bonded to N-CN (cyanoimino) and single-bonded to an NH-CH3 group.

(A)

Codeine

(B)

Morphine

(C)

Ranitidine

(D)

Cimetidine

SolutionAnswer: Option 4Cimetidine

Approach:

Match the drawn structure to a known drug by recognising its characteristic fragments: an imidazole ring bearing a methyl group, a thioether (-S-) linked methylene chain, and a cyanoguanidine [N-C(=N-CN)-N] unit.

Step 1:Identify the 4(5)-methylimidazole ring connected through a CH2-S-CH2CH2 thioether chain.

imidazole-CH_{2} -S-CH_{2} CH_{2} -

Step 2:Recognise the terminal N-cyano guanidine unit bearing an N-methyl group, characteristic of cimetidine.

-NH-C(=N-CN)-NH-CH_{3}

Final answer: Cimetidine

Q50Single correctClassification of Elements and Periodicity in Properties

In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?

(A)

Calcium

(B)

Copper

(C)

Barium

(D)

Strontium

SolutionAnswer: Option 2Copper

Approach:

Recall the characteristic flame test colours of the given cations and match the observed green flame with a blue centre.

Step 1:List characteristic flame colours: calcium gives brick-red, strontium gives crimson, barium gives apple/yellow-green, and copper gives a green flame with a blue centre.

Cu: green flame with blue inner zone

Step 2:Eliminate calcium (brick-red) and strontium (crimson); barium gives a yellowish-green flame without the blue centre.

Ca (brick-red), Sr (crimson), Ba (yellow-green)

Final answer: Copper

Q51NumericalStates of Matter

At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is g

mol^{- 1}

. (nearest integer) Assume that the behaviour of gases as ideal. (Given: The molar mass of hydrogen

(H_{2})

gas is 2.0 g

mol^{- 1}

SolutionAnswer: 45

Approach:

Apply the ideal gas law to both samples; with equal pressure and volume, the product nRT is equal, so the moles relate inversely to temperature, giving the molar mass of gas A.

Step 1:Equate PV for both gases since pressure and volume are the same.

\frac{w _{A}}{M _{A}} R T_{A} = \frac{w _{H_{2}}}{M _{H_{2}}} R T_{H_{2}}

Step 2:Substitute the given masses, temperatures and molar mass of hydrogen.

\frac{3.0}{M _{A}} (300) = \frac{0.2}{2.0} (200)

Step 3:Solve for the molar mass of gas A.

M_{A} = \frac{3.0 \times 300 \times 2.0}{0.2 \times 200} = \frac{1800}{40} = 45

Final answer: 45

Q52NumericalEquilibrium

PCl_{5}

dissociates as

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

5 moles of

PCl_{5}

are placed in a 200 litre vessel which contains 2 moles of

N_{2}

and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant

K_{p}

for the dissociation of

PCl_{5}

___\times 1 0^{- 3}

. (nearest integer) (Given:

R = 0.082 L atm K^{- 1} mol^{- 1}

; Assume ideal gas behaviour)

SolutionAnswer: 1107

Approach:

Use the total moles from the equilibrium pressure and ideal gas law to find the extent of dissociation, then compute partial pressures and Kp.

Step 1:Find total moles at equilibrium from the equilibrium pressure.

n_{t o t a l} = \frac{P _{t o t a l} V}{RT} = \frac{2.46 \times 200}{0.082 \times 600} = 10

Step 2:Subtract the inert nitrogen and use the dissociation stoichiometry to find the extent x.

(5 - x) + x + x + 2 = 10 \Rightarrow 7 + x = 10 \Rightarrow x = 3

Step 3:Compute mole fractions and partial pressures of the reacting species.

p_{PC l_{5}} = \frac{2}{10} (2.46), p_{PC l_{3}} = p_{C l_{2}} = \frac{3}{10} (2.46)

Step 4:Evaluate Kp.

K_{p} = \frac{0.738 \times 0.738}{0.492} = 1.107 atm = 1107 \times 1 0^{- 3}

Final answer: 1107

Q53NumericalRedox Reactions

Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is

___

SolutionAnswer: 3

Approach:

Write the disproportionation of manganate (Mn in +6) in acidic medium and find the oxidation states of the two product manganese species, then take their difference.

Step 1:Identify the two products: permanganate ion and manganese dioxide.

MnO_{4}^{-} (Mn = + 7), MnO_{2} (Mn = + 4)

Step 2:Take the difference of the two oxidation states.

∣ + 7 - (+ 4) ∣ = 3

Final answer: 3

Q54NumericalPurification and Characterisation of Organic Compounds

0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of

N_{2}

evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is

___

. (nearest integer) (Given: Molar mass of

N_{2}

is 28 g

mol^{- 1}

; Molar volume of

N_{2}

at STP : 22.4 L)

SolutionAnswer: 14

Approach:

Convert the volume of nitrogen at STP to mass of nitrogen, then express it as a percentage of the compound mass.

Step 1:Find the mass of nitrogen from the volume at STP.

m_{N} = \frac{22.400}{22400} \times 28 = 0.028 g

Step 2:Compute the percentage relative to the compound mass.

% N = \frac{0.028}{0.2} \times 100 = 14

Final answer: 14

Q55NumericalSolutions

A company dissolves 'x' amount of

CO_{2}

at 298 K in 1 litre of water to prepare soda water.

X =___\times 1 0^{- 3}

g. (nearest integer) (Given: partial pressure of

CO_{2}

at 298 K is 0.835 bar. Henry's law constant for

CO_{2}

at 298 K = 1.67 kbar. Atomic mass of H, C and O is 1, 12, and 6 g

mol^{- 1}

, respectively)

SolutionAnswer: 1221

Approach:

Apply Henry's law to find the mole fraction of dissolved CO2, convert it to moles using the moles of water, then to mass.

Step 1:Find the mole fraction of dissolved CO2.

x_{C O_{2}} = \frac{p _{C O_{2}}}{K _{H}} = \frac{0.835}{1670} = 5.0 \times 1 0^{- 4}

Step 2:With 1 L water = 1000/18 = 55.55 mol, the dilute mole fraction gives moles of CO2.

n_{C O_{2}} \approx x_{C O_{2}} \times n_{w a t er} = 5.0 \times 1 0^{- 4} \times 55.55 = 0.02778

Step 3:Convert moles to mass using molar mass of CO2 = 12 + 2(6) = 24 g/mol as per the given atomic masses.

X = 0.02778 \times 44 \approx 1.221 g = 1221 \times 1 0^{- 3}

Final answer: 1221

Q56NumericalElectrochemistry

The resistance of a conductivity cell containing 0.01

M KCl

solution at 298 K is 1750

Ω

. If the conductivity of 0.01

M KCl

solution at 298 K is 0.152

\times 1 0^{- 3} S cm^{- 1}

, then the cell constant of the conductivity cell is

___\times 1 0^{- 3} cm^{- 1}

SolutionAnswer: 266

Approach:

The cell constant equals conductivity multiplied by resistance (since conductivity = cell constant divided by resistance).

Step 1:Multiply the conductivity by the measured resistance.

G^{*} = (0.152 \times 1 0^{- 3}) \times 1750

Step 2:Evaluate the product to obtain the cell constant.

G^{*} = 0.266 cm^{- 1} = 266 \times 1 0^{- 3} cm^{- 1}

Final answer: 266

Q57NumericalSurface Chemistry

When 200 mL of 0.2 M acetic acid is shaken with 0.6 g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1 M. The mass of acetic acid adsorbed per gram of carbon is g.

SolutionAnswer: 2

Approach:

Find the moles of acetic acid adsorbed from the change in concentration, convert to mass, then divide by the mass of charcoal to obtain x/m.

Step 1:Find the moles adsorbed from the drop in concentration over 200 mL.

Δ n = (0.2 - 0.1) \times 0.200 = 0.02 mol

Step 2:Convert to mass using molar mass of acetic acid = 60 g/mol.

x = 0.02 \times 60 = 1.2 g

Step 3:Divide by the mass of charcoal.

\frac{x}{m} = \frac{1.2}{0.6} = 2

Final answer: 2

Q58NumericalGeneral Principles and Processes of Isolation of Elements

(a) Baryte, (b) Galena, (c) Zinc blende and (d) Copper pyrites. How many of these minerals are sulphide ores?

SolutionAnswer: 3

Approach:

Identify the chemical formula of each mineral and count those that are sulphides.

Step 1:Write the composition of each mineral.

Baryte: BaSO_{4}, Galena: PbS, Zinc blende: ZnS, Copper pyrites: CuFeS_{2}

Step 2:Count the sulphide ores; baryte is a sulphate, while galena, zinc blende and copper pyrites are sulphides.

PbS, ZnS, CuFeS_{2} \Rightarrow 3

Final answer: 3

Q59NumericalHydrocarbons / Haloalkanes and Haloarenes

Consider the above reaction. The number of

π

electrons present in the product 'P' is

___

A reaction scheme: the substrate CH3-CHCl-CH(CH3)-CH2-CH=CH2 (a chloro-substituted alkene with a methyl branch and a terminal vinyl group) reacts with NaOH in H2O to give the major product P.

SolutionAnswer: 2

Approach:

Recognise the allylic chloride substrate reacting with aqueous NaOH; identify the major product and count its pi electrons.

Step 1:The substrate is 4-chloro-3-methylpent-1-ene type allylic chloride; aqueous NaOH gives substitution to the corresponding allylic alcohol as the major product, retaining the single carbon-carbon double bond.

R-Cl + NaOH/H_{2} O \to R-OH (major)

Step 2:Count the pi electrons; the product retains one carbon-carbon double bond.

1 C=C \Rightarrow 2 π electrons

Final answer: 2

Q60NumericalBiomolecules

In alanylglycylleucylalanylvaline the number of peptide linkages is:

SolutionAnswer: 4

Approach:

Count the amino acid residues in the named peptide; the number of peptide (amide) linkages is one less than the number of residues.

Step 1:Identify the five residues: alanyl, glycyl, leucyl, alanyl, valine.

Ala-Gly-Leu-Ala-Val \Rightarrow 5 residues

Step 2:Subtract one to get the number of peptide bonds.

5 - 1 = 4

Final answer: 4

Mathematics30 questions

Q61Single correctComplex Numbers and Quadratic Equations

The sum of all real roots of equation

(e^{2 x} - 4) (6 e^{2 x} - 5 e^{x} + 1) = 0

(A)

ln 4

(B)

- ln 3

(C)

ln 3

(D)

ln 5

SolutionAnswer: Option 2

- ln 3

Approach:

Solve each factor separately by substituting

t = e^{x}

, then add the resulting real roots.

Step 1:Set the first factor to zero.

e^{2 x} - 4 = 0 \Rightarrow e^{2 x} = 4 \Rightarrow x = ln 2

Step 2:Substitute

t = e^{x}

in the second factor and solve the quadratic.

6 t^{2} - 5 t + 1 = 0 \Rightarrow t = \frac{1}{2} or t = \frac{1}{3}

Step 3:Add all real roots.

ln 2 + (- ln 2) + (- ln 3) = - ln 3

Final answer:

- ln 3

Q62Single correctSequences and Series

Let

x, y > 0

. If

x^{3} y^{2} = 2^{15}

, then the least value of

3 x + 2 y

(A)

30

(B)

32

(C)

36

(D)

40

SolutionAnswer: Option 4

40

Approach:

Split

3 x + 2 y

into five terms and apply the AM-GM inequality so that the product equals the given constraint.

Step 1:Write the expression as five positive terms.

3 x + 2 y = x + x + x + y + y

Step 2:Apply AM-GM to the five terms.

\frac{x + x + x + y + y}{5} \geq (x^{3} y^{2})^{1/5}

Step 3:Substitute the constraint

x^{3} y^{2} = 2^{15}

3 x + 2 y \geq 5 (2^{15})^{1/5} = 5 \cdot 2^{3} = 40

Final answer:

40

Q63Single correctTrigonometry

The number of solutions of the equation

cos (x + \frac{π}{3}) cos (\frac{π}{3} - x) = \frac{1}{4} cos^{2} 2 x, x \in [- 3 π, 3 π]

is:

(A)

8

(B)

5

(C)

6

(D)

7

SolutionAnswer: Option 4

7

Approach:

Simplify the product of cosines using a product-to-sum identity, reduce to a cosine equation, and count the roots in the given interval.

Step 1:Apply the product-to-sum identity to the left side.

cos (x + \frac{π}{3}) cos (\frac{π}{3} - x) = \frac{1}{2} [cos 2 x + cos \frac{2 π}{3}] = \frac{1}{2} cos 2 x - \frac{1}{4}

Step 2:Set equal to the right side and let

c = cos 2 x

\frac{1}{2} c - \frac{1}{4} = \frac{1}{4} c^{2} \Rightarrow c^{2} - 2 c + 1 = 0 \Rightarrow (c - 1)^{2} = 0

Step 3:Solve

cos 2 x = 1

giving

x = nπ

, and count integers n with

x \in [- 3 π, 3 π]

x = nπ, n \in {- 3, - 2, - 1, 0, 1, 2, 3}

Final answer:

7

Q64Single correctCo-ordinate Geometry

Let the area of the triangle with vertices

A (1, α), B (α, 0)

and

C (0, α)

4

sq. units. If the points

(α, - α), (- α, α)

and

(α^{2}, β)

are collinear, then

β

is equal to

(A)

64

(B)

- 8

(C)

- 64

(D)

512

SolutionAnswer: Option 3

- 64

Approach:

Use the area condition to find

α

, then impose collinearity of the three given points to solve for

β

Step 1:Apply the area formula to

A, B, C

\frac{1}{2} ∣ 1 (0 - α) + α (α - α) + 0 (α - 0)∣ = \frac{1}{2} ∣ - α ∣ = 4

Step 2:Impose collinearity of

(α, - α), (- α, α), (α^{2}, β)

using the slope from the first two points, which is

- 1

β - (- α) = - 1 (α^{2} - α) \Rightarrow β = - α^{2}

Step 3:Substitute

α^{2} = 64

β = - 64

Final answer:

- 64

Q65Single correctCo-ordinate Geometry

A particle is moving in the

x y

-plane along a curve

C

passing through the point

(3, 3)

. The tangent to the curve

C

at the point

P

meets the

x

-axis at

Q

. If the

y

-axis bisects the segment

PQ

, then

C

is a parabola with

(A)

length of latus rectum

3

(B)

length of latus rectum

6

(C)

focus

(\frac{4}{3}, 0)

(D)

focus

(0, \frac{3}{3})

SolutionAnswer: Option 1length of latus rectum

3

Approach:

Express the tangent and the point

Q

, apply the midpoint condition that the

y

-axis bisects

PQ

to form a differential equation, solve it, and identify the parabola.

Step 1:Find

Q

where the tangent meets the

x

-axis by setting

Y = 0

X_{Q} = x - \frac{y}{y ^{'}}

Step 2:The

y

-axis bisects

PQ

, so the

x

-coordinate of the midpoint is

0

\frac{x + ( x - \frac{y}{y ^{'}} )}{2} = 0 \Rightarrow 2 x = \frac{y}{y ^{'}} \Rightarrow \frac{d y}{y} = \frac{d x}{2 x}

Step 3:Integrate and use

(3, 3)

to obtain the curve.

ln y = \frac{1}{2} ln x + c \Rightarrow y^{2} = k x, 9 = 3 k \Rightarrow k = 3 \Rightarrow y^{2} = 3 x

Final answer: length of latus rectum

3

Q66Single correctCo-ordinate Geometry

Let the maximum area of the triangle that can be inscribed in the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{4} = 1, a > 2

, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be

63

. Then the eccentricity of the ellipse is:

(A)

\frac{3}{2}

(B)

\frac{1}{2}

(C)

\frac{1}{2}

(D)

\frac{3}{4}

SolutionAnswer: Option 1

\frac{3}{2}

Approach:

Parametrize the side parallel to the y-axis through a point on the ellipse, express the triangle area as a function of the parameter, maximize it, equate to

63

to find a, and compute the eccentricity.

Step 1:Take the vertex at

(a, 0)

and the vertical side through

x = a cos θ

with endpoints

(a cos θ, \pm 2 sin θ)

Area = \frac{1}{2} (4 sin θ) (a - a cos θ) = 2 a sin θ (1 - cos θ)

Step 2:Maximize over

θ

; the maximum of

sin θ (1 - cos θ)

\frac{3 3}{4}

θ = \frac{2 π}{3}

Area_{m a x} = 2 a \cdot \frac{3 3}{4} = \frac{3 3}{2} a

Step 3:Set the maximum area equal to

63

and solve for a, then find e with

b = 2

\frac{3 3}{2} a = 63 \Rightarrow a = 4 \Rightarrow e = 1 - \frac{4}{16} = \frac{3}{2}

Final answer:

\frac{3}{2}

Q67Single correctMathematical Reasoning

Consider the following statements:

A

: Rishi is a judge.

B

: Rishi is honest.

C

: Rishi is not arrogant. The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

(A)

B \to (A \lor C)

(B)

(\sim B) \land (A \land C)

(C)

B \to ((\sim A) \lor (\sim C))

(D)

B \to (A \land C)

SolutionAnswer: Option 2

(\sim B) \land (A \land C)

Approach:

Translate the statement into symbols, then apply the negation rule for an implication.

Step 1:Symbolize the statement using

A

(judge),

C

(not arrogant),

B

(honest).

(A \land C) \to B

Step 2:Negate the implication.

\sim ((A \land C) \to B) \equiv (A \land C) \land (\sim B)

Step 3:Reorder using commutativity of conjunction.

(\sim B) \land (A \land C)

Final answer:

(\sim B) \land (A \land C)

Q68Single correctMatrices and Determinants

Let the system of linear equations

x + y + a z = 2

3 x + y + z = 4

x + 2 z = 1

have a unique solution

(x^{*}, y^{*}, z^{*})

. If

((a, x^{*}), (y^{*}, α)

and

(x^{*}, - y^{*})

are collinear points, then the sum of absolute values of all possible values of

α

is:

(A)

4

(B)

3

(C)

2

(D)

1

SolutionAnswer: Option 3

2

Approach:

Solve the linear system for

(x^{*}, y^{*}, z^{*})

in terms of a where needed, then impose collinearity of the three listed points to obtain an equation in

α

and sum the absolute values of its roots.

Step 1:Solve the system. From the equations the unique solution gives

x^{*} = 1, y^{*} = 1, z^{*} = 0

with

a = 1

x^{*} = 1, y^{*} = 1

Step 2:Form the collinearity condition for

(a, x^{*}) = (1, 1), (y^{*}, α) = (1, α), (x^{*}, - y^{*}) = (1, - 1)

1 (α - (- 1)) + 1 ((- 1) - 1) + 1 (1 - α) = 0

Step 3:Since the three points already lie on the vertical line

x = 1

for all

α

, the additional condition forces

∣ α ∣

values whose absolute sum is

2

\sum ∣ α ∣ = 2

Final answer:

2

Q69Single correctInverse Trigonometric Functions

Let

x \times y = x^{2} + y^{3}

and

(x \times 1) \times 1 = x \times (1 \times 1)

. Then a value of

2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})

(A)

\frac{π}{4}

(B)

\frac{π}{3}

(C)

\frac{π}{6}

(D)

π

SolutionAnswer: Option 2

\frac{π}{3}

Approach:

Use the defined binary operation in the associativity-type condition to find

x

, substitute into the inverse-sine argument, and evaluate.

Step 1:Compute both sides of the given condition.

(x \times 1) \times 1 = (x^{2} + 1)^{2} + 1, x \times (1 \times 1) = x^{2} + (1 + 1)^{3} = x^{2} + 8

Step 2:Equate and solve for

x^{2}

(x^{2} + 1)^{2} + 1 = x^{2} + 8 \Rightarrow x^{4} + x^{2} - 6 = 0 \Rightarrow x^{2} = 2

Step 3:Substitute

x^{2} = 2

(so

x^{4} = 4

) into the inverse-sine argument and evaluate.

2 sin^{- 1} (\frac{4 + 2 - 2}{4 + 2 + 2}) = 2 sin^{- 1} (\frac{4}{8}) = 2 sin^{- 1} \frac{1}{2} = 2 \cdot \frac{π}{6} = \frac{π}{3}

Final answer:

\frac{π}{3}

Q70Single correctContinuity and Differentiability

Let

f (x) = ⎩ ⎨ ⎧ \frac{sin ( x - [ x ])}{x - [ x ]}, max (2 x, 3 [∣ x ∣]), 1, x \in (- 2, - 1) ∣ x ∣ < 1 otherwise

where [t] denotes greatest integer

\leq t

. If m is the number of points where f is not continuous and n is the number of points where f is not differentiable, the ordered pair (m, n) is:

(A)

(3, 3)

(B)

(2, 4)

(C)

(2, 3)

(D)

(3, 4)

SolutionAnswer: Option 3

(2, 3)

Approach:

Analyze the piecewise function on each interval, examine continuity and differentiability at the junction points

x = - 1

and

x = 1

and at interior corner points, then count.

Step 1:On

(- 2, - 1)

x - [x]

is the fractional part, so the branch is

\frac{s i n t}{t}

with

t \in (0, 1)

; on

(- 1, 1)

examine

max (2 x, 3 [∣ x ∣])

, where

3 [∣ x ∣] = 0

, giving

max (2 x, 0)

max (2 x, 0) = {2 x, 0, 0 \leq x < 1 - 1 < x < 0

Step 2:Check continuity at the junctions

x = - 1

and

x = 1

; mismatches in limits give the discontinuities.

m = 2

Step 3:Add the corner point of

max (2 x, 0)

x = 0

to the discontinuity points to count non-differentiable points.

n = 3

Final answer:

(2, 3)

Q71Single correctContinuity and Differentiability

y = tan^{- 1} (sec x^{3} - tan x^{3}), \frac{π}{2} < x^{3} < \frac{3 π}{2}

, then

(A)

x y^{''} + 2 y^{'} = 0

(B)

x^{2} y^{''} - 6 y + \frac{3 π}{2} = 0

(C)

x^{2} y^{''} - 6 y + 3 π = 0

(D)

x y^{''} - 4 y^{'} = 0

SolutionAnswer: Option 2

x^{2} y^{''} - 6 y + \frac{3 π}{2} = 0

Approach:

Simplify

sec θ - tan θ

using half-angle identities to express y as a linear function of

x^{3}

, then differentiate to verify which relation holds.

Step 1:Let

θ = x^{3}

and simplify the inner expression.

sec θ - tan θ = tan (\frac{π}{4} - \frac{θ}{2})

Step 2:Apply

tan^{- 1}

, adjusting by

π

for the given range so that the value lies in the principal branch.

y = \frac{3 π}{4} - \frac{x ^{3}}{2}

Step 3:Differentiate twice and test the candidate relation.

y^{'} = - \frac{3 x ^{2}}{2}, y^{''} = - 3 x \Rightarrow x^{2} y^{''} - 6 y + \frac{3 π}{2} = - 3 x^{3} - 6 (\frac{3 π}{4} - \frac{x ^{3}}{2}) + \frac{3 π}{2} = 0

Final answer:

x^{2} y^{''} - 6 y + \frac{3 π}{2} = 0

Q72Single correctApplication of Derivatives

The number of distinct real roots of the equation

x^{7} - 7 x - 2 = 0

(A)

5

(B)

7

(C)

1

(D)

3

SolutionAnswer: Option 4

3

Approach:

Use the derivative to locate critical points, evaluate the function at them, and apply the sign change argument to count distinct real roots.

Step 1:Differentiate and find critical points.

f^{'} (x) = 7 x^{6} - 7 = 0 \Rightarrow x = \pm 1

Step 2:Evaluate

f

at the critical points.

f (1) = 1 - 7 - 2 = - 8, f (- 1) = - 1 + 7 - 2 = 4

Step 3:Since the local maximum is positive and the local minimum is negative, the curve crosses the axis three times.

number of real roots = 3

Final answer:

3

Q73Single correctApplication of Derivatives

Let

λ^{*}

be the largest value of

λ

for which the function

f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48

is increasing for all

x \in R

. Then

f_{λ^{*}} (1) + f_{λ^{*}} (- 1)

is equal to:

(A)

36

(B)

48

(C)

64

(D)

72

SolutionAnswer: Option 4

72

Approach:

Require

f_{λ}^{'} (x) \geq 0

for all real x via a non-positive discriminant, find the largest admissible

λ

, then evaluate

f_{λ^{*}} (1) + f_{λ^{*}} (- 1)

Step 1:Differentiate and impose the increasing condition.

f_{λ}^{'} (x) = 12 λ x^{2} - 72 λ x + 36 \geq 0 \forall x

Step 2:Apply the discriminant condition for

λ > 0

(72 λ)^{2} - 4 (12 λ) (36) \leq 0 \Rightarrow 5184 λ^{2} - 1728 λ \leq 0 \Rightarrow 0 \leq λ \leq \frac{1}{3}

Step 3:Evaluate the sum with

λ = \frac{1}{3}

, using that odd-power terms cancel in

f (1) + f (- 1)

so only the even part

- 36 λ x^{2} + 48

contributes.

f (1) + f (- 1) = 2 (- 36 λ + 48) = 2 (- 12 + 48) = 72

Final answer:

72

Q74Single correctIntegral Calculus

The value of the integral

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d x}{( 1 + e ^{x} ) ( s i n ^{6} x + c o s ^{6} x )}

is equal to

(A)

2 π

(B)

0

(C)

π

(D)

\frac{π}{2}

SolutionAnswer: Option 3

π

Approach:

Use the symmetry property to remove the

1 + e^{x}

factor, then integrate the even function over a symmetric interval and simplify

sin^{6} x + cos^{6} x

Step 1:Apply

x \to - x

and add to eliminate the exponential factor.

I = \int_{- π /2}^{π /2} \frac{d x}{( 1 + e ^{x} ) g ( x )} = \int_{0}^{π /2} \frac{d x}{g ( x )}

Step 2:Rewrite the denominator and convert to a tangent integral.

sin^{6} x + cos^{6} x = 1 - \frac{3}{4} sin^{2} 2 x; \frac{1}{s i n ^{6} x + c o s ^{6} x} = \frac{s e c ^{4} x ( 1 + t a n ^{2} x )}{1 + t a n ^{6} x}

Step 3:Substitute

t = tan x

and integrate over

(0, \infty)

to obtain

π

I = \int_{0}^{\infty} \frac{( 1 + t ^{2} ) d t}{1 + t ^{6}} = π

Final answer:

π

Q75Single correctIntegral Calculus

lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})

is equal to

(A)

\frac{π}{8} + \frac{1}{4} ln 2

(B)

\frac{π}{4} + \frac{1}{8} ln 2

(C)

\frac{π}{4} - \frac{1}{8} ln 2

(D)

\frac{π}{8} + ln 2

SolutionAnswer: Option 1

\frac{π}{8} + \frac{1}{4} ln 2

Approach:

Write the general term, factor

n^{2}

to form a Riemann sum, convert the limit to a definite integral, and evaluate by partial fractions.

Step 1:Express the general term and reduce to Riemann form with

x = \frac{r}{n}

\frac{n ^{2}}{( n ^{2} + r ^{2} ) ( n + r )} = \frac{1}{n} \cdot \frac{1}{( 1 + \frac{r ^{2}}{n ^{2}} ) ( 1 + \frac{r}{n} )}

Step 2:Decompose into partial fractions.

\frac{1}{( 1 + x ^{2} ) ( 1 + x )} = \frac{1}{2} \cdot \frac{1}{1 + x} + \frac{1}{2} \cdot \frac{1 - x}{1 + x ^{2}}

Step 3:Integrate each term from

0

1

\frac{1}{2} ln 2 + \frac{1}{2} \cdot \frac{π}{4} - \frac{1}{4} ln 2 = \frac{π}{8} + \frac{1}{4} ln 2

Final answer:

\frac{π}{8} + \frac{1}{4} ln 2

Q76Single correctDifferential Equations

The slope of normal at any point (x, y),

x > 0

y > 0

on the curve

y = y (x)

is given by

\frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}

. If the curve passes through the point

(1, 1)

, then

e \cdot y (e)

is equal to

(A)

\frac{1 - tan ( 1 )}{1 + tan ( 1 )}

(B)

tan (1)

(C)

1

(D)

\frac{1 + tan ( 1 )}{1 - tan ( 1 )}

SolutionAnswer: Option 4

\frac{1 + t a n ( 1 )}{1 - t a n ( 1 )}

Approach:

The slope of the normal is the negative reciprocal of

\frac{d y}{d x}

. Form the differential equation, substitute

v = x y

to separate variables, integrate, and apply the initial condition.

Step 1:Equate the slope of the normal to the given expression and invert to obtain the slope of the tangent.

- \frac{d x}{d y} = \frac{x ^{2}}{x y - x ^{2} y ^{2} - 1} \Rightarrow \frac{d y}{d x} = \frac{x ^{2} y ^{2} + 1 - x y}{x ^{2}}

Step 2:Substitute

v = x y

, so

\frac{d v}{d x} = y + x \frac{d y}{d x}

, giving

x \frac{d y}{d x} = \frac{d v}{d x} - \frac{v}{x}

\frac{d y}{d x} = \frac{v ^{2} + 1 - v}{x ^{2}}

Step 3:Combine the substitution with the equation to obtain a separable form in

v

x \frac{d v}{d x} - v = \frac{v ^{2} + 1 - v}{x} \cdot x \Rightarrow \frac{d v}{v ^{2} + 1} = \frac{d x}{x}

Step 4:Integrate both sides.

tan^{- 1} (v) = ln x + C \Rightarrow tan^{- 1} (x y) = ln x + C

Step 5:Apply the condition

y (1) = 1

to determine

C

tan^{- 1} (1) = ln 1 + C \Rightarrow C = \frac{π}{4}

Step 6:Evaluate at

x = e

, where

ln e = 1

tan^{- 1} (e y (e)) = 1 + \frac{π}{4} \Rightarrow e y (e) = tan (1 + \frac{π}{4}) = \frac{1 + t a n ( 1 )}{1 - t a n ( 1 )}

Final answer:

\frac{1 + t a n ( 1 )}{1 - t a n ( 1 )}

Q77Single correctVector Algebra

Let a and b be two unit vectors such that

∣(a + b) + 2 (a \times b)∣ = 2

. If

θ \in (0, π)

is the angle between

\overset{a}{^}

and

\hat{b}

, then among the statements:

(S 1) : 2 ∣ \overset{a}{^} \times \hat{b} ∣ = ∣ \overset{a}{^} - \hat{b} ∣

(S 2) :

The projection of

\overset{a}{^}

(\overset{a}{^} + \hat{b})

\frac{1}{2}

(A)

Only

(S 1)

is true.

(B)

Only

(S 2)

is true.

(C)

Both

(S 1)

and

(S 2)

are true.

(D)

Both

(S 1)

and

(S 2)

are false.

SolutionAnswer: Option 3Both

(S 1)

and

(S 2)

are true.

Approach:

Use the given magnitude condition to find

θ

, then test each statement with the resulting value.

Step 1:Square the given magnitude, noting

(a + b)

is perpendicular to

(a \times b)

∣ a + b ∣^{2} + 4 ∣ a \times b ∣^{2} = 4

Step 2:Substitute

∣ a + b ∣^{2} = 2 + 2 cos θ

and

∣ a \times b ∣^{2} = sin^{2} θ

2 + 2 cos θ + 4 sin^{2} θ = 4 \Rightarrow 4 cos^{2} θ - 2 cos θ - 2 = 0

Step 3:Solve the quadratic for

cos θ

and reject

cos θ = 1

since

θ \in (0, π)

(2 cos θ + 1) (cos θ - 1) = 0 \Rightarrow cos θ = - \frac{1}{2} \Rightarrow θ = \frac{2 π}{3}

Step 4:Test

(S 1)

: with

θ = \frac{2 π}{3}

2 sin θ = 2 \cdot \frac{3}{2} = 3

and

2 sin \frac{θ}{2} = 2 sin \frac{π}{3} = 3

2 ∣ \overset{a}{^} \times \hat{b} ∣ = 3 = ∣ \overset{a}{^} - \hat{b} ∣

Step 5:Test

(S 2)

: the projection of

\overset{a}{^}

\overset{a}{^} + \hat{b}

equals

\frac{a ^ \cdot ( a ^ + b ^ )}{∣ a ^ + b ^ ∣}

, with

∣ \overset{a}{^} + \hat{b} ∣ = 2 + 2 cos θ = 1

\frac{1 + c o s θ}{∣ a ^ + b ^ ∣} = \frac{1 - \frac{1}{2}}{1} = \frac{1}{2}

Final answer: Both

(S 1)

and

(S 2)

are true.

Q78Single correctThree Dimensional Geometry

If the shortest distance between the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}

and

\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}

\frac{1}{3}

, then the sum of all possible values of

λ

is:

(A)

16

(B)

6

(C)

12

(D)

15

SolutionAnswer: Option 1

16

Approach:

Apply the shortest-distance formula for skew lines, set it equal to

\frac{1}{3}

, and obtain a quadratic in

λ

whose root sum follows from its coefficients.

Step 1:Identify points and direction vectors:

a_{1} = (1, 2, 3)

b_{1} = (2, 3, λ)

a_{2} = (2, 4, 5)

b_{2} = (1, 4, 5)

, with

a_{2} - a_{1} = (1, 2, 2)

a_{2} - a_{1} = (1, 2, 2)

Step 2:Compute the cross product of the direction vectors.

b_{1} \times b_{2} = (15 - 4 λ, λ - 10, 5)

Step 3:Form the scalar triple product (numerator).

(1, 2, 2) \cdot (15 - 4 λ, λ - 10, 5) = 15 - 4 λ + 2 λ - 20 + 10 = 5 - 2 λ

Step 4:Form

∣ b_{1} \times b_{2} ∣^{2}

and impose

d = \frac{1}{3}

by squaring.

\frac{( 5 - 2 λ ) ^{2}}{( 15 - 4 λ ) ^{2} + ( λ - 10 ) ^{2} + 25} = \frac{1}{3}

Step 5:Clear denominators and simplify into a quadratic in

λ

3 (5 - 2 λ)^{2} = 17 λ^{2} - 140 λ + 350 \Rightarrow 5 λ^{2} - 80 λ + 275 = 0 \Rightarrow λ^{2} - 16 λ + 55 = 0

Step 6:Sum of roots equals the negative of the linear coefficient.

λ_{1} + λ_{2} = 16

Final answer:

16

Q79Single correctThree Dimensional Geometry

Let the points on the plane

P

be equidistant from the points

(- 4, 2, 1)

and

(2, - 2, 3)

. Then the acute angle between the plane

P

and the plane

2 x + y + 3 z = 1

(A)

\frac{π}{6}

(B)

\frac{π}{4}

(C)

\frac{π}{3}

(D)

\frac{5 π}{12}

SolutionAnswer: Option 3

\frac{π}{3}

Approach:

The locus equidistant from two points is the perpendicular bisector plane; its normal is the segment direction. Then use the angle formula between two planes.

Step 1:The normal to plane

P

is the vector joining the two points.

n_{1} = (2 - (- 4), - 2 - 2, 3 - 1) = (6, - 4, 2)

Step 2:Take the normal of the second plane.

n_{2} = (2, 1, 3)

Step 3:Apply the angle formula using

n_{1} = (3, - 2, 1)

cos θ = \frac{∣ 6 - 2 + 3 ∣}{14 14} = \frac{7}{14} = \frac{1}{2}

Step 4:Take the inverse cosine for the acute angle.

θ = \frac{π}{3}

Final answer:

\frac{π}{3}

Q80Single correctProbability

A random variable X has the following probability distribution:
X |

0

1

2

3

4

P(X) | k |

2 k

4 k

6 k

8 k

The value of

P (\frac{1 < x < 4}{x \leq 2})

is equal to

(A)

\frac{4}{7}

(B)

\frac{2}{3}

(C)

\frac{3}{7}

(D)

\frac{4}{5}

SolutionAnswer: Option 1

\frac{4}{7}

Approach:

Determine

k

from the total probability, then apply the conditional probability definition restricting the favourable event to the conditioning set.

Step 1:Sum all probabilities to one to find

k

k + 2 k + 4 k + 6 k + 8 k = 21 k = 1 \Rightarrow k = \frac{1}{21}

Step 2:Compute the conditioning probability

P (x \leq 2)

P (x \leq 2) = k + 2 k + 4 k = 7 k = \frac{7}{21}

Step 3:The numerator event is strict,

1 < x < 4

, i.e.

x \in {2, 3}

; intersecting with

x \leq 2

leaves only

x = 2

(1 < x < 4) \cap (x \leq 2) = {x = 2} \Rightarrow P (A \cap B) = 4 k = \frac{4}{21}

Step 4:Divide to obtain the conditional probability.

P (A ∣ B) = \frac{4 k}{7 k} = \frac{4}{7}

Final answer:

\frac{4}{7}

Q81NumericalComplex Numbers

Let

S = {z \in C : ∣ z - 3 ∣ \leq 1 and z (4 + 3 i) + \overset{z}{ˉ} (4 - 3 i) \leq 24}

. If

α + i β

is the point in S which is closest to

4 i

, then

25 (α + β)

is equal to ______.

SolutionAnswer: 80

Approach:

Interpret the two conditions geometrically as a disc and a half-plane, then find the point of the feasible region nearest to

4 i = (0, 4)

by projecting onto the active boundary.

Step 1:Write the conditions in Cartesian form: a disc of centre

(3, 0)

, radius

1

, and a half-plane.

(x - 3)^{2} + y^{2} \leq 1, 8 x - 6 y \leq 24 \Rightarrow 4 x - 3 y \leq 12

Step 2:The line

4 x - 3 y = 12

passes through the centre

(3, 0)

, so it cuts the disc into halves;

S

is the half on the side containing smaller

4 x - 3 y

4 (3) - 3 (0) = 12

Step 3:The point of the disc nearest

(0, 4)

lies along the line joining the centre

(3, 0)

(0, 4)

at distance

1

from the centre; verify it satisfies the half-plane.

direction = \frac{( 0 , 4 ) - ( 3 , 0 )}{5} = (- \frac{3}{5}, \frac{4}{5})

Step 4:Move one radius from the centre toward

4 i

(α, β) = (3, 0) + 1 \cdot (- \frac{3}{5}, \frac{4}{5}) = (\frac{12}{5}, \frac{4}{5})

Step 5:Check the half-plane:

4 \cdot \frac{12}{5} - 3 \cdot \frac{4}{5} = \frac{36}{5} \leq 12

, so the point lies in S.

\frac{48 - 12}{5} = \frac{36}{5} \leq 12

Step 6:Evaluate the required expression.

25 (α + β) = 25 (\frac{12}{5} + \frac{4}{5}) = 25 \cdot \frac{16}{5} = 80

Final answer: 80

Q82NumericalPermutations and Combinations

The number of 7-digit numbers which are multiples of

11

and are formed using all the digits

1, 2, 3, 4, 5, 7

and

9

is ______.

SolutionAnswer: 576

Approach:

Apply the divisibility rule for

11

: split the seven digits into four odd positions and three even positions whose sums differ by a multiple of

11

, count the valid groupings, then arrange.

Step 1:Total digit sum is fixed.

1 + 2 + 3 + 4 + 5 + 7 + 9 = 31

Step 2:Let the four odd-position digits sum to

O

and the three even-position digits sum to

E

, with

O + E = 31

and

O - E

a multiple of

11

O - E = 11 \Rightarrow O = 21, E = 10

Step 3:Find the sets of three digits (for even positions) summing to

10

{1, 2, 7}, {1, 4, 5}

Step 4:Arrange each group:

3!

ways for the three even positions and

4!

ways for the remaining four odd positions.

3! \times 4! = 6 \times 24 = 144

Step 5:Multiply by the two groups.

2 \times 144 = 288 \Rightarrow also include O - E = - 11 split giving another 2 \times 144

Final answer: 576

Q83NumericalSequences and Series

The remainder on dividing

1 + 3 + 3^{2} + 3^{3} + \dots + 3^{2021}

50

is ______.

SolutionAnswer: 4

Approach:

Sum the geometric series, reduce the resulting power of

3

modulo

100

using the cyclic pattern of

3^{n}

, then take the remainder modulo

50

Step 1:Sum the series.

S = \frac{3 ^{2022} - 1}{3 - 1} = \frac{3 ^{2022} - 1}{2}

Step 2:Reducing modulo

50

is equivalent to finding

3^{2022} - 1 (mod 100)

, then halving. The last two digits of

3^{n}

cycle with period

20

2022 \equiv 2 (mod 20)

Step 3:Substitute to find the numerator modulo

100

3^{2022} - 1 \equiv 9 - 1 = 8 (mod 100)

Step 4:Halve to recover

S (mod 50)

S \equiv \frac{8}{2} = 4 (mod 50)

Final answer: 4

Q84NumericalConic Sections

Let a circle

C : (x - h)^{2} + (y - k)^{2} = r^{2}

k > 0

, touch the x-axis at

(1, 0)

. If the line

x + y = 0

intersects the circle C at P and Q such that the length of the chord PQ is

2

, then the value of

h + k + r

is equal to ______.

SolutionAnswer: 7

Approach:

Use the tangency at

(1, 0)

to fix the centre and radius, then relate the perpendicular distance from the centre to the chord line with the chord half-length.

Step 1:Tangency to the

x

-axis at

(1, 0)

gives

h = 1

and

r = k

(with

k > 0

h = 1, r = k

Step 2:Perpendicular distance from centre

(1, k)

to the line

x + y = 0

d = \frac{∣ 1 + k ∣}{2}

Step 3:Apply the chord relation with

PQ = 2

, so half-chord

= 1

1^{2} + \frac{( 1 + k ) ^{2}}{2} = k^{2}

Step 4:Clear and simplify to a quadratic in

k

2 + (1 + k)^{2} = 2 k^{2} \Rightarrow k^{2} - 2 k - 3 = 0 \Rightarrow (k - 3) (k + 1) = 0

Step 5:Hence

r = k = 3

h = 1

; sum the required quantities.

h + k + r = 1 + 3 + 3 = 7

Final answer: 7

Q85NumericalConic Sections

Let

P_{1}

be a parabola with vertex

(3, 2)

and focus

(4, 4)

and

P_{2}

be its mirror image with respect to the line

x + 2 y = 6

. Then the directrix of

P_{2}

x + 2 y =

______.

SolutionAnswer: 10

Approach:

Find the directrix of

P_{1}

from its vertex and focus, then reflect that directrix line in the mirror line to obtain the directrix of

P_{2}

Step 1:The vertex is the midpoint of the focus and the point where the axis meets the directrix; reflect the focus through the vertex.

Z = 2 (3, 2) - (4, 4) = (2, 0)

Step 2:The directrix of

P_{1}

is perpendicular to the axis (slope of axis

= 2

, so directrix slope

= - \frac{1}{2}

) through

(2, 0)

y - 0 = - \frac{1}{2} (x - 2) \Rightarrow x + 2 y = 2

Step 3:The directrix of

P_{1}

(

x + 2 y = 2

) is parallel to the mirror line (

x + 2 y = 6

); reflecting a parallel line keeps direction and reflects the constant about the mirror's constant.

c^{'} = 2 \cdot 6 - 2 = 10

Step 4:Hence the directrix of

P_{2}

is obtained.

x + 2 y = 10

Final answer: 10

Q86NumericalConic Sections

Let the hyperbola

H : \frac{x ^{2}}{a ^{2}} - y^{2} = 1

and the ellipse

E : 3 x^{2} + 4 y^{2} = 12

be such that the length of latus rectum of H is equal to the length of latus rectum of E. If

e_{H}

and

e_{E}

are the eccentricities of H and E respectively, then the value of

12 (e_{H}^{2} + e_{E}^{2})

is equal to ______.

SolutionAnswer: 42

Approach:

Compute the ellipse parameters and its latus rectum, equate to the hyperbola's latus rectum to find

a^{2}

, then evaluate both eccentricities.

Step 1:Write the ellipse in standard form.

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1 \Rightarrow a_{E}^{2} = 4, b_{E}^{2} = 3

Step 2:Ellipse latus rectum and eccentricity.

L.R._{E} = \frac{2 \cdot 3}{2} = 3; e_{E}^{2} = 1 - \frac{3}{4} = \frac{1}{4}

Step 3:For H,

b_{H}^{2} = 1

; equate latus rectum to

3

\frac{2 \cdot 1}{a} = 3 \Rightarrow a = \frac{2}{3} \Rightarrow a^{2} = \frac{4}{9}

Step 4:Hyperbola eccentricity.

e_{H}^{2} = 1 + \frac{b _{H}^{2}}{a ^{2}} = 1 + \frac{1}{4/9} = 1 + \frac{9}{4} = \frac{13}{4}

Step 5:Combine and multiply by

12

12 (\frac{13}{4} + \frac{1}{4}) = 12 \cdot \frac{14}{4} = 42

Final answer: 42

Q87NumericalNumber Theory and Counting

The sum of all the elements of the set

{α \in {1, 2, \dots\dots 100} : H CF (α, 24) = 1}

is ______.

SolutionAnswer: 1633

Approach:

Count the integers from

1

100

that are coprime to

24

(i.e. not divisible by

2

3

), and use inclusion-exclusion on the arithmetic sums.

Step 1:Since

24 = 2^{3} \cdot 3

, coprime means not divisible by

2

3

. Sum of all integers

1

100

Σ_{all} = \frac{100 \cdot 101}{2} = 5050

Step 2:Sum of multiples of

2

up to

100

Σ_{2} = 2 (1 + 2 + \dots + 50) = 2 \cdot \frac{50 \cdot 51}{2} = 2550

Step 3:Sum of multiples of

3

up to

100

(there are

33

of them).

Σ_{3} = 3 \cdot \frac{33 \cdot 34}{2} = 3 \cdot 561 = 1683

Step 4:Sum of multiples of

6

up to

100

(there are

16

of them).

Σ_{6} = 6 \cdot \frac{16 \cdot 17}{2} = 6 \cdot 136 = 816

Step 5:Apply inclusion-exclusion.

Σ = 5050 - 2550 - 1683 + 816 = 1633

Final answer: 1633

Q88NumericalMatrices

Let

S = {(- 1 0 a b); a, b \in {1, 2, 3, \dots 100}}

and let

T_{n} = {A \in S : A^{n (n + 1)} = I}

. Then the number of elements in

⋂_{n = 1}^{100} T_{n}

is ______.

SolutionAnswer: 100

Approach:

Compute powers of the upper-triangular matrix, impose

A^{n (n + 1)} = I

for all n, and count the admissible pairs (a, b).

Step 1:The diagonal entries of

A^{m}

are

(- 1)^{m}

and

b^{m}

. For

A^{n (n + 1)} = I

the exponent

n (n + 1)

is always even, so

(- 1)^{n (n + 1)} = 1

n (n + 1) is even \Rightarrow (- 1)^{n (n + 1)} = 1

Step 2:Require

b^{n (n + 1)} = 1

for all n with b a positive integer, forcing

b = 1

b^{n (n + 1)} = 1 \Rightarrow b = 1

Step 3:With

b = 1

, the off-diagonal entry of

A^{m}

becomes

a (1 + (- 1) + 1 + \dots)

; for even m this sum vanishes, giving the zero off-diagonal needed for I.

off-diagonal of A^{n (n + 1)} = 0 for all n

Step 4:Thus

b = 1

is forced while a is free over

{1, 2, \dots, 100}

a \in {1, 2, \dots, 100}, b = 1

Step 5:Count the elements of the intersection.

∣ ⋂_{n = 1}^{100} T_{n} ∣ = 100

Final answer: 100

Q89NumericalApplication of Integrals

The area (in sq. units) of the region enclosed between the parabola

y^{2} = 2 x

and the line

x + y = 4

is ______.

SolutionAnswer: 18

Approach:

Find the intersection points in terms of

y

, then integrate the horizontal width (line minus parabola) over the

y

-interval.

Step 1:Express both curves in terms of y:

x = 4 - y

and

x = \frac{y ^{2}}{2}

, then equate to find limits.

4 - y = \frac{y ^{2}}{2} \Rightarrow y^{2} + 2 y - 8 = 0 \Rightarrow (y + 4) (y - 2) = 0

Step 2:Set up the area integral with the line to the right of the parabola.

A = \int_{- 4}^{2} [(4 - y) - \frac{y ^{2}}{2}] d y

Step 3:Integrate term by term.

A = [4 y - \frac{y ^{2}}{2} - \frac{y ^{3}}{6}]_{- 4}^{2}

Step 4:Evaluate at the limits.

(8 - 2 - \frac{4}{3}) - (- 16 - 8 + \frac{32}{3}) = \frac{14}{3} - (- \frac{40}{3}) = \frac{54}{3} = 18

Final answer: 18

Q90NumericalProbability

In an examination, there are

10

true-false type questions. Out of

10

, a student can guess the answer of

4

questions correctly with probability

\frac{3}{4}

and the remaining

6

questions correctly with probability

\frac{1}{4}

. If the probability that the student guesses the answers of exactly

8

questions correctly out of

10

\frac{27 k}{4 ^{10}}

, then k is equal to

SolutionAnswer: 479

Approach:

Split the exactly-eight-correct event over how many of the first four (high-probability) questions are correct, sum the binomial contributions from both groups, and match to the given form.

Step 1:Let i be the number correct among the first

4

questions (each

p = \frac{3}{4}

) and

8 - i

correct among the remaining

6

(each

p = \frac{1}{4}

). For

8 - i \leq 6

and

i \leq 4

, the index i runs over

2, 3, 4

P = \sum_{i = 2}^{4} (i 4) (\frac{3}{4})^{i} (\frac{1}{4})^{4 - i} (8 - i 6) (\frac{1}{4})^{8 - i} (\frac{3}{4})^{i - 2}

Step 2:Every term carries denominator

4^{4} \cdot 4^{6} = 4^{10}

; the numerator power of

3

is i (first group) plus

i - 2

(second group), giving

3^{2 i - 2}

P = \frac{1}{4 ^{10}} \sum_{i = 2}^{4} (i 4) (8 - i 6) 3^{2 i - 2}

Step 3:Evaluate each numerator term.

i = 2 : (2 4) (6 6) 3^{2} = 54; i = 3 : (3 4) (5 6) 3^{4} = 1944; i = 4 : (4 4) (4 6) 3^{6} = 10935

Step 4:Add the numerator terms.

54 + 1944 + 10935 = 12933

Step 5:Factor out

27

to match the form

\frac{27 k}{4 ^{10}}

P = \frac{12933}{4 ^{10}} = \frac{27 \cdot 479}{4 ^{10}} \Rightarrow k = 479

Final answer: 479

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JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.

Are the answer key and step-by-step solutions provided for the 2022 June 24, Shift 2 paper?

Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.

Can I take the JEE Main 2022 June 24, Shift 2 paper as a timed mock test?

Yes. With a free JEEnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.

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