MHT CET 2026 18th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

For the vector

$\overrightarrow{𝐴}=2\widehat{𝑖}+3\widehat{𝑗},$

its magnitude is

$\left|\overrightarrow{𝐴}\right|=\sqrt{\surd 2^2+3^2}=\sqrt{\surd 13}.$

If $𝛼$ is the angle with the $𝑥$-axis, then

$\cos 𝛼=\frac{𝑥\text{-component}}{\left|\overrightarrow{𝐴}\right|}=\frac{2}{\sqrt{\surd 13}},$

so

$𝛼={\cos }^{-1}\left(\frac{2}{\sqrt{\surd 13}}\right).$

If $𝛽$ is the angle with the $𝑦$-axis, then

$\cos 𝛽=\frac{𝑦\text{-component}}{\left|\overrightarrow{𝐴}\right|}=\frac{3}{\sqrt{\surd 13}},$

so

$𝛽={\cos }^{-1}\left(\frac{3}{\sqrt{\surd 13}}\right).$

Hence, the correct answer is Option B.

For addition or subtraction, the maximum absolute errors are added.

For the sum:

${𝑅}_1+{𝑅}_2=350+140=490\Omega$

Absolute error:

$\Delta 𝑅=3+4=7\Omega$

Percentage error:

$\frac{7}{490}\times 100=1.43\%$

For the difference:

${𝑅}_1-{𝑅}_2=350-140=210\Omega$

The absolute error is again:

$\Delta 𝑅=3+4=7\Omega$

Percentage error:

$\frac{7}{210}\times 100=3.33\%$

Thus, the percentage errors in the sum and difference are respectively

$\overline{)1.43\%,3.33\%}$

Hence, Option A is correct.

To move from one end of a diameter to the other along the circular path, the body covers a semicircle.

The angular displacement for a semicircle is

$𝜃=𝜋\text{radian}$

Given time,

$𝑡=3\text{s}$

Angular speed is

$𝜔=\frac{𝜃}{𝑡}=\frac{𝜋}{3}{\text{rad s}}^{-1}$

Hence, the correct answer is Option C:

$\overline{)\frac{𝜋}{3}{\text{rad s}}^{-1}}$

The rope has mass $\frac{𝑀}{2}$. So, the total mass of the block-rope system is

$𝑀+\frac{𝑀}{2}=\frac{3𝑀}{2}.$

Hence, the common acceleration is

$𝑎=\frac{𝐹}{\frac{3𝑀}{2}}=\frac{2𝐹}{3𝑀}.$

To find the tension at the middle of the rope, consider the system consisting of the block and the half of the rope attached to it.

Its total mass is

$𝑀+\frac{1}{2}\left(\frac{𝑀}{2}\right)=𝑀+\frac{𝑀}{4}=\frac{5𝑀}{4}.$

The tension $𝑇$ at the middle accelerates this part of the system. Thus,

$𝑇=\frac{5𝑀}{4}\times \frac{2𝐹}{3𝑀}=\frac{5𝐹}{6}.$

So, the correct answer is

$\overline{)\frac{5𝐹}{6}}$

Option D.

For a rotating body, rotational kinetic energy is

$𝐾=\frac{1}{2}𝐿𝜔$

Since $𝜔=2𝜋𝑓$, tripling the frequency triples the angular velocity:

${𝜔}^{\prime }=3𝜔$

The new kinetic energy is one-third of the initial kinetic energy:

${𝐾}^{\prime }=\frac{𝐾}{3}$

Using ${𝐾}^{\prime }=\frac{1}{2}{𝐿}^{\prime }{𝜔}^{\prime }$,

$\frac{1}{2}{𝐿}^{\prime }\left(3𝜔\right)=\frac{1}{3}\left(\frac{1}{2}𝐿𝜔\right)$

Cancelling common terms gives

$3{𝐿}^{\prime }=\frac{𝐿}{3}$

Hence,

${𝐿}^{\prime }=\frac{𝐿}{9}$

So, the correct answer is Option D: $\frac{𝐿}{9}$.

For a body rolling without slipping down an inclined plane,

$𝑎=\frac{𝑔\sin 𝜃}{1+\frac{𝐼}{𝑀{𝑅}^2}}$

For a solid cylinder,

${𝐼}_{𝑐}=\frac{1}{2}𝑀{𝑅}^2$

So,

${𝑎}_{𝑐}=\frac{𝑔\sin 𝜃}{1+\frac{1}{2}}=\frac{2}{3}𝑔\sin 𝜃$

For a solid sphere,

${𝐼}_{𝑠}=\frac{2}{5}𝑀{𝑅}^2$

Hence,

${𝑎}_{𝑠}=\frac{𝑔\sin 𝜃}{1+\frac{2}{5}}=\frac{5}{7}𝑔\sin 𝜃$

Thus,

$\frac{{𝑎}_{𝑐}}{{𝑎}_{𝑠}}=\frac{\frac{2}{3}}{\frac{5}{7}}=\frac{14}{15}$

So the correct answer is Option A, $\overline{)\frac{14}{15}}$.

Here, pure rolling is assumed. A perfectly smooth plane would provide no friction, so both bodies would simply slide with the same acceleration.

For a hollow sphere, the moment of inertia about an axis passing through its center is:

${𝐼}_{𝐶𝑀}=\frac{2}{3}𝑀{𝑅}^2$

The axis $𝑋𝑌$ is parallel to this central axis and is at a distance $ℎ$ from the center. So, by the parallel axis theorem:

${𝐼}_{𝑋𝑌}={𝐼}_{𝐶𝑀}+𝑀{ℎ}^2$

Substitute the value of ${𝐼}_{𝐶𝑀}$:

${𝐼}_{𝑋𝑌}=\frac{2}{3}𝑀{𝑅}^2+𝑀{ℎ}^2$

The radius of gyration about the axis $𝑋𝑌$ is given as $𝐾=𝑅$. We know that moment of inertia can be written as:

${𝐼}_{𝑋𝑌}=𝑀{𝐾}^2=𝑀{𝑅}^2$

Now compare both values of ${𝐼}_{𝑋𝑌}$:

$𝑀{𝑅}^2=\frac{2}{3}𝑀{𝑅}^2+𝑀{ℎ}^2$

Cancel $𝑀$ from all terms:

${𝑅}^2=\frac{2}{3}{𝑅}^2+{ℎ}^2$

So,

${ℎ}^2={𝑅}^2-\frac{2}{3}{𝑅}^2$

${ℎ}^2=\frac{1}{3}{𝑅}^2$

Taking square root:

$ℎ=\frac{𝑅}{\sqrt{\surd 3}}$

For a simple pendulum,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

A seconds pendulum has the same time period at both places. Hence, $𝑙\propto 𝑔$.

On the Earth’s surface,

${𝑔}_{𝑠}=\frac{𝐺𝑀}{{𝑅}^2}$

At altitude $ℎ$,

${𝑔}_{ℎ}=\frac{𝐺𝑀}{(𝑅+ℎ{)}^2}$

Thus,

$\frac{{𝑙}_{𝑠}}{{𝑙}_{ℎ}}=\frac{{𝑔}_{𝑠}}{{𝑔}_{ℎ}}=\frac{(𝑅+ℎ{)}^2}{{𝑅}^2}$

Taking square roots,

$\frac{\sqrt{\surd {𝑙}_{𝑠}}}{\sqrt{\surd {𝑙}_{ℎ}}}=\frac{𝑅+ℎ}{𝑅}$

So,

$𝑅\sqrt{\surd {𝑙}_{𝑠}}=(𝑅+ℎ)\sqrt{\surd {𝑙}_{ℎ}}$

$𝑅(\sqrt{\surd {𝑙}_{𝑠}}-\sqrt{\surd {𝑙}_{ℎ}})=ℎ\sqrt{\surd {𝑙}_{ℎ}}$

Hence, the radius of the Earth is

$\overline{)𝑅=\frac{ℎ\sqrt{\surd {𝑙}_{ℎ}}}{\sqrt{\surd {𝑙}_{𝑠}}-\sqrt{\surd {𝑙}_{ℎ}}}}$

So, the correct answer is Option A.

For a simple pendulum,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

At a height $ℎ$ above Earth’s surface, acceleration due to gravity is

${𝑔}_{ℎ}=𝑔{\left(\frac{𝑅}{𝑅+ℎ}\right)}^2$

Here, $ℎ=2𝑅$. So,

${𝑔}_{ℎ}=𝑔{\left(\frac{𝑅}{𝑅+2𝑅}\right)}^2=𝑔{\left(\frac{1}{3}\right)}^2=\frac{𝑔}{9}$

Since $𝑇\propto \frac{1}{\sqrt{\surd 𝑔}}$,

$\frac{{𝑇}_2}{{𝑇}_1}=\sqrt{\surd \frac{𝑔}{{𝑔}_{ℎ}}}=\sqrt{\surd \frac{𝑔}{𝑔/9}}=3$

Thus,

$\overline{){𝑇}_1:{𝑇}_2=1:3}$

Hence, Option B is correct.

The pressure inside the lake includes atmospheric pressure and the pressure due to the water column.

At half the depth, pressure is

${𝑃}_{\text{half}}={𝑃}_0+𝜌𝑔\frac{ℎ}{2}$

At the bottom, pressure is

${𝑃}_{\text{bottom}}={𝑃}_0+𝜌𝑔ℎ$

Given that the pressure at half the depth is two-thirds of the pressure at the bottom,

${𝑃}_0+\frac{𝜌𝑔ℎ}{2}=\frac{2}{3}({𝑃}_0+𝜌𝑔ℎ)$

Multiplying by $6$,

$6{𝑃}_0+3𝜌𝑔ℎ=4{𝑃}_0+4𝜌𝑔ℎ$

Hence,

$𝜌𝑔ℎ=2{𝑃}_0$

So,

$\overline{)ℎ=\frac{2{𝑃}_0}{𝜌𝑔}}$

Thus, the correct answer is Option A.

At terminal velocity, the sphere moves with constant velocity, so its acceleration is zero. Hence, the net force on it must be zero.

The forces acting on the falling sphere are:

• Weight $𝑊$ acting downward

• Buoyant force ${𝐹}_{𝐵}$ acting upward

• Viscous force ${𝐹}_{𝑉}$ acting upward, opposite to the motion

Thus, at terminal velocity,

$𝑊={𝐹}_{𝐵}+{𝐹}_{𝑉}$

So, the buoyant force plus the viscous force becomes equal to the force due to gravity.

Correct option: D

Using Pascal’s law, the pressure is the same in both pistons:

$\frac{{𝐹}_1}{{𝐴}_1}=\frac{{𝐹}_2}{{𝐴}_2}$

The force exerted by the car is its weight:

${𝐹}_2=𝑚𝑔=3000\times 9.8=29400\text{N}$

So,

${𝐴}_2=\frac{{𝐹}_2{𝐴}_1}{{𝐹}_1}=\frac{29400\times 2.5\times {10}^{-2}}{250}=2.94{\text{m}}^2$

Hence, the correct answer is Option B: $\overline{)2.94{\text{m}}^2}$.

For a spherical black body, Stefan’s law gives the radiated power as

$𝑃=𝜎𝐴{𝑇}^4$

Since the surface area of a sphere is $𝐴=4𝜋{𝑅}^2$,

$𝑃=𝜎\left(4𝜋{𝑅}^2\right){𝑇}^4$

Both bodies radiate the same power, so

$4𝜋𝜎{𝑅}_2^1{𝑇}_4^1=4𝜋𝜎{𝑅}_2^2{𝑇}_4^2$

Cancelling common terms,

${𝑅}_2^1{𝑇}_4^1={𝑅}_2^2{𝑇}_4^2$

Thus,

${\left(\frac{{𝑅}_1}{{𝑅}_2}\right)}^2={\left(\frac{{𝑇}_2}{{𝑇}_1}\right)}^4$

Taking the square root,

$\overline{)\frac{{𝑅}_1}{{𝑅}_2}={\left(\frac{{𝑇}_2}{{𝑇}_1}\right)}^2}$

Hence, the correct answer is Option C.

Thermal resistance of a rod is

$𝑅=\frac{𝐿}{𝐾𝐴}$

Here, both rods have the same thermal resistance and the same area of cross-section. Hence,

$\frac{{𝐿}_1}{{𝐾}_1𝐴}=\frac{{𝐿}_2}{{𝐾}_2𝐴}$

Cancelling $𝐴$,

$\frac{{𝐿}_1}{{𝐿}_2}=\frac{{𝐾}_1}{{𝐾}_2}$

Given,

${𝐾}_1:{𝐾}_2=4:3$

So,

${𝐿}_1:{𝐿}_2=4:3$

Thus, the correct answer is Option B, $4:3$.

Option B is the wrong statement.

During free expansion, a gas expands into a vacuum. The process is very rapid and the gas does not remain in thermodynamic equilibrium during expansion. Hence, pressure and volume do not have well-defined equilibrium values at every intermediate stage.

So, the path of free expansion cannot be plotted on a $𝑝$-$𝑉$ diagram. Only the initial and final equilibrium states can be shown.

For an adiabatic free expansion:

• Heat exchanged: $𝑄=0$

• External pressure: ${𝑝}_{\text{ext}}=0$

• Work done:

$𝑊=\int {𝑝}_{\text{ext}}𝑑𝑉=0$

Thus, Options A, C and D are correct, while Option B is wrong.

An isochoric process means the volume of the gas remains constant. So, for this process, we write $𝑉=\text{constant}$.

In a pressure-volume graph, pressure is shown on the vertical axis and volume is shown on the horizontal axis.

If the volume is constant, the graph must stay at the same value of volume. So the line will be vertical, parallel to the pressure axis.

Among the given graphs, graph (B) is a vertical line. This shows that pressure changes, but volume remains constant.

So, the correct answer is (B).

At S.T.P., the initial temperature is

${𝑇}_1=273\text{K}$

At constant pressure, Charles’ law gives

$\frac{{𝑉}_1}{{𝑇}_1}=\frac{{𝑉}_2}{{𝑇}_2}$

Since the final volume is three times the initial volume,

${𝑉}_2=3{𝑉}_1$

Hence,

${𝑇}_2=3{𝑇}_1=3\times 273=819\text{K}$

The r.m.s. speed of a gas is proportional to the square root of its absolute temperature:

${𝑢}_{\text{rms}}\propto \sqrt{\surd 𝑇}$

Thus,

$\frac{{𝑢}_2}{𝑢}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}=\sqrt{\surd \frac{819}{273}}=\sqrt{\surd 3}$

So,

${𝑢}_2=\sqrt{\surd 3}𝑢\text{m/s}$

Hence, the correct answer is Option C:

$\overline{)819\text{K},\sqrt{\surd 3}𝑢\text{m/s}}$

At constant pressure, Charles’s law gives

$\frac{{𝑉}_1}{{𝑇}_1}=\frac{{𝑉}_2}{{𝑇}_2}$

Temperatures must be converted into kelvin:

${𝑇}_1=27+273=300\text{K}$

${𝑇}_2=627+273=900\text{K}$

Now,

${𝑉}_2={𝑉}_1\frac{{𝑇}_2}{{𝑇}_1}=4\times \frac{900}{300}=12{\text{m}}^3$

Hence, the new volume is

$\overline{)12{\text{m}}^3}$

Option A.

For figure (a), there is only one spring attached to the mass. So the equivalent spring constant is ${𝐾}_{𝑎}=𝐾$.

The time period of a spring-mass system is given by $𝑇=2𝜋\sqrt{\surd \frac{𝑚}{{𝐾}_{\text{eq}}}}$.

So, for figure (a), the time period is ${𝑇}_{𝑎}=2𝜋\sqrt{\surd \frac{𝑚}{𝐾}}$.

For figure (b), the two identical springs are connected in series. In series, the equivalent spring constant becomes smaller.

The equivalent spring constant is ${𝐾}_{𝑏}=\frac{𝐾\cdot 𝐾}{𝐾+𝐾}=\frac{𝐾}{2}$.

So the time period is ${𝑇}_{𝑏}=2𝜋\sqrt{\surd \frac{𝑚}{𝐾/2}}=2𝜋\sqrt{\surd \frac{2𝑚}{𝐾}}=\sqrt{\surd 2}\left(2𝜋\sqrt{\surd \frac{𝑚}{𝐾}}\right)=\sqrt{\surd 2}{𝑇}_{𝑎}$.

For figure (c), the two identical springs are connected in parallel. In parallel, the equivalent spring constant becomes the sum of the spring constants.

So, the equivalent spring constant is ${𝐾}_{𝑐}=𝐾+𝐾=2𝐾$.

The time period is ${𝑇}_{𝑐}=2𝜋\sqrt{\surd \frac{𝑚}{2𝐾}}=\frac{1}{\sqrt{\surd 2}}\left(2𝜋\sqrt{\surd \frac{𝑚}{𝐾}}\right)=\frac{{𝑇}_{𝑎}}{\sqrt{\surd 2}}$.

From this, ${𝑇}_{𝑎}=\sqrt{\surd 2}{𝑇}_{𝑐}$.

Also, ${𝑇}_{𝑏}=\sqrt{\surd 2}{𝑇}_{𝑎}$.

Substituting ${𝑇}_{𝑎}=\sqrt{\surd 2}{𝑇}_{𝑐}$ in this relation:

${𝑇}_{𝑏}=\sqrt{\surd 2}\left(\sqrt{\surd 2}{𝑇}_{𝑐}\right)=2{𝑇}_{𝑐}$.

For simple harmonic motion, the speed at displacement $𝑥$ is

$𝑣=𝜔\sqrt{\surd {𝑎}^2-{𝑥}^2}$

Here,

$𝜔=\frac{2𝜋}{𝑇},𝑥=\frac{𝑎}{3}$

Substituting,

$𝑣=\frac{2𝜋}{𝑇}\sqrt{\surd {𝑎}^2-{\left(\frac{𝑎}{3}\right)}^2}$

$𝑣=\frac{2𝜋}{𝑇}\sqrt{\surd \frac{8{𝑎}^2}{9}}=\frac{2𝜋}{𝑇}\cdot \frac{2\sqrt{\surd 2}𝑎}{3}$

Thus,

$\overline{)𝑣=\frac{4\sqrt{\surd 2}𝜋𝑎}{3𝑇}}$

Hence, the correct answer is Option C.

The angular frequencies are

${𝜔}_1=708𝜋,{𝜔}_2=700𝜋$

Using $𝜔=2𝜋𝑓$,

${𝑓}_1=\frac{708𝜋}{2𝜋}=354\text{Hz}$

and

${𝑓}_2=\frac{700𝜋}{2𝜋}=350\text{Hz}$

Hence, the number of beats heard per second is

${𝑓}_{\text{beat}}=|{𝑓}_1-{𝑓}_2|=|354-350|=4$

The amplitudes are ${𝑎}_1=4$ and ${𝑎}_2=3$.

During waxing, the waves reinforce each other:

${𝐴}_{max}={𝑎}_1+{𝑎}_2=7$

During waning, they oppose each other:

${𝐴}_{min}=|{𝑎}_1-{𝑎}_2|=1$

Since intensity is proportional to the square of amplitude,

${𝐼}_{max}:{𝐼}_{min}={𝐴}_2^{max}:{𝐴}_2^{min}=7^2:1^2=49:1$

Thus, the correct answer is

$\overline{)\text{Option D: }4,49:1}$

For an open pipe, all harmonics are present. Its fifth overtone is the sixth harmonic.

${𝑓}_{\text{open}}=\frac{6𝑣}{2{𝐿}_0}=\frac{3𝑣}{{𝐿}_0}$

For a pipe closed at one end, only odd harmonics are present:

$1,3,5,7,9,11,\dots$

Hence, its fifth overtone is the eleventh harmonic.

${𝑓}_{\text{closed}}=\frac{11𝑣}{4{𝐿}_{𝑐}}$

Since the two overtones are in unison, their frequencies are equal:

$\frac{3𝑣}{{𝐿}_0}=\frac{11𝑣}{4{𝐿}_{𝑐}}$

Cancelling $𝑣$ and cross-multiplying,

$12{𝐿}_{𝑐}=11{𝐿}_0$

Thus,

$\frac{{𝐿}_0}{{𝐿}_{𝑐}}=\frac{12}{11}$

So, the required ratio is

$\overline{){𝐿}_0:{𝐿}_{𝑐}=12:11}$

Option D.

For a pipe open at both ends, the fundamental frequency is

$𝑓=\frac{𝑣}{2𝑙}$

For the pipe of length $𝐿$,

${𝑓}_1=\frac{𝑣}{2𝐿}$

For the pipe of length $𝐿+{𝐿}_1$,

${𝑓}_2=\frac{𝑣}{2(𝐿+{𝐿}_1)}$

The beat frequency is the difference between the two frequencies:

${𝑓}_{𝑏}=|{𝑓}_1-{𝑓}_2|$

${𝑓}_{𝑏}=\frac{𝑣}{2𝐿}-\frac{𝑣}{2(𝐿+{𝐿}_1)}$

Taking the LCM,

${𝑓}_{𝑏}=\frac{𝑣}{2}\left(\frac{𝐿+{𝐿}_1-𝐿}{𝐿(𝐿+{𝐿}_1)}\right)=\frac{𝑣{𝐿}_1}{2𝐿(𝐿+{𝐿}_1)}$

Hence, the correct answer is

$\overline{)\frac{𝑣{𝐿}_1}{2𝐿(𝐿+{𝐿}_1)}}$

Option D.

The observer counts $54$ waves in $1$ minute.

Convert time into seconds:

$1\text{minute}=60\text{s}$

So, the frequency is

$𝜈=\frac{54}{60}=0.9\text{Hz}$

Using the wave relation,

$𝑣=𝜈𝜆$

where $𝜆=5\text{m}$. Thus,

$𝑣=0.9\times 5=4.5\text{m/s}$

Hence, the correct answer is Option A: $4.5\text{m/s}$.

For an electric dipole in a uniform electric field, the work done in rotating it from angle ${𝜃}_1$ to ${𝜃}_2$ is

$𝑊=𝑝𝐸(\cos {𝜃}_1-\cos {𝜃}_2)$

Initially, the dipole is parallel to the field, so ${𝜃}_1=0$.

For rotation through $𝜋$ radians:

$𝑊=𝑝𝐸(\cos 0^{\circ }-\cos {180}^{\circ })$

$𝑊=𝑝𝐸[1-(-1\left)\right]=2𝑝𝐸$

Thus,

$𝑝𝐸=\frac{𝑊}{2}$

For rotation through $\frac{𝜋}{3}$ radians:

${𝑊}^{\prime }=𝑝𝐸(\cos 0^{\circ }-\cos {60}^{\circ })$

${𝑊}^{\prime }=𝑝𝐸(1-\frac{1}{2})=\frac{𝑝𝐸}{2}$

Substituting $𝑝𝐸=\frac{𝑊}{2}$:

${𝑊}^{\prime }=\frac{1}{2}\times \frac{𝑊}{2}=\frac{𝑊}{4}$

Hence, the correct answer is Option A: $\overline{)\frac{𝑊}{4}}$.

Let the distance of every vertex from the centre $𝑂$ be $𝑟$.

The electric field at $𝑂$ due to a charge of magnitude $𝑞$ placed at any vertex is

$𝐸=\frac{1}{4𝜋{𝜖}_0}\frac{𝑞}{{𝑟}^2}$

In a regular hexagon, opposite vertices lie on the same straight line through the centre. So the opposite pairs are $(𝐴,𝐷)$, $(𝐵,𝐸)$, and $(𝐶,𝐹)$.

The charges at $𝐴$ and $𝐷$ are both $+𝑞$. Since they are equal charges at equal distances on opposite sides of $𝑂$, their electric fields at $𝑂$ are equal in magnitude and opposite in direction. So they cancel each other.

The charges at $𝐶$ and $𝐹$ are both $-𝑞$. Again, they are equal charges at equal distances on opposite sides of $𝑂$, so their electric fields at $𝑂$ also cancel each other.

So, among the five charges at $𝐴,𝐵,𝐶,𝐷$ and $𝐹$, only the charge $-𝑞$ at $𝐵$ gives a net electric field at the centre.

Hence, the magnitude of the electric field at $𝑂$ due to these five charges is

${𝐸}_1=\frac{1}{4𝜋{𝜖}_0}\frac{𝑞}{{𝑟}^2}$

The electric field at $𝑂$ due to the charge $+𝑄$ at $𝐸$ alone is

${𝐸}_2=\frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{{𝑟}^2}$

It is given that the electric field due to the five charges is three times the electric field due to $+𝑄$ alone.

${𝐸}_1=3{𝐸}_2$

$\frac{1}{4𝜋{𝜖}_0}\frac{𝑞}{{𝑟}^2}=3\left(\frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{{𝑟}^2}\right)$

Cancel the common terms from both sides.

$𝑞=3𝑄$

So,

$𝑄=\frac{𝑞}{3}$

The total energy of an oscillating LC circuit is equal to the maximum electric energy stored in the capacitor:

$𝑈=\frac{{𝑄}^2}{2𝐶}$

When the energy is shared equally between the electric and magnetic fields, the electric energy is half of the total energy.

If the charge at that instant is $𝑞$, then

$\frac{{𝑞}^2}{2𝐶}=\frac{1}{2}\left(\frac{{𝑄}^2}{2𝐶}\right)$

Cancelling common terms,

${𝑞}^2=\frac{{𝑄}^2}{2}$

Hence,

$𝑞=\frac{𝑄}{\sqrt{\surd 2}}$

So, the correct answer is Option A: $\frac{𝑄}{\sqrt{\surd 2}}$.

For a parallel-plate capacitor,

$𝐶=\frac{{𝜀}_0𝐴}{𝑑}$

For circular plates of radius $𝑟$,

$𝐴=𝜋{𝑟}^2$

Initially,

${𝐶}_1=\frac{{𝜀}_0𝜋{𝑟}^2}{𝑑}$

The new radius is $\sqrt{\surd 3}𝑟$, so the new area is

${𝐴}_2=𝜋(\sqrt{\surd 3}𝑟{)}^2=3𝜋{𝑟}^2$

The new separation is

${𝑑}_2=\frac{𝑑}{2}$

Thus,

${𝐶}_2=\frac{{𝜀}_0\left(3𝜋{𝑟}^2\right)}{𝑑/2}=6\frac{{𝜀}_0𝜋{𝑟}^2}{𝑑}=6{𝐶}_1$

Hence,

${𝐶}_1:{𝐶}_2=1:6$

So, the correct answer is Option C: $1:6$.

For a capacitor of fixed capacitance $𝐶$, the stored energy is

$𝑊=\frac{{𝑄}^2}{2𝐶}$

When the charge becomes $3𝑄$,

${𝑊}^{\prime }=\frac{(3𝑄{)}^2}{2𝐶}=\frac{9{𝑄}^2}{2𝐶}=9𝑊$

Hence, the energy stored becomes

$\overline{){𝑊}^{\prime }=9𝑊}$

So, Option A is correct.

The total resistance in the primary circuit includes the potentiometer wire, series resistance, and internal resistance of the cell:

${𝑅}_{\text{total}}=5+992+3=1000\Omega$

The current through the potentiometer wire is

$𝐼=\frac{𝐸}{{𝑅}_{\text{total}}}=\frac{4}{1000}=4\times {10}^{-3}\text{A}$

The potential difference across the $4\text{m}$ potentiometer wire is

$𝑉=𝐼𝑅=(4\times {10}^{-3})\left(5\right)=20\times {10}^{-3}\text{V}$

Hence, the potential gradient is

$𝑘=\frac{20\text{mV}}{4\text{m}}=5{\text{mV m}}^{-1}$

For a balancing length of $0.75\text{m}$, the e.m.f. is

${𝐸}^{\prime }=𝑘𝑙=\left(5\right)\left(0.75\right)=3.75\text{mV}$

Thus, the correct answer is

$\overline{)\text{Option C: }3.75\text{mV}}$

Let the resistance of the galvanometer be $𝐺$ and its full-scale deflection current be ${𝐼}_{𝑔}$.

The voltmeter range when $980\Omega$ is connected in series is

${𝑉}_1={𝐼}_{𝑔}(𝐺+980)$

With $470\Omega$ in series, the range becomes half:

${𝐼}_{𝑔}(𝐺+470)=\frac{1}{2}{𝐼}_{𝑔}(𝐺+980)$

Cancelling ${𝐼}_{𝑔}$,

$2(𝐺+470)=𝐺+980$

$2𝐺+940=𝐺+980$

$𝐺=40\Omega$

Hence, the resistance of the galvanometer is

$\overline{)40\Omega }$

Option B.

For a magnetic material, the relative permeability is

${𝜇}_{𝑟}=1+𝜒$

Without niobium, the magnetic field is

${𝐵}_0={𝜇}_0𝐻$

With niobium, it becomes

$𝐵={𝜇}_0(1+𝜒)𝐻$

Hence, the fractional increase is

$\frac{𝐵-{𝐵}_0}{{𝐵}_0}=𝜒$

So, the percentage increase is

$𝜒\times 100=2.6\times {10}^{-5}\times 100=2.6\times {10}^{-3}\%$

Thus, the correct answer is Option C: $2.6\times {10}^{-3}$.

The magnetic field at point $𝑂$ is due to three parts of the wire: the semi-infinite straight wire $𝐴𝐵$, the quarter-circular arc $𝐵𝐶$, and the semi-infinite straight wire $𝐶𝐷$.

For the semi-infinite straight wire $𝐴𝐵$, the perpendicular distance of point $𝑂$ from the wire is $𝑟$.

The magnetic field due to a semi-infinite straight wire at distance $𝑟$ is:

${𝐵}_1=\frac{{𝜇}_0𝐼}{4𝜋𝑟}$

Using the right-hand thumb rule, the direction of this magnetic field at $𝑂$ is out of the plane of the paper.

For the quarter-circular arc $𝐵𝐶$, the radius is $𝑟$ and the angle subtended at $𝑂$ is $\frac{𝜋}{2}$.

The magnetic field at the centre due to a circular arc is:

$𝐵=\frac{{𝜇}_0𝐼}{4𝜋𝑟}𝜃$

Here, $𝜃=\frac{𝜋}{2}$, so

${𝐵}_2=\frac{{𝜇}_0𝐼}{4𝜋𝑟}\left(\frac{𝜋}{2}\right)$

The current in the arc flows in the counter-clockwise direction around $𝑂$, so by the right-hand thumb rule, this magnetic field is also out of the plane of the paper.

For the semi-infinite straight wire $𝐶𝐷$, the perpendicular distance from $𝑂$ is again $𝑟$.

So, its magnetic field at $𝑂$ is:

${𝐵}_3=\frac{{𝜇}_0𝐼}{4𝜋𝑟}$

Its direction is also out of the plane of the paper.

Since all three magnetic fields are in the same direction, we add their magnitudes:

${𝐵}_{\text{net}}={𝐵}_1+{𝐵}_2+{𝐵}_3$

${𝐵}_{\text{net}}=\frac{{𝜇}_0𝐼}{4𝜋𝑟}+\frac{{𝜇}_0𝐼}{4𝜋𝑟}\left(\frac{𝜋}{2}\right)+\frac{{𝜇}_0𝐼}{4𝜋𝑟}$

${𝐵}_{\text{net}}=\frac{{𝜇}_0𝐼}{4𝜋𝑟}(1+\frac{𝜋}{2}+1)$

${𝐵}_{\text{net}}=\frac{{𝜇}_0𝐼}{4𝜋𝑟}(2+\frac{𝜋}{2})$

For a circular current-carrying loop of radius $𝑅$,

$𝐵=\frac{{𝜇}_0𝐼}{2𝑅}$

Its magnetic moment is

$𝑀=𝐼𝜋{𝑅}^2$

Thus, the ratio is

$\frac{𝐵}{𝑀}=\frac{{𝜇}_0𝐼}{2𝑅}\times \frac{1}{𝐼𝜋{𝑅}^2}=\frac{{𝜇}_0}{2𝜋{𝑅}^3}$

The ratio is independent of current and varies as $\frac{1}{{𝑅}^3}$.

When the radius is tripled,

$\frac{{𝐵}^{\prime }}{{𝑀}^{\prime }}=\frac{𝑥}{3^3}=\frac{𝑥}{27}$

Hence, the correct answer is Option A: $\overline{)\frac{𝑥}{27}}$.