MHT CET 2026 11th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

For a quantity of the form

$𝑃={𝑋}^{-2}{𝑌}^{-3/2}{𝑍}^{2/5}$

the maximum fractional error is found by multiplying each fractional error by the magnitude of its power and then adding them.

So,

$\frac{\Delta 𝑃}{𝑃}=|-2|\frac{\Delta 𝑋}{𝑋}+\mid -\frac{3}{2}\mid \frac{\Delta 𝑌}{𝑌}+\mid \frac{2}{5}\mid \frac{\Delta 𝑍}{𝑍}$

Given,

$\frac{\Delta 𝑋}{𝑋}=0.1,\frac{\Delta 𝑌}{𝑌}=0.2,\frac{\Delta 𝑍}{𝑍}=0.5$

Substitute the values:

$\frac{\Delta 𝑃}{𝑃}=2\left(0.1\right)+\frac{3}{2}\left(0.2\right)+\frac{2}{5}\left(0.5\right)$

$=0.2+0.3+0.2$

$=0.7$

So, the maximum fractional error in $𝑃$ is

$\overline{)0.7}$

Correct option: A.

For two vectors to be perpendicular, their dot product must be zero.

Here the vectors are $(\overrightarrow{𝐴}+\overrightarrow{𝐵})$ and $(\overrightarrow{𝐴}-\overrightarrow{𝐵})$.

So,

$(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot (\overrightarrow{𝐴}-\overrightarrow{𝐵})=0$

Expanding using distributive property of dot product:

$\overrightarrow{𝐴}\cdot \overrightarrow{𝐴}-\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}+\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}-\overrightarrow{𝐵}\cdot \overrightarrow{𝐵}=0$

Since dot product is commutative,

$\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}=\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}$

So the middle two terms cancel:

$|\overrightarrow{𝐴}{|}^2-|\overrightarrow{𝐵}{|}^2=0$

Hence,

$|\overrightarrow{𝐴}{|}^2=|\overrightarrow{𝐵}{|}^2$

So,

$\left|\overrightarrow{𝐴}\right|=\left|\overrightarrow{𝐵}\right|$

The correct option is:

$\overline{)\text{Option A }\left|\overrightarrow{𝐴}\right|=\left|\overrightarrow{𝐵}\right|}$

The minute hand completes one full rotation in $60$ minutes.

One full rotation means ${360}^{\circ }$.

Also,

$60\text{ minutes}=60\times 60=3600\text{ seconds}$

So, angular speed is

$\frac{{360}^{\circ }}{3600\text{ s}}={0.1}^{\circ }/\text{s}$

So the angular speed of the minute hand is

${0.1}^{\circ }/\text{s}$

Hence, the correct option is Option B: $0.1$.

In non-uniform circular motion, the acceleration has two components:

Radial acceleration is directed towards the centre of the circle:

${𝑎}_{𝑟}=\frac{{𝑉}^2}{𝑟}$

Tangential acceleration is along the tangent and is due to angular acceleration $𝛼$:

${𝑎}_{𝑡}=𝑟𝛼$

So, the ratio of radial acceleration to tangential acceleration is

$\frac{{𝑎}_{𝑟}}{{𝑎}_{𝑡}}=\frac{\frac{{𝑉}^2}{𝑟}}{𝑟𝛼}$

$\frac{{𝑎}_{𝑟}}{{𝑎}_{𝑡}}=\frac{{𝑉}^2}{{𝑟}^2𝛼}$

So the correct option is

$\overline{)\frac{{𝑉}^2}{{𝑟}^2𝛼}}$

Option B.

Since the two masses stick together after collision, they move with a common velocity. In such a collision, linear momentum is conserved.

Momentum of mass $𝑀$ moving along X-axis:

${\overrightarrow{𝑝}}_1=𝑀\left(𝑉\widehat{𝑖}\right)=𝑀𝑉\widehat{𝑖}$

Momentum of mass $2𝑀$ moving along Y-axis with velocity $3𝑉$:

${\overrightarrow{𝑝}}_2=2𝑀\left(3𝑉\widehat{𝑗}\right)=6𝑀𝑉\widehat{𝑗}$

Total initial momentum is

$\overrightarrow{𝑝}=𝑀𝑉\widehat{𝑖}+6𝑀𝑉\widehat{𝑗}$

After collision, total mass is

$𝑀+2𝑀=3𝑀$

If the common velocity is $\overrightarrow{𝑣}$, then

$3𝑀\overrightarrow{𝑣}=𝑀𝑉\widehat{𝑖}+6𝑀𝑉\widehat{𝑗}$

Dividing by $3𝑀$,

$\overrightarrow{𝑣}=\frac{𝑉}{3}\widehat{𝑖}+2𝑉\widehat{𝑗}$

So the correct option is

$\overline{)\text{Option B }\frac{𝑉}{3}\widehat{𝑖}+2𝑉\widehat{𝑗}}$

Since the second identical ring is gently placed on the first ring, we take it to be initially at rest.

Moment of inertia of first ring = $𝐼$

Initial angular velocity of first ring = $𝜔$

Initial angular momentum of the system is

${𝐿}_{𝑖}=𝐼𝜔$

The second ring is identical, so its moment of inertia is also $𝐼$.

After some time, due to friction between the rings, both rotate together with a common angular velocity, say ${𝜔}^{\prime }$.

Total moment of inertia after they rotate together is

$𝐼+𝐼=2𝐼$

Since there is no external torque about the axis, angular momentum is conserved.

$𝐼𝜔=2𝐼{𝜔}^{\prime }$

So,

${𝜔}^{\prime }=\frac{𝜔}{2}$

Initial kinetic energy is only of the first ring:

${𝐾}_{𝑖}=\frac{1}{2}𝐼{𝜔}^2$

Final kinetic energy of both rings together is

${𝐾}_{𝑓}=\frac{1}{2}\left(2𝐼\right){\left(\frac{𝜔}{2}\right)}^2$

${𝐾}_{𝑓}=𝐼\cdot \frac{{𝜔}^2}{4}$

${𝐾}_{𝑓}=\frac{𝐼{𝜔}^2}{4}$

Loss in kinetic energy is

${𝐾}_{𝑖}-{𝐾}_{𝑓}=\frac{1}{2}𝐼{𝜔}^2-\frac{1}{4}𝐼{𝜔}^2$

$=\frac{𝐼{𝜔}^2}{4}$

So, the loss in kinetic energy is

$\overline{)\frac{𝐼{𝜔}^2}{4}}$

Correct option: C.

For a body, radius of gyration $𝐾$ is related to moment of inertia by

$𝐼=𝑀{𝐾}^2$

So,

$𝐾=\sqrt{\surd \frac{𝐼}{𝑀}}$

For a thin uniform rod, when the axis passes through its centre and is perpendicular to its length,

${𝐼}_1=\frac{𝑀{𝐿}^2}{12}$

So the radius of gyration in the first case is

${𝐾}_1=\sqrt{\surd \frac{{𝐼}_1}{𝑀}}=\sqrt{\surd \frac{𝑀{𝐿}^2}{12𝑀}}=\frac{𝐿}{\sqrt{\surd 12}}$

When the axis passes through one end and is perpendicular to the length of the rod,

${𝐼}_2=\frac{𝑀{𝐿}^2}{3}$

So the radius of gyration in the second case is

${𝐾}_2=\sqrt{\surd \frac{{𝐼}_2}{𝑀}}=\sqrt{\surd \frac{𝑀{𝐿}^2}{3𝑀}}=\frac{𝐿}{\sqrt{\surd 3}}$

Now the required ratio is

$\frac{{𝐾}_1}{{𝐾}_2}=\frac{\frac{𝐿}{\sqrt{\surd 12}}}{\frac{𝐿}{\sqrt{\surd 3}}}$

$=\frac{\sqrt{\surd 3}}{\sqrt{\surd 12}}=\sqrt{\surd \frac{3}{12}}=\sqrt{\surd \frac{1}{4}}=\frac{1}{2}$

So the correct option is

$\overline{)\frac{1}{2}}$

Option B.

For a planet,

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

Also, mass of a spherical planet is

$𝑀=\frac{4}{3}𝜋{𝑅}^3𝜌$

Putting this value of $𝑀$ in the formula for $𝑔$,

$𝑔=\frac{𝐺\left(\frac{4}{3}𝜋{𝑅}^3𝜌\right)}{{𝑅}^2}$

$𝑔=\frac{4}{3}𝜋𝐺𝑅𝜌$

So, acceleration due to gravity is directly proportional to density and radius:

$𝑔\propto 𝜌𝑅$

Now,

$\frac{{𝑔}_{𝑝}}{{𝑔}_{𝑒}}=\frac{{𝜌}_{𝑝}}{{𝜌}_{𝑒}}\times \frac{{𝑅}_{𝑝}}{{𝑅}_{𝑒}}$

Given,

${𝜌}_{𝑝}=3{𝜌}_{𝑒}$

and

${𝑅}_{𝑝}=2.5{𝑅}_{𝑒}$

So,

$\frac{{𝑔}_{𝑝}}{{𝑔}_{𝑒}}=3\times 2.5=7.5$

Hence, the ratio ${𝑔}_{𝑝}:{𝑔}_{𝑒}$ is

$\overline{)7.5}$

Correct option: A.

Surface energy of a liquid drop is proportional to its surface area.

Suppose the radius of the original drop is $𝑅$.

Its surface area is

$4𝜋{𝑅}^2$

So its surface energy is given as

$𝐸$

Now the drop is sprayed into $512$ equal droplets. Let the radius of each small droplet be $𝑟$.

Since volume remains the same,

$\frac{4}{3}𝜋{𝑅}^3=512\times \frac{4}{3}𝜋{𝑟}^3$

Cancel common terms:

${𝑅}^3=512{𝑟}^3$

Since $512=8^3$,

${𝑅}^3=8^3{𝑟}^3$

So,

$𝑅=8𝑟$

Now final total surface area is

$512\times 4𝜋{𝑟}^2$

Using $𝑟=\frac{𝑅}{8}$,

$512\times 4𝜋{\left(\frac{𝑅}{8}\right)}^2$

$=512\times 4𝜋\frac{{𝑅}^2}{64}$

$=8\times 4𝜋{𝑅}^2$

So the final surface area becomes $8$ times the original surface area.

Hence final surface energy also becomes $8$ times:

${𝐸}_{𝑓}=8𝐸$

So the correct option is

$\overline{)8𝐸}$

Option B.

For water in a narrow capillary, we take the angle of contact as $0^{\circ }$, so $\cos 0^{\circ }=1$.

Here, water rises in the annular space between the capillary tube and the thin wire.

The surface tension pulls water upward along both contact boundaries:

• along the inner wall of the capillary: circumference $=2𝜋𝑅$

• along the outer surface of the wire: circumference $=2𝜋𝑟$

So, total upward force due to surface tension is

$𝐹=𝑇(2𝜋𝑅+2𝜋𝑟)$

$𝐹=2𝜋𝑇(𝑅+𝑟)$

The water column rises through the annular area:

$𝐴=𝜋{𝑅}^2-𝜋{𝑟}^2$

$𝐴=𝜋({𝑅}^2-{𝑟}^2)$

If the rise is $ℎ$, then weight of water column is

$𝑊=𝜌𝑔𝜋({𝑅}^2-{𝑟}^2)ℎ$

At equilibrium,

$2𝜋𝑇(𝑅+𝑟)=𝜌𝑔𝜋({𝑅}^2-{𝑟}^2)ℎ$

Cancel $𝜋$:

$2𝑇(𝑅+𝑟)=𝜌𝑔({𝑅}^2-{𝑟}^2)ℎ$

Since

${𝑅}^2-{𝑟}^2=(𝑅-𝑟)(𝑅+𝑟)$

we get

$ℎ=\frac{2𝑇(𝑅+𝑟)}{𝜌𝑔(𝑅-𝑟)(𝑅+𝑟)}$

$ℎ=\frac{2𝑇}{(𝑅-𝑟)𝜌𝑔}$

So the rise of water is

$\overline{)\frac{2𝑇}{(𝑅-𝑟)𝜌𝑔}}$

Correct option: B.

For a liquid drop, the excess pressure inside it due to surface tension is

$𝑃=\frac{2𝑇}{𝑟}$

So excess pressure is inversely proportional to the radius of the drop:

$𝑃\propto \frac{1}{𝑟}$

For each smaller drop, let the radius be $𝑟$ and excess pressure be

${𝑃}_{𝑠}=9\text{ units}$

Now $27$ smaller drops combine to form one bigger drop.

Volume is conserved, so

$27\times \frac{4}{3}𝜋{𝑟}^3=\frac{4}{3}𝜋{𝑅}^3$

Canceling common terms,

$27{𝑟}^3={𝑅}^3$

$𝑅=3𝑟$

So the bigger drop has radius $3𝑟$.

Now excess pressure inside the bigger drop is

${𝑃}_{𝑏}=\frac{2𝑇}{𝑅}$

Since $𝑅=3𝑟$,

${𝑃}_{𝑏}=\frac{2𝑇}{3𝑟}$

But

$\frac{2𝑇}{𝑟}=9$

So

${𝑃}_{𝑏}=\frac{9}{3}=3\text{ units}$

The excess pressure inside the bigger drop is

$\overline{)3\text{ units}}$

Correct option: A.

According to Newton’s law of cooling, the rate of cooling depends on the temperature difference between the body and its surroundings.

The surrounding temperature is ${10}^{\circ }\text{C}$.

So, instead of looking at the actual temperature, we look at the excess temperature above the surroundings.

When the body cools from ${60}^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$:

$60-10=50$

and

$40-10=30$

So the excess temperature changes from ${50}^{\circ }\text{C}$ to ${30}^{\circ }\text{C}$ in $7$ minutes.

That means the excess temperature becomes

$\frac{30}{50}=\frac{3}{5}$

of its initial value in $7$ minutes.

Now, when the body cools from ${40}^{\circ }\text{C}$ to ${28}^{\circ }\text{C}$:

$40-10=30$

and

$28-10=18$

So the excess temperature changes from ${30}^{\circ }\text{C}$ to ${18}^{\circ }\text{C}$.

Again,

$\frac{18}{30}=\frac{3}{5}$

So the excess temperature again becomes $\frac{3}{5}$ of its initial value.

Since the same fractional decrease happens, the time taken will also be the same.

Hence, the required time is

$7\text{ minutes}$

So the correct option is Option D, $7$.

For an adiabatic change of an ideal gas,

$𝑃{𝑉}^{𝛾}=\text{constant}$

Also, from ideal gas equation,

$𝑃𝑉=𝑛𝑅𝑇$

So,

$𝑉=\frac{𝑛𝑅𝑇}{𝑃}$

Substitute this in the adiabatic relation:

$𝑃{\left(\frac{𝑛𝑅𝑇}{𝑃}\right)}^{𝛾}=\text{constant}$

Since $𝑛𝑅$ is constant,

$𝑃\cdot \frac{{𝑇}^{𝛾}}{{𝑃}^{𝛾}}=\text{constant}$

${𝑇}^{𝛾}{𝑃}^{1-𝛾}=\text{constant}$

This gives

$𝑃\propto {𝑇}^{\frac{𝛾}{𝛾-1}}$

For a rigid diatomic gas, degrees of freedom are $5$, so

${𝐶}_{𝑉}=\frac{5}{2}𝑅,{𝐶}_{𝑃}=\frac{7}{2}𝑅$

Hence,

$𝛾=\frac{{𝐶}_{𝑃}}{{𝐶}_{𝑉}}=\frac{7}{5}$

So,

$𝑥=\frac{𝛾}{𝛾-1}=\frac{\frac{7}{5}}{\frac{7}{5}-1}=\frac{\frac{7}{5}}{\frac{2}{5}}=\frac{7}{2}=3.5$

So the correct option is Option C: $3.5$.

For a rigid diatomic gas, vibrational motion is not considered.

So the degrees of freedom are:

$𝑓=5$

For an ideal gas,

${𝐶}_{𝑉}=\frac{𝑓}{2}𝑅=\frac{5}{2}𝑅$

At constant pressure,

${𝐶}_{𝑃}={𝐶}_{𝑉}+𝑅=\frac{5}{2}𝑅+𝑅=\frac{7}{2}𝑅$

Now, for $𝑛$ moles and temperature change $\Delta 𝑇$:

Change in internal energy is

$\Delta 𝑈=𝑛{𝐶}_{𝑉}\Delta 𝑇=𝑛\cdot \frac{5}{2}𝑅\Delta 𝑇$

Heat supplied at constant pressure is

$\Delta 𝑄=𝑛{𝐶}_{𝑃}\Delta 𝑇=𝑛\cdot \frac{7}{2}𝑅\Delta 𝑇$

Work done by the gas at constant pressure is

$\Delta 𝑊=𝑃\Delta 𝑉=𝑛𝑅\Delta 𝑇$

So,

$\Delta 𝑊:\Delta 𝑈:\Delta 𝑄=𝑛𝑅\Delta 𝑇:\frac{5}{2}𝑛𝑅\Delta 𝑇:\frac{7}{2}𝑛𝑅\Delta 𝑇$

Canceling common terms,

$=1:\frac{5}{2}:\frac{7}{2}$

Multiplying by $2$,

$=2:5:7$

So the correct option is:

$\overline{)\text{Option A }2:5:7}$

The container and the gas are initially moving together with speed $𝑣$.

When the container is stopped suddenly, the gas still has kinetic energy due to its motion. Since the container is insulated, this kinetic energy gets converted into the internal energy of the gas, increasing its temperature.

For $𝑛$ moles of gas, total mass of gas is $𝑛𝑚$.

So the kinetic energy of the moving gas is

$\frac{1}{2}𝑛𝑚{𝑣}^2$

For a monoatomic ideal gas,

${𝐶}_{𝑉}=\frac{3}{2}𝑅$

Increase in internal energy is

$𝑛{𝐶}_{𝑉}\Delta 𝑇=𝑛\left(\frac{3}{2}𝑅\right)\Delta 𝑇$

Equating the loss of kinetic energy to the gain in internal energy,

$\frac{1}{2}𝑛𝑚{𝑣}^2=𝑛\left(\frac{3}{2}𝑅\right)\Delta 𝑇$

Cancel $𝑛$ and simplify:

$\frac{1}{2}𝑚{𝑣}^2=\frac{3}{2}𝑅\Delta 𝑇$

$\Delta 𝑇=\frac{𝑚{𝑣}^2}{3𝑅}$

So the change in temperature is

$\overline{)\frac{𝑚{𝑣}^2}{3𝑅}}$

Correct option: C.

For a Carnot engine, efficiency is

$𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}$

where ${𝑇}_1$ is the temperature of the hot reservoir and ${𝑇}_2$ is the temperature of the cold reservoir.

Given initial efficiency:

$0.2=1-\frac{{𝑇}_2}{{𝑇}_1}$

So,

$\frac{{𝑇}_2}{{𝑇}_1}=0.8$

${𝑇}_2=0.8{𝑇}_1$

Now, when ${𝑇}_2$ is lowered by $45$ K, the new cold reservoir temperature becomes

${𝑇}_2-45$

New efficiency is $0.5$, so

$0.5=1-\frac{{𝑇}_2-45}{{𝑇}_1}$

Thus,

$\frac{{𝑇}_2-45}{{𝑇}_1}=0.5$

${𝑇}_2-45=0.5{𝑇}_1$

Using ${𝑇}_2=0.8{𝑇}_1$,

$0.8{𝑇}_1-45=0.5{𝑇}_1$

$0.3{𝑇}_1=45$

${𝑇}_1=150\text{ K}$

Now,

${𝑇}_2=0.8\times 150=120\text{ K}$

So the temperatures are

${𝑇}_1=150\text{ K},{𝑇}_2=120\text{ K}$

Correct option: Option A.

Given,

$\frac{𝑅}{{𝐶}_{𝑣}}=0.4$

So,

${𝐶}_{𝑣}=\frac{𝑅}{0.4}$

Since $0.4=\frac{2}{5}$,

${𝐶}_{𝑣}=\frac{𝑅}{2/5}=\frac{5𝑅}{2}$

For an ideal gas, molar specific heat at constant volume is

${𝐶}_{𝑣}=\frac{𝑓}{2}𝑅$

where $𝑓$ is the number of degrees of freedom.

So,

$\frac{𝑓}{2}𝑅=\frac{5𝑅}{2}$

Hence,

$𝑓=5$

A molecule with $5$ degrees of freedom is a rigid diatomic molecule.

A rigid diatomic molecule has:

$3\text{ translational degrees of freedom}+2\text{ rotational degrees of freedom}=5$

So the correct option is:

$\overline{)\text{Option D: rigid diatomic}}$

For a simple pendulum,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{{𝑔}_{\text{eff}}}}$

where ${𝑔}_{\text{eff}}$ is the effective acceleration due to gravity.

When the lift is at rest,

${𝑇}_1=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

So,

${𝑇}_2^1=\frac{4{𝜋}^2𝑙}{𝑔}$

When the lift moves upward with acceleration $𝑎$, effective gravity becomes $𝑔+𝑎$.

${𝑇}_2=2𝜋\sqrt{\surd \frac{𝑙}{𝑔+𝑎}}$

So,

${𝑇}_2^2=\frac{4{𝜋}^2𝑙}{𝑔+𝑎}$

When the lift moves downward with acceleration $𝑎$, effective gravity becomes $𝑔-𝑎$.

${𝑇}_3=2𝜋\sqrt{\surd \frac{𝑙}{𝑔-𝑎}}$

So,

${𝑇}_2^3=\frac{4{𝜋}^2𝑙}{𝑔-𝑎}$

Now take reciprocals:

$\frac{1}{{𝑇}_2^2}=\frac{𝑔+𝑎}{4{𝜋}^2𝑙}$

and

$\frac{1}{{𝑇}_2^3}=\frac{𝑔-𝑎}{4{𝜋}^2𝑙}$

Adding both,

$\frac{1}{{𝑇}_2^2}+\frac{1}{{𝑇}_2^3}=\frac{𝑔+𝑎+𝑔-𝑎}{4{𝜋}^2𝑙}$

$\frac{1}{{𝑇}_2^2}+\frac{1}{{𝑇}_2^3}=\frac{2𝑔}{4{𝜋}^2𝑙}$

But

$\frac{1}{{𝑇}_2^1}=\frac{𝑔}{4{𝜋}^2𝑙}$

So,

$\frac{1}{{𝑇}_2^2}+\frac{1}{{𝑇}_2^3}=\frac{2}{{𝑇}_2^1}$

Hence,

$\frac{1}{{𝑇}_2^1}=\frac{1}{2}(\frac{1}{{𝑇}_2^2}+\frac{1}{{𝑇}_2^3})$

Simplifying,

${𝑇}_2^1=\frac{2{𝑇}_2^2{𝑇}_2^3}{{𝑇}_2^2+{𝑇}_2^3}$

Taking square root,

${𝑇}_1=\frac{\sqrt{\surd 2}{𝑇}_2{𝑇}_3}{\sqrt{\surd {𝑇}_2^2+{𝑇}_2^3}}$

So the correct option is

$\overline{)\text{Option B}}$

The total energy in S.H.M. remains constant.

At the given displacement,

$\text{K.E.}=2\text{ J}$

and

$\text{P.E.}=1.2\text{ J}$

So the total energy is

$𝐸=2+1.2=3.2\text{ J}$

For S.H.M., total energy is

$𝐸=\frac{1}{2}𝑚{𝜔}^2{𝐴}^2$

Given,

$𝑚=0.4\text{ kg}$

and frequency,

$𝑓=\frac{16}{𝜋}\text{ Hz}$

Angular frequency is

$𝜔=2𝜋𝑓$

$𝜔=2𝜋\times \frac{16}{𝜋}=32{\text{ rad s}}^{-1}$

Now put the values in the energy formula:

$3.2=\frac{1}{2}\times 0.4\times (32{)}^2\times {𝐴}^2$

$3.2=0.2\times 1024\times {𝐴}^2$

$3.2=204.8{𝐴}^2$

${𝐴}^2=\frac{3.2}{204.8}$

${𝐴}^2=\frac{1}{64}$

$𝐴=\frac{1}{8}=0.125\text{ m}$

So the amplitude of oscillation is

$\overline{)0.125\text{ m}}$

Correct option: Option A.

Given displacement is

$𝑥=3\times {10}^{-3}\cos 2𝜋𝑡$

For S.H.M., speed is maximum at the mean position, where displacement is zero.

So we put

$𝑥=0$

$3\times {10}^{-3}\cos 2𝜋𝑡=0$

This means

$\cos 2𝜋𝑡=0$

The first time cosine becomes zero is when

$2𝜋𝑡=\frac{𝜋}{2}$

So,

$𝑡=\frac{1}{4}\text{ s}$

Hence, the time taken to obtain maximum speed for the first time is

$\overline{)\frac{1}{4}\text{ s}}$

Correct option: C.

Take the original length of each pipe as $𝐿$ and speed of sound as $𝑣$.

For an open pipe, the first overtone frequency is

${𝑓}_{𝑜}=\frac{𝑣}{𝐿}$

For a closed pipe, the first overtone frequency is

${𝑓}_{𝑐}=\frac{3𝑣}{4𝐿}$

They produce $3$ beats, so

${𝑓}_{𝑜}-{𝑓}_{𝑐}=3$

$\frac{𝑣}{𝐿}-\frac{3𝑣}{4𝐿}=3$

$\frac{𝑣}{4𝐿}=3$

So,

$\frac{𝑣}{𝐿}=12$

Now the open pipe is made $\frac{1}{3}$ of its original length, so its new length is

$\frac{𝐿}{3}$

Its fundamental frequency becomes

${𝑓}_{\prime }^{𝑜}=\frac{𝑣}{2(𝐿/3)}=\frac{3𝑣}{2𝐿}$

Using $\frac{𝑣}{𝐿}=12$,

${𝑓}_{\prime }^{𝑜}=\frac{3}{2}\times 12=18$

The closed pipe is made $3$ times its original length, so its new length is

$3𝐿$

Its fundamental frequency becomes

${𝑓}_{\prime }^{𝑐}=\frac{𝑣}{4\left(3𝐿\right)}=\frac{𝑣}{12𝐿}$

${𝑓}_{\prime }^{𝑐}=\frac{1}{12}\times 12=1$

So the new number of beats is

$|18-1|=17$

Hence, the correct option is

$\overline{)17}$

Option C.

For a stretched wire, the frequency of the ${𝑛}^{\text{th}}$ harmonic is

${𝑓}_{𝑛}=\frac{𝑛}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}$

where $𝜇$ is the mass per unit length.

Since both wires are of the same material,

$𝜇\propto {𝑟}^2$

Given,

${𝑟}_1=2{𝑟}_2$

So,

$\frac{{𝜇}_1}{{𝜇}_2}=\frac{{𝑟}_2^1}{{𝑟}_2^2}=\frac{(2{𝑟}_2{)}^2}{{𝑟}_2^2}=4$

Hence,

${𝜇}_1=4{𝜇}_2$

The wave speed is

$𝑣=\sqrt{\surd \frac{𝑇}{𝜇}}$

Since tension is the same,

${𝑣}_1=\frac{{𝑣}_2}{2}$

Now, first overtone means second harmonic.

So for the first wire,

${𝑓}_1=\frac{2{𝑣}_1}{2{𝐿}_1}=\frac{{𝑣}_1}{{𝐿}_1}$

Second overtone means third harmonic.

So for the second wire,

${𝑓}_2=\frac{3{𝑣}_2}{2{𝐿}_2}$

Given that these frequencies are equal:

$\frac{{𝑣}_1}{{𝐿}_1}=\frac{3{𝑣}_2}{2{𝐿}_2}$

Substitute ${𝑣}_1=\frac{{𝑣}_2}{2}$:

$\frac{{𝑣}_2}{2{𝐿}_1}=\frac{3{𝑣}_2}{2{𝐿}_2}$

Cancel ${𝑣}_2$ and $2$:

$\frac{1}{{𝐿}_1}=\frac{3}{{𝐿}_2}$

So,

$\frac{{𝐿}_1}{{𝐿}_2}=\frac{1}{3}$

The correct option is

$\overline{)\frac{1}{3}}$

Option B.

Given wave equation is

$𝑌=3\sin [𝜋(\frac{𝑡}{3}-\frac{𝑥}{5})+\frac{𝜋}{4}]$

Compare it with the standard equation of a progressive wave travelling along positive $𝑥$-direction:

$𝑌=𝐴\sin (𝜔𝑡-𝑘𝑥+𝜙)$

Now rewrite the given equation:

$𝑌=3\sin (\frac{𝜋}{3}𝑡-\frac{𝜋}{5}𝑥+\frac{𝜋}{4})$

So,

$𝐴=3\text{ m}$

Amplitude is $3$ m, not $30$ cm.

So option B is incorrect.

Also,

$𝜔=\frac{𝜋}{3}$

and

$𝑘=\frac{𝜋}{5}$

We know,

$𝑘=\frac{2𝜋}{𝜆}$

So,

$\frac{𝜋}{5}=\frac{2𝜋}{𝜆}$

Cancel $𝜋$:

$\frac{1}{5}=\frac{2}{𝜆}$

Hence,

$𝜆=10\text{ m}$

So the correct option is:

$\overline{)\text{Option C: Wavelength }=10\text{ m}}$

For a stationary observer and a moving source, the apparent frequency is given by Doppler effect.

When the source moves towards the observer with speed ${𝑉}_{𝑠}$, the apparent frequency increases:

${𝑛}_1=𝑛\left(\frac{𝑉}{𝑉-{𝑉}_{𝑠}}\right)$

When the source moves away from the observer with speed ${𝑉}_{𝑠}$, the apparent frequency decreases:

${𝑛}_2=𝑛\left(\frac{𝑉}{𝑉+{𝑉}_{𝑠}}\right)$

The difference between the two apparent frequencies is

${𝑛}_1-{𝑛}_2=𝑛\left(\frac{𝑉}{𝑉-{𝑉}_{𝑠}}\right)-𝑛\left(\frac{𝑉}{𝑉+{𝑉}_{𝑠}}\right)$

Taking $𝑛𝑉$ common,

${𝑛}_1-{𝑛}_2=𝑛𝑉(\frac{1}{𝑉-{𝑉}_{𝑠}}-\frac{1}{𝑉+{𝑉}_{𝑠}})$

Now simplify the bracket:

$\frac{1}{𝑉-{𝑉}_{𝑠}}-\frac{1}{𝑉+{𝑉}_{𝑠}}=\frac{(𝑉+{𝑉}_{𝑠})-(𝑉-{𝑉}_{𝑠})}{(𝑉-{𝑉}_{𝑠})(𝑉+{𝑉}_{𝑠})}$

$=\frac{2{𝑉}_{𝑠}}{{𝑉}^2-{𝑉}_2^{𝑠}}$

So,

${𝑛}_1-{𝑛}_2=𝑛𝑉\left(\frac{2{𝑉}_{𝑠}}{{𝑉}^2-{𝑉}_2^{𝑠}}\right)$

$=\frac{2𝑛𝑉{𝑉}_{𝑠}}{{𝑉}^2-{𝑉}_2^{𝑠}}$

So the correct option is

$\overline{)\text{Option A }\frac{2𝑛𝑉{𝑉}_{𝑠}}{{𝑉}^2-{𝑉}_2^{𝑠}}}$

The correct option is B: remains unchanged.

According to Gauss’s law, the electric flux through a closed surface is

${\Phi }_{𝐸}=\frac{𝑞}{{𝜀}_0}$

where $𝑞$ is the charge enclosed by the Gaussian surface.

Here, the point charge is enclosed inside the spherical Gaussian surface. If we increase the radius of the sphere, the enclosed charge $𝑞$ remains the same.

Although the electric field becomes weaker as the radius increases, the surface area of the sphere increases. These two effects balance each other, so the total electric flux remains unchanged.

So, the electric flux depends only on the enclosed charge, not on the size of the Gaussian surface.

Hence, the answer is Option B: remains unchanged.

For an isolated conducting sphere of radius $𝑅$, its capacitance in free space is

$𝐶=4𝜋{𝜖}_0𝑅$

Now, the earth is assumed to be a sphere.

Its volume is

$𝑉=\frac{4}{3}𝜋{𝑅}^3$

and its surface area is

$𝐴=4𝜋{𝑅}^2$

So,

$\frac{𝑉}{𝐴}=\frac{\frac{4}{3}𝜋{𝑅}^3}{4𝜋{𝑅}^2}$

Cancel common terms:

$\frac{𝑉}{𝐴}=\frac{𝑅}{3}$

Hence,

$𝑅=\frac{3𝑉}{𝐴}$

Substitute this value of $𝑅$ in the capacitance formula:

$𝐶=4𝜋{𝜖}_0\left(\frac{3𝑉}{𝐴}\right)$

$𝐶=\frac{12𝜋{𝜖}_0𝑉}{𝐴}$

So, the correct option is

$\overline{)\text{Option D }\frac{12𝜋{𝜖}_0𝑉}{𝐴}}$

Suppose each capacitor has capacitance $𝐶$.

Initially, all $𝑛$ capacitors are connected in parallel and charged to potential $𝑉$.

In parallel, each capacitor gets the same potential difference $𝑉$.

So charge on each capacitor is

$𝑞=𝐶𝑉$

Energy stored in each capacitor is

$𝑈=\frac{1}{2}𝐶{𝑉}^2$

So total initial energy is

${𝑈}_{\text{initial}}=𝑛\times \frac{1}{2}𝐶{𝑉}^2$

${𝑈}_{\text{initial}}=\frac{1}{2}𝑛𝐶{𝑉}^2$

Now the capacitors are separated. Since they are isolated, the charge on each capacitor remains the same, that is $𝐶𝑉$.

When they are joined in series, each capacitor still has charge $𝐶𝑉$.

So potential difference across each capacitor is

$𝑉=\frac{𝑞}{𝐶}=\frac{𝐶𝑉}{𝐶}=𝑉$

Since there are $𝑛$ capacitors in series, the total potential difference becomes

${𝑉}_{\text{total}}=𝑉+𝑉+\cdots +𝑉$

${𝑉}_{\text{total}}=𝑛𝑉$

Energy stored in each capacitor is still

$\frac{1}{2}𝐶{𝑉}^2$

So total energy remains

${𝑈}_{\text{final}}=\frac{1}{2}𝑛𝐶{𝑉}^2$

Hence, the potential difference becomes $𝑛𝑉$ and the total energy remains the same.

The correct option is:

$\overline{)\text{Option C}}$

For a parallel plate air capacitor, capacitance is

$𝐶=\frac{{𝜖}_0𝐴}{𝑑}$

Here each capacitor has plate area $\frac{𝐴}{3}$.

For the first capacitor, separation is $𝑑$:

${𝐶}_1=\frac{{𝜖}_0\left(\frac{𝐴}{3}\right)}{𝑑}=\frac{{𝜖}_0𝐴}{3𝑑}$

For the second capacitor, separation is $2𝑑$:

${𝐶}_2=\frac{{𝜖}_0\left(\frac{𝐴}{3}\right)}{2𝑑}=\frac{{𝜖}_0𝐴}{6𝑑}$

For the third capacitor, separation is $3𝑑$:

${𝐶}_3=\frac{{𝜖}_0\left(\frac{𝐴}{3}\right)}{3𝑑}=\frac{{𝜖}_0𝐴}{9𝑑}$

Since the capacitors are connected in parallel, their capacitances add directly:

${𝐶}_{\text{eq}}={𝐶}_1+{𝐶}_2+{𝐶}_3$

${𝐶}_{\text{eq}}=\frac{{𝜖}_0𝐴}{3𝑑}+\frac{{𝜖}_0𝐴}{6𝑑}+\frac{{𝜖}_0𝐴}{9𝑑}$

Taking $\frac{{𝜖}_0𝐴}{𝑑}$ common:

${𝐶}_{\text{eq}}=\frac{{𝜖}_0𝐴}{𝑑}(\frac{1}{3}+\frac{1}{6}+\frac{1}{9})$

LCM of $3,6,9$ is $18$:

$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6+3+2}{18}=\frac{11}{18}$

So,

${𝐶}_{\text{eq}}=\frac{11{𝜖}_0𝐴}{18𝑑}$

Hence, the correct option is

$\overline{)\text{Option C }\frac{11{𝜖}_0𝐴}{18𝑑}}$

The balancing length will increase.

In a potentiometer, at null point,

$𝐸=𝑘𝑙$

where $𝐸$ is the emf of the cell, $𝑘$ is the potential gradient of the potentiometer wire, and $𝑙$ is the balancing length.

The cell is not changed, so $𝐸$ remains the same.

Now the length of the potentiometer wire is increased, but the driving source is not changed. So the same source has to maintain potential across a longer wire. Hence the potential drop per unit length decreases.

That means the potential gradient $𝑘$ decreases.

Since

$𝑙=\frac{𝐸}{𝑘}$

if $𝑘$ decreases and $𝐸$ remains the same, the balancing length $𝑙$ must increase.

So the correct option is:

$\overline{)\text{Option B: increase}}$

The given circuit is a Wheatstone bridge.

First, check whether the bridge is balanced or not.

For one side of the bridge:
$\frac{{𝑅}_{𝐴𝐵}}{{𝑅}_{𝐵𝐶}}=\frac{15}{3}=5$

For the other side of the bridge:
$\frac{{𝑅}_{𝐴𝐷}}{{𝑅}_{𝐷𝐶}}=\frac{20}{4}=5$

Since $\frac{{𝑅}_{𝐴𝐵}}{{𝑅}_{𝐵𝐶}}=\frac{{𝑅}_{𝐴𝐷}}{{𝑅}_{𝐷𝐶}}$, the bridge is balanced.

So, no current flows through the $6\Omega$ resistor in branch BD.

Now the circuit becomes two parallel branches.

Resistance of the upper branch is:
${𝑅}_1=15+3=18\Omega$

Resistance of the lower branch is:
${𝑅}_2=20+4=24\Omega$

The total current in the circuit is $𝐼=2.1\text{A}$.

The $15\Omega$ resistor is in the upper branch, so we need the current in the upper branch.

Using the current divider rule:
${𝐼}_{15\Omega }=𝐼\times \frac{{𝑅}_2}{{𝑅}_1+{𝑅}_2}$

Substitute the values:
${𝐼}_{15\Omega }=2.1\times \frac{24}{18+24}$

${𝐼}_{15\Omega }=2.1\times \frac{24}{42}=1.2\text{A}$