MHT CET 2026 15th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

For a cylindrical rod, the volume is

$𝑉=𝜋{𝑟}^2𝑙$

Since the diameter is measured by vernier callipers, it is better to write the formula in terms of diameter.

Radius is given as

$𝑟=2.0\text{cm}$

So diameter is

$𝑑=2𝑟=4.0\text{cm}$

Volume can be written as

$𝑉=\frac{𝜋{𝑑}^2𝑙}{4}$

For percentage error in a quantity involving powers, we add the relative errors with their powers.

So,

$\frac{\Delta 𝑉}{𝑉}=2\frac{\Delta 𝑑}{𝑑}+\frac{\Delta 𝑙}{𝑙}$

Given:

$𝑙=5.0\text{cm},\Delta 𝑙=0.1\text{cm}$

$𝑑=4.0\text{cm},\Delta 𝑑=0.01\text{cm}$

Now,

$\frac{\Delta 𝑙}{𝑙}\times 100=\frac{0.1}{5.0}\times 100=2\%$

and

$2\frac{\Delta 𝑑}{𝑑}\times 100=2\times \frac{0.01}{4.0}\times 100=0.5\%$

So total percentage error in volume is

$2\%+0.5\%=2.5\%$

Hence, the correct option is

$\overline{)2.5\%}$

Option B.

The two given vectors are

$(\widehat{𝑖}-2\widehat{𝑗}+2\widehat{𝑘})$

and

$(-2\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘})$

First add them:

$(\widehat{𝑖}-2\widehat{𝑗}+2\widehat{𝑘})+(-2\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘})$

Combine like components:

$=(1-2)\widehat{𝑖}+(-2+1)\widehat{𝑗}+(2-1)\widehat{𝑘}$

$=-\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘}$

Now, vector $\overrightarrow{𝐴}$ added to this sum gives a unit vector along the positive $𝑌$-axis.

A unit vector along positive $𝑌$-axis is $\widehat{𝑗}$.

So,

$\overrightarrow{𝐴}+(-\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘})=\widehat{𝑗}$

Hence,

$\overrightarrow{𝐴}=\widehat{𝑗}-(-\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘})$

$\overrightarrow{𝐴}=\widehat{𝑗}+\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘}$

$\overrightarrow{𝐴}=\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}$

Now find its magnitude:

$\left|\overrightarrow{𝐴}\right|=\sqrt{\surd 1^2+2^2+(-1{)}^2}$

$\left|\overrightarrow{𝐴}\right|=\sqrt{\surd 1+4+1}$

$\left|\overrightarrow{𝐴}\right|=\sqrt{\surd 6}$

So, the correct option is

$\overline{)\sqrt{\surd 6}}$

Option B.

Take the ground as the reference point.

The stone falling from the top starts from height $100$ m with initial velocity $0$.

So its height after time $𝑡$ is

${𝑦}_1=100-\frac{1}{2}𝑔{𝑡}^2$

The stone projected upwards from the ground has initial velocity $25$ m/s.

So its height after time $𝑡$ is

${𝑦}_2=25𝑡-\frac{1}{2}𝑔{𝑡}^2$

When the two stones meet, their heights from the ground are equal.

$100-\frac{1}{2}𝑔{𝑡}^2=25𝑡-\frac{1}{2}𝑔{𝑡}^2$

The term $\frac{1}{2}𝑔{𝑡}^2$ cancels from both sides because both stones have the same acceleration due to gravity.

$100=25𝑡$

$𝑡=\frac{100}{25}=4\text{ s}$

So, the two stones meet after

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

Correct option: Option C.

In uniform circular motion, tangential velocity is the distance covered along the circular path per unit time.

Radius is

$𝑟=\frac{𝜋}{2}\text{ m}$

Distance covered in one revolution is the circumference:

$2𝜋𝑟$

So,

$2𝜋𝑟=2𝜋\times \frac{𝜋}{2}={𝜋}^2$

If the particle makes $𝑥$ revolutions, total distance covered is

$𝑥\times {𝜋}^2={𝜋}^2𝑥$

Time taken is $𝑡$.

So tangential velocity is

$𝑣=\frac{\text{distance}}{\text{time}}=\frac{{𝜋}^2𝑥}{𝑡}$

Hence, the correct option is

$\overline{)\frac{{𝜋}^2𝑥}{𝑡}}$

Option B.

Take the direction of motion of ${𝑚}_1$ as positive.

So initially,

${𝑢}_1={𝑉}_1$

and since ${𝑚}_2$ is moving in the opposite direction,

${𝑢}_2=-{𝑉}_2$

After collision, the velocities are exchanged. So ${𝑚}_1$ moves with velocity ${𝑉}_2$ in the direction in which ${𝑚}_2$ was initially moving, and ${𝑚}_2$ moves with velocity ${𝑉}_1$ in the direction in which ${𝑚}_1$ was initially moving.

Thus,

${𝑣}_1=-{𝑉}_2$

and

${𝑣}_2={𝑉}_1$

Using conservation of linear momentum,

${𝑚}_1{𝑢}_1+{𝑚}_2{𝑢}_2={𝑚}_1{𝑣}_1+{𝑚}_2{𝑣}_2$

Substitute the values:

${𝑚}_1{𝑉}_1-{𝑚}_2{𝑉}_2=-{𝑚}_1{𝑉}_2+{𝑚}_2{𝑉}_1$

Rearrange:

${𝑚}_1{𝑉}_1+{𝑚}_1{𝑉}_2={𝑚}_2{𝑉}_1+{𝑚}_2{𝑉}_2$

${𝑚}_1({𝑉}_1+{𝑉}_2)={𝑚}_2({𝑉}_1+{𝑉}_2)$

Since ${𝑉}_1+{𝑉}_2\ne 0$,

${𝑚}_1={𝑚}_2$

So,

$\frac{{𝑚}_2}{{𝑚}_1}=1$

Hence, the correct option is:

$\overline{)1.0}$

Option D.

Power supplied to a rotating body is related to torque and angular velocity by

$𝑃=𝜏𝜔$

For a body rotating with angular acceleration $𝛼$, the torque is

$𝜏=𝐼𝛼$

Substitute this in the power formula:

$𝑃=\left(𝐼𝛼\right)𝜔$

So,

$𝜔=\frac{𝑃}{𝐼𝛼}$

Hence, the instantaneous angular velocity is

$\overline{)\frac{𝑃}{𝐼𝛼}}$

Correct option: D.

Since the earth is rotating about its own axis, its rotational kinetic energy is

$𝐾=\frac{1}{2}𝐼{𝜔}^2$

For a rotating sphere,

$𝐼=\frac{2}{5}𝑀{𝑅}^2$

So moment of inertia depends on ${𝑅}^2$.

The earth contracts to $\frac{1}{3}$ of its present radius, while its mass remains the same.

So,

${𝑅}^{\prime }=\frac{𝑅}{3}$

New moment of inertia will be

${𝐼}^{\prime }=\frac{2}{5}𝑀({𝑅}^{\prime }{)}^2$

${𝐼}^{\prime }=\frac{2}{5}𝑀{\left(\frac{𝑅}{3}\right)}^2$

${𝐼}^{\prime }=\frac{1}{9}\cdot \frac{2}{5}𝑀{𝑅}^2$

${𝐼}^{\prime }=\frac{𝐼}{9}$

Since no external torque is acting, angular momentum remains conserved:

$𝐼𝜔={𝐼}^{\prime }{𝜔}^{\prime }$

Rotational kinetic energy can also be written as

$𝐾=\frac{{𝐿}^2}{2𝐼}$

Here $𝐿$ is constant, so kinetic energy is inversely proportional to moment of inertia:

$𝐾\propto \frac{1}{𝐼}$

Hence,

$\frac{{𝐾}^{\prime }}{𝐾}=\frac{𝐼}{{𝐼}^{\prime }}$

$\frac{{𝐾}^{\prime }}{𝐾}=\frac{𝐼}{𝐼/9}=9$

So, the ratio of kinetic energy after and before contraction is

$\overline{)9}$

Correct option: A.

At the equator, the rotation of the earth produces a centrifugal effect away from the centre of the earth.

So the apparent weight becomes less than the actual gravitational weight.

If the angular velocity of earth is $𝜔$, then the centrifugal acceleration at the equator is

${𝜔}^2𝑅$

Hence effective acceleration due to gravity at the equator becomes

${𝑔}^{\prime }=𝑔-{𝜔}^2𝑅$

The weight is proportional to acceleration due to gravity, so if the weight becomes $\frac{3}{5}$ of its initial value, then

${𝑔}^{\prime }=\frac{3}{5}𝑔$

So,

$𝑔-{𝜔}^2𝑅=\frac{3}{5}𝑔$

Now solving,

${𝜔}^2𝑅=𝑔-\frac{3}{5}𝑔$

${𝜔}^2𝑅=\frac{2}{5}𝑔$

${𝜔}^2=\frac{2𝑔}{5𝑅}$

$𝜔={\left(\frac{2𝑔}{5𝑅}\right)}^{1/2}$

So the correct option is

$\overline{){\left(\frac{2𝑔}{5𝑅}\right)}^{1/2}}$

Option B.

For capillary rise, the height of liquid is

$ℎ=\frac{2𝑇\cos 𝜃}{𝑟𝜌𝑔}$

For the same liquid and same tube material, $𝑇$, $𝜃$, $𝜌$, and $𝑔$ remain constant.

So,

$ℎ\propto \frac{1}{𝑟}$

The cross-sectional area of the tube is

$𝐴=𝜋{𝑟}^2$

Given that the new area is reduced to $\frac{1}{3}$ of the initial area:

${𝐴}^{\prime }=\frac{𝐴}{3}$

So,

$𝜋{𝑟}^{\prime 2}=\frac{𝜋{𝑟}^2}{3}$

${𝑟}^{\prime 2}=\frac{{𝑟}^2}{3}$

${𝑟}^{\prime }=\frac{𝑟}{\sqrt{\surd 3}}$

Since height is inversely proportional to radius,

$\frac{{ℎ}^{\prime }}{ℎ}=\frac{𝑟}{{𝑟}^{\prime }}$

$\frac{{ℎ}^{\prime }}{3}=\frac{𝑟}{𝑟/\sqrt{\surd 3}}=\sqrt{\surd 3}$

${ℎ}^{\prime }=3\sqrt{\surd 3}\text{ cm}$

So, the correct option is Option A: $3\sqrt{\surd 3}$ cm.

The work required to blow a soap bubble is used in increasing its surface area.

A soap bubble has two surfaces, inner and outer. So its surface energy is

$𝑊=2\times 4𝜋{𝑅}^2\times 𝑇$

where $𝑇$ is the surface tension of the soap solution.

So,

$𝑊=8𝜋{𝑅}^2𝑇$

For the first bubble of radius $𝑅$,

${𝑊}_1=8𝜋{𝑅}^2𝑇$

After heating, the second bubble has radius $2𝑅$. If the surface tension after heating is ${𝑇}^{\prime }$, then

${𝑊}_2=8𝜋(2𝑅{)}^2{𝑇}^{\prime }$

${𝑊}_2=32𝜋{𝑅}^2{𝑇}^{\prime }$

If the temperature had not changed, the work would become $4$ times because area depends on ${𝑅}^2$.

But heating decreases the surface tension of a liquid, so

${𝑇}^{\prime }<𝑇$

Hence,

${𝑊}_2<4{𝑊}_1$

So the work required for the second bubble is slightly less than $4$ times the work required for the first bubble.

The correct option is:

$\overline{)\text{Option A: slightly less than }4\text{ times}}$

For a soap bubble, the excess pressure inside it is

$\Delta 𝑃=\frac{4𝑇}{𝑅}$

where $𝑇$ is surface tension and $𝑅$ is the radius of the bubble.

For the first bubble,

$\Delta {𝑃}_1=\frac{4𝑇}{{𝑅}_1}$

For the second bubble,

$\Delta {𝑃}_2=\frac{4𝑇}{{𝑅}_2}$

Given that the excess pressure inside the first bubble is three times that inside the second bubble,

$\Delta {𝑃}_1=3\Delta {𝑃}_2$

So,

$\frac{4𝑇}{{𝑅}_1}=3\left(\frac{4𝑇}{{𝑅}_2}\right)$

Cancel $4𝑇$ from both sides:

$\frac{1}{{𝑅}_1}=\frac{3}{{𝑅}_2}$

So,

${𝑅}_2=3{𝑅}_1$

Now, volume of a bubble is proportional to the cube of its radius:

$𝑉\propto {𝑅}^3$

So,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑅}_3^1}{{𝑅}_3^2}$

Since ${𝑅}_2=3{𝑅}_1$,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑅}_3^1}{(3{𝑅}_1{)}^3}$

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{1}{27}$

Hence, the ratio of volumes is

${𝑉}_1:{𝑉}_2=1:27$

So the correct option is Option C: $1:27$.

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

Room temperature is $𝜃$.

Initial temperature of the body is $4𝜃$.

So initial excess temperature is

$4𝜃-𝜃=3𝜃$

After $𝑡$ minutes, the temperature becomes $3𝜃$.

So the excess temperature after $𝑡$ minutes is

$3𝜃-𝜃=2𝜃$

Thus, in $𝑡$ minutes, the excess temperature becomes

$3𝜃\to 2𝜃$

So the fraction remaining is

$\frac{2𝜃}{3𝜃}=\frac{2}{3}$

For the next equal interval of $𝑡$ minutes, the excess temperature will again become $\frac{2}{3}$ of its previous value.

So new excess temperature is

$2𝜃\times \frac{2}{3}=\frac{4𝜃}{3}$

Hence, the actual temperature of the body after the next $𝑡$ minutes is

$𝜃+\frac{4𝜃}{3}=\frac{7𝜃}{3}$

So the correct option is

$\overline{)\frac{7𝜃}{3}}$

Option B.

For a black body, the total energy radiated per second is given by Stefan’s law:

$𝐸\propto 𝐴{𝑇}^4$

For a black disc of radius $𝑅$, its area is proportional to

$𝐴\propto {𝑅}^2$

So,

$𝐸\propto {𝑅}^2{𝑇}^4$

Now, the wavelength corresponding to maximum intensity is $𝜆$. By Wien’s displacement law,

$𝜆𝑇=\text{constant}$

So,

$𝑇\propto \frac{1}{𝜆}$

Hence,

${𝑇}^4\propto \frac{1}{{𝜆}^4}$

Putting this in the expression for $𝐸$:

$𝐸\propto {𝑅}^2{𝜆}^{-4}$

So the correct option is:

$\overline{)\text{Option B }{𝑅}^2,{𝜆}^{-4}}$

The wrong statement is:

Option A: For all wavelengths, power radiated is same

A black body emits radiation of all wavelengths, so Option D is correct.

But the power radiated is not the same for all wavelengths. At a particular temperature, the distribution of energy with wavelength follows a curve. The energy is maximum at a certain wavelength, called ${𝜆}_{𝑚}$, and it is less on either side of this wavelength.

So a black body does not radiate equal power at every wavelength.

Hence, the correct choice is:

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

A small note: Options B and C are written in a very general way. Strictly, black body radiation is not simply “more for all shorter wavelengths” and “less for all longer wavelengths”; it depends on the position relative to the wavelength of maximum emission. But among the given options, the clearly wrong statement is Option A.

In an isobaric process, the pressure remains constant.

For a gas, work done in an isobaric process is

$𝑊=𝑃\Delta 𝑉$

So, if the volume changes, work done is not zero.

Also, from the first law of thermodynamics,

$𝑄=\Delta 𝑈+𝑊$

In an isobaric process, heat supplied is generally used partly to increase internal energy and partly to do work.

For an ideal gas, internal energy depends on temperature. In an isobaric process, if the gas expands or contracts, its temperature changes, so the change in internal energy is generally not zero.

So the correct option is:

$\overline{)\text{Option C: change in internal energy is not zero}}$

Since the container is thermally insulated, heat exchange is not possible.

So the compression is treated as an adiabatic process. For an adiabatic process,

$𝑃{𝑉}^{𝛾}=\text{constant}$

For a monoatomic gas,

$𝛾=\frac{5}{3}$

So,

${𝑃}_1{𝑉}_{𝛾}^1={𝑃}_2{𝑉}_{𝛾}^2$

Hence,

$\frac{{𝑃}_2}{{𝑃}_1}={\left(\frac{{𝑉}_1}{{𝑉}_2}\right)}^{𝛾}$

The gas is compressed to $\left(\frac{1}{8}\right)$ of its initial volume, so

${𝑉}_2=\frac{{𝑉}_1}{8}$

Thus,

$\frac{{𝑉}_1}{{𝑉}_2}=8$

Now,

$\frac{{𝑃}_2}{{𝑃}_1}=8^{5/3}$

Since,

$8=2^3$

$8^{5/3}={\left(2^3\right)}^{5/3}=2^5=32$

So,

$\frac{{𝑃}_2}{{𝑃}_1}=32$

The ratio of final pressure to initial pressure is

$32:1$

So the correct option is Option C: $32:1$.

For an ideal gas in a closed container, the amount of gas and the volume of the container remain constant.

From the ideal gas equation,

$𝑃𝑉=𝑛𝑅𝑇$

Since $𝑛$, $𝑅$, and $𝑉$ are constant,

$𝑃\propto 𝑇$

The average force on the wall is related to pressure by

$𝑃=\frac{𝐹}{𝐴}$

So,

$𝐹=𝑃𝐴$

For a fixed container, the area $𝐴$ of the wall is also constant. Hence,

$𝐹\propto 𝑃$

Since $𝑃\propto 𝑇$,

$𝐹\propto 𝑇$

So the average force applied on the walls depends on temperature as

${𝑇}^1$

The correct option is Option C: ${𝑇}^1$.

For linear S.H.M., the speed of the particle at displacement $𝑥$ from mean position is

$𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$

and the magnitude of acceleration is

$𝑎={𝜔}^2𝑥$

Here,

$𝐴=4\text{ cm},𝑥=3\text{ cm}$

It is given that the magnitude of velocity and acceleration are equal at $𝑥=3\text{ cm}$.

So,

$𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}={𝜔}^2𝑥$

Cancel $𝜔$ from both sides:

$\sqrt{\surd {𝐴}^2-{𝑥}^2}=𝜔𝑥$

Substitute the values:

$\sqrt{\surd 4^2-3^2}=𝜔\left(3\right)$

$\sqrt{\surd 16-9}=3𝜔$

$\sqrt{\surd 7}=3𝜔$

So,

$𝜔=\frac{\sqrt{\surd 7}}{3}$

Time period of S.H.M. is

$𝑇=\frac{2𝜋}{𝜔}$

Hence,

$𝑇=\frac{2𝜋}{\sqrt{\surd 7}/3}$

$𝑇=\frac{6𝜋}{\sqrt{\surd 7}}\text{ s}$

So the correct option is

$\overline{)\frac{6𝜋}{\sqrt{\surd 7}}\text{ s}}$

Option B.

Since the particle starts from the mean position, we can write its displacement as

$𝑥=𝐴\sin 𝜔𝑡$

In S.H.M., total energy is

$𝐸=\frac{1}{2}𝑘{𝐴}^2$

and potential energy at displacement $𝑥$ is

$𝑈=\frac{1}{2}𝑘{𝑥}^2$

So kinetic energy is

$𝐾=𝐸-𝑈$

Using $𝑥=𝐴\sin 𝜔𝑡$,

$\frac{𝐾}{𝐸}=1-\frac{{𝑥}^2}{{𝐴}^2}$

$\frac{𝐾}{𝐸}=1-{\sin }^2𝜔𝑡$

$\frac{𝐾}{𝐸}={\cos }^2𝜔𝑡$

Given kinetic energy is $50\%$ of total energy,

$\frac{𝐾}{𝐸}=\frac{1}{2}$

So,

${\cos }^2𝜔𝑡=\frac{1}{2}$

$\cos 𝜔𝑡=\frac{1}{\sqrt{\surd 2}}$

Using $\cos {45}^{\circ }=\frac{1}{\sqrt{\surd 2}}$,

$𝜔𝑡={45}^{\circ }=\frac{𝜋}{4}$

Time period is $𝑇=6$ s, so

$𝜔=\frac{2𝜋}{𝑇}$

$𝜔=\frac{2𝜋}{6}=\frac{𝜋}{3}$

Now,

$\frac{𝜋}{3}𝑡=\frac{𝜋}{4}$

$𝑡=\frac{3}{4}$

$𝑡=0.75\text{ s}$

So the correct option is:

$\overline{)0.75\text{ s}}$

Option A.

For a small sphere moving inside a watch glass, the surface is like a part of a sphere of radius of curvature $𝑅$.

Given,

$𝑅=1.6\text{ m}$

$𝑔=10{\text{ m/s}}^2$

For small oscillations, the restoring acceleration along the curved surface is due to the component of weight.

If the sphere is displaced through a small angle $𝜃$, then

$\text{restoring acceleration}=-𝑔\sin 𝜃$

For small oscillations,

$\sin 𝜃\approx 𝜃$

Also, the displacement along the arc is

$𝑥=𝑅𝜃$

So,

$𝜃=\frac{𝑥}{𝑅}$

Hence,

$𝑎=-𝑔\frac{𝑥}{𝑅}$

This is of the form of SHM:

$𝑎=-{𝜔}^2𝑥$

So,

${𝜔}^2=\frac{𝑔}{𝑅}$

The time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝑅}{𝑔}}$

Substitute the values:

$𝑇=2𝜋\sqrt{\surd \frac{1.6}{10}}$

$𝑇=2𝜋\sqrt{\surd 0.16}$

$𝑇=2𝜋\times 0.4$

$𝑇=0.8𝜋\text{ s}$

So, the correct option is

$\overline{)0.8𝜋}$

Option C.

For a wave travelling on a stretched wire, the speed is

$𝑣=\sqrt{\surd \frac{𝑇}{𝜇}}$

where $𝑇$ is the tension and $𝜇$ is the mass per unit length of the wire.

Since the same wire is used, $𝜇$ remains constant. So,

$𝑣\propto \sqrt{\surd 𝑇}$

Initial tension is

${𝑇}_1=2\text{ kg wt}$

The new speed is thrice the original speed:

${𝑣}_2=3{𝑣}_1$

Using the relation,

$\frac{{𝑣}_2}{{𝑣}_1}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

So,

$3=\sqrt{\surd \frac{{𝑇}_2}{2}}$

Squaring both sides,

$9=\frac{{𝑇}_2}{2}$

${𝑇}_2=18\text{ kg wt}$

Increase in tension is

${𝑇}_2-{𝑇}_1=18-2=16\text{ kg wt}$

So, the increase in tension is

$\overline{)16\text{ kg wt}}$

Correct option: C.

For a pipe open at one end and closed at the other, the fundamental frequency is

$𝑓=\frac{𝑣}{4𝐿}$

Here,

$𝑣=320\text{m/s},𝐿=0.8\text{m}$

So,

$𝑓=\frac{320}{4\times 0.8}$

$𝑓=\frac{320}{3.2}=100\text{Hz}$

The string resonates with this frequency, so the frequency of the string in its first overtone is also

$100\text{Hz}$

For a stretched string, the frequency in first overtone is

$𝑓=\frac{1}{𝑙}\sqrt{\surd \frac{𝑇}{𝜇}}$

where $𝑙$ is the length of the string and $𝜇$ is mass per unit length.

Given,

$𝑙=0.5\text{m},𝑇=50\text{N}$

So,

$100=\frac{1}{0.5}\sqrt{\surd \frac{50}{𝜇}}$

$100=2\sqrt{\surd \frac{50}{𝜇}}$

$50=\sqrt{\surd \frac{50}{𝜇}}$

Squaring both sides,

$2500=\frac{50}{𝜇}$

$𝜇=\frac{50}{2500}=0.02\text{kg/m}$

Mass of the string is

$𝑚=𝜇𝑙$

$𝑚=0.02\times 0.5=0.01\text{kg}$

$𝑚=10\text{g}$

So, the mass of the string is

$\overline{)10\text{gram}}$

Correct option: C

For a sonometer wire fixed between two bridges, if it forms $𝑛$ antinodes, its frequency is

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

Here, the tuning fork is the same, so frequency $𝑓$ is the same.

The distance between the bridges is also the same, so $𝐿$ is constant. The wire is the same, so $𝜇$ is also constant.

So,

$𝑛\sqrt{\surd 𝑇}=\text{constant}$

In the first case,

${𝑛}_1=4,{𝑇}_1=1𝑔$

In the second case,

${𝑛}_2=2,{𝑇}_2=𝑀𝑔$

Since the frequency is same,

$4\sqrt{\surd 1𝑔}=2\sqrt{\surd 𝑀𝑔}$

Cancel $\sqrt{\surd 𝑔}$:

$4=2\sqrt{\surd 𝑀}$

$\sqrt{\surd 𝑀}=2$

$𝑀=4$

So, the required mass is

$\overline{)4\text{ kg}}$

Correct option: C.

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

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

Here tension $𝑇$ is kept the same, and the wire is the same, so $𝜇$ is also the same.

So frequency varies inversely with length:

$𝑓\propto \frac{1}{𝑙}$

The length is increased by $25\%$.

If the original length is $𝑙$, then the new length is

${𝑙}^{\prime }=𝑙+25\%\text{ of }𝑙$

${𝑙}^{\prime }=𝑙+\frac{25}{100}𝑙=1.25𝑙$

For the second harmonic also, frequency is inversely proportional to length.

So,

$\frac{{𝑓}_{\prime }^2}{{𝑓}_2}=\frac{𝑙}{{𝑙}^{\prime }}=\frac{𝑙}{1.25𝑙}$

$\frac{{𝑓}_{\prime }^2}{{𝑓}_2}=\frac{1}{1.25}=\frac{4}{5}=0.8$

So the new second harmonic frequency becomes $80\%$ of the original second harmonic frequency.

That means the frequency decreases by

$100\%-80\%=20\%$

So the frequency of the second harmonic decreases by $20\%$.

Correct option: B. decrease by $20\%$

The number of electric lines of force originating from a charge is taken as the electric flux due to that charge.

Using Gauss’s law in a medium,

$𝑁=\frac{𝑞}{𝜖}$

where

$𝜖=𝐾{𝜖}_0$

Given,

$𝑞=2\times 8.85\times {10}^{-9}\text{C}$

$𝐾=10$

${𝜖}_0=8.85\times {10}^{-12}$

So,

$𝜖=10\times 8.85\times {10}^{-12}$

Now,

$𝑁=\frac{2\times 8.85\times {10}^{-9}}{10\times 8.85\times {10}^{-12}}$

Cancel $8.85$:

$𝑁=\frac{2\times {10}^{-9}}{10\times {10}^{-12}}$

$𝑁=\frac{2}{10}\times {10}^3$

$𝑁=200$

So, the correct option is:

$\overline{)200}$

Option B.

For three identical capacitors, let each capacitance be $𝐶$.

When they are connected in parallel, the equivalent capacitance is

${𝐶}_{𝑝}=𝐶+𝐶+𝐶=3𝐶$

When they are connected in series, the equivalent capacitance is

$\frac{1}{{𝐶}_{𝑠}}=\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}$

So,

${𝐶}_{𝑠}=\frac{𝐶}{3}$

Energy stored in a capacitor combination is

$𝑈=\frac{1}{2}𝐶{𝑉}^2$

Let the potential difference applied in parallel combination be ${𝑉}_{𝑝}$, and in series combination be ${𝑉}_{𝑠}$.

Since the energy stored is the same,

$\frac{1}{2}{𝐶}_{𝑝}{𝑉}_2^{𝑝}=\frac{1}{2}{𝐶}_{𝑠}{𝑉}_2^{𝑠}$

Substitute ${𝐶}_{𝑝}=3𝐶$ and ${𝐶}_{𝑠}=\frac{𝐶}{3}$:

$\frac{1}{2}\left(3𝐶\right){𝑉}_2^{𝑝}=\frac{1}{2}\left(\frac{𝐶}{3}\right){𝑉}_2^{𝑠}$

Cancel common terms:

$3{𝑉}_2^{𝑝}=\frac{{𝑉}_2^{𝑠}}{3}$

$9{𝑉}_2^{𝑝}={𝑉}_2^{𝑠}$

Taking square root,

$3{𝑉}_{𝑝}={𝑉}_{𝑠}$

So,

$\frac{{𝑉}_{𝑝}}{{𝑉}_{𝑠}}=\frac{1}{3}$

Hence, the ratio of potential difference in parallel to series combination is

$\overline{)1:3}$

Correct option: A.

The correct option is Option C: reduce the effective potential on plates.

When a dielectric is placed between the plates of a capacitor, it gets polarised. Due to this polarisation, an electric field is produced inside the dielectric in the opposite direction to the original electric field between the plates.

So, the net electric field between the plates decreases.

Since potential difference is related to electric field by

$𝑉=𝐸𝑑$

a decrease in electric field $𝐸$ also decreases the potential difference $𝑉$ between the plates.

For the same charge $𝑄$, capacitance is

$𝐶=\frac{𝑄}{𝑉}$

So when $𝑉$ decreases, capacitance increases.

Hence, the function of a dielectric is to reduce the effective potential on the plates, thereby increasing the capacitance.

So the answer is:

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

In the circuit, capacitors ${𝐶}_1$, ${𝐶}_2$, and ${𝐶}_3$ are connected in series. This whole series branch is connected in parallel with capacitor ${𝐶}_4$ across the battery of voltage $𝑉$.

For capacitors in series, the equivalent capacitance ${𝐶}_{𝑠}$ is found using:

$\frac{1}{{𝐶}_{𝑠}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}+\frac{1}{{𝐶}_3}$

Here, ${𝐶}_1=𝐶$, ${𝐶}_2=2𝐶$, and ${𝐶}_3=3𝐶$.

$\frac{1}{{𝐶}_{𝑠}}=\frac{1}{𝐶}+\frac{1}{2𝐶}+\frac{1}{3𝐶}$

$\frac{1}{{𝐶}_{𝑠}}=\frac{6+3+2}{6𝐶}=\frac{11}{6𝐶}$

So,

${𝐶}_{𝑠}=\frac{6𝐶}{11}$

In a series combination, the charge on each capacitor is the same. So the charge on ${𝐶}_2$ will be equal to the charge on the whole series branch.

The voltage across the series branch is $𝑉$, because it is connected across the battery.

So, the charge on ${𝐶}_2$ is:

${𝑄}_2={𝐶}_{𝑠}𝑉=\frac{6𝐶}{11}\times 𝑉$

${𝑄}_2=\frac{6𝐶𝑉}{11}$

Capacitor ${𝐶}_4$ is directly connected across the battery, so the voltage across ${𝐶}_4$ is also $𝑉$.

Given ${𝐶}_4=4𝐶$, the charge on ${𝐶}_4$ is:

${𝑄}_4={𝐶}_4𝑉=4𝐶𝑉$

Now the required ratio is:

$\frac{{𝑄}_2}{{𝑄}_4}=\frac{\frac{6𝐶𝑉}{11}}{4𝐶𝑉}$

$\frac{{𝑄}_2}{{𝑄}_4}=\frac{6}{44}=\frac{3}{22}$

For a galvanometer to work as a voltmeter, a resistance is connected in series with it.

If the galvanometer current for full-scale deflection is ${𝐼}_{𝑔}$, then the voltage range is

$𝑉={𝐼}_{𝑔}(𝐺+𝑅)$

where $𝐺$ is the resistance of the galvanometer and $𝑅$ is the series resistance.

When $200\Omega$ is connected in series, the range is $𝑉$:

$𝑉={𝐼}_{𝑔}(𝐺+200)$

When $2000\Omega$ is connected in series, the range becomes triple, that is $3𝑉$:

$3𝑉={𝐼}_{𝑔}(𝐺+2000)$

Now divide the second equation by the first:

$\frac{3𝑉}{𝑉}=\frac{{𝐼}_{𝑔}(𝐺+2000)}{{𝐼}_{𝑔}(𝐺+200)}$

$3=\frac{𝐺+2000}{𝐺+200}$

So,

$𝐺+2000=3(𝐺+200)$

$𝐺+2000=3𝐺+600$

$1400=2𝐺$

$𝐺=700\Omega$

So, the value of $𝐺$ is

$\overline{)700\Omega }$

Correct option: C.

Let the resistance of the galvanometer be $𝐺$.

The shunt resistance is given as

$𝑆=\frac{𝐺}{𝑛-1}$

Since the galvanometer and shunt are connected in parallel, the same potential difference is across both.

If ${𝐼}_{𝑔}$ is the current through the galvanometer and ${𝐼}_{𝑠}$ is the current through the shunt, then

${𝐼}_{𝑔}𝐺={𝐼}_{𝑠}𝑆$

So,

$\frac{{𝐼}_{𝑠}}{{𝐼}_{𝑔}}=\frac{𝐺}{𝑆}$

Substitute $𝑆=\frac{𝐺}{𝑛-1}$:

$\frac{{𝐼}_{𝑠}}{{𝐼}_{𝑔}}=\frac{𝐺}{𝐺/(𝑛-1)}=𝑛-1$

So,

${𝐼}_{𝑠}=(𝑛-1){𝐼}_{𝑔}$

Total current is

$𝐼={𝐼}_{𝑔}+{𝐼}_{𝑠}$

$𝐼={𝐼}_{𝑔}+(𝑛-1){𝐼}_{𝑔}=𝑛{𝐼}_{𝑔}$

Hence, the fraction of total current passing through the galvanometer is

$\frac{{𝐼}_{𝑔}}{𝐼}=\frac{{𝐼}_{𝑔}}{𝑛{𝐼}_{𝑔}}=\frac{1}{𝑛}$

So the correct option is

$\overline{)\frac{1}{𝑛}}$

Option B.

For a paramagnetic substance, magnetic susceptibility $𝜒$ varies inversely with absolute temperature $𝑇$.

According to Curie’s law,

$𝜒=\frac{𝐶}{𝑇}$

where $𝐶$ is Curie’s constant.

Multiplying both sides by $𝑇$,

$𝜒𝑇=𝐶$

So, the product of magnetic susceptibility and absolute temperature remains constant for a paramagnetic substance.

Hence, the correct option is:

Option B: paramagnetic substance.

For a circular current-carrying coil, magnetic moment is

$𝑀=𝑁𝐼𝐴$

where $𝑁$ is the number of turns, $𝐼$ is current, and $𝐴$ is the area of one turn.

For the original coil, diameter is $𝑑$, so radius is $\frac{𝑑}{2}$.

$𝐴=𝜋{\left(\frac{𝑑}{2}\right)}^2=\frac{𝜋{𝑑}^2}{4}$

So original magnetic moment is

${𝑀}_1=𝑁𝐼\cdot \frac{𝜋{𝑑}^2}{4}$

Now the same wire is rewound into a new coil of diameter $2𝑑$.

Since the wire length remains the same,

$𝑁\cdot 𝜋𝑑={𝑁}^{\prime }\cdot 𝜋\left(2𝑑\right)$

So,

${𝑁}^{\prime }=\frac{𝑁}{2}$

The new radius is $𝑑$, so the new area is

${𝐴}^{\prime }=𝜋{𝑑}^2$

Now the new magnetic moment is

${𝑀}_2={𝑁}^{\prime }𝐼{𝐴}^{\prime }$

${𝑀}_2=\frac{𝑁}{2}\cdot 𝐼\cdot 𝜋{𝑑}^2$

Now compare with the original:

${𝑀}_1=𝑁𝐼\cdot \frac{𝜋{𝑑}^2}{4}$

So,

$\frac{{𝑀}_2}{{𝑀}_1}=\frac{\frac{𝑁}{2}𝐼𝜋{𝑑}^2}{𝑁𝐼\frac{𝜋{𝑑}^2}{4}}=2$

Hence, the ratio of magnetic moment of the new coil to the original coil is

$\overline{)2:1}$

Correct option: A.

The magnetic Lorentz force on a charged particle is

$\overrightarrow{𝐹}=𝑞(\overrightarrow{𝑉}\times \overrightarrow{𝐵})$

The direction of $\overrightarrow{𝑉}\times \overrightarrow{𝐵}$ is perpendicular to $\overrightarrow{𝑉}$.

Work done by a force depends on the component of force along the displacement. Since the particle’s displacement is along its velocity $\overrightarrow{𝑉}$, and the magnetic force is perpendicular to $\overrightarrow{𝑉}$, the magnetic force has no component along the displacement.

So,

$𝑊=0$

Also, the power delivered by the magnetic force is

$𝑃=\overrightarrow{𝐹}\cdot \overrightarrow{𝑉}$

$𝑃=𝑞(\overrightarrow{𝑉}\times \overrightarrow{𝐵})\cdot \overrightarrow{𝑉}=0$

Hence, the work done by the magnetic force is zero.

Correct option: Option A, zero.

The total resistance of the circuit is the resistance of the coil plus the external resistance.

Coil resistance $=𝑅$

External resistance $=\frac{𝑅}{4}$

So,

${𝑅}_{\text{total}}=𝑅+\frac{𝑅}{4}$

${𝑅}_{\text{total}}=\frac{5𝑅}{4}$

Now, the flux changes from ${\Phi }_1$ to ${\Phi }_2$ in time $𝑡$.

For a coil of $𝑛$ turns, the induced emf is

$𝑒=\frac{𝑛({\Phi }_1-{\Phi }_2)}{𝑡}$