MHT CET 2026 19th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

Given,

$\overrightarrow{𝑃}\cdot \overrightarrow{𝑄}=|\overrightarrow{𝑃}\times \overrightarrow{𝑄}|$

We know,

$\overrightarrow{𝑃}\cdot \overrightarrow{𝑄}=𝑃𝑄\cos 𝜃$

and

$|\overrightarrow{𝑃}\times \overrightarrow{𝑄}|=𝑃𝑄\sin 𝜃$

So,

$𝑃𝑄\cos 𝜃=𝑃𝑄\sin 𝜃$

Dividing by $𝑃𝑄$,

$\cos 𝜃=\sin 𝜃$

This happens when

$𝜃={45}^{\circ }$

Now, the magnitude of the resultant vector

$\overrightarrow{𝑅}=\overrightarrow{𝑃}+\overrightarrow{𝑄}$

is given by

$𝑅=\sqrt{\surd {𝑃}^2+{𝑄}^2+2𝑃𝑄\cos 𝜃}$

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

$𝑅=\sqrt{\surd {𝑃}^2+{𝑄}^2+2𝑃𝑄\cdot \frac{1}{\sqrt{\surd 2}}}$

$𝑅=\sqrt{\surd {𝑃}^2+{𝑄}^2+\sqrt{\surd 2}𝑃𝑄}$

So, the correct answer is

$\overline{)\sqrt{\surd {𝑃}^2+{𝑄}^2+\sqrt{\surd 2}𝑃𝑄}}$

Option D.

For circular motion, the centripetal force is

$𝐹=\frac{𝑚{𝑣}^2}{𝑟}$

For the lighter stone:

Mass $=𝑚$, radius $=𝑟$, speed $={𝑣}_{𝑙}$

${𝐹}_{𝑙}=\frac{𝑚{𝑣}_2^{𝑙}}{𝑟}$

For the heavier stone:

Mass $=3𝑚$, radius $=\frac{𝑟}{3}$, speed $={𝑣}_{ℎ}$

${𝐹}_{ℎ}=\frac{3𝑚{𝑣}_2^{ℎ}}{\frac{𝑟}{3}}$

Dividing by $\frac{𝑟}{3}$ is the same as multiplying by $\frac{3}{𝑟}$, so

${𝐹}_{ℎ}=\frac{9𝑚{𝑣}_2^{ℎ}}{𝑟}$

Both stones experience the same centripetal force, so

$\frac{𝑚{𝑣}_2^{𝑙}}{𝑟}=\frac{9𝑚{𝑣}_2^{ℎ}}{𝑟}$

Cancel $𝑚$ and $𝑟$:

${𝑣}_2^{𝑙}=9{𝑣}_2^{ℎ}$

Taking square root:

${𝑣}_{𝑙}=3{𝑣}_{ℎ}$

So the tangential speed of the lighter stone is $3$ times that of the heavier stone.

$𝑛=3$

Correct option: C $3$

For circular motion, angular velocity is related to time period by

$𝜔=\frac{2𝜋}{𝑇}$

Here, both particles have the same time period $𝑇$.

So for the first particle,

${𝜔}_1=\frac{2𝜋}{𝑇}$

and for the second particle,

${𝜔}_2=\frac{2𝜋}{𝑇}$

Thus,

$\frac{{𝜔}_1}{{𝜔}_2}=1$

So the ratio of angular velocity of the first particle to that of the second particle is

$1:1$

Correct option: Option A $1:1$

The correct option is Option B: A driver driving a bus which is moving with constant velocity.

An inertial frame of reference is a frame which is either at rest or moving with constant velocity. This means there is no acceleration.

In Option B, the bus is moving with constant velocity, so its acceleration is zero. Hence, the driver inside the bus is in an inertial frame of reference.

The other options are not inertial frames:

• A pilot in an aeroplane which is taking off is accelerating.

• A child revolving in a merry-go-round has centripetal acceleration because the direction of motion keeps changing.

• A man in a train which is slowing down has acceleration in the opposite direction of motion.

So, the answer is:

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

For a spring, the restoring force is given by Hooke’s law:

$𝐹=𝑘𝑥$

where $𝑘$ is the spring constant and $𝑥$ is the extension.

The work done in stretching a spring from extension $0$ to $𝑥$ is stored as elastic potential energy:

${𝑊}_1=\frac{1}{2}𝑘{𝑥}^2$

Now the spring is stretched further by length $𝑥$. So its extension changes from $𝑥$ to $2𝑥$.

The work done in stretching it from $𝑥$ to $2𝑥$ is

${𝑊}_2=\frac{1}{2}𝑘(2𝑥{)}^2-\frac{1}{2}𝑘{𝑥}^2$

${𝑊}_2=\frac{1}{2}𝑘\left(4{𝑥}^2\right)-\frac{1}{2}𝑘{𝑥}^2$

${𝑊}_2=\frac{1}{2}𝑘\left(3{𝑥}^2\right)$

${𝑊}_2=\frac{3}{2}𝑘{𝑥}^2$

But

${𝑊}_1=\frac{1}{2}𝑘{𝑥}^2$

So,

${𝑊}_2=3{𝑊}_1$

Hence, the correct option is

$\overline{)3{𝑊}_1}$

Option C.

When the rod falls from the vertical position to the ground, its centre of mass moves downward.

For a uniform rod, the centre of mass is at its middle, so its distance from the hinge $𝐴$ is

$\frac{𝑙}{2}$

Initially, the rod is vertical. When it reaches the ground, it becomes horizontal. So the centre of mass falls through a height

$\frac{𝑙}{2}$

The loss in gravitational potential energy is

$𝑚𝑔\cdot \frac{𝑙}{2}$

This becomes rotational kinetic energy of the rod about the hinge $𝐴$.

The moment of inertia of a uniform rod about one end is

$𝐼=\frac{1}{3}𝑚{𝑙}^2$

So,

$𝑚𝑔\frac{𝑙}{2}=\frac{1}{2}𝐼{𝜔}^2$

Substitute $𝐼=\frac{1}{3}𝑚{𝑙}^2$:

$𝑚𝑔\frac{𝑙}{2}=\frac{1}{2}\left(\frac{1}{3}𝑚{𝑙}^2\right){𝜔}^2$

$𝑚𝑔\frac{𝑙}{2}=\frac{1}{6}𝑚{𝑙}^2{𝜔}^2$

Cancel $𝑚$:

$𝑔\frac{𝑙}{2}=\frac{1}{6}{𝑙}^2{𝜔}^2$

$3𝑔𝑙={𝑙}^2{𝜔}^2$

${𝜔}^2=\frac{3𝑔}{𝑙}$

$𝜔=\sqrt{\surd \frac{3𝑔}{𝑙}}$

So, the correct option is

$\overline{)\sqrt{\surd \frac{3𝑔}{𝑙}}}$

Option B.

For a solid sphere, the moment of inertia about its diameter is

${𝐼}_{𝑑}=\frac{2}{5}𝑀{𝑅}^2$

Radius of gyration $𝐾$ is defined by

$𝐼=𝑀{𝐾}^2$

So, for the diameter,

$𝑀{𝐾}_2^{𝑑}=\frac{2}{5}𝑀{𝑅}^2$

${𝐾}_{𝑑}=𝑅\sqrt{\surd \frac{2}{5}}$

Now, a tangent to the sphere is parallel to a diameter and is at a distance $𝑅$ from the centre.

Using the parallel axis theorem,

${𝐼}_{𝑡}={𝐼}_{𝑑}+𝑀{𝑅}^2$

${𝐼}_{𝑡}=\frac{2}{5}𝑀{𝑅}^2+𝑀{𝑅}^2$

${𝐼}_{𝑡}=\frac{7}{5}𝑀{𝑅}^2$

So,

$𝑀{𝐾}_2^{𝑡}=\frac{7}{5}𝑀{𝑅}^2$

${𝐾}_{𝑡}=𝑅\sqrt{\surd \frac{7}{5}}$

Hence,

$\frac{{𝐾}_{𝑑}}{{𝐾}_{𝑡}}=\frac{𝑅\sqrt{\surd \frac{2}{5}}}{𝑅\sqrt{\surd \frac{7}{5}}}$

$\frac{{𝐾}_{𝑑}}{{𝐾}_{𝑡}}=\sqrt{\surd \frac{2}{7}}$

So the correct option is

$\overline{){\left(\frac{2}{7}\right)}^{1/2}}$

Option C.

For a satellite revolving around the earth, the square of its time period is proportional to the cube of its orbital radius:

${𝑇}^2\propto {𝑟}^3$

So for two satellites,

$\frac{{𝑇}_2}{{𝑇}_1}={\left(\frac{{𝑟}_2}{{𝑟}_1}\right)}^{3/2}$

The geostationary satellite has time period

${𝑇}_1=24\text{ hours}$

Its height above the earth’s surface is $4𝑅$, so its distance from the centre of the earth is

${𝑟}_1=𝑅+4𝑅=5𝑅$

For the second satellite, height above the surface is $1.5𝑅$, so its distance from the centre is

${𝑟}_2=𝑅+1.5𝑅=2.5𝑅=\frac{5𝑅}{2}$

Now,

$\frac{𝑇}{24}={\left(\frac{\frac{5𝑅}{2}}{5𝑅}\right)}^{3/2}$

$\frac{𝑇}{24}={\left(\frac{1}{2}\right)}^{3/2}$

$\frac{𝑇}{24}=\frac{1}{2\sqrt{\surd 2}}$

So,

$𝑇=\frac{24}{2\sqrt{\surd 2}}$

$𝑇=\frac{12}{\sqrt{\surd 2}}=6\sqrt{\surd 2}$

Hence, the time period is

$\overline{)6\sqrt{\surd 2}\text{ hours}}$

Correct option: B.

For a soap bubble, there are two surfaces: outer and inner.

So, surface energy of one soap bubble is

$\text{Surface energy}=𝑇\times 2\left(4𝜋{𝑟}^2\right)$

$=8𝜋𝑇{𝑟}^2$

Let the radius of the large bubble be $𝑅$.

Given diameter is $𝐷$, so

$𝑅=\frac{𝐷}{2}$

Initial surface energy is

${𝐸}_{𝑖}=8𝜋𝑇{𝑅}^2$

The large bubble breaks into $64$ smaller bubbles. Since air volume remains the same,

$\frac{4}{3}𝜋{𝑅}^3=64\times \frac{4}{3}𝜋{𝑟}^3$

So,

${𝑅}^3=64{𝑟}^3$

$𝑅=4𝑟$

$𝑟=\frac{𝑅}{4}$

Now, final surface energy of $64$ smaller bubbles is

${𝐸}_{𝑓}=64\times 8𝜋𝑇{𝑟}^2$

Substitute $𝑟=\frac{𝑅}{4}$:

${𝐸}_{𝑓}=64\times 8𝜋𝑇{\left(\frac{𝑅}{4}\right)}^2$

${𝐸}_{𝑓}=64\times 8𝜋𝑇\frac{{𝑅}^2}{16}$

${𝐸}_{𝑓}=32𝜋𝑇{𝑅}^2$

Change in surface energy is

$\Delta 𝐸={𝐸}_{𝑓}-{𝐸}_{𝑖}$

$\Delta 𝐸=32𝜋𝑇{𝑅}^2-8𝜋𝑇{𝑅}^2$

$\Delta 𝐸=24𝜋𝑇{𝑅}^2$

Since $𝑅=\frac{𝐷}{2}$,

$\Delta 𝐸=24𝜋𝑇{\left(\frac{𝐷}{2}\right)}^2$

$\Delta 𝐸=24𝜋𝑇\frac{{𝐷}^2}{4}$

$\Delta 𝐸=6𝜋𝑇{𝐷}^2$

So, the correct option is

$\overline{)6𝜋𝑇{𝐷}^2}$

Option C.

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.

Outside pressure is $1$ atmosphere.

For bubble A:

$\Delta {𝑃}_{𝐴}=1.02-1=0.02\text{ atmosphere}$

For bubble B:

$\Delta {𝑃}_{𝐵}=1.04-1=0.04\text{ atmosphere}$

Since

$\Delta 𝑃=\frac{4𝑇}{𝑟}$

for the same soap solution, $𝑇$ is same, so

$𝑟\propto \frac{1}{\Delta 𝑃}$

Thus,

$\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}=\frac{\Delta {𝑃}_{𝐵}}{\Delta {𝑃}_{𝐴}}$

$\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}=\frac{0.04}{0.02}=2$

So,

${𝑟}_{𝐴}:{𝑟}_{𝐵}=2:1$

Volume of a bubble is proportional to ${𝑟}^3$.

$\frac{{𝑉}_{𝐴}}{{𝑉}_{𝐵}}={\left(\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}\right)}^3$

$\frac{{𝑉}_{𝐴}}{{𝑉}_{𝐵}}=2^3=8$

So, the ratio of volumes is

${𝑉}_{𝐴}:{𝑉}_{𝐵}=8:1$

Correct option: Option B, $8:1$

For an incompressible fluid flowing steadily through a pipe, the equation of continuity is used:

${𝐴}_1{𝑣}_1={𝐴}_2{𝑣}_2$

Here, the cross-sectional area of a cylindrical pipe is

$𝐴=𝜋{𝑟}^2$

At point A, radius is $3𝑅$, so

${𝐴}_{𝐴}=𝜋(3𝑅{)}^2=9𝜋{𝑅}^2$

At point B, radius is $1.5𝑅$, so

${𝐴}_{𝐵}=𝜋(1.5𝑅{)}^2=2.25𝜋{𝑅}^2$

Using continuity,

${𝐴}_{𝐴}{𝑣}_{𝐴}={𝐴}_{𝐵}{𝑣}_{𝐵}$

Given ${𝑣}_{𝐴}=𝑣$,

$9𝜋{𝑅}^2\cdot 𝑣=2.25𝜋{𝑅}^2\cdot {𝑣}_{𝐵}$

Cancel $𝜋{𝑅}^2$:

$9𝑣=2.25{𝑣}_{𝐵}$

${𝑣}_{𝐵}=\frac{9}{2.25}𝑣=4𝑣$

So, the velocity at point B is

$\overline{)4𝑣}$

Correct option: Option D.

Incident energy on the surface is

$𝐸=170\text{ J}$

Reflected energy is

${𝐸}_{𝑟}=17\text{ J}$

Coefficient of absorption is

$𝑎=0.7$

The coefficient of absorption means the fraction of incident energy absorbed by the body.

So, absorbed energy is

${𝐸}_{𝑎}=𝑎𝐸$

${𝐸}_{𝑎}=0.7\times 170$

${𝐸}_{𝑎}=119\text{ J}$

Now, incident energy is divided into three parts:

$\text{Incident energy}=\text{Reflected energy}+\text{Absorbed energy}+\text{Transmitted energy}$

So,

$170=17+119+{𝐸}_{𝑡}$

${𝐸}_{𝑡}=170-136$

${𝐸}_{𝑡}=34\text{ J}$

So, the amount of energy transmitted is

$\overline{)34.0\text{ J}}$

Correct option: Option C.

For a black body, the rate of emission of energy is given by Stefan’s law:

$𝐸\propto {𝑇}^4$

Here, $𝑇$ must be the absolute temperature in kelvin, not in degree Celsius.

Given temperature:

${827}^{\circ }\text{C}$

Converting to kelvin:

$𝑇=827+273=1100\text{ K}$

So the rate at which it emits energy is proportional to

$(1100{)}^4$

So the correct option is:

$\overline{)\text{Option C }(1100{)}^4}$

For temperature differences, we use the relation between Celsius and Fahrenheit scales:

$\frac{\Delta 𝐶}{5}=\frac{\Delta 𝐹}{9}$

Here, the difference on the Fahrenheit scale is

$\Delta 𝐹={54}^{\circ }\text{F}$

So,

$\Delta 𝐶=\frac{5}{9}\times 54$

$\Delta 𝐶={30}^{\circ }\text{C}$

So, the difference in temperature on the Celsius scale is

${30}^{\circ }\text{C}$

Correct option: A ${30}^{\circ }\text{C}$

Since the gas is suddenly compressed, there is no time for heat exchange with the surroundings. So the process is adiabatic.

For an adiabatic process,

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

So,

${𝑇}_1{𝑉}_{𝛾-1}^1={𝑇}_2{𝑉}_{𝛾-1}^2$

Given:

${𝑉}_2=\frac{{𝑉}_1}{4}$

and

$𝛾=1.5$

So,

$𝛾-1=1.5-1=0.5$

Now,

$\frac{{𝑇}_2}{{𝑇}_1}={\left(\frac{{𝑉}_1}{{𝑉}_2}\right)}^{𝛾-1}$

$\frac{{𝑇}_2}{{𝑇}_1}={\left(\frac{{𝑉}_1}{{𝑉}_1/4}\right)}^{0.5}$

$\frac{{𝑇}_2}{{𝑇}_1}=4^{0.5}=2$

So,

${𝑇}_2=2{𝑇}_1$

At normal temperature,

${𝑇}_1=273\text{K}$

Hence,

${𝑇}_2=2\times 273=546\text{K}$

Increase in temperature:

$\Delta 𝑇={𝑇}_2-{𝑇}_1$

$\Delta 𝑇=546-273=273\text{K}$

So, the increase in temperature is

$\overline{)273\text{K}}$

Correct option: A.

The correct option is:

Option D: isochoric process

In thermodynamics, work done by a gas is given by

$𝑊=𝑃\Delta 𝑉$

where $\Delta 𝑉$ is the change in volume.

In an isochoric process, the volume remains constant. So,

$\Delta 𝑉=0$

Hence,

$𝑊=𝑃\times 0=0$

So, no work is done by the gas or on the gas in an isochoric process.

For an ideal gas, the kinetic energy of its molecules is directly proportional to the absolute temperature.

So,

$𝐸\propto 𝑇$

Temperature must be taken in kelvin.

Initial temperature:

${27}^{\circ }\text{C}=27+273=300\text{ K}$

Final temperature:

${177}^{\circ }\text{C}=177+273=450\text{ K}$

Since kinetic energy is proportional to temperature,

$\frac{𝐸}{{𝐸}_0}=\frac{450}{300}$

$\frac{𝐸}{{𝐸}_0}=\frac{3}{2}$

So,

$𝐸=\frac{3{𝐸}_0}{2}$

The correct option is Option C: $\frac{3{𝐸}_0}{2}$.

The time period of the SHM is

$𝑇=8\text{s}$

So the angular frequency is

$𝜔=\frac{2𝜋}{𝑇}=\frac{2𝜋}{8}=\frac{𝜋}{4}{\text{rad s}}^{-1}$

At $𝑡=0$, the particle is at the mean position. So we can write its displacement as

$𝑥=𝐴\sin 𝜔𝑡$

where $𝐴$ is the amplitude.

So,

$𝑥=𝐴\sin \left(\frac{𝜋}{4}𝑡\right)$

Distance travelled in the 1st second means from $𝑡=0$ to $𝑡=1$.

At $𝑡=0$,

${𝑥}_0=𝐴\sin 0=0$

At $𝑡=1$,

${𝑥}_1=𝐴\sin \frac{𝜋}{4}=𝐴\sin {45}^{\circ }=\frac{𝐴}{\sqrt{\surd 2}}$

So distance travelled in the 1st second is

${𝑠}_1=\frac{𝐴}{\sqrt{\surd 2}}-0=\frac{𝐴}{\sqrt{\surd 2}}$

Now distance travelled in the 2nd second means from $𝑡=1$ to $𝑡=2$.

At $𝑡=2$,

${𝑥}_2=𝐴\sin \frac{𝜋}{2}=𝐴\sin {90}^{\circ }=𝐴$

So distance travelled in the 2nd second is

${𝑠}_2=𝐴-\frac{𝐴}{\sqrt{\surd 2}}$

Now the required ratio is

$\frac{{𝑠}_2}{{𝑠}_1}=\frac{𝐴-\frac{𝐴}{\sqrt{\surd 2}}}{\frac{𝐴}{\sqrt{\surd 2}}}$

Taking $𝐴$ common in the numerator,

$\frac{{𝑠}_2}{{𝑠}_1}=\frac{𝐴(1-\frac{1}{\sqrt{\surd 2}})}{\frac{𝐴}{\sqrt{\surd 2}}}$

Cancel $𝐴$:

$\frac{{𝑠}_2}{{𝑠}_1}=\sqrt{\surd 2}(1-\frac{1}{\sqrt{\surd 2}})$

$\frac{{𝑠}_2}{{𝑠}_1}=\sqrt{\surd 2}-1$

So the correct option is

$\overline{)\sqrt{\surd 2}-1}$

Option C.

For a closed pipe, only odd harmonics are present.

So the overtones are:

• Fundamental: $1^{\text{st}}$ harmonic

• First overtone: $3^{\text{rd}}$ harmonic

• Second overtone: $5^{\text{th}}$ harmonic

• Third overtone: $7^{\text{th}}$ harmonic

So, frequency of third overtone of closed pipe is

${𝑓}_{𝑐}=\frac{7𝑣}{4{𝐿}_{𝑐}}$

For an open pipe, all harmonics are present.

So the third overtone is the $4^{\text{th}}$ harmonic.

Frequency of third overtone of open pipe is

${𝑓}_0=\frac{4𝑣}{2{𝐿}_0}=\frac{2𝑣}{{𝐿}_0}$

Given that both frequencies are equal,

$\frac{7𝑣}{4{𝐿}_{𝑐}}=\frac{2𝑣}{{𝐿}_0}$

Cancel $𝑣$:

$\frac{7}{4{𝐿}_{𝑐}}=\frac{2}{{𝐿}_0}$

Cross-multiply:

$7{𝐿}_0=8{𝐿}_{𝑐}$

So,

$\frac{{𝐿}_{𝑐}}{{𝐿}_0}=\frac{7}{8}$

Hence,

${𝐿}_{𝑐}:{𝐿}_0=7:8$

Correct option: B, $7:8$

For a sonometer wire, when tension and mass per unit length remain the same, the frequency is inversely proportional to the length:

$𝑓\propto \frac{1}{𝑙}$

So,

$𝑓=\frac{𝑘}{𝑙}$

where $𝑘$ is a constant.

For length $33$ cm, frequency of the wire is

${𝑓}_1=\frac{𝑘}{33}$

For length $32$ cm, frequency of the wire is

${𝑓}_2=\frac{𝑘}{32}$

When the wire is shortened, its frequency increases. So,

${𝑓}_2>{𝑓}_1$

The tuning fork gives $5$ beats per second with both lengths. This means the frequency of the tuning fork is $5$ Hz away from each of these two frequencies.

Since ${𝑓}_2$ is greater than ${𝑓}_1$, the fork frequency must lie exactly between them:

${𝑓}_1=𝑛-5$

and

${𝑓}_2=𝑛+5$

So,

${𝑓}_2-{𝑓}_1=10$

Now substitute the sonometer frequencies:

$\frac{𝑘}{32}-\frac{𝑘}{33}=10$

$𝑘\left(\frac{33-32}{32\times 33}\right)=10$

$\frac{𝑘}{1056}=10$

$𝑘=10560$

Now,

${𝑓}_1=\frac{𝑘}{33}$

${𝑓}_1=\frac{10560}{33}=320\text{ Hz}$

Since the fork frequency is $5$ Hz more than this,

$𝑛=320+5=325\text{ Hz}$

So, the frequency of the tuning fork is

$\overline{)325\text{ Hz}}$

Correct option: B.

Let the frequency of the first tuning fork be $𝑓$ Hz.

Since each fork produces $𝑥$ beats per second with the preceding fork, the difference between the frequencies of two consecutive forks is $𝑥$ Hz.

So the frequencies form an arithmetic progression:

$𝑓,𝑓+𝑥,𝑓+2𝑥,\dots ,𝑓+13𝑥$

There are $14$ forks, so the frequency of the last fork is

$𝑓+13𝑥$

Given that the last fork is an octave of the first fork, its frequency is double the first:

$𝑓+13𝑥=2𝑓$

So,

$13𝑥=𝑓$

Now, the seventh fork has frequency $114$ Hz.

The seventh fork is:

$𝑓+6𝑥$

So,

$𝑓+6𝑥=114$

Using $𝑓=13𝑥$,

$13𝑥+6𝑥=114$

$19𝑥=114$

$𝑥=6$

So the value of $𝑥$ is

$\overline{)6}$

Correct option: C.

The wrong statement is:

$\text{(D) In general male sound is sharper than that of a female.}$

Pitch tells us how sharp or shrill a sound is. A higher frequency gives a higher pitch.

In general, a female voice has a higher frequency than a male voice, so it is usually sharper or more shrill. A male voice is generally lower-pitched.

So the correct option is:

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

For each small spherical drop, let the radius be $𝑟$ and charge be $𝑞$.

The potential of a charged spherical drop is

$𝑉=\frac{1}{4𝜋{𝜀}_0}\frac{𝑞}{𝑟}$

When $𝑛$ identical drops coalesce, the total charge becomes

$𝑄=𝑛𝑞$

Also, volume is conserved. So if the radius of the big drop is $𝑅$,

$\frac{4}{3}𝜋{𝑅}^3=𝑛\left(\frac{4}{3}𝜋{𝑟}^3\right)$

So,

${𝑅}^3=𝑛{𝑟}^3$

$𝑅={𝑛}^{\frac{1}{3}}𝑟$

Now the potential of the big drop is

${𝑉}^{\prime }=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{𝑅}$

Substitute $𝑄=𝑛𝑞$ and $𝑅={𝑛}^{\frac{1}{3}}𝑟$:

${𝑉}^{\prime }=\frac{1}{4𝜋{𝜀}_0}\frac{𝑛𝑞}{{𝑛}^{\frac{1}{3}}𝑟}$

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

But

$\frac{1}{4𝜋{𝜀}_0}\frac{𝑞}{𝑟}=𝑉$

Hence,

${𝑉}^{\prime }={𝑛}^{\frac{2}{3}}𝑉$

So the correct option is

$\overline{)\text{Option D }\left({𝑛}^{\frac{2}{3}}\right)𝑉}$

For a point on the axial line of an electric dipole, the electric field is

$𝐸=\frac{1}{4𝜋{𝜀}_0}\frac{2𝑝}{{𝑟}^3}$

The two dipoles are placed in opposite directions, so at a point between them their electric fields are in opposite directions. Hence, for the net field to be zero, their magnitudes must be equal.

Let the distance between the dipoles be $24$ cm.

If the zero-field point is at a distance $𝑥$ from the dipole of moment $𝑃$, then its distance from the other dipole is

$24-𝑥$

So,

$\frac{2𝐾𝑃}{{𝑥}^3}=\frac{2𝐾\left(𝑛𝑃\right)}{(24-𝑥{)}^3}$

where $𝑛𝑃$ is the moment of the second dipole.

Cancel common terms:

$\frac{1}{{𝑥}^3}=\frac{𝑛}{(24-𝑥{)}^3}$

So,

$(24-𝑥{)}^3=𝑛{𝑥}^3$

Taking cube root:

$24-𝑥=\text{exception 25:}𝑥$

Thus,

$𝑥=\frac{24}{1+\text{exception 25:}}$

For the first case, $𝑛=8$:

${𝑥}_1=\frac{24}{1+\text{exception 25:}}$

${𝑥}_1=\frac{24}{1+2}=8\text{ cm}$

For the second case, $𝑛=27$:

${𝑥}_2=\frac{24}{1+\text{exception 25:}}$

${𝑥}_2=\frac{24}{1+3}=6\text{ cm}$

Now,

$|{𝑥}_2-{𝑥}_1|=|6-8|=2\text{ cm}$

So the correct option is

$\overline{)2\text{ cm}}$

Option A.

Charge on the first capacitor is

${𝑄}_1=𝐶𝑉$

Charge on the second capacitor is

${𝑄}_2=\left(3𝐶\right)\left(3𝑉\right)=9𝐶𝑉$

Now the batteries are disconnected, so charge is conserved.

The capacitors are connected in parallel with opposite polarity, because the positive terminal of one is connected to the negative terminal of the other.

So the effective charge left on the common plates is the difference of the two charges:

${𝑄}_{\text{net}}=9𝐶𝑉-𝐶𝑉=8𝐶𝑉$

In parallel, the equivalent capacitance is

${𝐶}_{\text{eq}}=𝐶+3𝐶=4𝐶$

So the final common potential is

${𝑉}_{𝑓}=\frac{{𝑄}_{\text{net}}}{{𝐶}_{\text{eq}}}$

${𝑉}_{𝑓}=\frac{8𝐶𝑉}{4𝐶}=2𝑉$

Final energy stored is

${𝑈}_{𝑓}=\frac{1}{2}{𝐶}_{\text{eq}}{𝑉}_2^{𝑓}$

${𝑈}_{𝑓}=\frac{1}{2}\left(4𝐶\right)(2𝑉{)}^2$

${𝑈}_{𝑓}=8𝐶{𝑉}^2$

So the final energy of the configuration is

$\overline{)8.0𝐶{𝑉}^2}$

Correct option: C.

The dielectric slab has thickness $\frac{4𝑑}{5}$, while the plate separation is $𝑑$.

So, an air gap is left of thickness

$𝑑-\frac{4𝑑}{5}=\frac{𝑑}{5}$

This arrangement behaves like two capacitors in series:

1. One capacitor with dielectric of thickness $\frac{4𝑑}{5}$

2. One capacitor with air of thickness $\frac{𝑑}{5}$

For such a combination, the effective capacitance is

$𝐶=\frac{{𝜀}_0𝐴}{\text{air thickness}+\frac{\text{dielectric thickness}}{𝐾}}$

So,

$𝐶=\frac{{𝜀}_0𝐴}{\frac{𝑑}{5}+\frac{1}{𝐾}\cdot \frac{4𝑑}{5}}$

$𝐶=\frac{{𝜀}_0𝐴}{\frac{𝑑}{5}+\frac{4𝑑}{5𝐾}}$

Taking common denominator,

$𝐶=\frac{{𝜀}_0𝐴}{\frac{𝑑(𝐾+4)}{5𝐾}}$

$𝐶=\frac{5𝐾{𝜀}_0𝐴}{𝑑(𝐾+4)}$

Without dielectric,

${𝐶}_0=\frac{{𝜀}_0𝐴}{𝑑}$

So,

$\frac{𝐶}{{𝐶}_0}=\frac{\frac{5𝐾{𝜀}_0𝐴}{𝑑(𝐾+4)}}{\frac{{𝜀}_0𝐴}{𝑑}}$

$\frac{𝐶}{{𝐶}_0}=\frac{5𝐾}{𝐾+4}$

So the correct option is

$\overline{)\frac{5𝐾}{𝐾+4}}$

Option C.

In a meter bridge, at balance point,

$\frac{\text{resistance in left gap}}{\text{resistance in right gap}}=\frac{\text{length of left wire}}{\text{length of right wire}}$

Initially, left gap resistance is $15\Omega$ and right gap resistance is $𝑅\Omega$.

So,

$\frac{15}{𝑅}=\frac{{𝑙}_1}{100-{𝑙}_1}$

Now, $𝑅$ is shunted with an equal resistance.

So the equivalent resistance in the right gap becomes

${𝑅}_{\text{eq}}=\frac{𝑅\times 𝑅}{𝑅+𝑅}=\frac{𝑅}{2}$

The new balance length is $1.6{𝑙}_1$.

So,

$\frac{15}{𝑅/2}=\frac{1.6{𝑙}_1}{100-1.6{𝑙}_1}$

This gives

$\frac{30}{𝑅}=\frac{1.6{𝑙}_1}{100-1.6{𝑙}_1}$

From the first equation,

$𝑅=\frac{15(100-{𝑙}_1)}{{𝑙}_1}$

Substitute this in the second equation:

$\frac{30}{\frac{15(100-{𝑙}_1)}{{𝑙}_1}}=\frac{1.6{𝑙}_1}{100-1.6{𝑙}_1}$

$\frac{2{𝑙}_1}{100-{𝑙}_1}=\frac{1.6{𝑙}_1}{100-1.6{𝑙}_1}$

Cancel ${𝑙}_1$:

$\frac{2}{100-{𝑙}_1}=\frac{1.6}{100-1.6{𝑙}_1}$

Cross multiply:

$2(100-1.6{𝑙}_1)=1.6(100-{𝑙}_1)$

$200-3.2{𝑙}_1=160-1.6{𝑙}_1$

$40=1.6{𝑙}_1$

${𝑙}_1=25\text{cm}$

Now,

$𝑅=\frac{15(100-25)}{25}$

$𝑅=\frac{15\times 75}{25}=45\Omega$

So, the resistance is

$\overline{)45\Omega }$

Correct option: C.

In a potentiometer, the balancing length is directly proportional to the potential difference being balanced.

When the two cells assist each other, their total e.m.f. is

$𝐸=1.5+1.2=2.7\text{ V}$

For this, the balancing length is

$𝑙=270\text{ cm}$

So the potential gradient is

$\frac{2.7}{270}=0.01\text{ V/cm}$

When the cells are connected in opposition, the effective e.m.f. is the difference of their e.m.f.s:

${𝐸}^{\prime }=1.5-1.2=0.3\text{ V}$

Using the same potential gradient,

${𝑙}^{\prime }=\frac{0.3}{0.01}=30\text{ cm}$

So the balancing length will be

$\overline{)30\text{ cm}}$

Correct option: D.

Power consumed by a lamp is given by

$𝑃=𝑉𝐼$

Here,

$𝑉=(6.0\pm 0.3)\text{V}$

and

$𝐼=(4.0\pm 0.1)\text{A}$

First find the power:

$𝑃=6.0\times 4.0=24.0\text{W}$

For multiplication, percentage errors are added.

Percentage error in voltage:

$\frac{0.3}{6.0}\times 100=5\%$

Percentage error in current:

$\frac{0.1}{4.0}\times 100=2.5\%$

So, percentage error in power:

$5\%+2.5\%=7.5\%$

Now absolute error in power:

$\Delta 𝑃=\frac{7.5}{100}\times 24.0$

$\Delta 𝑃=1.8\text{W}$

So the power consumed is

$𝑃=(24.0\pm 1.8)\text{W}$

Hence, the correct option is

$\overline{)\text{Option A }(24.0\pm 1.8)}$

A diamagnetic liquid is weakly repelled by a magnetic field.

So, when one arm of the U-tube is placed in an external magnetic field, the diamagnetic liquid tries to move away from the region of stronger magnetic field. Because of this repulsion, the liquid level in that arm decreases.

So the level of the liquid in that arm will fall.

Correct option: Option B — fall.

For a small magnet oscillating in a uniform magnetic field, the frequency of oscillation is

$𝑛=\frac{1}{2𝜋}\sqrt{\surd \frac{𝑀𝐵}{𝐼}}$

Here, $𝑀$ is the magnetic moment of the magnet and $𝐼$ is its moment of inertia.

For the same magnet, $𝑀$ and $𝐼$ remain constant. So the frequency depends only on the magnetic field as

$𝑛\propto \sqrt{\surd 𝐵}$

In the field $𝐵$, the frequency is $𝑛$.

In the field $𝑋$, the frequency becomes $\frac{𝑛}{3}$.

So,

$\frac{𝑛/3}{𝑛}=\sqrt{\surd \frac{𝑋}{𝐵}}$

$\frac{1}{3}=\sqrt{\surd \frac{𝑋}{𝐵}}$

Squaring both sides,

$\frac{1}{9}=\frac{𝑋}{𝐵}$

$𝑋=\frac{𝐵}{9}$

So the correct option is

$\overline{)\frac{𝐵}{9}}$

Option D.

For a circular coil of radius $𝑅$, carrying current $𝐼$, the magnetic field at a point on its axis at distance $𝑥$ from the centre is

$𝐵=\frac{{𝜇}_0𝐼{𝑅}^2}{2{({𝑅}^2+{𝑥}^2)}^{3/2}}$

For the bigger coil $𝑌$,

${𝑅}_{𝑌}=2𝑟,{𝑥}_{𝑌}=𝑑$

So,

${𝐵}_{𝑦}=\frac{{𝜇}_0𝐼(2𝑟{)}^2}{2{\left(\right(2𝑟{)}^2+{𝑑}^2)}^{3/2}}$

${𝐵}_{𝑦}=\frac{4{𝜇}_0𝐼{𝑟}^2}{2{(4{𝑟}^2+{𝑑}^2)}^{3/2}}$

${𝐵}_{𝑦}=\frac{2{𝜇}_0𝐼{𝑟}^2}{{({𝑑}^2+4{𝑟}^2)}^{3/2}}$

For the smaller coil $𝑋$,

${𝑅}_{𝑋}=𝑟,{𝑥}_{𝑋}=\frac{𝑑}{2}$

So,

${𝐵}_{𝑥}=\frac{{𝜇}_0𝐼{𝑟}^2}{2{({𝑟}^2+{\left(\frac{𝑑}{2}\right)}^2)}^{3/2}}$