MHT CET 2026 17th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Since the two vectors have the same direction, one vector must be a positive scalar multiple of the other.

Given,

$\overrightarrow{𝐴}=6\widehat{𝑖}+9\widehat{𝑗}-\widehat{𝑘}$

and

$\overrightarrow{𝐵}=2\widehat{𝑖}+3\widehat{𝑗}-𝑝\widehat{𝑘}$

Compare the corresponding components.

For $\widehat{𝑖}$ components:

$\frac{6}{2}=3$

For $\widehat{𝑗}$ components:

$\frac{9}{3}=3$

So $\overrightarrow{𝐴}$ is $3$ times $\overrightarrow{𝐵}$.

Now compare the $\widehat{𝑘}$ components:

$-1=3(-𝑝)$

$-1=-3𝑝$

$𝑝=\frac{1}{3}$

So the value of $𝑝$ is

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

Correct option: Option B.

Pressure is force per unit area.

For a square plate of side $𝐿$, the area is

$𝐴={𝐿}^2$

So pressure is

$𝑃=\frac{𝐹}{𝐴}=\frac{𝐹}{{𝐿}^2}$

For percentage error, when a quantity is multiplied or divided, percentage errors are added. Also, if a quantity has a power, its percentage error is multiplied by that power.

Here,

$𝑃=𝐹{𝐿}^{-2}$

So percentage error in pressure is

$\frac{\Delta 𝑃}{𝑃}\times 100=\frac{\Delta 𝐹}{𝐹}\times 100+2(\frac{\Delta 𝐿}{𝐿}\times 100)$

Given:

$\text{Percentage error in }𝐹=2\%$

$\text{Percentage error in }𝐿=3\%$

Hence,

$\text{Percentage error in }𝑃=2\%+2\left(3\%\right)$

$=2\%+6\%=8\%$

So the correct answer is

$\overline{)8\%}$

Option D.

For potential energy at the highest point, we use

$\text{P.E.}=𝑚𝑔ℎ$

Both balls have the same mass, so the ratio of their potential energies will be the same as the ratio of their maximum heights.

Ball P is projected at an angle of ${60}^{\circ }$ with the vertical.

So, its angle with the horizontal is

${90}^{\circ }-{60}^{\circ }={30}^{\circ }$

For a projectile, maximum height is

$𝐻=\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}$

For ball P,

${𝐻}_{𝑃}=\frac{{𝑢}^2{\sin }^2{30}^{\circ }}{2𝑔}$

Given,

$\sin {30}^{\circ }=0.5$

So,

${𝐻}_{𝑃}=\frac{{𝑢}^2(0.5{)}^2}{2𝑔}=\frac{{𝑢}^2}{8𝑔}$

Ball Q is projected vertically upwards with the same initial speed $𝑢$. Its maximum height is

${𝐻}_{𝑄}=\frac{{𝑢}^2}{2𝑔}$

Now,

$\frac{\text{P.E. of P}}{\text{P.E. of Q}}=\frac{𝑚𝑔{𝐻}_{𝑃}}{𝑚𝑔{𝐻}_{𝑄}}=\frac{{𝐻}_{𝑃}}{{𝐻}_{𝑄}}$

$=\frac{\frac{{𝑢}^2}{8𝑔}}{\frac{{𝑢}^2}{2𝑔}}=\frac{1}{4}$

So, the required ratio is

$1:4$

Correct option: A.

Mass of one bullet is

$𝑚=40𝑔=0.04𝑘𝑔$

Speed of bullet is

$𝑣=600𝑚{𝑠}^{-1}$

Momentum given to one bullet is

$𝑝=𝑚𝑣$

$𝑝=0.04\times 600=24𝑘𝑔𝑚{𝑠}^{-1}$

When the gun fires bullets, the bullets gain forward momentum. By Newton’s third law, the gun gets an equal backward effect, called recoil. The force on the person depends on the change of momentum per second.

If $𝑛$ bullets are fired per second, then total momentum per second is

$24𝑛$

So the average force on the person is

$𝐹=24𝑛$

The maximum force the person can bear is $168𝑁$, so

$24𝑛=168$

$𝑛=\frac{168}{24}=7$

So, the number of bullets that can be fired per second is

$\overline{)7}$

Correct option: B.

For a thin uniform circular disc of mass $𝑀$ and radius $𝑅$,

Moment of inertia about the axis passing through its centre and perpendicular to its plane is

${𝐼}_{𝑐}=\frac{1}{2}𝑀{𝑅}^2$

Radius of gyration is given by

$𝐼=𝑀{𝐾}^2$

So,

$𝑀{𝐾}_2^{𝑐}=\frac{1}{2}𝑀{𝑅}^2$

${𝐾}_{𝑐}=\frac{𝑅}{\sqrt{\surd 2}}$

Now, for the moment of inertia about a diameter of the disc, we use the perpendicular axis theorem.

For a plane lamina,

${𝐼}_{𝑧}={𝐼}_{𝑥}+{𝐼}_{𝑦}$

Here, the two diameters are identical axes, so

${𝐼}_{𝑥}={𝐼}_{𝑦}={𝐼}_{𝑑}$

Thus,

${𝐼}_{𝑐}=2{𝐼}_{𝑑}$

${𝐼}_{𝑑}=\frac{{𝐼}_{𝑐}}{2}$

${𝐼}_{𝑑}=\frac{1}{4}𝑀{𝑅}^2$

So,

$𝑀{𝐾}_2^{𝑑}=\frac{1}{4}𝑀{𝑅}^2$

${𝐾}_{𝑑}=\frac{𝑅}{2}$

Now the ratio is

${𝐾}_{𝑑}:{𝐾}_{𝑐}=\frac{𝑅}{2}:\frac{𝑅}{\sqrt{\surd 2}}$

${𝐾}_{𝑑}:{𝐾}_{𝑐}=1:\sqrt{\surd 2}$

So the correct option is

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

Option B.

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

${𝐼}_{\text{cm}}=\frac{2}{5}𝑀{𝑅}^2$

Here, the new axis is parallel to that diameter and is at a distance

$𝑑=\frac{𝑅}{3}$

from the centre.

Using the parallel axis theorem,

$𝐼={𝐼}_{\text{cm}}+𝑀{𝑑}^2$

So,

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

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

Taking LCM $45$,

$𝐼=(\frac{18}{45}+\frac{5}{45})𝑀{𝑅}^2$

$𝐼=\frac{23}{45}𝑀{𝑅}^2$

So the correct option is

$\overline{)\text{Option C }\frac{23}{45}𝑀{𝑅}^2}$

For a mass moving in a horizontal circle, the angular momentum is

$𝐿=𝐼𝜔$

For a point mass moving in a circle of radius $𝑟$,

$𝐼=𝑚{𝑟}^2$

So,

$𝐿=𝑚{𝑟}^2𝜔$

Here, the angular velocity $𝜔$ is kept the same.

If the length of the string is halved, the radius becomes

${𝑟}^{\prime }=\frac{𝑟}{2}$

So the new angular momentum is

${𝐿}^{\prime }=𝑚{\left(\frac{𝑟}{2}\right)}^2𝜔$

${𝐿}^{\prime }=𝑚\frac{{𝑟}^2}{4}𝜔$

${𝐿}^{\prime }=\frac{1}{4}𝑚{𝑟}^2𝜔$

Since $𝐿=𝑚{𝑟}^2𝜔$,

${𝐿}^{\prime }=\frac{𝐿}{4}$

So the angular momentum becomes

$\overline{)\frac{𝐿}{4}}$

Correct option: A.

For a planet orbiting around the sun, the orbital speed is given by

$𝑣=\sqrt{\surd \frac{𝐺𝑀}{𝑟}}$

Here, $𝐺$ is the gravitational constant, $𝑀$ is the mass of the sun, and $𝑟$ is the distance of the planet from the sun.

So, orbital speed is inversely proportional to the square root of the distance:

$𝑣\propto \frac{1}{\sqrt{\surd 𝑟}}$

For planets A and B,

$\frac{{𝑣}_{𝐵}}{{𝑣}_{𝐴}}=\sqrt{\surd \frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}}$

Given,

${𝑟}_{𝐵}=225{𝑟}_{𝐴}$

So,

$\frac{{𝑣}_{𝐵}}{𝑉}=\sqrt{\surd \frac{{𝑟}_{𝐴}}{225{𝑟}_{𝐴}}}$

$\frac{{𝑣}_{𝐵}}{𝑉}=\sqrt{\surd \frac{1}{225}}$

$\frac{{𝑣}_{𝐵}}{𝑉}=\frac{1}{15}$

Hence,

${𝑣}_{𝐵}=\frac{𝑉}{15}$

So the correct option is Option C:

$\overline{)\frac{𝑉}{15}}$

For capillary rise, the height to which water rises in a tube of radius $𝑟$ is

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

So, for the same liquid and same type of tube,

$ℎ\propto \frac{1}{𝑟}$

The mass of water in the capillary tube is

$𝑚=\text{volume}\times \text{density}$

The volume of water column is

$𝑉=𝜋{𝑟}^2ℎ$

So,

$𝑚=𝜌𝜋{𝑟}^2ℎ$

Since $ℎ\propto \frac{1}{𝑟}$,

$𝑚\propto {𝑟}^2\times \frac{1}{𝑟}$

$𝑚\propto 𝑟$

Now the new capillary tube has radius $𝑥𝑟$.

So the mass of water that rises will become $𝑥$ times the original mass:

${𝑚}^{\prime }=𝑥𝑚$

So the correct option is

$\overline{)𝑥\cdot 𝑚}$

Option A.

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.

Since both are soap bubbles of the same liquid, $𝑇$ is the same.

Given,

$\Delta {𝑃}_1=3\Delta {𝑃}_2$

Using $\Delta 𝑃\propto \frac{1}{𝑟}$,

$\frac{1}{{𝑟}_1}=3\cdot \frac{1}{{𝑟}_2}$

So,

${𝑟}_2=3{𝑟}_1$

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

$𝑉\propto {𝑟}^3$

So,

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

Putting ${𝑟}_2=3{𝑟}_1$,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑟}_3^1}{(3{𝑟}_1{)}^3}=\frac{1}{27}$

So the ratio of volumes is

${𝑉}_1:{𝑉}_2=1:27$

Correct option: D $1:27$

For a soap bubble, the excess pressure inside it is

$𝑃=\frac{4𝑇}{𝑟}$

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

Since both are soap bubbles of the same liquid, $𝑇$ is the same for both. So excess pressure is inversely proportional to radius:

$𝑃\propto \frac{1}{𝑟}$

Given,

${𝑟}_{𝐴}:{𝑟}_{𝐵}=3:2$

So,

${𝑃}_{𝐴}:{𝑃}_{𝐵}=\frac{1}{{𝑟}_{𝐴}}:\frac{1}{{𝑟}_{𝐵}}$

${𝑃}_{𝐴}:{𝑃}_{𝐵}=\frac{1}{3}:\frac{1}{2}$

${𝑃}_{𝐴}:{𝑃}_{𝐵}=2:3$

So the ratio of excess pressure inside bubble A to that inside bubble B is

$\overline{)2:3}$

Correct option: Option A.

For a black body, the wavelength corresponding to maximum intensity is related to temperature by Wien’s displacement law:

${𝜆}_{max}𝑇=\text{constant}$

So, if the temperature increases, the wavelength of maximum intensity decreases.

Initially,

${𝜆}_1=𝜆,{𝑇}_1=𝑇$

New temperature is

${𝑇}_2=2.5𝑇=\frac{5𝑇}{2}$

Using Wien’s law,

${𝜆}_1{𝑇}_1={𝜆}_2{𝑇}_2$

Substitute the values:

$𝜆\cdot 𝑇={𝜆}_2\cdot \frac{5𝑇}{2}$

Cancel $𝑇$ from both sides:

$𝜆={𝜆}_2\cdot \frac{5}{2}$

So,

${𝜆}_2=\frac{2𝜆}{5}$

Hence, the correct option is

$\overline{)\frac{2𝜆}{5}}$

Option B is correct.

For heat conduction through a composite slab in steady state, the same heat current passes through both materials.

Thermal resistance of a slab is

$𝑅=\frac{𝑙}{𝐾𝐴}$

For the first material:

• Thermal conductivity $=𝐾$

• Thickness $=𝑥$

So,

${𝑅}_1=\frac{𝑥}{𝐾𝐴}$

For the second material:

• Thermal conductivity $=2𝐾$

• Thickness $=4𝑥$

So,

${𝑅}_2=\frac{4𝑥}{\left(2𝐾\right)𝐴}$

${𝑅}_2=\frac{2𝑥}{𝐾𝐴}$

Since the two slabs are in series, total thermal resistance is

$𝑅={𝑅}_1+{𝑅}_2$

$𝑅=\frac{𝑥}{𝐾𝐴}+\frac{2𝑥}{𝐾𝐴}$

$𝑅=\frac{3𝑥}{𝐾𝐴}$

The rate of heat transfer is

$𝐻=\frac{{𝑇}_2-{𝑇}_1}{𝑅}$

$𝐻=\frac{{𝑇}_2-{𝑇}_1}{\frac{3𝑥}{𝐾𝐴}}$

$𝐻=\frac{𝐴𝐾({𝑇}_2-{𝑇}_1)}{3𝑥}$

The question gives the rate as

$\left[\frac{𝐴({𝑇}_2-{𝑇}_1)𝐾}{𝑥}\right]\cdot 𝑓$

Comparing,

$𝑓=\frac{1}{3}$

So the correct option is

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

Option D.

For a Carnot engine, efficiency is

$𝜂=1-\frac{{𝑇}_{𝐶}}{{𝑇}_{𝐻}}$

Initially, the efficiency is $\frac{1}{6}$.

So,

$\frac{1}{6}=1-\frac{{𝑇}_{𝐶}}{{𝑇}_{𝐻}}$

$\frac{{𝑇}_{𝐶}}{{𝑇}_{𝐻}}=\frac{5}{6}$

So,

${𝑇}_{𝐻}=\frac{6}{5}{𝑇}_{𝐶}$

Now the sink temperature is decreased by $64\text{K}$, so the new sink temperature is

${𝑇}_{𝐶}-64$

The new efficiency is $\frac{1}{3}$.

So,

$\frac{1}{3}=1-\frac{{𝑇}_{𝐶}-64}{{𝑇}_{𝐻}}$

$\frac{{𝑇}_{𝐶}-64}{{𝑇}_{𝐻}}=\frac{2}{3}$

Substitute ${𝑇}_{𝐻}=\frac{6}{5}{𝑇}_{𝐶}$:

$\frac{{𝑇}_{𝐶}-64}{\frac{6}{5}{𝑇}_{𝐶}}=\frac{2}{3}$

$\frac{5({𝑇}_{𝐶}-64)}{6{𝑇}_{𝐶}}=\frac{2}{3}$

Cross-multiply:

$15({𝑇}_{𝐶}-64)=12{𝑇}_{𝐶}$

$15{𝑇}_{𝐶}-960=12{𝑇}_{𝐶}$

$3{𝑇}_{𝐶}=960$

${𝑇}_{𝐶}=320\text{K}$

Now,

${𝑇}_{𝐻}=\frac{6}{5}\times 320$

${𝑇}_{𝐻}=384\text{K}$

So the temperatures ${𝑇}_{𝐻}$ and ${𝑇}_{𝐶}$ are respectively

$384\text{K},320\text{K}$

Correct option: Option B.

In an adiabatic process, no heat is exchanged with the surroundings.

So,

$𝑄=0$

When a thermodynamic system, like a gas, is compressed adiabatically, work is done on the system. Since no heat can escape, this work increases the internal energy of the system.

For a gas, internal energy is directly related to temperature. So when internal energy increases, temperature also increases.

Hence, during adiabatic compression, the temperature of the system increases.

Correct option: Option C — increases.

In cylinder B, the piston is fixed, so the volume remains constant.

For heating at constant volume,

$𝑄=𝑛{𝐶}_{𝑉}\Delta 𝑇$

Given that the rise in temperature in cylinder B is $63\text{K}$,

$𝑄=𝑛{𝐶}_{𝑉}\times 63$

In cylinder A, the piston is free to move, so the gas expands while being heated. The pressure remains constant.

For heating at constant pressure,

$𝑄=𝑛{𝐶}_{𝑃}\Delta {𝑇}_{𝐴}$

The same amount of heat is supplied to both cylinders, so

$𝑛{𝐶}_{𝑃}\Delta {𝑇}_{𝐴}=𝑛{𝐶}_{𝑉}\times 63$

For an ideal diatomic gas,

${𝐶}_{𝑉}=\frac{5}{2}𝑅$

and

${𝐶}_{𝑃}=\frac{7}{2}𝑅$

So,

$\Delta {𝑇}_{𝐴}=\frac{{𝐶}_{𝑉}}{{𝐶}_{𝑃}}\times 63$

$\Delta {𝑇}_{𝐴}=\frac{\frac{5}{2}𝑅}{\frac{7}{2}𝑅}\times 63$

$\Delta {𝑇}_{𝐴}=\frac{5}{7}\times 63$

$\Delta {𝑇}_{𝐴}=45\text{K}$

So, the rise in temperature of the gas in cylinder A is

$\overline{)45\text{K}}$

Option C.

For a gas, the r.m.s. velocity is

${𝑣}_{\text{rms}}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

So, for two gases,

${𝑣}_{\text{rms}}\propto \sqrt{\surd \frac{𝑇}{𝑀}}$

Here, hydrogen has molecular weight $2$, and oxygen has molecular weight $32$.

Temperature of oxygen is

${47}^{\circ }\text{C}=47+273=320\text{ K}$

It is given that the r.m.s. velocity of hydrogen is $4.5$ times that of oxygen:

$\frac{{𝑣}_{𝐻}}{{𝑣}_{𝑂}}=4.5$

Using the relation,

$\frac{{𝑣}_{𝐻}}{{𝑣}_{𝑂}}=\sqrt{\surd \frac{{𝑇}_{𝐻}/{𝑀}_{𝐻}}{{𝑇}_{𝑂}/{𝑀}_{𝑂}}}$

Substitute the values:

$4.5=\sqrt{\surd \frac{{𝑇}_{𝐻}/2}{320/32}}$

Squaring both sides,

$(4.5{)}^2=\frac{{𝑇}_{𝐻}/2}{320/32}$

$20.25=\frac{{𝑇}_{𝐻}/2}{10}$

$20.25=\frac{{𝑇}_{𝐻}}{20}$

So,

${𝑇}_{𝐻}=405\text{ K}$

Convert this into Celsius:

${𝑇}_{𝐻}=405-273={132}^{\circ }\text{C}$

So, the correct option is

$\overline{){132}^{\circ }\text{C}}$

Option B.

Given,

$𝑦=𝐴\cos \left[𝜋\right(𝑡+𝜙\left)\right]$

This is of the form

$𝑦=𝐴\cos (𝜔𝑡+\text{constant})$

So the angular frequency is

$𝜔=𝜋$

At $𝑡=0$, displacement is $2$ cm:

$2=𝐴\cos \left(𝜋𝜙\right)$

Now velocity is

$𝑣=\frac{𝑑𝑦}{𝑑𝑡}$

Differentiating,

$𝑣=-𝐴𝜋\sin \left[𝜋\right(𝑡+𝜙\left)\right]$

At $𝑡=0$, velocity is $2𝜋$ cm/s:

$2𝜋=-𝐴𝜋\sin \left(𝜋𝜙\right)$

Dividing by $𝜋$,

$2=-𝐴\sin \left(𝜋𝜙\right)$

So we have

$𝐴\cos \left(𝜋𝜙\right)=2$

and

$𝐴\sin \left(𝜋𝜙\right)=-2$

Now square and add both equations:

${𝐴}^2{\cos }^2\left(𝜋𝜙\right)+{𝐴}^2{\sin }^2\left(𝜋𝜙\right)=2^2+(-2{)}^2$

${𝐴}^2[{\cos }^2(𝜋𝜙)+{\sin }^2(𝜋𝜙\left)\right]=4+4$

Using

${\cos }^2𝜃+{\sin }^2𝜃=1$

we get

${𝐴}^2=8$

$𝐴=2\sqrt{\surd 2}$

So the amplitude is

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

Correct option: D.

For a mass attached to a light spring, the time period of small oscillations is

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}$

where $𝑚$ is the mass attached and $𝑘$ is the spring constant.

When only mass ${𝑚}_1$ is attached,

$𝑇=2𝜋\sqrt{\surd \frac{{𝑚}_1}{𝑘}}$

When mass ${𝑚}_2$ is added, the total mass becomes ${𝑚}_1+{𝑚}_2$.

So the new time period is

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{{𝑚}_1+{𝑚}_2}{𝑘}}$

Given that

${𝑇}^{\prime }=\frac{3}{2}𝑇$

Now divide the two time period formulas:

$\frac{{𝑇}^{\prime }}{𝑇}=\sqrt{\surd \frac{{𝑚}_1+{𝑚}_2}{{𝑚}_1}}$

Substitute ${𝑇}^{\prime }=\frac{3}{2}𝑇$:

$\frac{3}{2}=\sqrt{\surd \frac{{𝑚}_1+{𝑚}_2}{{𝑚}_1}}$

Squaring both sides,

$\frac{9}{4}=\frac{{𝑚}_1+{𝑚}_2}{{𝑚}_1}$

$\frac{9}{4}=1+\frac{{𝑚}_2}{{𝑚}_1}$

$\frac{{𝑚}_2}{{𝑚}_1}=\frac{9}{4}-1$

$\frac{{𝑚}_2}{{𝑚}_1}=\frac{5}{4}$

So,

$\frac{{𝑚}_1}{{𝑚}_2}=\frac{4}{5}$

Hence, the ratio is

${𝑚}_1:{𝑚}_2=4:5$

Correct option: C, $4:5$

For linear S.H.M., the total energy of the particle is

$𝐸=\frac{1}{2}𝑘{𝑟}^2$

where $𝑟$ is the amplitude.

At displacement $𝑥$ from the mean position, the potential energy is

$𝑈=\frac{1}{2}𝑘{𝑥}^2$

Given that the potential energy is $𝜆$ times the total energy,

$𝑈=𝜆𝐸$

So,

$\frac{1}{2}𝑘{𝑥}^2=𝜆\left(\frac{1}{2}𝑘{𝑟}^2\right)$

Cancel $\frac{1}{2}𝑘$ from both sides:

${𝑥}^2=𝜆{𝑟}^2$

Taking square root,

$𝑥=𝑟\sqrt{\surd 𝜆}$

So the displacement of the particle is

$\overline{)𝑟\sqrt{\surd 𝜆}}$

Correct option: C.

For a sonometer wire, the fundamental frequency is

$𝑛=\frac{1}{2𝑙}\sqrt{\surd \frac{𝑇}{𝜇}}$

Here, the length $𝑙$ and mass per unit length $𝜇$ of the wire remain the same.

So,

$𝑛\propto \sqrt{\surd 𝑇}$

or

${𝑛}^2\propto 𝑇$

When the bob is hanging in air, the tension in the wire is equal to its weight:

${𝑇}_1=𝑉𝜌𝑔$

where $𝜌$ is the density of the metal.

So,

${𝑛}_2^1\propto 𝑉𝜌𝑔$

When the bob is completely immersed in water, it loses weight due to upthrust. So the effective tension becomes

${𝑇}_2=𝑉𝜌𝑔-𝑉{𝜌}_{𝑤}𝑔$

${𝑇}_2=𝑉𝑔(𝜌-{𝜌}_{𝑤})$

So,

${𝑛}_2^2\propto 𝑉𝑔(𝜌-{𝜌}_{𝑤})$

Now taking ratio,

$\frac{{𝑛}_2^2}{{𝑛}_2^1}=\frac{𝑉𝑔(𝜌-{𝜌}_{𝑤})}{𝑉𝜌𝑔}$

$\frac{{𝑛}_2^2}{{𝑛}_2^1}=\frac{𝜌-{𝜌}_{𝑤}}{𝜌}$

$\frac{{𝑛}_2^2}{{𝑛}_2^1}=1-\frac{{𝜌}_{𝑤}}{𝜌}$

Relative density of the metal is

$𝑅=\frac{𝜌}{{𝜌}_{𝑤}}$

So,

$\frac{{𝜌}_{𝑤}}{𝜌}=\frac{1}{𝑅}$

Hence,

$\frac{{𝑛}_2^2}{{𝑛}_2^1}=1-\frac{1}{𝑅}$

$\frac{1}{𝑅}=1-\frac{{𝑛}_2^2}{{𝑛}_2^1}$

$\frac{1}{𝑅}=\frac{{𝑛}_2^1-{𝑛}_2^2}{{𝑛}_2^1}$

Therefore,

$𝑅=\frac{{𝑛}_2^1}{{𝑛}_2^1-{𝑛}_2^2}$

So the correct option is

$\overline{)\text{Option C }\frac{{𝑛}_2^1}{{𝑛}_2^1-{𝑛}_2^2}}$

Beats are produced when two sound waves of nearly equal frequencies superpose.

For beats to be heard clearly, the two waves should have almost the same amplitude and slightly different frequencies. Due to this small difference in frequency, the sound becomes alternately loud and soft.

So the correct option is:

Option C: same amplitude but slightly different frequencies.

In a pipe closed at one end, the closed end is a displacement node and the open end is a displacement antinode.

Only odd harmonics are present in a closed pipe:

$1^{\text{st}},3^{\text{rd}},5^{\text{th}},7^{\text{th}},9^{\text{th}},\dots$

The fourth overtone means:

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

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

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

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

• Fourth overtone = $9^{\text{th}}$ harmonic

For the $9^{\text{th}}$ harmonic in a closed pipe, the air column contains $9$ quarter-wavelength parts.

The pattern is:

$𝑁𝐴𝑁𝐴𝑁𝐴𝑁𝐴𝑁𝐴$

So, number of nodes $=5$ and number of antinodes $=5$.

Hence,

$𝑥=5,𝑦=5$

Correct option: D $5,5$

For sound waves in the same air at the same temperature, the velocity of sound is the same.

We know,

$𝑣=𝑓𝜆$

So,

$𝑓=\frac{𝑣}{𝜆}$

The two wavelengths are

${𝜆}_1=50\text{ cm}=0.50\text{ m}$

${𝜆}_2=50.5\text{ cm}=0.505\text{ m}$

Their frequencies differ by $6\text{ Hz}$, so

$\mid \frac{𝑣}{0.50}-\frac{𝑣}{0.505}\mid =6$

Take $𝑣$ common:

$𝑣(\frac{1}{0.50}-\frac{1}{0.505})=6$

Now,

$\frac{1}{0.50}=2$

and

$\frac{1}{0.505}\approx 1.9802$

So,

$𝑣(2-1.9802)=6$

$𝑣\left(0.0198\right)=6$

$𝑣=\frac{6}{0.0198}$

$𝑣\approx 303\text{ m/s}$

So, the velocity of sound in air is

$\overline{)303\text{ m/s}}$

Correct option: C.

For a system of point charges, the electrostatic potential energy is the sum of potential energies of all pairs of charges.

Since the charges are placed at the vertices of an equilateral triangle, all separations are equal. Let each side be $𝑎$.

The three pairs are:

$+𝑞$ and $+2𝑞$:

${𝑈}_1=\frac{𝑘\left(𝑞\right)\left(2𝑞\right)}{𝑎}$

$+𝑞$ and $+𝑄$:

${𝑈}_2=\frac{𝑘\left(𝑞\right)\left(𝑄\right)}{𝑎}$

$+2𝑞$ and $+𝑄$:

${𝑈}_3=\frac{𝑘\left(2𝑞\right)\left(𝑄\right)}{𝑎}$

Total potential energy is zero:

$\frac{𝑘}{𝑎}(2{𝑞}^2+𝑞𝑄+2𝑞𝑄)=0$

$\frac{𝑘}{𝑎}(2{𝑞}^2+3𝑞𝑄)=0$

Since $\frac{𝑘}{𝑎}$ is not zero,

$2{𝑞}^2+3𝑞𝑄=0$

Taking $𝑞$ common:

$𝑞(2𝑞+3𝑄)=0$

So,

$2𝑞+3𝑄=0$

$3𝑄=-2𝑞$

$𝑄=-\frac{2𝑞}{3}$

So the correct option is

$\overline{)-\frac{2𝑞}{3}}$

Option B.

The charge $+𝑞$ is placed between $+𝑒$ and $+4𝑒$.

Since all charges are positive, both fixed charges repel $+𝑞$.

• Force on $+𝑞$ due to $+𝑒$ is away from $+𝑒$.

• Force on $+𝑞$ due to $+4𝑒$ is away from $+4𝑒$.

As $+𝑞$ is between them, these two forces act in opposite directions. For equilibrium, their magnitudes must be equal.

Distance of $+𝑞$ from $+𝑒$ is $𝑥$.

So distance of $+𝑞$ from $+4𝑒$ is $𝑑-𝑥$.

Using Coulomb’s law,

${𝐹}_1=\frac{𝑘\left(𝑒\right)\left(𝑞\right)}{{𝑥}^2}$

and

${𝐹}_2=\frac{𝑘\left(4𝑒\right)\left(𝑞\right)}{(𝑑-𝑥{)}^2}$

For equilibrium,

$\frac{𝑘𝑒𝑞}{{𝑥}^2}=\frac{4𝑘𝑒𝑞}{(𝑑-𝑥{)}^2}$

Cancel $𝑘𝑒𝑞$ from both sides:

$\frac{1}{{𝑥}^2}=\frac{4}{(𝑑-𝑥{)}^2}$

Taking square root,

$\frac{1}{𝑥}=\frac{2}{𝑑-𝑥}$

So,

$𝑑-𝑥=2𝑥$

$𝑑=3𝑥$

$𝑥=\frac{𝑑}{3}$

So the correct option is:

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

Option A.

Let the two identical plates be brought close to form a parallel plate capacitor.

Since the plates are identical, the charge on the outer surfaces will be equal. Let this common charge be $𝑥$.

The facing inner surfaces form the capacitor. Let the charge on the inner surface of the plate having charge ${𝑞}_1$ be $+𝑄$, and on the inner surface of the other plate be $-𝑄$.

So, total charge on first plate is

$𝑥+𝑄={𝑞}_1$

and total charge on second plate is

$𝑥-𝑄={𝑞}_2$

Subtracting these two equations,

$(𝑥+𝑄)-(𝑥-𝑄)={𝑞}_1-{𝑞}_2$

$2𝑄={𝑞}_1-{𝑞}_2$

$𝑄=\frac{{𝑞}_1-{𝑞}_2}{2}$

For a capacitor,

$𝑉=\frac{𝑄}{𝐶}$

So,

$𝑉=\frac{{𝑞}_1-{𝑞}_2}{2𝐶}$

Hence the correct option is

$\overline{)\text{Option C }\frac{{𝑞}_1-{𝑞}_2}{2𝐶}}$

The capacitors $2𝐶$, $3𝐶$, and $4𝐶$ are connected in parallel.

For capacitors in parallel, the equivalent capacitance is the sum of all capacitances.

${𝐶}_{𝑝}=2𝐶+3𝐶+4𝐶=9𝐶$

So, the parallel combination has capacitance $9𝐶$.

This $9𝐶$ combination is connected in series with ${𝐶}^1$.

The total equivalent capacitance between A and B is given as $6𝐶$.

For capacitors in series, we use:

$\frac{1}{{𝐶}_{𝑒𝑞}}=\frac{1}{{𝐶}_{𝑝}}+\frac{1}{{𝐶}^1}$

Substitute the given values:

$\frac{1}{6𝐶}=\frac{1}{9𝐶}+\frac{1}{{𝐶}^1}$

Now subtract $\frac{1}{9𝐶}$ from both sides:

$\frac{1}{{𝐶}^1}=\frac{1}{6𝐶}-\frac{1}{9𝐶}$

Take LCM of $6𝐶$ and $9𝐶$, which is $18𝐶$:

$\frac{1}{{𝐶}^1}=\frac{3-2}{18𝐶}=\frac{1}{18𝐶}$

So,

${𝐶}^1=18𝐶$

The $9\text{ A}$ current enters at point P and has two possible paths to reach point R.

One path is P $\to$ Q $\to$ R. In this path, two wires of $3\Omega$ each are in series, so the resistance is

${𝑅}_1=3\Omega +3\Omega =6\Omega$

The current in this path is ${𝐼}_1$.

The other path is directly P $\to$ R. This path has only one wire of resistance $3\Omega$, so

${𝑅}_2=3\Omega$

The current in this path is ${𝐼}_2$.

These two paths are connected between the same points P and R, so they are in parallel. In parallel paths, the potential difference is the same.

${𝑉}_{𝑃𝑅}={𝐼}_1{𝑅}_1={𝐼}_2{𝑅}_2$

Substituting the resistances:

${𝐼}_1\left(6\right)={𝐼}_2\left(3\right)$

${𝐼}_2=2{𝐼}_1$

Now use Kirchhoff's current law at point P. The total current entering P is equal to the sum of currents in the two branches.

${𝐼}_1+{𝐼}_2=9$

Put ${𝐼}_2=2{𝐼}_1$:

${𝐼}_1+2{𝐼}_1=9$

$3{𝐼}_1=9$

${𝐼}_1=3\text{ A}$

Now find ${𝐼}_2$:

${𝐼}_2=2\left(3\right)=6\text{ A}$

So, the currents ${𝐼}_1$ and ${𝐼}_2$ are $3\text{ A}$ and $6\text{ A}$ respectively.

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

Given, current through galvanometer is $4\%$ of total current.

If total current is $𝐼$, then

${𝐼}_{𝑔}=\frac{4}{100}𝐼=\frac{𝐼}{25}$

So current through the shunt is

${𝐼}_{𝑠}=𝐼-{𝐼}_{𝑔}=𝐼-\frac{𝐼}{25}$

${𝐼}_{𝑠}=\frac{24𝐼}{25}$

Since galvanometer and shunt are in parallel,

${𝐼}_{𝑔}𝐺={𝐼}_{𝑠}𝑆$

So,

$𝑆=\frac{{𝐼}_{𝑔}𝐺}{{𝐼}_{𝑠}}$

Substitute the values:

$𝑆=\frac{\frac{𝐼}{25}𝐺}{\frac{24𝐼}{25}}$

$𝑆=\frac{𝐺}{24}$

So the shunt resistance is

$\overline{)\frac{𝐺}{24}}$

Correct option: C.

Let the length of each wire be $𝐿$ and the current in both loops be $𝐼$.

Magnetic moment of a current loop is

$𝑀=𝑁𝐼𝐴$

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

For the circular loop, the wire makes $2$ turns.

If radius of each circular turn is $𝑟$, then total length of wire is

$𝐿=2\left(2𝜋𝑟\right)=4𝜋𝑟$

So,

$𝑟=\frac{𝐿}{4𝜋}$

Magnetic moment of the circular loop is

${𝑀}_{𝑐}=2𝐼\left(𝜋{𝑟}^2\right)$

${𝑀}_{𝑐}=2𝐼𝜋{\left(\frac{𝐿}{4𝜋}\right)}^2$

${𝑀}_{𝑐}=\frac{𝐼{𝐿}^2}{8𝜋}$

For the square loop, let the side be $𝑎$.

Since it is one turn,

$𝐿=4𝑎$

So,

$𝑎=\frac{𝐿}{4}$

Magnetic moment of the square loop is

${𝑀}_{𝑠}=𝐼{𝑎}^2$

${𝑀}_{𝑠}=𝐼{\left(\frac{𝐿}{4}\right)}^2$

${𝑀}_{𝑠}=\frac{𝐼{𝐿}^2}{16}$

Now the ratio is

$\frac{{𝑀}_{𝑐}}{{𝑀}_{𝑠}}=\frac{\frac{𝐼{𝐿}^2}{8𝜋}}{\frac{𝐼{𝐿}^2}{16}}=\frac{2}{𝜋}$

So, the ratio is

$2:𝜋$

Correct option: D.

For a long straight current-carrying wire, the magnetic field at distance $𝑟$ is

$𝐵=\frac{{𝜇}_0𝐼}{2𝜋𝑟}$

So, the magnetic field is inversely proportional to the distance from the wire:

$𝐵\propto \frac{1}{𝑟}$

At distance $𝑟$, the field is given as

$𝐵=0.4\text{ T}$

At distance $2𝑟$, the distance becomes double, so the magnetic field becomes half:

${𝐵}^{\prime }=\frac{0.4}{2}=0.2\text{ T}$

So, the magnetic field at distance $2𝑟$ will be

$0.2\text{ tesla}$

Correct option: Option B, $0\cdot 2$ tesla