MHT CET 2026 16th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Density is given by

$𝜌=\frac{𝑚}{𝑉}$

For a cube of side length $𝑙$,

$𝑉={𝑙}^3$

So,

$𝜌=\frac{𝑚}{{𝑙}^3}$

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

So the percentage error in density is

$\frac{\Delta 𝜌}{𝜌}\times 100=\frac{\Delta 𝑚}{𝑚}\times 100+3(\frac{\Delta 𝑙}{𝑙}\times 100)$

Given:

$\text{Error in mass}=4\%$

$\text{Error in length}=3\%$

So,

$\text{Error in density}=4\%+3\left(3\%\right)$

$=4\%+9\%$

$=13\%$

So the correct option is

$\overline{)13\%}$

Option A.

Since the vectors are perpendicular, their dot product is zero.

The given vectors are

$𝑎\widehat{𝑖}+𝑏\widehat{𝑗}+\widehat{𝑘}$

and

$2\widehat{𝑖}-3\widehat{𝑗}+4\widehat{𝑘}$

So,

$\left(𝑎\right)\left(2\right)+\left(𝑏\right)(-3)+\left(1\right)\left(4\right)=0$

$2𝑎-3𝑏+4=0$

$2𝑎-3𝑏=-4\text{...(1)}$

Also given,

$3𝑎+2𝑏=7\text{...(2)}$

Now solve the two equations:

$2𝑎-3𝑏=-4$

$3𝑎+2𝑏=7$

Multiply the first equation by $3$:

$6𝑎-9𝑏=-12$

Multiply the second equation by $2$:

$6𝑎+4𝑏=14$

Subtract:

$13𝑏=26$

$𝑏=2$

Put $𝑏=2$ in $3𝑎+2𝑏=7$:

$3𝑎+2\left(2\right)=7$

$3𝑎+4=7$

$3𝑎=3$

$𝑎=1$

So,

$\frac{𝑎}{𝑏}=\frac{1}{2}$

Given that the ratio is $\frac{𝑥}{2}$,

$\frac{𝑥}{2}=\frac{1}{2}$

Hence,

$𝑥=1$

The correct option is Option B: $1$.

For a projectile projected with speed $𝑢$ at an angle $𝜃$ with the horizontal, the horizontal range is

$𝑅=\frac{{𝑢}^2\sin 2𝜃}{𝑔}$

For the first object projected at angle $𝛼$,

${𝑅}_1=\frac{{𝑢}^2\sin 2𝛼}{𝑔}$

For the second object projected at angle $𝛽$,

${𝑅}_2=\frac{{𝑢}^2\sin 2𝛽}{𝑔}$

Given,

$𝛼+𝛽={90}^{\circ }$

So,

$𝛽={90}^{\circ }-𝛼$

Now,

$2𝛽=2({90}^{\circ }-𝛼)={180}^{\circ }-2𝛼$

Using the given identity,

$\sin (𝜋-𝜃)=\sin 𝜃$

or in degrees,

$\sin ({180}^{\circ }-2𝛼)=\sin 2𝛼$

So,

$\sin 2𝛽=\sin 2𝛼$

Hence,

${𝑅}_1={𝑅}_2$

So the ratio of horizontal ranges is

${𝑅}_1:{𝑅}_2=1:1$

Correct option: C.

Since the body begins to move, its initial velocity is zero.

Given mass,

$𝑚=1\text{kg}$

Force is

$\overrightarrow{𝐹}=𝑡\widehat{𝑖}+3{𝑡}^2\widehat{𝑗}$

Using Newton’s second law,

$\overrightarrow{𝐹}=𝑚\overrightarrow{𝑎}$

Since $𝑚=1\text{kg}$,

$\overrightarrow{𝑎}=𝑡\widehat{𝑖}+3{𝑡}^2\widehat{𝑗}$

Now acceleration is the rate of change of velocity.

For the $𝑥$-direction,

${𝑎}_{𝑥}=𝑡$

So,

${𝑣}_{𝑥}=\int 𝑡𝑑𝑡=\frac{{𝑡}^2}{2}$

For the $𝑦$-direction,

${𝑎}_{𝑦}=3{𝑡}^2$

So,

${𝑣}_{𝑦}=\int 3{𝑡}^2𝑑𝑡={𝑡}^3$

Thus,

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

At $𝑡=2\text{s}$,

$\overrightarrow{𝐹}=2\widehat{𝑖}+3(2{)}^2\widehat{𝑗}$

$\overrightarrow{𝐹}=2\widehat{𝑖}+12\widehat{𝑗}$

Also,

$\overrightarrow{𝑣}=\frac{(2{)}^2}{2}\widehat{𝑖}+(2{)}^3\widehat{𝑗}$

$\overrightarrow{𝑣}=2\widehat{𝑖}+8\widehat{𝑗}$

Power is given by

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

So,

$𝑃=\left(2\right)\left(2\right)+\left(12\right)\left(8\right)$

$𝑃=4+96$

$𝑃=100\text{W}$

So, the correct option is

$\overline{)100}$

Option A.

For the disc, the tangent is in its plane.

This tangent is parallel to a diameter of the disc and is at a distance $𝑅$ from the centre.

Moment of inertia of a disc about a diameter is

${𝐼}_{\text{diameter}}=\frac{1}{4}𝑚{𝑅}^2$

Here mass of disc is $\frac{𝑀}{2}$, so

${𝐼}_{\text{diameter}}=\frac{1}{4}\cdot \frac{𝑀}{2}{𝑅}^2=\frac{𝑀{𝑅}^2}{8}$

Using parallel axis theorem for the tangent:

${𝐼}_{\text{disc tangent}}={𝐼}_{\text{diameter}}+𝑚{𝑅}^2$

${𝐼}_{\text{disc tangent}}=\frac{𝑀{𝑅}^2}{8}+\frac{𝑀}{2}{𝑅}^2$

${𝐼}_{\text{disc tangent}}=\frac{𝑀{𝑅}^2}{8}+\frac{4𝑀{𝑅}^2}{8}=\frac{5𝑀{𝑅}^2}{8}$

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

${𝐼}_{\text{diameter}}=\frac{2}{5}𝑀{𝑅}^2$

A tangent to the sphere is parallel to a diameter and is at a distance $𝑅$ from the centre. So by parallel axis theorem,

${𝐼}_{\text{sphere tangent}}=\frac{2}{5}𝑀{𝑅}^2+𝑀{𝑅}^2$

${𝐼}_{\text{sphere tangent}}=\frac{7}{5}𝑀{𝑅}^2$

Now the required ratio is

$\frac{{𝐼}_{\text{disc tangent}}}{{𝐼}_{\text{sphere tangent}}}=\frac{\frac{5}{8}𝑀{𝑅}^2}{\frac{7}{5}𝑀{𝑅}^2}$

$=\frac{5}{8}\times \frac{5}{7}=\frac{25}{56}$

So the ratio is

$\overline{)25:56}$

Correct option: B.

For a uniform solid cylinder, the moment of inertia about its own axis is

$𝐼=\frac{1}{2}𝑀{𝑅}^2$

For the original cylinder,

${𝐼}_1=\frac{1}{2}𝑀{𝑅}^2$

where $𝑀$ is its mass.

Now, the carved out cylinder has radius

$\frac{𝑅}{2}$

and length

$\frac{𝐿}{2}$

Since the material is uniform, mass is proportional to volume.

Volume of original cylinder:

$𝑉=𝜋{𝑅}^2𝐿$

Volume of carved out cylinder:

${𝑉}^{\prime }=𝜋{\left(\frac{𝑅}{2}\right)}^2\left(\frac{𝐿}{2}\right)$

${𝑉}^{\prime }=𝜋\cdot \frac{{𝑅}^2}{4}\cdot \frac{𝐿}{2}=\frac{1}{8}𝜋{𝑅}^2𝐿$

So the mass of the carved out portion is

${𝑀}^{\prime }=\frac{𝑀}{8}$

Moment of inertia of carved out portion about the same axis:

${𝐼}_2=\frac{1}{2}{𝑀}^{\prime }{\left(\frac{𝑅}{2}\right)}^2$

Substitute ${𝑀}^{\prime }=\frac{𝑀}{8}$:

${𝐼}_2=\frac{1}{2}\cdot \frac{𝑀}{8}\cdot \frac{{𝑅}^2}{4}$

${𝐼}_2=\frac{1}{32}\left(\frac{1}{2}𝑀{𝑅}^2\right)$

Since

${𝐼}_1=\frac{1}{2}𝑀{𝑅}^2$

we get

$\frac{{𝐼}_1}{{𝐼}_2}=32$

So,

${𝐼}_1:{𝐼}_2=32:1$

Correct option: D, $32:1$

For a solid sphere, the moment of inertia about an axis passing through its centre of mass is

${𝐼}_{𝑐𝑚}=\frac{2}{5}𝑀{𝑅}^2$

The given axis does not pass through the centre of the sphere. So we use the parallel axis theorem:

$𝐼={𝐼}_{𝑐𝑚}+𝑀{𝑑}^2$

Here, $𝑑$ is the distance between the centre of the sphere and the given axis.

Given $𝑅=5$ cm and $𝑑=10$ cm,

$𝐼=\frac{2}{5}𝑀(5{)}^2+𝑀(10{)}^2$

$𝐼=\frac{2}{5}𝑀\times 25+100𝑀$

$𝐼=10𝑀+100𝑀=110𝑀$

The radius of gyration $𝑘$ is defined by

$𝐼=𝑀{𝑘}^2$

So,

$𝑀{𝑘}^2=110𝑀$

${𝑘}^2=110$

$𝑘=\sqrt{\surd 110}$ cm

The radius of gyration is given as $\sqrt{\surd 𝑥}$ cm. So,

$𝑥=110$

The height is $ℎ=2𝑅$, so the body is taken from the surface of the earth to a distance

$𝑅+ℎ=𝑅+2𝑅=3𝑅$

from the centre of the earth.

For large heights, we cannot use $\Delta 𝑈=𝑚𝑔ℎ$ because $𝑔$ changes with height. We use gravitational potential energy:

$𝑈=-\frac{𝐺𝑀𝑚}{𝑟}$

At the surface of the earth, $𝑟=𝑅$:

${𝑈}_1=-\frac{𝐺𝑀𝑚}{𝑅}$

At height $2𝑅$, the distance from the centre is $3𝑅$:

${𝑈}_2=-\frac{𝐺𝑀𝑚}{3𝑅}$

Increase in potential energy is

$\Delta 𝑈={𝑈}_2-{𝑈}_1$

$\Delta 𝑈=-\frac{𝐺𝑀𝑚}{3𝑅}-(-\frac{𝐺𝑀𝑚}{𝑅})$

$\Delta 𝑈=\frac{𝐺𝑀𝑚}{𝑅}-\frac{𝐺𝑀𝑚}{3𝑅}$

$\Delta 𝑈=\frac{2𝐺𝑀𝑚}{3𝑅}$

Now, on the earth’s surface,

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

So,

$𝐺𝑀=𝑔{𝑅}^2$

Substitute this:

$\Delta 𝑈=\frac{2\left(𝑔{𝑅}^2\right)𝑚}{3𝑅}$

$\Delta 𝑈=\frac{2}{3}𝑚𝑔𝑅$

So the correct option is

$\overline{)\frac{2}{3}𝑚𝑔𝑅}$

Option A.

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

For one small drop of radius $𝑟$,

$\text{Surface area}=4𝜋{𝑟}^2$

So, for $1000$ small drops,

$\text{Total initial surface area}=1000\times 4𝜋{𝑟}^2$

When they combine to form one big drop, volume remains conserved.

So,

$1000\times \frac{4}{3}𝜋{𝑟}^3=\frac{4}{3}𝜋{𝑅}^3$

where $𝑅$ is the radius of the big drop.

Canceling common terms,

${𝑅}^3=1000{𝑟}^3$

$𝑅=10𝑟$

Now surface area of the big drop is

$4𝜋{𝑅}^2=4𝜋(10𝑟{)}^2$

$=400𝜋{𝑟}^2$

Initial total surface area is

$1000\times 4𝜋{𝑟}^2=4000𝜋{𝑟}^2$

Since surface energy is proportional to surface area,

$\frac{\text{Final surface energy}}{\text{Initial surface energy}}=\frac{400𝜋{𝑟}^2}{4000𝜋{𝑟}^2}=\frac{1}{10}$

So the ratio is

$1:10$

Correct option: B.

For a soap bubble, the excess pressure inside it is

$\Delta 𝑃=\frac{4𝑇}{𝑟}$

because a soap bubble has two liquid surfaces.

So, pressure of air inside a soap bubble of radius $𝑟$ is

$𝑃+\frac{4𝑇}{𝑟}$

where $𝑃$ is the outside pressure and $𝑇$ is the surface tension.

Since the process is isothermal, temperature remains constant. So we use

$𝑃𝑉=𝑛𝑅𝑇$

At constant temperature, the amount of gas is proportional to pressure $\times$ volume.

For the two initial bubbles,

${𝑛}_1+{𝑛}_2=𝑛$

So,

$(𝑃+\frac{4𝑇}{{𝑟}_1})\frac{4}{3}𝜋{𝑟}_3^1+(𝑃+\frac{4𝑇}{{𝑟}_2})\frac{4}{3}𝜋{𝑟}_3^2=(𝑃+\frac{4𝑇}{𝑅})\frac{4}{3}𝜋{𝑅}^3$

Cancel $\frac{4}{3}𝜋$ from all terms:

$(𝑃+\frac{4𝑇}{{𝑟}_1}){𝑟}_3^1+(𝑃+\frac{4𝑇}{{𝑟}_2}){𝑟}_3^2=(𝑃+\frac{4𝑇}{𝑅}){𝑅}^3$

Expand each term:

$𝑃{𝑟}_3^1+4𝑇{𝑟}_2^1+𝑃{𝑟}_3^2+4𝑇{𝑟}_2^2=𝑃{𝑅}^3+4𝑇{𝑅}^2$

Bring similar terms together:

$4𝑇({𝑟}_2^1+{𝑟}_2^2-{𝑅}^2)=𝑃({𝑅}^3-{𝑟}_3^1-{𝑟}_3^2)$

Hence,

$𝑇=\frac{𝑃({𝑅}^3-{𝑟}_3^1-{𝑟}_3^2)}{4({𝑟}_2^1+{𝑟}_2^2-{𝑅}^2)}$

So the correct option is

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

For a horizontal pipe, the height remains the same, so Bernoulli’s theorem gives

${𝑃}_1+\frac{1}{2}𝜚{𝑉}_2^1={𝑃}_2+\frac{1}{2}𝜚{𝑉}_2^2$

So,

${𝑃}_2={𝑃}_1+\frac{1}{2}𝜚({𝑉}_2^1-{𝑉}_2^2)$

Now use the equation of continuity:

${𝐴}_1{𝑉}_1={𝐴}_2{𝑉}_2$

Hence,

${𝑉}_2=\frac{{𝐴}_1{𝑉}_1}{{𝐴}_2}$

So,

${𝑉}_2^2=\frac{{𝐴}_2^1{𝑉}_2^1}{{𝐴}_2^2}$

Substitute this in Bernoulli’s equation:

${𝑃}_2={𝑃}_1+\frac{1}{2}𝜚({𝑉}_2^1-\frac{{𝐴}_2^1{𝑉}_2^1}{{𝐴}_2^2})$

Taking ${𝑉}_2^1$ common,

${𝑃}_2={𝑃}_1+\frac{1}{2}𝜚{𝑉}_2^1(1-\frac{{𝐴}_2^1}{{𝐴}_2^2})$

${𝑃}_2={𝑃}_1+\frac{𝜚{𝑉}_2^1}{2{𝐴}_2^2}({𝐴}_2^2-{𝐴}_2^1)$

So the correct option is

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

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

For a small temperature range, we use the average temperature of the body during cooling.

Room temperature $={22}^{\circ }\text{C}$

First case:

Water cools from ${98}^{\circ }\text{C}$ to ${86}^{\circ }\text{C}$ in $2$ minutes.

Average temperature of water:

$\frac{98+86}{2}={92}^{\circ }\text{C}$

Temperature difference from surroundings:

$92-22={70}^{\circ }\text{C}$

Rate of cooling:

$\frac{98-86}{2}=\frac{12}{2}=6^{\circ }\text{C/min}$

So,

$6=𝑘\left(70\right)$

Now for cooling from ${75}^{\circ }\text{C}$ to ${69}^{\circ }\text{C}$:

Average temperature:

$\frac{75+69}{2}={72}^{\circ }\text{C}$

Temperature difference from surroundings:

$72-22={50}^{\circ }\text{C}$

Let the required time be $𝑡$ minutes.

Rate of cooling:

$\frac{75-69}{𝑡}=\frac{6}{𝑡}$

Using Newton’s law:

$\frac{6}{𝑡}=𝑘\left(50\right)$

From the first case:

$𝑘=\frac{6}{70}$

So,

$\frac{6}{𝑡}=\frac{6}{70}\times 50$

$\frac{6}{𝑡}=\frac{30}{7}$

$𝑡=\frac{6\times 7}{30}=1.4\text{ minutes}$

So, the water will take

$\overline{)1.4\text{ minutes}}$

Correct option: Option B.

For a Carnot engine, efficiency is

$𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}$

where ${𝑇}_1$ is the source temperature and ${𝑇}_2$ is the sink temperature.

Given initial efficiency,

$𝜂=50\%=0.5$

and source temperature,

${𝑇}_1=600\text{K}$

So,

$0.5=1-\frac{{𝑇}_2}{600}$

$\frac{{𝑇}_2}{600}=0.5$

${𝑇}_2=300\text{K}$

So the sink temperature is $300\text{K}$.

Now the efficiency is increased by $20\%$, so the new efficiency becomes

$50\%+20\%=70\%=0.7$

The sink temperature remains the same, so ${𝑇}_2=300\text{K}$.

Using Carnot efficiency again,

$0.7=1-\frac{300}{{𝑇}_{\prime }^1}$

$\frac{300}{{𝑇}_{\prime }^1}=0.3$

${𝑇}_{\prime }^1=\frac{300}{0.3}$

${𝑇}_{\prime }^1=1000\text{K}$

So the new temperature of the source will be

$\overline{)1000\text{K}}$

Correct option: C.

During vaporisation, the heat supplied is used in two ways:

1. To do work in increasing the volume.

2. To increase the internal energy of the substance.

According to the first law of thermodynamics,

$𝑄=\Delta 𝑈+𝑊$

Here, the work done during expansion is

$𝑊=𝑃\Delta 𝑉$

Given,

$𝑃=3\times {10}^5\text{Pa}$

and

$\Delta 𝑉=1600{\text{cm}}^3$

Convert volume into SI unit:

$1600{\text{cm}}^3=1600\times {10}^{-6}{\text{m}}^3=1.6\times {10}^{-3}{\text{m}}^3$

So,

$𝑊=3\times {10}^5\times 1.6\times {10}^{-3}$

$𝑊=480\text{J}$

It is given that this work is $10\%$ of the heat supplied.

So,

$480=\frac{10}{100}𝑄$

$𝑄=4800\text{J}$

Now,

$𝑄=\Delta 𝑈+𝑊$

$4800=\Delta 𝑈+480$

$\Delta 𝑈=4800-480$

$\Delta 𝑈=4320\text{J}$

So, the increase in internal energy is

$\overline{)4320\text{J}}$

Correct option: C.

For a gas molecule, the average translational kinetic energy is

${𝐸}_1=\frac{3}{2}𝑘𝑇$

Using

$𝑘=\frac{𝑅}{𝑁}$

where $𝑁$ is Avogadro’s number, we get

${𝐸}_1=\frac{3}{2}\cdot \frac{𝑅}{𝑁}𝑇$

The kinetic energy gained by an electron accelerated from rest through a potential difference $𝑉$ is

${𝐸}_2=𝑒𝑉$

Given that

${𝐸}_1={𝐸}_2$

So,

$\frac{3𝑅𝑇}{2𝑁}=𝑒𝑉$

Now solve for $𝑇$:

$𝑇=\frac{2𝑁𝑒𝑉}{3𝑅}$

Hence, the required temperature is

$\overline{)\frac{2𝑉𝑁𝑒}{3𝑅}}$

So the correct option is Option A.

For a rigid diatomic gas, the molar heat capacity at constant volume is

${𝐶}_{𝑉}=\frac{5𝑅}{2}$

because a rigid diatomic molecule has $5$ degrees of freedom: $3$ translational and $2$ rotational.

Suppose the gas contains $𝑛$ moles.

Its total mass is

$𝑀=𝑛𝑚$

where $𝑚$ is the molar mass.

Since the container is moving with velocity $𝑉$, the gas also has the same bulk kinetic energy:

$𝐾=\frac{1}{2}𝑀{𝑉}^2$

$𝐾=\frac{1}{2}\left(𝑛𝑚\right){𝑉}^2$

When the insulated container is stopped suddenly, this kinetic energy is converted into internal energy of the gas. Since the container is insulated, heat is not exchanged with surroundings.

So,

$𝑛{𝐶}_{𝑉}\Delta 𝑇=\frac{1}{2}𝑛𝑚{𝑉}^2$

Substitute ${𝐶}_{𝑉}=\frac{5𝑅}{2}$:

$𝑛\left(\frac{5𝑅}{2}\right)\Delta 𝑇=\frac{1}{2}𝑛𝑚{𝑉}^2$

Cancel $𝑛$ and $\frac{1}{2}$:

$5𝑅\Delta 𝑇=𝑚{𝑉}^2$

So,

$\Delta 𝑇=\frac{𝑚{𝑉}^2}{5𝑅}$

The correct option is

$\overline{)\frac{𝑚{𝑉}^2}{5𝑅}}$

Option C.

For a gas, the r.m.s. velocity of molecules is given by

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

where $𝑅$ is the gas constant, $𝑇$ is the absolute temperature, and $𝑀$ is the molar mass of the gas.

In an isothermal compression, the temperature of the gas remains constant. Since ${𝑣}_{\text{rms}}$ depends only on temperature for a given gas, it does not change during isothermal compression.

So, the r.m.s. velocity remains the same.

Correct option: B. Remains the same

When ${𝑚}_1$ passes through the mean position, its speed is maximum.

For the original SHM,

$𝜔=\sqrt{\surd \frac{𝑘}{{𝑚}_1}}$

So maximum speed of ${𝑚}_1$ is

${𝑣}_{max}=𝜔𝐴$

${𝑣}_{max}=𝐴\sqrt{\surd \frac{𝑘}{{𝑚}_1}}$

At the mean position, mass ${𝑚}_2$ is placed on ${𝑚}_1$, and both move together. Since the spring force is zero at the mean position, horizontal momentum is conserved at that instant.

So,

${𝑚}_1{𝑣}_{max}=({𝑚}_1+{𝑚}_2){𝑣}^{\prime }$

${𝑣}^{\prime }=\frac{{𝑚}_1}{{𝑚}_1+{𝑚}_2}{𝑣}_{max}$

After this, the combined mass is ${𝑚}_1+{𝑚}_2$, so the new angular frequency is

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

Since the combined system starts from the mean position with speed ${𝑣}^{\prime }$, its new amplitude is

${𝐴}_1=\frac{{𝑣}^{\prime }}{{𝜔}^{\prime }}$

Substitute the values:

${𝐴}_1=\frac{\frac{{𝑚}_1}{{𝑚}_1+{𝑚}_2}𝐴\sqrt{\surd \frac{𝑘}{{𝑚}_1}}}{\sqrt{\surd \frac{𝑘}{{𝑚}_1+{𝑚}_2}}}$

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

So,

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

Hence, the correct option is

$\overline{){\left[\frac{{𝑚}_1+{𝑚}_2}{{𝑚}_1}\right]}^{\frac{1}{2}}}$

Option A.

The maximum potential energy in S.H.M. occurs at the extreme position, where displacement is $𝐴$.

So the total energy of the block is

$𝐸=\frac{1}{2}𝑘{𝐴}^2$

At displacement $\frac{𝐴}{2}$, the potential energy is

$𝑈=\frac{1}{2}𝑘{\left(\frac{𝐴}{2}\right)}^2$

$𝑈=\frac{1}{2}𝑘\cdot \frac{{𝐴}^2}{4}$

$𝑈=\frac{1}{4}\left(\frac{1}{2}𝑘{𝐴}^2\right)$

Since

$\frac{1}{2}𝑘{𝐴}^2=𝐸$

we get

$𝑈=\frac{𝐸}{4}$

Now, total energy in S.H.M. remains constant:

$\text{Total energy}=\text{Kinetic energy}+\text{Potential energy}$

So,

$𝐸=𝐾+\frac{𝐸}{4}$

$𝐾=𝐸-\frac{𝐸}{4}$

$𝐾=\frac{3𝐸}{4}$

So the kinetic energy of the block at displacement $\frac{𝐴}{2}$ is

$\overline{)\frac{3𝐸}{4}}$

Correct option: Option A.

For a simple pendulum, the time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

Here, $𝑙$ is the length of the pendulum and $𝑔$ is acceleration due to gravity.

At the earth’s surface,

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

The pendulum is taken to a height $𝑅$ above the earth’s surface. So its distance from the centre of the earth becomes

$𝑅+𝑅=2𝑅$

At this height, acceleration due to gravity becomes

${𝑔}^{\prime }=\frac{𝐺𝑀}{(2𝑅{)}^2}$

${𝑔}^{\prime }=\frac{𝐺𝑀}{4{𝑅}^2}$

Since

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

we get

${𝑔}^{\prime }=\frac{𝑔}{4}$

Now the new time period is

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{𝑙}{{𝑔}^{\prime }}}$

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{𝑙}{𝑔/4}}$

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{4𝑙}{𝑔}}$

${𝑇}^{\prime }=2\times 2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

${𝑇}^{\prime }=2𝑇$

So,

$𝑥𝑇=2𝑇$

$𝑥=2$

The correct option is C: $2$.

For a pipe closed at one end, resonance occurs only at odd multiples of $\frac{𝜆}{4}$.

First find the wavelength of the sound wave:

$𝑣=𝑓𝜆$

So,

$𝜆=\frac{𝑣}{𝑓}$

Given,

$𝑣=332\text{m/s},𝑓=415\text{Hz}$

$𝜆=\frac{332}{415}=0.8\text{m}$

$𝜆=80\text{cm}$

For a closed pipe, resonant lengths are:

$𝐿=\frac{𝜆}{4},\frac{3𝜆}{4},\frac{5𝜆}{4},\frac{7𝜆}{4},\dots$

So,

$\frac{𝜆}{4}=\frac{80}{4}=20\text{cm}$

The possible resonant lengths are:

$20\text{cm},60\text{cm},100\text{cm},\dots$

Among the given options:

• $20\text{cm}$ is resonant

• $60\text{cm}$ is resonant

• $100\text{cm}$ is resonant

• $40\text{cm}$ is not resonant

So the length not in resonance is:

$\overline{)40\text{cm}}$

Correct option: B.

For the two waves,

${𝑌}_1=0.25\sin 316𝑡$

and

${𝑌}_2=0.25\sin 310𝑡$

we compare them with the standard form

$𝑌=𝐴\sin 𝜔𝑡$

So their angular frequencies are

${𝜔}_1=316{\text{ rad s}}^{-1}$

and

${𝜔}_2=310{\text{ rad s}}^{-1}$

The number of beats produced per second is equal to the difference in ordinary frequencies:

$𝑛=|{𝑓}_1-{𝑓}_2|$

Since

$𝑓=\frac{𝜔}{2𝜋}$

we get

$𝑛=\frac{|{𝜔}_1-{𝜔}_2|}{2𝜋}$

Substituting the values,

$𝑛=\frac{|316-310|}{2𝜋}$

$𝑛=\frac{6}{2𝜋}$

$𝑛=\frac{3}{𝜋}$

So, the number of beats produced per second is

$\overline{)\frac{3}{𝜋}}$

Correct option: A.

For an open pipe of length $𝐿$, the fundamental frequency is

${𝑓}_{𝑜}=\frac{𝑣}{2𝐿}$

The second overtone of an open pipe is the third harmonic, so its frequency is

$3{𝑓}_{𝑜}$

Now the same pipe is closed at one end. Its fundamental frequency becomes

${𝑓}_{𝑐}=\frac{𝑣}{4𝐿}$

Since

${𝑓}_{𝑜}=\frac{𝑣}{2𝐿}$

we get

${𝑓}_{𝑐}=\frac{{𝑓}_{𝑜}}{2}$

In a closed pipe, only odd harmonics are present.

Third overtone of a closed pipe means the $7^{\text{th}}$ harmonic.

So its frequency is

$7{𝑓}_{𝑐}=7\left(\frac{{𝑓}_{𝑜}}{2}\right)=\frac{7{𝑓}_{𝑜}}{2}$

Given that this is higher than the second overtone of the open pipe by $150\text{Hz}$:

$\frac{7{𝑓}_{𝑜}}{2}-3{𝑓}_{𝑜}=150$

$\frac{7{𝑓}_{𝑜}-6{𝑓}_{𝑜}}{2}=150$

$\frac{{𝑓}_{𝑜}}{2}=150$

${𝑓}_{𝑜}=300\text{Hz}$

So, the fundamental frequency of the open pipe is

$\overline{)300\text{Hz}}$

Correct option: D.

For a stationary person at the station, train A is coming towards the person and train B is moving away from the person.

The speed of sound is

$𝑣=330\text{m/s}$

Speed of each train is

${𝑣}_{𝑠}=30\text{m/s}$

Original frequency of sound emitted by each train is

$𝑓=300\text{Hz}$

When the source of sound moves towards a stationary observer, the apparent frequency is higher:

${𝑓}_{𝐴}=𝑓\left(\frac{𝑣}{𝑣-{𝑣}_{𝑠}}\right)$

So for train A,

${𝑓}_{𝐴}=300\left(\frac{330}{330-30}\right)$

${𝑓}_{𝐴}=300\left(\frac{330}{300}\right)$

${𝑓}_{𝐴}=330\text{Hz}$

When the source of sound moves away from a stationary observer, the apparent frequency is lower:

${𝑓}_{𝐵}=𝑓\left(\frac{𝑣}{𝑣+{𝑣}_{𝑠}}\right)$

So for train B,

${𝑓}_{𝐵}=300\left(\frac{330}{330+30}\right)$

${𝑓}_{𝐵}=300\left(\frac{330}{360}\right)$

${𝑓}_{𝐵}=275\text{Hz}$

The difference in frequencies heard by the person is

${𝑓}_{𝐴}-{𝑓}_{𝐵}=330-275$

$=55\text{Hz}$

So the correct option is

$\overline{)55\text{Hz}}$

Option C.

For a hollow spherical conductor, charge spreads on its outer surface.

Surface charge density is

$𝜎=\frac{𝑄}{4𝜋{𝑅}^2}$

Both spheres have the same charge $𝑄$.

For the first sphere,

${𝜎}_1=\frac{𝑄}{4𝜋{𝑅}_2^1}$

For the second sphere,

${𝜎}_2=\frac{𝑄}{4𝜋{𝑅}_2^2}$

So,

$\frac{{𝜎}_1}{{𝜎}_2}=\frac{\frac{𝑄}{4𝜋{𝑅}_2^1}}{\frac{𝑄}{4𝜋{𝑅}_2^2}}$

Canceling common terms,

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

So the ratio is

${𝜎}_1:{𝜎}_2={𝑅}_2^2:{𝑅}_2^1$

Correct option: C.

The torque on an electric dipole in a uniform electric field is

$𝜏=𝑝𝐸\sin 𝜃$

where dipole moment is

$𝑝=𝑞𝑙$

Here, $𝑙$ is the length of the dipole.

So,

$𝜏=𝑞𝑙𝐸\sin 𝜃$

Given,

$𝜏=5\times {10}^{-9}\text{Nm}$

$𝑙=0.5𝜇\text{m}=0.5\times {10}^{-6}\text{m}$

$𝐸={10}^4\text{V/m}$

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

Now substitute the values:

$5\times {10}^{-9}=𝑞\times (0.5\times {10}^{-6})\times {10}^4\times 0.5$

$𝑞=\frac{5\times {10}^{-9}}{0.5\times {10}^{-6}\times {10}^4\times 0.5}$

$𝑞=\frac{5\times {10}^{-9}}{2.5\times {10}^{-3}}$

$𝑞=2\times {10}^{-6}\text{C}$

Since,

$1𝜇\text{C}={10}^{-6}\text{C}$

we get

$𝑞=2𝜇\text{C}$

So, the correct option is

$\overline{)\text{Option C }2𝜇\text{C}}$

The potential energy of charge ${𝑞}_3$ at point C is due to its interaction with both charges ${𝑞}_1$ and ${𝑞}_2$.

${𝑈}_{𝐶}=\frac{1}{4𝜋{𝜀}_0}(\frac{{𝑞}_1{𝑞}_3}{𝐴𝐶}+\frac{{𝑞}_2{𝑞}_3}{𝐵𝐶})$

From the figure, $𝐴𝐶=40\text{ cm}=0.4\text{ m}$ and $𝐴𝐵=30\text{ cm}=0.3\text{ m}$.

Triangle ABC is right-angled, so we use Pythagoras theorem to find $𝐵𝐶$.

$𝐵𝐶=\sqrt{\surd 𝐴{𝐵}^2+𝐴{𝐶}^2}=\sqrt{\surd (0.3{)}^2+(0.4{)}^2}=0.5\text{ m}$

So, the potential energy at C becomes:

${𝑈}_{𝐶}=\frac{{𝑞}_3}{4𝜋{𝜀}_0}(\frac{{𝑞}_1}{0.4}+\frac{{𝑞}_2}{0.5})$

Now ${𝑞}_3$ is moved from C to D along a circular path of radius $40\text{ cm}$ with centre at A. So the distance from A remains the same.

$𝐴𝐷=𝐴𝐶=40\text{ cm}=0.4\text{ m}$

From the figure, points A, B, and D lie on the same straight line. So the distance $𝐵𝐷$ is:

$𝐵𝐷=𝐴𝐷-𝐴𝐵=0.4\text{ m}-0.3\text{ m}=0.1\text{ m}$

The potential energy of charge ${𝑞}_3$ at point D is:

${𝑈}_{𝐷}=\frac{1}{4𝜋{𝜀}_0}(\frac{{𝑞}_1{𝑞}_3}{𝐴𝐷}+\frac{{𝑞}_2{𝑞}_3}{𝐵𝐷})=\frac{{𝑞}_3}{4𝜋{𝜀}_0}(\frac{{𝑞}_1}{0.4}+\frac{{𝑞}_2}{0.1})$

The change in potential energy is:

$\Delta 𝑈={𝑈}_{𝐷}-{𝑈}_{𝐶}$

$\Delta 𝑈=\frac{{𝑞}_3}{4𝜋{𝜀}_0}(\frac{{𝑞}_1}{0.4}+\frac{{𝑞}_2}{0.1}-\frac{{𝑞}_1}{0.4}-\frac{{𝑞}_2}{0.5})$