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

Given,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

To find the error in $𝑔$, first write the formula in terms of $𝑔$:

$𝑔=\frac{4{𝜋}^2𝑙}{{𝑇}^2}$

For percentage error, constants like $4{𝜋}^2$ do not contribute any error. Since $𝑙$ is in numerator and ${𝑇}^2$ is in denominator,

$\frac{\Delta 𝑔}{𝑔}\times 100=\frac{\Delta 𝑙}{𝑙}\times 100+2\frac{\Delta 𝑇}{𝑇}\times 100$

Now, length is about

$𝑙=80\text{ cm}$

and its accuracy is

$\Delta 𝑙=0.1\text{ cm}$

So,

$\frac{\Delta 𝑙}{𝑙}\times 100=\frac{0.1}{80}\times 100=0.125\%$

The period is about $1.5$ s. Time for $50$ oscillations is

$𝑡=50\times 1.5=75\text{ s}$

The least count of stopwatch is $0.1$ s, so

$\Delta 𝑡=0.1\text{ s}$

Since

$𝑇=\frac{𝑡}{50}$

the fractional error in $𝑇$ is same as the fractional error in $𝑡$:

$\frac{\Delta 𝑇}{𝑇}=\frac{\Delta 𝑡}{𝑡}=\frac{0.1}{75}$

So,

$2\frac{\Delta 𝑇}{𝑇}\times 100=2\times \frac{0.1}{75}\times 100=0.267\%$

Hence,

$\frac{\Delta 𝑔}{𝑔}\times 100=0.125+0.267=0.392\%$

Nearly,

$\overline{)0.4\%}$

So the correct option is Option B: $0.4$.

For motion in a straight line with uniform acceleration, we use the equation

${𝑣}^2-{𝑢}^2=2𝑎𝑠$

For distance $𝐴𝐵$:

Initial speed at $𝐴$, $𝑢=3\text{m/s}$

Final speed at $𝐵$, $𝑣=6\text{m/s}$

Distance $𝐴𝐵=60\text{m}$

So,

$6^2-3^2=2𝑎\left(60\right)$

$36-9=120𝑎$

$27=120𝑎$

$𝑎=\frac{27}{120}=\frac{9}{40}{\text{m/s}}^2$

Now for distance $𝐵𝐶$:

Initial speed at $𝐵$, $𝑢=6\text{m/s}$

Final speed at $𝐶$, $𝑣=9\text{m/s}$

Acceleration is the same, because the acceleration is uniform.

Again using

${𝑣}^2-{𝑢}^2=2𝑎𝑠$

$9^2-6^2=2\left(\frac{9}{40}\right)𝑠$

$81-36=\frac{18}{40}𝑠$

$45=\frac{9}{20}𝑠$

$𝑠=45\times \frac{20}{9}$

$𝑠=100\text{m}$

So, the distance $𝐵𝐶$ is

$\overline{)100\text{m}}$

Correct option: A.

For motion in a circle of constant radius $𝑟$, the centripetal acceleration is

${𝑎}_{𝑐}=\frac{{𝑣}^2}{𝑟}$

Given,

${𝑎}_{𝑐}={𝑘}^2𝑟{𝑡}^2$

So,

$\frac{{𝑣}^2}{𝑟}={𝑘}^2𝑟{𝑡}^2$

Multiplying by $𝑟$,

${𝑣}^2={𝑘}^2{𝑟}^2{𝑡}^2$

So the speed is

$𝑣=𝑘𝑟𝑡$

Since the speed is changing with time, there is a tangential acceleration also:

${𝑎}_{𝑡}=\frac{𝑑𝑣}{𝑑𝑡}$

Now,

$𝑣=𝑘𝑟𝑡$

So,

${𝑎}_{𝑡}=𝑘𝑟$

The centripetal force is always perpendicular to velocity, so it does no work. Only the tangential force delivers power.

Tangential force is

${𝐹}_{𝑡}=𝑚{𝑎}_{𝑡}=𝑚𝑘𝑟$

Power is

$𝑃={𝐹}_{𝑡}𝑣$

So,

$𝑃=\left(𝑚𝑘𝑟\right)\left(𝑘𝑟𝑡\right)$

$𝑃=𝑚{𝑘}^2{𝑟}^2𝑡$

Hence, the correct option is

$\overline{)𝑚{𝑘}^2{𝑟}^2𝑡}$

Option B.

The position vector is

$\overrightarrow{𝑟}=𝑎\cos 𝜔𝑡\widehat{𝑖}+𝑎\sin 𝜔𝑡\widehat{𝑗}$

Velocity is the rate of change of position vector with time:

$\overrightarrow{𝑣}=\frac{𝑑\overrightarrow{𝑟}}{𝑑𝑡}$

So,

$\overrightarrow{𝑣}=\frac{𝑑}{𝑑𝑡}(𝑎\cos 𝜔𝑡)\widehat{𝑖}+\frac{𝑑}{𝑑𝑡}(𝑎\sin 𝜔𝑡)\widehat{𝑗}$

$\overrightarrow{𝑣}=-𝑎𝜔\sin 𝜔𝑡\widehat{𝑖}+𝑎𝜔\cos 𝜔𝑡\widehat{𝑗}$

Now, to find the angle between $\overrightarrow{𝑟}$ and $\overrightarrow{𝑣}$, use dot product:

$\overrightarrow{𝑟}\cdot \overrightarrow{𝑣}=\left|\overrightarrow{𝑟}\right|\left|\overrightarrow{𝑣}\right|\cos 𝜃$

First calculate $\overrightarrow{𝑟}\cdot \overrightarrow{𝑣}$:

$\overrightarrow{𝑟}\cdot \overrightarrow{𝑣}=(𝑎\cos 𝜔𝑡)(-𝑎𝜔\sin 𝜔𝑡)+(𝑎\sin 𝜔𝑡)(𝑎𝜔\cos 𝜔𝑡)$

$=-{𝑎}^2𝜔\sin 𝜔𝑡\cos 𝜔𝑡+{𝑎}^2𝜔\sin 𝜔𝑡\cos 𝜔𝑡$

$=0$

So,

$\left|\overrightarrow{𝑟}\right|\left|\overrightarrow{𝑣}\right|\cos 𝜃=0$

Since $\left|\overrightarrow{𝑟}\right|$ and $\left|\overrightarrow{𝑣}\right|$ are not zero,

$\cos 𝜃=0$

We know that

$\cos {90}^{\circ }=0$

So,

$𝜃={90}^{\circ }$

The angle between the position vector and the velocity vector is

$\overline{){90}^{\circ }}$

Correct option: C.

When two springs are connected one after the other, they are in series.

Since the springs are massless and connected in a single chain, the same stretching force $𝐹$ acts through both springs.

For the first spring,

${𝑥}_1=\frac{𝐹}{{𝐾}_1}$

For the second spring,

${𝑥}_2=\frac{𝐹}{{𝐾}_2}$

The total extension is the sum of the extensions of both springs:

$𝑥={𝑥}_1+{𝑥}_2$

So,

$𝑥=\frac{𝐹}{{𝐾}_1}+\frac{𝐹}{{𝐾}_2}$

$𝑥=𝐹(\frac{1}{{𝐾}_1}+\frac{1}{{𝐾}_2})$

So the total extension produced is

$\overline{)𝐹(\frac{1}{{𝐾}_1}+\frac{1}{{𝐾}_2})}$

Correct option: D.

Initial angular speed of the wheel is $600$ rpm.

Since $1$ revolution $=2𝜋$ radian,

$600\text{rpm}=\frac{600}{60}\text{rev/s}$

$=10\text{rev/s}$

So,

$𝜔=10\times 2𝜋=20𝜋\text{rad/s}$

The wheel comes to rest in half a minute:

$𝑡=30\text{s}$

Final angular speed is

${𝜔}_0=0$

Using

${𝜔}_0=𝜔+𝛼𝑡$

we get

$0=20𝜋+𝛼\left(30\right)$

$𝛼=-\frac{20𝜋}{30}$

$𝛼=-\frac{2𝜋}{3}{\text{rad/s}}^2$

The negative sign shows retardation. So the magnitude of angular retardation is

$\frac{2𝜋}{3}{\text{rad/s}}^2$

So the correct option is Option A: $\frac{2𝜋}{3}$.

For a rotating flywheel, the kinetic energy is

$𝐾=\frac{1}{2}𝐼{𝜔}^2$

Here,

$𝐾=500\text{J}$

The frequency is given as

$𝑓=5\text{Hz}$

Angular velocity is related to frequency by

$𝜔=2𝜋𝑓$

So,

$𝜔=2𝜋\times 5=10𝜋{\text{rad s}}^{-1}$

Now substitute in the kinetic energy formula:

$500=\frac{1}{2}𝐼(10𝜋{)}^2$

$500=\frac{1}{2}𝐼\times 100{𝜋}^2$

Given,

${𝜋}^2=10$

So,

$500=\frac{1}{2}𝐼\times 1000$

$500=500𝐼$

$𝐼=1{\text{kg m}}^2$

So the correct option is:

$\overline{)1{\text{kg m}}^2}$

Option B.

Statement I is correct.

Acceleration due to gravity is maximum at the earth’s surface.

When we go to a height $ℎ$ above the earth’s surface,

${𝑔}_{ℎ}=𝑔{\left(\frac{𝑅}{𝑅+ℎ}\right)}^2$

So, as $ℎ$ increases, ${𝑔}_{ℎ}$ decreases.

When we go to a depth $𝑑$ below the earth’s surface, according to the NCERT result,

${𝑔}_{𝑑}=𝑔(1-\frac{𝑑}{𝑅})$

So, as $𝑑$ increases, ${𝑔}_{𝑑}$ also decreases.

Hence, gravity decreases both on going up and on going down from the earth’s surface.

Now check Statement II.

If $ℎ=𝑑$, then

${𝑔}_{ℎ}=𝑔{\left(\frac{𝑅}{𝑅+ℎ}\right)}^2$

and

${𝑔}_{𝑑}=𝑔(1-\frac{ℎ}{𝑅})$

These two expressions are not generally equal.

For small values of $ℎ$,

${𝑔}_{ℎ}\approx 𝑔(1-\frac{2ℎ}{𝑅})$

while

${𝑔}_{𝑑}=𝑔(1-\frac{ℎ}{𝑅})$

So even when $ℎ=𝑑$, the decrease in gravity at height $ℎ$ is not the same as the decrease at depth $𝑑$.

Thus, Statement II is incorrect.

The correct answer is:

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

Statement I is correct but Statement II is incorrect.

For a small spherical raindrop falling through air, terminal velocity is given by Stokes’ law:

$𝑉=\frac{2{𝑟}^2(𝜌-𝜎)𝑔}{9𝜂}$

Here, for the same liquid and same medium, all quantities except radius are constant.

So terminal velocity is proportional to the square of radius:

$𝑉\propto {𝑟}^2$

For the smaller drop of radius $𝑟$,

$𝑉\propto {𝑟}^2$

For the bigger drop of radius $𝑅$, suppose its terminal velocity is ${𝑉}^{\prime }$.

Then,

$\frac{{𝑉}^{\prime }}{𝑉}=\frac{{𝑅}^2}{{𝑟}^2}$

So,

${𝑉}^{\prime }=\frac{𝑉{𝑅}^2}{{𝑟}^2}$

Hence, the terminal velocity of the bigger drop is

$\overline{)\frac{𝑉{𝑅}^2}{{𝑟}^2}}$

Correct option: A.

At constant velocity, the ball has attained terminal velocity. So the net force on it is zero.

The forces are:

• Weight of the ball downward

• Buoyant force upward

• Viscous force upward

So, at terminal velocity,

$\text{Viscous force}=\text{Weight}-\text{Buoyant force}$

That is,

${𝐹}_{𝑣}=𝑉({𝜌}_{𝑠}-{𝜌}_{𝑙})𝑔$

Here,

$𝑟=1\text{ mm}={10}^{-3}\text{ m}$

${𝜌}_{𝑠}=10.5\text{ g/cc}=10500{\text{ kg/m}}^3$

${𝜌}_{𝑙}=1.5\text{ g/cc}=1500{\text{ kg/m}}^3$

So,

${𝜌}_{𝑠}-{𝜌}_{𝑙}=10500-1500=9000{\text{ kg/m}}^3$

Volume of the spherical ball is

$𝑉=\frac{4}{3}𝜋{𝑟}^3$

So,

${𝐹}_{𝑣}=\frac{4}{3}𝜋{𝑟}^3({𝜌}_{𝑠}-{𝜌}_{𝑙})𝑔$

Substituting the values,

${𝐹}_{𝑣}=\frac{4}{3}\times \frac{22}{7}\times ({10}^{-3}{)}^3\times 9000\times 9.8$

${𝐹}_{𝑣}=\frac{4}{3}\times \frac{22}{7}\times {10}^{-9}\times 88200$

${𝐹}_{𝑣}=3.696\times {10}^{-4}\text{ N}$

Now,

$3.696\times {10}^{-4}=3696\times {10}^{-7}$

So,

$𝑥=7$

Correct option: C

Let the total volume of the sphere be $𝑉$ and its density be $𝜌$.

When a body floats, its weight is equal to the total upthrust, or buoyant force, acting on it.

Weight of the sphere is:

$𝑊=𝑉𝜌𝑔$

Half of the sphere is in mercury and half is in oil. So, volume of sphere in mercury is $\frac{𝑉}{2}$ and volume of sphere in oil is $\frac{𝑉}{2}$.

The buoyant force due to mercury is:

${𝐹}_{𝐵,\text{mercury}}=\left(\frac{𝑉}{2}\right){𝜌}_{\text{Hg}}𝑔$

The buoyant force due to oil is:

${𝐹}_{𝐵,\text{oil}}=\left(\frac{𝑉}{2}\right){𝜌}_{\text{oil}}𝑔$

So, the total buoyant force is:

${𝐹}_{𝐵}=\left(\frac{𝑉}{2}\right){𝜌}_{\text{Hg}}𝑔+\left(\frac{𝑉}{2}\right){𝜌}_{\text{oil}}𝑔$

Since the sphere is floating,

$𝑉𝜌𝑔=\frac{𝑉}{2}{𝜌}_{\text{Hg}}𝑔+\frac{𝑉}{2}{𝜌}_{\text{oil}}𝑔$

Now divide both sides by $𝑉𝑔$:

$𝜌=\frac{{𝜌}_{\text{Hg}}+{𝜌}_{\text{oil}}}{2}$

Given, ${𝜌}_{\text{Hg}}=13.6{\text{ gm/cm}}^3$ and ${𝜌}_{\text{oil}}=0.8{\text{ gm/cm}}^3$.

Substitute these values:

$𝜌=\frac{13.6+0.8}{2}$

$𝜌=\frac{14.4}{2}$

$𝜌=7.2{\text{ gm/cm}}^3$

The density of the material of the sphere is $7.2{\text{ gm/cm}}^3$.

Let the length of each rod be $𝐿$ and the area of cross-section be $𝐴$.

Since the rods are connected in series and are insulated from outside, no heat is lost to the surroundings. So, in steady state, the same amount of heat flows through each rod per second.

For heat conduction, the rate of heat flow is proportional to thermal conductivity and temperature difference:

$\frac{𝑑𝑄}{𝑑𝑡}=\frac{2𝐾𝐴(3𝑇-{𝑇}_1)}{𝐿}=\frac{𝐾𝐴({𝑇}_1-{𝑇}_2)}{𝐿}=\frac{2𝐾𝐴({𝑇}_2-𝑇)}{𝐿}$

Here, the two side rods have thermal conductivity $2𝐾$, and the middle rod has thermal conductivity $𝐾$.

Now cancel the common factor $\frac{𝐾𝐴}{𝐿}$ from all parts:

$2(3𝑇-{𝑇}_1)={𝑇}_1-{𝑇}_2=2({𝑇}_2-𝑇)$

Compare the first and third parts:

$2(3𝑇-{𝑇}_1)=2({𝑇}_2-𝑇)$

$3𝑇-{𝑇}_1={𝑇}_2-𝑇$

${𝑇}_1+{𝑇}_2=4𝑇\dots \left(1\right)$

Now compare the middle and third parts:

${𝑇}_1-{𝑇}_2=2({𝑇}_2-𝑇)$

${𝑇}_1-{𝑇}_2=2{𝑇}_2-2𝑇$

${𝑇}_1-3{𝑇}_2=-2𝑇\dots \left(2\right)$

Subtract equation $\left(2\right)$ from equation $\left(1\right)$:

$({𝑇}_1+{𝑇}_2)-({𝑇}_1-3{𝑇}_2)=4𝑇-(-2𝑇)$

$4{𝑇}_2=6𝑇$

${𝑇}_2=\frac{3𝑇}{2}$

Put this value of ${𝑇}_2$ in equation $\left(1\right)$:

${𝑇}_1+\frac{3𝑇}{2}=4𝑇$

${𝑇}_1=\frac{5𝑇}{2}$

Now find the required ratio:

$\frac{{𝑇}_1}{{𝑇}_2}=\frac{\frac{5𝑇}{2}}{\frac{3𝑇}{2}}=\frac{5}{3}$

For a hot body, the power radiated is given by Stefan’s law:

$𝑃=𝑒𝜎𝐴{𝑇}^4$

Here, $𝑒$ is the emissivity of the wire.

The wire is cylindrical. Its length is $𝐿$ and diameter is $𝑑$.

The radiating surface area of the wire is mainly its curved surface area:

$𝐴=𝜋𝑑𝐿$

So,

$𝑃=𝑒𝜎\left(𝜋𝑑𝐿\right){𝑇}^4$

Now solve for $𝑒$:

$𝑒=\frac{𝑃}{𝜎{𝑇}^4𝜋𝑑𝐿}$

So the emissivity of the wire is

$\overline{)\frac{𝑃}{𝜎{𝑇}^4𝜋𝑑𝐿}}$

Correct option: A

For a black body, the wavelength at which energy is maximum is related to temperature by Wien’s displacement law:

${𝜆}_{𝑚}𝑇=\text{constant}$

Initially,

${𝜆}_0{𝑇}_0=\text{constant}$

After changing the temperature, the wavelength for maximum energy becomes

${𝜆}_{\prime }^{𝑚}=\frac{3{𝜆}_0}{2}$

So,

${𝜆}_0{𝑇}_0=\frac{3{𝜆}_0}{2}{𝑇}^{\prime }$

Cancel ${𝜆}_0$:

${𝑇}_0=\frac{3}{2}{𝑇}^{\prime }$

So,

${𝑇}^{\prime }=\frac{2{𝑇}_0}{3}$

Now, according to Stefan’s law, the power radiated by a black body is proportional to the fourth power of its absolute temperature:

$𝑃\propto {𝑇}^4$

So the ratio of new power to initial power is

$\frac{{𝑃}^{\prime }}{{𝑃}_0}={\left(\frac{{𝑇}^{\prime }}{{𝑇}_0}\right)}^4$

$\frac{{𝑃}^{\prime }}{{𝑃}_0}={\left(\frac{2}{3}\right)}^4$

$\frac{{𝑃}^{\prime }}{{𝑃}_0}=\frac{16}{81}$

So the ratio of power radiated is

$\overline{)\frac{16}{81}}$

Correct option: B.

In a cyclic process, the work done by the gas is equal to the area enclosed by the graph on the $𝑃-𝑉$ diagram.

Here, the path $𝐴𝐵𝐶𝐷𝐴$ makes a rectangle on the graph.

The change in volume is the width of the rectangle:

$\Delta 𝑉=5{𝑉}_0-3{𝑉}_0=2{𝑉}_0$

The change in pressure is the height of the rectangle:

$\Delta 𝑃=4{𝑃}_0-2{𝑃}_0=2{𝑃}_0$

So, the magnitude of work done is the area of this rectangle:

$\left|𝑊\right|=\Delta 𝑃\times \Delta 𝑉$

$\left|𝑊\right|=\left(2{𝑃}_0\right)\times \left(2{𝑉}_0\right)=4{𝑃}_0{𝑉}_0$

For sample A, the compression is isothermal.

In an isothermal process, temperature remains constant, so Boyle’s law applies:

$𝑃𝑉=\text{constant}$

Initially, pressure is $𝑃$ and volume is $𝑉$.

Finally, volume becomes $\frac{𝑉}{2}$, so

${𝑃}_{𝐴}\cdot \frac{𝑉}{2}=𝑃\cdot 𝑉$

${𝑃}_{𝐴}=2𝑃$

For sample B, the compression is adiabatic.

In an adiabatic process,

$𝑃{𝑉}^{𝛾}=\text{constant}$

where $𝛾>1$ for a gas.

So,

${𝑃}_{𝐵}{\left(\frac{𝑉}{2}\right)}^{𝛾}=𝑃{𝑉}^{𝛾}$

${𝑃}_{𝐵}=𝑃{\left(\frac{𝑉}{𝑉/2}\right)}^{𝛾}$

${𝑃}_{𝐵}=𝑃(2{)}^{𝛾}$

Since $𝛾>1$,

$2^{𝛾}>2$

So,

${𝑃}_{𝐵}>2𝑃$

But we found:

${𝑃}_{𝐴}=2𝑃$

Hence,

${𝑃}_{𝐴}<{𝑃}_{𝐵}$

So the correct option is:

$\overline{)\text{Option A }{𝑃}_{𝐴}<{𝑃}_{𝐵}}$

For hydrogen in the same vessel, the volume remains constant.

Using the ideal gas equation,

$𝑃𝑉=𝑛𝑅𝑇$

Since $𝑉$ and $𝑅$ are constant,

$𝑛\propto \frac{𝑃}{𝑇}$

Initial condition:

${𝑃}_1=𝑃,{𝑇}_1=500\text{ K},{𝑚}_1=6\text{ g}$

Final condition:

${𝑃}_2=\frac{𝑃}{2},{𝑇}_2=300\text{ K}$

So,

$\frac{{𝑛}_2}{{𝑛}_1}=\frac{{𝑃}_2/{𝑇}_2}{{𝑃}_1/{𝑇}_1}$

Substitute the values:

$\frac{{𝑛}_2}{{𝑛}_1}=\frac{\frac{𝑃}{2}/300}{𝑃/500}$

$\frac{{𝑛}_2}{{𝑛}_1}=\frac{𝑃}{600}\times \frac{500}{𝑃}=\frac{5}{6}$

So the final mass of hydrogen is also $\frac{5}{6}$ of the initial mass:

${𝑚}_2=\frac{5}{6}\times 6=5\text{ g}$

Hydrogen leaked out:

$6-5=1\text{ g}$

So, the correct option is:

$\overline{)\text{Option D: }1\text{ gram}}$

For SHM, we can write the displacement as

$𝑥=𝐴\sin 𝜔𝑡$

Take the particle to start from $𝑥=0$ and move towards $+𝐴$.

When the particle reaches $𝑥=\frac{𝐴}{2}$,

$\frac{𝐴}{2}=𝐴\sin 𝜔𝑡$

$\sin 𝜔𝑡=\frac{1}{2}$

$𝜔𝑡=\frac{𝜋}{6}$

Given this time is $2$ s,

$\frac{𝜋}{6𝜔}=2$

Now, for the motion from $𝑥=\frac{𝐴}{2}$ to $𝑥=𝐴$:

At $𝑥=\frac{𝐴}{2}$,

$𝜔𝑡=\frac{𝜋}{6}$

At $𝑥=𝐴$,

$\sin 𝜔𝑡=1$

so

$𝜔𝑡=\frac{𝜋}{2}$

Change in phase is

$\frac{𝜋}{2}-\frac{𝜋}{6}=\frac{𝜋}{3}$

Time taken is proportional to change in phase.

Since $\frac{𝜋}{6}$ phase takes $2$ s, then $\frac{𝜋}{3}$ phase will take

$2\times 2=4\text{ s}$

So, the time taken from $𝑥=\frac{𝐴}{2}$ to $𝑥=𝐴$ is

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

Correct option: D.

For a body performing SHM, the restoring force is

$𝐹=-𝐾𝑥$

Here, $𝐾$ is the force constant.

In SHM, the total mechanical energy is the sum of kinetic energy and potential energy.

At any displacement $𝑥$,

$\text{Potential energy}=\frac{1}{2}𝐾{𝑥}^2$

and

$\text{Kinetic energy}=\frac{1}{2}𝐾({𝑎}^2-{𝑥}^2)$

So total energy is

$𝐸=\frac{1}{2}𝐾{𝑥}^2+\frac{1}{2}𝐾({𝑎}^2-{𝑥}^2)$

$𝐸=\frac{1}{2}𝐾{𝑎}^2$

So the total energy depends only on the force constant $𝐾$ and the amplitude $𝑎$.

It does not depend on the displacement $𝑥$ or velocity $𝑣$ at a particular instant.

Hence, the correct option is:

$\overline{)\text{Option D: }𝐾,𝑎}$

When humidity increases, the amount of water vapour in air increases.

Water vapour is lighter than the main gases present in dry air, so humid air becomes less dense. Sound travels faster in a medium with lower density, if other conditions are nearly the same.

So, the speed of sound increases when humidity increases.

The correct answer is:

Option A: Speed of sound waves is more.

The beat frequency is equal to the difference between the frequencies of the two sources.

Let the frequency of the sonometer string be $𝑓$.

Given tuning fork frequency $=𝑛$ and beats per second $=𝑥$,

$|𝑓-𝑛|=𝑥$

So the string frequency could be either

$𝑓=𝑛+𝑥$

or

$𝑓=𝑛-𝑥$

Now, when the tension in a sonometer string is slightly increased, its frequency increases because

$𝑓\propto \sqrt{\surd 𝑇}$

It is given that after increasing the tension, the number of beats becomes lesser.

This means the string frequency must be moving closer to the tuning fork frequency $𝑛$.

That will happen only if the original frequency of the string was less than $𝑛$, i.e.

$𝑓=𝑛-𝑥$

If the original frequency were $𝑛+𝑥$, increasing the tension would increase the string frequency further, making the difference larger and beats more.

So the frequency of the string was

$\overline{)𝑛-𝑥}$

Correct option: B.

The linear mass density of the string is

$𝜇=\frac{𝑚}{𝐿}$

Given,

$𝑚={10}^{-3}\text{kg},𝐿=0.5\text{m}$

So,

$𝜇=\frac{{10}^{-3}}{0.5}=2\times {10}^{-3}{\text{kg m}}^{-1}$

The speed of a wave pulse on a stretched string is

$𝑣=\sqrt{\surd \frac{𝑇}{𝜇}}$

Here,

$𝑇=0.8\text{N}$

So,

$𝑣=\sqrt{\surd \frac{0.8}{2\times {10}^{-3}}}$

$𝑣=\sqrt{\surd 400}=20{\text{m s}}^{-1}$

For constructive interference between successive identical pulses, the first pulse should return to the starting end with the same phase.

At a fixed end, a pulse gets inverted after reflection.

So after one reflection, it is opposite in phase.

After two reflections, it comes back to the same phase.

The minimum time for this is the time for a round trip along the string:

$\Delta {𝑡}_{min}=\frac{2𝐿}{𝑣}$

$\Delta {𝑡}_{min}=\frac{2\left(0.5\right)}{20}$

$\Delta {𝑡}_{min}=\frac{1}{20}=0.05\text{s}$

So the minimum value of $\Delta 𝑡$ is

$\overline{)0.05\text{s}}$

Correct option: D.

The statement which is NOT TRUE is:

$\text{a. Work done to move a charge on an equipotential surface is not zero.}$

On an equipotential surface, the electric potential is the same at every point. So the potential difference between any two points on that surface is

$\Delta 𝑉=0$

Work done in moving a charge $𝑞$ is

$𝑊=𝑞\Delta 𝑉$

So,

$𝑊=𝑞\times 0=0$

This means the work done to move a charge on an equipotential surface is zero.

So statement (a) is false.

Hence, the correct option is:

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

For a metal sphere, charge is related to surface charge density by

$𝑄=𝜎\cdot 4𝜋{𝑟}^2$

Initially, both spheres have the same surface charge density $𝜎$.

For the smaller sphere of radius $𝑅$,

${𝑄}_1=𝜎\cdot 4𝜋{𝑅}^2$

For the bigger sphere of radius $3𝑅$,

${𝑄}_2=𝜎\cdot 4𝜋(3𝑅{)}^2$

${𝑄}_2=9𝜎\cdot 4𝜋{𝑅}^2$

So the bigger sphere initially has $9$ times the charge of the smaller sphere.

When the two metal spheres are brought in contact, charge flows until their potentials become equal.

For a conducting sphere,

$𝑉=\frac{𝑘𝑄}{𝑟}$

So after contact,

$\frac{𝑘{𝑄}_{\prime }^1}{𝑅}=\frac{𝑘{𝑄}_{\prime }^2}{3𝑅}$

Cancel $𝑘$ and $𝑅$:

${𝑄}_{\prime }^2=3{𝑄}_{\prime }^1$

Now surface charge density after separation is

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

and

${𝜎}_2=\frac{{𝑄}_{\prime }^2}{4𝜋(3𝑅{)}^2}$

Using ${𝑄}_{\prime }^2=3{𝑄}_{\prime }^1$,

${𝜎}_2=\frac{3{𝑄}_{\prime }^1}{4𝜋\cdot 9{𝑅}^2}$

${𝜎}_2=\frac{{𝑄}_{\prime }^1}{12𝜋{𝑅}^2}$

Also,

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

So,

$\frac{{𝜎}_1}{{𝜎}_2}=\frac{\frac{{𝑄}_{\prime }^1}{4𝜋{𝑅}^2}}{\frac{{𝑄}_{\prime }^1}{12𝜋{𝑅}^2}}$

$\frac{{𝜎}_1}{{𝜎}_2}=3$

So the correct option is

$\overline{)3}$

Option D.

Let us first fix the positions carefully. The three charges lie on a line whose total length is $𝑑$. The charge $𝑞$ is at one end, $𝑄$ is at a distance $\frac{𝑑}{3}$ from it, and $+4𝑞$ lies a further $\frac{2𝑑}{3}$ away. So, measuring everything from $𝑞$:

• distance of $𝑄$ from $𝑞$ is $\frac{𝑑}{3}$

• distance of $4𝑞$ from $𝑞$ is $\frac{𝑑}{3}+\frac{2𝑑}{3}=𝑑$

Now think about the directions of the two forces acting on $𝑞$.

The charge $+4𝑞$ is positive, so it pushes $𝑞$ away, that is, outward along the line. For the net force on $𝑞$ to vanish, the middle charge $𝑄$ must pull $𝑞$ inward. A pull is possible only if $𝑄$ is negative. So the sign of $𝑄$ is settled even before any calculation, and this is why all the options are negative.

For balance, the two magnitudes must be equal:

$\frac{1}{4𝜋{𝜀}_0}\frac{𝑞\left|𝑄\right|}{{\left(\frac{𝑑}{3}\right)}^2}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑞\left(4𝑞\right)}{{𝑑}^2}$

The constant $\frac{1}{4𝜋{𝜀}_0}$ and the charge $𝑞$ cancel from both sides:

$\frac{\left|𝑄\right|}{\frac{{𝑑}^2}{9}}=\frac{4𝑞}{{𝑑}^2}$

$\frac{9\left|𝑄\right|}{{𝑑}^2}=\frac{4𝑞}{{𝑑}^2}$

$\left|𝑄\right|=\frac{4𝑞}{9}$

Since $𝑄$ has to be negative,

$𝑄=-\frac{4𝑞}{9}$

So the correct choice is Option D.

The electric field and electric potential are related by $𝐸=-\frac{𝑑𝑉}{𝑑𝑥}$.

This means the electric field depends on the slope of the $𝑉-𝑥$ graph. For magnitude, we take the magnitude of the slope.

In regions 1 and 3, the potential $𝑉$ is constant. So the graph is horizontal there. A horizontal line has zero slope, so ${𝐸}_1=0$ and ${𝐸}_3=0$.

Thus, ${𝐸}_1={𝐸}_3$.

In region 2, the potential increases from ${𝑉}_{𝐵}$ to ${𝑉}_{𝐶}$. So the magnitude of the electric field is ${𝐸}_2=\mid \frac{\Delta 𝑉}{\Delta 𝑥}\mid =\frac{{𝑉}_{𝐶}-{𝑉}_{𝐵}}{{𝑥}_{𝐶}-{𝑥}_{𝐵}}$.

In region 4, the potential decreases from ${𝑉}_{𝐶}$ to $0$. So the magnitude of the electric field is ${𝐸}_4=\mid \frac{\Delta 𝑉}{\Delta 𝑥}\mid =\frac{{𝑉}_{𝐷}-{𝑉}_{𝐸}}{{𝑥}_{𝐸}-{𝑥}_{𝐷}}=\frac{{𝑉}_{𝐶}-0}{{𝑥}_{𝐸}-{𝑥}_{𝐷}}$.

From the graph, regions 2 and 4 have equal horizontal width. So ${𝑥}_{𝐶}-{𝑥}_{𝐵}={𝑥}_{𝐸}-{𝑥}_{𝐷}$.

Now compare the change in potential. In region 2, the change is ${𝑉}_{𝐶}-{𝑉}_{𝐵}$. In region 4, the change is ${𝑉}_{𝐶}-0={𝑉}_{𝐶}$.

Since ${𝑉}_{𝐵}>0$, we have ${𝑉}_{𝐶}>{𝑉}_{𝐶}-{𝑉}_{𝐵}$. So the change in potential is greater in region 4.

Because both regions have equal width, the slope is greater in region 4. Hence, ${𝐸}_4>{𝐸}_2$.

So the correct relation is ${𝐸}_1={𝐸}_3$ and ${𝐸}_2<{𝐸}_4$.

The dielectric slab has thickness $1\text{ mm}$, while the total plate separation is $2\text{ mm}$.

So the remaining space is air of thickness

$2\text{ mm}-1\text{ mm}=1\text{ mm}$

For a parallel plate capacitor partly filled with a dielectric slab along the separation, the effective separation is

${𝑑}_{\text{eff}}={𝑑}_{\text{air}}+\frac{{𝑑}_{\text{dielectric}}}{𝐾}$

Here,

$𝐴=40{\text{ cm}}^2=40\times {10}^{-4}{\text{ m}}^2=4\times {10}^{-3}{\text{ m}}^2$

${𝑑}_{\text{air}}=1\text{ mm}={10}^{-3}\text{ m}$

${𝑑}_{\text{dielectric}}=1\text{ mm}={10}^{-3}\text{ m}$

and

$𝐾=5$

So,

${𝑑}_{\text{eff}}={10}^{-3}+\frac{{10}^{-3}}{5}$

${𝑑}_{\text{eff}}={10}^{-3}+0.2\times {10}^{-3}$

${𝑑}_{\text{eff}}=1.2\times {10}^{-3}\text{ m}$

Now capacitance is

$𝐶=\frac{{𝜖}_0𝐴}{{𝑑}_{\text{eff}}}$

$𝐶=\frac{{𝜖}_0\times 4\times {10}^{-3}}{1.2\times {10}^{-3}}$

$𝐶=\frac{4}{1.2}{𝜖}_0$

$𝐶=\frac{10}{3}{𝜖}_0\text{ F}$

So the correct option is

$\overline{)\frac{10}{3}{𝜖}_0\text{ F}}$

Option C.

In a metre bridge, at balance condition,

$\frac{𝑅}{𝑆}=\frac{𝑙}{100-𝑙}$

Here the two resistances in the gaps are $2\Omega$ and $3\Omega$.

Taking $2\Omega$ in the left gap and $3\Omega$ in the right gap,

$\frac{2}{3}=\frac{𝑙}{100-𝑙}$

So,

$3𝑙=2(100-𝑙)$

$3𝑙=200-2𝑙$

$5𝑙=200$

$𝑙=40\text{ cm}$

Now a shunt of $𝑋\Omega$ is added across the $3\Omega$ resistor. So the effective resistance of that gap becomes

${𝑅}_{\text{eq}}=\frac{3𝑋}{3+𝑋}$

Since this is less than $3\Omega$, the balance point shifts towards the right.

The shift is $22.5\text{ cm}$, so the new balance length is

${𝑙}^{\prime }=40+22.5=62.5\text{ cm}$

Now,

$\frac{2}{{𝑅}_{\text{eq}}}=\frac{62.5}{100-62.5}$

$\frac{2}{{𝑅}_{\text{eq}}}=\frac{62.5}{37.5}$

$\frac{2}{{𝑅}_{\text{eq}}}=\frac{5}{3}$

So,

${𝑅}_{\text{eq}}=\frac{6}{5}=1.2\Omega$

Now,

$\frac{3𝑋}{3+𝑋}=1.2$

$3𝑋=1.2(3+𝑋)$

$3𝑋=3.6+1.2𝑋$

$1.8𝑋=3.6$

$𝑋=2\Omega$

So the correct option is

$\overline{)2\Omega }$

Option B.

For the galvanometer,

Full-scale current is

${𝐼}_{𝑔}=10\text{ mA}=0.01\text{ A}$

Voltage across the galvanometer at full-scale deflection is

${𝑉}_{𝑔}=0.75\text{ V}$

So the resistance of the galvanometer is

$𝐺=\frac{{𝑉}_{𝑔}}{{𝐼}_{𝑔}}$

$𝐺=\frac{0.75}{0.01}=75\Omega$

To convert it into an ammeter of range $10\text{ A}$, a shunt resistance $𝑆$ is connected in parallel with the galvanometer.

Total current required is

$𝐼=10\text{ A}$

Current through shunt will be

${𝐼}_{𝑠}=𝐼-{𝐼}_{𝑔}$

${𝐼}_{𝑠}=10-0.01=9.99\text{ A}$

Since galvanometer and shunt are in parallel, the voltage across both is the same:

${𝐼}_{𝑔}𝐺={𝐼}_{𝑠}𝑆$

So,

$𝑆=\frac{{𝐼}_{𝑔}𝐺}{{𝐼}_{𝑠}}$

$𝑆=\frac{0.01\times 75}{9.99}$

$𝑆=\frac{0.75}{9.99}$

Multiplying numerator and denominator by $100$,

$𝑆=\frac{75}{999}\Omega$

So the shunt resistance is

$\overline{)\frac{75}{999}\Omega }$

Correct option: D.