MHT CET 2026 18th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Since $\overrightarrow{𝑈}$ is a unit vector, its magnitude must be $1$.

Given,

$\overrightarrow{𝑈}=0.9\widehat{𝑖}-0.2\widehat{𝑗}+𝑚\widehat{𝑘}$

Magnitude of $\overrightarrow{𝑈}$ is

$\left|\overrightarrow{𝑈}\right|=\sqrt{\surd (0.9{)}^2+(-0.2{)}^2+{𝑚}^2}$

Since it is a unit vector,

$\sqrt{\surd (0.9{)}^2+(-0.2{)}^2+{𝑚}^2}=1$

Squaring both sides,

$(0.9{)}^2+(-0.2{)}^2+{𝑚}^2=1$

$0.81+0.04+{𝑚}^2=1$

$0.85+{𝑚}^2=1$

${𝑚}^2=0.15$

So,

$𝑚=\sqrt{\surd 0.15}$

Hence, the correct option is

$\overline{)\sqrt{\surd 0.15}}$

Option B.

For a uniform cube, density is

$𝜌=\frac{𝑚}{𝑉}$

Since the cube has side length $𝑙$,

$𝑉={𝑙}^3$

So,

$𝜌=\frac{𝑚}{{𝑙}^3}$

Rearranging,

${𝑙}^3=\frac{𝑚}{𝜌}$

Taking percentage errors, for multiplication and division, percentage errors add:

$\frac{\Delta \left({𝑙}^3\right)}{{𝑙}^3}=\frac{\Delta 𝑚}{𝑚}+\frac{\Delta 𝜌}{𝜌}$

Given:

$\frac{\Delta 𝑚}{𝑚}=2\%$

and

$\frac{\Delta 𝜌}{𝜌}=8\%$

So,

$\frac{\Delta \left({𝑙}^3\right)}{{𝑙}^3}=2\%+8\%=10\%$

Now, since ${𝑙}^3$ contains the power $3$,

$3\frac{\Delta 𝑙}{𝑙}=10\%$

Hence,

$\frac{\Delta 𝑙}{𝑙}=\frac{10}{3}\%=3.33\%$

The nearest option is

$\overline{)4\%}$

So, the correct option is Option B.

For projectile motion,

$𝑅=\frac{{𝑢}^2\sin 2𝜃}{𝑔}$

and maximum height is

$𝐻=\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}$

Given that range is $𝑛$ times the maximum height,

$𝑅=𝑛𝐻$

So,

$\frac{𝑅}{𝐻}=𝑛$

Now substitute the formulas:

$\frac{𝑅}{𝐻}=\frac{\frac{{𝑢}^2\sin 2𝜃}{𝑔}}{\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}}$

Cancel common terms:

$\frac{𝑅}{𝐻}=\frac{2\sin 2𝜃}{{\sin }^2𝜃}$

Using $\sin 2𝜃=2\sin 𝜃\cos 𝜃$,

$\frac{𝑅}{𝐻}=\frac{2(2\sin 𝜃\cos 𝜃)}{{\sin }^2𝜃}$

$\frac{𝑅}{𝐻}=4\cot 𝜃$

Since $\frac{𝑅}{𝐻}=𝑛$,

$𝑛=4\cot 𝜃$

So,

$\tan 𝜃=\frac{4}{𝑛}$

Hence,

$𝜃={\tan }^{-1}\left(\frac{4}{𝑛}\right)$

Correct option: C

$\overline{){\tan }^{-1}\left(\frac{4}{𝑛}\right)}$

For two masses connected by a light inextensible string over a smooth pulley, the acceleration of the system is

$𝑎=\frac{({𝑚}_2-{𝑚}_1)𝑔}{{𝑚}_1+{𝑚}_2}$

Here ${𝑚}_2>{𝑚}_1$, so ${𝑚}_2$ moves downward and ${𝑚}_1$ moves upward.

Given,

$𝑎=\frac{𝑔}{𝑛}$

So,

$\frac{({𝑚}_2-{𝑚}_1)𝑔}{{𝑚}_1+{𝑚}_2}=\frac{𝑔}{𝑛}$

Cancel $𝑔$ from both sides:

$\frac{{𝑚}_2-{𝑚}_1}{{𝑚}_1+{𝑚}_2}=\frac{1}{𝑛}$

Now let

$\frac{{𝑚}_2}{{𝑚}_1}=𝑟$

So ${𝑚}_2=𝑟{𝑚}_1$.

Substitute this:

$\frac{𝑟{𝑚}_1-{𝑚}_1}{{𝑚}_1+𝑟{𝑚}_1}=\frac{1}{𝑛}$

Take ${𝑚}_1$ common and cancel it:

$\frac{𝑟-1}{𝑟+1}=\frac{1}{𝑛}$

Cross-multiply:

$𝑛(𝑟-1)=𝑟+1$

$𝑛𝑟-𝑛=𝑟+1$

$𝑛𝑟-𝑟=𝑛+1$

$𝑟(𝑛-1)=𝑛+1$

$𝑟=\frac{𝑛+1}{𝑛-1}$

So,

$\frac{{𝑚}_2}{{𝑚}_1}=\frac{𝑛+1}{𝑛-1}$

Hence, the correct option is

$\overline{)\text{Option C }\frac{𝑛+1}{𝑛-1}}$

The central ring has its centre on the given axis. For a thin ring, the moment of inertia about an axis through its centre and perpendicular to its plane is:

${𝐼}_0=𝑀{𝑅}^2$

Now look at the six outer rings. Each outer ring touches the central ring, so the distance between their centres is:

$𝑑=𝑅+𝑅=2𝑅$

For one outer ring, its own moment of inertia about an axis through its centre and perpendicular to its plane is:

${𝐼}_{𝑐𝑚}=𝑀{𝑅}^2$

But the given axis passes through the centre of the central ring, not through the centre of the outer ring. So we use the parallel axis theorem:

${𝐼}_1={𝐼}_{𝑐𝑚}+𝑀{𝑑}^2$

${𝐼}_1=𝑀{𝑅}^2+𝑀(2𝑅{)}^2$

${𝐼}_1=𝑀{𝑅}^2+4𝑀{𝑅}^2$

${𝐼}_1=5𝑀{𝑅}^2$

There are six such outer rings, so their total moment of inertia is:

${𝐼}_{\text{surrounding}}=6\times 5𝑀{𝑅}^2=30𝑀{𝑅}^2$

The total moment of inertia of all seven rings is:

${𝐼}_{\text{total}}={𝐼}_0+{𝐼}_{\text{surrounding}}$

${𝐼}_{\text{total}}=𝑀{𝑅}^2+30𝑀{𝑅}^2$

${𝐼}_{\text{total}}=31𝑀{𝑅}^2$

Initial kinetic energy of rotating dancer is

$𝐾=\frac{1}{2}𝐼{𝜔}^2$

where $𝐼$ is the initial moment of inertia and $𝜔$ is the initial angular velocity.

Now the dancer stretches his arms.

New moment of inertia becomes

${𝐼}^{\prime }=3𝐼$

New angular velocity becomes

${𝜔}^{\prime }=\frac{𝜔}{3}$

Rotational kinetic energy now is

${𝐾}^{\prime }=\frac{1}{2}{𝐼}^{\prime }{{𝜔}^{\prime }}^2$

Substitute the new values:

${𝐾}^{\prime }=\frac{1}{2}\left(3𝐼\right){\left(\frac{𝜔}{3}\right)}^2$

${𝐾}^{\prime }=\frac{1}{2}\left(3𝐼\right)\frac{{𝜔}^2}{9}$

${𝐾}^{\prime }=\frac{1}{3}\left(\frac{1}{2}𝐼{𝜔}^2\right)$

Since

$\frac{1}{2}𝐼{𝜔}^2=𝐾$

we get

${𝐾}^{\prime }=\frac{𝐾}{3}$

So the kinetic energy of the dancer now is

$\overline{)\frac{𝐾}{3}}$

Correct option: B.

For a solid sphere rolling without slipping,

$𝑣=𝑅𝜔$

The total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy:

${𝐾}_{\text{total}}={𝐾}_{\text{trans}}+{𝐾}_{\text{rot}}$

For a solid sphere,

$𝐼=\frac{2}{5}𝑀{𝑅}^2$

Rotational kinetic energy is

${𝐾}_{\text{rot}}=\frac{1}{2}𝐼{𝜔}^2$

Substitute $𝐼=\frac{2}{5}𝑀{𝑅}^2$:

${𝐾}_{\text{rot}}=\frac{1}{2}\times \frac{2}{5}𝑀{𝑅}^2{𝜔}^2$

${𝐾}_{\text{rot}}=\frac{1}{5}𝑀{𝑅}^2{𝜔}^2$

Using $𝑣=𝑅𝜔$, we get ${𝑅}^2{𝜔}^2={𝑣}^2$:

${𝐾}_{\text{rot}}=\frac{1}{5}𝑀{𝑣}^2$

Translational kinetic energy is

${𝐾}_{\text{trans}}=\frac{1}{2}𝑀{𝑣}^2$

So,

${𝐾}_{\text{total}}=\frac{1}{2}𝑀{𝑣}^2+\frac{1}{5}𝑀{𝑣}^2$

${𝐾}_{\text{total}}=\frac{7}{10}𝑀{𝑣}^2$

Now the required ratio is

$\frac{{𝐾}_{\text{total}}}{{𝐾}_{\text{rot}}}=\frac{\frac{7}{10}𝑀{𝑣}^2}{\frac{1}{5}𝑀{𝑣}^2}$

$=\frac{7}{10}\times 5$

$=\frac{7}{2}$

So the correct option is

$\overline{)\frac{7}{2}}$

Option B.

At a height $ℎ$ above the earth’s surface, the value of acceleration due to gravity is

${𝑔}_{ℎ}=𝑔(1-\frac{2ℎ}{𝑅})$

where $𝑅$ is the radius of the earth.

At a depth $𝑑$ below the earth’s surface, the value of acceleration due to gravity is

${𝑔}_{𝑑}=𝑔(1-\frac{𝑑}{𝑅})$

Here, the depth is

$𝑑=90\text{ km}$

It is given that acceleration due to gravity at height $ℎ$ is equal to acceleration due to gravity at depth $90$ km.

So,

$𝑔(1-\frac{2ℎ}{𝑅})=𝑔(1-\frac{90}{𝑅})$

Cancel $𝑔$ from both sides:

$1-\frac{2ℎ}{𝑅}=1-\frac{90}{𝑅}$

So,

$\frac{2ℎ}{𝑅}=\frac{90}{𝑅}$

Cancel $𝑅$:

$2ℎ=90$

$ℎ=45\text{ km}$

So, the value of $ℎ$ is

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

Correct option: D.

From the figure, end P is narrower than end Q. So the cross-sectional area at P is smaller than the cross-sectional area at Q.

So, we write:

${𝐴}_{𝑃}<{𝐴}_{𝑄}$

For an incompressible fluid, the equation of continuity says that area multiplied by velocity remains constant:

${𝐴}_{𝑃}{𝑣}_{𝑃}={𝐴}_{𝑄}{𝑣}_{𝑄}$

This means that where the tube is narrow, the fluid moves faster. Where the tube is wider, the fluid moves slower.

Since ${𝐴}_{𝑃}<{𝐴}_{𝑄}$, the velocity at P is greater than the velocity at Q:

${𝑣}_{𝑃}>{𝑣}_{𝑄}$

The kinetic energy per unit volume of a fluid is:

$𝐾=\frac{1}{2}𝜌{𝑣}^2$

Here, $𝜌$ is the density of the fluid. Since the fluid is incompressible, $𝜌$ remains the same at P and Q.

Now, because ${𝑣}_{𝑃}>{𝑣}_{𝑄}$, we also have:

${𝑣}_2^{𝑃}>{𝑣}_2^{𝑄}$

So,

$\frac{1}{2}𝜌{𝑣}_2^{𝑃}>\frac{1}{2}𝜌{𝑣}_2^{𝑄}$

Hence, the kinetic energy per unit volume at P is greater than that at Q.

The flow of a fluid becomes unsteady or turbulent when the Reynolds number is large.

According to NCERT, for Reynolds number ${𝑅}_{𝑛}$:

• If ${𝑅}_{𝑛}<1000$, the flow is generally steady or streamline.

• If $1000<{𝑅}_{𝑛}<2000$, the flow is unstable or in transition.

• If ${𝑅}_{𝑛}>2000$, the flow becomes turbulent or unsteady.

So, the correct option is:

$\overline{)\text{Option D: greater than 2000}}$

The correct option is C.

For a curved liquid surface, surface tension creates a pressure difference on the two sides of the surface.

The pressure is always greater on the concave side than on the convex side.

For a single curved liquid surface, the excess pressure is

$\Delta 𝑃=\frac{2𝑇}{𝑅}$

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

So,

${𝑃}_{\text{concave}}>{𝑃}_{\text{convex}}$

Hence, the correct answer is:

$\overline{)\text{Option C: the pressure on the concave side is greater than that on the convex side.}}$

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

${𝜆}_{𝑚}𝑇=\text{constant}$

Initially,

$𝜆{𝑇}_1=\text{constant}$

After changing the temperature, maximum energy is radiated at wavelength

$\frac{2𝜆}{3}$

So,

$\frac{2𝜆}{3}\times {𝑇}_2=𝜆{𝑇}_1$

Cancel $𝜆$:

$\frac{2}{3}{𝑇}_2={𝑇}_1$

So,

${𝑇}_2=\frac{3}{2}{𝑇}_1$

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

$𝐸\propto {𝑇}^4$

So,

$\frac{{𝐸}_2}{𝐸}={\left(\frac{{𝑇}_2}{{𝑇}_1}\right)}^4$

Substitute ${𝑇}_2=\frac{3}{2}{𝑇}_1$:

$\frac{{𝐸}_2}{𝐸}={\left(\frac{3}{2}\right)}^4$

$\frac{{𝐸}_2}{𝐸}=\frac{81}{16}$

Hence,

${𝐸}_2=\frac{81}{16}𝐸$

So the correct option is:

$\overline{)\text{Option B }\frac{81}{16}𝐸}$

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

Room temperature is $𝜃$.

Initial temperature of the body is $3𝜃$, so initial excess temperature is

$3𝜃-𝜃=2𝜃$

After 10 minutes, the body cools to $2𝜃$, so the excess temperature becomes

$2𝜃-𝜃=𝜃$

So, in 10 minutes, the excess temperature becomes half:

$2𝜃\to 𝜃$

Since Newton’s law of cooling is applicable, the same fraction of excess temperature will be lost in the next 10 minutes also.

So after the next 10 minutes, the excess temperature will again become half:

$𝜃\to \frac{𝜃}{2}$

Now temperature of the body will be:

$\text{Room temperature}+\text{excess temperature}$

$𝑥=𝜃+\frac{𝜃}{2}$

$𝑥=\frac{3𝜃}{2}$

So the correct option is

$\overline{)\frac{3}{2}𝜃}$

Option C.

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

Here, the cycle ABCDA forms a rectangle.

Area of the rectangle = change in volume $\times$ change in pressure

So,

Area = $({𝑉}_{𝐵}-{𝑉}_{𝐴})\times ({𝑃}_{𝐶}-{𝑃}_{𝐵})$

From the graph, ${𝑉}_{𝐵}=3{𝑉}_0$, ${𝑉}_{𝐴}={𝑉}_0$, ${𝑃}_{𝐶}=2{𝑃}_0$, and ${𝑃}_{𝐵}={𝑃}_0$.

Putting these values,

Area = $(3{𝑉}_0-{𝑉}_0)\times (2{𝑃}_0-{𝑃}_0)$

Area = $\left(2{𝑉}_0\right)\times \left({𝑃}_0\right)=2{𝑃}_0{𝑉}_0$

The cycle ABCDA is in the counter-clockwise direction. For a counter-clockwise cycle, the work done by the gas is negative.

So, the work done during the cycle is $-2{𝑃}_0{𝑉}_0$.

For the isothermal expansion, temperature remains constant.

For an ideal gas in an isothermal process,

$𝑃𝑉=\text{constant}$

Initially,

${𝑃}_1=𝑃,{𝑉}_1=𝑉$

Finally,

${𝑃}_{𝑖}=?,{𝑉}_2=2𝑉$

So,

$𝑃𝑉={𝑃}_{𝑖}\left(2𝑉\right)$

${𝑃}_{𝑖}=\frac{𝑃}{2}$

Now for the adiabatic expansion,

$𝑃{𝑉}^{𝛾}=\text{constant}$

So,

$𝑃{𝑉}^{𝛾}={𝑃}_{𝑎}(2𝑉{)}^{𝛾}$

${𝑃}_{𝑎}=\frac{𝑃}{2^{𝛾}}$

Now the required ratio is

$\frac{{𝑃}_{𝑖}}{{𝑃}_{𝑎}}=\frac{\frac{𝑃}{2}}{\frac{𝑃}{2^{𝛾}}}$

$\frac{{𝑃}_{𝑖}}{{𝑃}_{𝑎}}=\frac{𝑃}{2}\times \frac{2^{𝛾}}{𝑃}$

$\frac{{𝑃}_{𝑖}}{{𝑃}_{𝑎}}=2^{𝛾-1}$

So the correct option is

$\overline{)2^{𝛾-1}}$

Option C.

For an ideal gas, the equation of state is

$𝑃𝑉=𝑛𝑅𝑇$

where $𝑛$ is the number of moles.

Given mass of oxygen $=7\text{ g}$

Molecular weight of oxygen $=32{\text{ g mol}}^{-1}$

So,

$𝑛=\frac{\text{given mass}}{\text{molecular mass}}$

$𝑛=\frac{7}{32}$

Putting this in the ideal gas equation,

$𝑃𝑉=\frac{7}{32}𝑅𝑇$

So, the correct option is

$\overline{)\text{Option D }𝑃𝑉=\left(\frac{7}{32}\right)𝑅𝑇}$

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

$𝑣=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

Here, $𝑅$ is the gas constant and $𝑀$ is the molar mass of the gas.

For the same gas, $𝑅$ and $𝑀$ are constants. So,

$𝑣\propto \sqrt{\surd 𝑇}$

Squaring both sides,

${𝑣}^2\propto 𝑇$

So,

$\frac{{𝑣}^2}{𝑇}=\text{constant}$

Hence, the true relation is

$\overline{)\frac{{𝑣}^2}{𝑇}=\text{constant}}$

So, the correct option is Option D.

Let the spring constant be $𝑘$.

For a mass attached to a spring, the time period is given by

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}$

When the $500\text{ g}$ mass is removed, the masses left on the spring are $100\text{ g}$ and $300\text{ g}$.

So the remaining mass is

${𝑚}_1=100\text{ g}+300\text{ g}=400\text{ g}$

The time period for this mass is given as

${𝑇}_1=3\text{ s}$

When the $300\text{ g}$ mass is also removed, only the $100\text{ g}$ mass remains.

So the new mass is

${𝑚}_2=100\text{ g}$

Since the same spring is used, $𝑘$ remains the same. So we can compare the two time periods using

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

Substitute the values:

$\frac{{𝑇}_2}{3}=\sqrt{\surd \frac{100}{400}}$

$\frac{{𝑇}_2}{3}=\sqrt{\surd \frac{1}{4}}$

$\frac{{𝑇}_2}{3}=\frac{1}{2}$

So,

${𝑇}_2=\frac{3}{2}=1.5\text{ s}$

For a particle performing SHM, the maximum velocity is

${𝑣}_{max}=𝑎𝜔$

where $𝑎$ is the amplitude and $𝜔$ is the angular frequency.

For a mass attached to a spring,

$𝜔=\sqrt{\surd \frac{𝐾}{𝑚}}$

The masses of particles A and B are equal.

So,

${𝑣}_{max,𝐴}={𝑎}_{𝐴}\sqrt{\surd \frac{{𝐾}_{𝐴}}{𝑚}}$

and

${𝑣}_{max,𝐵}={𝑎}_{𝐵}\sqrt{\surd \frac{{𝐾}_{𝐵}}{𝑚}}$

Given that their maximum velocities are equal,

${𝑎}_{𝐴}\sqrt{\surd \frac{{𝐾}_{𝐴}}{𝑚}}={𝑎}_{𝐵}\sqrt{\surd \frac{{𝐾}_{𝐵}}{𝑚}}$

Since $𝑚$ is same for both,

${𝑎}_{𝐴}\sqrt{\surd {𝐾}_{𝐴}}={𝑎}_{𝐵}\sqrt{\surd {𝐾}_{𝐵}}$

So,

$\frac{{𝑎}_{𝐵}}{{𝑎}_{𝐴}}=\sqrt{\surd \frac{{𝐾}_{𝐴}}{{𝐾}_{𝐵}}}$

Hence, the ratio of amplitude of B to amplitude of A is

${𝑎}_{𝐵}:{𝑎}_{𝐴}=\sqrt{\surd {𝐾}_{𝐴}}:\sqrt{\surd {𝐾}_{𝐵}}$

So the correct option is

$\overline{)\text{Option B }\sqrt{\surd {𝐾}_{𝐴}}:\sqrt{\surd {𝐾}_{𝐵}}}$

For a particle executing S.H.M., the potential energy at displacement $𝑥$ is

$𝑈=\frac{1}{2}𝑚{𝜔}^2{𝑥}^2$

Here, the given frequency is

$𝑓=\sqrt{\surd \frac{𝐾𝑎}{𝜋𝑚}}$

In S.H.M., angular frequency $𝜔$ is related to frequency $𝑓$ by

$𝜔=2𝜋𝑓$

So,

${𝜔}^2=(2𝜋𝑓{)}^2=4{𝜋}^2{𝑓}^2$

Substitute the value of ${𝑓}^2$:

${𝑓}^2=\frac{𝐾𝑎}{𝜋𝑚}$

Hence,

${𝜔}^2=4{𝜋}^2\times \frac{𝐾𝑎}{𝜋𝑚}$

${𝜔}^2=\frac{4𝜋𝐾𝑎}{𝑚}$

Now substitute this in the potential energy formula:

$𝑈=\frac{1}{2}𝑚{𝜔}^2{𝑥}^2$

$𝑈=\frac{1}{2}𝑚\times \frac{4𝜋𝐾𝑎}{𝑚}\times {𝑥}^2$

$𝑈=2𝜋𝐾𝑎{𝑥}^2$

So, the potential energy of the particle is

$\overline{)2𝜋𝐾𝑎{𝑥}^2}$

Correct option: Option C.

For the original organ pipe closed at one end, if its length is $𝐿$, then its fundamental frequency is

$𝜈=\frac{𝑣}{4𝐿}$

where $𝑣$ is the speed of sound in air.

Now the pipe is cut into two equal parts, so each new pipe has length

$\frac{𝐿}{2}$

One part will remain closed at one end and open at the other.

For this closed pipe of length $\frac{𝐿}{2}$, the fundamental frequency is

${𝜈}_1=\frac{𝑣}{4\left(\frac{𝐿}{2}\right)}$

${𝜈}_1=\frac{𝑣}{2𝐿}=2𝜈$

The other part will be open at both ends.

For an open pipe of length $\frac{𝐿}{2}$, the fundamental frequency is

${𝜈}_2=\frac{𝑣}{2\left(\frac{𝐿}{2}\right)}$

${𝜈}_2=\frac{𝑣}{𝐿}=4𝜈$

So the fundamental frequencies of the two pipes are

$2𝜈,4𝜈$

Correct option: C.

The fundamental frequency is given as

$𝑛=288\text{Hz}$

For resonance in the harmonic series, the possible resonant frequencies are integral multiples of the fundamental frequency:

$𝑛,2𝑛,3𝑛,4𝑛,\dots$

So,

$𝑛=288\text{Hz}$

$2𝑛=2\times 288=576\text{Hz}$

$3𝑛=3\times 288=864\text{Hz}$

So $288$ Hz, $576$ Hz, and $864$ Hz are possible resonant frequencies.

But

$844\text{Hz}$

is not an integral multiple of $288$ Hz.

So it will not resonate with

$\overline{)844\text{Hz}}$

Correct option: C.

For a wire stretched between two rigid supports, the frequency of the ${𝑛}^{\text{th}}$ harmonic is

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

Here $𝐿$, $𝑇$, and $𝑚$ are fixed, so the frequency is directly proportional to the harmonic number $𝑛$.

The wire resonates in the ${𝑃}^{\text{th}}$ harmonic at $320\text{Hz}$.

So,

${𝑓}_{𝑃}=320$

The next higher frequency is in the next successive mode, so it will be the $(𝑃+1{)}^{\text{th}}$ harmonic.

${𝑓}_{𝑃+1}=400$

Since frequency is proportional to harmonic number,

$\frac{{𝑓}_{𝑃+1}}{{𝑓}_{𝑃}}=\frac{𝑃+1}{𝑃}$

Substitute the given frequencies:

$\frac{400}{320}=\frac{𝑃+1}{𝑃}$

$\frac{5}{4}=\frac{𝑃+1}{𝑃}$

Now solve:

$5𝑃=4(𝑃+1)$

$5𝑃=4𝑃+4$

$𝑃=4$

So, the value of $𝑃$ is

$\overline{)4}$

Correct option: B.

For an echo, sound travels to the cliff and comes back.

So the total distance travelled by sound is twice the distance between the person and that cliff.

Using

$\text{Distance}=\text{Speed}\times \text{Time}$

For the first echo after $1$ second:

$2{𝑑}_1=330\times 1$

${𝑑}_1=\frac{330}{2}=165\text{ m}$

So the person is $165$ m from one cliff.

For the second echo after $3$ seconds:

$2{𝑑}_2=330\times 3$

${𝑑}_2=\frac{990}{2}=495\text{ m}$

So the person is $495$ m from the other cliff.

Since the person is standing between the two parallel cliffs, the distance between the cliffs is

${𝑑}_1+{𝑑}_2=165+495=660\text{ m}$

So the correct option is:

$\overline{)660\text{ m}}$

Option B.

The total electrostatic potential energy of three charges is the sum of the potential energies of each pair of charges.

Here, the charges are $𝑄$, $+2𝑞$, and $+𝑞$.

From the figure, the triangle is a right-angled isosceles triangle. The two equal sides are each $\sqrt{\surd 2}𝑎$.

So, the distances are:

Distance between $𝑄$ and $2𝑞$ is $\sqrt{\surd 2}𝑎$.

Distance between $𝑄$ and $𝑞$ is $\sqrt{\surd 2}𝑎$.

The distance between $2𝑞$ and $𝑞$ is the hypotenuse:

$\sqrt{\surd (\sqrt{\surd 2}𝑎{)}^2+(\sqrt{\surd 2}𝑎{)}^2}=\sqrt{\surd 2{𝑎}^2+2{𝑎}^2}=\sqrt{\surd 4{𝑎}^2}=2𝑎$

The total potential energy is:

$𝑈=\frac{1}{4𝜋{𝜖}_0}(\frac{𝑄\left(2𝑞\right)}{\sqrt{\surd 2}𝑎}+\frac{𝑄\left(𝑞\right)}{\sqrt{\surd 2}𝑎}+\frac{\left(2𝑞\right)\left(𝑞\right)}{2𝑎})$

It is given that the net electrostatic potential energy is zero, so $𝑈=0$.

Since $\frac{1}{4𝜋{𝜖}_0}$ is a common non-zero factor, we can write:

$\frac{2𝑞𝑄}{\sqrt{\surd 2}𝑎}+\frac{𝑞𝑄}{\sqrt{\surd 2}𝑎}+\frac{2{𝑞}^2}{2𝑎}=0$

Simplify the last term:

$\frac{2𝑞𝑄}{\sqrt{\surd 2}𝑎}+\frac{𝑞𝑄}{\sqrt{\surd 2}𝑎}+\frac{{𝑞}^2}{𝑎}=0$

Now divide the whole equation by $\frac{𝑞}{𝑎}$:

$\frac{2𝑄}{\sqrt{\surd 2}}+\frac{𝑄}{\sqrt{\surd 2}}+𝑞=0$

Combine the first two terms:

$\frac{3𝑄}{\sqrt{\surd 2}}+𝑞=0$

$\frac{3𝑄}{\sqrt{\surd 2}}=-𝑞$

$𝑄=-\frac{\sqrt{\surd 2}}{3}𝑞$

For two equal point charges in air, Coulomb’s force is

$𝐹=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑑}^2}$

In a medium of dielectric constant $𝐾$, the force becomes

$𝐹=\frac{1}{4𝜋{𝜀}_0𝐾}\frac{{𝑞}^2}{{𝐷}^2}$

The force is given to be the same in both cases, so

$\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑑}^2}=\frac{1}{4𝜋{𝜀}_0𝐾}\frac{{𝑞}^2}{{𝐷}^2}$

Cancel common terms:

$\frac{1}{{𝑑}^2}=\frac{1}{𝐾{𝐷}^2}$

So,

$𝐾{𝐷}^2={𝑑}^2$

${𝐷}^2=\frac{{𝑑}^2}{𝐾}$

$𝐷=\frac{𝑑}{\sqrt{\surd 𝐾}}$

So the correct option is

$\overline{)\text{Option A }\frac{𝑑}{\sqrt{\surd 𝐾}}}$

Let the positive terminal of the battery be connected to node A and the negative terminal to node C. The top wire is node B.

Now check which capacitors are connected between the same two nodes. Capacitors connected between the same two nodes are in parallel.

The two $3𝐶$ capacitors are both connected between nodes A and B. So, they are in parallel.

Their equivalent capacitance is:

${𝐶}_{𝐴𝐵}=3𝐶+3𝐶=6𝐶$

The $1𝐶$ capacitor and the right-side $2𝐶$ capacitor are both connected between nodes B and C. So, they are also in parallel.

Their equivalent capacitance is:

${𝐶}_{𝐵𝐶}=1𝐶+2𝐶=3𝐶$

Now the capacitor ${𝐶}_{𝐴𝐵}=6𝐶$ is between A and B, and ${𝐶}_{𝐵𝐶}=3𝐶$ is between B and C. These two are in series.

The equivalent capacitance of these two in series is:

${𝐶}_{𝑠𝑒𝑟𝑖𝑒𝑠}=\frac{{𝐶}_{𝐴𝐵}\times {𝐶}_{𝐵𝐶}}{{𝐶}_{𝐴𝐵}+{𝐶}_{𝐵𝐶}}=\frac{6𝐶\times 3𝐶}{6𝐶+3𝐶}=2𝐶$

This series combination is connected between A and C. The middle $2𝐶$ capacitor is also connected between A and C.

So, the series combination and the middle $2𝐶$ capacitor are in parallel.

The total equivalent capacitance is:

${𝐶}_{𝑒𝑞}={𝐶}_{𝑠𝑒𝑟𝑖𝑒𝑠}+2𝐶=2𝐶+2𝐶=4𝐶$

The energy stored in a capacitor network is given by:

$𝑈=\frac{1}{2}{𝐶}_{𝑒𝑞}{𝑉}^2$

Putting ${𝐶}_{𝑒𝑞}=4𝐶$, we get:

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

For capacitors connected to the same source, the potential difference $𝑉$ is the same in both cases.

Energy stored in a capacitor combination is

$𝑈=\frac{1}{2}{𝐶}_{\text{eq}}{𝑉}^2$

In series, the net capacitance is given as

${𝐶}_{𝑠}=𝑥$

Since the three capacitors are identical, suppose each capacitor has capacitance $𝐶$.

For three identical capacitors in series,

${𝐶}_{𝑠}=\frac{𝐶}{3}$

So,

$\frac{𝐶}{3}=𝑥$

$𝐶=3𝑥$

Now, when the same three capacitors are connected in parallel,

${𝐶}_{𝑝}=𝐶+𝐶+𝐶=3𝐶$

${𝐶}_{𝑝}=3\left(3𝑥\right)=9𝑥$

Since both combinations are connected to the same source, $𝑉$ is the same.

So the ratio of energies is

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

$\frac{{𝑈}_{𝑠}}{{𝑈}_{𝑝}}=\frac{{𝐶}_{𝑠}}{{𝐶}_{𝑝}}$

$\frac{{𝑈}_{𝑠}}{{𝑈}_{𝑝}}=\frac{𝑥}{9𝑥}=\frac{1}{9}$

So, the ratio of energy stored in series to parallel is

$\overline{)1:9}$

Correct option: C.

For a potentiometer, the balancing length is directly proportional to the E.M.F. being balanced, if the potential gradient of the wire remains the same.

Let the potential gradient of the potentiometer wire be $𝑘$.

When the cell of E.M.F. ${𝐸}_1$ is connected alone, the balancing length is ${𝑙}_1$.

So,

${𝐸}_1=𝑘{𝑙}_1$

Now, the second cell of E.M.F. ${𝐸}_2$ is connected with ${𝐸}_1$ such that the two cells oppose each other.

Since ${𝐸}_1>{𝐸}_2$, the effective E.M.F. of the combination is

${𝐸}_1-{𝐸}_2$

The balancing length for this opposing combination is ${𝑙}_2$.

So,

${𝐸}_1-{𝐸}_2=𝑘{𝑙}_2$

Using ${𝐸}_1=𝑘{𝑙}_1$,

$𝑘{𝑙}_1-{𝐸}_2=𝑘{𝑙}_2$

${𝐸}_2=𝑘{𝑙}_1-𝑘{𝑙}_2$

${𝐸}_2=𝑘({𝑙}_1-{𝑙}_2)$

Now,

${𝐸}_1:{𝐸}_2=𝑘{𝑙}_1:𝑘({𝑙}_1-{𝑙}_2)$

Cancelling $𝑘$,

${𝐸}_1:{𝐸}_2={𝑙}_1:({𝑙}_1-{𝑙}_2)$

So, the correct option is

$\overline{)\text{Option B }{𝑙}_1:({𝑙}_1-{𝑙}_2)}$

Use Kirchhoff's Current Law at each junction. According to KCL, the total current entering a junction is equal to the total current leaving it.

At the top-left node, ${𝐼}_5$ enters the node and ${𝐼}_6$ leaves the node. So,
${𝐼}_6={𝐼}_5$

Given ${𝐼}_5=0.4\text{A}$, we get
${𝐼}_6=0.4\text{A}$

At the bottom-left node, ${𝐼}_3$ and ${𝐼}_4$ enter the node, and ${𝐼}_5$ leaves the node. So,
${𝐼}_3+{𝐼}_4={𝐼}_5$

Substitute ${𝐼}_4=1\text{A}$ and ${𝐼}_5=0.4\text{A}$:
${𝐼}_3+1=0.4$
${𝐼}_3=-0.6\text{A}$

The negative sign means the actual direction of ${𝐼}_3$ is opposite to the direction shown in the diagram.

At the bottom-right node, ${𝐼}_1$ and ${𝐼}_2$ enter the node, and ${𝐼}_4$ leaves the node. So,
${𝐼}_1+{𝐼}_2={𝐼}_4$

Substitute ${𝐼}_1=-0.4\text{A}$ and ${𝐼}_4=1\text{A}$:
$-0.4+{𝐼}_2=1$
${𝐼}_2=1.4\text{A}$

So, the values of ${𝐼}_2$, ${𝐼}_3$, and ${𝐼}_6$ are respectively
$1.4\text{A}$ , $-0.6\text{A}$ , $0.4\text{A}$

For a charge moving in a circular path, the motion of charge is like a current loop.

The time period of revolution is

$𝑇=\frac{2𝜋}{𝜔}$

So the current is

$𝐼=\frac{𝑞}{𝑇}=\frac{𝑞𝜔}{2𝜋}$

The area of the circular orbit is

$𝐴=𝜋{𝑟}^2$

So the magnetic moment is

$𝜇=𝐼𝐴$

$𝜇=\frac{𝑞𝜔}{2𝜋}\times 𝜋{𝑟}^2$

$𝜇=\frac{𝑞𝜔{𝑟}^2}{2}$

The angular momentum of the particle is

$𝐿=𝑚𝑣𝑟$

Since $𝑣=𝜔𝑟$,

$𝐿=𝑚𝜔{𝑟}^2$

Now the ratio is

$\frac{𝜇}{𝐿}=\frac{\frac{𝑞𝜔{𝑟}^2}{2}}{𝑚𝜔{𝑟}^2}$

$\frac{𝜇}{𝐿}=\frac{𝑞}{2𝑚}$

So the ratio depends only on $𝑞$ and $𝑚$.

Hence, the correct option is

$\overline{)\text{Option A: }𝑞\text{ and }𝑚}$

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

$𝐵=\frac{{𝜇}_0𝐼}{2𝜋𝑟}$

Here, the current $𝐼$ is the same in both cases, so the magnetic field is inversely proportional to the distance:

$𝐵\propto \frac{1}{𝑟}$

At distance $2\text{ cm}$,

${𝐵}_1\propto \frac{1}{2}$