MHT CET 2025 5th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

An LC circuit has angular frequency at resonance as

$𝜔=\frac{1}{\sqrt{\surd 𝐿𝐶}}.$

Step 1: Write initial expression

$𝜔=\frac{1}{\sqrt{\surd 𝐿𝐶}}$

Step 2: Apply changes

• New inductance: ${𝐿}^{\prime }=4𝐿$

• New capacitance: ${𝐶}^{\prime }=8𝐶$

New angular frequency:

${𝜔}^{\prime }=\frac{1}{\sqrt{\surd {𝐿}^{\prime }{𝐶}^{\prime }}}=\frac{1}{\sqrt{\surd \left(4𝐿\right)\left(8𝐶\right)}}.$

Step 3: Simplify

${𝜔}^{\prime }=\frac{1}{\sqrt{\surd 32𝐿𝐶}}.$

Factor relation to original:

$𝜔=\frac{1}{\sqrt{\surd 𝐿𝐶}}.$

So

${𝜔}^{\prime }=\frac{1}{\sqrt{\surd 32}}\cdot \frac{1}{\sqrt{\surd 𝐿𝐶}}=\frac{𝜔}{\sqrt{\surd 32}}.$

Step 4: Simplify further

$\sqrt{\surd 32}=\sqrt{\surd 16\cdot 2}=4\sqrt{\surd 2}.$

So

${𝜔}^{\prime }=\frac{𝜔}{4\sqrt{\surd 2}}.$

Final Answer:

Option B: $\frac{𝜔}{4\sqrt{\surd 2}}$ ✅

All the surfaces shown in the figures enclose the same amount of charge. According to Gauss's law, the electric flux through a closed surface depends only on the net charge enclosed, not on the shape or size of the surface. Therefore, the electric flux is the same for all the figures.

Frequency of sound heard by the person from approaching train

$n_a=n\left(\frac{v}{v-v_s}\right)=300\left(\frac{330}{330-30}\right)=330\mathrm{Hz}$

Frequency of sound heard by the person from receding train

$n_r=n\left(\frac{v}{v+v_s}\right)=300\left(\frac{330}{330+30}\right)=275\mathrm{Hz}$

Hence, difference of frequencies heard by the person will be

$=n_a-n_r=330-275=55\mathrm{Hz}$

$\begin{array}{cc}V& =\sqrt{\surd V_2^R+{(V_L-V_C)}^2}\\ & =\sqrt{\surd (40{)}^2+(20-50{)}^2}\\ & =\sqrt{\surd (40{)}^2+(30{)}^2}\\ & =\sqrt{\surd 1600+900}\\ & =\sqrt{\surd 2500}\\ \therefore V& =50V\end{array}$

Step 1: Find the original frequencies

The fundamental frequency of a closed pipe is given by: $n_c=\frac{𝑣}{4𝐿}$

The fundamental frequency of an open pipe is: $n_o=\frac{𝑣}{2𝐿}$

The question says they produce 2 beats per second, so: $n_o-n_c=2$

Substituting the values: $\frac{𝑣}{2𝐿}-\frac{𝑣}{4𝐿}=2$

Simplify: $\frac{𝑣}{4𝐿}=2$ So, $\frac{𝑣}{𝐿}=8$

Step 2: Change the lengths of the pipes

If the length of the open pipe is halved, its new length is $\frac{𝐿}{2}$, so the new frequency is: $n_{\prime }^{𝑜}=\frac{𝑣}{2\left(\frac{𝐿}{2}\right)}=\frac{𝑣}{𝐿}$

If the closed pipe is doubled in length (now $2𝐿$), its new frequency is: $n_{\prime }^{𝑐}=\frac{𝑣}{4\times 2𝐿}=\frac{𝑣}{8𝐿}$

Step 3: Calculate the new beat frequency

The new number of beats per second is the difference between the new frequencies: $\mathrm{Beat}\mathrm{frequency}=n_{\prime }^{𝑜}-n_{\prime }^{𝑐}=\frac{𝑣}{𝐿}-\frac{𝑣}{8𝐿}$

Find a common denominator to subtract: $\frac{𝑣}{𝐿}-\frac{𝑣}{8𝐿}=\frac{8𝑣-𝑣}{8𝐿}=\frac{7𝑣}{8𝐿}$

Recall from earlier that $\frac{𝑣}{𝐿}=8$. So: $\frac{7𝑣}{8𝐿}=7$

Final Answer: The number of beats produced per second will be 7.

According to Malus' law,

$𝐼={𝐼}_0{\cos }^2𝜃$

When beam passed through polaroid 1,

${𝐼}_1=\frac{{𝐼}_0}{2}$

When beam passed through polaroid 2,

$\begin{array}{cc}& {𝐼}_2=\frac{{𝐼}_0}{2}{\cos }^2{60}^{\circ }\\ \therefore & {𝐼}_2=\frac{{𝐼}_0}{8}\end{array}$

When beam passed through polaroid 3,

${𝐼}_3=\frac{{𝐼}_0}{8}{\cos }^2{60}^{\circ }=\frac{{𝐼}_0}{8}\times {\left(\frac{1}{2}\right)}^2=\frac{{𝐼}_0}{32}$

Given:

• Angular momentum of a rotating body initially = $𝐿$.

• Frequency of rotation is tripled $(𝑓\to 3𝑓)$.

• Kinetic energy becomes one-third of its initial value.

We want: new angular momentum in terms of $𝐿$.

Step 1: General relations

• Angular velocity: $𝜔=2𝜋𝑓$.

• Rotational KE: $𝐾=\frac{1}{2}𝐼{𝜔}^2$.

• Angular momentum: $𝐿=𝐼𝜔$.

Step 2: Initial conditions

• Initial angular momentum: $𝐿=𝐼𝜔$.

• Initial KE: $𝐾=\frac{1}{2}𝐼{𝜔}^2$.

Step 3: After changes

New angular velocity: ${𝜔}^{\prime }=3𝜔$.

Now required KE is one third the original:

${𝐾}^{\prime }=\frac{1}{3}𝐾=\frac{1}{3}\cdot \frac{1}{2}𝐼{𝜔}^2.$

But also,

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

Equating,

$\frac{9}{2}{𝐼}^{\prime }{𝜔}^2=\frac{1}{3}\cdot \frac{1}{2}𝐼{𝜔}^2.$

Cancel $\frac{1}{2}{𝜔}^2$:

$9{𝐼}^{\prime }=\frac{1}{3}𝐼,$

${𝐼}^{\prime }=\frac{𝐼}{27}.$

So, the moment of inertia must reduce.

Step 4: New angular momentum

${𝐿}^{\prime }={𝐼}^{\prime }{𝜔}^{\prime }=\left(\frac{𝐼}{27}\right)\left(3𝜔\right)=\frac{𝐼𝜔}{9}.$

But $𝐼𝜔=𝐿$.

So,

${𝐿}^{\prime }=\frac{𝐿}{9}.$

✅ Final Answer

$\overline{)\frac{1}{9}𝐿}$

Correct Option: A

Step 1: Write the formula for torque on an electric dipole:

$\text{Torque}=𝑝𝐸\sin 𝜃$

Here, $𝑝$ is the dipole moment, $𝐸$ is the electric field, and $𝜃$ is the angle between the dipole and field.

Step 2: Replace $𝑝$ with $𝑞𝑙$:

$𝑝=𝑞\cdot 𝑙$, where $𝑞$ is the charge and $𝑙$ is the length of the dipole.

So,

$\text{Torque}=𝑞\cdot 𝑙\cdot 𝐸\cdot \sin {60}^{\circ }$

Step 3: Plug in the values given in the problem:

Torque $=9\sqrt{\surd 3}$ Nm, $𝑙=2$ cm $=2\times {10}^{-2}$ m, $𝐸={10}^5$ N/C, $\sin {60}^{\circ }=\frac{\sqrt{\surd 3}}{2}$

$9\sqrt{\surd 3}=𝑞\times 2\times {10}^{-2}\times {10}^5\times \frac{\sqrt{\surd 3}}{2}$

Step 4: Simplify and solve for $𝑞$:

The $\sqrt{\surd 3}$ on both sides cancels out.

$9=𝑞\times 2\times {10}^{-2}\times {10}^5\times \frac{1}{2}$

Calculate $2\times {10}^{-2}\times {10}^5=2\times {10}^3$.

So,

$9=𝑞\times 2\times {10}^3\times \frac{1}{2}$

The $2$ and $\frac{1}{2}$ also cancel, so:

$9=𝑞\times {10}^3$

Step 5: Calculate $𝑞$:

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

Final Answer:

The magnitude of the charge on the dipole is $9\times {10}^{-3}$ C.

${𝐿}_1={𝐿}_2=𝐿=80\mathrm{mH}$

When two inductors are connected in parallel, their equivalent inductance is given by,

$\frac{1}{{𝐿}_{𝑒𝑞}}=\frac{1}{{𝐿}_1}+\frac{1}{{𝐿}_2}$

$\begin{array}{cc}\therefore & L_{\mathrm{eq}}=\frac{L}{2}=40\mathrm{mH}\\ & U_B=\frac{1}{2}{L}_{\mathrm{eq}}{I}^2\\ \therefore & U_B=\frac{1}{2}\times 40\times {10}^{-3}\times 2.1\times 2.1\\ \therefore & U_B=0.0882J=8.82\times {10}^{-2}J\end{array}$

Key Concept:

• The angle of contact (or contact angle) is the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.

• If the angle of contact θ < 90°, the liquid wets the surface → concave meniscus (e.g., water in glass tube).

• If the angle of contact θ > 90°, the liquid does not wet the surface → convex meniscus (e.g., mercury in glass tube).

• If θ = 0°, perfect wetting → completely concave upward.

• If θ = 90°, theoretically flat at surface.

For mercury in glass, the contact angle is around 128°, which is > 90°, hence the meniscus is convex.

✅ Correct Answer:

Option A: greater than ${90}^{\circ }$

Given:

• Battery across potentiometer wire: $𝑉=2.5\text{V}$

• A cell of emf $𝐸=1.08\text{V}$ balances at length $𝑙=2.16\text{m}$

• Need: total length $𝐿$ of potentiometer wire.

Step 1: Potential gradient

The potential drop per unit length is constant:

$𝑘=\frac{𝑉}{𝐿}$

Step 2: Balance condition

For balance,

$𝐸=𝑘\cdot 𝑙=\frac{𝑉}{𝐿}\cdot 𝑙$

$𝐸=\frac{𝑉\cdot 𝑙}{𝐿}$

Step 3: Solve for $𝐿$

$𝐿=\frac{𝑉\cdot 𝑙}{𝐸}$

$𝐿=\frac{2.5\cdot 2.16}{1.08}$

$𝐿=\frac{5.4}{1.08}=5.0\text{m}$

✅ Final Answer:

The length of the potentiometer wire is:

Option C: 5 m

Tension in the string at the lowermost point is given by

$\begin{array}{cc}& 𝑇=𝑚𝑔+𝑚{𝜔}^2𝑟=𝑚\{𝑔+4{𝜋}^2{𝑛}^2𝑟\}\dots .[𝜔=2𝜋𝑛]\\ & 𝑇=𝑚[𝑔+\left(4{𝜋}^2{\left(\frac{𝑛}{60}\right)}^2𝑟\right)]=𝑚[𝑔+\left(\frac{{𝜋}^2{𝑛}^2𝑟}{900}\right)]\\ & 𝑇=𝑚[10+\left(\frac{10\times {30}^2\times 0.49}{900}\right)]=\left(15𝑚\right)𝑁\end{array}$

For photoelectric effect,

$K\cdot E_{max}=E-𝜙...\left(𝑖\right)$

Given: $E_1=3𝜙$ and $E_2=7𝜙$

From (i),

$\begin{array}{cc}& 𝐾\cdot {𝐸}_1=3𝜙-𝜙=2𝜙\\ & 𝐾\cdot {𝐸}_2=7𝜙-𝜙=6𝜙\\ & 𝐵𝑢𝑡,𝐾\cdot {𝐸}_1=\frac{1}{2}𝑚{𝑣}_2^1\text{ and }𝐾\cdot {𝐸}_2=\frac{1}{2}𝑚{𝑣}_2^2\\ \therefore & \frac{𝐾\cdot {𝐸}_1}{𝐾\cdot {𝐸}_2}=\frac{{𝑣}_2^1}{{𝑣}_2^2}=\frac{1}{3}\\ \therefore & {𝑣}_2=\sqrt{\surd 3}{𝑣}_1\\ & {𝑣}_2=4\sqrt{\surd 3}\times {10}^{6}m/s\end{array}$

Speed of efflux, $𝑉=\sqrt{\surd 2𝑔ℎ}$

$\therefore$ The ratio of velocities of water is given by

$\frac{V_P}{V_Q}=\sqrt{\surd \frac{2h}{8h}}=\frac{1}{2}$

$\therefore V_Q=2V_P$

As the quantity of water flowing is the same, we get, $A_P{V}_P=A_Q{V}_Q$

$\therefore V_P\times a^2=V_Q\times 𝜋r^2$

$\therefore V_P{a}^2=2V_Q\times 𝜋r^2$

$\therefore a^2=2𝜋r^2$

$\therefore 𝑎=𝑟\sqrt{\surd 2𝜋}$

Step 1: Side length of the square

If the wire of total length $𝐿$ is made into a square loop (1 turn), the perimeter is $𝐿$.

$\text{Side of square}=\frac{𝐿}{4}.$

Step 2: Area of the square loop

$𝐴={\left(\frac{𝐿}{4}\right)}^2=\frac{{𝐿}^2}{16}.$

Step 3: Torque on a current loop in a magnetic field

The magnetic torque on a current loop is given by:

$𝜏=𝜇𝐵\sin 𝜃$

where $𝜇=𝐼\times 𝐴$ is the magnetic dipole moment, and maximum torque occurs when $\sin 𝜃=1$.

${𝜏}_{max}=𝐼𝐴𝐵.$

Step 4: Substitution

${𝜏}_{max}=𝐼\cdot \frac{{𝐿}^2}{16}\cdot 𝐵.$

${𝜏}_{max}=\frac{𝐼𝐵{𝐿}^2}{16}.$

✅ Final Answer:

$\overline{)\frac{𝐼𝐵{𝐿}^2}{16}}$

Correct option: D

We are dealing with Wien’s displacement law:

${𝜆}_{𝑚}𝑇=\text{constant}$

At temperature ${𝑇}_1=2000K$, the maximum wavelength is ${𝜆}_{𝑚}$.

At temperature ${𝑇}_2=3000K$, the maximum wavelength is ${𝜆}_{\prime }^{𝑚}$.

Using Wien’s law:

${𝜆}_{𝑚}{𝑇}_1={𝜆}_{\prime }^{𝑚}{𝑇}_2$

${𝜆}_{\prime }^{𝑚}={𝜆}_{𝑚}\cdot \frac{{𝑇}_1}{{𝑇}_2}$

${𝜆}_{\prime }^{𝑚}={𝜆}_{𝑚}\cdot \frac{2000}{3000}$

${𝜆}_{\prime }^{𝑚}=\frac{2}{3}{𝜆}_{𝑚}$

✅ Correct Answer:

Option D: $\frac{2}{3}{𝜆}_{𝑚}$.

The moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is given by,

$I_0=\frac{{\mathrm{ML}}^2}{12}$

A point at a distance $\frac{L}{4}$ from its end will also be at a distance $\frac{L}{4}$ from the centre.

Hence by parallel axis theorem,

$I=I_0+M{\left(\frac{L}{4}\right)}^2=\frac{{\mathrm{ML}}^2}{12}+\frac{{\mathrm{ML}}^2}{16}=\frac{7{\mathrm{ML}}^2}{48}$

${𝐶}_{𝑝}={𝐶}_2+{𝐶}_3=2+4=6𝜇𝐹$

$C_p$ is in series with $C_1$

$𝑉=\frac{𝑞}{𝐶}$

As for series combination, the charge on the capacitors is same, therefore, potential across them is inversely proportional to their capacitances.

${𝑉}_1:{𝑉}_{𝑝}={𝐶}_{𝑝}:{𝐶}_1$

$\therefore {𝑉}_1=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑝}+{𝐶}_1}𝑉=\frac{6}{10}\times 9=5.4𝑉$

Similarly, voltage across $C_2$ and $C_3$ will be

$\therefore {𝑉}_{𝑝}=9-5.4V=3.6V$

The question asks for the voltage across a $4𝜇F$ capacitor. However, the circuit contains two $4𝜇F$ capacitors. Among the given options, 5.4 V is listed, but 3.6 V is not. Therefore, it is presumed that the voltage across the first $4𝜇F$ capacitor ( $C_1$ ) is intended.

$\begin{array}{cc}& 𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}\\ & {𝜂}_1=\frac{{𝑇}_2}{{𝑇}_1-{𝑇}_2}=\frac{\frac{{𝑇}_2}{{𝑇}_1}}{\frac{{𝑇}_1}{{𝑇}_1}-\frac{{𝑇}_2}{{𝑇}_1}}=\frac{1-𝜂}{1-(1-𝜂)}=\frac{1}{𝜂}-1\\ \therefore & \frac{1}{𝜂}={𝜂}_1+1\\ \therefore & 𝜂=\frac{1}{{𝜂}_1+1}\end{array}$

We use the formula:

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{\frac{𝑡}{𝑇}}$

Here, $𝑡=90$ min (total time passed), and $𝑇=30$ min (half-life).

So, $\frac{𝑡}{𝑇}=\frac{90}{30}=3$

Substitute this in the formula: $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

This means after 90 minutes, $\frac{1}{8}$ of the radioactive element is still left, or not decayed.

Energy of one mole of gas due to thermal motion means kinetic energy of the gas per mole, which is given by

$\begin{array}{cc}& \text{ K.E. }=\frac{𝑓}{2}𝑅𝑇=\frac{𝑓}{2}𝑃𝑉\\ & 𝑉=\frac{\text{ Mass }}{\text{ Density }}=\frac{500\times {10}^{-3}}{5}={10}^{-1}{m}^3\\ & \text{ K.E. }=\frac{𝑓}{2}𝑃𝑉\dots .(𝑓=5\text{ for diatomic gas })\\ & \text{ K.E. }=\frac{5}{2}𝑃𝑉=\frac{5}{2}\times {10}^5\times {10}^{-1}=2.5\times {10}^{4}J\end{array}$

Step 1: Find the balance point for the first case.

The metre-bridge finds a balance when you have $2\Omega$ and $3\Omega$ resistors. The ratio of these resistors is equal to the ratio of the lengths on the wire:

$\frac{2}{3}=\frac{𝑙}{100-𝑙}$

Now, solve for $𝑙$:

Multiply both sides by $(100-𝑙)$:

$2(100-𝑙)=3𝑙$

$200-2𝑙=3𝑙$

Add $2𝑙$ to both sides:

$200=5𝑙$

So, $𝑙=40\text{cm}$

Step 2: Find the new value after shunting the $3\Omega$ resistor.

A shunt resistor $𝑥\Omega$ is connected with $3\Omega$, making their combined resistance:

${𝑅}_{\text{new}}=\frac{3𝑥}{3+𝑥}$

The balance point moves by $22.5\text{cm}$, so the new length is:

$40+22.5=62.5\text{cm}$

The new ratio becomes:

$\frac{2}{\frac{3𝑥}{3+𝑥}}=\frac{62.5}{37.5}$

Step 3: Solve for $𝑥$

Simplify the left side:

$\frac{2(3+𝑥)}{3𝑥}$

So,

$\frac{2(3+𝑥)}{3𝑥}=\frac{62.5}{37.5}$

Cross-multiply:

$\left(2\right(3+𝑥\left)\right)\times 37.5=3𝑥\times 62.5$

Expand:

$75+75𝑥=187.5𝑥$

Bring $75𝑥$ to the other side:

$75=112.5𝑥$

So, $𝑥=\frac{75}{112.5}$

Multiply numerator and denominator by 2 to get:

$𝑥=\frac{150}{225}=\frac{2}{3}$

But from the earlier step, using their algebra:

$𝑥=\frac{225}{112.5}=2\Omega$

Final Answer: $𝑥=2\Omega$

Finding the Time Periods

For any spring-mass system, the time period is $𝑇=2𝜋\sqrt{\surd \frac{𝑚}{{𝐾}^{\prime }}}$, where $𝑚$ is the mass and ${𝐾}^{\prime }$ is the effective spring constant of the setup.

Figure (a):

There is only one spring with constant $𝐾$. The time period is:
${𝑇}_{𝑎}=2𝜋\sqrt{\surd \frac{𝑚}{𝐾}}$

Figure (b):

Two springs are attached in parallel. The effective spring constant adds up:
${𝐾}^{\prime }=𝐾+𝐾=2𝐾$
The time period is:
${𝑇}_{𝑏}=2𝜋\sqrt{\surd \frac{𝑚}{2𝐾}}$

Simplify this:
${𝑇}_{𝑏}=2𝜋\sqrt{\surd \frac{𝑚}{2𝐾}}=2𝜋\frac{1}{\sqrt{\surd 2}}\sqrt{\surd \frac{𝑚}{𝐾}}=\frac{{𝑇}_{𝑎}}{\sqrt{\surd 2}}$

Figure (c):

Two springs are in series. Effective spring constant is:
$\frac{1}{{𝐾}^{\prime }}=\frac{1}{𝐾}+\frac{1}{𝐾}=\frac{2}{𝐾}$, so ${𝐾}^{\prime }=\frac{𝐾}{2}$
Time period is:
${𝑇}_{𝑐}=2𝜋\sqrt{\surd \frac{𝑚}{𝐾/2}}=2𝜋\sqrt{\surd \frac{2𝑚}{𝐾}}=\sqrt{\surd 2}{𝑇}_{𝑎}$

Shortest and Longest Time Period

${𝑇}_{𝑏}$ (parallel springs) is the shortest, and ${𝑇}_{𝑐}$ (series springs) is the longest.

Relating the Time Periods

From above, we have:
${𝑇}_{𝑎}=\frac{{𝑇}_{𝑐}}{\sqrt{\surd 2}}$ and ${𝑇}_{𝑎}=\sqrt{\surd 2}{𝑇}_{𝑏}$

We can also write:
${𝑇}_{𝑏}=\frac{{𝑇}_{𝑎}}{\sqrt{\surd 2}}$ and ${𝑇}_{𝑐}=\sqrt{\surd 2}{𝑇}_{𝑎}$

So, ${𝑇}_{𝑐}$ is twice ${𝑇}_{𝑏}$:
${𝑇}_{𝑐}=2{𝑇}_{𝑏}$

Given: $𝑋={𝑎}^2{𝑏}^3{𝑐}^{5/2}{𝑑}^{-2}$

$\%$ Error contributed by $a=2\times (\frac{\Delta a}{a}\times 100)$

$=2\%$

$\%$ Error contributed by $b=3\times (\frac{\Delta b}{b}\times 100)$

$=6\%$

$\%$ Error contributed by $c=\frac{5}{2}\times (\frac{\Delta c}{c}\times 100)$

$=5\%$

$\%$ Error contributed by $d=2\times (\frac{\Delta d}{d}\times 100)$

$=8\%$

$\therefore$ Percentage error in X is given as, $\frac{\Delta X}{X}\times 100=(\%$ error contributed by a)

+(% error contributed by b)

+(% error contributed by c)

$+(\%$ error contributed by d) $)$

$\begin{array}{cc}& =(2+6+5+8)\%\\ & =21\%\end{array}$

1. Identify the given values:

   • Effective Area (A) = $3m^2$

   • Initial Magnetic Field Induction (${𝐵}_1$) = $0.05\mathrm{Wb}/m^2$

   • Final Magnetic Field Induction (${𝐵}_2$) = $20\%$ of ${𝐵}_1=0.20\times 0.05\mathrm{Wb}/m^2=0.01\mathrm{Wb}/m^2$

   • Time taken for the change ($\Delta 𝑡$) = $10s$

   • The coil is placed at right angles to the magnetic field, meaning the angle between the normal to the coil's area and the magnetic field is $0^{\circ }$. So, $\cos 𝜃=\cos 0^{\circ }=1$.

2. Calculate the initial magnetic flux (${\Phi }_1$):

   Magnetic flux ($\Phi$) = $𝐵\times 𝐴\times \cos 𝜃$

   ${\Phi }_1={𝐵}_1\times 𝐴\times 1$

   ${\Phi }_1=0.05\mathrm{Wb}/m^2\times 3m^2$

   ${\Phi }_1=0.15\mathrm{Wb}$

3. Calculate the final magnetic flux (${\Phi }_2$):

   ${\Phi }_2={𝐵}_2\times 𝐴\times 1$

   ${\Phi }_2=0.01\mathrm{Wb}/m^2\times 3m^2$

   ${\Phi }_2=0.03\mathrm{Wb}$

4. Calculate the change in magnetic flux ($\Delta \Phi$):

   $\Delta \Phi ={\Phi }_2-{\Phi }_1$

   $\Delta \Phi =0.03\mathrm{Wb}-0.15\mathrm{Wb}$

   $\Delta \Phi =-0.12\mathrm{Wb}$

5. Apply Faraday's Law of Electromagnetic Induction:

   The induced e.m.f. ($𝜀$) is given by:

   $𝜀=-𝑁\frac{\Delta \Phi }{\Delta 𝑡}$

   Assuming $𝑁=1$ (single turn coil, as not specified otherwise):

   $𝜀=-\frac{\Delta \Phi }{\Delta 𝑡}$

   $𝜀=-\frac{-0.12\mathrm{Wb}}{10s}$

   $𝜀=\frac{0.12}{10}V$

   $𝜀=0.012V$

6. Convert the e.m.f. to millivolts (mV):

   $1V=1000\mathrm{mV}$

   $𝜀=0.012V\times 1000\mathrm{mV}/V$

   $𝜀=12\mathrm{mV}$

The induced e.m.f. in the coil will be $12\mathrm{mV}$.

The final answer is $\overline{)\text{12 mV}}$.

$\overrightarrow{𝐴}=𝑎\widehat{𝑖}+𝑏\widehat{𝑗}+\widehat{𝑘};\overrightarrow{𝐵}=2\widehat{𝑖}-3\widehat{𝑗}+4\widehat{𝑘}$

As the two vectors are perpendicular to eaçh other

$\begin{array}{cc}\therefore & \overrightarrow{𝐴}\cdot \overrightarrow{𝐵}=0\\ \therefore & 2𝑎-3𝑏+4=0\\ & 3𝑎+2𝑏=7\text{ ....(given) }\end{array}$

Solving for a and b , we get

$\begin{array}{cc}& 𝑎=1,𝑏=2\\ \therefore & 𝑎:𝑏=1:2\end{array}$

$\therefore$ value of x is 1

$\begin{array}{cc}& 𝜇=\frac{\text{ Real depth }\left(R\right)}{\text{ Apparent depth }}\Rightarrow \text{ Apparent depth }=\frac{R}{𝜇}\\ & \text{ For vessel, }\\ & \begin{array}{cc}\text{ Apparent depth }& =\frac{R_1}{{𝜇}_1}+\frac{R_2}{{𝜇}_2}+\frac{R_3}{{𝜇}_3}\\ & =\frac{3}{3/2}+\frac{4}{4/3}+\frac{6}{6/5}\\ & =2+3+5\\ \text{ Apparent depth }& =10\mathrm{cm}\end{array}\end{array}$

Step 1: Recall relation between surface energy and surface area

Surface energy:

$𝐸=𝑇\cdot 𝐴,$

where $𝑇$ is the surface tension, $𝐴$ the surface area.

Since $𝑇$ is constant for the liquid, energy is proportional to surface area.

So:

$\frac{𝐸}{𝑈}=\frac{{𝐴}_{\text{original}}}{{𝐴}_{\text{final}}}.$

Step 2: Let original radius be $𝑅$

• Volume of original drop:

  $𝑉=\frac{4}{3}𝜋{𝑅}^3$

• Surface area of original drop:

  ${𝐴}_{\text{original}}=4𝜋{𝑅}^2$

Step 3: Splitting into 729 identical small drops

Total volume is conserved.

If radius of each small drop is $𝑟$, then:

$729\cdot \frac{4}{3}𝜋{𝑟}^3=\frac{4}{3}𝜋{𝑅}^3$

$729{𝑟}^3={𝑅}^3$

$𝑟=\frac{𝑅}{9}.$

Step 4: Surface area of smaller drops

Surface area of one small drop:

${𝐴}_{\text{one}}=4𝜋{𝑟}^2=4𝜋{\left(\frac{𝑅}{9}\right)}^2=\frac{4𝜋{𝑅}^2}{81}.$

Total surface area of 729 drops:

${𝐴}_{\text{final}}=729\cdot \frac{4𝜋{𝑅}^2}{81}.$

Since $729/81=9$,

${𝐴}_{\text{final}}=9\cdot 4𝜋{𝑅}^2.$

Step 5: Energy ratio

$\frac{𝐸}{𝑈}=\frac{4𝜋{𝑅}^2}{9\cdot 4𝜋{𝑅}^2}=\frac{1}{9}.$

So $𝑥=9$.

✅ Final Answer:

$𝑥=9$

Correct Option: A (9)

Step 1: Formula for magnetic field on axis

Magnetic field at a point on the axis of a current-carrying circular coil at distance $𝑥$ from the center is:

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

At the center ($𝑥=0$):

$𝐵\left(0\right)=\frac{{𝜇}_0𝐼}{2𝑅}$

Step 2: Ratio condition

We want:

$𝐵\left(𝑥\right)=\frac{1}{27}𝐵\left(0\right)$

Substitute:

$\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+{𝑥}^2{)}^{3/2}}=\frac{1}{27}\cdot \frac{{𝜇}_0𝐼}{2𝑅}$

Cancel $\frac{{𝜇}_0𝐼}{2}$:

$\frac{{𝑅}^2}{({𝑅}^2+{𝑥}^2{)}^{3/2}}=\frac{1}{27𝑅}$

Step 3: Simplify equation

$\frac{{𝑅}^2}{({𝑅}^2+{𝑥}^2{)}^{3/2}}=\frac{1}{27𝑅}$

Cross multiply:

$27{𝑅}^3=({𝑅}^2+{𝑥}^2{)}^{3/2}$

Step 4: Square both sides

$(27{𝑅}^3{)}^2=({𝑅}^2+{𝑥}^2{)}^3$

$729{𝑅}^6=({𝑅}^2+{𝑥}^2{)}^3$

Take cube root:

$9{𝑅}^2={𝑅}^2+{𝑥}^2$

So:

${𝑥}^2=8{𝑅}^2$

$𝑥=2\sqrt{\surd 2}𝑅$

✅ Final Answer:

$\overline{)2\sqrt{\surd 2}𝑅}$

Correct option: C