MHT CET 2026 20th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

The area of the parallelogram formed by two vectors is the magnitude of their cross product:

$\text{Area}=\mid \overrightarrow{𝑃}\times \overrightarrow{𝑄}\mid$

Given,

$\overrightarrow{𝑃}=2\widehat{𝑖}-\widehat{𝑗}+5\widehat{𝑘},\overrightarrow{𝑄}=3\widehat{𝑖}-2\widehat{𝑗}+4\widehat{𝑘}$

Their cross product is

$\overrightarrow{𝑃}\times \overrightarrow{𝑄}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 2& -1& 5\\ 3& -2& 4\end{array}\mid$

$=6\widehat{𝑖}+7\widehat{𝑗}-\widehat{𝑘}$

Hence,

$\mid \overrightarrow{𝑃}\times \overrightarrow{𝑄}\mid =\sqrt{\surd 6^2+7^2+(-1{)}^2}=\sqrt{\surd 36+49+1}=\sqrt{\surd 86}$

So, the correct answer is Option B: $\sqrt{\surd 86}$.

Momentum is

$𝑝=𝑚𝑣$

For a product, percentage errors are added:

$\frac{\Delta 𝑝}{𝑝}\times 100=0.73\%+1.84\%=2.57\%$

Kinetic energy is

$𝐾=\frac{1}{2}𝑚{𝑣}^2$

The constant $\frac{1}{2}$ does not contribute to error. Since speed has power $2$, its percentage error is multiplied by $2$:

$\frac{\Delta 𝐾}{𝐾}\times 100=0.73\%+2\left(1.84\%\right)$

$=0.73\%+3.68\%=4.41\%$

Thus, the percentage errors in momentum and kinetic energy are respectively

$\overline{)2.57\%,4.41\%}$

Hence, Option B is correct.

To cross directly along $𝐴𝐵$, the boat must cancel the river current.

Let the boat be directed at an angle $𝜃$ upstream from $𝐴𝐵$. Its upstream velocity component must equal the speed of water:

${𝑉}_{𝐵}\sin 𝜃={𝑉}_{𝑊}$

Given ${𝑉}_{𝐵}=2{𝑉}_{𝑊}$,

$\sin 𝜃=\frac{{𝑉}_{𝑊}}{{𝑉}_{𝐵}}=\frac{1}{2}$

Hence,

$𝜃={30}^{\circ }$

The effective velocity of the boat along $𝐴𝐵$ is

${𝑉}_{𝐵}\cos {30}^{\circ }={𝑉}_{𝐵}\left(\frac{\sqrt{\surd 3}}{2}\right)$

Since distance across the river is $𝐷$,

$𝑡=\frac{𝐷}{{𝑉}_{𝐵}\cos {30}^{\circ }}=\frac{𝐷}{{𝑉}_{𝐵}\left(\frac{\sqrt{\surd 3}}{2}\right)}=\frac{2𝐷}{{𝑉}_{𝐵}\sqrt{\surd 3}}$

Thus, the correct answer is

$\overline{)\frac{2𝐷}{{𝑉}_{𝐵}\sqrt{\surd 3}}}$

Option C.

When the lift accelerates upward, the apparent weight is

$𝑚(𝑔+𝑎)=660$

When it accelerates downward with the same acceleration,

$𝑚(𝑔-𝑎)=380$

Adding both equations,

$2𝑚𝑔=660+380=1040$

$𝑚𝑔=520\text{N}$

Hence, the real weight of the man is

$\overline{)520\text{N}}$

So, Option C is correct.

For a solid sphere rolling without slipping,

$𝐼=\frac{2}{5}𝑚{𝑅}^2,𝜔=\frac{{𝑉}_1}{𝑅}$

Using conservation of mechanical energy,

$𝑚𝑔ℎ=\frac{1}{2}𝑚{𝑉}_2^1+\frac{1}{2}𝐼{𝜔}^2$

$𝑚𝑔ℎ=\frac{1}{2}𝑚{𝑉}_2^1+\frac{1}{2}\left(\frac{2}{5}𝑚{𝑅}^2\right){\left(\frac{{𝑉}_1}{𝑅}\right)}^2$

$𝑚𝑔ℎ=\frac{7}{10}𝑚{𝑉}_2^1$

Hence,

${𝑉}_1=\sqrt{\surd \frac{10𝑔ℎ}{7}}$

When the sphere slides without friction, its potential energy changes only into translational kinetic energy:

$𝑚𝑔ℎ=\frac{1}{2}𝑚{𝑉}_2^2$

So,

${𝑉}_2=\sqrt{\surd 2𝑔ℎ}$

Thus,

$\frac{{𝑉}_1}{{𝑉}_2}=\sqrt{\surd \frac{10𝑔ℎ/7}{2𝑔ℎ}}=\sqrt{\surd \frac{5}{7}}$

Therefore,

$\overline{){𝑉}_1:{𝑉}_2=\sqrt{\surd 5}:\sqrt{\surd 7}}$

Hence, Option D is correct.

For a ring of mass $𝑀$ and radius $𝑅$, the moment of inertia about its diameter is

${𝐼}_{𝐴}=\frac{1}{2}𝑀{𝑅}^2$

For ring B, the tangential axis lies in its plane and is parallel to a diameter. Using the parallel axis theorem,

${𝐼}_{𝐵}={𝐼}_{𝐴}+𝑀{𝑅}^2$

${𝐼}_{𝐵}=\frac{1}{2}𝑀{𝑅}^2+𝑀{𝑅}^2=\frac{3}{2}𝑀{𝑅}^2$

Both rings have the same rotational kinetic energy:

$\frac{1}{2}{𝐼}_{𝐴}{𝜔}_2^{𝐴}=\frac{1}{2}{𝐼}_{𝐵}{𝜔}_2^{𝐵}$

So,

$\frac{{𝜔}_{𝐵}}{{𝜔}_{𝐴}}=\sqrt{\surd \frac{{𝐼}_{𝐴}}{{𝐼}_{𝐵}}}=\sqrt{\surd \frac{\frac{1}{2}𝑀{𝑅}^2}{\frac{3}{2}𝑀{𝑅}^2}}=\frac{1}{\sqrt{\surd 3}}$

Hence,

$\overline{){𝜔}_{𝐵}:{𝜔}_{𝐴}=1:\sqrt{\surd 3}}$

So, the correct answer is Option B.

For a rotating body, rotational kinetic energy is

$𝐾=\frac{{𝐿}^2}{2𝐼}$

Here, $𝐾=𝑥$ and $𝐿=𝑦$. So,

$𝑥=\frac{{𝑦}^2}{2𝐼}$

Rearranging,

$𝐼=\frac{{𝑦}^2}{2𝑥}$

Hence, the correct answer is Option D:

$\overline{)\frac{{𝑦}^2}{2𝑥}}$

The weight of a body at the Earth’s surface is

$𝑊=\frac{𝐺𝑀𝑚}{{𝑅}^2}$

At a height $ℎ$, its weight becomes

${𝑊}_{ℎ}=\frac{𝐺𝑀𝑚}{(𝑅+ℎ{)}^2}$

For a small height compared with the Earth’s radius, the fractional decrease in weight is approximately

$\frac{\Delta 𝑊}{𝑊}=\frac{2ℎ}{𝑅}$

Substituting $ℎ=48\text{km}$ and $𝑅=6400\text{km}$,

$\text{Percentage decrease}=\frac{2\times 48}{6400}\times 100=1.5\%$

Hence, the correct answer is

$\overline{)1.5\%}$

Option B.

For a soap bubble, there are two liquid surfaces. The work done in blowing it is equal to the increase in surface energy:

$𝑊=2𝑇\left(4𝜋{𝑟}^2\right)=8𝜋𝑇{𝑟}^2$

Thus, $𝑊$ is proportional to ${𝑟}^2$.

Given that the work required to produce a bubble of radius $𝑟$ is

${𝑊}_{𝑟}=450\text{erg}$

The total work required for radius $3𝑟$ is

${𝑊}_{3𝑟}=450{\left(\frac{3𝑟}{𝑟}\right)}^2=450\times 9=4050\text{erg}$

Hence, the additional work needed to increase the radius from $𝑟$ to $3𝑟$ is

${𝑊}_{\text{additional}}=4050-450=3600\text{erg}$

So, the correct answer is $\overline{)\text{Option C: }3600\text{erg}}$.

For a soap bubble, there are two liquid surfaces. Hence, the excess pressure is

$\Delta 𝑃=\frac{4𝑇}{𝑟}$

Given:

$\Delta 𝑃=48{\text{dyne/cm}}^2,𝑟=2.5\text{cm}$

So,

$𝑇=\frac{\Delta 𝑃𝑟}{4}=\frac{48\times 2.5}{4}=30\text{dyne/cm}$

Thus, the surface tension is

$\overline{)30\text{dyne/cm}}$

Option B.

For liquid flow:

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

• If ${𝑅}_{𝑛}>2000$, the flow is turbulent.

• If $1000<{𝑅}_{𝑛}<2000$, the flow may change between streamline and turbulent flow.

Since

${𝑅}_{𝑛}=1450,$

it lies between $1000$ and $2000$. Hence, the flow may sometimes be turbulent and sometimes streamline.

Correct option: D

According to Stefan’s law of cooling, the rate of loss of heat by radiation is proportional to

${𝑇}^4-{𝑇}_4^0$

where $𝑇$ is the temperature of the sphere and ${𝑇}_0$ is the temperature of the surroundings.

Initially,

$𝑅\propto {600}^4-{200}^4$

When the sphere reaches $400\text{K}$,

${𝑅}^{\prime }\propto {400}^4-{200}^4$

Hence,

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{{400}^4-{200}^4}{{600}^4-{200}^4}$

Taking ${200}^4$ common,

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{2^4-1}{3^4-1}=\frac{16-1}{81-1}=\frac{15}{80}=\frac{3}{16}$

Thus,

$\overline{){𝑅}^{\prime }=\frac{3}{16}𝑅}$

So, the correct answer is Option D.

Let the thermal resistance of each rod be $𝑅$.

In figure (1), the two rods are connected in parallel. So the equivalent thermal resistance is:

${𝑅}_{𝑝}=\frac{𝑅\times 𝑅}{𝑅+𝑅}=\frac{𝑅}{2}$

Heat transferred in time $𝑡$ is given by:

$𝑄=\frac{\Delta 𝑇}{{𝑅}_{𝑝}}𝑡=\frac{2\Delta 𝑇}{𝑅}𝑡$

In figure (2), the two rods are connected in series. So the equivalent thermal resistance becomes:

${𝑅}_{𝑠}=𝑅+𝑅=2𝑅$

Now let the time taken in this case to transfer the same heat $𝑄$ under the same temperature difference $\Delta 𝑇$ be ${𝑡}^{\prime }$.

Then,

$𝑄=\frac{\Delta 𝑇}{{𝑅}_{𝑠}}{𝑡}^{\prime }=\frac{\Delta 𝑇}{2𝑅}{𝑡}^{\prime }$

Since the amount of heat transferred is the same in both cases, we equate the two expressions of $𝑄$:

$\frac{2\Delta 𝑇}{𝑅}𝑡=\frac{\Delta 𝑇}{2𝑅}{𝑡}^{\prime }$

Cancel $\Delta 𝑇$ and $𝑅$ from both sides:

$2𝑡=\frac{{𝑡}^{\prime }}{2}$

So,

${𝑡}^{\prime }=4𝑡$

For an adiabatic process,

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

So,

${𝑇}_{𝑖}{𝑉}_{𝛾-1}^{𝑖}={𝑇}_{𝑓}{𝑉}_{𝛾-1}^{𝑓}$

Given

$𝛾=\frac{7}{5},{𝑉}_{𝑓}=\frac{{𝑉}_{𝑖}}{32}$

Thus,

$\frac{{𝑇}_{𝑓}}{{𝑇}_{𝑖}}={\left(\frac{{𝑉}_{𝑖}}{{𝑉}_{𝑓}}\right)}^{𝛾-1}={\left(32\right)}^{\frac{7}{5}-1}={32}^{\frac{2}{5}}$

Since $32=2^5$,

$\frac{{𝑇}_{𝑓}}{{𝑇}_{𝑖}}=(2^5{)}^{2/5}=2^2=4$

Hence,

${𝑇}_{𝑓}=4{𝑇}_{𝑖}$

So, $\overline{)𝑥=4}$, which is Option C.

For an ideal gas undergoing an adiabatic change, the relation between temperature and volume is

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

Comparing it with

$𝑇{𝑉}^{𝑥}=\text{constant},$

we get

$𝑥=𝛾-1.$

Given $𝛾=\frac{7}{5}$,

$𝑥=\frac{7}{5}-1=\frac{2}{5}=0.40.$

Hence, the correct answer is Option C: $0.40$.

Using the ideal gas equation,

$𝑃𝑉=𝑁{𝑘}_B𝑇$

where $𝑁$ is the number of molecules.

For jar A,

${𝑁}_{𝐴}=\frac{𝑃𝑉}{{𝑘}_B𝑇}$

For jar B, pressure is $2𝑃$, volume is $\frac{𝑉}{4}$, and temperature is $2𝑇$:

${𝑁}_{𝐵}=\frac{\left(2𝑃\right)\left(\frac{𝑉}{4}\right)}{{𝑘}_B\left(2𝑇\right)}=\frac{𝑃𝑉}{4{𝑘}_B𝑇}$

Thus,

$\frac{{𝑁}_{𝐵}}{{𝑁}_{𝐴}}=\frac{𝑃𝑉/\left(4{𝑘}_B𝑇\right)}{𝑃𝑉/\left({𝑘}_B𝑇\right)}=\frac{1}{4}$

Hence, the required ratio is

$\overline{){𝑁}_{𝐵}:{𝑁}_{𝐴}=1:4}$

So, Option B is correct.

For a simple pendulum,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

Since both pendulums are at the same place, $𝑔$ is constant. Hence,

$𝑙\propto {𝑇}^2$

For the pendulum of length $({𝑙}_1-{𝑙}_2)$, let its time period be $𝑇$. Then,

${𝑇}^2={𝑇}_2^1-{𝑇}_2^2$

Substituting ${𝑇}_1=2.4\text{s}$ and ${𝑇}_2=1.8\text{s}$:

$𝑇=\sqrt{\surd (2.4{)}^2-(1.8{)}^2}$

$𝑇=\sqrt{\surd 5.76-3.24}=\sqrt{\surd 2.52}\approx 1.6\text{s}$

Hence, the correct answer is Option C: $1.6\text{s}$.

When the mass $𝑚$ is displaced by a small distance $𝑥$ from its equilibrium position, one spring is stretched by $𝑥$ and the other spring is compressed by $𝑥$.

In both cases, the spring force acts towards the equilibrium position. So both springs provide restoring force in the same direction, opposite to the displacement.

The restoring force due to the spring of force constant $2𝑘$ is $-2𝑘𝑥$, and the restoring force due to the spring of force constant $𝑘$ is $-𝑘𝑥$.

So, the total restoring force on the mass is:

$𝐹=-2𝑘𝑥-𝑘𝑥=-3𝑘𝑥$

This matches the form $𝐹=-{𝑘}_{𝑒𝑞}𝑥$. So the equivalent spring constant is:

${𝑘}_{𝑒𝑞}=3𝑘$

For a mass-spring system, the frequency of oscillation is:

$𝑓=\frac{1}{2𝜋}\sqrt{\surd \frac{{𝑘}_{𝑒𝑞}}{𝑚}}=\frac{1}{2𝜋}\sqrt{\surd \frac{3𝑘}{𝑚}}$

The pendulums are initially in phase.

Their frequencies are

${𝑓}_1=\frac{1}{𝑇},{𝑓}_2=\frac{1}{4𝑇/3}=\frac{3}{4𝑇}$

They will again be in phase when their phase difference becomes $2𝜋$. Thus,

$2𝜋({𝑓}_1-{𝑓}_2)𝑡=2𝜋$

$(\frac{1}{𝑇}-\frac{3}{4𝑇})𝑡=1$

$\frac{𝑡}{4𝑇}=1$

Hence,

$𝑡=4𝑇$

In this time, the first pendulum completes $4$ oscillations and the second completes $3$ oscillations.

$\overline{)\text{Option B: }4𝑇}$

The fundamental frequency of a stretched wire is

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

where $𝜇$ is the mass per unit length:

$𝜇=𝜌𝐴=𝜌\frac{𝜋{𝑑}^2}{4}$

For the second wire:

• Length: ${𝐿}_{𝐵}=2𝐿$

• Tension: ${𝑇}_{𝐵}=2𝑇$

• Density: ${𝜌}_{𝐵}=2𝜌$

• Diameter: ${𝑑}_{𝐵}=3𝑑$

Since area is proportional to the square of diameter,

$\frac{{𝜇}_{𝐵}}{{𝜇}_{𝐴}}=\frac{\left(2𝜌\right)(3𝑑{)}^2}{𝜌{𝑑}^2}=18$

Thus,

$\frac{{𝑛}_{𝐵}}{{𝑛}_{𝐴}}=\frac{𝐿}{2𝐿}\sqrt{\surd \frac{2𝑇}{𝑇}\cdot \frac{{𝜇}_{𝐴}}{{𝜇}_{𝐵}}}$

$\frac{{𝑛}_{𝐵}}{{𝑛}_{𝐴}}=\frac{1}{2}\sqrt{\surd \frac{2}{18}}=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

Hence,

$\overline{){𝑛}_{𝐵}:{𝑛}_{𝐴}=1:6}$

So, Option C is correct.

Option A: intensity of sound and frequency of sound.

• Loudness is the human perception of the intensity of sound. Greater intensity makes a sound seem louder.

• Pitch is the human perception of the frequency of sound. A higher frequency produces a higher-pitched sound.

The tube acts as a closed organ pipe because the water surface forms the closed end, while the other end is open.

For the two shortest resonant lengths,

${𝑙}_1=\frac{𝜆}{4}$

and

${𝑙}_2=\frac{3𝜆}{4}$

Subtracting,

${𝑙}_2-{𝑙}_1=\frac{3𝜆}{4}-\frac{𝜆}{4}=\frac{𝜆}{2}$

So,

$𝜆=2({𝑙}_2-{𝑙}_1)$

Since the speed of sound is $𝑣=𝑛𝜆$,

$𝑣=2𝑛({𝑙}_2-{𝑙}_1)$

Hence, the correct answer is

$\overline{)2𝑛({𝑙}_2-{𝑙}_1)}$

Option A.

The speed of sound in a gas is

$𝑣=\sqrt{\surd \frac{𝛾𝑅𝑇}{𝑀}}$

Since helium and nitrogen are at the same temperature,

$\frac{{𝑣}_{\text{He}}}{{𝑣}_{{\text{N}}_2}}=\sqrt{\surd \frac{{𝛾}_{\text{He}}/{𝑀}_{\text{He}}}{{𝛾}_{{\text{N}}_2}/{𝑀}_{{\text{N}}_2}}}$

Substituting the given values,

$\frac{{𝑣}_{\text{He}}}{{𝑣}_{{\text{N}}_2}}=\sqrt{\surd \frac{\frac{5}{3}/4}{\frac{7}{5}/28}}$

$=\sqrt{\surd \frac{5}{12}\times \frac{28\times 5}{7}}=\sqrt{\surd \frac{25}{3}}=\frac{5}{\sqrt{\surd 3}}$

Hence, the correct answer is

$\overline{)\frac{5}{\sqrt{\surd 3}}}$

Option D.

For a source moving towards a stationary observer, the apparent frequency is

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

Here,

$𝑓=450\text{Hz},𝑣=340\text{m/s},{𝑣}_{𝑠}=34\text{m/s}$

Substituting,

${𝑓}^{\prime }=450\left(\frac{340}{340-34}\right)$

${𝑓}^{\prime }=450\left(\frac{340}{306}\right)=450\times \frac{10}{9}=500\text{Hz}$

The frequency heard by the observer is $\overline{)500\text{Hz}}$.

Option C.

The two negative charges are at adjacent corners of the square. Their midpoint lies on the side joining them.

Since the side of the square is $2𝑟$, the distance of this midpoint from each negative charge is

$𝑟.$

The distance from the midpoint to each positive charge is

$\sqrt{\surd (2𝑟{)}^2+{𝑟}^2}=𝑟\sqrt{\surd 5}.$

Electric potential is a scalar quantity, so the potentials due to all four charges are added algebraically:

$𝑉=\frac{1}{4𝜋{𝜖}_0}[\frac{𝑞}{𝑟\sqrt{\surd 5}}+\frac{𝑞}{𝑟\sqrt{\surd 5}}-\frac{𝑞}{𝑟}-\frac{𝑞}{𝑟}].$

Thus,

$𝑉=\frac{1}{4𝜋{𝜖}_0}\frac{2𝑞}{𝑟}(\frac{1}{\sqrt{\surd 5}}-1).$

Hence, the correct answer is

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

Using Coulomb’s law, the initial force is

$𝐹=𝑘\frac{\left|{𝑞}_1{𝑞}_2\right|}{{𝑙}^2}$

One charge is doubled and the distance is halved. The new force is

${𝐹}^{\prime }=𝑘\frac{\left|\right(2{𝑞}_1\left){𝑞}_2\right|}{{\left(\frac{𝑙}{2}\right)}^2}$

Since ${\left(\frac{𝑙}{2}\right)}^2=\frac{{𝑙}^2}{4}$,

${𝐹}^{\prime }=𝑘\frac{2\left|{𝑞}_1{𝑞}_2\right|}{{𝑙}^2/4}=8𝑘\frac{\left|{𝑞}_1{𝑞}_2\right|}{{𝑙}^2}=8𝐹$

Thus, the force becomes $8$ times.

$\overline{)𝑛=8}$

So, Option C is correct.

For the air capacitor,

${𝐶}_0=\frac{{𝜀}_0𝐴}{𝑑}$

After inserting a slab of thickness $𝑡$ and dielectric constant $𝐾$, the remaining air gap is $(𝑑-𝑡)$. The effective separation becomes

$(𝑑-𝑡)+\frac{𝑡}{𝐾}$

Hence,

$𝐶=\frac{{𝜀}_0𝐴}{𝑑-𝑡+\frac{𝑡}{𝐾}}$

Given that the new capacitance is one-third of the original capacitance,

$\frac{{𝜀}_0𝐴}{𝑑-𝑡+\frac{𝑡}{𝐾}}=\frac{1}{3}\frac{{𝜀}_0𝐴}{𝑑}$

So,

$𝑑-𝑡+\frac{𝑡}{𝐾}=3𝑑$

$\frac{𝑡}{𝐾}=2𝑑+𝑡$

Thus,

$𝐾=\frac{𝑡}{2𝑑+𝑡}$

Hence, the expected answer is

$\overline{)\text{Option B: }\frac{𝑡}{2𝑑+𝑡}}$

For a parallel plate air capacitor, the energy density of the electric field is

$𝑢=\frac{1}{2}{𝜖}_0{𝐸}^2$

The volume of the space between the plates is

$𝑉=𝐴𝑑$

Hence, the total energy stored is

$𝑈=𝑢𝑉=\frac{1}{2}{𝜖}_0{𝐸}^2\left(𝐴𝑑\right)=\frac{1}{2}𝐴{𝜖}_0{𝐸}^2𝑑$

So, the correct answer is

$\overline{)\text{Option C: }\frac{1}{2}𝐴{𝜖}_0{𝐸}^2𝑑}$

For a balanced meter bridge, the ratio of resistances in the two gaps is equal to the ratio of the corresponding wire lengths.

Initially,

$\frac{𝑋}{𝑌}=\frac{2}{3}$

So, take

$𝑋=2𝑘,𝑌=3𝑘$

After increasing each resistance by $30\Omega$,

$\frac{𝑋+30}{𝑌+30}=\frac{5}{6}$

Substituting $𝑋=2𝑘$ and $𝑌=3𝑘$:

$\frac{2𝑘+30}{3𝑘+30}=\frac{5}{6}$

$6(2𝑘+30)=5(3𝑘+30)$

$12𝑘+180=15𝑘+150$

$𝑘=10$

Hence,

$𝑋=2𝑘=20\Omega ,𝑌=3𝑘=30\Omega$

So, the correct choice is Option A: $20\Omega ,30\Omega$.

From the circuit, the potential difference between points A and B is only due to the cell ${𝐸}_1$. So,

${𝑉}_{𝐴𝐵}={𝐸}_1$

For a potentiometer, potential difference is proportional to balancing length. If $𝑘$ is the potential gradient, then for balancing length $412\text{ cm}$,

${𝐸}_1=412𝑘$

The two cells are connected in opposition, so between points A and C, the effective potential difference is the difference of their emfs.

${𝑉}_{𝐴𝐶}={𝐸}_1-{𝐸}_2$

For balancing length $103\text{ cm}$,

${𝐸}_1-{𝐸}_2=103𝑘$

Now divide the two equations:

$\frac{{𝐸}_1}{{𝐸}_1-{𝐸}_2}=\frac{412}{103}=4$

So,

${𝐸}_1=4({𝐸}_1-{𝐸}_2)$

${𝐸}_1=4{𝐸}_1-4{𝐸}_2$

$3{𝐸}_1=4{𝐸}_2$

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{4}{3}$

So the ratio is

${𝐸}_1:{𝐸}_2=4:3$

For a magnet vibrating in a uniform magnetic field, the time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝐼}{𝑀𝐵}}$

where $𝐼$ is the moment of inertia and $𝑀$ is the magnetic moment.

For the new magnet,

${𝐼}^{\prime }=3𝐼,{𝑀}^{\prime }=\frac{𝑀}{3}$

Thus,

$\frac{{𝑇}^{\prime }}{{𝑇}_0}=\sqrt{\surd \frac{{𝐼}^{\prime }/{𝑀}^{\prime }}{𝐼/𝑀}}=\sqrt{\surd \frac{3𝐼}{𝑀/3}\times \frac{𝑀}{𝐼}}=\sqrt{\surd 9}=3$

Hence,

${𝑇}^{\prime }=3{𝑇}_0$

So, the correct answer is Option C: $3{𝑇}_0$.

The magnetic field made by the long straight wire $𝑋𝑌$ at a distance $𝑟$ is

$𝐵=\frac{{𝜇}_0{𝐼}_2}{2𝜋𝑟}$

This field is directed into the plane of the paper at the position of the loop.

Side $𝐴𝐵$ is nearer to the straight wire. Its distance from the wire is $𝑟=\frac{𝐿}{3}$. The current in $𝐴𝐵$ is in the same direction as the current in $𝑋𝑌$, so the force is attractive.

${𝐹}_{𝐴𝐵}={𝐼}_1𝐿{𝐵}_{𝐴𝐵}={𝐼}_1𝐿\left(\frac{{𝜇}_0{𝐼}_2}{2𝜋(𝐿/3)}\right)=\frac{3{𝜇}_0{𝐼}_1{𝐼}_2}{2𝜋}$

Side $𝐶𝐷$ is farther from the straight wire. Its distance is

$𝑟=\frac{𝐿}{3}+𝐿=\frac{4𝐿}{3}$

The current in $𝐶𝐷$ is opposite to the current in $𝑋𝑌$, so the force is repulsive.

${𝐹}_{𝐶𝐷}={𝐼}_1𝐿{𝐵}_{𝐶𝐷}={𝐼}_1𝐿\left(\frac{{𝜇}_0{𝐼}_2}{2𝜋(4𝐿/3)}\right)=\frac{3{𝜇}_0{𝐼}_1{𝐼}_2}{8𝜋}$

The forces on sides $𝐵𝐶$ and $𝐷𝐴$ are equal and opposite, so they cancel each other.

So the net force on the loop is only because of sides $𝐴𝐵$ and $𝐶𝐷$.

${𝐹}_{𝑛𝑒𝑡}={𝐹}_{𝐴𝐵}-{𝐹}_{𝐶𝐷}=\frac{3{𝜇}_0{𝐼}_1{𝐼}_2}{2𝜋}-\frac{3{𝜇}_0{𝐼}_1{𝐼}_2}{8𝜋}$

${𝐹}_{𝑛𝑒𝑡}=\frac{12{𝜇}_0{𝐼}_1{𝐼}_2-3{𝜇}_0{𝐼}_1{𝐼}_2}{8𝜋}=\frac{9{𝜇}_0{𝐼}_1{𝐼}_2}{8𝜋}$

Answer: $\frac{9{𝜇}_0{𝐼}_1{𝐼}_2}{8𝜋}$

For a long solenoid, the magnetic field along its axis is

$𝐵={𝜇}_0𝑛𝐼$

where $𝑛$ is the number of turns per unit length and $𝐼$ is the current.

The current is tripled:

${𝐼}^{\prime }=3𝐼$

The number of turns per cm is halved:

${𝑛}^{\prime }=\frac{𝑛}{2}$

So, the new magnetic field is

${𝐵}^{\prime }={𝜇}_0{𝑛}^{\prime }{𝐼}^{\prime }={𝜇}_0\left(\frac{𝑛}{2}\right)\left(3𝐼\right)=\frac{3}{2}{𝜇}_0𝑛𝐼$

Since ${𝜇}_0𝑛𝐼=𝐵$,

$\overline{){𝐵}^{\prime }=\frac{3}{2}𝐵}$

Hence, Option C is correct.

For a solenoid, self-inductance is directly proportional to the permeability of the core:

$𝐿=\frac{{𝜇}_0{𝜇}_{𝑟}{𝑁}^2𝐴}{𝑙}$

For an air-core solenoid, ${𝜇}_{𝑟}\approx 1$. Hence,

${𝜇}_{𝑟}=\frac{{𝐿}_{\text{iron}}}{{𝐿}_{\text{air}}}$

Substituting the values:

${𝜇}_{𝑟}=\frac{30\text{mH}}{0.03\text{mH}}=1000={10}^3$

Thus, the relative permeability of the iron core is

$\overline{){10}^3}$

So, Option D is correct.

The effective area of the coil is $3{\text{m}}^2$. Since the coil is placed at right angles to the magnetic field, the magnetic flux is maximum.

Initial magnetic field:

${𝐵}_{𝑖}=0.05{\text{Wb m}}^{-2}$

Final magnetic field is $20\%$ of the initial value:

${𝐵}_{𝑓}=\frac{20}{100}\times 0.05=0.01{\text{Wb m}}^{-2}$

Change in magnetic field:

$\Delta 𝐵=0.05-0.01=0.04{\text{Wb m}}^{-2}$

Using Faraday’s law, the magnitude of induced e.m.f. is

$E=\frac{{𝐴}_{\text{eff}}\Delta 𝐵}{\Delta 𝑡}$

$E=\frac{3\times 0.04}{10}=0.012\text{V}$

Since $1\text{V}=1000\text{mV}$,

$E=12\text{mV}$

Hence, the correct answer is Option B: $12\text{mV}$.

At resonance in a series LCR circuit,

${𝑋}_{𝐿}={𝑋}_{𝐶}=\frac{1}{𝜔𝐶}$