MHT CET 2026 19th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Given

$𝑉=𝑙\times 𝑙\times \sqrt{\surd 𝑙}={𝑙}^{5/2}$

For a quantity raised to a power, the relative error is multiplied by that power:

$\frac{\Delta 𝑉}{𝑉}=\frac{5}{2}\frac{\Delta 𝑙}{𝑙}$

Hence, the percentage error in $𝑙$ is

$\frac{\Delta 𝑙}{𝑙}\times 100=\frac{2}{5}\times 6.25\%=2.5\%$

The least count of the meter scale is

$\Delta 𝑙=1\text{mm}=0.1\text{cm}$

Now,

$\frac{0.1}{𝑙}\times 100=2.5$

So,

$𝑙=\frac{0.1\times 100}{2.5}=4\text{cm}$

Thus, the relative percentage error in length is $2.5\%$ and the length is $4\text{cm}$.

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

In half the period of revolution, the particle completes half a circle and reaches the diametrically opposite point.

The displacement is the straight-line distance between the initial and final positions, which is the diameter:

$\text{Displacement}=2𝑅$

The distance covered is the length of the semicircular path:

$\text{Distance}=\frac{1}{2}\left(2𝜋𝑅\right)=𝜋𝑅$

Hence, the correct answer is Option B: $2𝑅,𝜋𝑅$.

The standard equation of the trajectory of a projectile is

$𝑦=𝑥\tan 𝜃-\frac{𝑔{𝑥}^2}{2{𝑢}^2{\cos }^2𝜃}$

The given equation is

$𝑦=𝑥-\frac{𝑔{𝑥}^2}{2}$

Comparing the coefficient of $𝑥$,

$\tan 𝜃=1$

Hence,

$𝜃={45}^{\circ }$

Now compare the coefficients of ${𝑥}^2$:

$\frac{𝑔}{2{𝑢}^2{\cos }^2𝜃}=\frac{𝑔}{2}$

So,

${𝑢}^2{\cos }^2𝜃=1$

Using $\cos {45}^{\circ }=\frac{1}{\sqrt{\surd 2}}$,

${𝑢}^2{\left(\frac{1}{\sqrt{\surd 2}}\right)}^2=1$

$\frac{{𝑢}^2}{2}=1$

$𝑢=\sqrt{\surd 2}\text{m/s}$

Hence, the correct answer is Option C: $\sqrt{\surd 2}\text{m/s}$.

Given:

$𝑚=40\text{g}=0.04\text{kg},𝑟=50\text{cm}=0.5\text{m}$

The frequency is

$𝑓=\frac{30}{60}=0.5\text{revolutions per second}$

Hence, the angular speed is

$𝜔=2𝜋𝑓=2𝜋\left(0.5\right)=𝜋\text{rad/s}$

At the lowest point, tension acts upwards while weight acts downwards. Their resultant provides the centripetal force:

$𝑇-𝑚𝑔=𝑚{𝜔}^2𝑟$

Thus,

$𝑇=𝑚𝑔+𝑚{𝜔}^2𝑟$

$𝑇=\left(0.04\right)\left(10\right)+\left(0.04\right)\left({𝜋}^2\right)\left(0.5\right)$

Using ${𝜋}^2=10$,

$𝑇=0.4+0.2=0.6\text{N}$

Hence, the correct answer is Option C: $\overline{)0.6\text{N}}$.

For a body just completing a vertical circle, the tension at the highest point is zero.

At the top,

$\frac{𝑚{𝑣}_2^{𝑡}}{𝑟}=𝑚𝑔$

So,

${𝑣}_2^{𝑡}=𝑔𝑟$

Using conservation of mechanical energy between the top and the lowest point,

$\frac{1}{2}𝑚{𝑣}_2^{𝑏}=\frac{1}{2}𝑚{𝑣}_2^{𝑡}+𝑚𝑔\left(2𝑟\right)$

Substituting ${𝑣}_2^{𝑡}=𝑔𝑟$,

${𝑣}_2^{𝑏}=5𝑔𝑟$

At the lowest point, the tension $𝑇$ acts upwards and weight $𝑚𝑔$ acts downwards:

$𝑇-𝑚𝑔=\frac{𝑚{𝑣}_2^{𝑏}}{𝑟}$

$𝑇-𝑚𝑔=5𝑚𝑔$

Hence,

$𝑇=6𝑚𝑔$

The apparent weight is equal to the tension in the string.

$\overline{)6𝑚𝑔}$

So, Option A is correct.

Take point A as the origin.

The $2\text{g}$ mass is at ${𝑥}_1=0$ and the $4\text{g}$ mass is at ${𝑥}_2=9\text{cm}$.

The position of the centre of mass is

${𝑥}_{\text{CM}}=\frac{{𝑚}_1{𝑥}_1+{𝑚}_2{𝑥}_2}{{𝑚}_1+{𝑚}_2}$

${𝑥}_{\text{CM}}=\frac{2\left(0\right)+4\left(9\right)}{2+4}=\frac{36}{6}=6\text{cm}$

Thus, the centre of mass is $6\text{cm}$ from A.

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

Torque is given by the cross product of the position vector and force:

$\overrightarrow{𝜏}=\overrightarrow{𝑟}\times \overrightarrow{𝐹}$

Here,

$\overrightarrow{𝑟}=\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘},\overrightarrow{𝐹}=3\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}$

So,

$\overrightarrow{𝜏}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& 1& -1\\ 3& 2& -1\end{array}\mid$

$\overrightarrow{𝜏}=\widehat{𝑖}\left(1\right)-\widehat{𝑗}\left(2\right)+\widehat{𝑘}(-1)=\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘}$

Its magnitude is

$\left|\overrightarrow{𝜏}\right|=\sqrt{\surd 1^2+(-2{)}^2+(-1{)}^2}=\sqrt{\surd 6}\text{N m}$

Hence, the correct answer is Option C: $\sqrt{\surd 6}\text{N m}$.

For a satellite revolving close to the planet’s surface, the orbital radius is approximately $𝑅$.

The time period of a satellite in a circular orbit is

$𝑇=2𝜋\sqrt{\surd \frac{{𝑅}^3}{𝐺𝑀}}$

If $𝜌$ is the mean density of the planet, its mass is

$𝑀=\frac{4}{3}𝜋{𝑅}^3𝜌$

Substituting this value of $𝑀$,

$𝑇=2𝜋\sqrt{\surd \frac{{𝑅}^3}{𝐺\left(\frac{4}{3}𝜋{𝑅}^3𝜌\right)}}$

The factor ${𝑅}^3$ cancels:

$𝑇=2𝜋\sqrt{\surd \frac{3}{4𝜋𝐺𝜌}}=\sqrt{\surd \frac{3𝜋}{𝜌𝐺}}$

Hence, the correct answer is

$\overline{)\text{Option D: }\sqrt{\surd \frac{3𝜋}{𝜌𝐺}}}$

The wire is supported by surface tension acting along its two sides.

For the maximum possible upward force, the surface tension on each side acts vertically upward.

Hence, total upward force is

$𝐹=2𝑇𝐿$

If the radius of the wire is $𝑟$, its volume is

$𝑉=𝜋{𝑟}^2𝐿$

So, the weight of the wire is

$𝑊=𝑉𝑑𝑔=𝜋{𝑟}^2𝐿𝑑𝑔$

At the limiting condition, when the wire is just about to sink,

$𝜋{𝑟}^2𝐿𝑑𝑔=2𝑇𝐿$

Cancelling $𝐿$,

${𝑟}^2=\frac{2𝑇}{𝜋𝑑𝑔}$

Thus, the maximum radius is

$\overline{)𝑟=\sqrt{\surd \frac{2𝑇}{𝜋𝑑𝑔}}}$

Hence, Option B is correct.

The volume of the liquid remains unchanged when the large drop breaks into smaller droplets.

Hence,

$\frac{4}{3}𝜋{𝑅}^3=𝑛\left(\frac{4}{3}𝜋{𝑟}^3\right)$

So,

$𝑛=\frac{{𝑅}^3}{{𝑟}^3}$

Initial surface area of the large drop is

${𝐴}_{𝑖}=4𝜋{𝑅}^2$

Final surface area of $𝑛$ small droplets is

${𝐴}_{𝑓}=𝑛\left(4𝜋{𝑟}^2\right)=4𝜋{𝑟}^2\left(\frac{{𝑅}^3}{{𝑟}^3}\right)=\frac{4𝜋{𝑅}^3}{𝑟}$

The energy needed equals surface tension multiplied by the increase in surface area:

$𝐸=𝑇({𝐴}_{𝑓}-{𝐴}_{𝑖})$

$𝐸=𝑇(\frac{4𝜋{𝑅}^3}{𝑟}-4𝜋{𝑅}^2)$

$\overline{)𝐸=4𝜋𝑇{𝑅}^2(\frac{𝑅}{𝑟}-1)}$

Hence, the correct answer is Option A.

Option A: decreases

As temperature rises, the kinetic energy of liquid molecules increases. This weakens the effective cohesive forces between molecules at the surface. Hence, the surface tension of most liquids decreases with an increase in temperature.

A black body emits radiation of all wavelengths, but the intensity is not the same at every wavelength. At a fixed temperature, intensity varies with wavelength and reaches a maximum at a particular wavelength.

Hence, the statement in Option B is not true.

The water and vessel together have an effective water mass of $60\text{g}$.

First, check whether the ice melts completely.

Heat released by the water and vessel on cooling from ${40}^{\circ }\text{C}$ to $0^{\circ }\text{C}$ is

$𝑄=60\times 1\times 40=2400\text{cal}$

Heat needed to melt $15\text{g}$ of ice is

${𝑄}_{\text{melt}}=15\times 80=1200\text{cal}$

Since $2400>1200$, all the ice melts and the final temperature is above $0^{\circ }\text{C}$.

Let the final temperature be $𝑇$.

Heat lost by the water and vessel:

$60(40-𝑇)$

Heat gained by the ice in melting and then warming to $𝑇$:

$15\times 80+15\times 1\times 𝑇$

Using heat lost $=$ heat gained,

$60(40-𝑇)=1200+15𝑇$

$2400-60𝑇=1200+15𝑇$

$1200=75𝑇$

$𝑇={16}^{\circ }\text{C}$

Hence, the correct answer is Option C: $\overline{){16}^{\circ }\text{C}}$.

Using the first law of thermodynamics,

$𝑑𝑄=𝑑𝑈+𝑑𝑊$

Given $𝑑𝑊=0$ and $𝑑𝑄<0$,

$𝑑𝑈=𝑑𝑄<0$

For an ideal gas, internal energy depends only on temperature. Since internal energy decreases, the temperature must decrease.

Also, $𝑑𝑊=𝑃𝑑𝑉=0$. Since $𝑃\ne 0$, $𝑑𝑉=0$, so the volume remains constant. At constant volume, $𝑃\propto 𝑇$, so pressure also decreases.

Hence, the correct answer is:

$\overline{)\text{Option C: }𝑇\text{ will decrease}}$

Using the ideal gas equation,

$𝑃𝑉=𝑛𝑅𝑇$

Initially, the number of moles in the vessel is

${𝑛}_{𝑖}=\frac{𝑃𝑉}{𝑅𝑇}$

After leakage, the volume and temperature remain unchanged, while the pressure becomes ${𝑃}^1$. So, the number of moles left is

${𝑛}_{𝑓}=\frac{{𝑃}^1𝑉}{𝑅𝑇}$

Hence, the number of moles that leaked is

${𝑛}_{\text{leaked}}={𝑛}_{𝑖}-{𝑛}_{𝑓}$

${𝑛}_{\text{leaked}}=\frac{𝑃𝑉}{𝑅𝑇}-\frac{{𝑃}^1𝑉}{𝑅𝑇}=\frac{𝑉}{𝑅𝑇}(𝑃-{𝑃}^1)$

Thus, the correct answer is Option B:

$\overline{)\frac{𝑉}{𝑅𝑇}(𝑃-{𝑃}^1)}$

From the kinetic theory of gases,

$𝑃=\frac{1}{3}𝜌\overline{{𝑣}^2}$

The kinetic energy per unit volume of the gas is

$𝑢=\frac{1}{2}𝜌\overline{{𝑣}^2}$

Comparing these expressions,

$𝑃=\frac{2}{3}𝑢$

Thus, pressure is $\frac{2}{3}$ of the kinetic energy per unit volume of the gas.

Correct option: A

For an ideal gas, the root mean square velocity is

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

Thus, ${𝑣}_{\text{rms}}\propto \sqrt{\surd 𝑇}$.

If the rms velocity becomes half,

$\frac{{𝑣}_2}{{𝑣}_1}=\frac{1}{2}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

Squaring,

$\frac{{𝑇}_2}{{𝑇}_1}=\frac{1}{4}$

For an adiabatic process,

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

So,

${𝑇}_1{𝑉}_{𝛾-1}^1={𝑇}_2{𝑉}_{𝛾-1}^2$

Given $𝛾=1.5$, we have $𝛾-1=0.5$. Hence,

$\frac{{𝑇}_1}{{𝑇}_2}={\left(\frac{{𝑉}_2}{{𝑉}_1}\right)}^{0.5}$

$4={\left(\frac{{𝑉}_2}{{𝑉}_1}\right)}^{1/2}$

Squaring both sides,

$\frac{{𝑉}_2}{{𝑉}_1}=16$

The gas should be expanded to $16$ times its initial volume.

$\overline{)\text{Option B: }16\text{ times}}$

At the equilibrium position, the speed of the mass is maximum. For a spring–mass system,

${𝑣}_{max}=𝜔𝐴$

where

$𝜔=\sqrt{\surd \frac{𝑘}{𝑚}}$

Given:

$𝑘=180\text{N/m},𝑚=0.2\text{kg},𝐴=4\text{cm}=0.04\text{m}$

So,

$𝜔=\sqrt{\surd \frac{180}{0.2}}=\sqrt{\surd 900}=30\text{rad/s}$

Hence,

${𝑣}_{max}=30\times 0.04=1.2\text{m/s}$

Thus, the velocity at the equilibrium position is

$\overline{)1.2\text{m/s}}$

Option C.

For a particle performing linear S.H.M., the velocity at displacement $𝑥$ from the mean position is

${𝑣}^2={𝜔}^2({𝐴}^2-{𝑥}^2)$

At distances ${𝑥}_1$ and ${𝑥}_2$:

${𝑉}_2^1={𝜔}^2({𝐴}^2-{𝑥}_2^1)$

${𝑉}_2^2={𝜔}^2({𝐴}^2-{𝑥}_2^2)$

Subtracting the first equation from the second:

${𝑉}_2^2-{𝑉}_2^1={𝜔}^2({𝑥}_2^1-{𝑥}_2^2)$

Hence,

${𝜔}^2=\frac{{𝑉}_2^2-{𝑉}_2^1}{{𝑥}_2^1-{𝑥}_2^2}$

So,

$\overline{)𝜔=\sqrt{\surd \frac{{𝑉}_2^2-{𝑉}_2^1}{{𝑥}_2^1-{𝑥}_2^2}}}$

Thus, the correct answer is Option A.

For a resonance tube closed at one end, the first resonance condition is

${𝐿}_1+𝑒=\frac{𝜆}{4}$

Here,

${𝐿}_1=18\text{ cm},𝑒=1.1\text{ cm}$

So,

$\frac{𝜆}{4}=18+1.1=19.1\text{ cm}$

For the next resonance,

${𝐿}_2+𝑒=\frac{3𝜆}{4}=3\left(19.1\right)$

Thus,

${𝐿}_2=57.3-1.1=56.2\text{ cm}$

Hence, the next resonating length is

$\overline{)56.2\text{ cm}}$

Option D.

For the stationary wave,

$𝑦=12\cos \left(\frac{𝜋𝑥}{10}\right)\sin \left(36𝜋𝑡\right)$

Comparing the position part with $\cos \left(𝑘𝑥\right)$,

$𝑘=\frac{𝜋}{10}{\text{cm}}^{-1}$

Using $𝑘=\frac{2𝜋}{𝜆}$,

$𝜆=\frac{2𝜋}{𝑘}=\frac{2𝜋}{𝜋/10}=20\text{cm}$

The distance between a node and the successive antinode in a stationary wave is $\frac{𝜆}{4}$.

$\frac{𝜆}{4}=\frac{20}{4}=5\text{cm}$

Hence, the correct answer is Option D: $5\text{cm}$.

When two waves of the same frequency and speed travel in opposite directions, they form a stationary wave.

The distance between two consecutive antinodes is half the wavelength:

$𝑑=\frac{𝜆}{2}$

Using the wave relation,

$𝑣=𝑛𝜆$

we get

$𝜆=\frac{𝑣}{𝑛}=\frac{18}{𝑛}$

Hence,

$𝑑=\frac{𝜆}{2}=\frac{1}{2}\times \frac{18}{𝑛}=\frac{9}{𝑛}\text{m}$

So, the correct answer is Option C: $\frac{9}{𝑛}$.

Given,

$𝑦=6\cos (100𝑡-4𝑥)$

Comparing it with the progressive wave equation

$𝑦=𝐴\cos (𝜔𝑡-𝑘𝑥),$

we get

$𝐴=6𝜇\text{m}=6\times {10}^{-6}\text{ m},𝜔=100{\text{ rad s}}^{-1},𝑘=4{\text{ rad m}}^{-1}.$

The maximum particle velocity is

${𝑣}_{\text{particle,max}}=𝐴𝜔$

$=(6\times {10}^{-6})\left(100\right)=6\times {10}^{-4}{\text{ m s}}^{-1}.$

The wave velocity is

${𝑣}_{\text{wave}}=\frac{𝜔}{𝑘}=\frac{100}{4}=25{\text{ m s}}^{-1}.$

Thus,

$\frac{{𝑣}_{\text{particle,max}}}{{𝑣}_{\text{wave}}}=\frac{6\times {10}^{-4}}{25}=2.4\times {10}^{-5}.$

Hence, the correct answer is $\overline{)\text{Option C }2.4\times {10}^{-5}}$.

The frequency of a wave depends on the source, so it remains the same in both media.

Using $𝑣=𝑓𝜆$,

$\frac{{𝑣}_{𝑄}}{{𝑣}_{𝑃}}=\frac{{𝜆}_{𝑄}}{{𝜆}_{𝑃}}$

Given:

${𝜆}_{𝑃}=150\text{mm}=0.15\text{m}$

${𝜆}_{𝑄}=0.30\text{m},{𝑣}_{𝑃}=80{\text{cm s}}^{-1}=0.80{\text{m s}}^{-1}$

So,

${𝑣}_{𝑄}=0.80\times \frac{0.30}{0.15}=0.80\times 2=1.6{\text{m s}}^{-1}$

Hence, the correct answer is Option D: $1.6{\text{m s}}^{-1}$.

The charged spherical shell pushes outward because like charges repel each other. For a uniformly charged conducting thin spherical shell, the electrostatic pressure on its surface is

$𝑃=\frac{{𝜎}^2}{2{𝜖}_0}$

Take a small area $𝑑𝐴$ on one hemisphere. The outward force on this small area is

$𝑑𝐹=𝑃𝑑𝐴=\frac{{𝜎}^2}{2{𝜖}_0}𝑑𝐴$

This force acts normally outward from the surface. But for separating the two hemispheres, only the component along the axis of the hemisphere is useful.

If $𝜃$ is the angle between the outward normal and this axis, then the required component is

$𝑑{𝐹}_{⟂}=𝑑𝐹\cos 𝜃=𝑃𝑑𝐴\cos 𝜃$

Now add this component over the whole curved surface of one hemisphere:

${𝐹}_{𝑟𝑒𝑝}=\int 𝑃𝑑𝐴\cos 𝜃=𝑃\int 𝑑𝐴\cos 𝜃$

The quantity $\int 𝑑𝐴\cos 𝜃$ is the projected area of the hemisphere on its circular base. This projected area is just the area of a circle of radius $𝑅$.

$\int 𝑑𝐴\cos 𝜃=𝜋{𝑅}^2$

So the repulsive force between the two hemispheres is

${𝐹}_{𝑟𝑒𝑝}=\left(\frac{{𝜎}^2}{2{𝜖}_0}\right)\left(𝜋{𝑅}^2\right)$

${𝐹}_{𝑟𝑒𝑝}=\frac{𝜋{𝜎}^2{𝑅}^2}{2{𝜖}_0}$

To keep the two hemispheres together, the applied force $𝐹$ must balance this repulsive force.

$𝐹=\frac{𝜋{𝜎}^2{𝑅}^2}{2{𝜖}_0}$

Hence, $𝐹$ is proportional to

$\frac{1}{{𝜖}_0}{𝜎}^2{𝑅}^2$

Since $𝐶$ lies on $𝐴𝐵$ and $𝐴𝐵=3𝑥$,

$𝐴𝐶=𝑥,𝐶𝐵=𝐴𝐵-𝐴𝐶=2𝑥$

The electrostatic potential energy of three charges is the sum of the potential energies of all charge pairs:

$𝑈=\frac{1}{4𝜋{𝜖}_0}(\frac{{𝑞}_{𝐴}{𝑞}_{𝐵}}{𝐴𝐵}+\frac{{𝑞}_{𝐴}{𝑞}_{𝐶}}{𝐴𝐶}+\frac{{𝑞}_{𝐵}{𝑞}_{𝐶}}{𝐵𝐶})$

Substituting ${𝑞}_{𝐴}={𝑞}_{𝐵}=𝑞$ and ${𝑞}_{𝐶}=-3𝑞$,

$𝑈=\frac{1}{4𝜋{𝜖}_0}[\frac{{𝑞}^2}{3𝑥}-\frac{3{𝑞}^2}{𝑥}-\frac{3{𝑞}^2}{2𝑥}]$

$𝑈=\frac{{𝑞}^2}{4𝜋{𝜖}_0𝑥}(\frac{1}{3}-3-\frac{3}{2})=-\frac{25}{6}\frac{{𝑞}^2}{4𝜋{𝜖}_0𝑥}$

Since

$-\frac{25}{6}\approx -4.17\approx -4,$

the potential energy is nearly

$\overline{)-\frac{4{𝑞}^2}{4𝜋{𝜖}_0𝑥}}$

Hence, Option D is correct.

The electric field does work on the particle while moving it through distance $𝐷$.

Hence, the kinetic energy gained is

$𝐾=𝑞𝐸𝐷$

As the particle approaches the fixed charge $𝑄$ of the same sign, it experiences repulsion. At the distance of closest approach $𝑟$, its speed becomes zero. Its kinetic energy is completely converted into electrostatic potential energy:

$𝑞𝐸𝐷=\frac{1}{4𝜋{𝜖}_0}\frac{𝑄𝑞}{𝑟}$

Cancelling $𝑞$ and solving for $𝑟$,

$𝑟=\frac{𝑄}{4𝜋{𝜖}_0𝐸𝐷}$

Thus, the distance of closest approach is

$\overline{)\frac{𝑄}{4𝜋{𝜖}_0𝐸𝐷}}$

Hence, Option B is correct.

The total energy of an oscillating LC circuit is

$𝑈=\frac{{𝑄}^2}{2𝐶}$

When the charge on the capacitor is $𝑞$, the electric energy stored in it is

${𝑈}_{𝐸}=\frac{{𝑞}^2}{2𝐶}$

If the energy is shared equally between the electric and magnetic fields, then the electric energy is half of the total energy:

$\frac{{𝑞}^2}{2𝐶}=\frac{1}{2}\left(\frac{{𝑄}^2}{2𝐶}\right)$

Cancelling common terms,

${𝑞}^2=\frac{{𝑄}^2}{2}$

Hence,

$𝑞=\frac{𝑄}{\sqrt{\surd 2}}$

So, the correct answer is Option C: $\frac{𝑄}{\sqrt{\surd 2}}$.

When the gap is completely filled with a dielectric of dielectric constant $𝐾$, the capacitance becomes

${𝐶}^{\prime }=𝐾𝐶$

Assuming the charged capacitor is disconnected from the battery, its charge $𝑄$ remains constant.

The new potential difference is

${𝑉}^{\prime }=\frac{𝑄}{{𝐶}^{\prime }}=\frac{𝑄}{𝐾𝐶}=\frac{𝑉}{𝐾}$

So, the potential decreases.

The energy stored is

${𝐸}^{\prime }=\frac{{𝑄}^2}{2{𝐶}^{\prime }}=\frac{{𝑄}^2}{2𝐾𝐶}=\frac{𝐸}{𝐾}$

So, the energy also decreases.

Hence, the correct answer is:

$\overline{)\text{Option D: Both }𝑉\text{ and }𝐸\text{ decrease.}}$

Let the resistance of the galvanometer be $𝐺$ and its full-scale current be ${𝐼}_{𝑔}$.

When a shunt resistance $𝑆$ is connected, the current capacity becomes $𝑛{𝐼}_{𝑔}$. Hence, the current through the shunt is

${𝐼}_{𝑠}=(𝑛-1){𝐼}_{𝑔}$

Since the galvanometer and shunt are connected in parallel, the potential difference across them is equal:

${𝐼}_{𝑔}𝐺=(𝑛-1){𝐼}_{𝑔}𝑆$

Thus,

$𝐺=(𝑛-1)𝑆$

Similarly, for shunt resistance ${𝑆}^1$ and current capacity ${𝑛}^1{𝐼}_{𝑔}$,

$𝐺=({𝑛}^1-1){𝑆}^1$

Equating both expressions for $𝐺$:

$(𝑛-1)𝑆=({𝑛}^1-1){𝑆}^1$

So,

$𝑛-1=\frac{{𝑆}^1({𝑛}^1-1)}{𝑆}$

Therefore,

$\overline{)𝑛=\frac{𝑆+{𝑆}^1({𝑛}^1-1)}{𝑆}}$

Hence, the correct answer is Option D.

For a balanced meter bridge,

$\frac{𝑅}{𝑆}=\frac{𝑙}{100-𝑙}$

Initially, the resistances are $𝑋$ and $3𝑋$:

$\frac{𝑋}{3𝑋}=\frac{𝑙}{100-𝑙}$

$\frac{1}{3}=\frac{𝑙}{100-𝑙}$

This gives

$𝑙=25\text{cm}$

After interchanging the resistances, the balance point becomes

${𝑙}^{\prime }=100-25=75\text{cm}$

Hence, the shift in balance point is

${𝑙}^{\prime }-𝑙=75-25=50\text{cm}$

So, the correct answer is Option C: $50\text{cm}$.

The magnetic field at an axial point at distance $𝑥$ from the centre of a circular coil is

${𝐵}_1=\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+{𝑥}^2{)}^{3/2}}.$

At the centre of the coil, $𝑥=0$, so

${𝐵}_2=\frac{{𝜇}_0𝐼}{2𝑅}.$

Given that ${𝐵}_2=8{𝐵}_1$,

$\frac{{𝐵}_2}{{𝐵}_1}=8.$

Substituting the expressions,

$\frac{({𝑅}^2+{𝑥}^2{)}^{3/2}}{{𝑅}^3}=8.$

Raising both sides to the power $\frac{2}{3}$,

${𝑅}^2+{𝑥}^2=4{𝑅}^2.$

Thus,

${𝑥}^2=3{𝑅}^2$

and hence,

$\overline{)𝑥=\sqrt{\surd 3}𝑅}.$

So, the correct answer is Option B.

For an ideal solenoid, the magnetic induction inside it is

$𝐵=𝜇𝑛𝐼$

where:

• $𝑛$ is the number of turns per unit length,

• $𝐼$ is the current flowing through the solenoid,

• $𝜇$ is the permeability of the medium.

The magnetic field does not depend on the radius of the wire.

Hence, it depends on quantities (a), (c), and (d).

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

The current in the primary coil is

${𝐼}_{𝑝}=12\sin \left(50𝜋𝑡\right)$

So, the angular frequency is

$𝜔=50𝜋{\text{rad s}}^{-1}$

The maximum induced emf in the secondary coil is

${𝐸}_{max}=𝑀𝜔{𝐼}_{𝑝,max}$

Given,

$𝑀=𝜋\text{mH}=𝜋\times {10}^{-3}\text{H}$

Thus,

${𝐸}_{max}=(𝜋\times {10}^{-3})\left(50𝜋\right)\left(12\right)=0.6{𝜋}^2$

Using ${𝜋}^2=10$,

${𝐸}_{max}=6\text{V}$

For the secondary coil,

$𝐿=\frac{60}{𝜋}\text{mH}=\frac{60}{𝜋}\times {10}^{-3}\text{H}$

Its inductive reactance is

${𝑋}_{𝐿}=𝜔𝐿=\left(50𝜋\right)(\frac{60}{𝜋}\times {10}^{-3})=3\Omega$

The impedance of the secondary coil is

$𝑍=\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐿}}=\sqrt{\surd 4^2+3^2}=5\Omega$

Hence, the maximum induced current is

${𝐼}_{max}=\frac{{𝐸}_{max}}{𝑍}=\frac{6}{5}=1.2\text{A}$

So, the correct answer is $\overline{)1.2\text{A}}$, Option D.

When the iron core is inserted, the magnetic flux linked with the solenoid increases. According to Lenz’s law, an induced emf is produced that opposes this increase in magnetic flux.

This induced emf acts against the battery emf, so the current in the coil decreases during the insertion.

Hence, the correct answer is:

Option C — will decrease

After the iron core is fully inserted and the magnetic flux becomes steady, the induced emf becomes zero and the current returns to its original steady value.

The magnetic energy stored in an inductor is

$𝑈=\frac{1}{2}𝐿{𝐼}^2$

Given $𝐿=4\text{H}$ and $𝐼=1.5\text{A}$,

$𝑈=\frac{1}{2}\times 4\times (1.5{)}^2$

$𝑈=2\times 2.25=4.5\text{J}$

Since $1\text{J}=1000\text{mJ}$,

$4.5\text{J}=4500\text{mJ}$

Hence, the correct answer is Option D: $4500\text{mJ}$.

In an AC circuit with a pure capacitor, the current does not stay in phase with the emf.

For a pure capacitor, the alternating current leads the alternating emf by a phase angle of $\frac{𝜋}{2}$ or ${90}^{\circ }$.

If the emf across the capacitor is

${𝑒}_{𝑐}={𝐸}_0\sin \left(𝜔𝑡\right)$,

then the current is

${𝐼}_{𝑐}={𝐼}_0\sin (𝜔𝑡+\frac{𝜋}{2})$.

The term $+\frac{𝜋}{2}$ shows that the current reaches its maximum value earlier than the emf. So, current leads emf by ${90}^{\circ }$.

In the phasor diagram, this means the phasor for current ${𝐼}_{𝑐}$ should be ${90}^{\circ }$ ahead of the phasor for emf ${𝑒}_{𝑐}$ in the anticlockwise direction.

Now check the given options:

(A) ${𝐼}_{𝑐}$ and ${𝑒}_{𝑐}$ are in the same phase. This is the case for a pure resistor.

(B) ${𝐼}_{𝑐}$ leads ${𝑒}_{𝑐}$ by ${90}^{\circ }$. This is the correct case for a pure capacitor.

(C) ${𝐼}_{𝑐}$ lags behind ${𝑒}_{𝑐}$ by ${90}^{\circ }$. This is the case for a pure inductor.

(D) ${𝐼}_{𝑐}$ and ${𝑒}_{𝑐}$ differ in phase by ${180}^{\circ }$.

So, figure (B) correctly shows the phase relation for a pure capacitor.

Answer: (B)