MHT CET 2024 4th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

Total normal electric induction is given by,

T.N.E.I. $=\sum q_{\text{enclosed }}$

$\therefore$ T.N.E.I. for $𝐴=(+2𝑞-𝑞)=+𝑞$

T.N.E.I. for $𝐵=(+3𝑞-𝑞)=+2𝑞$

Current is given by, $I=I_0\sin 𝜔t\text{.... (i)}$

Given, $𝐼=100\sin (50𝜋t$)

Comparing with (i),

$𝜔=50𝜋$

$\therefore 2𝜋f=50𝜋$

$\therefore f=25\mathrm{Hz}\Rightarrow 25$ cycles per second Current becomes zero twice in one cycle

$\therefore$ Current becomes zero, $25\times 2=50$ times

According to Stefan-Boltzmann's law, the emissive power of a body is given by:

$\frac{𝑑𝑄}{𝑑𝑡}=𝑒\left(𝜎𝐴{𝑇}^4\right)\text{..... (i)}$

Here, $𝑑𝑄/𝑑𝑡$ represents the rate of emission of radiant energy, $𝑒$ is the emissivity, $𝜎$ is the Stefan-Boltzmann constant, $𝐴$ is the surface area, and $𝑇$ is the temperature in Kelvin.

Given that the dimensions of the two bodies, X and Y, are identical:

${𝐴}_1={𝐴}_2\text{..... (ii)}$

It is also stated that the emissive powers of both bodies are the same. Therefore, we have:

$\begin{array}{cc}& \frac{𝑑{𝑄}_1}{𝑑𝑡}=\frac{𝑑{𝑄}_2}{𝑑𝑡}\\ \Rightarrow & {𝑒}_1{𝑇}_4^1={𝑒}_2{𝑇}_4^2\text{... [From equations (i) and (ii)]}\\ \Rightarrow & {\left(\frac{{𝑇}_1}{{𝑇}_2}\right)}^4=\frac{{𝑒}_2}{{𝑒}_1}\end{array}$

This equation describes the relationship between the temperatures ${𝑇}_1$ and ${𝑇}_2$ of the two bodies, given their equal emissive powers.

The magnetic field at the centre of a loop is given by

$B=\frac{{𝜇}_0\mathrm{NI}}{2R}\text{.... (i)}$

$\therefore$ Magnetic field produced by ring $𝐴,{𝐵}_{𝐴}=\frac{{𝜇}_0𝐼}{2{𝑟}_1}$

$\therefore$ Magnetic flux produced in ring B due to $B_A$,

$\begin{array}{ccc}\therefore & {𝜙}_{𝐵}={𝐵}_{𝐴}{𝐴}_{𝐵}& \dots \left({𝐴}_{𝐵}\text{ is area }\right)\\ \therefore & {𝜙}_{𝐵}=\frac{{𝜋}_0𝐼}{2{𝑟}_1}\times 𝜋{𝑟}_2^2=\frac{{𝜇}_0𝜋{𝑟}_2^2}{2{𝑟}_1}𝐼& \dots \left[\text{ From(i}\right)]\end{array}$

Mutual Inductance $𝑀=\frac{𝜙}{𝐼}$

$\therefore 𝑀=\frac{{𝜙}_B}{I}=\frac{{𝜇}_0𝜋{𝑟}_2^{2}I}{2r_{1}I}=\frac{{𝜇}_0𝜋{𝑟}_2^2}{2{𝑟}_1}$

According to Gauss’s law, the net electric flux through any closed surface is given by

${\Phi }_{𝐸}=\frac{{𝑞}_{\text{enc}}}{{𝜀}_0}$

where:

${\Phi }_{𝐸}$ is the electric flux,

${𝑞}_{\text{enc}}$ is the total enclosed charge, and

${𝜀}_0$ is the permittivity of free space.

In our case, the balloon carries a total charge $𝑄$ which is uniformly distributed over its surface. When you blow up the balloon, its size increases, but the total charge $𝑄$ remains the same.

Because the net flux depends only on the total enclosed charge and not on the size or shape of the surface, the total electric flux will remain unchanged.

Thus, the correct answer is:

Option D: remains unchanged.

$\begin{array}{cc}& \text{ For balancing the bridge, }\\ & \begin{array}{cc}\frac{P}{Q}& =\frac{R}{S}\\ S& =\frac{S_1{S}_2}{S_1+S_2}\dots .(\because S_1,S_2\text{ are in parallel })\\ \therefore \frac{P}{Q}& =\frac{R(S_1+S_2)}{S_1{S}_2}\end{array}\end{array}$

Mass $=$ Volume $\times$ Density

Mass of spheres $𝐴$ and $𝐵$ is,

$M_A=\frac{4}{3}𝜋R_3^A{𝜌}_A$ and $M_B=\frac{4}{3}𝜋R_3^B{𝜌}_B$

If masses are equal,

$M_A=M_B$

$\begin{array}{cc}\therefore & \frac{4}{3}𝜋R_3^A{𝜌}_A=\frac{4}{3}𝜋R_3^B{𝜌}_{𝐵}\\ & \frac{R_{𝐵}}{R_A}={\left(\frac{{𝜌}_A}{{𝜌}_{𝐵}}\right)}^{\frac{1}{3}}\text{..... (i)}\end{array}$

Moment of inertia of sphere about its diameter is given by, $I=\frac{2}{5}{\mathrm{MR}}^2$

${𝐼}_{𝐴}=\frac{2}{5}{𝑀}_{𝐴}{𝑅}_2^{𝐴}$ and ${𝐼}_{𝐵}=\frac{2}{5}{𝑀}_{𝐵}{𝑅}_2^{𝐵}$

$\begin{array}{cc}\therefore & \frac{{𝐼}_{𝐵}}{{𝐼}_{𝐴}}=\frac{{𝑅}_2^{𝐵}}{{𝑅}_2^{𝐴}}\\ & \frac{{𝐼}_{𝐵}}{{𝐼}_{𝐴}}={\left(\frac{{𝜌}_{𝐴}}{{𝜌}_{𝐵}}\right)}^{\frac{2}{3}}\text{.... [From (i)]}\end{array}$

To determine the distance between two successive antinodes in a stationary wave, consider the given wave equation:

$𝑦=12\cos \left(\frac{𝜋}{6}𝑥\right)\sin \left(8𝜋𝑡\right)$

This can be compared to the standard form of a stationary wave equation:

$𝑦=2𝐴\sin \left(\frac{2𝜋𝑥}{𝜆}\right)\cos \left(\frac{2𝜋𝑡}{𝜆}\right)$

From the comparison, we identify:

$\frac{𝜋}{6}=\frac{2𝜋}{𝜆}$

Solving for $𝜆$:

$𝜆=12\text{cm}$

The distance between two successive antinodes is given by $\frac{𝜆}{2}$. Therefore, this distance is:

$\frac{𝜆}{2}=\frac{12\text{cm}}{2}=6\text{cm}$

Net upward zone force on the bottom of the liquid = weight of the liquid displaced by cylinder + thrust force on upper surface of the cylinder due to h column of liquid

$\begin{array}{cc}{F}_{\text{net }}& =𝜌\mathrm{Vg}+𝜌\mathrm{gh}\times 𝜋r^2\\ & =𝜌𝑔(𝑉+𝜋r^2ℎ)\end{array}$

Given that, $25\%$ of total intensity of incident light is reflected from upper surface. This implies, if intensity of incident light is $I_0$, the intensity of light reaching the lower surface of plate will be $\frac{3}{4}{I}_0$.

As $50\%$ of this intensity is reflected, the final intensity of light emerging from glass plate will be $\frac{3}{8}{I}_0$.

$\begin{array}{cc}\therefore & I_1=\frac{I_0}{4}\\ & I_2=\frac{3}{8}{I}_0\end{array}$

Now, $\frac{{𝐼}_{max}}{{𝐼}_{min}}=\frac{{(\sqrt{\surd {𝐼}_1}+\sqrt{\surd {𝐼}_2})}^2}{{(\sqrt{\surd {𝐼}_1}-\sqrt{\surd {𝐼}_2})}^2}={\left(\frac{\frac{1}{2}+\sqrt{\surd \frac{3}{8}}}{\frac{1}{2}-\sqrt{\surd \frac{3}{8}}}\right)}^2$

To determine the new power of a convex lens when it is placed in a liquid, we use the lens maker's formula. Originally, the lens has a refractive index of $\frac{3}{2}$ and a power of 2.5 D. When immersed in a liquid with a refractive index of 2, the new power of the lens can be calculated as follows:

The lens maker’s formula is given by:

$\frac{1}{𝑓}=𝑃=(𝜇-1)(\frac{1}{{𝑅}_1}-\frac{1}{{𝑅}_2})$

Where $𝑃$ is the power of the lens, $𝜇$ is the refractive index, and ${𝑅}_1$ and ${𝑅}_2$ are the radii of curvature of the lens surfaces.

When the lens is placed in a liquid, the power is recalculated using:

${𝑃}^{\prime }=({𝜇}^{\prime }-1)(\frac{1}{{𝑅}_1}-\frac{1}{{𝑅}_2})$

To find the refractive index of the lens in the liquid, ${𝜇}^{\prime }$, we have:

${𝜇}^{\prime }=\frac{𝜇}{\text{Refractive index of the liquid}}=\frac{\frac{3}{2}}{2}=\frac{3}{4}$

We calculate the ratio of the new power, ${𝑃}^{\prime }$, to the original power, $𝑃$, as follows:

$\frac{{𝑃}^{\prime }}{𝑃}=\frac{{𝜇}^{\prime }-1}{𝜇-1}=\frac{\frac{3}{4}-1}{\frac{3}{2}-1}=\frac{-\frac{1}{4}}{\frac{1}{2}}=-\frac{1}{2}$

Thus, the new power, ${𝑃}^{\prime }$, is:

${𝑃}^{\prime }=-\frac{1}{2}\times 𝑃=-\frac{1}{2}\times 2.5=-1.25\text{D}$

The lens's new power in the liquid is therefore -1.25 D.

From the logic circuit,

$Y=\overline{\left(\stackrel{–}{A+B}\right)\cdot (A\cdot B)}$

Using De-Morgan's law, $\overline{A+B}=\bar{A}\cdot \bar{B}$

$\begin{array}{cc}& Y=\overline{(\stackrel{–}{A}\cdot \stackrel{–}{B})\cdot (A\cdot B)}\\ & Y=\overline{(\stackrel{–}{A}\cdot A)\cdot (\stackrel{–}{B}\cdot B)}\\ & Y=\overline{0.0}=1\end{array}$

$\therefore$ Output is always 1 .

To find the width of the diffraction pattern when the slit is partially covered, we start with the formula for the width of the diffraction pattern:

$W=\frac{2𝜆D}{d}\text{(i)}$

Given:

The initial wavelength, $𝜆=400\text{nm}$

The initial slit width, $d$

The initial width of the diffraction pattern is $Y$

Now, when half of the slit is covered, the effective width of the slit becomes:

$d^{\prime }=\frac{d}{2}$

The new wavelength is ${𝜆}^{\prime }=600\text{nm}$.

Substituting these values into our formula, the new width of the diffraction pattern is:

$\begin{array}{cc}{W}^{\prime }& =\frac{2{𝜆}^{\prime }D}{d^{\prime }}\\ & =\frac{2\cdot 600\text{nm}\cdot D}{d/2}\text{(ii)}\end{array}$

So, comparing the new width to the original:

$\begin{array}{cc}\frac{W^{\prime }}{W}& =\frac{2{𝜆}^{\prime }D}{𝜆D}\\ & =\frac{2\times 600}{400}\\ & =3\end{array}$

Thus, the new width of the diffraction pattern, $W^{\prime }$, is:

$W^{\prime }=3W=3Y$

Therefore, the width of the diffraction pattern, when half the slit is covered and illuminated by yellow light, becomes three times the initial width, or $3𝑌$.

Angular momentum $=I𝜔$

By conservation of angular momentum,

$\begin{array}{cc}& I_1{𝜔}_1=I_2{𝜔}_2\\ & \text{ Here, }{I}_1=\frac{{\mathrm{MR}}^2}{2},I_2=\frac{(M+M/3)}{2}{R}^2=\frac{2{\mathrm{MR}}^2}{3}\\ \therefore & \frac{{\mathrm{MR}}^2}{2}{𝜔}_1=\frac{2{\mathrm{MR}}^2}{3}{𝜔}_2\dots \left[\text{ From }\right(i\left)\right]\\ \therefore & {𝜔}_2=\frac{3}{4}{𝜔}_1=\frac{3}{4}𝜔\dots (\because {𝜔}_1=𝜔)\end{array}$

Charged particle moving in a circular path acts like a current carrying coil.

Let $𝐼$ be the current due to the particle current is,

$𝐼=\frac{𝑞}{𝑡}=\frac{100\times 𝑒}{1}=100e$

Magnetic field at the centre due to current carrying coil,

$\begin{array}{cc}{B}_{\text{centre }}& =\frac{{𝜇}_0}{4𝜋}\frac{2𝜋I}{r}=\frac{{𝜇}_0}{4𝜋}\frac{2𝜋\times 100e}{r}\\ & =\frac{{𝜇}_0}{4}\frac{2\times 100(1.6\times {10}^{-19})}{0.8}={10}^{-17}{𝜇}_0\end{array}$

Density $(𝜌=𝑀/𝑉)$ of carpet remains the same after it is unrolled,

${𝜌}_1={𝜌}_2$

If $M_2$ and $V_2$ are mass and volume respectively of unrolled carpet,

$\begin{array}{cc}& \frac{M}{V}=\frac{M_2}{V_2}\\ & M_2=\frac{M}{𝜋R^2𝑙}\times 𝜋R_2𝑙=\frac{M}{𝜋R^2𝑙}\times 𝜋\left(\frac{R^2}{4}\right)𝑙\dots (\because R_2=R/2)\\ & M_2=\frac{M}{4}\text{.... (i)}\end{array}$

Potential energy of rolled carpet,

$U_1=\mathrm{MgR}$

Potential energy of unrolled carpet,

$U_2=M_2{\mathrm{gR}}_2=\left(\frac{M}{4}\right)g\left(\frac{R}{2}\right)\text{.... [From (i)]}$

Change in potential energy,

$\begin{array}{cc}& \Delta 𝑈={𝑈}_1-{𝑈}_2\\ & \Delta 𝑈=𝑀𝑔𝑅-\left(\frac{𝑀}{4}\right)𝑔\left(\frac{𝑅}{2}\right)\\ & \Delta 𝑈=\frac{7}{8}𝑀𝑔𝑅\end{array}$

Surface area of film,

$A_1=𝑙\times d=10\times 0.5=5{\mathrm{cm}}^2$

The distance between wires ' $𝑑$ ' is increased by

$1\mathrm{mm}=0.1\mathrm{cm}$

Surface area of film becomes,

$A_2=10\times (0.5+0.1)=6{\mathrm{cm}}^2$

Increase in the surface area of the film is,

$\begin{array}{cc}& \Delta A=A_2-A_1\\ & \Delta A=6-5=1{\mathrm{cm}}^2={10}^{-4}{m}^2\end{array}$

Work done,

$W=2T.\Delta A\dots$ (where T is surface tension)

$\begin{array}{cc}& W=2\times 72\times {10}^{-4}J\\ & W=1.44\times {10}^{-2}J\end{array}$

When iron rod is inserted, the coil magnetizes the iron rod increasing the magnetic field.

$\therefore$ Inductance ( L ) of coil increases.

$V=L\frac{\mathrm{di}}{\mathrm{dt}}$

$\therefore$ Voltage drop across the coil increases

i.e. voltage drop at bulb decreases.

$\therefore$ Brightness of bulb decreases.

The wavelength of the wave is given by,

$𝜆=\frac{v}{n}=\frac{30}{250}=0.12m$

Also, the phase difference ' $𝜙$ ' is given by,

$𝜙=\frac{2𝜋}{𝜆}\times$ path difference

$\begin{array}{cc}\therefore & 𝜙=\frac{2𝜋\left(0.1\right)}{𝜆}\text{... (path difference = 10 cm = 0.1 m)}\\ \therefore & 𝜙=\frac{2𝜋}{0.12}\times 0.1\\ \therefore & 𝜙={\left(\frac{5𝜋}{3}\right)}^{𝑐}\end{array}$

To determine the increase in temperature when a lead bullet moving with velocity $𝑣$ strikes a wall and stops, given that 50% of its energy is converted into heat, follow these calculations:

Initial Kinetic Energy (K.E.) of the Bullet:

$\text{K.E.}=\frac{1}{2}𝑚{𝑣}^2$

Energy Converted to Heat:

$\text{Energy converted to heat}=50\%\text{ of K.E.}=\frac{1}{2}\times \frac{1}{2}𝑚{𝑣}^2=\frac{1}{4}𝑚{𝑣}^2$

Heat Absorbed by the Bullet:

Let $\Delta 𝑇$ be the increase in temperature, $𝑠$ the specific heat capacity of lead, and $𝑚$ the mass. Then the heat absorbed $𝑄$ is:

$𝑄=𝑚𝑠\Delta 𝑇$

Equating Mechanical Energy to Thermal Energy:

Using the formula for the mechanical equivalent of heat $𝐽$:

$𝐽=\frac{\text{Work}}{\text{Heat}}=\frac{\frac{1}{4}𝑚{𝑣}^2}{𝑚𝑠\Delta 𝑇}$

Solving for Increase in Temperature:

Rearrange to find $\Delta 𝑇$:

$\Delta 𝑇=\frac{{𝑣}^2}{4𝐽𝑠}$

Thus, the increase in temperature $\Delta 𝑇$ of the bullet is expressed as $\frac{{𝑣}^2}{4𝐽𝑠}$.

The frequency of the pendulum at the surface is given as $f=\frac{1}{2𝜋}\sqrt{\surd \frac{g}{𝑙}}$

At depth the formula for gravitational acceleration is $g_{\mathrm{cff}}=g(1-\frac{d}{R})$

For $d=\frac{R}{4},g_{\mathrm{eff}}=g(1-\frac{1}{4})=\frac{3}{4}𝑔$

The frequency at depth $d=\frac{R}{4}$

$f_d=\frac{1}{2𝜋}\sqrt{\surd \frac{\frac{3}{4}g}{𝑙}}=\frac{1}{2𝜋}\sqrt{\surd \frac{3g}{4𝑙}}$

Take the ratio of both frequencies

$\begin{array}{cc}& \frac{{𝑓}_{𝑑}}{𝑓}=\frac{\sqrt{\surd 3}}{2}\\ & {𝑓}_{𝑑}=\frac{\sqrt{\surd 3}}{2}𝑓=\frac{\sqrt{\surd 3}𝑛}{2}\dots (\because 𝑓=𝑛)\end{array}$

Wavelength, $\frac{1}{𝜆}=R(\frac{1}{n^2}-\frac{1}{m^2})$

For shortest wavelength in Lyman series:

$\begin{array}{cc}& 𝑛=1,𝑚=\infty \\ & \frac{1}{{𝜆}_{𝐿}}=𝑅(\frac{1}{1^2}-\frac{1}{{\infty }^2})=𝑅\text{.... (i)}\end{array}$

For shortest wavelength in Balmer series:

$\begin{array}{cc}& n=2,m=\infty \\ & \frac{1}{{𝜆}_B}=R(\frac{1}{2^2}-\frac{1}{{\infty }^2})=\frac{R}{4}\text{.... (ii)}\end{array}$

Dividing equation (ii) by equation (i),

$\frac{{𝜆}_1}{{𝜆}_{𝐵}}=\frac{R/4}{R}=\frac{1}{4}=0.25$

Due to the $10𝜇C$ change, potential at all the four corners will be the same

$\therefore$ work done in moving $2𝜇C$ change from point A to $𝐵$ will be zero.

To understand how to derive the expression for ${𝐶}_{𝑝}$, the molar specific heat of an ideal gas at constant pressure, let's go through the following steps:

We start with the given relation for the heat capacity ratio (also known as the adiabatic index), $𝛾$, which is defined as:

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}$

This implies:

$\frac{{𝐶}_{𝑣}}{{𝐶}_{𝑝}}=\frac{1}{𝛾}$

We can express the difference in specific heats as:

$\frac{{𝐶}_{𝑣}-{𝐶}_{𝑝}}{{𝐶}_{𝑝}}=\frac{1-𝛾}{𝛾}$

Rearranging the terms gives:

$\frac{-({𝐶}_{𝑝}-{𝐶}_{𝑣})}{{𝐶}_{𝑝}}=\frac{1-𝛾}{𝛾}$

Since the relation between the specific heats is ${𝐶}_{𝑝}-{𝐶}_{𝑣}=𝑅$ (where $𝑅$ is the universal gas constant), we substitute this into the equation:

$\frac{-𝑅}{{𝐶}_{𝑝}}=\frac{1-𝛾}{𝛾}$

Simplifying, we have:

$\frac{𝑅}{{𝐶}_{𝑝}}=\frac{𝛾-1}{𝛾}$

Finally, solving for ${𝐶}_{𝑝}$, we find:

${𝐶}_{𝑝}=\frac{𝛾𝑅}{𝛾-1}$

Thus, the molar specific heat at constant pressure ${𝐶}_{𝑝}$ for an ideal gas is given by the expression $\frac{𝛾𝑅}{𝛾-1}$.

Using Lenz law,

Induced e.m.f. is given by, $𝑒=-𝑁\frac{d𝜙}{\mathrm{dt}}\text{.... (i)}$

Change in magnetic flux $=\frac{d𝜙}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}}A\cos 𝜃$

From (i),

$\begin{array}{cc}& 𝑒=-𝑁\frac{𝑑𝐵}{𝑑𝑡}𝐴\cos 𝜃\\ & \therefore 𝑒=-\frac{𝑁({𝐵}_2-{𝐵}_1)𝐴\cos 𝜃}{𝑡}\\ & \therefore 𝑡=\frac{-\left(50\right)\times (0-2\times {10}^{-2})\times (100\times {10}^{-4})\times \cos 0^{\circ }}{0.1}\\ & \therefore 𝑡=0.1s\end{array}$

Let $𝑟$ be radius,

Circumference $=2𝜋$,

Half the circle is covered,

$\therefore$ Distance covered $=𝜋r=15r$

... (given, $r=15\mathrm{cm}$)

Speed $=\frac{\text{ distance }}{\text{ time }}=\frac{15𝜋}{3}=5𝜋$

Angular speed, $𝜃=\frac{𝑣}{𝑟}=\frac{5𝜋}{15}=\frac{𝜋}{3}\mathrm{rad}/s$

Kinetic Energy, $𝐸=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\frac{(\mathrm{mv}{)}^2}{m}=\frac{p^2}{2m}$

$\therefore p=\sqrt{\surd 2\mathrm{mE}}\text{..... (i)}$

De-Broglie wavelength,

$\begin{array}{cc}& 𝜆=\frac{h}{p}=\frac{h}{\sqrt{\surd 2\mathrm{mE}}}\text{.... [from (i)]}\\ & \Rightarrow 𝜆\propto \frac{1}{\sqrt{\surd E}}\end{array}$

$\therefore$ When kinetic energy is increased by 2 times, $E_2=E_1+2E_1=3E_1$

$\therefore 𝜆$ changes by factor of $\frac{1}{\sqrt{\surd 3}}$

The change in the internal energy, ΔU, of a gas when its volume changes from 'V' to '2V' at constant pressure 'P' can be determined using the following derivation.

The formula for the change in internal energy is given by:

$\Delta 𝑈=𝑛{𝐶}_{𝑉}\Delta 𝑇\text{.... (i)}$

We are given that the ratio of specific heats is:

$\frac{{𝐶}_{𝑃}}{{𝐶}_{𝑉}}=𝛾$

and the relation between them is:

${𝐶}_{𝑃}-{𝐶}_{𝑉}=𝑅$

This leads to:

$1+\frac{𝑅}{{𝐶}_{𝑉}}=𝛾$

From this, we derive:

${𝐶}_{𝑉}=\frac{𝑅}{𝛾-1}$

Substitute ${𝐶}_{𝑉}$ into equation (i):

$\Delta 𝑈=𝑛\left(\frac{𝑅}{𝛾-1}\right)\Delta 𝑇$

Under constant pressure, we use the equation:

$𝑃\Delta 𝑉=𝑛𝑅\Delta 𝑇$

Thus, substituting for $\Delta 𝑇$ from above:

$\Delta 𝑈=\frac{𝑃\Delta 𝑉}{𝛾-1}=\frac{𝑃(2𝑉-𝑉)}{𝛾-1}=\frac{𝑃𝑉}{𝛾-1}$

This indicates that the change in internal energy is $\frac{𝑃𝑉}{𝛾-1}$.

Potential difference across $1200\Omega$ resistor $V_{1200\Omega }=15V$

$\therefore$ current across $1200\Omega$ resistor,

$I_{1200\Omega }=\frac{15}{1200}A=12.5\mathrm{mA}$

Potential difference across $300\Omega$ resistor,

$V_{300\Omega }=24-15=9V$

$\therefore$ Current across $300\Omega$ resistor,

$I_{300\Omega }=\frac{9}{300}A=30\mathrm{mA}$

$\therefore$ Current through zener diode,

$I_z=30-12.5=17.5\mathrm{mA}.$

We can solve this using Gauss's Law, which states that:

$\oint 𝐄\cdot 𝑑𝐀=\frac{{𝑞}_{\text{enc}}}{{𝜀}_0}$

Here's a step-by-step explanation:

The flux leaving the surface contributes positively with a value of ${𝜙}_2,$ while the flux entering the surface contributes negatively with a value of ${𝜙}_1.$ Thus, the net electric flux is:

${𝜙}_{\text{net}}={𝜙}_2-{𝜙}_1$

Applying Gauss's Law:

${𝜙}_{\text{net}}=\frac{{𝑞}_{\text{enc}}}{{𝜀}_0}$

Substituting for the net flux:

${𝜙}_2-{𝜙}_1=\frac{{𝑞}_{\text{enc}}}{{𝜀}_0}$

Solving for the enclosed charge ${𝑞}_{\text{enc}}$:

${𝑞}_{\text{enc}}={𝜀}_0({𝜙}_2-{𝜙}_1)$

Thus, the correct answer is:

Option D

${𝜀}_0({𝜙}_2-{𝜙}_1)$

Refractive index of a medium is given by,

$n=$ Real depth/ Apparent depth ...(i)

Given that, height of container $=30\mathrm{cm}$

Apparent depth $=15\mathrm{cm}\dots$ (ii)

Real depth $=n\times$ Apparent depth $\dots [$ From (i)]

Given that, ${𝑛}_{𝑤}=\frac{4}{3}$ ...(iii)

Real depth $=20\mathrm{cm}\dots$ [From (ii) and (iii)]

$\therefore x=20\mathrm{cm}$

From the law of conservation of energy, we have

Potential energy = Translational kinetic energy + Rotational kinetic energy

or $\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}I{𝜔}^2$

or $\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2{𝜔}^2+\frac{1}{2}\left(\frac{1}{2}{\mathrm{mr}}^2\right){𝜔}^2=\frac{3}{4}{𝑚{𝑟}^2{𝜔}^2}^2$

or ${𝜔}^2=\frac{4\mathrm{gh}}{3r^2}$

Now the rotational kinetic energy $=\frac{1}{2}I{𝜔}^2$

$\therefore$ Substituting for ${𝜔}^2$ and I, we have,

Rotational kinetic energy $=\frac{1}{2}\left(\frac{1}{2}{\mathrm{mr}}^2\right)\frac{4\mathrm{gh}}{3r^2}$

$=\frac{\mathrm{Mgh}}{3}\dots (\because 𝑀=𝑚)$

For S.H.M.,

Maximum velocity, $𝛼=A𝜔$

$𝜔=\frac{𝛼}{𝐴}\text{.... (i)}$

Maximum acceleration, $𝛽=A{𝜔}^2$

$\begin{array}{cc}& 𝛽=𝐴{\left(\frac{𝛼}{𝐴}\right)}^2=\frac{{𝛼}^2}{𝐴}\\ & \Rightarrow 𝐴=\frac{{𝛼}^2}{𝛽}\end{array}$

$\therefore$ Path length $=2𝐴=\frac{2{𝛼}^2}{𝛽}$

When a galvanometer is shunted with a resistance ' $𝑠$ ', its current handling capacity increases by a factor of ' $𝑛$ '. If this galvanometer is then shunted with a different resistance ' ${𝑠}_1$ ', its capacity is enhanced to ' ${𝑛}_1$ ' times the original current capacity.

Derivation

Shunt Resistance Formula:

The formula for shunt resistance, when applied to a galvanometer, is:

$𝑠=𝐺\left(\frac{1}{1-\frac{𝐼}{{𝐼}_{𝑔}}}\right)$

Here, $𝐺$ represents the galvanometer constant, $𝐼$ is the total current, and ${𝐼}_{𝑔}$ is the maximum current the galvanometer can handle initially.

Current Capacity Factor:

The current capacity factor $𝑛$ is expressed as:

$𝑛=\frac{1}{{𝐼}_{𝑔}}$

Therefore, using the relationship for shunt resistance:

$𝑠=𝐺\left(\frac{1}{1-𝑛}\right)$

Rearranging the terms provides:

$𝐺=𝑠(𝑛-1)\text{(Equation i)}$

Shunted with Another Resistance ${𝑠}_1$:

When the galvanometer is shunted with another resistance ${𝑠}_1$, the new formula becomes:

${𝑠}_1=𝐺\left(\frac{1}{1-{𝑛}_1}\right)$

Solving for ${𝑛}_1$:

${𝑛}_1=\frac{𝐺+{𝑠}_1}{{𝑠}_1}$

Substituting Equation (i) into this:

${𝑛}_1=\frac{𝑠(𝑛-1)+{𝑠}_1}{{𝑠}_1}$

This expression gives you the new current capacity factor ' ${𝑛}_1$ ' when the galvanometer is shunted with the resistance ' ${𝑠}_1$ '.