MHT CET 2024 10th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

In a semiconductor at room temperature, some electrons from the valence band are thermally excited across the band gap into the conduction band. Therefore:

The valence band is not completely filled (i.e., it is partially empty).

The conduction band is not empty (i.e., it is partially filled).

Hence, the correct statement is:

(D) The valence band is partially empty and the conduction band partially filled.

$\begin{array}{cc}& V_z=8V\text{.... (given)}\\ & P=V_{z}I\\ \therefore & 1.6=8I\\ \therefore & I=0.2A\end{array}$

Voltage drop across Zener is 8 V . Hence the voltage drop across R will be $10-8=2V$.

$R=\frac{V}{I}=\frac{2}{0.2}=10\Omega$

Given: $R=\frac{l}{𝜔C}=X_c$.

$\begin{array}{cc}\therefore 𝑍& =\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐶}}=\sqrt{\surd 2}𝑅\\ & {𝐼}_0=\frac{{𝑉}_0}{𝑍}=\frac{{𝑉}_0}{\sqrt{\surd 2}𝑅}\\ & \Rightarrow \frac{{𝐼}_0}{\sqrt{\surd 2}}=\frac{{𝑉}_0}{2𝑅}\text{.... (i)}\end{array}$

When $𝜔$ becomes $\frac{1}{\sqrt{\surd 3}}$ times, $X_C$ will become $\sqrt{\surd 3}\text{ times, i.e., }\sqrt{\surd 3}𝑅.$

$\begin{array}{cc}\therefore & Z^{\prime }=\sqrt{\surd R^2+(\sqrt{\surd 3}𝑅{)}^2}=2𝑅\\ \therefore & I_{\prime }^0=\frac{V_0}{Z^{\prime }}=\frac{V_0}{2R}=\frac{I_0}{\sqrt{\surd 2}}\text{...[From(i)]}\end{array}$

The black discs $x$, $y$, and $z$ have radii of $1\text{m}$, $2\text{m}$, and $3\text{m}$, respectively. The wavelengths corresponding to maximum intensity are $200\text{nm}$, $300\text{nm}$, and $400\text{nm}$, respectively. Here, we explore the relationship between the emissive power ${𝐸}_{𝑥}$, ${𝐸}_{𝑦}$, and ${𝐸}_{𝑧}$ for these discs.

Emissive Power

The formula for emissive power is given by:

$𝐸=𝜎𝐴{𝑇}^4\Rightarrow 𝐸\propto 𝐴{𝑇}^4$

Where:

$𝐴$ is the area of the disc.

$𝑇$ is the absolute temperature.

$𝜎$ is the Stefan-Boltzmann constant.

Area Calculation

The area $𝐴$ of a disc is given by $𝐴=𝜋{𝑅}^2$. Therefore:

$𝐴\propto {𝑅}^2$

Given:

${𝑅}_1=1\text{m}$

${𝑅}_2=2\text{m}$

${𝑅}_3=3\text{m}$

The areas are proportional as follows:

${𝐴}_{𝑥}:{𝐴}_{𝑦}:{𝐴}_{𝑧}\text{proportional to}1:4:9\text{... (i)}$

Temperature Analysis Using Wien's Displacement Law

By Wien's displacement law, we have:

${𝜆}_{max}𝑇=\text{constant}\Rightarrow 𝑇\propto \frac{1}{𝜆}$

Given maximum wavelengths:

${𝜆}_{max1}=200\text{nm}$

${𝜆}_{max2}=300\text{nm}$

${𝜆}_{max3}=400\text{nm}$

The wavelengths are proportional as follows:

${𝜆}_{𝑥}:{𝜆}_{𝑦}:{𝜆}_{𝑧}\text{proportional to}2:3:4$

Thus the temperatures are:

$\frac{1}{{𝑇}_{𝑥}}:\frac{1}{{𝑇}_{𝑦}}:\frac{1}{{𝑇}_{𝑧}}\text{proportional to}2:3:4$

Therefore, the temperatures are proportional as:

${𝑇}_{𝑥}:{𝑇}_{𝑦}:{𝑇}_{𝑧}\text{proportional to}\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$

Simplified, this gives:

${𝑇}_{𝑥}:{𝑇}_{𝑦}:{𝑇}_{𝑧}\text{proportional to}6:4:3\text{... (ii)}$

Comparison of Emissive Powers

We compare the product $𝐴{𝑇}^4$ for the discs:

For disc $x$: ${𝐴}_{𝑥}{𝑇}_4^{𝑥}=1\times (6{)}^4=1296$

For disc $y$: ${𝐴}_{𝑦}{𝑇}_4^{𝑦}=4\times (4{)}^4=1024$

For disc $z$: ${𝐴}_{𝑧}{𝑇}_4^{𝑧}=9\times (3{)}^4=729$

Thus, the relationship between the emissive powers is:

${𝐸}_{𝑥}>{𝐸}_{𝑦}>{𝐸}_{𝑧}$

Since, $𝑔=\frac{𝐺𝑀}{R^2}$

$g_h=\frac{\mathrm{GM}}{(R+h{)}^2}$

When value of $𝑔$ is reduced by $64\%$, $g_h=36\%$ of g

$\begin{array}{cc}\therefore & \frac{{𝑔}_{ℎ}}{𝑔}=\frac{{𝑅}^2}{(𝑅+ℎ{)}^2}=\frac{36}{100}\\ \therefore & \frac{𝑅}{(𝑅+ℎ)}=\frac{6}{10}\\ \therefore & 4𝑅=6ℎ\\ \therefore & ℎ=\frac{4}{6}𝑅=\frac{2}{3}𝑅\end{array}$

For a sonometer wire, the fundamental frequency ($𝑓$) of vibration is inversely proportional to the length ($𝐿$) of the vibrating segment. Thus, the frequency is given by:

$𝑓\propto \frac{1}{𝐿}$

Given that the fundamental frequencies of the three segments are in the ratio $1:2:3$, we need to determine the lengths of these segments such that their inverses also follow the ratio $1:2:3$:

Let the lengths of the segments be ${𝐿}_1$, ${𝐿}_2$, and ${𝐿}_3$. The ratio of their fundamental frequencies being $1:2:3$ implies:

$\frac{1}{{𝐿}_1}:\frac{1}{{𝐿}_2}:\frac{1}{{𝐿}_3}=1:2:3$

Inverting the terms in each ratio, we have:

${𝐿}_1:{𝐿}_2:{𝐿}_3=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}$

To simplify this ratio, find a common multiple of the denominators (2 and 3), which is 6. Therefore, multiplying each term by 6 gives:

${𝐿}_1:{𝐿}_2:{𝐿}_3=6:3:2$

Since the total length of the wire is 110 cm, we can write:

${𝐿}_1+{𝐿}_2+{𝐿}_3=110$

Substituting the ratio terms:

$6𝑥+3𝑥+2𝑥=110$

Solving for $𝑥$:

$11𝑥=110$

$𝑥=10$

Thus, the lengths of the segments are:

${𝐿}_1=6𝑥=6\times 10=60$ cm

${𝐿}_2=3𝑥=3\times 10=30$ cm

${𝐿}_3=2𝑥=2\times 10=20$ cm

Therefore, the positions of the bridges from point $𝐴$ should be at:

First bridge at 60 cm (end of the first segment)

Second bridge at 60 cm + 30 cm = 90 cm (end of the second segment)

Hence, the placement of the bridges should be at 60 cm and 90 cm.

Answer: Option A: 60 cm and 90 cm

When a charge is moved from one point to another, the work done by the electric force is given by

$𝑊=𝑞\Delta 𝑉$

where:

$𝑞$ is the charge,

$\Delta 𝑉$ is the difference in electric potential between the two points.

On an equipotential surface, all points have the same potential. This means that

$\Delta 𝑉=0.$

Thus, for a unit charge (where $𝑞=1$),

$𝑊=1\times 0=0.$

So, no work is required to move a unit charge along an equipotential surface. This corresponds to Option D:

work done to move a charge is zero.

The pressure inside a soap bubble is given by the formula:

${𝑃}_{\text{inside}}={𝑃}_0+\frac{4𝑇}{𝑟}$

where:

${𝑃}_0$ is the external pressure,

$𝑇$ is the surface tension,

$𝑟$ is the radius of the soap bubble.

For two soap bubbles with radii ${𝑟}_1$ and ${𝑟}_2$, and inside pressures ${𝑃}_1$ and ${𝑃}_2$ respectively, we can write:

${𝑃}_1={𝑃}_0+\frac{4𝑇}{{𝑟}_1}$

${𝑃}_2={𝑃}_0+\frac{4𝑇}{{𝑟}_2}$

Rearranging for ${𝑟}_1$ and ${𝑟}_2$, we have:

${𝑟}_1=\frac{4𝑇}{{𝑃}_1-{𝑃}_0}$

${𝑟}_2=\frac{4𝑇}{{𝑃}_2-{𝑃}_0}$

The volume $𝑉$ of a sphere (soap bubble) is given by:

$𝑉=\frac{4}{3}𝜋{𝑟}^3$

Thus, the volumes of the two soap bubbles are:

${𝑉}_1=\frac{4}{3}𝜋{𝑟}_3^1=\frac{4}{3}𝜋{\left(\frac{4𝑇}{{𝑃}_1-{𝑃}_0}\right)}^3$

${𝑉}_2=\frac{4}{3}𝜋{𝑟}_3^2=\frac{4}{3}𝜋{\left(\frac{4𝑇}{{𝑃}_2-{𝑃}_0}\right)}^3$

The ratio of their volumes is:

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{\left(\frac{4𝑇}{{𝑃}_1-{𝑃}_0}\right)}^3}{{\left(\frac{4𝑇}{{𝑃}_2-{𝑃}_0}\right)}^3}={\left(\frac{{𝑃}_2-{𝑃}_0}{{𝑃}_1-{𝑃}_0}\right)}^3$

Thus, the correct option for the ratio of their volumes is:

Option C

$\frac{{({𝑃}_2-{𝑃}_0)}^3}{{({𝑃}_1-{𝑃}_0)}^3}$

To find the number of turns in the secondary coil of a transformer, use the formula for the voltage ratio in a transformer:

$\frac{{𝑉}_{𝑠}}{{𝑉}_{𝑝}}=\frac{{𝑁}_{𝑠}}{{𝑁}_{𝑝}}$

where:

${𝑉}_{𝑠}$ is the secondary voltage,

${𝑉}_{𝑝}$ is the primary voltage,

${𝑁}_{𝑠}$ is the number of turns in the secondary coil,

${𝑁}_{𝑝}$ is the number of turns in the primary coil.

Given:

${𝑉}_{𝑠}=560\text{V}$

${𝑁}_{𝑝}=120$ turns

The primary voltage ${𝑉}_{𝑝}$ can be found using the power formula:

$𝑃={𝑉}_{𝑝}\cdot {𝐼}_{𝑝}$

where:

$𝑃=1000\text{W}$ (1 kilowatt is the input power)

${𝐼}_{𝑝}=5\text{A}$ (current in the primary coil)

Solving for ${𝑉}_{𝑝}$:

${𝑉}_{𝑝}=\frac{𝑃}{{𝐼}_{𝑝}}=\frac{1000\text{W}}{5\text{A}}=200\text{V}$

Now use the voltage ratio formula:

$\frac{560\text{V}}{200\text{V}}=\frac{{𝑁}_{𝑠}}{120}$

Solving for ${𝑁}_{𝑠}$:

${𝑁}_{𝑠}=\frac{560}{200}\cdot 120=2.8\cdot 120=336$

Therefore, the number of turns in the secondary coil will be 336.

Option C: 336

Average velocity, $V_{\mathrm{avg}}=\frac{2V_1{V}_2}{V_1+V_2}$

Given $V_1=V$ and $V_2=\frac{V}{3}$

$\therefore V_{\text{avg }}=\frac{2\times V\times \frac{V}{3}}{V+\frac{V}{3}}=\frac{2V^2}{3}\times \frac{3}{4V}=\frac{V}{2}$

In uniform circular motion, the centripetal force required to maintain the motion is given by the formula:

$𝐹=\frac{𝑚{𝑣}^2}{𝑟}$

Let's denote the initial values of mass, velocity, and radius as $𝑚$, $𝑣$, and $𝑟$, respectively. The initial force is:

${𝐹}_{\text{initial}}=\frac{𝑚{𝑣}^2}{𝑟}$

Now, if mass, velocity, and radius each increase by 20%, the new values will be:

New mass = ${𝑚}^{\prime }=1.2𝑚$

New velocity = ${𝑣}^{\prime }=1.2𝑣$

New radius = ${𝑟}^{\prime }=1.2𝑟$

The new force required, ${𝐹}_{\text{new}}$, is:

${𝐹}_{\text{new}}=\frac{{𝑚}^{\prime }({𝑣}^{\prime }{)}^2}{{𝑟}^{\prime }}=\frac{\left(1.2𝑚\right)(1.2𝑣{)}^2}{1.2𝑟}$

Calculating the new force, we have:

${𝐹}_{\text{new}}=\frac{1.2𝑚\times \left(1.44{𝑣}^2\right)}{1.2𝑟}$

Simplifying the expression:

${𝐹}_{\text{new}}=\frac{1.2\times 1.44\times 𝑚\times {𝑣}^2}{1.2\times 𝑟}$

${𝐹}_{\text{new}}=\frac{1.44𝑚{𝑣}^2}{𝑟}$

The change in force can be calculated as:

$\text{Change in force}={𝐹}_{\text{new}}-{𝐹}_{\text{initial}}$

Substituting the expressions for ${𝐹}_{\text{new}}$ and ${𝐹}_{\text{initial}}$:

$\text{Change in force}=\frac{1.44𝑚{𝑣}^2}{𝑟}-\frac{𝑚{𝑣}^2}{𝑟}$

Simplifying the change in force expression:

$\text{Change in force}=(1.44-1)\frac{𝑚{𝑣}^2}{𝑟}$

$\text{Change in force}=0.44\frac{𝑚{𝑣}^2}{𝑟}$

The percentage change in force is:

$\text{Percentage change}=\left(\frac{\text{Change in force}}{{𝐹}_{\text{initial}}}\right)\times 100\%$

$\text{Percentage change}=\left(\frac{0.44\frac{𝑚{𝑣}^2}{𝑟}}{\frac{𝑚{𝑣}^2}{𝑟}}\right)\times 100\%$

$\text{Percentage change}=0.44\times 100\%$

$\text{Percentage change}=44\%$

Therefore, the necessary change in force required is 44%. The correct answer is Option B: $44\%$.

$\begin{array}{cc}{𝑆}_2𝑃& ={({𝐷}^2+{𝑑}^2)}^{1/2}\\ & =𝐷{[1+\frac{{𝑑}^2}{{𝐷}^2}]}^{1/2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& S_{2}P=D{[1+\frac{1}{2}\frac{d^2}{D^2}]}^{1/2}=D+\frac{d^2}{2D}\\ & \Rightarrow \text{ Path difference }=\frac{d^2}{2D}\end{array}\\ & \text{ For dark fringe, }\frac{d^2}{2D}=(2n-1)\frac{𝜆}{2}\\ & \begin{array}{cc}\therefore & \frac{{𝑑}^2}{2𝐷}=(2𝑛-1)\frac{{𝑑}^2}{6𝐷}\\ \therefore & 2𝑛-1=3\\ \therefore & 𝑛=2\end{array}\end{array}$

For SHM,

$\begin{array}{cc}& \text{ P.E. }=\frac{1}{2}𝑚{𝜔}^2{𝑥}^2\text{.... (i)}\\ & \text{ and K.E. }=\frac{1}{2}𝑚{𝜔}^2({𝐴}^2-{𝑥}^2)\text{.... (ii)}\end{array}$

Given $t=\frac{T}{12}$

$\begin{array}{cc}\therefore 𝑥& =𝐴\sin 𝜔𝑡=𝐴\sin \frac{2𝜋}{𝑇}\cdot 𝑡\\ \therefore 𝑥& =𝐴\sin \frac{2𝜋}{𝑇}\times \frac{𝑇}{12}\\ & =𝐴\sin \left(\frac{𝜋}{6}\right)\\ & =\frac{𝐴}{2}\text{.... (iii)}\end{array}$

Substituting equation (iii) in equation (i),

$\begin{array}{cc}\text{ P.E. }& =\frac{1}{2}𝑚{𝜔}^2\left(\frac{A^2}{4}\right)\\ & =\frac{𝑚{𝜔}^2{𝐴}^2}{8}\end{array}$

Substituting equation (iii) in equation (ii),

$\begin{array}{cc}\text{ K.E. }& =\frac{1}{2}m{𝜔}^2(A^2-\frac{A^2}{4})\\ & =\frac{1}{2}m{𝜔}^2\left(\frac{3A^2}{4}\right)\\ \therefore \frac{\text{ P.E. }}{\text{ K.E. }}& =\frac{\frac{1}{2}𝑚{𝜔}^2{𝐴}^2\times \frac{1}{4}}{\frac{1}{2}𝑚{𝜔}^2{𝐴}^2\times \frac{3}{4}}=\frac{1}{3}\end{array}$

To find the power of a combination of lenses, you can use the formula:

$𝑃=\frac{1}{𝑓}=\frac{1}{{𝑓}_1}+\frac{1}{{𝑓}_2}$

Here, ${𝑓}_1=0.4\text{m}$ (the focal length of the convex lens) and ${𝑓}_2=-0.25\text{m}$ (the focal length of the concave lens, which is negative because it's concave).

Let's calculate the power:

$𝑃=\frac{1}{0.4}+\frac{1}{-0.25}$

Convert the focal lengths to have a common denominator:

For the convex lens: $\frac{1}{0.4}=\frac{10}{4}=2.5$

For the concave lens: $\frac{1}{-0.25}=\frac{-4}{1}=-4$

Now, add these values:

$𝑃=2.5-4=-1.5\text{D}$

Therefore, the power of the lens combination is $-1.5\text{D}$.

We have $𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$

Substituting the given values in the above equation we get,

$8=𝜔\sqrt{\surd A^2-9}$ $\dots$ (i) $(\because 𝑥=3\mathrm{cm},v=8{\mathrm{cms}}^{-1})$

$6=𝜔\sqrt{\surd {𝐴}^2-16}$ ...(ii) $(\because 𝑥=4\mathrm{cm},v=6{\mathrm{cms}}^{-1})$

Diving equation (i) and (ii),

$\begin{array}{cc}& \frac{8}{6}=\frac{\sqrt{\surd A^2-9}}{\sqrt{\surd A^2-16}}\\ & \frac{4}{3}=\frac{\sqrt{\surd A^2-9}}{\sqrt{\surd A^2-16}}\\ \therefore & \frac{16}{9}=\frac{A^2-9}{A^2-16}\\ \therefore & 16A^2-256=9A^2-81\\ & 7A^2=175\\ \therefore & A^2=\frac{175}{7}\end{array}$

Substituting this value of ${𝐴}^2$ in equation (ii) and solving we get

$\begin{array}{cc}6& =𝜔\sqrt{\surd \frac{175}{7}-16}\\ 6& =𝜔\sqrt{\surd \frac{175-112}{7}}=𝜔\sqrt{\surd 9}\\ 6& =3𝜔\\ 𝜔& =2\\ \therefore 𝜔& =\frac{2𝜋}{𝑇}=2\\ \therefore 𝑇& =𝜋\text{ seconds }\end{array}$

$\begin{array}{cc}& E_1=\frac{\mathrm{hc}}{𝜆}-𝜙\text{.... (i)}\\ & E_2=\frac{\mathrm{hc}}{𝜆/2}-𝜙=\frac{2\mathrm{hc}}{𝜆}-𝜙\text{.... (ii)}\end{array}$

Given, $E_1=\frac{E_2}{4}$

$\therefore 4E_1=E_2\text{.... (iii)}$

From (ii) and (iii),

$\begin{array}{cc}& 4E_1=\frac{2\mathrm{hc}}{𝜆}-𝜙\\ & 4(\frac{\mathrm{hc}}{𝜆}-𝜙)=\frac{2\mathrm{hc}}{𝜆}-𝜙\text{.... [Substituting from (i)]}\end{array}$

$\begin{array}{cc}\therefore & \frac{4ℎ𝑐}{𝜆}-4𝜙=\frac{2ℎ𝑐}{𝜆}-𝜙\\ \therefore & 3𝜙=\frac{2ℎ𝑐}{𝜆}\\ \therefore & 𝜙=\frac{2ℎ𝑐}{3𝜆}\end{array}$

Magnetic field due to current carrying arc is,

$𝐵=\frac{{𝜇}_0𝐼}{4𝜋r}\times 𝜃=\frac{{𝜇}_{0}I}{4𝜋r}\times \frac{𝜋}{2}=\frac{{𝜇}_{0}I}{8r}$

Magnetic field due to semi-infinite current carrying straight wires AB and CD is,

$\frac{2{𝜇}_0𝐼}{4𝜋r}=\frac{{𝜇}_0𝐼}{2𝜋r}$

$\therefore$ Total magnetic field at 'O' will be,

$\begin{array}{cc}\frac{{𝜇}_{0}I}{2𝜋r}+\frac{{𝜇}_{0}I}{8r}& =\frac{{𝜇}_{0}I}{2r}(\frac{1}{𝜋}+\frac{1}{4})\\ & =\frac{{𝜇}_0}{4𝜋}\frac{1}{r}(\frac{2𝜋}{𝜋}+\frac{2𝜋}{4})\\ & =\frac{{𝜇}_0}{4𝜋}\frac{1}{𝑟}(2+\frac{𝜋}{2})\end{array}$

$\begin{array}{cc}& \text{ Radius for }{n}^{\text{th }}\text{ orbit, }{r}_n=a_0\times n^2\\ & \text{ For third orbit, }\\ & r_3=a_0\times 3^2\\ & =9a_0\\ & \text{ Also, }\mathrm{mvr}=\frac{\mathrm{nh}}{2𝜋}\\ & \Rightarrow \mathrm{mv}=\frac{\mathrm{nh}}{2𝜋r}=\frac{3h}{2𝜋\times 9a_0}\\ & 𝜆=\frac{h}{\mathrm{mv}}=\frac{h}{3h}\times 2𝜋\times 9a_0=6𝜋a_0\end{array}$

Given: $\frac{𝑅}{{𝐶}_{𝑣}}=0.4$

To determine the type of gas molecules, we start with the given ratio:

Finding ${𝐶}_{𝑣}$:

${𝐶}_{𝑣}=\frac{𝑅}{0.4}=\frac{5𝑅}{2}$

Calculate ${𝐶}_{𝑝}$:

${𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅=\frac{5𝑅}{2}+𝑅=\frac{7𝑅}{2}$

Determine $𝛾$ (the heat capacity ratio):

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{\frac{7𝑅}{2}}{\frac{5𝑅}{2}}=\frac{7}{5}$

The value of $𝛾=\frac{7}{5}$ is characteristic of gases made up of rigid diatomic molecules. Hence, the gas consists of rigid diatomic molecules.

Option C, Gravitational force, is 'NOT' a contact force.

Contact forces require physical contact between objects, including:

Force of friction: This is a contact force that opposes the relative motion or tendency of such motion of two surfaces in contact.

Normal reaction: A contact force exerted perpendicular to the surfaces in contact, often experienced as the support force from a surface.

Viscous force: Another contact force, occurring between layers of fluid in motion relative to each other.

In contrast, a gravitational force is a non-contact force that acts at a distance, pulling two masses toward each other without needing any direct physical interaction. The mathematical expression for gravitational force between two masses ${𝑚}_1$ and ${𝑚}_2$ separated by a distance $𝑟$ is given by:

$𝐹=\frac{𝐺{𝑚}_1{𝑚}_2}{{𝑟}^2}$

where $𝐺$ is the gravitational constant. This formula shows how gravitational force operates over a distance, not requiring objects to touch or interact directly.

In an isothermal process, the change in internal energy, $\Delta 𝑈$, is zero. According to the first law of thermodynamics, the relationship is given by:

$\Delta 𝑄=\Delta 𝑈+𝑊$

For an adiabatic process, where no heat is exchanged with the surroundings, the heat transfer, $\Delta 𝑄$, is zero. Therefore, the equation simplifies to:

$\Delta 𝑈=-𝑊$

To find the energy stored in the coil, we can use the formula for the energy stored in a magnetic field of an inductor:

$𝐸=\frac{1}{2}𝐿{𝐼}^2$

where $𝐸$ is the energy, $𝐿$ is the inductance, and $𝐼$ is the current.

First, we need to determine the inductance $𝐿$ of the coil. The relationship between magnetic flux ($\Phi$), inductance ($𝐿$), and current ($𝐼$) is given by:

$\Phi =𝐿\cdot 𝐼$

Given:

$\Phi =3\times {10}^{-3}\text{Wb}$,

$𝐼=4\text{A}$,

Number of turns ($𝑁$) = 400.

The total flux linkage is $𝑁\cdot \Phi$:

$𝑁\cdot \Phi =𝐿\cdot 𝐼$

Plug in the values:

$400\times 3\times {10}^{-3}=𝐿\times 4$

Calculate $𝐿$:

$𝐿=\frac{400\times 3\times {10}^{-3}}{4}=\frac{1.2}{4}=0.3\text{H}$

Now, substitute $𝐿=0.3\text{H}$ and $𝐼=4\text{A}$ into the energy formula:

$𝐸=\frac{1}{2}\times 0.3\times 4^2=\frac{1}{2}\times 0.3\times 16=0.15\times 16=2.4\text{J}$

Therefore, the energy stored in the coil is 2.4 J.

The correct answer is Option B: 2.4 J.

$\begin{array}{cc}& \text{ (K.E. }{)}_A=(\text{ K.E. }{)}_B\\ & \frac{1}{2}{I}_1{𝜔}_2^1=\frac{1}{2}{I}_2{𝜔}_2^2\\ \therefore & \frac{{𝜔}_2^2}{{𝜔}_2^1}=\frac{{𝐼}_1}{{𝐼}_2}\text{.... (i)}\end{array}$

$\begin{array}{cc}& \text{ Also, K.E }=\frac{1}{2}L𝜔\\ \therefore & \frac{1}{2}{L}_1{𝜔}_1=\frac{1}{2}{L}_2{𝜔}_2\\ \therefore & \frac{L_1}{L_2}=\frac{{𝜔}_2}{{𝜔}_1}=\frac{1}{\sqrt{\surd 3}}\text{.... (ii)}\\ \therefore & \frac{I_1}{I_2}=\frac{1}{3}\text{... [From (i) and (ii)]}\\ \therefore & I_2=3I_1\end{array}$

According to law of conservation of energy,

$\frac{1}{2}{\mathrm{mv}}^2=-\frac{\mathrm{GMm}}{R_0}-\left(\frac{\mathrm{GMm}}{R}\right)=\mathrm{GMm}(\frac{1}{R}-\frac{1}{R_0})$

$\therefore V^2=2\mathrm{GM}(\frac{1}{R}-\frac{1}{R_0})$

$\therefore$ The velocity acquired by the body when it reaches the surface of earth is:

$v=\sqrt{\surd 2\mathrm{GM}(\frac{1}{R}-\frac{1}{R_0})}$

Terminal velocity is given by

$𝑣=\frac{2{𝑟}^2(𝜌-𝜎)𝑔}{9𝜂}$

As the velocity is the same,

$\begin{array}{cc}\therefore & {𝑟}_2^{𝐴}({𝜌}_{𝐴}-𝜎)={𝑟}_2^{𝐵}({𝜌}_{𝐵}-𝜎)\\ \therefore & \frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}=\sqrt{\surd \frac{{𝜌}_{𝐵}-𝜎}{{𝜌}_{𝐴}-𝜎}}\end{array}$

Substituting the given values, we get

$\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}=\sqrt{\surd \frac{8.5\times {10}^3-2.5\times {10}^3}{11.5\times {10}^3-2.5\times {10}^3}}=\sqrt{\surd \frac{6}{9}}=\sqrt{\surd \frac{2}{3}}$

In an LCR circuit

$\begin{array}{cc}𝐼=\frac{𝐸}{𝑍}& =\frac{𝐸}{\sqrt{\surd {𝑅}^2+{({𝑋}_{𝐿}-{𝑋}_{𝑐})}^2}}\\ & =\frac{𝐸}{\sqrt{\surd {𝑅}^2+{(2𝜋\mathrm{fL}-\frac{1}{2𝜋\mathrm{fC}})}^2}}\end{array}$

To keep I constant, if f is doubled both L and C should be halved.

The potential energy between two charges is defined by the formula:

$𝑈=\frac{𝐾{𝑞}_1{𝑞}_2}{𝑟}$

Calculating the Increase in Potential Energy:

Initial Potential Energy (${𝑈}_{𝑖}$):

The initial distance between the charges is $𝑑=10\text{cm}=0.1\text{m}$.

Initial potential energy ${𝑈}_{𝑖}=\frac{𝐾{𝑞}_1{𝑞}_2}{𝑑}$

Final Potential Energy (${𝑈}_{𝑓}$):

The charge ${𝑞}_2$ is moved towards ${𝑞}_1$ by $𝑥=2\text{cm}=0.02\text{m}$.

The new separation becomes $𝑑-𝑥=0.1\text{m}-0.02\text{m}=0.08\text{m}$.

Final potential energy ${𝑈}_{𝑓}=\frac{𝐾{𝑞}_1{𝑞}_2}{𝑑-𝑥}$

Increase in Potential Energy ($\Delta 𝑈$):

The increase in potential energy is given by the difference:

$\Delta 𝑈={𝑈}_{𝑓}-{𝑈}_{𝑖}=𝐾{𝑞}_1{𝑞}_2(\frac{1}{𝑑-𝑥}-\frac{1}{𝑑})$

$\Delta 𝑈=\frac{𝐾{𝑞}_1{𝑞}_2𝑥}{𝑑(𝑑-𝑥)}$

Substituting Given Values:

$𝐾=9\times {10}^9{\text{N m}}^2/{\text{C}}^2$

${𝑞}_1=6𝜇\text{C}=6\times {10}^{-6}\text{C}$

${𝑞}_2=4𝜇\text{C}=4\times {10}^{-6}\text{C}$

$𝑥=0.02\text{m}$

$𝑑=0.1\text{m}$

Performing the calculation:

$\Delta 𝑈=\frac{(9\times {10}^9)\times (6\times {10}^{-6})\times (4\times {10}^{-6})\times 0.02}{0.1\times 0.08}$

Simplifying:

$\Delta 𝑈=\frac{4.32\times {10}^{-3}}{0.008}=0.54\text{J}$

Therefore, the increase in potential energy of the system is $0.54\text{J}$.

To understand how the phase difference between two light waves is determined, consider the following:

Initial Conditions: Two light rays with the same wavelength, denoted as 'λ', are initially in phase.

Traveling through Different Media:

The first ray travels a distance 'L₁' through a medium with a refractive index 'μ₁'.

The second ray travels a distance 'L₂' through a medium with a refractive index 'μ₂'.

Optical Path Difference: When light travels through a medium, the distance is effectively multiplied by the medium's refractive index to obtain the optical path length. Hence, the optical path for the first ray is 'μ₁L₁', and for the second ray, it is 'μ₂L₂'.

Phase Difference Formula:

$\text{Optical Path Difference}={𝜇}_1{𝐿}_1-{𝜇}_2{𝐿}_2$

The phase difference (Δφ) between the two waves is directly related to their optical path difference and is given by:

$\Delta 𝜙=\frac{2𝜋}{𝜆}\times \left(\text{Optical Path Difference}\right)$

Simplifying, we have:

$\Delta 𝜙=\frac{2𝜋}{𝜆}[{𝜇}_1{𝐿}_1-{𝜇}_2{𝐿}_2]$

This formula helps determine how the difference in paths affects the phase relationship of the two waves when they recombine, thus influencing interference patterns.

Let's break down the problem step by step:

The phrase "distance between five consecutive crests" means the distance covering four wavelengths (since there are four intervals between five points). Thus, if we denote the wavelength by $𝜆$, then:

$4𝜆=0.8\text{m}$

Solve for $𝜆$:

$𝜆=\frac{0.8\text{m}}{4}=0.2\text{m}$

The relationship between wave velocity, frequency, and wavelength is given by:

$𝑣=𝑓\times 𝜆$

where

$𝑣$ is the velocity (56 m/s)

$𝜆$ is the wavelength (0.2 m)

$𝑓$ is the frequency we are finding.

Rearrange the formula to solve for frequency:

$𝑓=\frac{𝑣}{𝜆}$

Substitute the given values:

$𝑓=\frac{56\text{m/s}}{0.2\text{m}}=280\text{Hz}$

So, the frequency of the tuning fork is $280\text{Hz}$, which corresponds to Option B.