MHT CET 2024 3rd May Evening Shift (Physics & Chemistry) Question Paper with Solutions

To analyze how the temperature of a gas changes during expansion, given that the pressure $𝑃$ and volume $𝑉$ satisfy the condition ${\mathrm{PV}}^2=\text{constant}$, we start with the ideal gas equation:

$\mathrm{PV}=\mathrm{nRT}$

Using the given condition:

${\mathrm{PV}}^2=\text{constant}$

From the ideal gas equation, we can express $𝑃$ as:

$𝑃=\frac{𝑛𝑅𝑇}{𝑉}$

Substituting this into the given condition:

$\left(\frac{𝑛𝑅𝑇}{𝑉}\right){𝑉}^2=\text{constant}$

This simplifies to:

$𝑛𝑅𝑇\cdot 𝑉=\text{constant}$

Therefore, we find that:

$𝑇𝑉=\text{constant}$

From this relationship, we observe that:

$𝑇\propto \frac{1}{𝑉}$

Thus, if the gas expands (meaning $𝑉$ increases), the temperature $𝑇$ must decrease to maintain the equation $𝑇𝑉=\text{constant}$. Therefore, during expansion under this condition, the temperature of the gas decreases.

The magnetic field at the center of a rotating charged ring can be derived using the concept of a current loop. The magnetic field due to a circular loop of current $𝐼$ at its center is given by:

$𝐵=\frac{{𝜇}_0𝐼}{2𝑅}$

where ${𝜇}_0$ is the permeability of free space and $𝑅$ is the radius of the loop.

In this scenario, the ring is carrying a charge $𝑄$ and is rotating at $𝑁$ revolutions per second (r.p.s.). The equivalent current $𝐼$ due to this rotation can be calculated using the relation:

$𝐼=𝑄𝑁$

since in one full rotation, all the charge completes a full cycle, effectively forming a current loop.

Substitute $𝐼=𝑄𝑁$ into the equation for the magnetic field:

$𝐵=\frac{{𝜇}_0𝑄𝑁}{2𝑅}$

Rearranging for the charge $𝑄$:

$𝑄=\frac{2𝑅𝐵}{{𝜇}_0𝑁}$

Thus, the charge on the ring is given by option D:

$\frac{2𝑅𝐵}{{𝜇}_{0}N}$

The root mean square (r.m.s.) velocity of gas molecules is given by the equation:

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

where ${𝑣}_{\text{rms}}$ is the r.m.s. velocity, $𝑅$ is the universal gas constant, $𝑇$ is the temperature in Kelvin, and $𝑀$ is the molar mass of the gas in kg/mol.

Given:

${𝑣}_{\text{rms}}=61\text{m/s}$

$𝑇={27}^{\circ }\text{C}=27+273=300\text{K}$

$\text{R = 8.31 text{J/(mol cdot K)}}$

To find the molar mass $𝑀$, rearrange the equation:

$𝑀=\frac{3𝑅𝑇}{{𝑣}_2^{\text{rms}}}$

Substitute the given values into the equation:

$𝑀=\frac{3\times 8.31\text{J/(mol}\cdot \text{K)}\times 300\text{K}}{(61\text{m/s}{)}^2}$

Calculating:

Square the velocity:

${61}^2=3721{\text{m}}^2/{\text{s}}^2$

Multiply the constants:

$3\times 8.31\times 300=7479\text{J/mol}$

Find $𝑀$:

$𝑀=\frac{7479}{3721}\text{kg/mol}$

Calculate the value:

$𝑀\approx 2.0105\text{kg/mol}$

Since the molecular weight should be expressed in terms of grams per mole for standard usage:

$𝑀\approx 2.0105\times 1000\text{g/mol}=2.0105\text{g/mol}$

Therefore, the molecular weight of the gas is nearly 2. The correct option is:

Option A: 2

$\begin{array}{cc}& \text{ Potential at the centre of hexagon }\\ & \begin{array}{cc}& \text{ Potential at }\\ & {𝑉}_0={𝑉}_{𝐴}+{𝑉}_{𝐵}+{𝑉}_{𝐶}+{𝑉}_{𝐷}+{𝑉}_{𝐸}+{𝑉}_{𝐹}\\ & =\frac{1}{4𝜋{𝜀}_0}[\frac{𝑞}{𝑂𝐴}+\frac{𝑞}{𝑂𝐵}+\frac{𝑞}{𝑂𝐶}+\frac{𝑞}{𝑂𝐷}+\frac{𝑞}{𝑂𝐸}+\frac{𝑞}{𝑂𝐹}]\end{array}\end{array}$

For a hexagon, distance from centre to vertices is equal to length of side.

$\begin{array}{cc}& \therefore \mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{OD}=\mathrm{OE}=\mathrm{OF}=10\mathrm{cm}\\ & =0.1m\\ & \therefore V_0=\frac{1}{4𝜋{𝜀}_0}\frac{q}{0.1}\left(6\right)\\ & =9\times {10}^9\times \frac{(1\times {10}^{-6})6}{0.1}\\ & =5.4\times {10}^5\text{ volt }\end{array}$

Distance travelled by the wheel in half revolution $=\frac{C}{2}=\frac{2𝜋r}{2}=𝜋r$

Where C is the circumference of the wheel.

$\therefore$ From figure,

Displacement of initial point of contact after half revolution $=\mathrm{AB}$

$\begin{array}{cc}& \therefore {\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{DB}}^2\\ & {\mathrm{AB}}^2=(𝜋𝑟{)}^2+(2r{)}^2=r^2({𝜋}^2+4)\\ & \therefore \mathrm{AB}=r\sqrt{\surd ({𝜋}^2+4)}\\ & \therefore \mathrm{AB}=\sqrt{\surd ({𝜋}^2+4)}\text{.... (given)}\end{array}$

Let the shortest resonating length be denoted as ${𝑙}_1=15.0\text{cm}$.

The end correction is $𝑒=1.0\text{cm}$.

We can express the shortest resonating length with the tuning fork as:

${𝑙}_1+𝑒=\frac{𝜆}{4}$

where $𝜆$ is the wavelength.

For the next higher resonating length, it will correspond to $\frac{3𝜆}{4}$, so:

${𝑙}_2+𝑒=\frac{3𝜆}{4}$

To find the ratio of these expressions:

$\begin{array}{cc}\therefore & \frac{{𝑙}_1+𝑒}{{𝑙}_2+𝑒}=\frac{1}{3}\\ \therefore & \frac{15+1}{{𝑙}_2+1}=\frac{1}{3}\\ \therefore & {𝑙}_2+1=48\\ \therefore & {𝑙}_2=47\text{cm}\end{array}$

Thus, the next resonating length is $47\text{cm}$.

When 80 V dc is applied,

$\begin{array}{cc}& 𝑍=\sqrt{\surd {𝑅}^2+(𝜔𝐿{)}^2}=𝑅\dots (\because 𝜔=0\text{ in dc circuit })\\ & 𝑍=𝑅=\frac{𝑉}{𝐼}=\frac{80}{0.8}=100\Omega \dots \text{ (i) }\end{array}$

When 80 V ac is applied,

$\begin{array}{cc}& 𝑍=\frac{𝑉}{𝐼}=\frac{100}{0.5}=200\Omega \text{.... (ii)}\\ & 𝑍=\sqrt{\surd {𝑅}^2+(𝜔𝐿{)}^2}\\ & \Rightarrow (2𝜋𝑓𝐿{)}^2={𝑍}^2-{𝑅}^2\\ & 𝐿=\sqrt{\surd \frac{{200}^2-{100}^2}{(2𝜋\times 50{)}^2}}=0.55H\text{.... [From (i) and (ii)]}\end{array}$

The angular momentum of an object rotating about an axis is given by the formula:

$𝐿=𝐼𝜔$

where $𝐿$ is the angular momentum, $𝐼$ is the moment of inertia of the object, and $𝜔$ is the angular velocity.

For a solid sphere, the moment of inertia $𝐼$ about an axis through its center is:

$𝐼=\frac{2}{5}𝑀{𝑅}^2$

where $𝑀$ is the mass and $𝑅$ is the radius of the sphere.

The angular velocity $𝜔$ can be expressed in terms of the period of rotation $𝑇$ as:

$𝜔=\frac{2𝜋}{𝑇}$

Plugging these expressions into the angular momentum formula, we have:

$𝐿=\left(\frac{2}{5}𝑀{𝑅}^2\right)\left(\frac{2𝜋}{𝑇}\right)$

Simplifying, we get:

$𝐿=\frac{4𝜋𝑀{𝑅}^2}{5𝑇}$

Therefore, the angular momentum of the Earth about its axis of rotation is given by:

Option B:

$\frac{4𝜋{\mathrm{MR}}^2}{5T}$

Let ${𝐼}_{𝑔}$ be the current capacity of the galvanometer without any shunt. When a shunt resistance $𝑆$ is connected, the current capacity increases to $𝑛{𝐼}_{𝑔}$. The fraction of current passing through the galvanometer is given by:

$\frac{{𝐼}_{𝑔}}{𝑛{𝐼}_{𝑔}}=\frac{𝑆}{𝐺+𝑆}$

where $𝐺$ is the galvanometer resistance. This simplifies to:

$\frac{1}{𝑛}=\frac{𝑆}{𝐺+𝑆}$

Solving for $𝐺$:

$𝐺+𝑆=𝑛𝑆$

$𝐺=𝑛𝑆-𝑆=𝑆(𝑛-1)$

Now, if the galvanometer is shunted by another resistance ${𝑆}^{\prime }$, the current capacity increases to ${𝑛}^{\prime }{𝐼}_{𝑔}$. Using the same principle:

$\frac{{𝐼}_{𝑔}}{{𝑛}^{\prime }{𝐼}_{𝑔}}=\frac{{𝑆}^{\prime }}{𝐺+{𝑆}^{\prime }}$

$\frac{1}{{𝑛}^{\prime }}=\frac{{𝑆}^{\prime }}{𝐺+{𝑆}^{\prime }}$

Substitute the expression for $𝐺$ we derived earlier:

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

Solving for ${𝑛}^{\prime }$:

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

Therefore, the value of ${𝑛}^{\prime }$ in terms of $𝑛$, $𝑆$, and ${𝑆}^{\prime }$ is:

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

Thus, the correct option is D.

Considering $m_1$ and $m_2$ together as system. The only external force acting on the system is

$\begin{array}{cc}& F_{\mathrm{ext}}=(m_1+m_2)g\\ & \frac{\Delta 𝑃}{\Delta 𝑡}=\frac{[({𝑚}_1\overline{{𝑉}_1}+{𝑚}_2\overline{{𝑉}_{\prime }^2})-({𝑚}_1\overline{{𝑉}_1}+{𝑚}_2\overline{{𝑉}_2})]}{2t-0}\\ & \because F_{\mathrm{ext}}=\frac{\Delta P}{\Delta t}\\ & \therefore (m_1+m_2)g=\frac{[(m_1\overline{V_{\prime }^1}+m_2\overline{V_{\prime }^2})-(m_1{\stackrel{\to \leftarrow }{V}}_1+m_2\overline{V_2})]}{2t}\\ & \therefore [(m_1\overline{V_{\prime }^1}+m_2\overline{V_{\prime }^2})-(m_1{\bar{V}}_1+m_2\overline{V_2})]=2(m_1+m_2)\mathrm{gt}\end{array}$

The magnetic susceptibility, denoted as ${𝜒}_{𝑚}$, is related to the relative permeability ${𝜇}_{𝑟}$ by the equation:

${𝜇}_{𝑟}=1+{𝜒}_{𝑚}$

Given that the magnetic susceptibility ${𝜒}_{𝑚}$ of the material is 599, the relative permeability ${𝜇}_{𝑟}$ can be calculated as:

${𝜇}_{𝑟}=1+599=600$

The absolute permeability $𝜇$ is then given by the product of the relative permeability and the permeability of free space (${𝜇}_0$):

$𝜇={𝜇}_{𝑟}\cdot {𝜇}_0$

Substituting the given values:

$𝜇=600\cdot 4𝜋\times {10}^{-7}$

Calculating this gives:

$𝜇=2400𝜋\times {10}^{-7}$

Which simplifies to:

$𝜇=2.4𝜋\times {10}^{-4}$

Thus, the absolute permeability of the material of the rod is $2.4𝜋\times {10}^{-4}$, which corresponds to Option B.

For an open organ pipe, the effective length of the pipe, taking end correction into account, is calculated as follows:

$𝐿=𝑙+2𝑒=𝑙+2\times 0.6𝑟$

Since $𝑒=0.6𝑟$ for an open pipe, this simplifies to:

$𝐿=𝑙+1.2𝑟$

Therefore, the fundamental frequency of the open pipe is expressed as:

$𝑓=\frac{𝑣}{2𝐿}=\frac{𝑣}{2(𝑙+1.2𝑟)}$

Here, $𝑣$ is the velocity of sound in air, $𝑙$ is the length of the pipe, and $𝑟$ is the internal radius of the pipe.

The magnetic force acting on a charged particle moving through a magnetic field is given by the expression:

$𝐅=𝑞(𝐯\times 𝐁)$

where:

$𝐅$ is the magnetic force,

$𝑞$ is the charge of the particle,

$𝐯$ is the velocity vector of the particle,

$𝐁$ is the magnetic field vector,

$\times$ denotes the cross product.

When a charged particle moves along a magnetic field line, the velocity of the particle ($𝐯$) is parallel to the magnetic field ($𝐁$). In this scenario, the angle $𝜃$ between the velocity vector and the magnetic field vector is $0^{\circ }$. Therefore, the magnitude of the magnetic force can be computed as:

$𝐹=𝑞𝑣𝐵\sin 𝜃=𝑞𝑣𝐵\sin 0^{\circ }=𝑞𝑣𝐵\times 0=0$

Thus, the magnetic force acting on the particle is zero. The correct answer is:

Option C: Zero.

For a diatomic gas undergoing an adiabatic process, the relationship between pressure ($𝑃$) and temperature ($𝑇$) can be derived using the adiabatic condition and the gas law.

The adiabatic condition for an ideal gas is given by:

$𝑃{𝑉}^{𝛾}=\text{constant}$

where $𝑃$ is pressure, $𝑉$ is volume, and $𝛾$ (gamma) is the adiabatic index or heat capacity ratio given by:

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}$

For a diatomic gas, the degrees of freedom $𝑓$ are 5, so:

${𝐶}_{𝑣}=\frac{𝑓}{2}𝑅=\frac{5}{2}𝑅$

${𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅=\frac{7}{2}𝑅$

Thus,

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

The ideal gas law is given by:

$𝑃𝑉=𝑛𝑅𝑇$

During an adiabatic process, we also have:

$𝑃\propto {𝑉}^{-𝛾}$

and

$𝑇\propto {𝑉}^{𝛾-1}$

Substituting the relationship for temperature:

From the ideal gas law $𝑃𝑉=𝑛𝑅𝑇$, we have $𝑃=\frac{𝑛𝑅𝑇}{𝑉}$.

We can substitute $𝑇\propto {𝑉}^{𝛾-1}$ into the ideal gas law:

$𝑃\propto {\left({𝑉}^{𝛾-1}\right)}^{\frac{1}{𝑉}}\Rightarrow 𝑃\propto \left({𝑉}^{1-𝛾}\right)$

Hence, $𝑃\propto {𝑇}^{\frac{𝛾}{𝛾-1}}$.

Plugging in the value of γ for a diatomic gas:

$𝛾=\frac{7}{5}$

$\frac{𝛾}{𝛾-1}=\frac{\frac{7}{5}}{\frac{7}{5}-1}=\frac{\frac{7}{5}}{\frac{2}{5}}=\frac{7}{2}$

Thus, for a diatomic gas, the relationship is:

$𝑃\propto {𝑇}^{\frac{7}{2}}$

Therefore, the value of $𝑥$ is $3.5$.

Answer: Option A, $𝑥=3.5$.

To determine the height at which the weight of a body becomes $\frac{1}{16}$ of its weight on the surface of the Earth, we start by using the formula for gravitational force:

$𝐹=\frac{𝐺𝑀𝑚}{{𝑟}^2}$

Where:

$𝐹$ is the gravitational force (weight),

$𝐺$ is the gravitational constant,

$𝑀$ is the mass of the Earth,

$𝑚$ is the mass of the body,

$𝑟$ is the distance from the center of the Earth.

On the surface of the Earth, the weight ${𝑊}_0$ is given by:

${𝑊}_0=\frac{𝐺𝑀𝑚}{{𝑅}^2}$

At a height $ℎ$ above the Earth's surface, the new distance from the center of the Earth becomes $𝑅+ℎ$. The weight at this height, ${𝑊}_{ℎ}$, is:

${𝑊}_{ℎ}=\frac{𝐺𝑀𝑚}{(𝑅+ℎ{)}^2}$

According to the problem, ${𝑊}_{ℎ}=\frac{1}{16}{𝑊}_0$. Plugging in the expressions for ${𝑊}_0$ and ${𝑊}_{ℎ}$:

$\frac{𝐺𝑀𝑚}{(𝑅+ℎ{)}^2}=\frac{1}{16}\frac{𝐺𝑀𝑚}{{𝑅}^2}$

Canceling $𝐺𝑀𝑚$ from both sides and solving for $ℎ$:

$\frac{1}{(𝑅+ℎ{)}^2}=\frac{1}{16{𝑅}^2}$

Taking the reciprocal gives:

$(𝑅+ℎ{)}^2=16{𝑅}^2$

Taking the square root of both sides:

$𝑅+ℎ=4𝑅$

Solving for $ℎ$:

$ℎ=4𝑅-𝑅=3𝑅$

Thus, the height at which the weight becomes $\frac{1}{16}$ of its weight on the surface is:

Option B: 3R

Number of turns of in primary and secondary coil be $N_p$ and $N_s$,

For ideal transformer,

$\begin{array}{cc}& \frac{{𝑁}_{𝑠}}{{𝑁}_{𝑝}}=\frac{{𝑉}_{𝑠}}{{𝑉}_{𝑝}}\\ & {𝑁}_{𝑠}=\frac{4.4\times {10}^3}{220}\times 1000=2\times {10}^4\end{array}$

Power supply at primary coil is given as,

$\begin{array}{cc}& 𝑃={𝐼}_{𝑝}{𝑉}_{𝑝}=6.6\times {10}^{3}V\\ & {𝐼}_{𝑝}=\frac{6.6\times {10}^3𝑉}{220}=30A\end{array}$

For ideal transformer,

$\begin{array}{cc}& \frac{{𝐼}_{𝑠}}{{𝐼}_{𝑝}}=\frac{{𝑁}_{𝑝}}{{𝑁}_{𝑠}}\\ & {𝐼}_{𝑠}=\frac{{10}^3}{2\times {10}^4}\times 30=1.5A\end{array}$

For a monoatomic gas, when heated at constant pressure, the heat added to the system can be used to do work and to change the internal energy.

The specific heat at constant pressure is related to the specific heat at constant volume by:

${𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅$

where $𝑅$ is the ideal gas constant. For a monoatomic gas, ${𝐶}_{𝑣}=\frac{3}{2}𝑅$, so:

${𝐶}_{𝑝}=\frac{3}{2}𝑅+𝑅=\frac{5}{2}𝑅$

The work done by the gas during expansion at constant pressure can be represented by:

$𝑊=𝑃\Delta 𝑉=𝑛𝑅\Delta 𝑇$

The total heat supplied at constant pressure is:

$𝑄=𝑛{𝐶}_{𝑝}\Delta 𝑇=𝑛\left(\frac{5}{2}𝑅\right)\Delta 𝑇$

The percentage of heat used to do work is given by the ratio:

$\text{Percentage of heat used for work}=\frac{𝑊}{𝑄}\times 100=\frac{𝑛𝑅\Delta 𝑇}{𝑛\left(\frac{5}{2}𝑅\right)\Delta 𝑇}\times 100$

This simplifies to:

$\frac{2}{5}\times 100=40\%$

Thus, the percentage of total heat used for doing external work is 40%.

Option B (40%) is the correct answer.

Option B is NOT correct.

In a parallel plate capacitor, the charge $𝑄$ remains constant after it is isolated, as stated in Option A. When the separation $𝑑$ between the plates is increased, the capacitance $𝐶$ of the capacitor is given by the formula:

$𝐶=\frac{𝜀𝐴}{𝑑}$

where $𝜀$ is the permittivity of the dielectric material between the plates, and $𝐴$ is the area of the plates. Since $𝐶$ is inversely proportional to $𝑑$, increasing the separation decreases the capacitance, as described in Option D.

The potential difference $𝑉$ across the plates is related to charge and capacitance by

$𝑉=\frac{𝑄}{𝐶}$

Since the charge $𝑄$ remains constant and the capacitance $𝐶$ decreases when the separation is increased, the potential difference $𝑉$ must increase, which aligns with Option C. Therefore, the statement that the potential difference decreases (Option B) is incorrect.

The radius of the $𝑛$-th Bohr orbit for an electron in a hydrogen atom is given by:

${𝑟}_{𝑛}={𝑛}^2\cdot {𝑟}_1$

where ${𝑟}_1$ is the radius of the first Bohr orbit (also known as the Bohr radius).

For the first orbit ($𝑛=1$):

${𝑟}_1=1^2\cdot {𝑟}_1={𝑟}_1$

For the second orbit ($𝑛=2$):

${𝑟}_2=2^2\cdot {𝑟}_1=4{𝑟}_1$

The ratio of the radius of the first Bohr orbit to that of the second Bohr orbit is:

$\frac{{𝑟}_1}{{𝑟}_2}=\frac{{𝑟}_1}{4{𝑟}_1}=\frac{1}{4}$

Thus, the correct answer is:

Option C

$1:4$

$i_c={\sin }^{-1}\left(\frac{1}{n}\right)\text{ and }n\propto \frac{1}{𝜆}$

Violet, indigo, blue have smaller wavelength than green, so n will be more for these rays, consequently critical angle for these rays will be low, hence if green is just totally internally reflected then red, orange and yellow rays will emerge out and violet, indigo and blue rays will be reflected inside.

The moment of inertia of a loop made from uniform wire about its center is given by:

$𝐼=𝑚{𝑅}^2$

where $𝑚$ is the mass of the wire, and $𝑅$ is the radius of the loop.

Assuming that the wire is uniform, both loops $𝐴$ and $𝐵$ will have the same linear mass density. This means the mass of a loop is proportional to its circumference:

$𝑚=𝜆\cdot 2𝜋𝑅$

where $𝜆$ is the linear mass density.

Therefore, the moments of inertia for loops $𝐴$ and $𝐵$ are:

${𝐼}_{𝐴}=(𝜆\cdot 2𝜋{𝑅}_1){𝑅}_2^1=𝜆\cdot 2𝜋{𝑅}_3^1$

${𝐼}_{𝐵}=(𝜆\cdot 2𝜋{𝑅}_2){𝑅}_2^2=𝜆\cdot 2𝜋{𝑅}_3^2$

Given that $\frac{{𝐼}_{𝐴}}{{𝐼}_{𝐵}}=27$, we have:

$\frac{𝜆\cdot 2𝜋{𝑅}_3^1}{𝜆\cdot 2𝜋{𝑅}_3^2}=27$

This simplifies to:

$\frac{{𝑅}_3^1}{{𝑅}_3^2}=27$

Taking the cube root of both sides:

$\frac{{𝑅}_1}{{𝑅}_2}=3$

Thus, the ratio ${𝑅}_2/{𝑅}_1$ is:

$\frac{{𝑅}_2}{{𝑅}_1}=\frac{1}{3}$

Hence, the correct option is:

Option C $1:3$

In a series LCR circuit at resonance, the impedance is purely resistive, and the resonance condition is achieved when the reactive components of the impedance, i.e., the inductive reactance (${𝑋}_{𝐿}=𝜔𝐿$) and capacitive reactance (${𝑋}_{𝐶}=\frac{1}{𝜔𝐶}$), are equal.

At resonance, the voltage across the inductor can be related to the voltage across the resistor using their reactance and resistance. The voltage across the inductor ${𝑉}_{𝐿}$ can be expressed as:

${𝑉}_{𝐿}=𝐼\cdot 𝜔𝐿$

And the voltage across the resistor ${𝑉}_{𝑅}$ is:

${𝑉}_{𝑅}=𝐼\cdot 𝑅$

where $𝐼$ is the current through the circuit.

Since at resonance the same current $𝐼$ flows through both the resistor and the inductor:

$𝐼=\frac{{𝑉}_{𝑅}}{𝑅}$

Substitute $𝐼$ into the expression for ${𝑉}_{𝐿}$:

${𝑉}_{𝐿}=\frac{{𝑉}_{𝑅}}{𝑅}\cdot 𝜔𝐿$

From this, rearrange to find the inductance $𝐿$:

$𝐿=\frac{{𝑉}_{𝐿}𝑅}{{𝑉}_{𝑅}𝜔}$

Thus, the correct option is Option D:

$\frac{V_{L}R}{V_R𝜔}$

The period of a simple pendulum is given by the formula:

$𝑇=2𝜋\sqrt{\surd \frac{𝐿}{𝑔}}$

where:

$𝑇$ is the period of the pendulum,

$𝐿$ is the length of the pendulum,

$𝑔$ is the acceleration due to gravity.

To determine when the period $𝑇$ gets doubled, consider the scenario where ${𝑇}^{\prime }=2𝑇$. Using the period formula, we have:

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{{𝐿}^{\prime }}{𝑔}}=2\cdot 2𝜋\sqrt{\surd \frac{𝐿}{𝑔}}$

This simplifies to:

$\sqrt{\surd \frac{{𝐿}^{\prime }}{𝑔}}=2\sqrt{\surd \frac{𝐿}{𝑔}}$

Squaring both sides gives:

$\frac{{𝐿}^{\prime }}{𝑔}=4\cdot \frac{𝐿}{𝑔}$

which implies:

${𝐿}^{\prime }=4𝐿$

Therefore, the period of a simple pendulum gets doubled when its length is made four times. The correct option is Option B: its length is made four times.

The Boolean expression for an XOR (exclusive OR) gate, where the output is true if and only if exactly one of the inputs is true, is given by:

$𝐶=(𝐴\cdot \overline{𝐵})+(\overline{𝐴}\cdot 𝐵)$

This expression represents the XOR operation between two variables $𝐴$ and $𝐵$. It is equivalent to saying that $𝐶$ is true when either $𝐴$ is true and $𝐵$ is false, or $𝐴$ is false and $𝐵$ is true.

Thus, the correct option corresponding to this expression is:

Option D

$(\bar{A}\cdot B)+(A\cdot \bar{B})$

Force due to gravity,

$\begin{array}{cc}& F=\frac{\mathrm{GMM}}{(2R{)}^2}\\ & F=\frac{{\mathrm{GM}}^2}{4R^2}\text{.... (i)}\end{array}$

Density, $𝜌=\frac{𝑀}{𝑉}$

$𝑀=𝜌\times \frac{4}{3}𝜋{𝑅}^3$

Substituting in (i),

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

$\therefore 𝐹\propto {𝑅}^4{𝜌}^2$

The frequency of a vibrating string is given by the formula:

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

where:

$𝑛$ is the frequency,

$𝐿$ is the length of the string,

$𝑇$ is the tension in the string,

$𝜇$ is the mass per unit length of the string.

Given that the initial frequency is ${𝑛}_1$ with tension $𝑇$, when the tension is increased by $44\%$, the new tension becomes:

${𝑇}^{\prime }=𝑇+0.44𝑇=1.44𝑇$

The new frequency ${𝑛}_2$ when the tension is ${𝑇}^{\prime }$ is:

${𝑛}_2=\frac{1}{2𝐿}\sqrt{\surd \frac{1.44𝑇}{𝜇}}$

We can simplify this to:

${𝑛}_2=\frac{1}{2𝐿}\cdot \sqrt{\surd \frac{1.44𝑇}{𝜇}}=\frac{1}{2𝐿}\cdot \sqrt{\surd 1.44}\cdot \sqrt{\surd \frac{𝑇}{𝜇}}$

Recognizing that $\sqrt{\surd 1.44}=1.2$, we have:

${𝑛}_2=1.2\cdot \frac{1}{2𝐿}\cdot \sqrt{\surd \frac{𝑇}{𝜇}}$

Since we know ${𝑛}_1=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}$, we can substitute:

${𝑛}_2=1.2\cdot {𝑛}_1$

The ratio of ${𝑛}_2$ to ${𝑛}_1$ is:

$\frac{{𝑛}_2}{{𝑛}_1}=1.2$

Expressing $1.2$ as a ratio, we have:

$\frac{{𝑛}_2}{{𝑛}_1}=\frac{6}{5}$

Therefore, the correct answer is:

Option C: $6:5$

$\begin{array}{cc}& \text{ We know }\\ & 𝜙=ℎ{𝑣}_0\Rightarrow 𝜙\propto {𝑣}_0\\ & \text{ Also, }\\ & {𝑣}_0<{𝑣}_{\prime }^0\text{... (From graph)}\\ \therefore & {𝜙}_{𝑥}<{𝜙}_{𝑦}\end{array}$

The terminal speed of a sphere moving through a viscous fluid can be described by Stokes' Law, which relates terminal speed to the radius of the sphere. The terminal velocity ${𝑣}_{𝑡}$ is given by:

${𝑣}_{𝑡}=\frac{2}{9}\cdot \frac{{𝑟}^2(𝜌-𝜎)𝑔}{𝜂}$

where:

$𝑟$ is the radius of the ball,

$𝜌$ is the density of the ball,

$𝜎$ is the density of the fluid,

$𝑔$ is the acceleration due to gravity,

$𝜂$ is the viscosity of the liquid.

The terminal speed is directly proportional to the square of the radius:

${𝑣}_{𝑡}\propto {𝑟}^2$

For a ball with a radius of 6 mm, the terminal speed is $12\text{cm/s}$. Let the terminal speed for a ball of radius 3 mm be ${𝑣}_{\prime }^{𝑡}$.

Using the proportionality:

$\frac{{𝑣}_{\prime }^{𝑡}}{{𝑣}_{𝑡}}={\left(\frac{{𝑟}^{\prime }}{𝑟}\right)}^2$

Substitute the known values:

$\frac{{𝑣}_{\prime }^{𝑡}}{12}={\left(\frac{3}{6}\right)}^2$

This simplifies to:

$\frac{{𝑣}_{\prime }^{𝑡}}{12}={\left(\frac{1}{2}\right)}^2$

$\frac{{𝑣}_{\prime }^{𝑡}}{12}=\frac{1}{4}$

Solving for ${𝑣}_{\prime }^{𝑡}$:

${𝑣}_{\prime }^{𝑡}=12\cdot \frac{1}{4}$

${𝑣}_{\prime }^{𝑡}=3\text{cm/s}$

Therefore, the terminal speed of the steel ball with a radius of 3 mm is $\mathrm{𝟑}\text{cm/s}$, which corresponds to option D.

When analyzing the scenario of two coils placed near each other, the mutual inductance ($𝑀$) plays a key role in determining the electromotive force (e.m.f.) induced in one coil due to a change in current of the other. The e.m.f. ($𝜀$) induced due to the rate of change of current in the second coil is given by:

$𝜀=-𝑀\frac{𝑑{𝐼}_2}{𝑑𝑡}$

Given that $𝜀=20\text{mV}=20\times {10}^{-3}\text{V}$ and the rate of change of current in the second coil $\frac{𝑑{𝐼}_2}{𝑑𝑡}=10\text{A/s}$, we can rearrange the formula to solve for the mutual inductance $𝑀$:

$𝑀=-\frac{𝜀}{\frac{𝑑{𝐼}_2}{𝑑𝑡}}=-\frac{20\times {10}^{-3}}{10}=2\times {10}^{-3}\text{H}$

Now, consider the situation where no current passes through the second coil, and a current of 3.6 A passes through the first coil. The flux linkage (${\Phi }_{21}$) in the second coil due to the current in the first coil can be found using:

${\Phi }_{21}=𝑀{𝐼}_1$

Substitute $𝑀=2\times {10}^{-3}\text{H}$ and ${𝐼}_1=3.6\text{A}$:

${\Phi }_{21}=(2\times {10}^{-3}\text{H})\left(3.6\text{A}\right)=7.2\times {10}^{-3}\text{Wb}$

Therefore, the correct answer is:

Option D: $7.2\times {10}^{-3}\text{Wb}$

When two rods of different materials are connected and increase by the same amount in length due to temperature change, the expansion can be described as follows:

The change in length for the copper rod is given by:

$\Delta {𝐿}_1={𝐿}_1{𝛼}_{𝑐}𝑡$

The change in length for the iron rod is:

$\Delta {𝐿}_2={𝐿}_2{𝛼}_{𝑖}𝑡$

Given that these changes in lengths are equal:

${𝐿}_1{𝛼}_{𝑐}𝑡={𝐿}_2{𝛼}_{𝑖}𝑡$

By simplifying, we have:

${𝐿}_1{𝛼}_{𝑐}={𝐿}_2{𝛼}_{𝑖}$

From this, we can write:

$\frac{{𝐿}_1}{{𝐿}_2}=\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}$

We need to find the ratio:

$\frac{{𝐿}_1-{𝐿}_2}{{𝐿}_1+{𝐿}_2}$

Start by expressing ${𝐿}_1$ and ${𝐿}_2$ in terms of a common variable:

Let ${𝐿}_2=𝑥$, then ${𝐿}_1=\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}𝑥$.

Substitute these into $\frac{{𝐿}_1-{𝐿}_2}{{𝐿}_1+{𝐿}_2}$:

$=\frac{\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}𝑥-𝑥}{\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}𝑥+𝑥}$

Factor out $𝑥$ from the numerator and the denominator:

$=\frac{𝑥(\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}-1)}{𝑥(\frac{{𝛼}_{𝑖}}{{𝛼}_{𝑐}}+1)}$

Cancel out $𝑥$:

$=\frac{\frac{{𝛼}_{𝑖}-{𝛼}_{𝑐}}{{𝛼}_{𝑐}}}{\frac{{𝛼}_{𝑖}+{𝛼}_{𝑐}}{{𝛼}_{𝑐}}}$

Simplify further:

$=\frac{{𝛼}_{𝑖}-{𝛼}_{𝑐}}{{𝛼}_{𝑖}+{𝛼}_{𝑐}}$

Thus, the correct ratio is:

Option C

$\frac{{𝛼}_{𝑖}-{𝛼}_{𝑐}}{{𝛼}_{𝑐}+{𝛼}_{𝑖}}$

To find the maximum speed of the particle in simple harmonic motion (SHM) with frequency $10\text{Hz}$, we will use the relationship between angular frequency, frequency, and maximum speed.

Determine Angular Frequency ($𝜔$):

The angular frequency $𝜔$ is related to the frequency $𝑓$ by the formula:

$𝜔=2𝜋𝑓$

Given $𝑓=10\text{Hz}$, we have:

$𝜔=2𝜋\times 10=20𝜋\text{rad/s}$

Maximum Speed Formula in SHM:

The maximum speed ${𝑣}_{\text{max}}$ of a particle in SHM is given by:

${𝑣}_{\text{max}}=𝐴𝜔$

where $𝐴$ is the amplitude of the oscillation.

Determine Amplitude ($𝐴$):

At the highest point, the spring is unstretched, meaning the elongation due to the particle's weight is equal to the amplitude of oscillation.

Use the equilibrium condition $𝑚𝑔=𝑘𝐴$, and since $𝜔=\sqrt{\surd \frac{𝑘}{𝑚}}$, $𝑘=𝑚{𝜔}^2$.

Therefore, at maximum elongation, the net force provides the centripetal force for the circular motion:

$𝑚𝑔=𝑚{𝜔}^2𝐴$

Consequently, $𝐴=\frac{𝑔}{{𝜔}^2}$.

Substitute Values to Find ${𝑣}_{\text{max}}$:

Substituting for amplitude:

$𝐴=\frac{10}{(20𝜋{)}^2}=\frac{10}{400{𝜋}^2}=\frac{1}{40{𝜋}^2}$

Now, substitute $𝐴$ and $𝜔$ into ${𝑣}_{\text{max}}=𝐴𝜔$:

${𝑣}_{\text{max}}=\left(\frac{1}{40{𝜋}^2}\right)\times 20𝜋=\frac{20𝜋}{40{𝜋}^2}$