MHT CET 2024 16th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

In magnetism:

Diamagnetic materials have a negative magnetic susceptibility ($𝜒<0$).

Paramagnetic and ferromagnetic materials both have a positive magnetic susceptibility ($𝜒>0$), with ferromagnets having a much larger (and nonlinear) susceptibility.

Hence, the materials with negative magnetic susceptibility are diamagnetic.

Answer: (C) diamagnetic.

The surface tension of a liquid generally decreases with an increase in temperature.

As the temperature rises, the kinetic energy of the molecules also increases. This increased molecular motion results in a reduction in the cohesive forces between the molecules at the surface of the liquid. Consequently, the surface tension, which is a measure of these cohesive forces, decreases.

Option C: decreases.

When two capacitors, A and B, are connected in series, they will have the same charge initially due to the series configuration. Consider the capacitors connected to a battery of voltage $𝐸$.

Initial Configuration with Dielectric in Capacitor B:

The capacitance of capacitor B with a dielectric is ${𝐶}_{\prime }^{𝐵}=𝐾𝐶$, where $𝐶$ is the capacitance of capacitor A, and $𝐾$ is the dielectric constant.

For capacitors in series,

$\frac{1}{{𝐶}_{\text{eq}}}=\frac{1}{𝐶}+\frac{1}{𝐾𝐶}=\frac{𝐾+1}{𝐾𝐶}.$

${𝐶}_{\text{eq}}=\frac{𝐾𝐶}{𝐾+1}.$

The charge on the capacitors in series is given by:

$𝑄={𝐶}_{\text{eq}}\cdot 𝐸=\frac{𝐾𝐶}{𝐾+1}\cdot 𝐸.$

Charges on Each Capacitor:

Since ${𝑄}_{𝐴}={𝑄}_{𝐵}=𝑄$ (initially when the dielectric is present in B),

${𝑄}_{𝐴}={𝑄}_{𝐵}=\frac{𝐾𝐶}{𝐾+1}\cdot 𝐸.$

After Removing the Dielectric from Capacitor B:

Without the dielectric, the capacitance of B becomes $𝐶$. The equivalent capacitance is now:

$\frac{1}{{𝐶}_{\prime }^{\text{eq}}}=\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{2}{𝐶}.$

${𝐶}_{\prime }^{\text{eq}}=\frac{𝐶}{2}.$

The charge on the new configuration is:

${𝑄}^{\prime }={𝐶}_{\prime }^{\text{eq}}\cdot 𝐸=\frac{𝐶}{2}\cdot 𝐸.$

Since the capacitors are still in series, the charges on A and B are equal:

${𝑄}_{\prime }^{𝐴}={𝑄}_{\prime }^{𝐵}={𝑄}^{\prime }.$

Comparing the Charges:

For capacitor A:

$\frac{{𝑄}_{\prime }^{𝐴}}{{𝑄}_{𝐴}}=\frac{{𝑄}^{\prime }}{{𝑄}_{𝐴}}=\frac{\frac{𝐶}{2}\cdot 𝐸}{\frac{𝐾𝐶}{𝐾+1}\cdot 𝐸}=\frac{𝐾+1}{2𝐾}.$

For capacitor B:

$\frac{{𝑄}_{\prime }^{𝐵}}{{𝑄}_{𝐵}}=\frac{\frac{𝐶}{2}\cdot 𝐸}{\frac{𝐾𝐶}{𝐾+1}\cdot 𝐸}=\frac{𝐾+1}{2𝐾}.$

The correct option in this context would be:

Option D:

$\frac{Q_{\prime }^B}{Q_B}=\frac{K+1}{2K}$

Number of protons $=8$

Number of neutrons $=17-8=9$

Binding energy,

$\begin{array}{cc}𝐵𝐸& =({𝑀}_{\text{nucleus }}-{𝑀}_{\text{nucleons }}){𝐶}^2\\ \\ & =({𝑀}_{𝑜}-8{𝑀}_{𝑝}-9{𝑀}_{𝑁}){𝐶}^2\end{array}$

To determine the point on the x-axis where the net electric field due to two charges $+8𝑞$ and $-2𝑞$ is zero, let's denote this point as $𝐴$. Suppose $𝐴$ is located at a distance $𝑟$ from the charge $-2𝑞$ which is at $𝑥=𝐿$, and thus at a distance $𝐿+𝑟$ from the charge $+8𝑞$ located at $𝑥=0$.

To find the condition where the electric fields from both charges cancel each other, set the magnitudes of the electric fields equal to each other:

${𝐸}_1={𝐸}_2$

Given by:

$\frac{𝐾\cdot 8𝑞}{(𝐿+𝑟{)}^2}=\frac{𝐾\cdot 2𝑞}{{𝑟}^2}$

Simplifying, we have:

$\frac{4}{(𝐿+𝑟{)}^2}=\frac{1}{{𝑟}^2}$

Solving the equation, we find:

$2𝑟=𝐿+𝑟$

$𝑟=𝐿$

Therefore, the position on the x-axis from the origin where the electric fields cancel is:

$𝐿+𝑟=𝐿+𝐿=2𝐿$

Hence, the point is located at $2𝐿$ from the origin.

For an electron in the hydrogen atom, according to Bohr’s model,

$2𝜋{𝑟}_{𝑛}=𝑛𝜆$

where

• ${𝑟}_{𝑛}$ = radius of the ${𝑛}^{\text{th}}$ orbit

• $𝜆$ = de-Broglie wavelength

• $𝑛$ = principal quantum number

Also, for hydrogen atom,

${𝑟}_{𝑛}\propto {𝑛}^2$

So,

$𝜆=\frac{2𝜋{𝑟}_{𝑛}}{𝑛}$

Since ${𝑟}_{𝑛}\propto {𝑛}^2$,

$𝜆\propto \frac{{𝑛}^2}{𝑛}=𝑛$

This means the de-Broglie wavelength increases with increase in orbit number $𝑛$.

When the electron jumps from ground state ($𝑛=1$) to a higher excited state ($𝑛=2,3,\dots$), its de-Broglie wavelength increases.

Therefore, the correct option is:

$\overline{)\text{Option D: will increase}}$

Flux, $𝜙=$ LI

Comparing the above equation with $y=\mathrm{mx}$, $L=𝜙/I$, which is the slope of the graph.

Since line S has least slope, it indicates the smallest value of self-inductance.

The magnetic induction along the axis of a toroidal solenoid can be described by the formula:

$𝐵={𝜇}_0\cdot 𝑛\cdot 𝐼\text{where}𝑛=\frac{𝑁}{2𝜋𝑟}$

In this equation:

${𝜇}_0$ represents the permeability of the medium.

$𝑛$ is the number of turns per unit length.

$𝐼$ is the current passing through the solenoid.

$𝑁$ is the total number of turns.

$𝑟$ is the radius of the toroidal solenoid.

From this relationship, it's evident that the magnetic induction $𝐵$ does not depend on the radius $𝑟$ of the toroidal solenoid.

To determine the equivalent optical path in another medium, we can use the concept that the optical paths are equal when the product of the refractive index and the distance traveled in both media are the same.

Given:

Refractive index of flint glass, ${𝜇}_g=1.6$

Distance traveled in flint glass, $x_g=3\text{ cm}$

Refractive index of the other medium, ${𝜇}_m=1.25$

Distance traveled in the other medium, $x_m=𝑥\text{ cm}$

The relationship for equal optical paths is given by:

${𝜇}_g\cdot x_g={𝜇}_m\cdot x_m$

Solving for $x_m$:

$x_m=\frac{{𝜇}_g\cdot x_g}{{𝜇}_m}=\frac{1.6\times 3}{1.25}=3.84\text{ cm}$

Therefore, the distance 'x' that the light travels in the other medium is 3.84 cm.

To determine the mass of the string, we can follow these steps:

Calculate the Fundamental Frequency:

The fundamental frequency is the difference between the two consecutive resonant frequencies:

$\text{Fundamental frequency}={𝑓}_2-{𝑓}_1=450\text{Hz}-375\text{Hz}=75\text{Hz}$

Use the Formula for Frequency:

The frequency $𝑓$ of a vibrating string is given by:

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

where $𝑇$ is the tension in the string, $𝜇$ is the mass per unit length, and $𝐿$ is the length of the string.

Rewrite the Equation and Solve for $𝐿$:

Plug in the known values:

$75=\frac{1}{2𝐿}\sqrt{\surd \frac{180}{2\times {10}^{-3}}}$

Solving for $𝐿$:

$𝐿=2\text{m}$

Find the Mass ($𝑚$) of the String:

Using the relationship for mass per unit length:

$𝜇=\frac{𝑚}{𝐿}$

Rearrange to find $𝑚$:

$𝑚=𝜇\times 𝐿=(2\times {10}^{-3}\text{Kg/m})\times 2\text{m}=4\times {10}^{-3}\text{Kg}=4\text{grams}$

Therefore, the mass of the string is 4 grams.

Let the radii of the thin spherical shell and the solid sphere be ${𝑅}_1$ and ${𝑅}_2$, respectively. Then, the moment of inertia of the shell about its diameter is given by,

$I_{\text{shell }}=\frac{2}{3}{\mathrm{MR}}_2^1\text{.... (i)}$

And the moment of inertia of the solid sphere is given by,

$I_{\text{sphere }}=\frac{2}{5}{\mathrm{MR}}_2^2\text{...(ii)}$

Given that, the masses and moment of inertia for both the bodies are equal, then from equations (i) and (ii),

$\begin{array}{cc}& \frac{2}{3}{\mathrm{MR}}_2^1=\frac{2}{5}{\mathrm{MR}}_2^2\\ \therefore & \frac{R_2^1}{R_2^2}=\frac{3}{5}\\ \therefore & R_1:R_2=\sqrt{\surd 3}:\sqrt{\surd 5}\end{array}$

The pressure exerted by a gas on the walls of a container is derived from the kinetic theory of gases. It is represented by the equation:

$𝑃=\frac{1}{3}𝜌{𝑣}^2$

where $𝜌$ is the density of the gas, and $𝑣$ is the root mean square (r.m.s.) speed of the gas molecules. We can further express this by noting that density ($𝜌$) is mass ($𝑀$) per unit volume ($𝑉$):

$𝑃=\frac{1}{3}\left(\frac{𝑀}{𝑉}\right){𝑣}^2$

Next, we multiply and divide the equation by $2$, to reframe it in terms of kinetic energy:

$𝑃=\frac{2}{3}\cdot \frac{1}{2}\left(\frac{𝑀}{𝑉}\right){𝑣}^2$

Considering that the kinetic energy ($K.E.$) of the gas per unit volume is given by:

$K.E.=\frac{1}{2}𝑀{𝑣}^2$

The expression for pressure can thus be rewritten as:

$𝑃=\frac{2}{3}\left(\frac{K.E.}{𝑉}\right)$

This equation shows that the pressure ($𝑃$) is $\frac{2}{3}$ of the kinetic energy per unit volume of the gas.

Particle velocities are

$\begin{array}{cc}& {𝑉}_2^1={𝜔}^2({𝐴}^2-{𝑥}_2^1)\text{.... (i)}\\ & {𝑉}_2^2={𝜔}^2({𝐴}^2-{𝑥}_2^2)\text{.... (ii)}\end{array}$

Subtracting (i) from (ii),

$\begin{array}{cc}& {𝑉}_2^2-{𝑉}_2^1={𝜔}^2({𝑥}_2^1-{𝑥}_2^2)\\ & 𝜔=\sqrt{\surd \frac{{𝑉}_2^2-{𝑉}_2^1}{{𝑥}_2^1-{𝑥}_2^2}}\end{array}$

As $𝜔=2𝜋𝑓$ we get,

$f=\frac{1}{2𝜋}\sqrt{\surd \frac{V_2^2-V_2^1}{x_2^1-x_2^2}}$

Snell's law states that:

$𝜇=\frac{\sin 𝑖}{\sin 𝑟}$

where $𝑖$ is the angle of incidence and $𝑟$ is the angle of refraction.

Given that the angle of incidence $𝑖$ is twice the angle of refraction $𝑟$, or $𝑖=2𝑟$, we can express $𝑟$ as:

$𝑟=\frac{𝑖}{2}$

Substituting this into Snell's law gives:

$𝜇=\frac{\sin 𝑖}{\sin \left(\frac{𝑖}{2}\right)}$

Using the trigonometric identity $\sin 2𝐴=2\sin 𝐴\cos 𝐴$, we can express $\sin 𝑖$ as $2\sin \left(\frac{𝑖}{2}\right)\cos \left(\frac{𝑖}{2}\right)$. Thus:

$𝜇=\frac{2\sin \left(\frac{𝑖}{2}\right)\cos \left(\frac{𝑖}{2}\right)}{\sin \left(\frac{𝑖}{2}\right)}$

Simplifying, we find:

$𝜇=2\cos \left(\frac{𝑖}{2}\right)$

This implies:

$𝑖=2{\cos }^{-1}\left(\frac{𝜇}{2}\right)$

$\begin{array}{cc}& \begin{array}{cc}& {𝑉}_{𝑖}=2\cos (12𝑡+\frac{𝜋}{3})\text{ ... (given)}\\ & \text{ Voltage gain }=\frac{{𝑉}_0}{{𝑉}_{𝑖}}\\ & \begin{array}{cc}{𝑉}_0& =126\times 2\cos (12𝑡+\frac{𝜋}{3})\\ & =252\cos (12𝑡+\frac{𝜋}{3})\end{array}\end{array}\end{array}$

In CE amplifier, $V_0$ and $V_i$ have a phase difference of $𝜋$ between them.

$\begin{array}{cc}{𝑉}_0& =252\cos (12𝑡+\frac{𝜋}{3}+𝜋)\\ & =252\cos (12𝑡+\frac{4𝜋}{3})\end{array}$

Since the stopping potential is same, they all have same frequency i.e., ${𝑓}_{𝑎}={𝑓}_{𝑏}={𝑓}_{𝑐}={𝑓}_{𝑑}$

From figure Photocurrent is highest for $𝑑$ followed by $c,b$ and a.

$\therefore I_a<I_b<I_c<I_d$

Given:

First scenario with a resistance of $100\Omega$ in series leads to a range $𝑉$.

Second scenario, for doubling the range, a resistance of $1000\Omega$ is connected.

The governing equation for resistance in series to reach the required voltage range is:

${𝑅}_{𝑠}=\frac{𝑉}{{𝐼}_{𝑠}}-𝐺$

For the first scenario:

$100=\frac{𝑉}{{𝐼}_{𝑠}}-𝐺\text{(Equation 1)}$

For the second scenario, where the range is doubled:

$1000=\frac{2𝑉}{{𝐼}_{𝑠}}-𝐺\text{(Equation 2)}$

Subtract Equation 1 from Equation 2:

$1000-100=(\frac{2𝑉}{{𝐼}_{𝑠}}-\frac{𝑉}{{𝐼}_{𝑠}})$

Solving the above, we find:

$900=\frac{𝑉}{{𝐼}_{𝑠}}\text{(Equation 3)}$

Using Equation 3 in Equation 1:

$100=900-𝐺$

Solving for $𝐺$, we get:

$𝐺=800\Omega$

Therefore, the resistance of the galvanometer is $800\Omega$.

The angular width of the central maximum in a Fraunhofer diffraction pattern is given by:

$𝜃=\frac{2𝜆}{𝑎}$

This implies that:

$𝜃\propto 𝜆$

Therefore, if the angle changes when a different wavelength is used, we get:

$\frac{{𝜃}_2}{{𝜃}_1}=\frac{{𝜆}_2}{{𝜆}_1}$

Given that the angular separation decreases by 20% when a new wavelength is used, we have:

${𝜃}_2={𝜃}_1-\frac{20}{100}{𝜃}_1=0.8{𝜃}_1$

Thus:

${𝜆}_2=0.8{𝜆}_1=0.8\times 6000=4800\stackrel{𝑜}{𝐴}$

Therefore, the wavelength of the light used is $4800\stackrel{𝑜}{𝐴}$.

Using kinematic equation,

$\begin{array}{cc}& 𝜃={𝜔}_0𝑡+\frac{1}{2}𝛼{𝑡}^2\\ & 𝜃=\frac{1}{2}𝛼{𝑡}^2\dots (\because \text{ particle was at rest, }{𝜔}_0=0)\\ \therefore & 𝑡={\left(\frac{2𝜃}{𝛼}\right)}^{1/2}\text{.... (i)}\end{array}$

Angular displacement of the particle $=𝑟𝜃\text{.... (iii)}$

$\begin{array}{cc}& \therefore \text{ Average velocity }=\frac{\text{ Angular displacement }}{\text{ time }}\\ & v_{\text{avenge }}=\frac{r𝜃}{t}=\frac{r𝜃}{{\left(\frac{2𝜃}{𝛼}\right)}^{1/2}}=\frac{r}{\sqrt{\surd 2}}\sqrt{\surd 𝛼𝜃}\end{array}$

When heat energy $\Delta 𝑄$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta 𝑈$, and the work done by the gas is $\Delta 𝑊$. We need to find the ratio $\Delta 𝑊:\Delta 𝑈:\Delta 𝑄$.

The fraction of heat energy used to perform external work is determined by the formula:

$\left(\frac{\Delta 𝑊}{\Delta 𝑄}\right)=(1-\frac{1}{𝛾})$

Here, $𝛾$ (gamma) for a diatomic gas is $\frac{7}{5}$. Thus, substituting into the formula:

$\frac{\Delta 𝑊}{\Delta 𝑄}=1-\frac{1}{\left(\frac{7}{5}\right)}=1-\frac{5}{7}=\frac{2}{7}\text{.....(i)}$

Next, the fraction of heat energy used to increase the internal energy is:

$\frac{\Delta 𝑈}{\Delta 𝑄}=\frac{1}{𝛾}=\frac{5}{7}\text{.....(ii)}$

Therefore, combining equations (i) and (ii), the ratios are:

$\Delta 𝑊:\Delta 𝑈:\Delta 𝑄=2:5:7$

Loudness of two sounds are given as $L_2=60\mathrm{dB}$ and $L_1=30\mathrm{dB}$

Loudness of sound $L=10{\log }_{10}\frac{I}{I_0}$

$\begin{array}{cc}\therefore & {𝐿}_2-{𝐿}_1=10{\log }_{10}\frac{{𝐼}_2}{{𝐼}_1}\\ & 60-30=10{\log }_{10}\frac{{𝐼}_2}{{𝐼}_1}\\ \therefore & {\log }_{10}\frac{{𝐼}_2}{{𝐼}_1}=3\\ & \frac{{𝐼}_2}{{𝐼}_1}={10}^3\Rightarrow {𝐼}_2=1000{𝐼}_1\end{array}$

Two loops made from same material

$I={\mathrm{MR}}^2\text{...(i)}$

$M=𝜎.2𝜋R$ ...($𝜎$ is mass per unit length)

Substituting in (i),

$I=𝜎.2𝜋R^3\text{.... (ii)}$

$\therefore \frac{I_p}{I_Q}=\frac{𝜎\cdot 2𝜋R_3^1}{𝜎\cdot 2𝜋R_3^2}\text{...[From(ii)]}$

$\begin{array}{cc}\frac{I_P}{I_Q}& ={\left(\frac{R_1}{R_2}\right)}^3\\ \frac{27}{1}& ={\left(\frac{R_1}{R_2}\right)}^3\dots (\text{ given, }\frac{I_P}{I_Q}=27)\\ \therefore \frac{R_1}{R_2}& =\frac{3}{1}\end{array}$

According to Wien's Law, the product of the wavelength at which a black body radiates maximum energy (${𝜆}_{𝑚}$) and its temperature (T) is a constant:

${𝜆}_{𝑚}𝑇=\text{constant}$

If initially the black body radiates maximum energy at wavelength ${𝜆}_0$ and temperature ${𝑇}_1$, and then it changes to radiate maximum energy at $\frac{{𝜆}_0}{2}$, the new temperature ${𝑇}_2$ can be found as:

${𝜆}_{{𝑚}_1}{𝑇}_1={𝜆}_{{𝑚}_2}{𝑇}_2$

Substituting the given values:

${𝑇}_2=\frac{{𝜆}_{{𝑚}_1}}{{𝜆}_{{𝑚}_2}}{𝑇}_1=\frac{{𝜆}_0}{\frac{{𝜆}_0}{2}}{𝑇}_1=2{𝑇}_1\text{... (i)}$

According to the Stefan-Boltzmann Law, the power radiated by a black body is proportional to the fourth power of its temperature:

$𝑃\propto {𝑇}^4$

For the new power ${𝑃}_2$ at temperature ${𝑇}_2$, we have:

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

Substituting for ${𝑇}_2$ from equation (i):

$\frac{{𝑃}_2}{{𝑃}_1}={\left(\frac{2{𝑇}_1}{{𝑇}_1}\right)}^4=16$

Thus, the power radiated by the black body increases by a factor of 16.

To determine the magnitude of the magnetic field at the common center of the two coils, we begin by considering the contributions from each coil separately.

Magnetic Field due to Coil P:

Coil P, with radius $𝑅$ and current $𝐼$, produces a magnetic field at its center given by:

${𝐵}_{𝑃}=\frac{{𝜇}_0𝐼}{2𝑅}\text{..... (i)}$

Magnetic Field due to Coil Q:

Coil Q, also with radius $𝑅$ but carrying a current of $\sqrt{\surd 8}\times 𝐼$, produces a magnetic field at its center:

${𝐵}_{𝑄}=\frac{{𝜇}_0\sqrt{\surd 8}𝐼}{2𝑅}\text{.... (ii)}$

Resultant Magnetic Field:

Since coils P and Q lie in perpendicular planes, their magnetic fields are orthogonal to each other. Thus, the resultant magnetic field ${𝐵}_{\text{net}}$ at the common center is found using the Pythagorean theorem:

${𝐵}_{\text{net}}=\sqrt{\surd {𝐵}_2^{𝑃}+{𝐵}_2^{𝑄}}$

Substituting the expressions from (i) and (ii):

${𝐵}_{\text{net}}=\sqrt{\surd {\left(\frac{{𝜇}_0𝐼}{2𝑅}\right)}^2+{\left(\frac{{𝜇}_0\sqrt{\surd 8}𝐼}{2𝑅}\right)}^2}$

Simplifying:

${𝐵}_{\text{net}}=\sqrt{\surd {\left(\frac{{𝜇}_0𝐼}{2𝑅}\right)}^2+{\left(\frac{2\sqrt{\surd 2}{𝜇}_0𝐼}{2𝑅}\right)}^2}$

$=\frac{{𝜇}_0𝐼}{2𝑅}\sqrt{\surd 1+8}$

$=\frac{{𝜇}_0𝐼}{2𝑅}\times 3$

$=\frac{3{𝜇}_0𝐼}{2𝑅}$

Thus, the magnitude of the magnetic field at the center is $\frac{3{𝜇}_0𝐼}{2𝑅}$.

For any point in interference pattern,

$\begin{array}{cc}& I=I_{max}{\cos }^2\frac{𝜙}{2}\\ \therefore & \frac{I_{max}}{4}=I_{max}{\cos }^2\frac{𝜙}{2}\\ \therefore & {\cos }^2\frac{𝜙}{2}=\frac{1}{4}\\ \therefore & \cos \frac{𝜙}{2}=\frac{1}{2}\\ \therefore & \frac{𝜙}{2}={60}^{\circ }=\frac{𝜋}{3}\\ \therefore & 𝜙=\frac{2𝜋}{3}\end{array}$

We know that,

$𝜙=\left(\frac{2𝜋}{𝜆}\right)\Delta 𝑥$, where $\Delta 𝑥$ is path difference.

$\begin{array}{cc}& \text{ and }\Delta 𝑥=𝑑\sin 𝜃\\ \therefore & \frac{2𝜋}{3}=\frac{2𝜋}{𝜆}(d\sin 𝜃)\\ \therefore & \frac{𝜆}{3d}=\sin 𝜃\\ \therefore & 𝜃={\sin }^{-1}\left(\frac{𝜆}{3d}\right)\end{array}$

The coefficient of coupling ($𝑘$) between two coils can be calculated using the mutual inductance ($𝑀$) and the self-inductances (${𝐿}_1$ and ${𝐿}_2$) of the coils. The formula for the coefficient of coupling is:

$𝑘=\frac{𝑀}{\sqrt{\surd {𝐿}_1\cdot {𝐿}_2}}$

Given:

$𝑀=45\text{mH}$

${𝐿}_1=75\text{mH}$

${𝐿}_2=48\text{mH}$

Substitute the given values into the formula:

$𝑘=\frac{45\text{mH}}{\sqrt{\surd 75\text{mH}\cdot 48\text{mH}}}$

Calculate the product inside the square root:

$75\cdot 48=3600$

So:

$\sqrt{\surd 3600}=60$

Now, plug the values back into the formula:

$𝑘=\frac{45}{60}=0.75$

Thus, the coefficient of coupling is 0.75. The correct option is C: 0.75.

Wavelength of Balmer series is given as,

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

For series limit, $𝑚=\infty$

$\therefore \frac{1}{𝜆}=R(\frac{1}{4}-\frac{1}{\infty })=R\left(\frac{1}{4}\right)$

Now, $𝜈=\frac{C}{𝜆}$

$\therefore 𝜈=\frac{𝑅𝐶}{4}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑋=𝛼{𝑡}^3\widehat{𝑖}+𝛽{𝑡}^3\widehat{𝑗}\\ & 𝑉=\frac{𝑑𝑋}{𝑑𝑡}=3𝛼{𝑡}^2\widehat{𝑖}+3𝛽{𝑡}^2\widehat{𝑗}\end{array}\\ & \begin{array}{cc}\text{ Magnitude of }𝑉& =\sqrt{\surd {\left(3𝛼{𝑡}^2\right)}^2+{\left(3𝛽{𝑡}^2\right)}^2}\\ & =\sqrt{\surd 9{𝛼}^2{𝑡}^4+9{𝛽}^2{𝑡}^4}\\ & =\sqrt{\surd 9{𝑡}^4({𝛼}^2+{𝛽}^2)}\\ & =3{𝑡}^2\sqrt{\surd {𝛼}^2+{𝛽}^2}\end{array}\end{array}$

According to Newton's law of cooling, the rate at which an object cools is proportional to the difference between its average temperature and the ambient temperature.

Thus, the rate of cooling can be expressed as:

$\text{Rate of cooling}\propto \text{Mean temperature difference}$

This relationship can be mathematically represented as:

$\frac{\text{Fall in temperature}}{\text{Time}\left(𝑡\right)}\propto (\frac{{𝜃}_1+{𝜃}_2}{2}-{𝜃}_0)$

Where:

${𝜃}_1$ and ${𝜃}_2$ are the initial and final temperatures, respectively.

${𝜃}_0$ is the ambient room temperature.

Let's evaluate each cooling interval:

Case 1: From ${75}^{\circ }\text{C}$ to ${70}^{\circ }\text{C}$

$\text{Mean temperature}=\left(\frac{75+70}{2}\right)=72.5$

Case 2: From ${70}^{\circ }\text{C}$ to ${65}^{\circ }\text{C}$

$\text{Mean temperature}=\left(\frac{70+65}{2}\right)=67.5$

Case 3: From ${65}^{\circ }\text{C}$ to ${60}^{\circ }\text{C}$

$\text{Mean temperature}=\left(\frac{65+60}{2}\right)=62.5$

From these calculations, we observe:

$72.5>67.5>62.5$

Since the mean temperature decreases in each subsequent case, the cooling rate is higher initially and decreases over time. This implies:

${𝑡}_1<{𝑡}_2<{𝑡}_3$

$\begin{array}{cc}& \text{ We know, }\\ & V=\sqrt{\surd 2\mathrm{gh}}\text{... (where ’ }ℎ\text{ ’ is depth from the surface)}\\ & \\ & h_1<h_2<h_3\text{...(From figure)}\\ \therefore & V_1<V_2<V_3\end{array}$

$𝑒=160\sin \left(100𝑡\right)$ volt and

$i=250\sin (100t+\frac{𝜋}{3})\mathrm{mA}$

Comparing given equations with the standard forms,

$𝑒={𝑒}_0\sin 𝜔𝑡$ and $i=i_0\sin (𝜔t+𝜙)$ we get,

$\begin{array}{cc}{e}_0=& 160V,I_0=250\mathrm{mA}\\ \therefore \text{ Power }& =\frac{e_0}{\sqrt{\surd 2}}\cdot \frac{I_0}{\sqrt{\surd 2}}\cos 𝜙\\ & =\frac{160\times 250\times {10}^{-3}}{2}\times \cos \left(\frac{𝜋}{3}\right)\\ & =\frac{40}{2}\times \frac{1}{2}\\ & =10W\end{array}$

$\begin{array}{cc}& \text{ Power }=\frac{\text{ Work done }}{\text{ time }}\\ & P=\frac{\text{ Torque }\times \text{ angular displacement }}{\text{ time }}=𝜏\times 𝜔\end{array}$

$\dots [\because 𝜔=\frac{𝜃}{𝑡}]\dots (\because 𝜏=𝐼𝛼)$

$\begin{array}{cc}\therefore P& =I𝛼𝜔\\ \therefore 𝜔& =\frac{P}{I𝛼}\\ & =P(I𝛼{)}^{-1}\end{array}$

Dipole moment will be in same direction as the external field. The collective effect of dipole moment produces a field that opposes the electric field inside the dielectric and hence, the net electric field inside the dielectric is less than external electric field.