MHT CET 2023 14th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

For sphere: M.I. $\mathrm{I}_{\mathrm{S}}=\frac{2}{5} \mathrm{MR}^2$

For cylinder: M.I. $\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{MR}^2$

Acceleration: $\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{MR}^2}}$

$$\begin{aligned} & \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{1+\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2}}{1+\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{MR}^2}}=\frac{\mathrm{MR}^2+\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2+\mathrm{I}_{\mathrm{c}}}=\frac{\mathrm{MR}^2+\frac{2}{5} \mathrm{MR}^2}{\mathrm{MR}^2+\frac{1}{2} \mathrm{MR}^2} \\ & \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{7}{5} \times \frac{2}{3}=\frac{14}{15} \end{aligned}$$

First maximum in single slit diffraction pattern is obtained at, $x=\frac{3 \lambda D}{2 d}$

$$\begin{array}{ll} \therefore & \Delta \mathrm{x}=\frac{3 \mathrm{D} \times \Delta \lambda}{2 \mathrm{~d}}=\frac{3 \times 1.5}{2 \times 2 \times 10^{-6}}(596-590) \\ \therefore & \Delta \mathrm{x}=6.75 \mathrm{~mm} \end{array}$$

The correct answer is A) decreases surface tension of water

• Water molecules are attracted to each other due to cohesive forces.


• At the surface of the water, there's an imbalance of force because there are no water molecules above to create attraction.


• This results in a net inward force, causing the surface to behave like a stretched membrane – this is surface tension.

• Soap molecules have a hydrophilic (water-loving) head and a hydrophobic (water-hating) tail.


• When soap is added to water, the hydrophobic tails disrupt the cohesive forces between water molecules at the surface.


• This reduces the net inward force, decreasing the surface tension.

• Spraying creates tiny droplets of water, which increases the surface area significantly.


• High surface tension would require extra energy to overcome the cohesive forces and create those droplets.


• Soap, by reducing surface tension, makes this process easier, requiring less energy for spraying.

• B) increases surface tension of water: This would make it harder to spray.


• C) makes surface tension of water zero: While surfactants reduce surface tension significantly, they don't reduce it to zero.


• D) increases its density: Density isn't directly related to ease of spraying in this context.

Given data: $\mu_{\mathrm{g}}=1.5, \mathrm{R}_1=\mathrm{R}_2=20 \mathrm{~cm}, \mu_l=1.25$

$$\begin{aligned} \mathrm{P}_1=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) & =(1.5-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ \therefore \quad \mathrm{P}_2=\left(\frac{\mu_{\mathrm{g}}}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) & =\left(\frac{1.5}{1.25}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & =(1.2-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \end{aligned}$$

Taking ratio,

$\frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{1.2-1}{1.5-1}=\frac{2}{5}$

Magnetic field inside the solenoid is, $B=\frac{\mu_0 \mathrm{NI}}{l}$

Flux inside the coil, $\phi=\mathrm{N}(\mathrm{BA})$

$\therefore \quad \phi=\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}$

Also, self inductance, $\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}$

$\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$

$$\begin{array}{ll} & \mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2 \\ \therefore \quad & \mathrm{n}_1 \times 49=\mathrm{n}_2 \times 48 \\ \therefore \quad & \mathrm{n}_2=\frac{49}{48} \mathrm{n}_1 \end{array}$$

Given that, beat frequency, $\left|\mathrm{n}_1-\mathrm{n}_2\right|=6$

$$\begin{aligned} & \therefore \quad\left|\mathrm{n}_1-\frac{49}{48} \mathrm{n}_1\right|=6 \\ & \therefore \quad \frac{\mathrm{n}_1}{48}=6 \\ & \therefore \quad \mathrm{n}_1=288 \mathrm{~Hz} \end{aligned}$$

To find when the acceleration of the particle is zero, we first need to understand how acceleration relates to position mathematically. Acceleration is the second derivative of position with respect to time. Given the position function

$x = a t^2 - b t^3,$

let's first find the velocity of the particle, which is the first derivative of position with respect to time ($\frac{dx}{dt}$), and then the acceleration, which is the second derivative of position with respect to time ($\frac{d^2x}{dt^2}$).

The first derivative (velocity) is:

$v = \frac{dx}{dt} = \frac{d}{dt}(a t^2 - b t^3) = 2at - 3bt^2.$

Now, let's take the second derivative (acceleration):

$a(t) = \frac{d^2x}{dt^2} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt.$

To find when the acceleration is zero, set the acceleration function to zero and solve for $t$:

$2a - 6bt = 0.$

We can solve for $t$ as follows:

$2a = 6bt$

$t = \frac{2a}{6b}$

$t = \frac{a}{3b}.$

Therefore, the acceleration of the particle will be zero at $t = \frac{a}{3b}$.

This corresponds to option C, which is the correct answer.

The surface charge density is given by,

$\sigma=\frac{\mathrm{q}}{\mathrm{A}}=\frac{17.7 \times 10^{-4}}{200}=8.85 \times 10^{-6} \mathrm{C} / \mathrm{m}^2$

The electric field intensity at a distance of $20 \mathrm{~cm}$ in air is,

$\mathrm{E}=\frac{\sigma}{2 \varepsilon_0}=\frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}=5 \times 10^5 \mathrm{~N} / \mathrm{C}$

Hence, option (A).

The acceleration due to gravity on the surface of the Earth varies with latitude due to the centrifugal force resulting from the Earth's rotation. The effective acceleration due to gravity at a latitude $\phi$, denoted as $\mathrm{g}_\phi$, takes into account the centrifugal force and is less than the acceleration due to gravity that would be experienced if the Earth were not rotating (denoted as $g_0$). At the equator ($\phi = 0^{\circ}$), this effect is maximal because the velocity due to Earth's rotation is maximal. As we go to the poles, the effect of rotation becomes minimal.

The effective acceleration due to gravity at latitude $\phi$ is given by:

$g_{\phi} = g_{0} - \omega^2 \mathrm{R} \cos^2 \phi$

Where:

• $g_0$ is the acceleration due to gravity without the Earth's rotation, at the equator
• $\omega$ is the angular velocity of the Earth's rotation
• $\mathrm{R}$ is the radius of the Earth
• $\phi$ is the latitude

At the equator, the effective acceleration due to gravity, $g$, is:

$g = g_{0} - \omega^2 \mathrm{R}$

because $\cos 0^{\circ} = 1$.

At latitude $30^{\circ}$, the effective acceleration due to gravity, $\mathrm{g}_\phi$, using $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, is:

$g_{\phi} = g_{0} - \omega^2 \mathrm{R} \left(\frac{\sqrt{3}}{2}\right)^2$

$g_{\phi} = g_{0} - \omega^2 \mathrm{R} \times \frac{3}{4}$

To find $\left|g-g_\phi\right|$, we subtract the above two equations:

$\left|g-g_\phi\right| = \left|\left(g_{0} - \omega^2 \mathrm{R}\right) - \left(g_{0} - \omega^2 \mathrm{R} \times \frac{3}{4}\right)\right|$

$\left|g-g_\phi\right| = \omega^2 \mathrm{R} - \omega^2 \mathrm{R} \times \frac{3}{4}$

$\left|g-g_\phi\right| = \omega^2 \mathrm{R} \left(1 - \frac{3}{4}\right)$

$\left|g-g_\phi\right| = \omega^2 \mathrm{R} \times \frac{1}{4}$

So the value of $\left|g-g_\phi\right|$ is given by the option A, which is:

$\frac{1}{4} \omega^2 \mathrm{R}$

Wavelength for Lyman series is,

$\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}\right]$

For the longest wavelength, $\lambda=\lambda_{\max }$ and $\mathrm{n}=2$

$\frac{1}{\lambda_{\max (\mathrm{L})}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}$

Wavelength for Balmer series is,

$\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$

For the longest wavelength, $\mathrm{n}=3$

$$\begin{aligned} & \frac{1}{\lambda_{\max (\mathrm{B})}}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5}{36} \\ & \frac{\lambda_{\max (\mathrm{L})}}{\lambda_{\max (\mathrm{B})}}=\frac{4}{3} \times \frac{5}{36}=\frac{5}{27} \end{aligned}$$

To solve this problem, we need to understand the relationship between the surface energy of a drop and its radius. The surface energy $E$ of a drop is proportional to its surface area $A$. The surface area of a sphere (which is the shape of a water drop) can be calculated using the formula:

where $r$ is the radius of the sphere. The surface energy is then given by:

where $\sigma$ is the surface tension of the liquid. When $1000$ small drops combine to form one large drop, the volume of the large drop is equal to the sum of the volumes of the small drops. The volume of a sphere is given by

If the radius of each small drop is $r$, the total volume of the small drops is $1000 \times \dfrac{4}{3}\pi r^3$. Let the radius of the large drop be $R$. Then,

Canceling common terms and taking the cube root of both sides, we get

Since $1000 = 10^3$, the cube root of $1000$ is $10$.

So the radius of the large drop is ten times the radius of each small drop.

Now we can compare the initial surface energy of all small drops with the final surface energy of the large drop. Let's denote the initial surface energy as $E_{\text{initial}}$ and the final surface energy as $E_{\text{final}}$.

Substituting $R = 10r$ into the final energy equation, we get:

Now, we compare the final energy to the initial energy:

Hence, the ratio of the final surface energy to the total initial surface energy is $1 : 10$. Therefore, the correct answer is:

To determine the time taken by the wave pulse to travel along the wire from point P to R, we first need to calculate the speed of the wave in each segment of the wire (PQ and QR) and then find the time taken to travel through each segment.

We have two wires PQ and QR with different lengths and masses but the same radius and therefore the same cross-sectional area A. Both wires are under the same tension T. Given that the tension and cross-sectional area are the same for both wires, we can assume that the linear density (mass per unit length) is different because they have different masses and lengths.

To find the linear density (μ) for each wire, we use the formula:

$\mu = \frac{m}{L}$

where

m is the mass of the wire, and

L is the length of the wire.

For wire PQ:

$\mu_{PQ} = \frac{m_{PQ}}{L_{PQ}} = \frac{0.06\ kg}{4.8\ m} = 0.0125\ kg/m$

For wire QR:

$\mu_{QR} = \frac{m_{QR}}{L_{QR}} = \frac{0.2\ kg}{2.56\ m} = 0.078125\ kg/m$

The speed of a wave in a stretched string or wire is given by the formula:

$v = \sqrt{\frac{T}{\mu}}$

where

v is the speed of the wave, and

T is the tension in the wire.

Using this formula, we can calculate the speed of the wave in each section of the wire.

For wire PQ:

$v_{PQ} = \sqrt{\frac{T}{\mu_{PQ}}} = \sqrt{\frac{80\ N}{0.0125\ kg/m}} = \sqrt{6400\ m^2/s^2} = 80\ m/s$

For wire QR:

$v_{QR} = \sqrt{\frac{T}{\mu_{QR}}} = \sqrt{\frac{80\ N}{0.078125\ kg/m}} = \sqrt{1024\ m^2/s^2} = 32\ m/s$

Now we will find the time taken for the wave to travel through each section.

Time is distance over speed, so for wire PQ:

$t_{PQ} = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8\ m}{80\ m/s} = 0.06\ s$

And for wire QR:

$t_{QR} = \frac{L_{QR}}{v_{QR}} = \frac{2.56\ m}{32\ m/s} = 0.08\ s$

To find the total time taken by the wave pulse to travel from P to R, we add the times for PQ and QR:

$t_{total} = t_{PQ} + t_{QR} = 0.06\ s + 0.08\ s = 0.14\ s$

The correct answer is:

Option C: 0.14 s

When light travels from one medium to another, it undergoes refraction twice in the case of a prism: first when entering the prism and second when exiting it. The path of light inside the prism is parallel to the base, which means that the angle of incidence at the first surface is equal to the angle of emergence at the second surface.

Since we are dealing with an equilateral prism, the angles at each corner are $60^{\circ}$. Let's denote the angle of the prism as $A$. When the ray inside the prism is parallel to the base, the angle of refraction at the first surface (inside the prism) will be equal to the angle of incidence at the second surface. Let's denote this angle as $r$ (which is also the angle of refraction at the second surface), then according to Snell's Law, where the refractive index of glass is denoted by $n$:

and

Since the refracted ray in glass is parallel to the base of the prism, the deviation of the ray inside the prism is minimal, and this occurs when $r$ equals $A/2$ for an equilateral prism, so

Since the values of $\sin(i)$ and $\sin(e)$ are both obtained by multiplying $\sin(r)$ with the refractive index $n$, we have

Because the sine function is increasing for angles between $0^\circ$ and $90^\circ$, if $\sin(i)$ equals $\sin(e)$, the corresponding angles must also be equal, thus

Hence, the correct option is:

The two springs are connected in parallel. So, the effective spring constant is, $\mathrm{k}_{\mathrm{eff}}=2 \mathrm{k}$

Time period of the spring system is,

$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\ & \therefore \quad \mathrm{T}^2=4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{k}} \\ & \therefore \quad \mathrm{k}=4 \pi^2 \times \frac{\mathrm{m}}{2 \mathrm{~T}^2}=4 \times 10 \times \frac{12}{2 \times 4}=60 \mathrm{~N} / \mathrm{m} \end{aligned}$$

For semicircular arc the magnetic field is given as $B=\frac{\mu_0 i}{4 R}$

The equivalent magnetic field at the centre is

$$\begin{aligned} & B_{\text {eq }}=\frac{\mu_0 I}{4 R_1}+\frac{\mu_0 I}{4 R_2} \\ & B_{\text {eq }}=\frac{\mu_0 I}{4}\left(\frac{1}{R_1}+\frac{1}{R_2}\right) \end{aligned}$$

The half-life of a radioactive substance is the time it takes for half of the substance to decay. The remaining fraction of the original substance after a certain number of half-lives can be calculated using the formula:

$N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

Where:

$N$ is the remaining quantity of the substance,

$N_0$ is the initial quantity of the substance,

$t$ is the elapsed time,

$T_{1/2}$ is the half-life of the substance.

Given that the half-life $T_{1/2}$ is 1600 years, and we want to find the amount of substance left after 6400 years, we can use the above formula:

$\frac{t}{T_{1/2}} = \frac{6400}{1600} = 4$ half-lives

Therefore,

$N = N_0 \left(\frac{1}{2}\right)^4$

$N = N_0 \left(\frac{1}{2^4}\right)$

$N = N_0 \left(\frac{1}{16}\right)$

The fraction of the original sample remaining undecayed after 6400 years is $\frac{1}{16}$.

Thus, the correct answer is:

Option A $\frac{1}{16}$

Retentivity or remanence is the ability of a magnetic substance to retain magnetism even in the absence of a magnetising field.

$\therefore \quad$ Retentivity corresponds to the value of B when value of $\mathrm{H}$ is zero.

According to law of conservation of energy,

$$\begin{aligned} & \frac{1}{2} m v^2=-\frac{G M m}{R_0}-\left(\frac{G M m}{R}\right)=G M m\left(\frac{1}{R}-\frac{1}{R_0}\right) \\ \therefore \quad & v^2=2 G M\left(\frac{1}{R}-\frac{1}{R_0}\right) \end{aligned}$$

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

$\mathrm{V}=\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$

Amongst the given options, only option (B) dimensionally equates to velocity.

According to Newton's law of cooling,

$$\begin{aligned} & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\theta_1-\theta_0}{\theta_2-\theta_0} \\ \therefore \quad & \frac{4}{1}=\frac{90-\theta_0}{30-\theta_0} \\ \therefore \quad & 120-90=4 \theta_0-\theta_0 \\ \therefore \quad & 3 \theta_0=30 \end{aligned}$$

$\therefore \quad$ Temperature of the surroundings is,

$\theta_0=10^{\circ} \mathrm{C}$

To solve this problem, we need to understand the electric field intensity due to a short dipole at two specific points: on the axial line and on the equatorial line.

For a short dipole, the expressions for the electric field intensity are as follows:

1. On the axial line (line extending along the dipole axis):

$E_a=\frac{1}{4𝜋{𝜖}_0}\cdot \frac{2𝑝}{{𝑟}^3}$

where $𝑝$ is the dipole moment, $𝑟$ is the distance from the center of the dipole, and ${𝜖}_0$ is the permittivity of free space.

2. On the equatorial line (line perpendicular to the dipole axis at the center):

$E_q=\frac{1}{4𝜋{𝜖}_0}\cdot \frac{𝑝}{{𝑟}^3}$

Notice the difference in the coefficients. The electric field on the axial line is twice as strong as that on the equatorial line at the same distance $𝑟$.

Thus, the relationship between $E_a$ and $E_q$ can be derived:

Given:

$E_a=2E_q$

Therefore, the correct option is:

Option D
$E_a=2E_q$

The equation for the shift in the fringe pattern is given as,

$\mathrm{n} \lambda=(\mu-1) \mathrm{t}$

$\therefore \quad$ Refractive index of mica is:

$$\begin{aligned} & \mu=\frac{\mathrm{n} \lambda}{\mathrm{t}}+1 \\ \therefore \quad \mu & =\frac{1 \times 6 \times 10^{-5}}{12 \times 10^{-5}}+1=0.5+1 \\ \therefore \quad \mu & =1.5 \end{aligned}$$

In an n-p-n transistor, the collector current (Ic) is primarily due to the flow of electrons from the emitter to the collector. However, not all the electrons emitted by the emitter reach the collector; a small percentage of them recombine in the base. This percentage is given by the common base current gain ($\alpha$), which in this case is 80% or 0.80.

The base current (I_b) can be calculated by knowing that the emitter current (I_e) is the sum of the collector current (I_c) and the base current (I_b). The relationship is given by the equation:

$I_e = I_c + I_b$

Given that $\alpha$, which is the common base current gain, is the ratio of the collector current to the emitter current:

$\alpha = \frac{I_c}{I_e}$

Let's solve for $I_e$:

$I_e = \frac{I_c}{\alpha} = \frac{28 \mathrm{~mA}}{0.80} = 35 \mathrm{~mA}$

Now, we can find the base current using the initial relation between $I_e$, $I_c$, and $I_b$:

$I_b = I_e - I_c = 35 \mathrm{~mA} - 28 \mathrm{~mA}$

$I_b = 7 \mathrm{~mA}$

Hence, the base current $\mathbf{I_b}$ is $\mathbf{7 \mathrm{~mA}}$, which corresponds to Option D.

Flux linked with each turn of the solenoid is, $\phi=2.8 \times 10^{-3} \mathrm{~Wb}$

$\therefore \quad$ Total magnetic flux of the solenoid is,

$$\begin{aligned} & \phi_{\text {net }}=\mathrm{N} \phi \\ & \phi_{\text {net }}=1500 \times 2.8 \times 10^{-3}=4.2 \mathrm{~Wb} \end{aligned}$$

Flux can also be given by, $\phi=\mathrm{LI}$

$\therefore \quad$ The self-inductance of the solenoid is,

$$\begin{aligned} & L=\frac{\phi}{I}=\frac{4.2}{3.5} \\ & L=1.2 \mathrm{~H} \end{aligned}$$

The correct answer is Option B, intrinsic.

Intrinsic semiconductors are materials where the electrical conductivity is derived purely from the semiconductor itself, without any external doping. The conductivity in intrinsic semiconductors occurs because of the breaking of covalent bonds at higher temperatures, which creates electron-hole pairs. These electron-hole pairs contribute to the electrical conductivity.

Here's a brief explanation of each term to clarify the concepts:

Intrinsic Semiconductors (Option B): This is the pure form of a semiconductor, such as silicon or germanium, with no impurities added. The charge carriers (electrons and holes) are generated by thermal energy breaking the covalent bonds. Electrons are excited from the valence band to the conduction band, leaving behind holes in the valence band. Both the excited electrons and the holes contribute to the charge transport in intrinsic semiconductors.

Extrinsic Semiconductors (Option D): These are impure semiconductors intentionally 'doped' with impurities to alter their electrical properties. The addition of a small amount of dopant to a semiconductor can greatly increase its conductivity. There are two types of extrinsic semiconductors:

• Donor (n-type) semiconductors (Option A): These have been doped with elements that have more valence electrons than the semiconductor, typically five-valence electrons such as phosphorus or arsenic when doping silicon. These extra electrons can easily move to the conduction band and significantly increase the conductivity of the material.


• Acceptor (p-type) semiconductors (Option C): These are doped with elements that have fewer valence electrons (typically three valence electrons) than the semiconductor, such as boron or aluminum when doping silicon. These create holes (positive charge carriers) in the valence band by accepting electrons that jump from the valence band, thus increasing the conductivity.

To summarize, the answer is Option B, intrinsic, because this is the term used to describe a pure semiconductor where conductivity is due solely to its own lattice structure and not influenced by impurities or added elements.

Series resistance is $\mathrm{n}$ times of galvanometer resistance: $\mathrm{R}_{\mathrm{s}}=\mathrm{nR}_{\mathrm{G}}$

Relation between voltage and current:

$$\begin{aligned} & \therefore \quad \mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{s}}+\mathrm{R}_{\mathrm{G}}\right) \\ & \therefore \quad \mathrm{V}=\mathrm{I}\left(\mathrm{nR}_{\mathrm{G}}+\mathrm{R}_{\mathrm{G}}\right) \\ & \therefore \quad \mathrm{V}=\mathrm{IR}_{\mathrm{G}}(\mathrm{n}+1) \\ & \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2}=\frac{8+4}{8-4} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{12}{4} \\ & \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=3 \\ & \end{aligned}$$

$$\begin{array}{ll} & \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}} \\ \therefore \quad & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{M}_1}{\mathrm{M}_2}} \\ & \text { Given that }\left(\mathrm{v}_{\mathrm{rms}}\right)_2=4\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ \therefore \quad & \sqrt{\frac{32}{\mathrm{M}_2}}=4 \\ \therefore \quad & \frac{32}{\mathrm{M}_2}=16 \\ \therefore \quad & \mathrm{M}_2=2 \end{array}$$

The two condensers are in parallel. Their equivalent capacitance is $\mathrm{C}_{\mathrm{c}}=\mathrm{C}+\frac{\mathrm{C}}{2}=\frac{3 \mathrm{C}}{2}$

The formula for work done is

$$\begin{aligned} & \mathrm{W}=\frac{1}{2} \mathrm{C}_{\mathrm{e}} \mathrm{V}^2 \\ & \mathrm{~W}=\frac{1}{2}\left(\frac{3 \mathrm{C}}{2}\right) \mathrm{V}^2 \\ & \mathrm{~W}=\frac{3}{4} \mathrm{CV}^2 \end{aligned}$$

For adiabatic process, $\mathrm{PV}^\gamma=$ a constant

$\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\left(\frac{\mathrm{V}^{\prime}}{\mathrm{V}}\right)^\gamma=\left(\frac{\rho^{\prime}}{\rho}\right)^\gamma$

Given that, $\frac{\rho^{\prime}}{\rho}=32$ and $\gamma=\frac{7}{5}$

$\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=(32)^{\frac{7}{5}}=128$

The power factor of an R-L circuit is given as,

$$\begin{aligned} & \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}} \\ & \therefore \quad\frac{R}{\sqrt{R^2+X_L^2}}=\frac{1}{\sqrt{2}} \\ & \therefore \quad\frac{1}{\sqrt{1+\left(\frac{X_L}{R}\right)^2}}=\frac{1}{\sqrt{2}} \\ & \therefore \quad\frac{1}{\sqrt{1+\left(\frac{\omega L}{R}\right)^2}}=\frac{1}{\sqrt{2}} \\ & \therefore \quad\left(\frac{\omega L}{R}\right)^2+1=2 \\ & \therefore \quad\frac{\omega L}{R}=1 \\ \end{aligned}$$

So, when the AC frequency is doubled,

$$\begin{aligned} & \quad\frac{\omega L}{R}=2 \\ & \therefore \quad\cos \phi=\frac{1}{\sqrt{1+\left(\frac{\omega L}{R}\right)^2}}=\frac{1}{\sqrt{1+(2)^2}} \\ & \therefore \quad\cos \phi=\frac{1}{\sqrt{5}} \end{aligned}$$