MHT CET 2021 24th September Evening Shift (Physics & Chemistry) Question Paper with Solutions

When temperature is $80^{\circ} \mathrm{C}$, by Newton's law of cooling, we have

$1.5=\mathrm{k}\left(\frac{80+30}{2}-30\right)=\mathrm{k}(55-30)=25 \mathrm{~K}$

When temperature is $50^{\circ} \mathrm{C}$, let $\mathrm{r}$ be the rate of cooling.

Then

$$\begin{aligned} & \text { Then } \\ & r=k\left(\frac{50+30}{2}-30\right)=k(40-30)=10 \mathrm{~K} \\ & \therefore \quad \frac{r}{1.5}=\frac{10}{25} \quad \therefore r=\frac{10}{25} \times 1.5=\frac{3}{5}=0.6^{\circ} \mathrm{C} / \mathrm{min} \end{aligned}$$

The moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is given by

$\mathrm{I}_0=\frac{\mathrm{ML}^2}{12}$

A point at a distance $\frac{\mathrm{L}}{4}$ from its end will also be at a distance $\frac{\mathrm{L}}{4}$ from the centre.

Hence by parallel axis theorem,

$\mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48}$

To determine the depth from the surface of the Earth at which the acceleration due to gravity is 60% of its value on the surface, we can use the concept of the gravitational force inside the Earth.

Let $g_0$ be the acceleration due to gravity at the surface of the Earth, and let the radius of the Earth be $R$. When at a depth $d$ from the surface, the gravity is determined by the distance from the center of the Earth, which is $R - d$.

The gravity inside the Earth at a depth $d$ is given by the formula:

$g_d = g_0 \left(1 - \frac{d}{R}\right)$

Given that at depth $d$, the gravity is 60% of $g_0$, we get:

$g_d = 0.60 \cdot g_0$

Substitute this into the formula:

$0.60 \cdot g_0 = g_0 \left(1 - \frac{d}{R}\right)$

Divide both sides by $g_0$:

$0.60 = 1 - \frac{d}{R}$

Rearrange the equation to solve for $d$:

$\frac{d}{R} = 1 - 0.60 = 0.40$

Hence:

$d = 0.40 R$

Thus, the depth $d$ is:

$d = 0.40 R$

To find the depth from the surface, use:

$d = R - \frac{d}{R}$

Rearranging for $d$ we get:

$d = 0.40 R$

Therefore, the depth from the surface is:

$d = 0.40 R$

Which corresponds to the option:

Option B

$\frac{2 R}{5}$

To determine the mutual inductance between the two coils, we can use Faraday's Law of Induction. According to Faraday's Law, the electromotive force (emf) induced in a coil is given by:

$\mathcal{E} = -M \frac{dI}{dt}$

where:

• $\mathcal{E}$ is the induced emf,
• $M$ is the mutual inductance,
• $I$ is the current in the primary coil.

Given the current in the primary coil:

$I = 10 \sin (100 \pi t) \, \text{A}$

First, we need to compute the derivative of the current with respect to time:

$\frac{dI}{dt} = \frac{d}{dt} (10 \sin (100 \pi t)) = 10 \cdot 100 \pi \cos (100 \pi t) = 1000 \pi \cos (100 \pi t)$

We are given that the maximum emf induced in the neighbouring coil is:

$\mathcal{E}_{\text{max}} = 5 \pi \, \text{V}$

The maximum value of cosine function is 1. Therefore, the maximum emf is induced when:

$\mathcal{E}_{\text{max}} = M \cdot (1000 \pi)$

Setting the given maximum emf equal to this expression, we get:

$5 \pi = M \cdot 1000 \pi$

Solving for $M$:

$M = \frac{5 \pi}{1000 \pi} = \frac{5}{1000} = 0.005 \, \text{H}$

Converting Henry to milliHenry (1 H = 1000 mH):

$M = 0.005 \, \text{H} = 5 \, \text{mH}$

Therefore, the mutual inductance of the two coils is:

Option A: $5 \mathrm{~mH}$

The given circuit can be redrawn as shown in the figure. It is a balanced Wheatstone bridge. No current will flow through $5 \Omega$ resistance, and hence it can be removed from the circuit. $3 \Omega$ and $2 \Omega$ resistances are in series. Hence we have two branches with $5 \Omega$ resistances each, connected in parallel. Their equivalent resistance is $2.5 \Omega$

$\therefore \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{2.5}=2.4 \mathrm{~A}$

Energy stored in the capacitor $\mathrm{U}=\frac{1}{2} q \mathrm{~V}$

Work done by the battery $\mathrm{W}=\mathrm{qV}$

$\therefore \frac{\mathrm{U}}{\mathrm{W}}=\frac{1}{2}$

To find the frequency of the sound wave given the relationship between the path difference and phase difference, let's start by understanding the relationship between these quantities.

The path difference (x) and the phase difference ($n \pi$) for a wave can be related via the wavelength ($\lambda$) of the wave. The path difference corresponding to a phase difference of $2\pi$ radians is one wavelength, i.e., $x = \lambda$ when the phase difference is $2\pi$.

Given that the path difference $x$ is equivalent to a phase difference of $n \pi$, we can express the path difference as:

$x = \frac{n \pi}{2\pi} \lambda$

Then, simplifying this expression, we get:

$x = \frac{n \lambda}{2}$

Now, rearranging this equation to solve for $\lambda$ (wavelength), we have:

$\lambda = \frac{2 x}{n}$

Next, the frequency ($f$) of a wave is related to its wavelength ($\lambda$) and the velocity ($V$) of the wave via the equation:

$V = f \lambda$

Solving for the frequency, we get:

$f = \frac{V}{\lambda}$

Substituting the expression for $\lambda$, we obtain:

$f = \frac{V}{\frac{2 x}{n}}$

Which simplifies to:

$f = \frac{V n}{2 x}$

Thus, the correct option is:

Option C: $\frac{V n}{2 x}$

In a half-wave rectifier, only one half of the input AC voltage is allowed through during each cycle. Since the input frequency is $60 \ \text{Hz}$, the output frequency will also be $60 \ \text{Hz}$ because it passes for one half of each cycle.

For a full-wave rectifier, both halves of the input AC voltage are used, effectively doubling the frequency of the output waveform. As a result, the output frequency will be $2 \times 60 \ \text{Hz} = 120 \ \text{Hz}$.

Therefore, the correct answer is:

Option B: 60 Hz, 120 Hz.

$$\begin{aligned} & \text { Volume }=\frac{4}{3} \pi r^3 n=\frac{4}{3} \pi R^3 \\ & \therefore R=n^{1 / 3} r \quad \therefore \frac{R}{r}=n^{\frac{1}{3}} \end{aligned}$$

Terminal velocity $\mathrm{v} \propto \mathrm{r}^2$

$$\begin{aligned} & \frac{v_2}{v_1}=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{R}{r}\right)^2=n^{2 / 3} \\ & \therefore v_2=n^{2 / 3} v_1=5 \mathrm{n}^{2 / 3} \mathrm{~m} / \mathrm{s} \\ & \end{aligned}$$

When $40 \%$ nuclei decay, $60 \%$ nuclei remain undecayed

Let this number be $\mathrm{N}_0$

When $85 \%$ nuclei decay, $15 \%$ nuclei remain undecayed This number will be $\mathrm{N}=\frac{\mathrm{N}_0}{4}$

The number of nuclei will become one-fourth in two half lives i.e. 60 minutes.

To determine the ratio of the magnitude of the magnetic moment to the angular momentum for a particle of charge 'q' and mass 'm' moving in a circular orbit of radius 'r' with angular speed $\omega$, we need to calculate each quantity separately and then find their ratio.

Magnetic Moment (M):

The magnetic moment for a charged particle moving in a circular orbit is given by:

$M = \frac{q}{2m} \cdot L$

where $L$ is the angular momentum of the particle.

Angular Momentum (L):

Angular momentum for a particle moving in a circle is:

$L = m \cdot r^2 \cdot \omega$

Substituting $L$ into the Magnetic Moment formula:

$M = \frac{q}{2m} \cdot (m \cdot r^2 \cdot \omega)$

$M = \frac{q \cdot r^2 \cdot \omega}{2}$

Ratio of Magnetic Moment to Angular Momentum:

We need to find the ratio $\frac{M}{L}$.

$\frac{M}{L} = \frac{\frac{q \cdot r^2 \cdot \omega}{2}}{m \cdot r^2 \cdot \omega}$

$\frac{M}{L} = \frac{q}{2m}$

From this ratio, it is clear that the ratio of the magnitude of its magnetic moment to that of its angular momentum depends only on the charge 'q' and mass 'm' of the particle.

Thus, the correct option is:

Option C

q and m

$\frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{I}}{4 \mathrm{I}}=\frac{1}{4}=\frac{\mathrm{a}_1^2}{\mathrm{a}_2^2}$

where $a_1$ and $a_2$ are the amplitudes

$$\begin{aligned} & \therefore \frac{a_1}{a_2}=\frac{1}{2} \quad \therefore \frac{a_{\max }}{a_{\min }}=\frac{a_1+a_2}{a_1-a_2}=\frac{1.2}{1-2}=\frac{3}{-1} \\ & \therefore \frac{I_{\max }}{I_{\min }}=\frac{a_{\max }^2}{a_{\min }^2}=\frac{9}{1} \end{aligned}$$

If Vis the volume then we have

$$\begin{aligned} V & =\frac{4}{3} \pi r^3 \\ V & \propto r^3 \quad \therefore \frac{V_2}{V_1}=\left(\frac{r_2}{r_1}\right)^3 \\ \therefore \quad \frac{r_2}{r_1} & =\left(\frac{v_2}{v_1}\right)^{1 / 3}=(2)^{\frac{1}{3}} \\ W_1 & =8 \pi \pi_1^2 \cdot T \text { and } W_2=8 \pi \pi_2^2 T \\ \therefore \quad \frac{W_2}{W_1} & =\left(\frac{r_2}{r_1}\right)^2=(2)^{\frac{2}{3}}=(4)^{\frac{1}{3}} \\ \therefore \quad W_2 & =(4)^{1 / 3} W_1 \end{aligned}$$

Here's how to determine the ratio of volumes of the soap bubbles:

Understanding the Concept

The pressure inside a soap bubble is slightly higher than the pressure outside due to the surface tension of the soap film. The excess pressure inside a soap bubble is inversely proportional to the radius of the bubble. This relationship can be expressed as:

$P_{inside} - P_{outside} = \frac{4T}{r}$

Where:

• $P_{inside}$ is the pressure inside the bubble
• $P_{outside}$ is the pressure outside the bubble
• $T$ is the surface tension of the soap film
• $r$ is the radius of the bubble

Applying the Concept to the Problem

Let's denote the radii of the two bubbles as $r_1$ and $r_2$. We are given:

• $P_{inside1} = 1.01 \ atm$
• $P_{inside2} = 1.03 \ atm$
• $P_{outside} = 1 \ atm$

Using the formula, we can write:

• $1.01 - 1 = \frac{4T}{r_1}$
• $1.03 - 1 = \frac{4T}{r_2}$

Dividing the first equation by the second equation, we get:

$\frac{1.01 - 1}{1.03 - 1} = \frac{r_2}{r_1}$

Simplifying the equation:

$\frac{0.01}{0.03} = \frac{r_2}{r_1}$

$\frac{1}{3} = \frac{r_2}{r_1}$

Calculating the Volume Ratio

The volume of a sphere is given by:

$V = \frac{4}{3}πr^3$

Therefore, the ratio of volumes is:

$\frac{V_2}{V_1} = \frac{\frac{4}{3}πr_2^3}{\frac{4}{3}πr_1^3} = \left(\frac{r_2}{r_1}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

Since we want the ratio of $V_1$ to $V_2$, we invert the above result:

$\frac{V_1}{V_2} = \boxed{27:1}$

Therefore, the correct answer is Option B: 27 : 1.

Centripetal force $\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$

$\mathrm{F_1=\frac{m v_1^2}{r_1}}$ and $\mathrm{F_2=\frac{m v_2^2}{r_2}}$

$$\begin{aligned} & \therefore \frac{F_2}{F_1}=\frac{v_2^2}{v_1^2} \cdot \frac{r_1}{r_2} \\ & v_2=\frac{v_1}{2} \text { and } r_2=3 r_1 \\ & \therefore \frac{F_2}{F_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \therefore F_2=\frac{F_1}{12} \quad \therefore F_2 < F_1 \\ & \therefore F_1-F_2=F_1-\frac{F_1}{12}=\frac{11}{12} F_1 \end{aligned}$$

A rejector circuit, also known as a notch filter or band-stop filter, is a resonant circuit designed to reject or significantly attenuate a specific frequency while allowing others to pass. The correct configuration for a rejector circuit is when the elements $\mathrm{L}$ (inductor), $\mathrm{C}$ (capacitor), and $\mathrm{R}$ (resistor) are connected in parallel.

Thus, the correct option is:

Option A $\mathrm{L}-\mathrm{C}-\mathrm{R}$ are connected in parallel.

Let $\mathrm{E_1}$ be the energy of the first energy level Then $\mathrm{E_2=\frac{E_1}{4}}$

Highest level $\mathrm{E_{\infty}}=0$

$$\begin{aligned} & \therefore \quad E_2-E_1=\frac{E_1}{4}-E_1=-\frac{3}{4} E_1 \\ & E_{\infty}-E_2=0-\frac{E_1}{4}=-\frac{E_1}{4} \\ & \therefore \quad \frac{E_2-E_1}{E_{\infty}-E_2}=3 \\ & \end{aligned}$$

$$\begin{aligned} & \mathrm{I}_4=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3=1+2+3=6 \mathrm{~A} \\ & \mathrm{I}_6=\mathrm{I}_4-\mathrm{I}_5=6-1.2=4.8 \mathrm{~A} \\ & \mathrm{I}_8=\mathrm{I}_6+\mathrm{I}_7=4.8+0.8=5.6 \mathrm{~A} \\ & \mathrm{I}_{10}=\mathrm{I}_8-\mathrm{I}_9=5.6-0.5=5.1 \mathrm{~A} \\ & \mathrm{I}=\mathrm{I}_{10}-\mathrm{I}_{11}=5.1-1.7=3.4 \mathrm{~A} \end{aligned}$$

If k is the kinetic energy and p is the momentum then

$\mathrm{k}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \quad \therefore \mathrm{p}^2=2 \mathrm{mk}$

If $\mathrm{k}$ is constant, then $\mathrm{p}^2 \propto \mathrm{m}$

At resonance, the net reactance of the circuit is zero and the impedance is equal to the resistance.

$\therefore I_0=\frac{E_0}{R}$

To determine the ratio of the particle's acceleration after and before the change in speed, let's start with the formula for centripetal acceleration.

The centripetal acceleration of a particle moving with a constant speed $v$ along a circular path of radius $r$ is given by:

$a = \frac{v^2}{r}$

If the speed of the particle is doubled, the new speed $v'$ will be:

$v' = 2v$

The new centripetal acceleration $a'$ for this doubled speed is:

$a' = \frac{{v'}^2}{r}$

Substituting $v' = 2v$ into the equation for $a'$:

$a' = \frac{{(2v)}^2}{r} = \frac{4v^2}{r}$

The initial centripetal acceleration $a$ is:

$a = \frac{v^2}{r}$

Now, the ratio of the new centripetal acceleration $a'$ to the initial centripetal acceleration $a$ is:

$\frac{a'}{a} = \frac{\frac{4v^2}{r}}{\frac{v^2}{r}} = \frac{4v^2}{v^2} = 4$

Thus, the ratio of the particle's acceleration after doubling the speed to the acceleration before the speed change is:

$4: 1$

Therefore, the correct option is:

Option D: 4: 1

Let’s analyze the potential energy of a magnetic dipole in a magnetic field. The potential energy $U$ of a magnetic dipole with a magnetic moment $\mathrm{M}$ in a magnetic field of induction $\mathrm{B}$ is given by:

$U = -\mathrm{M} \cdot \mathrm{B} \cdot \cos{\theta}$

where $\theta$ is the angle between the magnetic moment vector $\mathrm{M}$ and the magnetic field vector $\mathrm{B}$. The potential energy varies with the angle $\theta$.

To find the minimum and maximum values of the potential energy, let's examine the extreme values of $\cos{\theta}$, which are $-1$ and $1$ respectively.

1. When $\theta = 0$ degrees (or $0$ radians),

$\cos(0) = 1$

$U_{\text{min}} = -\mathrm{M} \cdot \mathrm{B} \cdot 1 = -\mathrm{MB}$

So, the minimum potential energy is $-\mathrm{MB}$.

2. When $\theta = 180$ degrees (or $\pi$ radians),

$\cos(180^\circ) = -1$

$U_{\text{max}} = -\mathrm{M} \cdot \mathrm{B} \cdot (-1) = \mathrm{MB}$

So, the maximum potential energy is $\mathrm{MB}$.

Hence, the correct option is:

Option A

$-\mathrm{MB}, +\mathrm{MB}$

The interference pattern formed by light waves is characterized by alternating bright and dark bands (fringes). The path difference between two interfering waves determines whether we get a bright or dark band at a specific point. The criterion for these bands based on the path difference can be stated as follows:

For the bright band (constructive interference):

$\Delta x = m\lambda$

where $\Delta x$ is the path difference, $m$ is any integer (0, 1, 2, 3, ...), and $\lambda$ is the wavelength of light.

For the dark band (destructive interference):

$\Delta x = (m + \frac{1}{2}) \lambda$

where $m$ is any integer (0, 1, 2, 3, ...).

Given that the path difference is:

$\Delta x = \frac{57}{2} \lambda$

We observe that:

$\frac{57}{2} = 28.5$

This suggests that the path difference is equal to $(28.5) \lambda$. Hence, this can be expressed in terms of the dark band condition:

$\Delta x = (28 + \frac{1}{2}) \lambda$

This matches the formula for the dark band, with $m = 28$. Therefore, the point corresponds to the 29^th dark band (since we start counting from m = 0).

Hence, the correct answer is:

Option D: 29^th dark band

Consider mass $\mathrm{m}$ at $\mathrm{A}$.

The forces exerted on it by the other two masses are given by

$F_1=G \frac{\mathrm{m}^2}{\mathrm{~L}^2}=\mathrm{F}_2$

The angle between the two forces is $60^{\circ}$.

Hence the resultant force

$$\begin{aligned} & \quad F=\sqrt{\mathrm{F}_1^2+\mathrm{F}_1^2+2 \mathrm{~F}_1^2 \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}_1 \\ & \therefore \quad F=\sqrt{3} \cdot G \frac{\mathrm{m}^2}{\mathrm{~L}^2} \end{aligned}$$

Mass $\mathrm{m}$ rotates around the centre $\mathrm{G}$. The radius of the circular motion is $\mathrm{AG}$.

$$\begin{aligned} \mathrm{AG} & =\frac{2}{3} \mathrm{AD} \\ \mathrm{AD} & =\mathrm{AC} \sin 60^{\circ}=\mathrm{L} \sin 60^{\circ}=\frac{\sqrt{3}}{2} \mathrm{~L} \\ \therefore \quad \mathrm{AG} & =\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \mathrm{~L}=\frac{\mathrm{L}}{\sqrt{3}} \\ \therefore \text { radius } \mathrm{r} & =\frac{\mathrm{L}}{\sqrt{3}} \end{aligned}$$

For uniform circular motion, the gravitational force provides the centripetal force.

$$\begin{aligned} & \therefore \mathrm{mr} \frac{\mathrm{L}}{\sqrt{3}} \cdot \omega^2=\sqrt{3} \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2} \\ & \therefore \omega^2=3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3} \\ & \therefore \quad \omega=\left(3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3}\right)^{1 / 2} \\ & \therefore \frac{2 \pi}{T}=\left(\frac{3 G \mathrm{~m}}{\mathrm{~L}^3}\right)^{1 / 2} \\ & \therefore T=2 \pi\left(\frac{\mathrm{L}^3}{3 \mathrm{Gm}}\right)^{1 / 2} \\ & \therefore T \propto \mathrm{L}^{3 / 2} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}=\mu_0 \mathrm{nI} \\ & \mathrm{n}=\frac{\mathrm{N}}{\mathrm{L}}=\frac{4 \times 1000}{2}=2000 / \mathrm{m}, \mathrm{I}=5 \mathrm{~A} \\ & \begin{aligned} \therefore \quad B & =4 \pi \times 10^{-7} \times 2000 \times 5 \\ & =4 \pi \times 10^{-3} \mathrm{~T} \end{aligned} \end{aligned}$$

If $\mathrm{g}$ is the acceleration due to gravity on earth's surface and $\mathrm{g}^{\prime}$ on the planet, then

$$\begin{aligned} \mathrm{g} & =\frac{\mathrm{GM}}{\mathrm{r}^2} \text { and } \mathrm{g}^{\prime}=\frac{\mathrm{G} \times 1.5 \mathrm{M}}{(1.5)^2 \mathrm{r}^2}=\frac{1}{1.5} \frac{\mathrm{GM}}{\mathrm{r}^2} \\ \therefore \quad \mathrm{g}^{\prime} & =\frac{\mathrm{g}}{1.5} . \end{aligned}$$

For second's pendulum $\mathrm{T}=2 \mathrm{~s}$

$$\begin{aligned} & \therefore \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}=2 \pi \sqrt{\frac{\ell^{\prime}}{\mathrm{g}^{\prime}}} \\ & \therefore \frac{\ell}{\mathrm{g}}=\frac{\ell^{\prime}}{\mathrm{g}^{\prime}} \quad \therefore \ell^{\prime}=\ell \cdot \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1 \times \frac{1}{1.5}=\frac{2}{3}=0.67 \mathrm{~m} \end{aligned}$$

To find the wavelength range of white light when it passes through a medium with a refractive index, we need to understand how wavelength changes as light enters a different medium. The refractive index $n$ is related to the wavelength in the medium by the formula:

$\lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n}$

where $\lambda_{\text{medium}}$ is the wavelength of light in the medium (glass in this case), $\lambda_{\text{vacuum}}$ is the wavelength of light in a vacuum (or air, for practical purposes), and $n$ is the refractive index of the medium.

Given that the refractive index of glass is $1.6$, we will now calculate the new wavelength range of the light inside the glass:

The given range of wavelengths in the air is from $480 \mathrm{~nm}$ to $672 \mathrm{~nm}$.

For the lower end of the range:

$\lambda_{\text{glass, min}} = \frac{480 \mathrm{~nm}}{1.6} = 300 \mathrm{~nm}$

For the upper end of the range:

$\lambda_{\text{glass, max}} = \frac{672 \mathrm{~nm}}{1.6} = 420 \mathrm{~nm}$

Thus, when white light passes through glass with a refractive index of $1.6$, the wavelength range becomes $300 \mathrm{~nm} - 420 \mathrm{~nm}$.

The correct option is:

Option C

$300 \mathrm{~nm} - 420 \mathrm{~nm}$

Fundamenetal frequency of closed pipe $\mathrm{n}_{\mathrm{c}}=\frac{\mathrm{V}}{4 \mathrm{~L}}$

Fundamental frequency of open pipe $\mathrm{n}_0=\frac{\mathrm{v}}{2 \mathrm{~L}}$

They produce 2 beats per second

$$\begin{aligned} & \therefore n_0-n_c=2, \quad \therefore \frac{v}{2 L}-\frac{v}{4 L}=2 \\ & \therefore \frac{v}{4 L}=2 \text { or } \frac{v}{L}=8 \end{aligned}$$

When length of open pipe is halved

$\mathrm{n}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left(\frac{\mathrm{L}}{2}\right)}=\frac{\mathrm{v}}{\mathrm{L}}$

When length of closed pipe is doubled

$\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{4 \times 2 \mathrm{~L}}=\frac{\mathrm{v}}{8 \mathrm{~L}}$

New beat frequency $=\mathrm{n}_{\mathrm{o}}^{\prime}-\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{\mathrm{L}}-\frac{\mathrm{v}}{8 \mathrm{~L}}=\frac{7 \mathrm{v}}{8 \mathrm{~L}}=\frac{7}{8} \times 8=7$

For a monoatomic gas undergoing adiabatic compression, we use the adiabatic relation:

$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$

Given that the gas is compressed to $(1/8)$ of its initial volume, we have:

$V_2 = \frac{V_1}{8}$

We need to find the ratio of the final pressure $(P_2)$ to the initial pressure $(P_1)$. Substituting the volumes into the adiabatic relation, we get:

$P_1 V_1^{\gamma} = P_2 \left(\frac{V_1}{8}\right)^{\gamma}$

Let's solve for $(P_2 / P_1)$:

$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma}$

Since $(V_2 = V_1 / 8)$, we have:

$\frac{V_1}{V_2} = 8$

Therefore:

$P_2 = P_1 \cdot 8^{\gamma}$

Given $(\gamma = 5/3)$, we need to calculate $(8^{\gamma})$:

$8^{5/3} = (2^3)^{5/3} = 2^5 = 32$

Thus, the ratio of the final pressure to the initial pressure is:

$\frac{P_2}{P_1} = 32$

The correct answer is Option A: 32.

Kinetic energy is given by

$\text { (K.E. })_{\max }=\mathrm{h} v-\mathrm{h} v_0$

Comparing $y=m x+c$

We get $x$-intercept when

$\text { (K.E. })_{\max }=0 \text {, i.e. } \mathrm{h} v-\mathrm{h} v_0=0$

or $v=v_0=A$

We get $y$-intercept when $v=0$

$$\begin{aligned} & \therefore \text { (K.E. })_{\max }=-\mathrm{h} v_0=\mathrm{B} \\ & \therefore\left|\frac{\mathrm{B}}{\mathrm{A}}\right|=\frac{\mathrm{h} v_0}{v_0}=\mathrm{h} \end{aligned}$$

The emergent ray just grazes the second face.

Hence angle of emergence $\mathrm{e}=90^{\circ}$

$$\begin{aligned} & \mu=\frac{\sin e}{\sin r_2}=\frac{\sin 90^{\circ}}{\sin r_2}=\frac{1}{\sin r_2} \\ & \therefore \quad \frac{1}{\sin r_2}=\sqrt{2} \quad \text { or } \quad \sin r_2=\frac{1}{\sqrt{2}} . \\ & \therefore \quad r_2=45^{\circ} ; \quad A=r_1+r_2 \quad \therefore r_1=A-r_2=60-45=15^{\circ} \\ & \text { Also, } \frac{\sin i}{\sin r_1}=\mu \quad \therefore \sin i=\mu \sin r_1=\sqrt{2} \sin 15^{\circ} \\ & \therefore \quad i=\sin ^{-1}\left(\sqrt{2} \sin 15^{\circ}\right) \end{aligned}$$