MHT CET 2023 10th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

The truth table for given configuration is as shown below,

Considering case (I), in order to have output (Y) as $1, C_1$ has to be 1. For input values, $\mathrm{A}=0$ and $\mathrm{B}=0$, if $\mathrm{C}_1$ is to be high, the gate $\mathrm{G}$ could be either NAND or NOR.

Considering case (II), in order to have output (Y) as $0, C_2$ has to be 0. For input values, $\mathrm{A}=0$ and $\mathrm{B}=1$. If $\mathrm{C}_2$ is to be 0 , the gate must be NOR.

Frequency of first harmonic $=\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

Frequency of second harmonic $=2 \mathrm{n}=\frac{1}{l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ ..... (i)

For the first overtone, $\lambda=l$

$\therefore \quad$ Frequency of first overtone $\mathrm{n}_1=\frac{1}{l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ ...... (ii)

Comparing (i) and (ii), $\mathrm{n}_1=2 \mathrm{n}$

M.I. of thin circular disc through the tangent in the plane of the disc is $\mathrm{I}=\frac{5}{4} \mathrm{MR}^2$

$\Rightarrow \mathrm{MR}^2=\frac{4}{5} \mathrm{I}$

$\therefore \quad$ M.I. of thin circular disc about an axis perpendicular to plane and passing through its

$\text { centre }=\frac{\mathrm{MR}^2}{2}=\frac{\left(\frac{4}{5} \mathrm{I}\right)}{2}=\frac{2 \mathrm{I}}{5}$

For electron, de Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{V}}$

Given: $\lambda_2=\lambda_1+0.5 \lambda_1=1.5 \lambda_1$

$$\begin{array}{ll} \therefore \quad & \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\ & \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4} \end{array}$$

From $\phi=\mathrm{LI}$

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

Since line A has highest slope, it indicates the largest value of self-inductance.

$\mathrm{n}_{\text {open }}=\frac{\mathrm{v}}{2 \mathrm{~L}}$ ...... (i)

When dipped in water, pipe becomes closed at one end and open at the other.

Length available for resonance is

$$\begin{aligned} l_1 & =25 \% \times \mathrm{L} \\ & =\mathrm{L} \times \frac{25}{100} \\ & =\mathrm{L} / 4 \end{aligned}$$

$\therefore \quad \mathrm{n}_{\text {closed }}=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{L}} \quad \text{..... (ii)}$

Comparing (i) and (ii),

$\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{L}}\right)}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}$

$$\begin{aligned} &\begin{aligned} \text { Dipole moment } \mathrm{p} & =\mathrm{q} \times 2 \mathrm{l}=2 \times 10^{-6} \times 3 \times 10^{-2} \\ & =6 \times 10^{-8} \mathrm{Cm} \end{aligned}\\ &\begin{aligned} \therefore \quad \tau_{\max } & =\rho E \sin \theta=\mathrm{pE} \sin 90^{\circ} \\ & =6 \times 10^{-8} \times 2 \times 10^5 \times 1 \\ & =12 \times 10^{-3} \mathrm{Nm} \end{aligned} \end{aligned}$$

Using Rydberg's formula,

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

where $R_H$ is the Rydberg's constant

For calculating the shortest wavelength in the Balmer series of hydrogen atom, $\mathrm{n}=2, \mathrm{~m}=\infty$ and $\mathrm{Z}=1$

$\therefore \quad \lambda_1=\frac{4}{\mathrm{R}_{\mathrm{H}}}$

For hydrogen like atom, the shortest wavelength is given by,

In Brackett series

$$\begin{aligned} & \mathrm{n}=4, \mathrm{~m}=\infty \\ \therefore \quad & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2\left(\frac{1}{16}\right) \\ \therefore \quad \lambda_2 & =\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \end{aligned}$$

Given: $\lambda_1=\lambda_2$,

$$\begin{aligned} & \therefore \quad \frac{4}{R_H}=\frac{16}{R_H \cdot Z^2} \Rightarrow Z^2=4 \\ & \therefore \quad Z=2 \end{aligned}$$

Let the initial capacitance be $\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$

Let the charge on the capacitor be $Q_{\text {initial }}=C_0 V_0$

Plate separation is increased by 3 times i.e., $\mathrm{d}^{\prime}=3 \mathrm{~d}$

$\mathrm{C}_{\text {final }}=\frac{\varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{1}{3}\left(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=\frac{\mathrm{C}_0}{3}$

Let $\mathrm{Q}_{\text {final }}$ be the final charge on the capacitor and $V_{\text {final }}$ be the final potential on the capacitor.

$\therefore \quad \mathrm{Q}_{\text {final }}=\mathrm{C}_{\text {final }} \mathrm{V}_{\text {final }}=\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {final }}$

As the capacitor is isolated,

$$\begin{aligned} & \mathrm{Q}_{\text {final }}=\mathrm{Q}_{\text {initial }} \text {, } \\ & \mathrm{C}_0 \mathrm{~V}_0=\frac{1}{3} \mathrm{C}_0 \mathrm{~V}_{\text {final }} \\ & \therefore \quad \mathrm{V}_{\text {final }}=3 \mathrm{~V}_0 \\ & \text { Work done }=\text { Final P.E }- \text { Initial P.E } \\ & =\frac{1}{2} \mathrm{C}_{\text {final }}\left(\mathrm{V}_{\text {final }}\right)^2-\frac{1}{2} \mathrm{C}_0\left(\mathrm{~V}_0\right)^2 \\ & =\frac{1}{2} \frac{\mathrm{C}_0}{3} 9 \mathrm{~V}_0^2-\frac{1}{2} \mathrm{C}_0 \mathrm{~V}_0^2 \\ & =\mathrm{C}_0 \mathrm{~V}_0^2\left(\frac{3}{2}-\frac{1}{2}\right) \\ & =\mathrm{C}_0 \mathrm{~V}_0^2 \\ & =\frac{\varepsilon_0 \mathrm{AV}_0^2}{\mathrm{~d}} \quad \ldots\left(\because \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\right) \\ \end{aligned}$$

M.I of a circular disc $=\frac{\mathrm{MR}^2}{2}$

Using parallel axis theorem, M.I. about origin $\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+6 \mathrm{I}$

where, $\mathrm{I_{c m}=M. I}$ of the central disc

$\mathrm{I}^{\prime}=\mathrm{M}$. I of the each disc about the given axis.

$$\begin{aligned} \therefore \quad \mathrm{I} & =\frac{\mathrm{MR}^2}{2}+6\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{MD}^2\right) \\ & =\frac{\mathrm{MR}^2}{2}+6\left(\frac{\mathrm{MR}^2}{2}+4 \mathrm{MR}^2\right) \quad \ldots(\because \mathrm{D}=2 \mathrm{R}) \\ & =\frac{\mathrm{MR}^2}{2}+6\left(\frac{\mathrm{MR}^2+8 \mathrm{MR}^2}{2}\right) \\ & =\frac{55 \mathrm{MR}^2}{2} \end{aligned}$$

$$\begin{aligned} & \mathrm{C}_1=3 \mu \mathrm{F} \text { and } \mathrm{C}_2=2 \mu \mathrm{F} \\ \therefore \quad & \mathrm{C}_{\text {series }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{6}{5} \mu \mathrm{F} \\ & \text { Also, } \mathrm{Q}=\mathrm{CV} \\ \mathrm{Q} & =\mathrm{C}_{\text {series }} \times \mathrm{V} \\ & =\frac{6}{5} \times 100=120 \mu \mathrm{C} \end{aligned}$$

Q will be the same across both the capacitors as they are in series.

$\therefore \quad$ Potential across capacitors,

$$\begin{aligned} & \mathrm{V}_1=\frac{\mathrm{Q}}{\mathrm{C}_1}=\frac{120}{3}=40 \mathrm{~V} \\ & \mathrm{~V}_2=\frac{\mathrm{Q}}{\mathrm{C}_2}=\frac{120}{2}=60 \mathrm{~V} \\ \therefore \quad \mathrm{V}_2 & : \mathrm{V}_1=60: 40=3: 2 \end{aligned}$$

Given: area of square loop $=25 \mathrm{~cm}$

$$\begin{array}{ll} \therefore \quad l=\sqrt{25}=5 \mathrm{~cm} \Rightarrow 0.05 \mathrm{~m} \\ \quad \mathrm{R}=10 \Omega, \mathrm{t}=1 \mathrm{sec}, \mathrm{B}=40 \mathrm{~T} \\ \therefore \quad \text { Velocity } \mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s} \\ \quad \text { Motional emf } \varepsilon_{\max }=\mathrm{B} / \mathrm{v} \\ \therefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}} \\ \therefore \quad \\ \quad \mathrm{I}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A} \end{array}$$

We know, Force acting on loop

$$\begin{aligned} & \therefore \quad \mathrm{F}=\mathrm{BI} l=40 \times 0.01 \times 0.05 \\ & =0.02 \mathrm{~N} \\ & \text { Using } \mathrm{W}=\mathrm{F} . \mathrm{s} \text {, } \\ & \text { Work done } \mathrm{W}=\mathrm{BI} l \times l \\ & =0.02 \times 0.05 \\ & =1 \times 10^{-3} \mathrm{~J} \\ \end{aligned}$$

Let the initial length and tension be $l$ and $T$ respectively.

After shortening,

The new length $l_{\text {new }}=l-\frac{40}{100} l=\frac{3}{5} l$

After increase in tension,

the new tension $\mathrm{T}_{\text {new }}=\mathrm{T}+\frac{44}{100} \mathrm{~T}=\frac{144 \mathrm{~T}}{100}$

Fundamental frequency of a vibrating string is given by

$$\begin{aligned} \mathrm{n} & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ \therefore \quad \mathrm{n}_1 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ \mathrm{n}_2 & =\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{\text {new }}}{\mathrm{m}}} \\ \therefore \quad \frac{\mathrm{n}_1}{\mathrm{n}_2} & =\frac{l_{\text {new }}}{l} \times \frac{\sqrt{\mathrm{T}}}{\sqrt{\mathrm{T}_{\text {new }}}} \\ & =\frac{\frac{3}{5} l}{l} \times \sqrt{\frac{100 \mathrm{~T}}{144 \mathrm{~T}}} \\ & =\frac{3}{5} \times \frac{10}{12}=\frac{1}{2} \\ \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1} & =\frac{2}{1} \end{aligned}$$

The correct answer for the description of electronic bands in insulators is:

Option B: conduction band is empty and valence band is completely filled with electrons.

In insulators, there is a large energy gap between the valence band and the conduction band. The valence band is the band of electron orbitals that electrons occupy at absolute zero temperature, and it's usually filled with electrons in an insulator. The conduction band is the band above the valence band where an electron may be excited to in order to conduct electricity. In insulators, this conduction band is empty because the energy gap (band gap) between the valence band and the conduction band is so large that electrons in the valence band do not normally have enough energy to jump into the conduction band under room temperatures or normal conditions. Hence, there is no electrical conductivity in insulators under these conditions.

Let's break down the incorrect options:

Option A: In insulators, the valence band is not empty. It is filled with electrons, which is in direct contradiction to what this option states.

Option C: In insulators, the valence band is completely filled, not partially filled. It is the complete filling of the valence band which contributes to their insulating properties, combined with the large energy gap to the conduction band.

Option D: The conduction band in insulators is not partially filled; it is empty due to the energy required to move electrons from the valence band to the conduction band being prohibitively large for thermal excitation.

To understand the conditions for a stable circular orbit around Earth, we need to know the meanings of the critical velocity ($\mathrm{V}_{\mathrm{c}}$) and escape velocity ($\mathrm{V}_{\mathrm{e}}$).

Critical Velocity ($\mathrm{V}_{\mathrm{c}}$) is the necessary orbital velocity a body must have to be in a stable circular orbit around Earth without the need for propulsion. It depends on the gravitational force providing the necessary centripetal force to keep the satellite in orbit.

Escape Velocity ($\mathrm{V}_{\mathrm{e}}$) is the minimum velocity an object must have to break free from the gravitational attraction of a celestial body, like earth, without further propulsion.

For a satellite to maintain a stable orbit, its horizontal velocity ($\mathrm{V}_{\mathrm{H}}$) must be equal to the critical velocity ($\mathrm{V}_{\mathrm{c}}$). If the horizontal velocity is less than critical velocity, the gravitational pull will cause the satellite to spiral downwards towards Earth. If the horizontal velocity is greater than critical velocity but less than escape velocity, the satellite will enter an elliptical orbit. And if the horizontal velocity equals or exceeds escape velocity, the satellite will leave Earth's orbit entirely.

In simple terms, for a stable, circular orbit, the option is:

Option C : $V_H=V_c$

Options A and D would not result in a stable circular orbit. Option A would lead to a decaying orbit, and Option D would cause the satellite to leave orbit entirely. Option B implies that the satellite would be on an escape path and thus not in a stable orbit around the Earth.

$\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_0$ ....... (i)

Given: $\lambda=\frac{\lambda}{3}$ and $\mathrm{E}=4 \mathrm{E}$

$$\begin{aligned} & 4 \mathrm{E}=\frac{\mathrm{hc}}{\lambda / 3}-\phi_0 \quad \text{..... (ii)} \\ & =\frac{3 \mathrm{hc}}{\lambda}-\phi_0 \\ & 4\left(\frac{\mathrm{hc}}{\lambda}-\phi_0\right)=\frac{3 \mathrm{hc}}{\lambda}-\phi_0 \quad \text{.....(From (i))}\\ & \frac{4 \mathrm{hc}}{\lambda}-4 \phi_0=\frac{3 \mathrm{hc}}{\lambda}-\phi_0 \\ & \frac{\mathrm{hc}}{\lambda}=3 \phi_0 \\ & \phi_0=\frac{\mathrm{hc}}{3 \lambda} \\ \end{aligned}$$

Let the body be at $\mathrm{x}$ from the top after $\frac{\mathrm{t}}{2} \mathrm{~s}$.

$$\begin{aligned} \therefore \quad \mathrm{x} & =\frac{1}{2} \mathrm{~g} \frac{\mathrm{t}^2}{4}=\frac{g \mathrm{t}^2}{8} \quad \text{.... (i)} \\ \mathrm{H} & =\frac{1}{2} \mathrm{gt}^2 \quad \text{.... (ii)} \end{aligned}$$

Eliminating $\mathrm{t}$ from (i) and (ii), we get

$\frac{8 \mathrm{x}}{\mathrm{g}}=\frac{2 \mathrm{H}}{\mathrm{g}} \Rightarrow \mathrm{x}=\frac{\mathrm{H}}{4}$

$\therefore \quad$ Height of the body from the ground

$=\mathrm{H}-\frac{\mathrm{H}}{4}=\frac{3 \mathrm{H}}{4}$ metres

Let the normal length be $\mathrm{L}$ and the extension be $\mathrm{x}$.

$\therefore \quad$ Restoring Force $=$ Centripetal Force

$$\begin{aligned} & \mathrm{kx}=\mathrm{m}(\mathrm{L}+\mathrm{x}) \omega^2 \\ & 12.5 \mathrm{x}=0.2(\mathrm{~L}+\mathrm{x}) 25 \quad \ldots(\because \omega=5 \mathrm{~rad} / \mathrm{s}) \end{aligned}$$

$$\begin{aligned} & 12.5 x=5(L+x) \\ & 7.5 x=5 L \\ \therefore \quad & \frac{x}{L}=\frac{5}{7.5}=\frac{2}{3}=2: 3 \end{aligned}$$

From Wien's Displacement law,

$\lambda_{\max }=\frac{b}{T} \Rightarrow T=\frac{b}{\lambda_{\max }}$

From Stefan-Boltzmann law

$E=\sigma T^4=\sigma\left(\frac{b}{\lambda_{\max }}\right)^4$

Let the new emissive power be $\mathrm{E}^{\prime}$.

$$\begin{aligned} \therefore \quad & E^{\prime}=\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^4 \\ & E^{\prime}=\frac{81}{16} E \end{aligned}$$

$$\begin{aligned} & \text { Given }: \mathrm{K} \cdot \mathrm{E}=\mathrm{P} \cdot \mathrm{E}+\frac{25}{100} \cdot \mathrm{P} \cdot \mathrm{E} \\ & \mathrm{K} \cdot \mathrm{E}=\mathrm{P} \cdot \mathrm{E}+\frac{1}{4} \mathrm{P} \cdot \mathrm{E} \\ & \text { We know K.E }=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \text { and P.E }=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 \\ & \therefore \quad \mathrm{K} \cdot \mathrm{E}=\frac{5}{4} \mathrm{P} \cdot \mathrm{E} \\ & \frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right)=\frac{5}{4}\left(\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2\right) \\ & \mathrm{A}^2-\mathrm{x}^2=\frac{5}{4} \mathrm{x}^2 \\ & \mathrm{~A}^2=\frac{9}{4} \mathrm{x}^2 \\ \therefore \quad & \mathrm{~A}=\frac{3}{2} \mathrm{x} \\ \therefore \quad & \mathrm{x}=\mathrm{A} \times \frac{2}{3}=2 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & T_H=600 \mathrm{k} \\ & \eta=1-\frac{T_C}{T_H} \\ & \text { but, } \eta=\frac{1}{2} \quad ....(\text{Given} : \eta=50 \%)\\ & \Rightarrow \frac{1}{2}=1-\frac{T_C}{600} \\ & \therefore \quad T_C=300 K \end{aligned}$$

With $\mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}$, the efficiency is increased to $70 \%$

$\therefore \quad$ New temperature of the source will be $\mathrm{T}_{\mathrm{H}_{\text {new }}}$

$$\begin{array}{ll} \therefore \quad \eta=1-\frac{300}{\mathrm{~T}_{\mathrm{H}_{\text {new }}}} \\ \frac{300}{\mathrm{~T}_{\mathrm{H}_{\text {new }}}}=1-\frac{7}{10} \quad \ldots .(\because \eta=70 \%) \\ \therefore \quad \mathrm{T}_{\mathrm{H}_{\text {new }}}=\frac{3000}{3}=1000 \mathrm{~K} \end{array}$$

For Paschen series, Longest wavelength corresponds to

$$\begin{aligned} & \mathrm{n}_1=3, \mathrm{n}_2=\mathrm{n}_1+1=4 \\ & \lambda_{\max }=\frac{144}{7 \mathrm{R}} \end{aligned}$$

Shortest wavelength corresponds to $\mathrm{n}_1=3, \mathrm{n}_2=\infty$

$$\begin{aligned} \lambda_{\min } & =\frac{9}{R} \\ \therefore \quad \frac{\lambda_{\max }}{\lambda_{\min }} & =\frac{\left(\frac{144}{7 R}\right)}{\left(\frac{9}{R}\right)}=\frac{144}{63} \end{aligned}$$

$$\begin{aligned} X_L & =\omega L=2 \pi f L \\ f_{\text {new }} & =2 f \text { and } L_{\text {new }}=3 L \\ \therefore \quad X_L^{\prime} & =2 \pi(2 f) 3 L \\ & =6 \times 2 \pi f L=6 X_L \end{aligned}$$

Resolving Power $R=\frac{\text { N.A }}{0.61 \lambda}$

N.A $=2 \mathrm{n} \sin \alpha$

$\therefore$ By increasing the refractive index of the medium between the subject and the objective lens, the resolving power can be increased.

[Note: Since mass of the metal is not mentioned in the question, it is not possible to arrive at the answer.]

Mutual inductance of two concentric coplanar circular coils,

$$\begin{aligned} & \quad M=\frac{\mu_0 \pi N_1 N_2 R_2^2}{2 R_1} \\ & \therefore \quad M \propto \frac{R_2^2}{R_1} \end{aligned}$$

Given: $I_g=\frac{1}{100} I$

$$\begin{aligned} & \therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=100 \quad \text{..... (i)} \\ & \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1} \quad \text{..... (ii)} \end{aligned}$$

Substituting, (i) into (ii),

$\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$

Internal resistance,

$$\begin{aligned} r & =\left[\frac{E}{V}-1\right] R \\ & =\left[\frac{3}{2.5}-1\right] 150 \\ & =30 \Omega \end{aligned}$$

$$\begin{aligned} & \text { K.E }=\frac{G M m}{2 r} \\ & P . E=-\frac{G M m}{r} \\ \therefore \quad & \frac{K . E}{|P . E|}=\frac{G M m}{2 r} \times \frac{r}{G M m} \\ \therefore \quad \frac{P . E}{K . E} & =\frac{2}{1}=2 \end{aligned}$$

$\mu_{\text {water }}=1.33 \text { and } \mu_{\text {air }}=1$

As air is optically rarer, $\mu_{\text {air }}<\mu_{\text {glass, }}$ converging nature of the convex lens changes to diverging. Thus it behaves as divergent lens.

The option that is NOT involved in the formation of a secondary rainbow is Option C: Interference.

Rainbows are formed from the optical processes that occur when sunlight interacts with raindrops. A secondary rainbow, which appears outside the primary rainbow and is fainter and has its colors reversed, is created through three processes :

• Refraction: When sunlight enters a raindrop, it is bent (refracted) due to the change in medium from air to water. This bending of light is essential in separating the white sunlight into its component colors.


• Angular dispersion: This term refers to the separation of white light into its colors due to the different degrees of bending (refraction) for different wavelengths/colors. Violet light bends more than red light, resulting in dispersion.


• Total internal reflection: After refraction, the light rays strike the other side of the raindrop and reflect. In the secondary rainbow, light reflects twice inside the raindrop before emerging out. This double reflection causes the colors to be reversed as compared to the primary rainbow, and contributes to the formation of the secondary rainbow.

Interference, on the other hand, is the phenomenon where waves overlap and combine with each other, leading to patterns of constructive and destructive interference. While this process is crucial in some optical phenomena, like the thin film interference seen in soap bubbles, it is not a key process in the formation of the typical primary or secondary rainbow. In rainbows, each individual raindrop contributes to the rainbow separately, and interference between waves from different raindrops does not play a role in their formation.

For observer moving towards a stationary source,

$\mathrm{n}_1=\mathrm{n}_0\left[\frac{\mathrm{v}+\mathrm{v}_{\mathrm{L}}}{\mathrm{v}}\right]$

For source moving towards a stationary observer,

$\mathrm{n}_2=\mathrm{n}_0\left[\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right]$

Substituting the values for $\mathrm{v}, \mathrm{v}_{\mathrm{L}}$ and $\mathrm{v}_{\mathrm{S}}$ in the equations above

$$\begin{aligned} & \mathrm{n}_1=\mathrm{n}_0\left[\frac{330+50}{330}\right]=1.15 \mathrm{~n}_0 \\ & \mathrm{n}_2=\mathrm{n}_0\left[\frac{330}{330-50}\right]=1.17 \mathrm{~n}_0 \\ & \Rightarrow \mathrm{n}_2>\mathrm{n}_1 \end{aligned}$$

$\therefore \quad$ The frequency heard will be more in the second case than in the first case.

Fringe width, $\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{6000 \times 10^{-10} \times 1.2}{0.6 \times 10^{-3}}=1.2 \mathrm{~mm}$

$\therefore \quad$ Number of fringes $(\mathrm{n})=\frac{3}{1.2}=2.5$

$\therefore \quad$ Phase difference,

$\Delta \phi=2 \mathrm{n} \pi=2 \times 2.5 \pi=5 \pi$

Using the formula for adiabatic expansion,

$$\begin{aligned} & \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\ & \mathrm{~T}_1 \mathrm{~V}_1{ }^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1{ }^{1 / 2}\right) \quad ..... (\text{Given}: \gamma=\frac{3}{2})\\ \therefore \quad & \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \\ \therefore \quad & \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}} \end{aligned}$$