MHT CET 2023 12th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

$\therefore \quad$ There are two NOR gates, one NOT gate, and one NAND gate.

$\therefore \quad$ Output of NAND gate: $\overline{A \cdot B}$

Output of NOT and NOR Gate: $\overline{\bar{A}+B}$

Final output: $\overline{(\overline{\mathrm{A} \cdot \mathrm{B}})+(\overline{\overline{\mathrm{A}}+\mathrm{B}})}$

So, the output $\mathrm{Y}$ is 1 only if the input $\mathrm{A}$ and $\mathrm{B}$ is 1.

$$\begin{aligned} & A=1 \\ & B=1 \\ & Y=\overline{\overline{1 \cdot 1}+\overline{1}+1} \\ & Y=1 \end{aligned}$$

Here, the normal shift is $x$

The formula for the normal shift is

$$\begin{aligned} x & =t\left(1-\frac{1}{\mu}\right) \\ t & =\frac{x}{\left(1-\frac{1}{\mu}\right)} \\ x & =t\left(1-\frac{1}{\mu}\right) \\ t & =\frac{x}{\left(1-\frac{1}{\mu}\right)} \\ \therefore \quad t & =\frac{x \mu}{(\mu-1)} \end{aligned}$$

At both points, the momentum will be $\mathrm{mv}$, but will be in opposite directions.

$$\begin{array}{rlrl} \therefore & \Delta \mathrm{P} =\mathrm{mv}-(-\mathrm{mv}) \\ \therefore & \Delta \mathrm{P}=2 \mathrm{~mv} \end{array}$$

The momentum changes by $2 \mathrm{~mv}$.

The tension at the lowest point is $\mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}}+\mathrm{mg}$

To complete the vertical circle, the minimum velocity should be $\mathrm{v}=\sqrt{5 \mathrm{gr}}$

$$\begin{aligned} \therefore \quad \mathrm{T} & =\frac{\mathrm{m}(\sqrt{5 \mathrm{gr}})^2}{\mathrm{r}}+\mathrm{mg} \\ \mathrm{R} & =5 \mathrm{mg}+\mathrm{mg} \\ \therefore \quad \mathrm{T} & =6 \mathrm{mg} \end{aligned}$$

$$\begin{array}{ll} & \text { For an adiabatic process } \\ & \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\ \therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \end{array}$$

For a monoatomic gas, $r=\frac{5}{3}$

$$\begin{array}{cl} & \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\ & V_1=A L_1 \text { and } V_2=A L_2 \\ \therefore \quad & \frac{T_2}{T_1}=\left[\frac{A L_1}{A L_2}\right]^{2 / 3}=\left[\frac{L_1}{L_2}\right]^{2 / 3} \\ & \gamma-1=\frac{2}{3} \end{array}$$

For an adiabatic process,

$$\begin{aligned} & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\frac{2}{3}} \\ & \mathrm{~V} \propto \mathrm{L} \\ & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}} \end{aligned}$$

The frequency of a vibrating wire $f=\frac{1}{2 l} \sqrt{\frac{T}{m}}$

$\therefore \quad \mathrm{f} \propto \frac{1}{l} \Rightarrow \mathrm{fl}=$ constant

Let the frequency of the fork be $f$ and the initial and final frequencies of the wire be $f_1$ and $f_2$.

The number of beats heard before decreasing the length is $f-f_1=3$ .... (i)

The number of beats after decreasing the length is $\mathrm{f}_2-\mathrm{f}=3$ .... (ii)

$\therefore \quad \mathrm{f}_1 l_1=\mathrm{f}_2 l_2$

$\therefore (\mathrm{f}-3) 50=(\mathrm{f}+3) 49 \quad$ ... from (i) and (ii)

$\quad 50 \mathrm{f}-49 \mathrm{f}=147+150$

$\therefore \quad \mathrm{f}=297 \mathrm{~Hz}$

In streamlined flow, the velocity of a fluid at a specific point is always constant because each particle of the fluid follows a well-defined path and there is no mixing or crossing of paths between particles.

At resonance condition, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$

The impedance is given as:

$$\begin{aligned} \mathrm{Z} & =\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \\ \therefore \quad \mathrm{Z} & =\mathrm{R}=2 \Omega \\ \qquad\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{e}_{\mathrm{ms}}}{\mathrm{R}} \\ \mathrm{I}_{\mathrm{rms}} & =\frac{0.1}{2}=0.05 \mathrm{~A} \\ \omega & =\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{10^{-4} \times 4 \times 10^{-10}}} \\ \omega & =5 \times 10^6 \end{aligned}$$

$\therefore \quad$ The voltage-drop across the inductor is,

$$\begin{aligned} & V=I_{\mathrm{rms}} \times X_L=I_{\mathrm{rms}} \times L \omega \\ & V=0.05 \times 10^{-4} \times 5 \times 10^6 \\ & V=25 \mathrm{~V} \end{aligned}$$

$$\begin{aligned} & \omega=\frac{2 \pi}{T} \\ & I=\frac{q \omega}{2 \pi}=\frac{1.6 \times 10^{-19} \times 2 \pi}{2 \pi} \quad \ldots .\left(\because I=\frac{q}{t}\right) \\ & I=1.6 \times 10^{-19} \mathrm{~A} \end{aligned}$$

$\therefore \quad$ The magnetic field at the centre of the circle is:

$$\begin{aligned} & \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\ & \mathrm{~B}=4 \pi \times 10^{-26} \mathrm{~T} \end{aligned}$$

The heat loss is given as $\mathrm{R}=\operatorname{e} \sigma \mathrm{A}\left(\mathrm{T}^4-\mathrm{T}_0^4\right)$

Let surrounding temperature be denoted as $\mathrm{T}$

Heat loss from the metal sphere at a temperature $\mathrm{T}_1$

$\mathrm{R}_1=\operatorname{e\sigma A}\left(\mathrm{T}_1^4-\mathrm{T}_0^4\right)$

Heat loss from the metal sphere at a temperature $\mathrm{T}_2$

$\mathrm{R}_2=\operatorname{e} \sigma \mathrm{A}\left(\mathrm{T}_2^4-\mathrm{T}^4\right)$

$$\begin{aligned} \therefore \quad \frac{R_1}{R_2} & =\frac{T_1^4-T^4}{T_2^4-T^4}=\frac{900^4-300^4}{600^4-300^4} \\ \frac{R_1}{R_2} & =5.3 \end{aligned}$$

Net electrostatic potential energy of the system is,

$$\begin{aligned} & \quad U=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q q}{a}+\frac{2 Q q}{a}+\frac{2 q q}{\sqrt{2} a}\right)=0 \\ & Q+2 Q+\frac{2 q}{\sqrt{2}}=0 \\ & 3 Q+\sqrt{2} q=0 \\ & \therefore \quad Q=\frac{-\sqrt{2} q}{3} \end{aligned}$$

$\therefore \quad$ Emissive power of black body is, From Wien's Law:

$$\begin{aligned} \mathrm{E} & =\sigma \mathrm{T}^4 \\ \lambda & =\frac{\mathrm{b}}{\mathrm{T}} \\ \therefore \quad \mathrm{E} & =\frac{\sigma \mathrm{b}^4}{\lambda^4} \\ \therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\lambda_1^4}{\lambda_2^4}=\frac{\lambda^4}{\left(\frac{\lambda}{3}\right)^4}=\frac{81}{1} \end{aligned}$$

$\therefore \quad$ Dispersive Power of Prism:

$$\begin{aligned} & \omega=\frac{\delta_v-\delta_r}{\delta_y} \\ & \delta_y=\frac{\delta_v+\delta_r}{2}=\frac{11+9}{2} \end{aligned}$$

$\therefore \quad$ For first Prism,

$$\begin{aligned} \omega_1 & =\frac{2(11-9)}{11+9} \\ & =\frac{1}{5} \end{aligned}$$

For Second Prism:

$$\begin{aligned} \omega_2 & =\frac{2(13-11)}{13+11} \\ & =\frac{1}{6} \end{aligned}$$

Ratio:

$\frac{\omega_2}{\omega_1}=\frac{\frac{1}{6}}{\frac{1}{5}}=\frac{5}{6}$

$\therefore \quad$ The ratio is $5: 6$.

Time period of pendulum: $T=2 \pi \sqrt{\frac{L}{g}}$

When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,

$$\begin{aligned} & g=g-\frac{g}{4} \\ & g=\frac{3 g}{4} \end{aligned}$$

$\therefore\quad$ New Time period will be

$$\begin{aligned} & T_1=2 \pi \sqrt{\frac{4 L}{3 g}} \\ & T_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{L}{g}} \\ & T_1=\frac{2}{\sqrt{3}} T \end{aligned}$$

$\therefore \quad$ Activity is given as:

$$\begin{aligned} \therefore \quad \mathrm{A} & =\lambda \mathrm{N} \\ \lambda & =\frac{\ln 2}{\mathrm{~T}} \end{aligned}$$

$\therefore \quad$ Ratio of two different radioactive elements will be:

$$\begin{aligned} & \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\lambda_1 \mathrm{~N}_1}{\lambda_2 \mathrm{~N}_2} \\ & \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\frac{\ln 2}{\mathrm{~T}_1} \mathrm{~N}_1}{\frac{\ln 2}{\mathrm{~T}_2} \mathrm{~N}_2} \\ & \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{N}_1 \mathrm{~T}_2}{\mathrm{~N}_2 \mathrm{~T}_1} \end{aligned}$$

The statement that is WRONG regarding LED is:

Option C: Brightness of light emitted by LED cannot be controlled.

This statement is incorrect because the brightness of LEDs can indeed be controlled. The brightness of an LED is directly proportional to the current passing through it. Thus, by adjusting the current, we can dim or brighten an LED. This can be achieved by various methods such as pulse-width modulation (PWM), which rapidly turns the LED on and off to control the average amount of light being emitted without changing its color, or by using a current limiting resistor or a constant current source to directly adjust the current. Advanced lighting systems also make use of digital controls to regulate the intensity of LED lighting.

Let's briefly evaluate the other options to understand why they are considered correct:

Option A: LEDs are energy efficient - This is true. LEDs use much less energy compared to traditional incandescent bulbs for the same amount of light output, making them a more energy-efficient lighting solution.

Option B: LEDs have long life if properly manufactured - This statement is correct as well. LEDs can have a very long lifespan, often up to 25,000-50,000 hours, provided they are well-manufactured and used within their specified operating conditions.

Option D: Colours produced by LED do not fade out - LEDs maintain their color throughout their lifespan. Unlike some traditional lighting options that can change color as the filament wears out or the gas in the bulb is depleted, LEDs do not suffer from this issue and thus the colors remain consistent.

The potential energy between two charges is given as $U=\frac{k q_1 q_2}{r}$

Initial potential energy is $U_{\mathrm{f}}=\frac{k q_1 q_2}{r}$

When charge $q_2$ moves towards the $q_1$ the separation between the charges becomes $\mathrm{d}-\mathrm{x}$

The final potential energy is $U_{\mathrm{f}}=\frac{\mathrm{kq}_1 \mathrm{q}_2}{(\mathrm{~d}-\mathrm{x})}$

The increase in potential energy is

$$\begin{aligned} & \therefore \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{f}} \\ & \therefore \quad \Delta U=\frac{k q_1 q_2}{d}-\frac{k_1 q_2}{(d-x)} \\ & \therefore \quad \Delta U=\mathrm{kq}_1 \mathrm{q}_2\left(\frac{1}{\mathrm{~d}}-\frac{1}{\mathrm{~d}-\mathrm{x}}\right) \\ & \therefore \quad \Delta U=\frac{-\mathrm{kq}_1 \mathrm{q}_2 \mathrm{x}}{\mathrm{d}(\mathrm{d}-\mathrm{x})} \\ \end{aligned}$$

Energy stored in a charged capacitor is

$\mathrm{U}=\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad \Rightarrow \mathrm{U} \propto \mathrm{Q}^2$

As per given condition, when, $\mathrm{Q}_2=\left(\mathrm{Q}_1+3\right)$,

$$\begin{array}{ll} \therefore & \mathrm{U}_2=44 \% \text { of } \mathrm{U}_1=\frac{144}{100} \mathrm{U}_1 \\ \therefore & \frac{\mathrm{Q}_2^2}{2 \mathrm{C}}=\frac{144}{100} \times \frac{\mathrm{Q}_1^2}{2 \mathrm{C}} \\ \therefore & \mathrm{Q}_1=\frac{10}{12} \mathrm{Q}_2 \\ \therefore & \mathrm{Q}_1=\frac{5}{6}\left(\mathrm{Q}_1+3\right) \\ \therefore & 6 \mathrm{Q}_1=5 \mathrm{Q}_1+15 \\ \therefore & \mathrm{Q}_1=15 \mathrm{C} \end{array}$$

Option C is the incorrect statement for an isothermal process. An isothermal process is one in which the temperature (T) of a system remains constant throughout the process. In thermodynamics, the equation of state for an ideal gas is given by the ideal gas law, which states:

$PV = nRT$

where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature.

In an isothermal process, because the temperature T is constant, the product of P and V must also remain constant for a given amount of gas with a fixed number of moles (n). This is because R is a constant and T is held constant in the process, and thus, the product PV does not change even as pressure and volume do. Therefore,

$PV = constant$

This directly contradicts Option C, which states that the equation of state PV is not constant. Since in an isothermal process for an ideal gas, PV always remains constant if the temperature is constant, Option C is the wrong statement.

Let's verify the other options for an ideal gas undergoing an isothermal process:

Option A is correct because in an isothermal expansion or compression, the energy exchanged with the surroundings is indeed used to do work. No change in internal energy occurs (Option D), due to the fact that internal energy for an ideal gas is a function of temperature alone, and in an isothermal process, temperature does not change. The energy needed to change the volume of the gas comes from or goes into the work done by or on the gas.

Option B is a somewhat ambiguous statement. It can be interpreted as saying the system is in a state of thermal equilibrium with the environment because the temperature remains constant. However, maintaining a perfect thermal equilibrium throughout the process would mean the surroundings are also at the same temperature and capable of supplying or absorbing energy without changing temperature themselves, which is a theoretical idealization.

Nevertheless, Option C is clearly the wrong statement out of the four provided when considering the definition and properties of an isothermal process.

$$\begin{aligned} \therefore & \quad \mathrm{n}_3=\frac{3 \mathrm{v}}{2 l}=\frac{3 \mathrm{v}}{2(l+2 \mathrm{e})} \\ & \lambda_3=\frac{2(l+2 \mathrm{e})}{3} \quad \ldots .\left(\because \mathrm{v}=\mathrm{n}_3 \lambda\right) \end{aligned}$$

For polyatomic gases, molar-specific heat at constant volume is $\mathrm{C}_{\mathrm{V}}=(3+\mathrm{f}) \mathrm{R}$ and Molar-specific heat at constant pressure is $C_P=(4+f) R$

$\therefore \quad \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{V}}}=\frac{4+\mathrm{f}}{3+\mathrm{f}}$

For spectral series, $\frac{1}{\lambda}=R Z^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$

For the Balmer series, $\mathrm{n}_1=2$

The wavelength for $2^{\text {nd }}$ line of the Balmer series is

$$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ & \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ & \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{3}{16}\right) \Rightarrow \lambda_1=\left[\frac{16}{3}\right] \end{aligned}$$

For the Paschen series, $\mathrm{n}_1=3$

The wavelength for $1^{\text {st }}$ line of the Paschen series is

$$\begin{aligned} \frac{1}{\lambda_2} & =R Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ \frac{1}{\lambda_2} & =R Z^2\left(\frac{1}{9}-\frac{1}{16}\right) \\ \lambda_2 & =\frac{144}{7} \\ \frac{1}{\lambda_2} & =R Z^2\left(\frac{7}{144}\right) \\ \therefore \quad \frac{\lambda_1}{\lambda_2} & =\frac{16}{3} \times \frac{7}{144}=\frac{7}{27} \end{aligned}$$

We know,

$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$

The total time required for the ball to go up and reach the ground is $t=t_1+t_2$, and the total displacement is zero.

$$\begin{aligned} & \therefore \quad 0=u\left(t_1+t_2\right)+\frac{1}{2} g\left(t_1+t_2\right)^2 \\ & \therefore \quad u=\frac{1}{2} g\left(t_1+t_2\right) \end{aligned}$$

The displacement in time $t_1$ is

$$\begin{aligned} h & =\frac{1}{2} g\left(t_1+t_2\right) t_1-\frac{1}{2} g t_1^2 \\ h & =\frac{1}{2} g t_1\left(t_1+t_2-t_1\right) \\ \therefore \quad h & =\frac{1}{2} g t_1 t_2 \end{aligned}$$

de Broglie wavelength $\lambda=\frac{h}{p}$

$$\begin{aligned} & \therefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}} \\ & \therefore \lambda \sqrt{\mathrm{v}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq}}} \Rightarrow \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{v}}}\right)}=\frac{1}{\sqrt{2 \mathrm{mq}}} \end{aligned}$$

$\therefore \quad$ Slope of the graph $=\frac{1}{\sqrt{2 \mathrm{mq}}}$

The slope will be maximum for minimum mass,

$\therefore \quad \mathrm{m}_4$ is minimum and $\mathrm{m}_1$ will be minimum.

Given data: $\mathrm{E}=15 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}, \mathrm{I}=0.5 \mathrm{~A},\phi=\frac{\pi}{3} \mathrm{~rad}$

Impedance is given as $Z=\frac{E}{I}=\frac{15}{0.5}=30 \Omega$

$\begin{gathered} \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ \therefore \quad \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R} \end{gathered}$

The formula for impedance is

$$\begin{aligned} Z & =\sqrt{R^2+X_L^2} \\ Z & =\sqrt{R^2+(\sqrt{3} R)^2} \\ Z & =\sqrt{4 R^2} \\ \therefore \quad 2 R & =Z \\ \therefore \quad R & =\frac{Z}{2}=\frac{30}{2}=15 \Omega \end{aligned}$$

The gravitational acceleration at depth is given as $g_d=g\left[1-\frac{d}{R}\right]$

Given $\mathrm{g}_{\mathrm{d}}=\frac{\mathrm{g}}{2 \mathrm{n}}$

$$\begin{aligned} \therefore \quad \frac{g}{2 n} & =g\left[1-\frac{d}{R}\right] \\ & \frac{d}{R}=1-\frac{1}{2 n} \\ d & =\left[\frac{2 n-1}{2 n}\right] R \end{aligned}$$

$v=\frac{2 r^2(\rho-\sigma) g}{9 \Delta}$

$\rho, \sigma$ and $\eta$ are constant $\quad$ .... (given)

$\Rightarrow \mathrm{v} \propto \mathrm{r}^2$

The ratio of terminal velocities is

$$\begin{aligned} \frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2} \\ \frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}^2}{\left(\frac{\mathrm{r}}{3}\right)^2} \\ \frac{\mathrm{v}_1}{\mathrm{v}_2} & =9 \\ \therefore \quad \mathrm{v}_2 & =\frac{\mathrm{v}}{9} \end{aligned}$$

The inductance long solenoid is

$$\begin{aligned} & \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\ & \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}\right)^2 \times \pi \times \frac{\mathrm{d}^2}{4} \end{aligned}$$

$\therefore \quad$ The inductance per unit length near the middle of a long solenoid is:

$\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)$

The correct option is:

Explanation:

Option A is incorrect because gravitational force is actually much weaker than the electrostatic force. The gravitational force between two particles is given by Newton's law of gravitation:

where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the particles, and $r$ is the distance between their centers. On the other hand, the electrostatic force is given by Coulomb's law:

where $k$ is Coulomb's constant, and $q_1$ and $q_2$ are the electric charges of the particles. The electrostatic force is much stronger than the gravitational force because the gravitational constant $G$ is a much smaller number than Coulomb's constant $k$.

Option B is incorrect because the electrostatic force can be either attractive or repulsive, depending on whether the charges are opposite (attractive) or like charges (repulsive). However, the gravitational force is always attractive, as it always acts to pull masses together.

Option C is correct. Both gravitational and electrostatic forces act along the straight line joining the centers of the two masses or charges, respectively. This line is known as the line of action for these forces.

Option D is incorrect because the electrostatic force does indeed obey an inverse square law, just like the gravitational force. Both forces diminish in strength with the square of the distance between the two objects, as is reflected in the equations for both forces (Newton's law of gravitation and Coulomb's law).

$\therefore \quad$ Moment of inertia of thin wire:

$$\begin{aligned} I & =\frac{M^2}{2} \\ M & =v \times m \text { and } L=2 \pi R \\ R & =\frac{L}{2 \pi} \\ \therefore \quad I & =\frac{L m}{2}\left(\frac{L}{2 \pi}\right)^2 \\ I & =\frac{m L^3}{8 \pi^2} \end{aligned}$$

$\therefore \quad$ Using Parallel axis theorem:

$$\begin{aligned} & I^{\prime}=I+M R^2 \\ & I=\frac{m L^3}{8 \pi^2}+L m\left(\frac{L}{2 \pi}\right)^2 \\ & I=\frac{3 m L^3}{8 \pi^2} \end{aligned}$$

Let $\mathrm{R}$ be the radius of the coalesced drop.

$$\begin{aligned} \therefore \quad & \frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \\ & R^3=8 \pi r^3 \\ & R=2 r \end{aligned}$$