MHT CET 2021 23rd September Evening Shift (Physics & Chemistry) Question Paper with Solutions

To find the inductance value in the L-C-R circuit at resonance, we can use the given information about the angular frequency ($\omega$), the voltage across the resistance, and the voltage across the inductor at resonance.

The given parameters are:

• Resistance, $R = 120 ~\Omega$
• Angular frequency, $\omega = 4 \times 10^5 ~\text{rad} ~\text{s}^{-1}$
• Voltage across resistance, $V_R = 60 ~\text{V}$
• Voltage across inductor, $V_L = 40 ~\text{V}$

At resonance in an L-C-R circuit, the impedance is purely resistive, i.e., the impedance of the circuit is equal to the resistance $R$.

The voltage across the inductor $V_L$ at resonance is given by the equation:

$V_L = \omega L I$

where $I$ is the current in the circuit, which can be found using Ohm's law considering the voltage across the resistance:

$V_R = I \cdot R$

Rearrange to solve for $I$:

$I = \frac{V_R}{R}$

Substitute the given values:

$I = \frac{60 ~\text{V}}{120 ~\Omega} = 0.5 ~\text{A}$

Now, use the value of current to solve for inductance $L$:

$V_L = \omega L \cdot I$

$40 ~\text{V} = \left(4 \times 10^5 ~\text{rad} ~\text{s}^{-1}\right) \cdot L \cdot 0.5 ~\text{A}$

Rearrange to solve for $L$:

$L = \frac{40 ~\text{V}}{(4 \times 10^5 ~\text{rad} ~\text{s}^{-1}) \cdot 0.5 ~\text{A}}$

Simplify the expression:

$L = \frac{40}{2 \times 10^5}$

$L = 2 \times 10^{-4} ~\text{H} = 0.2 ~\text{mH}$

Therefore, the value of the inductance is $0.2 ~\text{mH}$, which corresponds to option A.

Due to spherical aberration, marginal and paraxial rays converge at different points along the principal axis. Between the marginal focus and the paraxial focus, the cross-section of the converging beam reaches a minimum diameter, forming a circular patch of minimum size known as the circle of least confusion.

Hence, the correct option is (D).

To find the acceleration of the particle performing uniform circular motion (U.C.M.), we will use the formula for centripetal acceleration:

$a = \omega^2 r$

where $\omega$ is the angular velocity and $r$ is the radius of the circle.

First, we need to determine the angular velocity $\omega$. The formula for angular velocity in terms of frequency $f$ is:

$\omega = 2\pi f$

Given the frequency $f = 2 \mathrm{~Hz}$, we have:

$\omega = 2\pi \times 2 = 4\pi \mathrm{~rad/s}$

Next, we need to find the radius $r$ of the circle. Given the diameter as $50 \mathrm{~cm}$, the radius $r$ is:

$r = \frac{50 \mathrm{~cm}}{2} = 25 \mathrm{~cm} = 0.25 \mathrm{~m}$

Now we can calculate the centripetal acceleration:

$a = \omega^2 r = (4\pi)^2 \times 0.25 = 16\pi^2 \times 0.25 = 4\pi^2 \mathrm{~m/s}^2$

Therefore, the acceleration of the particle is:

Option B

$4 \pi^2 \mathrm{~m/s}^2$

The output of a two-input AND gate is given by the Boolean expression:

$𝑌=𝐴\cdot 𝐵$

Evaluating the output for each entry:

For Entry 1: $𝐴=0$, $𝐵=1⟹𝑌=0\cdot 1=0$, which matches the table.

For Entry 2: $𝐴=1$, $𝐵=0⟹𝑌=1\cdot 0=0$, which matches the table.

For Entry 3: $𝐴=1$, $𝐵=1⟹𝑌=1\cdot 1=1$, which matches the table.

For Entry 4: $𝐴=0$, $𝐵=0⟹𝑌=0\cdot 0=0$, whereas the table gives $𝑌=1$.

Therefore, only entries 1, 2, and 3 are correct.

Hence, the correct option is (B).

To determine the final pressure of the gas after undergoing an isothermal expansion followed by an adiabatic compression, let's break down the problem step-by-step.

Step 1: Isothermal Expansion

During an isothermal process, the temperature remains constant. For an ideal gas, the relationship between pressure and volume is given by:

$P_1 V_1 = P_2 V_2$

Here, the initial pressure is $\mathrm{P}$ and the initial volume is $\mathrm{V}$. After the isothermal expansion, the volume becomes $9 \mathrm{~V}$ and the pressure becomes $\mathrm{P_2}$.

So we have:

$\mathrm{P} \cdot \mathrm{V} = \mathrm{P_2} \cdot 9 \mathrm{V}$

Solving for $\mathrm{P_2}$:

$\mathrm{P_2} = \frac{\mathrm{P}}{9}$

Step 2: Adiabatic Compression

During an adiabatic process, no heat is exchanged with the surroundings. The relationship between pressure and volume for an adiabatic process is given by:

$P_2 V_2^\gamma = P_3 V_3^\gamma$

Here, $\gamma$ is the adiabatic index, which is the ratio of specific heat at constant pressure to constant volume ($=\frac{3}{2} = 1.5$). The final volume $V_3$ is equal to the initial volume $\mathrm{V}$.

Given:

$\mathrm{P_2} = \frac{\mathrm{P}}{9}, \quad \mathrm{V_2} = 9 \mathrm{V}, \quad \mathrm{V_3} = \mathrm{V}$

Thus, we have:

$\left(\frac{\mathrm{P}}{9}\right) \cdot (9 \mathrm{V})^\gamma = \mathrm{P_3} \cdot \mathrm{V}^\gamma$

Substituting $\gamma = 1.5$:

$\left(\frac{\mathrm{P}}{9}\right) \cdot (9 \mathrm{V})^{1.5} = \mathrm{P_3} \cdot \mathrm{V}^{1.5}$

Since $\gamma = 1.5$, we have:

$(9 \mathrm{V})^{1.5} = 9^{1.5} \cdot \mathrm{V}^{1.5} = 27 \cdot \mathrm{V}^{1.5}$

Thus:

$\left(\frac{\mathrm{P}}{9}\right) \cdot 27 \cdot \mathrm{V}^{1.5} = \mathrm{P_3} \cdot \mathrm{V}^{1.5}$

Dividing both sides by $\mathrm{V}^{1.5}$:

$\left(\frac{\mathrm{P}}{9}\right) \cdot 27 = \mathrm{P_3}$

Simplifying, we get:

$\mathrm{P_3} = 3 \mathrm{P}$

So, the final pressure of the gas after the adiabatic compression is 3P.

Therefore, the correct answer is:

Option C: 3 P

The angular velocity

$$\begin{aligned} \omega & =1 \mathrm{~rad} / \mathrm{s} \\ I & =P \cdot\left(\frac{1}{2} \mathrm{I} \omega^2\right) \\ 1 & =\mathrm{P} \cdot \frac{1}{2} \cdot 1 \\ \therefore \quad P & =2 \end{aligned}$$

$$\begin{aligned} & e=\frac{d \phi}{d t}, i=\frac{e}{R}=\frac{1}{R} \cdot \frac{d \phi}{d t} \\ & \frac{d q}{d t}=\frac{1}{R} \cdot \frac{d \phi}{d t} \\ & \therefore d q=\frac{d \phi}{R} \end{aligned}$$

The speed of sound in a medium is influenced by the properties of the medium itself. Specifically, it depends on the medium's density and elasticity. The formula for the speed of sound $v$ in a medium is given by:

$v = \sqrt{\frac{E}{\rho}}$

Here, $E$ represents the modulus of elasticity (such as Bulk modulus for gases, Young's modulus for solids), and $\rho$ represents the density of the medium.

For gases, we can further refine this relationship using the ideal gas law. For example, in an ideal gas, the speed of sound can be expressed as:

$v = \sqrt{\frac{\gamma P}{\rho}}$

where:

• $\gamma$ is the adiabatic index (ratio of specific heats $C_p/C_v$).
• $P$ is the pressure of the gas.
• $\rho$ is the density of the gas.

At constant temperature, the density ($\rho$) of a gas is directly proportional to its pressure ($P$) according to the ideal gas law:

$P \propto \rho \implies \rho \propto P$

Thus, if pressure $P$ is doubled (keeping temperature constant), the density $\rho$ will also double. Substituting this into the speed of sound equation gives:

$v = \sqrt{\frac{\gamma P}{\rho}}$

If $P$ is doubled and $\rho$ is also doubled, the effect on the speed of sound is:

$v = \sqrt{\frac{\gamma (2P)}{2\rho}} = \sqrt{\frac{2\gamma P}{2\rho}} = \sqrt{\frac{\gamma P}{\rho}}$

We see that the expression for the speed of sound remains unchanged. Therefore, doubling the pressure at constant temperature does not affect the speed of sound in the medium.

Therefore, the correct option is:

Option A: Remains same

Distance of $6^{\text {th }}$ bright band $=\frac{6 \lambda_1 \mathrm{D}}{\mathrm{d}}$

Distance of $7^{\text {th }}$ dark band $=\frac{6.5 \lambda_2 \mathrm{D}}{\mathrm{d}}$

$\therefore \frac{3 \lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{6.5 \lambda_2 \mathrm{D}}{\mathrm{d}}, \quad \therefore \frac{\lambda_1}{\lambda_2}=\frac{6.5}{6}=\frac{13}{12}$

$\mathrm{u}_{\mathrm{x}}=\mathrm{a}=$ Horizontal component of the velocity

$\mathrm{u}_{\mathrm{y}}=\mathrm{b}=$ Vertical component of the velocity

Maximum height $\mathrm{H}=\frac{\mathrm{u}_{\mathrm{y}}^2}{2 \mathrm{~g}}=\frac{\mathrm{b}^2}{2 \mathrm{~g}}$

$$\begin{aligned} \text { Range } \mathrm{R} & =\frac{2 \mathrm{u}_{\mathrm{y}} \mathrm{u}_{\mathrm{x}}}{\mathrm{g}}=\frac{2 \mathrm{ba}}{\mathrm{g}} \\ \mathrm{R} & =2 \mathrm{H} \quad \therefore \frac{2 \mathrm{ba}}{\mathrm{g}}=\frac{2 \mathrm{~b}^2}{2 \mathrm{~g}} \\ \therefore \quad \mathrm{b} & =2 \mathrm{a} \end{aligned}$$

$$\begin{aligned} \mathrm{r} & =0.05 \mathrm{~nm}=0.05 \times 10^{-9} \mathrm{~m}=5 \times 10^{-11} \mathrm{~m} \\ \mathrm{f} & =10^{16} \text { r.p.s } \\ \mathrm{M} & =\mathrm{IA}=\mathrm{ef} \pi \mathrm{r}^2 \\ & =1.6 \times 10^{-19} \times 10^{16} \times 3.14 \times\left(5 \times 10^{-11}\right)^2 \\ & =1.26 \times 10^{-23} \mathrm{~A}-\mathrm{m}^2 \end{aligned}$$

$$\begin{aligned} & \mathrm{f}=220 \mathrm{~Hz}, v=330 \mathrm{~m} / \mathrm{s} \\ & v=\mathrm{f} \lambda \\ & \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{220}=\frac{3}{2} \mathrm{~m} \end{aligned}$$

Distance travel by 80 vibrations is

$80 \times \frac{3}{2}=120 \mathrm{~m}$

In a semiconductor, the behavior of resistance and electric conductivity with the change in temperature is distinct from that in conductors. As the temperature of a semiconductor is increased, more electrons gain enough thermal energy to jump from the valence band to the conduction band. This increase in charge carriers (electrons and holes) enhances the material's conductivity.

Consequently, the resistance of the semiconductor decreases because resistance is inversely proportional to the number of charge carriers. Since conductivity is the reciprocal of resistance, the increase in the number of charge carriers due to the higher temperature leads to an increase in conductivity.

Therefore, when the temperature of a semiconductor is increased:

• Resistance decreases
• Electric conductivity increases

So, the correct answer is:

Option D: decreases and increases

Let's analyze the given problem systematically. The magnetic field inside a long solenoid is determined by the formula:

${ B = \mu_0 n I }$

where:

- $\mu_0$ is the permeability of free space,

- $n$ is the number of turns per unit length,

- $I$ is the current passing through the solenoid.

Initially, the solenoid produces a magnetic field given by:

${ B_1 = \mu_0 n_1 I_1 }$

When the current is reduced to $20\%$ of its original value, the new current becomes:

${ I_2 = 0.20 I_1 }$

Additionally, if the number of turns per centimeter is increased five times, the new number of turns per unit length becomes:

${ n_2 = 5 n_1 }$

Now, using the formula for the magnetic field in a solenoid, the new magnetic field $B_2$ can be calculated as:

${ B_2 = \mu_0 n_2 I_2 }$

Substitute the values of $n_2$ and $I_2$ into the equation:

${ B_2 = \mu_0 (5 n_1) (0.20 I_1) }$

Simplifying this, we get:

${ B_2 = \mu_0 \cdot 5 n_1 \cdot 0.20 I_1 }$

${ B_2 = \mu_0 n_1 I_1 \cdot 1.0 }$

${ B_2 = B_1 }$

Therefore, the new magnetic field $B_2$ is equal to the original magnetic field $B_1$. Hence, the correct answer is:

Option A: $\mathrm{B}_1$

Initial energy $\mathrm{W}_1=\frac{1}{2} \mathrm{CV}_1^2$

Final energy $\mathrm{W}_2=\frac{1}{2} \mathrm{CV}_2^2$

$\therefore \quad \frac{W_2}{W_1}=\left(\frac{V_2}{V_1}\right)^2=\left(\frac{15}{5}\right)^2=(3)^2=9$

In a Common Emitter (CE) transistor configuration, the relationship between the emitter current ($I_E$), collector current ($I_C$), and base current ($I_B$) can be described using the following equations:

$I_E = I_B + I_C$

$I_C = \beta I_B$

Given in the problem, the change in emitter current ($\Delta I_E$) is $8.0 \mathrm{~mA}$ and the change in collector current ($\Delta I_C$) is $7.8 \mathrm{~mA}$.

Since, in CE configuration, the emitter current is the sum of the base and collector currents, we can express the change in emitter current as:

$\Delta I_E = \Delta I_B + \Delta I_C$

We need to find the change in the base current ($\Delta I_B$). Let's rearrange the above equation to solve for $\Delta I_B$:

$\Delta I_B = \Delta I_E - \Delta I_C$

Substituting the given values:

$\Delta I_B = 8.0 \mathrm{~mA} - 7.8 \mathrm{~mA} = 0.2 \mathrm{~mA}$

Recall that $1 \mathrm{~mA} = 1000 \mu \mathrm{A}$:

$0.2 \mathrm{~mA} = 0.2 \times 1000 \mu \mathrm{~A} = 200 \mu \mathrm{~A}$

Therefore, the change in the base current necessary to produce the same change in the collector current is:

Option A: $200 \mu \mathrm{A}$

$$\begin{aligned} \therefore \quad \mathrm{T}_1=\mathrm{K}_1 \mathrm{x} & =2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \quad \mathrm{F}_2=\mathrm{K}_2 \mathrm{x}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \quad \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{x} \\ \therefore \quad \mathrm{T}_1^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1} \quad \mathrm{~T}_2^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_2} \\ \mathrm{~T}^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \\ \therefore \quad \frac{1}{\mathrm{~T}^2} & =\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \\ & =\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2} \\ & =\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \\ \therefore \quad \mathrm{T}^2 & =\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2} \\ \therefore \mathrm{T} & =\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}} \end{aligned}$$

$$\begin{aligned} & \lambda_1=5.0 \mathrm{~m} \text { and } \lambda_2=5.5 \mathrm{~m}, \mathrm{v}=300 \mathrm{~m} / \mathrm{s} \\ & \mathrm{n}_1=\frac{\mathrm{V}}{\lambda_1}=\frac{300}{5}=60 \mathrm{~Hz} \\ & \mathrm{n}_2=\frac{\mathrm{V}}{\lambda_2}=\frac{300}{5.5}=54.5 \mathrm{~Hz} \simeq 54 \mathrm{~Hz} \end{aligned}$$

Number of beats $=\mathrm{n}_1-\mathrm{n}_2=60-54=6 \mathrm{~Hz}$

$$\begin{aligned} & E \propto\frac{1}{n^2} \\ & \therefore \frac{E_3}{E_2}=\frac{4}{9} \quad \therefore E_3=\frac{4}{9} E_2 \end{aligned}$$

$$\begin{aligned} & \mathrm{T}^2 \alpha \mathrm{r}^3 \\ & \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{r}_2^3}{\mathrm{r}_1^3}=(4)^3=64 \end{aligned}$$

$\therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=8$

or $\mathrm{T}_2=8 \mathrm{~T}_1=8 \times 5=40$ hours.

$$\begin{aligned} & \text { Kinetic energy }=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\ & \text { At } x=\frac{A}{2}, \text { K.E. }=\frac{1}{2} m \omega^2\left(\frac{3}{4} A^2\right)=\frac{3}{8} m \omega^2 A^2 \\ & =\frac{3}{8} \mathrm{~m}\left(\frac{2 \pi}{\mathrm{T}}\right)^2 \cdot \mathrm{A}^2 \\ & =\frac{3}{8} \mathrm{~m} \cdot \frac{4 \pi^2}{\mathrm{~T}^2} \cdot \mathrm{A}^2 \\ & =\frac{3}{2} \cdot \frac{m \pi^2 A^2}{T^2} \\ & \end{aligned}$$

In Young's double slit experiment, the position of dark and bright fringes on the screen is a result of constructive and destructive interference of light waves coming from the two slits.

The distance between two consecutive bright or dark fringes is called the fringe width, denoted by $\beta$. The fringe width is given by:

$\beta = \dfrac{\lambda D}{d}$

where:

• $\lambda$ is the wavelength of light used.
• $D$ is the distance between the slits and the screen.
• $d$ is the separation between the two slits.

The position of the nth dark band from the central bright band is given by the condition for destructive interference, which occurs at:

$y_n = \left(n - \dfrac{1}{2}\right) \beta$

where $n$ is the order of the dark fringe.

Therefore, the distance of the $\mathrm{n}^{\text{th}}$ dark band from the central bright band is:

Option C: $(\mathrm{n}-0.5) \beta$

$\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{\left(\mu_1-1\right)}{R}$ and

$$\begin{aligned} & \frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{\mu_2-1}{-R} \\ & \frac{1}{\mathrm{f}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{\mu_1-1}{R}-\frac{\mu_2-1}{R}=\frac{1}{R}\left[\mu_1-1-\mu_2+1\right]=\frac{\mu_1-\mu_2}{R} \end{aligned}$$

$$\begin{aligned} & \mathrm{r}=\mathrm{R}\left(\frac{\ell_1}{\ell_2}-1\right)=\mathrm{R}^{\prime}\left(\frac{\ell_1}{\ell_2^{\prime}}-1\right) \\ & \ell_2=1 \mathrm{~m}, \ell_2^{\prime}=1.5 \mathrm{~m} \mathrm{R}=3 \Omega, \mathrm{R}^{\prime}=6 \Omega \\ & \therefore 3\left(\frac{\ell_1}{1}-1\right)=6\left(\frac{\ell_1}{1.5}-1\right) \end{aligned}$$

Solving we get $\ell_1=3 \mathrm{~m}$

$\therefore \quad r=3\left(\frac{3}{1}-1\right)=3 \times 2=6 \Omega$

To solve this problem, let's start by understanding how the moment of inertia works for a circular loop. For a circular loop made from a uniform wire of radius $R$ and mass $M$, the moment of inertia about its own axis can be given by:

$\mathrm{I} = M R^2$

Given two loops P and Q with radii $R_1$ and $R_2$ respectively, and their moments of inertia $\mathrm{I_P}$ and $\mathrm{I_Q}$, we have the following relationship:

$\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I_Q}} = \frac{1}{8}$

Using the formula for the moment of inertia, we can write:

$\mathrm{I_P} = M_1 R_1^2$

$\mathrm{I_Q} = M_2 R_2^2$

Since the loops are made of the same uniform wire, the mass (which is proportional to the length of the wire used) will be different for different radii. The mass of a loop is proportional to the circumference so:

$M_1 \propto 2 \pi R_1$

$M_2 \propto 2 \pi R_2$

We can express the masses in terms of a common unit length mass $\lambda$:

$M_1 = \lambda (2 \pi R_1)$

$M_2 = \lambda (2 \pi R_2)$

Substitute these into the moments of inertia:

$\mathrm{I_P} = \lambda (2 \pi R_1) R_1^2 = 2 \pi \lambda R_1^3$

$\mathrm{I_Q} = \lambda (2 \pi R_2) R_2^2 = 2 \pi \lambda R_2^3$

Using the given ratio:

$\frac{ \mathrm{I_P} }{ \mathrm{I_Q} } = \frac{ 2 \pi \lambda R_1^3 }{ 2 \pi \lambda R_2^3 } = \frac{R_1^3}{R_2^3}$

Given that:

$\frac{R_1^3}{R_2^3} = \frac{1}{8}$

We take the cube root of both sides:

$\frac{R_1}{R_2} = \sqrt[3]{\frac{1}{8}}$

Simplify the cube root:

$\frac{R_1}{R_2} = \frac{1}{2}$

Therefore:

$\frac{R_2}{R_1} = 2$

So the correct answer is Option C: 2.

Since the some current flows through the second wire, the magnetic field at the same distance will be same. $\therefore B_1=B_2$

$$\begin{aligned} & \ell=31.4 \mathrm{~cm}=0.314 \mathrm{~m} \\ & \mathrm{~A}=10^{-3} \mathrm{~m}^2, \mathrm{~N}=500 \\ & \text { Self induction } \mathrm{L}=\mu_0 \mathrm{n}^2 \ell \mathrm{A} \\ & =\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\ell} \quad \text { where } \mathrm{n}=\frac{\mathrm{N}}{\ell} \\ & =\frac{4 \pi \times 10^{-7} \times(500)^2 \times 10^{-3}}{0.314} \\ & =\frac{4 \times 3.14 \times 25 \times 10^{-6}}{0.314} \\ & =10^{-3} \mathrm{H} \\ & \end{aligned}$$

The force due to surface tension at the wall of the capillary is given by $\mathrm{f}_{\mathrm{T}}$ = ( surface tension) $\times$ (length in contact)

$=\mathrm{T} \times 2 \pi \mathrm{r}_2$

The vertical component of this force is

$\mathrm{fV}_{\mathrm{V}}=\mathrm{T} \times 2 \pi \mathrm{r}_2 \cos \theta$

where $\theta$ is the angle of contact.

Similarly the vertical component of the force at the surface of the rod is

$\mathrm{f}_{\mathrm{V}}^{\prime}=\mathrm{T} \times 2 \pi \pi_1 \cos \theta$

Total force $\mathrm{F=f_V+f_V}^{\prime}$

$\mathrm{F}=\left(\mathrm{r}_2+\mathrm{r}_1\right) 2 \pi \mathrm{T} \cos \theta$

Weight of the liquid in the capillary

$\mathrm{W}=\pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}$

This is balanced by the vertical component of the face due to the surface tension

$\therefore \pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho g=\left(\mathrm{r}_2+\mathrm{r}_1\right) \times 2 \pi \mathrm{T} \cos \theta$

Simplifying and solving for h we get,

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}=\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}$

(For pure water, $\theta=0^{\circ}, \cos \theta=1$)

$x=1$ cm

$\therefore$ Volume of the cube $v=x^3=1 \mathrm{~cm}^3$, volume of drop = volume of cube.

$\frac{4}{3} \pi \mathrm{r}^3=\mathrm{x}^3=1 \mathrm{~cm}^3$

$$\begin{aligned} & \therefore r^3=\frac{3}{4 \pi} \quad \text { or } r=\left(\frac{3}{4 \pi}\right)^{\frac{1}{3}} \\ & \therefore r^2=\left(\frac{9}{16 \pi^2}\right)^{\frac{1}{3}} \end{aligned}$$

$$\begin{aligned} \text { Surface area of drop } & =4 \pi \mathrm{r}^2=4 \pi\left(\frac{9}{16 \pi \mathrm{r}^2}\right)^{\frac{1}{3}} \\ & =\left(64 \pi^3 \times \frac{9}{16 \pi^2}\right)^{\frac{1}{3}} \\ & =(36 \pi)^{\frac{1}{3}} \end{aligned}$$