JEE Main 2023 11th April Evening Shift Question Paper with Solutions

The conservation of angular momentum states that the angular momentum (L) remains constant. The relation between kinetic energy (KE), angular momentum (L), and moment of inertia (I) is given by:

$\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}$

Using this relation, we can find the ratio of the final kinetic energy ($\mathrm{KE}_{\text{final}}$) to the initial kinetic energy ($\mathrm{KE}_{\text{initial}}$ or E):

$\frac{\mathrm{KE}_{\text{final}}}{\mathrm{KE}_{\text{initial}}}=\frac{\mathrm{I}_{\text{initial}}}{\mathrm{I}_{\text{final}}}$

Since the moment of inertia triples, we have $\mathrm{I}_{\text{final}} = 3\mathrm{I}_{\text{initial}}$. Therefore,

$\frac{\mathrm{KE}_{\text{final}}}{\mathrm{E}}=\frac{\mathrm{I}_{\text{initial}}}{3\mathrm{I}_{\text{initial}}}=\frac{1}{3}$

This means that the final kinetic energy of the system is:

$\mathrm{KE}_{\text{final}}=\frac{E}{3}$

So, the value of $x$ is 3.

The electric field inside a parallel plate capacitor is uniform and given by $\mathbf{E}=\frac{V}{d} \hat{\mathbf{j}}$ where $d$ is the separation between the plates.

The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by $\Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}$.

Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by $\Phi_E = E A$, where $A$ is the area of the plate.

The area of the positive plate is $A = \frac{Q}{\varepsilon_0 V}$, where $Q = C V$ is the charge on the positive plate. Therefore, the electric flux through the surface is given by:

$\Phi_E = \frac{Q}{\varepsilon_0 V} \frac{V}{d} = \frac{Q}{\varepsilon_0 d} = \frac{C V}{\varepsilon_0 d}$

Thus, the answer is $\frac{C V}{\varepsilon_0}$.

$$\begin{aligned} & q E=e V B \\ & E=V B \end{aligned}$$

$$\begin{aligned} \Delta V & =(E . d) \\ & =(V B) \times 0.15 \\ & =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\ & =0.15 \mathrm{~V}=150 \mathrm{mV} \end{aligned}$$

When a light ray is reflected from a plane mirror, the angle of incidence (i) is equal to the angle of reflection (r). In this case, the angle of reflection is given as $30^{\circ}$, so the angle of incidence is also $30^{\circ}$.

The angle of deviation (D) is the angle between the incident ray and the reflected ray. To find this angle, consider the fact that the angle between the incident ray and the normal to the mirror and the angle between the reflected ray and the normal add up to $180^{\circ}$, since they are supplementary angles.

Thus, we have:

$i + r + D = 180^{\circ}$

Since $i = r$, we can rewrite the equation as:

$2i + D = 180^{\circ}$

Substitute the value of the angle of incidence:

$2(30^{\circ}) + D = 180^{\circ}$

$60^{\circ} + D = 180^{\circ}$

Solve for the angle of deviation (D):

$D = 180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the angle of deviation of the ray after reflection is $120^{\circ}$.

In satellite communication, the uplink frequency band refers to the range of frequencies used to transmit signals from the Earth to the satellite. The specific frequency bands used for uplink communication can vary depending on the type of satellite and the service it provides. However, one of the most common frequency bands used for satellite uplinks in geostationary satellite communication is the C-band.

$5.925-6.425 \mathrm{~GHz}$, is the typical frequency range used for uplink communication in the C-band.

To find the root mean square speed of molecules of nitrogen gas, we can use the formula:

$v_{rms} = \sqrt{\frac{3k_BT}{m}}$

where $v_{rms}$ is the root mean square speed, $k_B$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a nitrogen molecule.

First, we need to convert the temperature from Celsius to Kelvin:

$T = 27^{\circ}\mathrm{C} + 273 = 300\,\mathrm{K}$

Now, substitute the given values of $k_B$, $T$, and $m$ into the formula:

$v_{rms} = \sqrt{\frac{3(1.4 \times 10^{-23}\, \mathrm{JK}^{-1})(300\,\mathrm{K})}{4.6 \times 10^{-26}\,\mathrm{kg}}}$

Simplify and calculate the root mean square speed:

$v_{rms} = 523\,\mathrm{m/s}$

The root mean square speed of molecules of nitrogen gas at $27^{\circ}\mathrm{C}$ is 523 m/s.

We know that force (F) has dimensions of $\mathrm{MLT}^{-2}$, velocity (V) has dimensions of $\mathrm{LT}^{-1}$, and time (T) has dimensions of $\mathrm{T}$. To express density ($\rho$), which has dimensions of $\mathrm{ML}^{-3}$, in terms of F, V, and T, we need to find the exponents of F, V, and T.

Let the dimensions of density be expressed as:

$\mathrm{[M]}^a \mathrm{[L]}^b \mathrm{[T]}^c$

Substituting the dimensions of F, V, and T:

$\mathrm{[F]}^a \mathrm{[V]}^b \mathrm{[T]}^c = (\mathrm{MLT}^{-2})^a (\mathrm{LT}^{-1})^b (\mathrm{T})^c$

Since density has dimensions of $\mathrm{ML}^{-3}$, we can set up the following equations:

$a = 1$ (for the mass term M)

$a + b = -3$ (for the length term L)

$-2a - b + c = 0$ (for the time term T)

Solving these equations, we get:

$a = 1$

$b = -4$

$c = -2$

Thus, the dimensional formula of density in terms of F, V, and T is:

$\mathrm{FV}^{-4} \mathrm{T}^{-2}$

In free space, the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:

$B = \frac{E}{c}$

where c is the speed of light in a vacuum, approximately equal to $3 \times 10^8 \mathrm{~m} / \mathrm{s}$. We are given that the electric field is $\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}$. To find the magnetic field, we can first calculate the magnitude of B:

$B = \frac{6.6 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} = 2.2 \times 10^{-8} \mathrm{~T}$

Now, we need to find the direction of the magnetic field. Since the electromagnetic wave propagates in the x-direction, and the electric field is in the y-direction (j), the magnetic field should be in the z-direction (k) to satisfy the right-hand rule for electromagnetic waves. The direction of the magnetic field will be positive k (counter-clockwise rotation from x to y).

So, $\overrightarrow{\mathrm{B}} = 2.2 \times 10^{-8} \hat{k} \mathrm{T}$.

To find the velocity of the projectile at t = 2 s, we need to find the horizontal and vertical components of the velocity at that time.

The initial horizontal component of the velocity is constant and is given by:

$v_{0x} = v_0 \cos\theta = 40\,\mathrm{ms}^{-1} \cos(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\mathrm{ms}^{-1}$

The initial vertical component of the velocity is:

$v_{0y} = v_0 \sin\theta = 40\,\mathrm{ms}^{-1} \sin(30^{\circ}) = 40\,\mathrm{ms}^{-1} \cdot \frac{1}{2} = 20\,\mathrm{ms}^{-1}$

To find the vertical component of the velocity at t = 2 s, we use the equation:

$v_y = v_{0y} - gt = 20\,\mathrm{ms}^{-1} - (10\,\mathrm{ms}^{-2})(2\,\mathrm{s}) = 20\,\mathrm{ms}^{-1} - 20\,\mathrm{ms}^{-1} = 0\,\mathrm{ms}^{-1}$

At t = 2 s, the horizontal component of the velocity is still $20\sqrt{3}\,\mathrm{ms}^{-1}$, and the vertical component is 0. The overall velocity at t = 2 s is:

$\vec{v} = 20\sqrt{3}\,\mathrm{ms}^{-1} \hat{i} + 0\,\mathrm{ms}^{-1} \hat{j} = 20\sqrt{3}\,\mathrm{ms}^{-1}$

So the correct answer is: $20\sqrt{3}\,\mathrm{ms}^{-1}$

The magnetic field inside a solenoid is uniform and parallel to the axis of the solenoid. When an electron moves with constant velocity along the axis of the solenoid, the angle between its velocity vector and the magnetic field is 0°.

The magnetic force experienced by a moving charge is given by the Lorentz force formula:

$\vec{F} = q(\vec{v} \times \vec{B})$

Since the angle between the velocity vector and the magnetic field is 0°, the cross product term becomes zero:

$\vec{v} \times \vec{B} = 0$

Therefore, the magnetic force experienced by the electron is also zero:

$\vec{F} = 0$

As a result, the electron will not experience any magnetic force (Option B) and will continue to move along the axis of the solenoid with constant velocity (Option C).

Thus, the correct answer is:

B and C only

If the temperature (T) remains constant, the internal energy (U) also remains constant, since the internal energy of an ideal gas depends only on its temperature.

In this case, the thermodynamic process in which the internal energy of the system remains constant is an isothermal process. Isothermal processes occur at constant temperature, and for an ideal gas, this means that the internal energy remains constant as well.

To calculate the work done to increase the radius of a soap bubble, we can use the formula:

$W = T \Delta A$

where W is the work done, T is the surface tension, and ΔA is the change in surface area.

For a soap bubble, we need to consider both the inner and outer surfaces, so the surface area is doubled. The surface area of a sphere is:

$A = 4\pi r^2$

The initial surface area of the soap bubble is:

$A_1 = 2 \cdot 4\pi (0.1\,\mathrm{m})^2 = 8\pi (0.1\,\mathrm{m})^2$

The final surface area of the soap bubble is:

$A_2 = 2 \cdot 4\pi (0.2\,\mathrm{m})^2 = 8\pi (0.2\,\mathrm{m})^2$

The change in surface area is:

$\Delta A = A_2 - A_1 = 8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2$

Now, we can calculate the work done:

$W = T \Delta A = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2]$

Using the given value of π:

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.2\,\mathrm{m})^2 - 8(22/7)(0.1\,\mathrm{m})^2]$

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.04\,\mathrm{m^2}) - 8(22/7)(0.01\,\mathrm{m^2})]$

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.03\,\mathrm{m^2})]$

$$\begin{aligned} & W=2 \times 1.32 \times 10^{-2} \\ &W =2 \times 132 \times 10^{-4} \mathrm{~J} \\ & W=264 \times 10^{-4} \mathrm{~J} \end{aligned}$$

The work done to increase the radius of the soap bubble is 264 × 10⁻⁴ J.

The de Broglie wavelength (λ) of a particle can be found using the formula:

$\lambda = \frac{h}{p}$

where h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:

$p = \sqrt{2mK}$

Combining these two equations, we get:

$\lambda = \frac{h}{\sqrt{2mK}}$

Now, we are given that the kinetic energy of the proton and electron is the same. Let's denote the masses of the proton and electron as $m_p$ and $m_e$, respectively. We are given the relationship between the two masses:

$m_p = 1849 \times m_e$

Let's find the ratio of the de Broglie wavelengths of the proton ($λ_p$) and the electron ($λ_e$):

$\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}}$

Simplifying the expression, we get:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m_eK}}{\sqrt{2m_pK}}$ The 2K terms cancel out:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}}$

Substitute the given relationship between the masses:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 \times m_e}}$

Further simplification:

$\frac{\lambda_p}{\lambda_e} = \frac{1}{\sqrt{1849}}$

Since 1849 is equal to $43^2$:

$\frac{\lambda_p}{\lambda_e} = \frac{1}{43}$

Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.

To find the energy stored in the magnetic field after the current has built up to its equilibrium value, we first need to find the steady-state current in the coil.

When the current reaches its equilibrium value, the coil behaves like a resistor because the back-emf induced by the changing magnetic field is zero. Ohm's law can be applied:

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

where

• $I$ is the current
• $V$ is the voltage across the coil (10 V)
• $R$ is the resistance of the coil (4 Ω)

Plugging in the values:

$I = \frac{10}{4}$ $I = 2.5 \mathrm{~A}$

Now that we have the steady-state current, we can find the energy stored in the magnetic field using the formula:

$W = \frac{1}{2}LI^2$

where

• $W$ is the energy stored in the magnetic field
• $L$ is the inductance of the coil (2 H)
• $I$ is the steady-state current (2.5 A)

Plugging in the values:

$W = \frac{1}{2}(2)(2.5)^2$

$W = 1(6.25)$

$W = 6.25 \mathrm{~J}$

To express this in terms of $10^{-2} \mathrm{~J}$, divide by $10^{-2}$:

$6.25 \div 10^{-2} = 625$

Therefore, the energy stored in the magnetic field after the current has built up to its equilibrium value is 625$\times 10^{-2} \mathrm{~J}$.

Assertion A is true because a bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with the same geometry and mass. This is due to the effect of the magnet's magnetic field on the metallic pipe.

Reason R is also true because when the magnetic bar moves through the metallic pipe, it induces a changing magnetic field in the pipe. This changing magnetic field, in turn, induces Eddy currents in the pipe. According to Lenz's law, these Eddy currents produce their own magnetic field, which opposes the motion of the magnetic bar, causing it to fall more slowly through the pipe.

Since both Assertion A and Reason R are true and R provides the correct explanation for A, the correct answer is Both A and R are true and R is the correct explanation of A.

To find the power delivered by the pulling force at t = 10 s, we first need to find the work done by the force. The work done is given by the product of force and displacement, and the power is the rate of work done.

Calculate the velocity (v) at t = 10 s:

Since the block starts from rest and is pulled up with an effective acceleration of 1 m/s², we can use the equation of motion to find the velocity (v) at t = 10 s:

$v = u + at$

Here, u = 0 (initial velocity) and a = 1 m/s² (acceleration). Plugging in the values:

$v_{10} = 0 + 1(10) = 10 \mathrm{~m/s}$

1. Calculate the net force acting on the block ($F_{net}$):
   
   The net force acting on the block along the incline plane is the difference between the pulling force (F) and the gravitational force component acting parallel to the incline (mgsinθ):

$F_\text{net} = F - mgsinθ$

Since F_net = ma, we can write:

$F = ma + mgsinθ$

Plugging in the values (m = 5 kg, a = 1 m/s², g = 10 m/s², and θ = 30°):

$F = 5(1) + 5(10)(\sin 30°) = 5 + 25 = 30 \mathrm{~N}$

Calculate the power (P) at t = 10 s:

The power (P) can be calculated as the product of force (F) and velocity (v):

$P_{10} = Fv = 30(10) = 300 \mathrm{~W}$

So, the power delivered by the pulling force at t = 10 s from the start is 300 W.

$\mathrm{R}_{\mathrm{eq}}=2 \Omega$

$\text { I through battery } \frac{8}{2}=4 \mathrm{~A}$

$I \text { through } C D=$$\frac{2 \times 3}{3+6}=\frac{2}{3} A$

$I \text { through } R_2=$$\frac{2 / 3}{2}=\frac{1}{3} \mathrm{~A}$

Given that when vector $\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ is subtracted from vector $\overrightarrow{\mathrm{B}}$, it gives a vector equal to $2 \hat{j}$.

We can write this as:

$\vec{B} - \vec{A} = 2 \hat{j}$

Now, let's express the vector $\overrightarrow{\mathrm{B}}$ in terms of its components:

$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

Subtract vector $\vec{A}$ from vector $\vec{B}$:

$(B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) - (2 \hat{i}+3 \hat{j}+2 \hat{k}) = 2 \hat{j}$

Comparing the components, we get:

$B_x - 2 = 0 \\ B_y - 3 = 2 \\ B_z - 2 = 0$

Solving these equations, we find the components of vector $\vec{B}$:

$B_x = 2 \\ B_y = 5 \\ B_z = 2$

Now, we can find the magnitude of vector $\vec{B}$:

$|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33}$

Therefore, the magnitude of vector $\overrightarrow{\mathrm{B}}$ is $\sqrt{33}$

To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity. The formula for orbital velocity (v) is given by:

$v = \sqrt{\frac{GM}{r}}$

where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth to the spaceship (which is the sum of the Earth's radius and the altitude of the spaceship's orbit).

In this problem, the spaceship is orbiting close to Earth's surface, so we can approximate r as the Earth's radius. Given that g = 10 m/s² and Earth's radius R = 6400 km, we can relate the gravitational constant G and the mass of Earth M through the formula:

$g = \frac{GM}{R^2}$

Now, we can find the orbital velocity:

$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$

Converting the Earth's radius to meters:

$R = 6400 \times 10^3 \mathrm{~m}$

Plugging in the values:

$v = \sqrt{10 \times 6400 \times 10^3} = 8 \times 10^3 \mathrm{~m/s}$

Now, we need to find the additional velocity required to overcome the gravitational pull. To do this, we can use the formula for escape velocity:

$v\text{escape} = \sqrt{2} \times v\text{orbital}$

Finding the additional velocity required:

$\Delta v = v\text{escape} - v\text{orbital} = (\sqrt{2} - 1) \times v_\text{orbital}$

Plugging in the values:

$\Delta v = (\sqrt{2} - 1) \times 8 \times 10^3 \mathrm{~m/s}$

Converting the velocity to km/s:

$\Delta v = (\sqrt{2} - 1) \times 8 \mathrm{~km/s}$

Thus, the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit is:

$8(\sqrt{2}-1) \mathrm{~km/s}$

Let the internal resistance of each cell be $r$.

Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $2r$. The total EMF of the series combination of cells is $1.5\,\text{V} + 1.5\,\text{V} = 3\,\text{V}$.

Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:

$\text{EMF}_{total} - I(R + 2r) = 0$

We are given that the voltage across the $10\,\Omega$ resistor, as measured by the ideal voltmeter, is $1.5\,\text{V}$. According to Ohm's law, the current in the circuit can be determined as:

$I = \frac{V}{R} = \frac{1.5\,\text{V}}{10\,\Omega} = 0.15\,\text{A}$

Now, substitute the given values into the KVL equation:

$3\,\text{V} - 0.15\,\text{A}(10\,\Omega + 2r) = 0$

Solve for $2r$:

$3\,\text{V} - 1.5\,\text{V} = 0.15\,\text{A} \cdot 2r$

$1.5\,\text{V} = 0.3\,\text{A} \cdot r$

Now, solve for the internal resistance $r$:

$r = \frac{1.5\,\text{V}}{0.3\,\text{A}} = 5\,\Omega$

So, the internal resistance of each cell is $5\,\Omega$.

$$\begin{aligned} Z =\overline{(A+B) \cdot(A \cdot B)} \\ Y =\bar{Z}=(A+B) \cdot(A \cdot B) \\ \end{aligned}$$
$Y=A \cdot(A B)+B \cdot(A B)$

$$\begin{aligned} & =A B+A B \\ & =(A B) \end{aligned}$$

$\therefore$ It is an AND gate.

$$\begin{aligned} & \mathrm{C}_{\mathrm{eq}}=0.5 \mu \mathrm{F} \\ & \mathrm{Q}=0.5 \times 10=5 \mu \mathrm{C} \\ & \mathrm{Q}^{\prime}=\frac{5 \mu \mathrm{C} \times 0.8}{0.8+0.2}=4 \mu \mathrm{C} \end{aligned}$$

The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:

$V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right)$

At the surface of the sphere (r = R), the gravitational potential is:

$V = \frac{GM}{R}$

Now, let's find the gravitational potential at the center of the sphere (r = 0):

$V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R}$

The gravitational potential at the center of the sphere is $\frac{3}{2}$ times the potential at the surface of the sphere. Therefore:

$V(0) = \frac{3V}{2}$

The energy levels of a one-electron ion can be described by the formula:

$E_n = -\frac{Z^2}{n^2} \times E_0$

where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13.6 eV).

For the $\mathrm{He}^{+}$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:

$E_2 = -\frac{2^2}{2^2} \times (-13.6 ~\mathrm{eV}) = -13.6 ~\mathrm{eV}$

So, the energy of the $\mathrm{He}^{+}$ ion in its first excited state is $-13.6 ~\mathrm{eV}$.

Using the Doppler effect formula:

$f = f_0\left(\frac{c + v_0}{c + v_s}\right)$

where

• $f$ is the observed frequency (frequency heard by the passenger in car Q)
• $f_0$ is the source frequency (400 Hz)
• $c$ is the speed of sound (360 m/s)
• $v_0$ is the velocity of the observer (passenger in car Q) relative to the medium (air) in the direction of the source (40 m/s)
• $v_s$ is the velocity of the source (car P) relative to the medium (air) in the direction of the observer (20 m/s, since it's moving in the same direction as car Q)

Plugging in the values:

$f = 400\left(\frac{360 + 40}{360 + 20}\right)$

$f = 400\left(\frac{400}{380}\right)$

$f = 421$

So, the observed frequency is 421 Hz.

$\text { Apparent depth of } \mathrm{O}=\frac{\mathrm{d}}{\mu}$$=\frac{8}{\frac{4}{3}}=6 \mathrm{~cm}$

Distance of object from mirror = 42 + 6 = 48 m

$\text { Distance between } \mathrm{O} \text { and } \mathrm{I}_2=48+50=98 \mathrm{~cm}$

In this problem, we need to consider the terminal velocity of the droplets, which is reached when the gravitational force is balanced by the drag force acting on the droplet. Terminal velocity is related to the square of the droplet's radius.

The relationship between the terminal velocity (v) and the radius (r) of the droplet is given by:

$v \propto r^2$

Initially, there are 8 equal drops of water, each with radius r and velocity 10 cm/s. When these droplets coalesce, they form a single droplet with a larger radius R. The volume of the new droplet should be equal to the total volume of the 8 smaller droplets.

Using the volume formula for spheres, we can write the relationship between the radii as:

$8 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$

Solving for R, we get:

$R = 2r$

Now, we can use the relationship between the terminal velocities and radii of the droplets:

$\frac{v_1}{v_2} = \left(\frac{r}{R}\right)^2$

Given the initial terminal velocity of 10 cm/s for the smaller droplets ($v_1$) and the relationship between r and R:

$\frac{10}{v_2} = \left(\frac{1}{2}\right)^2$

Solving for the new terminal velocity ($v_2$):

$v_2 = 40 \mathrm{~cm} / \mathrm{s}$

The new terminal velocity after the droplets coalesce is 40 cm/s.

Let the masses of the two nuclear parts be $m_1$ and $m_2$, and their respective speeds be $v_1$ and $v_2$. According to the problem, the ratio of their nuclear sizes is $1 : 2^{1/3}$. Since the nuclear size is proportional to the cube root of the mass, we can write:

$\frac{m_1}{m_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2}$

Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:

$m_1 v_1 = m_2 v_2$

From the problem statement, the ratio of their respective speeds is $n : 1$, so we can write:

$v_1 = n \cdot v_2$

Substitute the expression for $v_1$ into the momentum conservation equation:

$m_1 (n \cdot v_2) = m_2 v_2$

We know the mass ratio, so substitute that into the equation:

$\frac{1}{2} m_2 (n \cdot v_2) = m_2 v_2$

Divide both sides by $m_2 v_2$:

$\frac{1}{2} n = 1$

Now, solve for $n$:

$n = 2$

Thus, the value of $n$ is 2.

To determine the fundamental frequency of the vibrating wire, we need to first find the tension (T) in the wire and the wave velocity (V) in the wire.

1. Tension in the wire (T): We used Young's modulus (Y) to relate the stress and strain in the wire. The formula for stress is:

$\text{stress} = Y \times \text{strain}$

Here, the strain is the extension ($\Delta L$) divided by the original length (L):

$\text{strain} = \frac{\Delta L}{L}$

Now, the tension (T) in the wire is the product of stress and cross-sectional area (A):

$T = \text{stress} \times A$

Combining the above equations, we get the expression for tension:

$T = \frac{Y \Delta L}{L} \times A$

1. Wave velocity in the wire (V): The linear mass density ($\mu$) of the wire is given by:

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

We need to find the ratio $\frac{T}{\mu}$, which represents the square of the wave velocity. Using the expressions for tension and linear mass density, we get:

$\frac{T}{\mu} = \frac{Y \Delta L}{L} \times \frac{A}{m} = \frac{Y \Delta L}{L} \times \frac{1}{\rho}$

Here, $\rho$ is the density of the wire material. Plugging in the given values, we find the value of $\frac{T}{\mu}$, which is:

$\frac{T}{\mu} = 6.4 \times 10^3$

Now, we find the wave velocity (V) by taking the square root of $\frac{T}{\mu}$:

$V = \sqrt{T/\mu} = 80 \mathrm{~m/s}$

1. Fundamental frequency (f): Finally, we find the fundamental frequency of the vibrating wire using the formula:

$f = \frac{V}{2L}$

Plugging in the values, we get the fundamental frequency (f) as:

$f = 80 \mathrm{~Hz}$

So, the fundamental frequency of vibration in the wire is 80 Hz.

Given that the velocity of the body varies with displacement x according to the relation:

$v = 10\sqrt{x}\,\mathrm{ms}^{-1}$

To find the force acting on the body, we first need to find its acceleration, which can be obtained by differentiating the velocity with respect to time. However, we don't have the velocity expressed as a function of time, but rather as a function of displacement. To work around this, we will use the chain rule:

$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$

Now, differentiate the velocity with respect to displacement:

$\frac{dv}{dx} = \frac{1}{2} \cdot 10 \cdot x^{-1/2} = 5x^{-1/2}$

Recall that $\frac{dx}{dt}$ is the velocity, so we have:

$\frac{dv}{dt} = 5x^{-1/2} \cdot 10\sqrt{x} = 50$

Thus, the acceleration is constant and equal to 50 m/s². Now we can find the force acting on the body using Newton's second law:

$F = ma$

First, convert the mass from grams to kilograms:

$m = \frac{500\,\mathrm{g}}{1000} = 0.5\,\mathrm{kg}$

Now, calculate the force:

$F = (0.5\,\mathrm{kg})(50\,\mathrm{ms}^{-2}) = 25\,\mathrm{N}$

The force acting on the body is 25 N.

A. $\mathrm{E_{\text {cell }}}$ is an intensive parameter.

This statement is correct. The standard cell potential ($\mathrm{E}^{\ominus}_{\text{cell}}$) is an intensive parameter because it depends only on the nature of the reactants and the products, and not on the size or shape of the electrodes or the amount of material present.

B. A negative $\mathrm{E}^{\ominus}$ means that the redox couple is a stronger reducing agent than the $\mathrm{H}^{+} / \mathrm{H}_{2}$ couple.

This statement is also correct. A negative standard reduction potential ($\mathrm{E}^{\ominus}$) for a redox couple indicates that the couple is a stronger reducing agent than the standard hydrogen electrode ($\mathrm{H}^{+} / \mathrm{H}_{2}$), which has a standard reduction potential of 0 V.

C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

This statement is correct. The amount of electricity required for a particular oxidation or reduction reaction depends on the number of electrons involved in the reaction, which in turn depends on the stoichiometry of the electrode reaction.

D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

This statement is also correct. Faraday's laws of electrolysis state that the amount of chemical reaction at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

Therefore, all of the statements A, B, C, and D are correct.