WBJEE 2024 Paper 2 (Physics & Chemistry) Question Paper with Solutions

To find the values of $m$ and $n$ through dimensional analysis, we must ensure that the dimensions on both sides of the proportionality equation are the same. The fundamental dimensions involved here are mass ($\mathrm{M}$), length ($\mathrm{L}$), and time ($\mathrm{T}$).

Let's start with the dimensional analysis of each quantity:

The power $\mathrm{P}$ has the dimensions of energy per unit time, which is:

$[\mathrm{P}] = [\mathrm{ML}^2\mathrm{T}^{-3}]$

The charge $\mathrm{q}$ has the dimensions of electric current ($I$) times time ($\mathrm{T}$), giving us:

$[\mathrm{q}] = [\mathrm{IT}]$

The acceleration $\mathrm{a}$ is the change in velocity per unit time, where velocity ($\mathrm{v}$) has dimensions of length per unit time:

$[\mathrm{a}] = [\mathrm{LT}^{-2}]$

The speed of light $\mathrm{c}$ has the dimensions of velocity, which are:

$[\mathrm{c}] = [\mathrm{LT}^{-1}]$

Given the proportionality relationship:

$\mathrm{P} \propto \frac{(q \mathrm{a})^{\mathrm{m}}}{\mathrm{c}^{\mathrm{n}}}$

The dimensions of the right-hand side are:

$\left[\frac{(q \mathrm{a})^{\mathrm{m}}}{\mathrm{c}^{\mathrm{n}}}\right] = \left[\frac{(\mathrm{IT \cdot LT^{-2}})^{\mathrm{m}}}{(\mathrm{LT^{-1}})^{\mathrm{n}}}\right]$

Simplifying further:

$\left[\frac{(\mathrm{I T \cdot L T^{-2}})^{\mathrm{m}}}{\mathrm{L^{n} T^{-n}}}\right] = \left[\frac{(\mathrm{I L T^{-1}})^{\mathrm{m}}}{\mathrm{L^{n} T^{-n}}}\right]$

Combining the powers:

$\left[\frac{\mathrm{I^m L^m T^{-m}}}{\mathrm{L^{n} T^{-n}}}\right] = \left[\mathrm{I^m L^{m-n} T^{-m+n}}\right]$

We want this to match the dimensions of power:

$[\mathrm{ML}^2\mathrm{T}^{-3}]$

Therefore, we equate the dimensions of each term on both sides:

For the current ($\mathrm{I}$):

$\mathrm{m} = 0 \quad \text{(since there is no current dimension in the power)}$

For the length ($\mathrm{L}$):

$m - n = 2$

For the time ($\mathrm{T}$):

$-m + n = -3$

Now, we solve these two simultaneous equations:

1. $m - n = 2$

2. $-m + n = -3$

Adding these two equations:

$(m - n) + (-m + n) = 2 - 3$

$0 = -1$

So we have:

$m - n = 2$

$-m + n = -3$

Solving these, we get:

$m = 2$

$n = 3$

Thus, the correct values are:

Option B

$m=2, n=3$

$\mathrm{F} \cos 60^{\circ}<\mu \mathrm{F} \sin 60^{\circ}$

Block does not move

To determine which of the options is correct, it helps to understand the properties of longitudinal waves. Longitudinal waves are waves in which the particle displacement is parallel to the direction of wave propagation. Sound waves in air are a common example of longitudinal waves.

Let's analyze each option:

Option A: Have a unique wavelength

Longitudinal waves, like any other type of waves, have a wavelength which is the distance between successive compressions or rarefactions. Therefore, longitudinal waves can have a unique wavelength.

Option B: Have a unique wave velocity

The wave velocity in a longitudinal wave is determined by the medium in which the wave is traveling. If the medium is uniform, the wave will have a unique velocity. Therefore, longitudinal waves can have a unique wave velocity.

Option C: Transmit energy

Waves, in general, transmit energy through the medium. This is true for both longitudinal and transverse waves. Longitudinal waves can transmit energy.

Option D: Be polarized

Polarization refers to the orientation of the oscillations in a wave. Only transverse waves can be polarized because their oscillations are perpendicular to the direction of wave propagation. Since the oscillations in longitudinal waves are parallel to the direction of wave propagation, they cannot be polarized.

Therefore, the correct answer is:

Option D: Be polarized

To determine the direction of the vector expression

$\overrightarrow{\mathrm{B}} - \mathrm{B} \sin \theta \hat{\mathrm{a}}_{\perp}$, let's break it down step-by-step:

1. Understanding the given vectors:

Let $\vec{A}$ and $\vec{B}$ be two vectors with an angle $\theta$ between them. The unit vector $\hat{a}_{\perp}$ is perpendicular to $\vec{A}$.

2. Projection of $\vec{B}$:

The term $\mathrm{B} \sin \theta$ is equivalent to the magnitude of the component of $\vec{B}$ that is perpendicular to $\vec{A}$. Thus, multiplying it with $\hat{a}_{\perp}$ gives a vector component of $\vec{B}$ perpendicular to $\vec{A}$, which can be written as:

$\mathrm{B} \sin \theta \hat{\mathrm{a}}_{\perp}$

3. Subtracting the perpendicular component from $\vec{B}$:

When we subtract this perpendicular component from $\vec{B}$, we get:

$\overrightarrow{\mathrm{B}} - \mathrm{B} \sin \theta \hat{\mathrm{a}}_{\perp}$

This operation essentially removes the part of $\vec{B}$ that is perpendicular to $\vec{A}$, leaving only the component of $\vec{B}$ that is parallel to $\vec{A}$.

4. Conclusion:

The resulting vector will be pointing in the direction along $\vec{A}$ because the perpendicular component has been removed. Thus, the direction of

$\overrightarrow{\mathrm{B}} - \mathrm{B} \sin \theta \hat{\mathrm{a}}_{\perp}$ is along $\vec{A}$.

Therefore, the correct answer is:

Option C: along $\vec{A}$.

Here's how to determine the kinetic energy (K) and total energy (E) of the satellite:

1. Angular Momentum and Velocity:

The angular momentum (J) of a satellite in circular orbit is given by:

$J = I\omega = mR^2 \omega$

where:

• m is the mass of the satellite
• R is the orbital radius
• ω is the angular velocity

We can relate angular velocity (ω) to linear velocity (v) using:

$v = Rω$

2. Kinetic Energy:

The kinetic energy of the satellite is:

$K = \frac{1}{2}mv^2$

Substituting v = Rω and J = mR²ω, we get:

$K = \frac{1}{2}m(Rω)^2 = \frac{1}{2}m(J^2 / m^2R^4) = \frac{J^2}{2mR^2}$

3. Total Energy:

The total energy (E) of the satellite is the sum of its kinetic energy (K) and potential energy (U). The potential energy in a gravitational field is:

$U = -\frac{GMm}{R}$

where G is the gravitational constant and M is the mass of the Earth.

The total energy is then:

$E = K + U = \frac{J^2}{2mR^2} - \frac{GMm}{R}$

We can express the gravitational constant (G) and the mass of the Earth (M) in terms of the acceleration due to gravity (g) at the Earth's surface:

$GM = gR^2$

Substituting this into the equation for total energy:

$E = \frac{J^2}{2mR^2} - \frac{gR^2m}{R} = \frac{J^2}{2mR^2} - gRm$

Now, using the centripetal acceleration equation, g = v²/R = (Rω)²/R = Rω² and substituting J = mR²ω, we get:

$gRm = Rω²m = \frac{J^2}{mR^2}$.

Therefore, the total energy is:

$E = \frac{J^2}{2mR^2} - \frac{J^2}{mR^2} = -\frac{J^2}{2mR^2}$

Conclusion:

The correct answer is Option B:

$K = \frac{J^2}{2mR^2}, E = -\frac{J^2}{2mR^2}$

To determine which quantity has the dimension of length, we need to analyze the dimensions of each option provided. We have:

- $h$ as Planck's constant
- $m$ as the mass of the electron
- $c$ as the velocity of light

First, we need to recall the dimensions of these constants:

- Planck's constant ($h$) has the dimension of action, which is energy multiplied by time:

$[h] = [E] [T]$

where $[E]$ is the dimension of energy and $[T]$ is the dimension of time.

We can calculate the dimensions for each option:

Option A:

$\frac{\mathrm{hc}}{\mathrm{m}}$

Substituting the known dimensions:

$[\frac{\mathrm{hc}}{\mathrm{m}}] = \frac{[E][T][L]/[T]}{[M]} = \frac{[M][L]^2/[T]^2[L]/[T]}{[M]} = [L]^3$

Option B:

$\frac{\mathrm{h}}{\mathrm{mc}^2}$

Substituting the known dimensions:

$[\frac{\mathrm{h}}{\mathrm{mc}^2}] = \frac{[E][T]}{[M][L]^2/[T]^2} = \frac{[M][L]^2/[T]^2[T]}{[M][L]^2 / [T]^2} = [T] = [1 / c]^{-1}[L]$

Option C:

$\frac{\mathrm{h}^2}{\mathrm{mc}^2}$

Substituting the known dimensions:

$[\frac{\mathrm{h}^2}{\mathrm{mc}^2}] = \frac{([E][T])^2}{[M][L]^2/[T]^2} = \frac{[M]^2[L]^4/[T]^4[T]^2}{[M][L]^2/[T]^2} = [M][L]^2$

Option D:

$\frac{\mathrm{h}}{\mathrm{mc}}$

Substituting the known dimensions:

$[\frac{\mathrm{h}}{\mathrm{mc}}] = \frac{[E][T]}{[M][L]/[T]} = \frac{[M][L]^2/[T]^2[T]}{[M][L]/[T]} = [L]$

Thus, the quantity that has the dimension of length is Option D:

$\frac{\mathrm{h}}{\mathrm{mc}}$

$I=\frac{2}{10}=0.2 \mathrm{~A}$

To determine the angular frequency of the oscillation of a particle under the given potential energy function, we first need to find the effective force acting on the particle by taking the negative gradient of the potential energy:

$F(x) = - \frac{dU(x)}{dx}$

Given the potential energy function:

$U(x) = -\frac{\alpha x}{x^2 + \beta^2}$

we can compute the force as follows:

$F(x) = - \frac{d}{dx} \left( -\frac{\alpha x}{x^2 + \beta^2} \right)$

Performing the differentiation using the quotient rule for $\frac{u}{v}$, where $u = \alpha x$ and $v = x^2 + \beta^2$:

$\frac{d}{dx} \left( \frac{\alpha x}{x^2 + \beta^2} \right) = \frac{(\alpha)(x^2 + \beta^2) - (\alpha x)(2x)}{(x^2 + \beta^2)^2}$

Simplifying the numerator:

$\alpha x^2 + \alpha \beta^2 - 2\alpha x^2 = \alpha \beta^2 - \alpha x^2$

So the force is:

$F(x) = -\frac{\alpha \beta^2 - \alpha x^2}{(x^2 + \beta^2)^2}$

In simple harmonic motion, near the equilibrium position at $x = 0$, the potential energy function can be approximated as a quadratic function of $x$. That is:

$U(x) \approx \frac{1}{2} k x^2$

where $k$ is the effective spring constant. The force near the equilibrium can also be approximated as:

$F(x) \approx -kx$

Here, we can determine $k$ by comparing the second derivative of the potential energy function at $x = 0$.

Taking the second derivative of $U(x)$ with respect to $x$:

$\frac{d^2U(x)}{dx^2} = \frac{d}{dx} \left( \frac{d U(x)}{dx} \right)$

We previously found the first derivative as:

$\frac{dU(x)}{dx} = \frac{\alpha \beta^2 - \alpha x^2}{(x^2 + \beta^2)^2}$

Evaluating the second derivative at $x = 0$:

$\left. \frac{d^2U(x)}{dx^2} \right|_{x=0} = \left. \frac{d}{dx} \left( \frac{\alpha \beta^2}{(\beta^2)^2} \right) \right|_{x=0}$

Since the potential energy is symmetric around $x = 0$ and the first derivative at $x=0$ is zero, we get:

$\left. \frac{d^2U(x)}{dx^2} \right|_{x=0} = \left. \frac{d}{dx} \left( \frac{\alpha}{\beta^2} \right) \right|_{x=0}$

So, the effective spring constant $k$ is:

$k = \frac{\alpha}{\beta^2}$

The angular frequency $\omega$ of the oscillation for a mass $m$ attached to a spring constant $k$ is given by:

$\omega = \sqrt{\frac{k}{m}}$

Substituting the value of $k$:

$\omega = \sqrt{\frac{\alpha}{m \beta^2}}$

This shows that the angular frequency is proportional to:

$\sqrt{\frac{\alpha}{m \beta^3}}$

Thus, the correct option is:

Option C: $\sqrt{\frac{\alpha}{\mathrm{m} \beta^3}}$

To determine the magnetic field $\vec{B}$, we need to apply the Lorentz force equation. The force experienced by a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by:

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

Here, $\vec{v}$ is the velocity of the particle and $\vec{B}$ is the magnetic field. Given:

$\vec{v} = v_1 \hat{i} + v_2 \hat{j}$

and

$\vec{F} = F_1 \hat{i} + F_2 \hat{j}$

We need to calculate $\vec{v} \times \vec{B}$ in different scenarios and equate to $\vec{F}$ to determine the appropriate $\vec{B}$.

Let's consider each option:

Option A: $\vec{B} = B_1 \hat{i} + B_2 \hat{j}$, where $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

Calculating the cross product:

$\vec{v} \times \vec{B} = (v_1 \hat{i} + v_2 \hat{j}) \times (B_1 \hat{i} + B_2 \hat{j})$

Since

$\hat{i} \times \hat{i} = 0$,

$\hat{j} \times \hat{j} = 0$,

$\hat{i} \times \hat{j} = \hat{k}$, and $\hat{j} \times \hat{i} = -\hat{k}$,

The cross product will be zero:

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

This means that the force would be zero, which contradicts the given condition. Thus, Option A is invalid.

Option B: $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$, with $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

Calculating the cross product:

$\vec{v} \times \vec{B} = (v_1 \hat{i} + v_2 \hat{j}) \times (B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k})$

$= v_1 (\hat{i} \times B_3 \hat{k}) + v_2 (\hat{j} \times B_3 \hat{k})$

Which yields:

$= v_1 B_3 (-\hat{j}) + v_2 B_3 \hat{i}$

$= B_3 (v_2 \hat{i} - v_1 \hat{j})$

This produces a force in the $\hat{i}$ and $\hat{j}$ directions, aligned correctly for constants $F_1$ and $F_2$. Therefore, Option B is valid under the given conditions.

Option C: $\vec{B} = \mathrm{B}_3 \hat{j}$, where $B_1 = B_2 = 0$.

Calculating the cross product:

$\vec{v} \times \vec{B} = (v_1 \hat{i} + v_2 \hat{j}) \times (0 \hat{i} + \mathrm{B}_3 \hat{j})$

The resultant force will be :

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

Thus, Option C is invalid.

Option D: $\vec{B} = B_1 \hat{j} + B_2 \hat{k}$, with $\frac{B_1}{B_2} = \frac{v_1}{v_2}$.

Calculating the cross product:

$\vec{v} \times \vec{B} = (v_1 \hat{i} + v_2 \hat{j}) \times (B_1 \hat{j} + B_2 \hat{k})$

The resultant force will be:

$v_1 \hat{i} \times B_1 \hat{j} = v_1 B_1 \hat{k}$

$v_2 \hat{j} \times B_2 \hat{k} = v_2 B_2 \hat{i}$

Therefore:

$\vec{v} \times \vec{B} = v_2 B_2 \hat{i} + v_1 B_1 \hat{k}$

This produces a force in both directions, but does not align with the given force vectors. Therefore, Option D is not valid.

The correct and valid option is:

Option B: $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ with $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

The elastic potential energy stored in a strained body can be derived from the relationship between stress, strain, and the deformation energy per unit volume. This relationship is often represented using the formula for the elastic potential energy density in a material. The correct option is:

Option D: $\frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

Explanation:

Elastic potential energy is the energy stored in an object when it is deformed but not permanently. When a material follows Hooke's Law, the stress (force per unit area) and strain (deformation per unit length) are linearly related. The formula for elastic potential energy $U$ in a body is given by:

$U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

Where:

- Stress is the internal force per unit area within the material.

• Strain is the relative deformation, i.e., the change in length divided by the original length.


• Volume is the measure of the space occupied by the body where the deformation occurs.

The factor $\frac{1}{2}$ comes from the integration of the linear stress-strain relationship, considering that stress increases linearly with strain from 0 to its maximum value. Thus, the correct representation of the elastic potential energy is given by Option D.

To find the total flux of the electric field through the six surfaces of the cube, we can use Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as:

$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$

where:

• $\Phi_E$ is the electric flux
• $Q_{\text{enc}}$ is the charge enclosed within the surface
• $\epsilon_0$ is the permittivity of free space

In this scenario, the charge $Q$ is placed at the center of the cube. The cube is a closed surface, and according to Gauss's Law, the total flux through this surface only depends on the enclosed charge $Q$ and not on the dimensions of the cube.

Therefore, the total flux $\Phi_E$ through the six surfaces of the cube is:

$\Phi_E = \frac{Q}{\epsilon_0}$

So, the correct answer is:

Option C

$\mathrm{Q} / \epsilon_0$

Let's analyze each of the statements in detail to determine their correctness in respect to nuclear binding energy.

(i) The mass energy of a nucleus is larger than the total mass energy of its individual protons and neutrons.

This statement is actually false. Due to the binding energy, the mass of a nucleus is less than the sum of the masses of its individual protons and neutrons. This difference in mass is converted into binding energy, which is given by Einstein's mass-energy equivalence principle: $E = mc^2$. Therefore, the mass energy of a nucleus is smaller, not larger, than the total mass energy of its constituent nucleons.

(ii) If a nucleus could be separated into its nucleons, an energy equal to the binding energy would have to be transferred to the particles during the separating process.

This statement is true. The binding energy is precisely the energy required to separate a nucleus into its individual protons and neutrons. Without this energy, the nucleons would remain bound within the nucleus.

(iii) The binding energy is a measure of how well the nucleons in a nucleus are held together.

This statement is true. The binding energy quantifies the stability of a nucleus. A higher binding energy indicates that the nucleons are held together more tightly.

(iv) The nuclear fission is somehow related to acquiring higher binding energy.

This statement is true. In nuclear fission, a large nucleus splits into two or more smaller nuclei, and the total binding energy of the resulting nuclei is higher than that of the original nucleus. This release of energy is what drives the process of nuclear fission.

Given the analysis above, we find that statements (ii), (iii), and (iv) are true, while statement (i) is false. Thus, the correct answer is:

Option B

Statements (ii), (iii), and (iv) are true.

We know that for a system, the internal energy (U) is a function of entropy (S) and volume (V). We can write this as:

$dU = TdS - PdV$

Where:

• T is the temperature
• P is the pressure

Now, let's differentiate the given expression for internal energy:

$dU = \frac{4}{3}as^{1/3}V^\alpha dS + a\alpha s^{4/3}V^{\alpha-1}dV$

Comparing the coefficients of dS and dV in both expressions, we get:

$T = \frac{4}{3}as^{1/3}V^\alpha$

$P = -a\alpha s^{4/3}V^{\alpha-1}$

Now, let's find the relation between T, P, V, and S using the following thermodynamic relations:

$T = \left( \frac{\partial U}{\partial S} \right)_V$

$P = -\left( \frac{\partial U}{\partial V} \right)_S$

From the expression for T, we have:

$T = \frac{4}{3}as^{1/3}V^\alpha$

From the expression for P, we have:

$P = -a\alpha s^{4/3}V^{\alpha-1}$

Now, let's substitute the expressions for T and P into the relation $T = \left( \frac{\partial U}{\partial S} \right)_V$ and $P = -\left( \frac{\partial U}{\partial V} \right)_S$:

For $T = \left( \frac{\partial U}{\partial S} \right)_V$:

$\frac{4}{3}as^{1/3}V^\alpha = \frac{4}{3}as^{1/3}V^\alpha$

For $P = -\left( \frac{\partial U}{\partial V} \right)_S$:

$-a\alpha s^{4/3}V^{\alpha-1} = -a\alpha s^{4/3}V^{\alpha-1}$

We can see that both equations are consistent. Therefore, the value of $\alpha$ can be any number. However, in the context of thermodynamics, the internal energy is often expressed as a function of entropy and volume, where the exponent of volume is often -1/3. This leads to the following expression for internal energy:

$U = as^{4/3}V^{-1/3}$

Therefore, the correct answer is D. $-\frac{1}{3}$

Area $=\frac{1}{2}\left(2 x \tan \frac{\theta}{2}\right) x$

$\begin{aligned} & \text { flux, } \varphi=\left(x^2 \tan \frac{\theta}{2}\right) B \\ & \frac{d \varphi}{d t}=\left(2 x \frac{d x}{d t} \tan \frac{\theta}{2}\right) B=2 x v B \tan \frac{\theta}{2}=2(v t) v B \tan \frac{\theta}{2} \\ & \text { At } t=1 \mathrm{sec}, \quad \varepsilon=2 v^2 B \tan \frac{\theta}{2}\end{aligned}$

To determine the height $z$ for which the range $R$ of the emerging water through the hole will be maximum, we need to analyze the problem using the principles of fluid dynamics and projectile motion.

When fluid flows out of a hole in a tank, the speed at which it exits can be described by Torricelli's theorem, which states:

$v = \sqrt{2 g (H - z)}$

where:

• $v$ is the velocity of the water exiting the hole,
• $g$ is the acceleration due to gravity,
• $H$ is the height of the water column above the hole, and
• $z$ is the height of the hole from the bottom of the vessel.

Next, we consider that the water emerging from the hole would follow a parabolic trajectory due to gravity. The range $R$ (horizontal distance) is given by:

$R = v \cdot t$

where:

• $v$ is the horizontal component of the initial velocity (since it exits horizontally), and
• $t$ is the time of flight, given by the time it takes to fall from height $z$ to the ground.

Time of flight $t$ can be determined from the kinematic equation:

$z = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2z}{g}}$

So, the range $R$ can be expressed as:

$R = \sqrt{2 g (H - z)} \cdot \sqrt{\frac{2z}{g}}$

$R = 2 \sqrt{z (H - z)}$

To maximize $R$, we need to find the value of $z$ that maximizes the function $R = 2 \sqrt{z (H - z)}$. We can do this by taking the derivative with respect to $z$ and setting it to zero:

$\frac{d}{dz} \left( 2 \sqrt{z (H - z)} \right) = 0$

Let:

$R = 2 \sqrt{z (H - z)}$

$\frac{dR}{dz} = 2 \cdot \frac{1}{2} (z (H - z))^{-\frac{1}{2}} \cdot (H - 2z) = \frac{H - 2z}{\sqrt{z (H - z)}}$

Setting the derivative to zero:

$\frac{H - 2z}{\sqrt{z (H - z)}} = 0$

$H - 2z = 0$

$z = \frac{H}{2}$

Therefore, the height $z$ for which the range $R$ of the emerging water will be maximum is:

Option B: $\mathrm{z=\frac{H}{2}}$

$$\begin{aligned} & 3 M \times x+M \times \frac{a}{2}=4 M \times 0 \\ & \Rightarrow x=-\frac{a}{6} \end{aligned}$$

Similarly, $y=-\frac{b}{6}$

To determine from which metals electrons will be emitted, we need to determine if the energy of the incident monochromatic light is greater than the work function of the metal. The energy of the incident light can be calculated using the formula:

$E = \dfrac{hc}{\lambda}$

where:

• $h$ is Planck's constant: $6.626 \times 10^{-34} \, \mathrm{Js}$
• $c$ is the speed of light: $3 \times 10^8 \, \mathrm{m/s}$
• $\lambda$ is the wavelength of the light in meters

Given that the wavelength $\lambda = 4770 \mathop A\limits^o$ (which is equivalent to $4770 \times 10^{-10} \, \mathrm{m}$), we can substitute into the formula:

$E = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4770 \times 10^{-10}}$

By calculating this, we get:

$E \approx 4.155 \times 10^{-19} \, \mathrm{J}$

To convert this energy from joules (J) to electron volts (eV), we use the conversion factor of $1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}$:

$E \approx \dfrac{4.155 \times 10^{-19}}{1.602 \times 10^{-19}} \, \mathrm{eV}$

This simplifies to:

$E \approx 2.59 \, \mathrm{eV}$

Now, we compare this energy with the work functions of the metals A, B, C, and D:

• A: $4.2 \, \mathrm{eV}$
• B: $3.7 \, \mathrm{eV}$
• C: $3.2 \, \mathrm{eV}$
• D: $2.3 \, \mathrm{eV}$

Since the energy of the incident light ($2.59 \, \mathrm{eV}$) is less than the work functions of metals A ($4.2 \, \mathrm{eV}$), B ($3.7 \, \mathrm{eV}$), and C ($3.2 \, \mathrm{eV}$), electrons will not be emitted from these metals. However, the energy is greater than the work function of metal D ($2.3 \, \mathrm{eV}$), so electrons will be emitted from metal D.

Therefore, the correct answer is:

$\text{Option D: D only}$