JEE Main 2026 4th April Evening Shift Question Paper with Solutions

We recall the basic facts of photo-electric effect:
  • $E = h\nu$, where $h$ is Planck’s constant.
  • Einstein’s equation $h\nu = \phi + eV_s$, where
$\phi$ = work function (minimum energy required to eject an electron),
    $V_s$ = stopping potential, and
    $e$ = magnitude of electronic charge.
  • Threshold frequency $\nu_0$ satisfies $h\nu_0 = \phi$.

Let us evaluate the dimensions of each physical quantity.

Planck’s constant $h$
From $E = h\nu$ we have $h = \dfrac{E}{\nu}$.
Energy has dimensions $[ML^2T^{-2}]$ and frequency has $[T^{-1}]$.
Therefore
$\left[h\right] = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$, which is set (III).

Stopping potential $V_s$
Potential (voltage) is work per unit charge: $V = \dfrac{W}{Q}$.
Work/energy has $[ML^2T^{-2}]$ and charge has $[AT]$ (since current $I$ has $[A]$ and $Q = IT$).
Hence
$\left[V_s\right] = \dfrac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$, which is set (IV).

Work function $\phi$
It is an energy, so
$\left[\phi\right] = [ML^2T^{-2}]$, which is set (I).

Threshold frequency $\nu_0$
Frequency is reciprocal of time, so
$\left[\nu_0\right] = [T^{-1}]$, which is set (II).

Collecting the matches:
A) Planck’s constant → (III) $[ML^2T^{-1}]$
B) Stopping potential → (IV) $[ML^2T^{-3}A^{-1}]$
C) Work function → (I) $[ML^2T^{-2}]$
D) Threshold frequency → (II) $[T^{-1}]$

This corresponds to Option A: A-III, B-IV, C-I, D-II.

Let the speeds with respect to the ground be:

Car A : $u_A = 100$ km/h
Car B : $u_B = 80$ km/h

A passenger sitting in car B throws the stone straight ahead with speed $v$ relative to car B. Hence the speed of the stone with respect to the ground is

$u_s = u_B + v \quad$ (because both the car and the throw are in the same direction).

The stone must hit car A with a relative speed (speed of approach) of $5$ m/s. Therefore, in ground frame, the relative speed between the stone and car A is

$u_{\text{rel}} = u_s - u_A = (u_B + v) - u_A.$

This relative speed should equal $5$ m/s. First convert $5$ m/s to km/h:

$5\;\text{m/s} = 5 \times \frac{18}{5} = 18 \;\text{km/h}.$

Set up the equation

$u_{\text{rel}} = 18$
$\Longrightarrow (u_B + v) - u_A = 18$
$\Longrightarrow (80 + v) - 100 = 18$

Solving for $v$:

$v - 20 = 18 \quad\Longrightarrow\quad v = 38 \;\text{km/h}.$

Hence the required throwing speed is

Option C which is: 38 km/h.

The mass of the body is given in grams, so first convert it into SI units:
$m = 100\,\text{g} = 0.1\,\text{kg}$

The constant force acting on the body is
$\mathbf{F} = 5\hat{i} + 10\hat{j}\,\text{N}$

Using Newton’s second law $\mathbf{F}=m\mathbf{a}$, the acceleration components are

$a_x = \frac{5}{0.1} = 50\,\text{m\,s}^{-2}$
$a_y = \frac{10}{0.1} = 100\,\text{m\,s}^{-2}$

The question does not state any initial velocity, so we take the body to start from rest at $t = 0$:
$u_x = 0,\; u_y = 0$.

For motion with constant acceleration, the displacement after time $t$ in each direction is
$s = ut + \tfrac12 a t^2$.

Time interval: $t = 2\,\text{s}$.

Along the $x$-axis:
Displacement is given as $2x$, so
$2x = \tfrac12 a_x t^2$
$2x = \tfrac12 \times 50 \times (2)^2$
$2x = 0.5 \times 50 \times 4 = 100$
$x = 50$.

Along the $y$-axis:
Displacement is given as $5y$, so
$5y = \tfrac12 a_y t^2$
$5y = \tfrac12 \times 100 \times (2)^2$
$5y = 0.5 \times 100 \times 4 = 200$
$y = 40$.

Therefore, the required ratio is
$x : y = 50 : 40 = 5 : 4$.

Hence the correct choice is:
Option D which is: $5:4$

The position of the projectile is given as a function of time:

$x = 24t, \quad y = 43.6t - 4.9t^{2}$

The instantaneous velocity components are obtained by differentiating the position coordinates with respect to time.

Horizontal component: $v_x = \frac{dx}{dt} = \frac{d(24t)}{dt} = 24 \text{ m\,s}^{-1}$

Vertical component: $v_y = \frac{dy}{dt} = \frac{d\bigl(43.6t - 4.9t^{2}\bigr)}{dt} = 43.6 - 9.8t \text{ m\,s}^{-1}$

At $t = 2$ s, substitute in the expressions above:

$v_x = 24 \text{ m\,s}^{-1}$ (constant for all $t$)
$v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 \text{ m\,s}^{-1}$

The angle $\theta$ that the velocity vector makes with the horizontal is given by

$\tan\theta = \frac{v_y}{v_x} = \frac{24}{24} = 1$

Therefore, $\theta = 45^{\circ}$.

Hence, the correct choice is:
Option B which is: $45°$

Acceleration due to gravity at a height $h$ above Earth’s surface is given by
$g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^{2}}$

The question states that this value becomes $\dfrac{g}{9}$, so set
$\frac{g}{\left(1 + \frac{h}{R}\right)^{2}} = \frac{g}{9}$

Cancel $g$ from both sides:
$\left(1 + \frac{h}{R}\right)^{2} = 9$

Take the positive square root (height is positive):
$1 + \frac{h}{R} = 3$

Solve for $h$:
$\frac{h}{R} = 3 - 1 = 2$
$h = 2R$

The required height is $2R$.

Option C which is: $2R$

Both strings have the same cross-sectional area, call it $A$. The masses of the strings themselves are negligible, so the only forces are the weights of the attached masses.

For string B
Only the mass $2M$ is hanging from its lower end, so the tension throughout string B is
$F_B = 2Mg$

For string A
String A has to support both masses: the mass $M$ fixed to its lower end and the mass $2M$ hanging further below. Hence the tension in string A is
$F_A = (M + 2M)g = 3Mg$

The elongation of a string of length $L$ under tension $F$ is given by
$\Delta L = \frac{F\,L}{A\,Y}$
where $Y$ is Young’s modulus.

Let the original lengths be $L_A$ and $L_B$. The data give
$\frac{L_A}{L_B} = 2 \quad\Rightarrow\quad L_A = 2L_B$
and
$\frac{Y_A}{Y_B} = 0.5 \quad\Rightarrow\quad Y_B = 2Y_A$

Now write the ratio of elongations:

$\frac{\Delta L_A}{\Delta L_B} = \frac{\dfrac{F_A L_A}{A Y_A}} {\dfrac{F_B L_B}{A Y_B}} = \frac{F_A L_A Y_B}{F_B L_B Y_A}$

Insert the known values:

$\frac{\Delta L_A}{\Delta L_B} = \frac{(3Mg)\,(2L_B)\,(2Y_A)}{(2Mg)\,(L_B)\,(Y_A)} = \frac{3 \times 2 \times 2}{2} = 6$

Therefore, the ratio of elongations is $\mathbf{6}$.

Option D which is: $6$

The volume flow rate of an incompressible fluid remains constant along a streamline. This is the equation of continuity: $Q = A_1 v_1 = A_2 v_2$, where $Q$ is the discharge (volume per time), $A$ is area and $v$ is speed.

Inlet (hose)
Cross-sectional area: $30\text{ cm}^2 = 30 \times 10^{-4}\text{ m}^2 = 3.0 \times 10^{-3}\text{ m}^2$
Speed of water: $50\text{ cm/s} = 0.50\text{ m/s}$

Volume flow rate at inlet:
$Q = A_{\text{in}} v_{\text{in}} = (3.0 \times 10^{-3}) (0.50) = 1.5 \times 10^{-3}\text{ m}^3\!\!/\text{s}$

Outlet (10 perforations)
Area of one perforation: $15\text{ mm}^2 = 15 \times 10^{-6}\text{ m}^2$
Total outlet area: $A_{\text{out}} = 10 \times 15 \times 10^{-6} = 1.5 \times 10^{-4}\text{ m}^2$

Let $v_{\text{out}}$ be the speed of water emerging from each perforation (all equal). Using continuity: $Q = A_{\text{out}} v_{\text{out}}$

Therefore,
$v_{\text{out}} = \frac{Q}{A_{\text{out}}} = \frac{1.5 \times 10^{-3}}{1.5 \times 10^{-4}} = 10\text{ m/s}$

Hence, the speed of water leaving each perforation is $10\text{ m/s}$.

Option B which is: $10\text{ m/s}$

Average translational kinetic energy per molecule in an ideal gas is given by
$\overline{E_k} = \frac{3}{2}kT$
where $k$ is Boltzmann’s constant and $T$ is the absolute temperature.

Thus, when $\overline{E_k}$ of any two gases is equal, we must have the same temperature $T$ for both. Therefore, the Assertion (A) is true.

The rms speed of molecules of an ideal gas is
$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$
where $m$ is the mass of one molecule (or molar mass $M$ in suitable units). For two different gases at the same temperature:
$\frac{v_{\text{rms}(1)}}{v_{\text{rms}(2)}} = \sqrt{\frac{m_2}{m_1}}$

Because $m_{H_2} \neq m_{O_2}$, their rms speeds cannot be equal at the same temperature. Hence the Reason (R) is false.

Assertion is true while Reason is false ⇒ Option C.

Let the original length of the strip at temperature $T_1$ be $L_0$. For a solid with constant linear expansion coefficient $\alpha$, the change in length for a temperature change $\Delta T$ is

$\Delta L = L \,\alpha\,\Delta T \quad -(1)$

Elongation from $T_1$ to $T_2$
The temperature rise is $T_2-T_1 = \Delta T$. Using $(1)$ with the initial length $L_0$, we have

$\Delta L_1 = L_0 \,\alpha\,\Delta T \quad -(2)$

After this heating, the new length at $T_2$ becomes

$L_{T_2} = L_0 + \Delta L_1 = L_0\bigl(1+\alpha\Delta T\bigr) \quad -(3)$

Elongation from $T_2$ to $T_3$
We are told $T_3+T_1 = 2T_2$. Re-arranging, $T_3 - T_2 = T_2 - T_1 = \Delta T \quad -(4)$ Thus the second heating also raises the temperature by $\Delta T$.

Now use $(1)$ again, but the initial length for this stage is $L_{T_2}$ from $(3)$:

$\Delta L_2 = L_{T_2}\,\alpha\,(T_3-T_2) = L_0\bigl(1+\alpha\Delta T\bigr)\alpha\Delta T \quad -(5)$

Compare $(5)$ with $(2)$ (note that $\Delta L_1 = L_0\alpha\Delta T$):

$\Delta L_2 = \Delta L_1\bigl(1+\alpha\Delta T\bigr)$

Hence

$\boxed{\;\Delta L_2 = \Delta L_1[\,1+\alpha\Delta T\,]\;}$

Therefore the correct option is
Option D which is: $\Delta L_1[1 + \alpha\Delta T]$.

For small-angle oscillations about a fixed horizontal axis, any rigid body behaves as a physical pendulum. Its time period is

$T = 2\pi\sqrt{\frac{I_A}{Mgd}} \quad -(1)$
where
  • $I_A$ is the moment of inertia about the axis of suspension A,
  • $M$ is the mass of the body,
  • $d$ is the distance between the axis A and the centre of mass (C.O.M.),
  • $g$ is the acceleration due to gravity.

Moment of inertia about axis A

The disc is uniform, has radius $R$ and mass $M$. The axis A passes through a point on the rim and is perpendicular to the plane of the disc (a tangential axis). Using the parallel-axis theorem,

$I_A = I_{\text{cm dia}} + Md^{2} \quad -(2)$
Here,
  • $I_{\text{cm dia}}$ for a disc about its central axis (parallel to the plane) is $\frac{1}{4}MR^{2}$.
  • The distance from the rim to the centre is $d = R$.

Substitute in $(2):$

$I_A = \frac{1}{4}MR^{2} \;+\; M R^{2} \;=\; \frac{5}{4} M R^{2} \quad -(3)$

Distance $d$ between axis and C.O.M.

The C.O.M. of a uniform disc is at its centre, so $d = R$.

Time period

Insert $I_A$ from $(3)$ and $d = R$ into the physical-pendulum formula $(1):$

$T = 2\pi\sqrt{\frac{\frac{5}{4}MR^{2}}{M g R}} = 2\pi\sqrt{\frac{5}{4}\,\frac{R}{g}} = 2\pi\sqrt{\frac{5R}{4g}}$

Thus the time period of oscillation is

$T = 2\pi\sqrt{\frac{5R}{4g}}$

Option A which is: $2\pi\sqrt{\frac{5R}{4g}}$

The restoring torque on a rigid electric dipole of moment magnitude $p$ kept in a uniform field $\vec E$ is $\tau = -pE\sin\theta$, where $\theta$ is the small angular displacement from the equilibrium direction.

For small oscillations $\sin\theta \approx \theta$, so the equation of rotational SHM is

$I\frac{d^{2}\theta}{dt^{2}} = -pE\,\theta$

Here $I$ is the moment of inertia of the dipole about its axis of rotation. Comparing with $\ddot\theta + \omega^{2}\theta = 0$, the angular frequency is

$\omega = \sqrt{\frac{pE}{I}}$

Case 1: Only the original field $\vec{E}_1 = E_0\hat x$ is present.
Magnitude $E_1 = E_0$

$\omega_1 = \sqrt{\frac{pE_0}{I}}$

Case 2: The additional field $\vec{E}_2 = 2E_0(\hat y + \hat z)$ is switched on.

The net field is the vector sum

$\vec E_{\text{net}} = E_0\hat x + 2E_0\hat y + 2E_0\hat z$

Magnitude

$E_{\text{net}} = E_0\sqrt{1^{2} + 2^{2} + 2^{2}} = E_0\sqrt{1 + 4 + 4} = 3E_0$

The new angular frequency is therefore

$\omega_2 = \sqrt{\frac{pE_{\text{net}}}{I}} = \sqrt{\frac{p(3E_0)}{I}} = \sqrt{3}\,\omega_1$

The ordinary (cyclic) frequency is proportional to angular frequency, so the same factor applies:

$f_2 = \sqrt{3}\,f_1$

Percentage change in frequency:

$\frac{f_2 - f_1}{f_1}\times 100\% = (\sqrt{3}-1)\times 100\% \approx (1.732 - 1)\times 100\% \approx 0.732 \times 100\% \approx 73\%$

Hence, the frequency increases by approximately $73\%$.

Option A which is: $73\%$

Let the node between $C_1$ and $C_2$ be P. 
The node between $C_2$ and $C_3$ is A (since that is shorted)
The two capacitors $C_1 , C_2$ are connected end-to-end between the terminals A and P, so they are in parallel.
The simplified circuit looks like 

$(C_1 \parallel C_2) \text{ in series with } C_4$
$\text{and the above eq. cap in parallel with } C_3$

$C_{eq} = \frac{(C_1 + C_2)C_4}{(C_1 + C_2) + C_4} + C_3$

$= \frac{(1 + 1)(2)}{1 + 1 + 2} + 1$

$= \frac{4}{4} + 1$

$= 2\mu F$

Option C which is: $2,\mu F$

The assertion states: “A conductor has no net charge inside it.”

In electrostatic equilibrium, the electric field $\mathbf{E}$ inside a conductor is zero. By Gauss’s law, $\oint \mathbf{E}\cdot d\mathbf{A}= \frac{Q_{\text{encl}}}{\varepsilon_0}$. Since $\mathbf{E}=0$ everywhere on the Gaussian surface drawn wholly inside the conductor, the left side is zero, giving $Q_{\text{encl}} = 0$. Hence no net charge can reside in the interior; all excess charge moves to the outer surface. Therefore the Assertion (A) is true.

The reason says: “Inside a parallel plate capacitor (without dielectric), free charges experience a force and drift.” Between the plates of an isolated parallel-plate capacitor there exists a uniform electric field $\mathbf{E} = \frac{\sigma}{\varepsilon_0}$ (directed from positive to negative plate). Any free charge placed in this region will indeed feel the force $\mathbf{F}=q\mathbf{E}$ and will drift toward one of the plates. Hence the Reason (R) is also true.

However, the phenomenon of charge drift inside the gap of a capacitor explains the behaviour of charges in a region where a finite electric field already exists; it does not account for the absence of net charge inside a conductor, which follows from zero electric field and Gauss’s law as shown above. Thus R is not the correct explanation of A.

Both A and R are true, but R is NOT the correct explanation of A. Hence the correct choice is:

Option B which is: Both A and R are true, but R is NOT the correct explanation of A

The magnetic flux density, magnetic field strength and permeability are related by
$B = \mu\,H = \mu_{0}\,\mu_{r}\,H$

Hence
$H = \frac{B}{\mu_{0}\,\mu_{r}}$

Given data:
$B = 1.0\ \text{T},\qquad \mu_{r} = 400,\qquad \mu_{0} = 4\pi \times 10^{-7}\ \text{H m}^{-1}$

Substituting the values:
$H = \frac{1.0}{\bigl(4\pi \times 10^{-7}\bigr)\,(400)}$

Simplify the denominator:
$4\pi \times 10^{-7} \times 400 = 1600\pi \times 10^{-7} = 1.6\times10^{-4}\,\pi$

Therefore
$H = \frac{1}{1.6\times10^{-4}\,\pi}\ \text{A m}^{-1}$

Working out the numerical factor:
$\frac{1}{1.6\times10^{-4}} = \frac{1}{1.6}\times10^{4} = 0.625\times10^{4} = 6.25\times10^{3}$

Hence
$H = \frac{6.25\times10^{3}}{\pi}\ \text{A m}^{-1}$

Express this in the form $H = \alpha \times 10^{5}\ \text{A m}^{-1}$:
$\frac{6.25\times10^{3}}{\pi} = \frac{0.0625\times10^{5}}{\pi}$

Thus
$\alpha = \frac{0.0625}{\pi} = \frac{1}{16\pi}$

Therefore, the correct choice is:
Option B which is: $\frac{1}{16\pi}$

The given magnetic field of the electromagnetic (EM) wave is
$\vec{B}=B_0\sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\,\hat{j}$

Step 1 — Identify the direction of propagation.
For a plane EM wave, the phase $\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)$ has the form $\omega t-kx$, which corresponds to a wave travelling along the $+x$-axis. Hence, the propagation vector $\vec{k}$ is along $+\hat{i}$.

Step 2 — Use the orthogonality of $\vec{E},\;\vec{B}$ and $\vec{k}$.
In a uniform EM wave, $\vec{E}\perp\vec{B}\perp\vec{k}$ and
$\vec{E}\times\vec{B} = \frac{1}{\mu_0}\,\vec{S}$ points in the propagation direction $\vec{k}$(i.e. $+\hat{i}$ here).

Step 3 — Determine the unit vector for $\vec{E}$.
We already have $\vec{k}=+\hat{i}$ and $\vec{B}$ along $+\hat{j}$. Let $\vec{E}$ be along $\hat{n}$. We need
$\hat{n}\times\hat{j}=+\hat{i}$.
Using the right-hand rule: $(-\hat{k})\times\hat{j}=+\hat{i}$. Therefore $\hat{n}=-\hat{k}$ and $\vec{E}$ must be along $-\hat{k}$.

Step 4 — Relate the magnitudes $E_0$ and $B_0$.
In free space, $E_0 = cB_0$ where $c$ is the speed of light. Using $c=\nu\lambda$, we get
$E_0 = \nu\lambda\,B_0$.

Step 5 — Write the complete electric field.
Keeping the same phase factor as $\vec{B}$, the electric field is
$\vec{E}= -\nu\lambda B_0 \sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\hat{k}$.

Thus the correct option is:
Option A which is: $-\nu\lambda B_0 \sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\hat{k}$.

The focal length of a thin lens kept in air is obtained from the Lens-Maker’s formula

$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where
 • $n$ is the refractive index of the lens material with respect to air,
 • $R_1$ is the radius of curvature of the first surface (sign taken according to the Cartesian convention),
 • $R_2$ is the radius of curvature of the second surface.

The lens is given as a symmetric bi-convex lens, so the magnitudes of the radii are equal: $|R_1| = |R_2| = R$.

For light traveling from left to right:
 • The centre of curvature of the first surface lies to the right of that surface, hence $R_1 = +R$.
 • The centre of curvature of the second surface lies to the left of that surface, hence $R_2 = -R$.

Substituting these signs:

$\frac{1}{f}=(n-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right)=2(n-1)\frac{1}{R}$

Rearranging for $f$:

$f=\frac{R}{2(n-1)}$

The given refractive index is $n = 1.4$, so $n-1 = 0.4$.

Therefore:

$f=\frac{R}{2\times 0.4}=\frac{R}{0.8}=1.25\,R$

Hence

$\frac{f}{R}=1.25$

Option D which is: $1.25$

Let the intensity of the incident unpolarised light be $I_0$. When unpolarised light passes through a polariser, its intensity is reduced to half.

Case 1: Only the original two polarisers (axes at $30^{\circ}$ and $90^{\circ}$)
After the first polariser (axis $30^{\circ}$):
$I_1 = \frac{I_0}{2}$

The second polariser’s axis is at $90^{\circ}$, i.e. it is at an angle $\theta = 90^{\circ}-30^{\circ}=60^{\circ}$ to the first axis.
By Malus’ law, transmitted intensity is
$I_{\text{without}} = I_1 \cos^{2} \theta = \frac{I_0}{2}\cos^{2}60^{\circ} = \frac{I_0}{2}\left(\frac{1}{2}\right)^{2} = \frac{I_0}{8}$

Case 2: A third polariser (axis $60^{\circ}$) inserted between the first two
Order of axes: $30^{\circ} \rightarrow 60^{\circ} \rightarrow 90^{\circ}$.

Step-1 First polariser (axis $30^{\circ}$):
$I_1 = \frac{I_0}{2}$

Step-2 Second polariser (axis $60^{\circ}$) is at $30^{\circ}$ to the first.
$I_2 = I_1 \cos^{2}30^{\circ} = \frac{I_0}{2}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{I_0}{2}\left(\frac{3}{4}\right) = \frac{3I_0}{8}$

Step-3 Third polariser (axis $90^{\circ}$) is at $30^{\circ}$ to the second.
$I_{\text{with}} = I_2 \cos^{2}30^{\circ} = \frac{3I_0}{8}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3I_0}{8}\left(\frac{3}{4}\right) = \frac{9I_0}{32}$

Required ratio
$\frac{I_{\text{with}}}{I_{\text{without}}} = \frac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \frac{9}{32}\times\frac{8}{1} = \frac{9}{4}$

Hence, the intensity becomes $\dfrac{9}{4}$ times larger when the third polariser at $60^{\circ}$ is inserted.

Option C which is: $\frac{9}{4}$

The question asks why, in Rutherford’s gold-foil experiment, only a very small fraction of incident $\alpha$-particles are scattered backward (rebound through angles $\gt 90^{\circ}$).
To answer, recall two key ideas of Rutherford scattering:

1. The nuclear size is extremely small compared with the atomic size.
2. Large deflection (up to $180^{\circ}$) occurs only when the $\alpha$-particle passes very close to the nucleus, i.e. the impact parameter $b$ is extremely small. Such trajectories are often called “head-on” or “near head-on” collisions.

Using these ideas, examine each statement:

Statement A: “The gold nucleus is very small compared to the atom.”
Because the nucleus occupies only about $10^{-15}\text{ m}$ while the atom is about $10^{-10}\text{ m}$, the geometrical area that can give large deflection is tiny. Hence very few $\alpha$-particles come close enough to rebound. Statement A is correct.

Statement B: “$\alpha$-particle and gold nucleus have equal charge.”
The charges are $+2e$ for an $\alpha$-particle and $+79e$ for a gold nucleus; they are not equal. Therefore Statement B is wrong.

Statement C: “Impact parameter is minimum for very few particles.”
The distribution of $\alpha$-particles over the foil is uniform, so only a minute fraction possess a sufficiently small impact parameter $b$ ($b \approx 0$) required for large-angle scattering. Hence Statement C is correct.

Statement D: “Very few $\alpha$-particles have very high kinetic energy.”
All $\alpha$-particles in Rutherford’s beam are emitted from the same radioactive source and have nearly the same kinetic energy. The fraction with “very high” energy is not the reason for rare back-scattering. Statement D is wrong.

Statement E: “Only few $\alpha$-particles undergo head-on collision.”
A head-on collision corresponds to $b \approx 0$ and produces a $180^{\circ}$ reversal. Since $b \approx 0$ events are rare, only a few $\alpha$-particles rebound. Statement E is correct.

Thus the correct set of reasons is A, C, and E.

Therefore, the correct option is
Option D which is: A, C, E Only.

The de Broglie wavelength of any particle is given by $\lambda=\frac{h}{p}$, where $h$ is Planck’s constant and $p$ is the linear momentum of the particle.

When a charged particle of charge magnitude $e$ is accelerated from rest through a potential difference $V$, the gain in kinetic energy is $eV$. Hence for a particle of mass $m$:

$\text{K.E.}=eV=\frac{p^{2}}{2m}\; \Longrightarrow\; p=\sqrt{2meV}$

Substituting this in the de Broglie formula:

$\lambda=\frac{h}{\sqrt{2meV}}$ $-(1)$

Both the electron (mass $m_e$) and the proton (mass $m_p$) are accelerated through the same potential difference $V$, so the factor $\sqrt{2eV}$ is common. Writing their wavelengths from $(1)$:

$\lambda_e=\frac{h}{\sqrt{2m_e eV}}, \qquad \lambda_p=\frac{h}{\sqrt{2m_p eV}}$

Taking the ratio:

$\frac{\lambda_e}{\lambda_p} =\frac{h}{\sqrt{2m_e eV}}\;\Big/\;\frac{h}{\sqrt{2m_p eV}} =\frac{1}{\sqrt{m_e}}\times\sqrt{m_p} =\sqrt{\frac{m_p}{m_e}}$

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

Option A which is: $\sqrt{\frac{m_p}{m_e}}$

In a $p$-$n$ junction diode the biasing conditions decide the direction and magnitude of current.
Let the applied voltage be $V$ with the convention that forward bias is positive and reverse bias is negative.

Behaviour under reverse bias
  • When the junction is reverse-biased, the depletion region widens and the majority carriers are repelled away from the junction.
  • The only carriers that are able to cross the junction are the thermally generated minority carriers (electrons in the $p$-region and holes in the $n$-region).
  • Their concentration is fixed by temperature, not by the magnitude of the reverse voltage (as long as the electric field is below the breakdown limit).
  • Hence the reverse current remains almost constant and is called the reverse saturation current $I_S$.

When the reverse voltage magnitude becomes large enough, the electric field in the depletion region exceeds a critical value. Avalanche multiplication (or Zener tunnelling, depending on doping) starts, causing a sudden, very large increase in reverse current. This point is called the breakdown voltage $V_{BR}$.

Verification of Assertion (A)
The statement in (A) says: “A reverse-biased diode gives a small current independent of voltage until a critical limit is reached.”
This is exactly the description of the constant reverse saturation current up to the breakdown voltage. Therefore Assertion (A) is true.

Verification of Reason (R)
Reason (R) states: “Below the critical voltage, only majority carriers flow.”
This is incorrect because, under reverse bias, majority carriers are blocked by the widened potential barrier. The small reverse current is due to minority carriers, not majority carriers. Hence the Reason (R) is false.

Relationship between A and R
Since (A) is true and (R) is false, (R) cannot be the correct explanation of (A).

Therefore, the correct choice is:
Option C which is: A is true but R is false.