JEE Main 2025 8th April Evening Shift Question Paper with Solutions

Electrostatic force is a conservative force, so the work $W$ done by (or against) the electric field in taking a test charge $q_0$ from point $A$ to point $B$ is related to the potential difference by
$W = q_0\,(V_B - V_A)\,\,\,\,\,\,\,\, -(1)$

For a uniformly charged thin spherical shell of radius $R$, the magnitude of the electric field $E$ at any point whose distance from the centre is $r$ satisfies
$E = 0 \quad \text{for} \; r \lt R$
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^{2}} \quad \text{for} \; r \gt R$

Since $E = 0$ everywhere inside the shell, the line integral of $\mathbf{E}\cdot d\mathbf{l}$ between any two interior points is zero. Therefore the electrostatic potential throughout the interior is the same constant value, equal to the potential on the surface:
$V_{\text{inside}} = V_{\text{surface}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\,\,\,\,\,\,\,\, -(2)$

Using $(2)$ in $(1)$, for any two interior points $A$ and $B$ we get
$V_B = V_A \;\; \Longrightarrow \;\; W = q_0\,(V_B - V_A) = 0$

This result does not depend on the path followed because the electric force is conservative. Hence:

Case 1 : Assertion A
"Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen."
We have just proven $W = 0$, so Assertion A is true.

Case 2 : Reason R
"Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell."
Equation $(2)$ confirms this, so Reason R is also true.

Because the constancy of potential (Reason R) directly leads to zero potential difference, which in turn makes the work done zero (Assertion A), R is the correct explanation of A.

Therefore the correct choice is Option B: Both A and R are true and R is the correct explanation of A.

The straight rod is bent until its two ends meet, so the whole length of the rod becomes the circumference of a circle.

Hence, $\text{circumference}=2\pi R=L$ $-(1)$ which gives $R=\frac{L}{2\pi}$ $-(2)$

Mass of the ring: linear mass density $\lambda$ means $m=\lambda\times(\text{length})=\lambda L$ $-(3)$

For a thin ring of mass $m$ and radius $R$:

• Moment of inertia about the axis perpendicular to its plane and passing through the centre is $I_{z}=mR^{2}$. • For any diameter (say $x$-axis) lying in the plane of the ring, use the perpendicular-axis theorem:   $I_{x}+I_{y}=I_{z}$. Because of symmetry, $I_{x}=I_{y}=I_{d}$ (moment of inertia about any diameter), so

$2I_{d}=mR^{2} \;\; \Longrightarrow \;\; I_{d}=\frac{mR^{2}}{2}$ $-(4)$

Substitute $m$ from $(3)$ and $R$ from $(2)$ into $(4)$:

$I_{d}=\frac{\lambda L}{2}\left(\frac{L}{2\pi}\right)^{2} =\frac{\lambda L}{2}\cdot\frac{L^{2}}{4\pi^{2}} =\frac{\lambda L^{3}}{8\pi^{2}}$

Thus, the moment of inertia of the ring about any diameter is $\boxed{\displaystyle \frac{\lambda L^{3}}{8\pi^{2}}}$, which corresponds to Option D.

The extension in a wire under a load is related to Young’s modulus $Y$ by the formula
$Y = \frac{F\,L}{A\,\Delta L}$
where
  $F$ = stretching force,
  $L$ = original length of the wire,
  $A$ = cross-sectional area, and
  $\Delta L$ = extension produced.

First, find each quantity from the data given:

Force
The load is a mass of $50\ \text{kg}$, so the force is its weight:
$F = m g = 50 \times 3\pi = 150\pi\ \text{N}$

Original length
$L = 3\ \text{m}$

Extension
$\Delta L = 0.1\ \text{mm} = 0.1 \times 10^{-3}\ \text{m} = 1 \times 10^{-4}\ \text{m}$

Cross-sectional area
The wire has radius $r = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$.
Area $A = \pi r^{2} = \pi \left(3 \times 10^{-3}\right)^{2} = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6}\ \text{m}^{2}$

Substitute into the formula for $Y$
$Y = \frac{150\pi \times 3}{9\pi \times 10^{-6} \times 1 \times 10^{-4}}$

Simplify step by step:

• Multiply force and length:
$150\pi \times 3 = 450\pi$

• Multiply area and extension:
$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$

• Divide:
$Y = \frac{450\pi}{9\pi \times 10^{-10}} = \frac{450}{9} \times 10^{10} = 50 \times 10^{10}\ \text{N m}^{-2}$

• Write in the required power-of-ten form:
$50 \times 10^{10} = 5 \times 10^{11}\ \text{N m}^{-2}$

Hence the Young’s modulus is $P \times 10^{11}\ \text{N m}^{-2}$ with $P = 5$.

Therefore, the correct option is Option A.

The surface charge density at any point on a conductor is inversely proportional to its local radius of curvature ($R$): $\sigma \propto \frac{1}{R}$

$\sigma_1$ is located at the sharpest point (pointed tip) at the top. This region has the absolute smallest radius of curvature ($R_1$).

$\sigma_3$ is located at the curved base at the bottom, which has a distinct curvature but is less sharp than the tip ($R_3 > R_1$).

$\sigma_2$ and $\sigma_4$ are situated on the symmetrical side flanks, which are relatively flat and identical in curvature ($R_2 = R_4 > R_3 > R_1$).

$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$

When water falls freely, its gravitational potential energy is converted into internal energy (heat) on reaching the pool.
Ignoring any heat loss to the surroundings, the gain in internal energy raises the water’s temperature.

For a mass $m$ of water falling through height $h$, the potential energy lost is
$\text{Potential energy} = m g h$

This energy appears as heat: $m c \Delta T$, where $c$ is the specific heat capacity and $\Delta T$ is the rise in temperature.

Equating the two energies:
$m g h = m c \Delta T$

Cancel the common factor $m$ (mass of water):
$g h = c \Delta T$

Hence the temperature rise is
$\Delta T = \frac{g h}{c}$

Insert the given values: $g = 10 \text{ m s}^{-2}$, $h = 200 \text{ m}$, $c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$.

$\Delta T = \frac{10 \times 200}{4200}$

$\Delta T = \frac{2000}{4200} = 0.476 \text{ K} \approx 0.48 \text{ K}$

Therefore, the rise in temperature of the water is about $0.48 \text{ K}$.

Correct option: Option D (0.48 K)

$\frac{1}{F} = \frac{1}{f_{\text{liquid}}} + \frac{1}{f_{\text{glass}}}$ (Net focal length)

$\frac{1}{f} = (\mu_{\text{lens}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ (individual focal length)

$\frac{1}{f_{\text{glass}}} = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{-20} \right)$

$\frac{1}{f_{\text{glass}}} = 0.5 \left( -\frac{1}{30} + \frac{1}{20} \right) = 0.5 \left( \frac{-2 + 3}{60} \right) = 0.5 \times \frac{1}{60} = \frac{1}{120}\text{ cm}^{-1}$

$\frac{1}{f_{\text{liquid}}} = (1.3 - 1) \left( \frac{1}{\infty} - \frac{1}{-30} \right)$

$\frac{1}{f_{\text{liquid}}} = 0.3 \left( 0 + \frac{1}{30} \right) = \frac{0.3}{30} = \frac{1}{100}\text{ cm}^{-1}$

$\frac{1}{F} = \frac{1}{100} + \frac{1}{120}$

$F = \frac{600}{11}\text{ cm}$

Since the wire passes through two corners of the cube that are maximally separated, it passes along the body diagonal of the cube.

The side of the cube is  $a=\sqrt{3}\,cm$

Therefore, the length of the wire inside the cube is the body diagonal:  $l=a\sqrt{3}$

$l=\sqrt{3}\times\sqrt{3}\,cm=3\,cm$

$l=3\times10^{-2}\,m$

Hence, the charge enclosed by the cube is  $q=\lambda l$

Given,  $\lambda=2\times10^{-9}\,C\,m^{-1}$

Therefore,  $q=(2\times10^{-9})(3\times10^{-2})$

$q=6\times10^{-11}\,C$

By Gauss's law,  $\Phi=\frac{q}{\epsilon_0}$

Using  $\frac{1}{4\pi\epsilon_0}=9\times10^9$

we get  $\frac{1}{\epsilon_0}=4\pi(9\times10^9)$

Therefore,  $\Phi=6\times10^{-11}\times4\pi\times9\times10^9$

$\Phi=2.16\pi\,Nm^2C^{-1}$

Hence,  ${x=2.16\pi}$

Hence, the correct option is D.

Both diodes have their p-sides connected to the central junction node ($V_{\text{out}}$), which is pulled up toward $+5\text{ V}$ via a resistor.

If $V_{\text{out}}$ rises above $0\text{ V}$, the anode of $D_2$ becomes more positive than its cathode ($0\text{ V}$). This makes diode $D_2$ forward-biased.

For $D_1$ to be forward-biased, $V_{\text{out}}$ would have to exceed $+5\text{ V}$, which is impossible since the maximum supply voltage is $+5\text{ V}$. Thus, diode $D_1$ remains reverse-biased (open circuit).

Since ideal diode $D_2$ is forward-biased, it acts as a perfect closed switch (short circuit) directly to the ground node:

$V_{\text{out}} = V_{\text{ground}} = 0\text{ V}$

The entire voltage drop of $5\text{ V}$ occurs across the pull-up resistor, clamping the output node securely to $0\text{ V}$.

Let the radii of the two conducting spheres be $R$ (smaller) and $3R$ (bigger).
Both spheres initially have the same surface charge density $\sigma$.

Initial charges
Charge on the smaller sphere: $Q_{s}= \sigma \times 4\pi R^{2}$
Charge on the bigger sphere: $Q_{b}= \sigma \times 4\pi (3R)^{2}= \sigma \times 4\pi \times 9R^{2}= 9\sigma \,4\pi R^{2}$

Total initial charge on the system: $Q_{\text{total}} = Q_{s}+Q_{b} = \sigma \,4\pi R^{2}(1+9)=10\sigma \,4\pi R^{2}$

When the spheres are connected by a conducting wire, they come to the same potential. For an isolated charged sphere of radius $r$, the electrostatic potential is

$V = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r}$

After contact, let the final charges be $Q'_s$ (smaller) and $Q'_b$ (bigger).

Equality of potentials gives
$\frac{Q'_s}{R} = \frac{Q'_b}{3R} \;\;\Longrightarrow\;\; Q'_s = \frac{Q'_b}{3} \quad -(1)$

Charge conservation gives
$Q'_s + Q'_b = Q_{\text{total}} = 10\sigma \,4\pi R^{2} \quad -(2)$

Substituting $Q'_s$ from $(1)$ into $(2)$:

$\frac{Q'_b}{3} + Q'_b = 10\sigma \,4\pi R^{2}$
$\frac{4}{3}Q'_b = 10\sigma \,4\pi R^{2}$
$Q'_b = \frac{30}{4}\,\sigma \,4\pi R^{2}= \frac{15}{2}\,\sigma \,4\pi R^{2}$

Using $(1)$,
$Q'_s = \frac{Q'_b}{3}= \frac{15}{2}\times \frac{1}{3}\,\sigma \,4\pi R^{2}= \frac{5}{2}\,\sigma \,4\pi R^{2}$

Final surface charge densities

Smaller sphere:
$\sigma_1 = \frac{Q'_s}{4\pi R^{2}} = \frac{5}{2}\sigma$

Bigger sphere:
$\sigma_2 = \frac{Q'_b}{4\pi (3R)^{2}} = \frac{\frac{15}{2}\,\sigma \,4\pi R^{2}}{4\pi \times 9R^{2}} = \frac{15}{2}\times\frac{1}{9}\,\sigma = \frac{5}{6}\sigma$

Required ratio

$\frac{\sigma_1}{\sigma_2} = \frac{\frac{5}{2}\sigma}{\frac{5}{6}\sigma} = \frac{1}{2}\times\frac{6}{1} = 3$

The ratio $\dfrac{\sigma_1}{\sigma_2}$ equals 3, which matches Option D.

The quantity is $Q = X^{-2}\,Y^{-\,\frac{3}{2}}\,Z^{\frac{2}{5}}$.

For a physical quantity of the form $Q = X^{a}\,Y^{b}\,Z^{c}$, the maximum fractional (relative) error is obtained by adding the absolute contributions of each factor:
$\frac{\Delta Q}{Q}\Big|_{\max}=|a|\;\frac{\Delta X}{X}+|b|\;\frac{\Delta Y}{Y}+|c|\;\frac{\Delta Z}{Z}$.

Here $a=-2,\;b=-\frac{3}{2},\;c=\frac{2}{5}$.
Their absolute values are $|a|=2,\;|b|=\frac{3}{2}=1.5,\;|c|=\frac{2}{5}=0.4$.

The given fractional errors are
$\frac{\Delta X}{X}=0.1,\qquad \frac{\Delta Y}{Y}=0.2,\qquad \frac{\Delta Z}{Z}=0.5$.

Substituting these values:
$\frac{\Delta Q}{Q}\Big|_{\max}=2(0.1)+1.5(0.2)+0.4(0.5)$

Calculate each term:
$2(0.1)=0.2,\qquad 1.5(0.2)=0.3,\qquad 0.4(0.5)=0.2$

Add them:
$0.2+0.3+0.2 = 0.7$

Therefore, the maximum fractional error in $Q$ is $0.7$.

Option C is correct.

The process is sudden and the container is thermally insulated, so no heat enters or leaves the gas. Such a process is adiabatic.

For an adiabatic change in an ideal gas, the equation is
$P V^{\gamma} = \text{constant}$

Let the initial state be $(P_1 , V_1)$ and the final state be $(P_2 , V_2)$. Writing the adiabatic relation for the two states gives
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$

Re-arrange to obtain the pressure ratio:
$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{\gamma}$ $-(1)$

The gas is compressed to $\left(\frac{1}{8}\right)^{\text{th}}$ of its original volume, so
$V_2 = \frac{V_1}{8}$ ⇒ $\frac{V_1}{V_2} = 8$

Given $\gamma = \frac{5}{3}$ for a monoatomic gas, substitute these values in $(1)$:
$\frac{P_2}{P_1} = 8^{\,\frac{5}{3}}$

Write 8 as $2^3$ and simplify:
$8^{\,\frac{5}{3}} = \left(2^3\right)^{\frac{5}{3}} = 2^{\,3 \times \frac{5}{3}} = 2^5 = 32$

Hence the ratio of the final pressure to the initial pressure is $32$.

Therefore, Option C is correct.

For two thin lenses kept in contact, the equivalent focal length $F$ is given by the formula
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

The convex lens has $f_1 = +30\ \text{cm}$ and the concave lens has $f_2 = -20\ \text{cm}$ (negative because it is concave).

Substituting,
$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = \frac{2}{60} - \frac{3}{60} = -\frac{1}{60}$
Hence,
$F = -60\ \text{cm}$

The combination therefore behaves like a single concave lens of focal length $60\ \text{cm}$.

Using the lens formula for a thin lens,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad -(1)$
where
• $u$ = object distance (negative when the object is on the left of the lens)
• $v$ = image distance (positive to the right, negative to the left)
• $F$ = focal length of the equivalent lens

The object is placed $20\ \text{cm}$ to the left, so $u = -20\ \text{cm}$. The focal length is $F = -60\ \text{cm}$. Substituting in $(1)$:

$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60}$
$\frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$

Move the second term to the right side:
$\frac{1}{v} = -\frac{1}{60} - \frac{1}{20}$

Convert to a common denominator of $60$:
$\frac{1}{v} = -\frac{1}{60} - \frac{3}{60} = -\frac{4}{60} = -\frac{1}{15}$

Therefore,
$v = -15\ \text{cm}$

The negative sign shows the image forms on the same side as the object (to the left of the lens system). The distance from the lens is the magnitude:

Image distance = $15\ \text{cm}$.

Hence, the correct option is Option D (15 cm).

The speed of a transverse wave on a stretched string is given by the fundamental relation
$v = \sqrt{\dfrac{T}{\mu}}$ where $T$ is the tension and $\mu$ is the mass per unit length of the string.

Because both strings are made of the same material and carry the same tension $T$, the only quantity that changes from one string to the other is $\mu$.

For a string of density $\rho$ (mass per unit volume) and circular cross-section of radius $R$, the cross-sectional area is
$A = \pi R^{2}$.
Hence,
$\mu = \rho A = \rho \pi R^{2} \;-\!(1)$

Case 1: Radius $R$
Using $(1)$, $\mu_1 = \rho \pi R^{2}$.
Therefore,
$v_1 = \sqrt{\dfrac{T}{\mu_1}} = \sqrt{\dfrac{T}{\rho \pi R^{2}}} \;-\!(2)$

Case 2: Radius $\dfrac{R}{2}$
New area: $A_2 = \pi \left(\dfrac{R}{2}\right)^{2} = \dfrac{\pi R^{2}}{4}$.
Thus,
$\mu_2 = \rho A_2 = \rho \dfrac{\pi R^{2}}{4} = \dfrac{\mu_1}{4} \;-\!(3)$

Wave speed in the second string:
$v_2 = \sqrt{\dfrac{T}{\mu_2}}$.
Substitute $\mu_2$ from $(3)$:
$v_2 = \sqrt{\dfrac{T}{\mu_1/4}} = \sqrt{\dfrac{4T}{\mu_1}} = 2\sqrt{\dfrac{T}{\mu_1}} \;-\!(4)$

Compare $(4)$ with $(2)$: $v_1 = \sqrt{\dfrac{T}{\mu_1}}$.
Therefore,
$\dfrac{v_2}{v_1} = \dfrac{2\sqrt{T/\mu_1}}{\sqrt{T/\mu_1}} = 2$.

Hence $\dfrac{v_2}{v_1} = 2$.
Correct option: Option B.

Resistance of $ABC = r$

Resistance of $ADC = 2r$

$i_1 = I \times \frac{2r}{r + 2r} = \frac{2}{3}I$

$i_2 = I \times \frac{r}{r + 2r} = \frac{1}{3}I$

$B_0 = \frac{\mu_0 i}{4\pi \left(\frac{a}{2}\right)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a}\left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2}\mu_0 i}{2\pi a}$

$B_{ABC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_1}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{2}{3}I\right)}{\pi a} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} \quad (\odot)$

$B_{ADC} = 2 \times \left(\frac{\sqrt{2}\mu_0 i_2}{2\pi a}\right) = \frac{\sqrt{2}\mu_0 \left(\frac{1}{3}I\right)}{\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a} \quad (\otimes)$

$B_{\text{net}} = B_{ABC} - B_{ADC} = \frac{2\sqrt{2}\mu_0 I}{3\pi a} - \frac{\sqrt{2}\mu_0 I}{3\pi a} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$

Initial velocity of the body is given as $\vec v_{in}=3\hat i+4\hat j$ ms$^{-1}$, so

$u_x = 3$ ms$^{-1},\; u_y = 4$ ms$^{-1},\; u_z = 0$ ms$^{-1}$.

The constant force acting on the mass is $\vec F = 6\hat k$ N (along +z-axis).

Using Newton’s second law, $\vec F = m\vec a \Rightarrow \vec a = \frac{\vec F}{m}$.

The mass of the body is $m = 2$ kg, therefore

$\vec a = \frac{6\hat k}{2} = 3\hat k$ ms$^{-2}$.

The body stays in this field for $t = \frac{5}{3}$ s. The kinematics relation $\vec v = \vec u + \vec a t$ gives the final velocity:

Along x-axis: $v_x = u_x + a_x t = 3 + 0 \times t = 3$ ms$^{-1}$ (no force in x).
Along y-axis: $v_y = u_y + a_y t = 4 + 0 \times t = 4$ ms$^{-1}$ (no force in y).
Along z-axis: $v_z = u_z + a_z t = 0 + 3 \times \frac{5}{3} = 5$ ms$^{-1}$.

Hence the velocity vector when the body emerges is

$\vec v_{out}=3\hat i + 4\hat j + 5\hat k$ ms$^{-1}$.

Therefore, the correct option is Option B.

Let the common initial speed of both balls be $u$ and let the projection angles with the horizontal be $\theta_1$ and $\theta_2$ for the first and the second ball, respectively.

Step 1 - Write the expression for maximum height.
For a projectile launched with speed $u$ at an angle $\theta$, the maximum height reached is

$H=\frac{u^{2}\sin^{2}\theta}{2g}$

Step 2 - Set up the given height ratio.
Given that the first ball rises eight times higher than the second,

$\frac{H_1}{H_2}=8$

Using the formula for $H$,

$\frac{\dfrac{u^{2}\sin^{2}\theta_1}{2g}}{\dfrac{u^{2}\sin^{2}\theta_2}{2g}}=8$

Simplifying (the factors $u^{2}$ and $2g$ cancel),

$\frac{\sin^{2}\theta_1}{\sin^{2}\theta_2}=8$

Taking square root on both sides,

$\frac{\sin\theta_1}{\sin\theta_2}=2\sqrt{2}$ $-(1)$

Step 3 - Write the expression for time of flight.
For the same projectile, the total time of flight is

$T=\frac{2u\sin\theta}{g}$

Step 4 - Form the required ratio of flight times.
Taking the ratio of times for the two balls,

$\frac{T_1}{T_2}=\frac{\dfrac{2u\sin\theta_1}{g}}{\dfrac{2u\sin\theta_2}{g}} =\frac{\sin\theta_1}{\sin\theta_2}$

Using result $-(1)$,

$\frac{T_1}{T_2}=2\sqrt{2}$

Final ratio.
Hence

$T_1:T_2 = 2\sqrt{2}:1$

So the correct choice is Option A.

Let the two waves be
$y_1(x,t)=4\sin\bigl(kx-\omega t\bigr)$
$y_2(x,t)=2\sin\bigl(kx-\omega t+\tfrac{2\pi}{3}\bigr)$

Both waves have the same angular argument $\theta=kx-\omega t$, but different amplitudes and a phase difference $\phi=\tfrac{2\pi}{3}$.
Write each wave in the compact form $A\sin\theta$ and $B\sin(\theta+\phi)$, where
$A=4,\; B=2,\; \phi=\tfrac{2\pi}{3}$.

The superposition principle gives
$y=y_1+y_2=A\sin\theta+B\sin(\theta+\phi)$

For two sine functions with the same frequency, the resultant is again a sine function:
$A\sin\theta+B\sin(\theta+\phi)=R\sin(\theta+\delta)$
where

Amplitude formula:
$R^2=A^2+B^2+2AB\cos\phi\quad -(1)$

Phase formula:
$\tan\delta=\frac{B\sin\phi}{A+B\cos\phi}\quad -(2)$

Compute the trigonometric values of $\phi=\tfrac{2\pi}{3}$:
$\cos\phi=\cos\!\left(\tfrac{2\pi}{3}\right)=-\tfrac12,\qquad\sin\phi=\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}$

Substitute into $(1)$:
$$\begin{aligned}R^2&=4^2+2^2+2\cdot4\cdot2\cos\!\left(\tfrac{2\pi}{3}\right)\\&=16+4+16\left(-\tfrac12\right)\\&=20-8\\&=12\end{aligned}$$
$\Rightarrow\;R=\sqrt{12}=2\sqrt3$

Substitute into $(2)$:
$$\begin{aligned}\tan\delta&=\frac{2\cdot\left(\tfrac{\sqrt3}{2}\right)}{4+2\left(-\tfrac12\right)}\\&=\frac{\sqrt3}{4-1}\\&=\frac{\sqrt3}{3}=\frac1{\sqrt3}\end{aligned}$$
$\therefore\;\delta=\tfrac{\pi}{6}$ since $\tan\left(\tfrac{\pi}{6}\right)=\tfrac1{\sqrt3}$ and $0\lt\delta\lt\pi$.

Hence, the resultant wave is
$y(x,t)=2\sqrt3\;\sin\!\bigl(kx-\omega t+\tfrac{\pi}{6}\bigr)$

Amplitude  =  $2\sqrt3$, Phase  =  $\tfrac{\pi}{6}$.

The correct choice is Option D.

For sustained interference in Young’s double-slit experiment, the two light waves reaching a point on the screen must satisfy two basic conditions:
  • They must have the same (or almost the same) frequency / wavelength.
  • They must maintain a constant phase difference.

If the source is white light and both slits are uncovered, every wavelength from the same source reaches both slits, so corresponding components of each wavelength are coherent and an interference pattern is obtained (central white fringe, coloured side fringes).

In the present situation one slit is covered with a red filter while the other is covered with a green filter. Hence:

  • The light emerging from slit $S_1$ is almost monochromatic red with wavelength say $\lambda_r$.
  • The light emerging from slit $S_2$ is almost monochromatic green with wavelength say $\lambda_g$, where $\lambda_g \neq \lambda_r$.

Therefore the two beams that superpose on the screen have different frequencies $\nu_r$ and $\nu_g$. The phase difference between them changes continuously with time because

$\text{Phase difference} = 2\pi\bigl(\nu_r - \nu_g\bigr)t + \text{(constant path term)}$

Since $\nu_r \neq \nu_g$, the term $2\pi\bigl(\nu_r - \nu_g\bigr)t$ varies rapidly; the phase relation is not stable. Time-averaged (detector) intensity at any point becomes simply the sum of individual intensities:

$I = I_r + I_g$

No position on the screen receives a consistently larger or smaller resultant amplitude, so bright and dark bands cannot form. Each slit only produces its own single-slit diffraction envelope; the two envelopes just overlap without producing any alternating maxima and minima.

Hence no interference fringes are observed.

Correct option: Option B

The nucleus is assumed to be a uniform solid sphere.

Radius-mass number relation:
Experimental data give $R = R_0\,A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}\,\text{m}$.

Volume of the nucleus:
$V = \frac{4}{3}\pi R^{3}$

Substitute $R = R_0\,A^{1/3}$:
$V = \frac{4}{3}\pi \left(R_0\,A^{1/3}\right)^{3} = \frac{4}{3}\pi R_0^{3}\,A$ $-(1)$

Mass of the nucleus:
Each nucleon (proton or neutron) has nearly the same mass $m_N$, so for mass number $A$ the nuclear mass is approximately
$M \approx A\,m_N$ $-(2)$

Nuclear mass density $\rho$ is defined as mass per unit volume:
$\rho = \frac{M}{V}$

Insert $(1)$ and $(2)$:
$\rho = \frac{A\,m_N}{\dfrac{4}{3}\pi R_0^{3}\,A} = \frac{3\,m_N}{4\pi R_0^{3}}$

The factor $A$ cancels out; therefore $\rho$ does not depend on the mass number $A$.