JEE Main 2025 April 8 shift 2

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 $R$: \bullet Moment of inertia about the axis perpendicular to its plane and passing through the centre is $I_{z}=mR^{2}$. \bullet 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: \bullet Multiply force and length: $150\pi \times 3 = 450\pi$ \bullet Multiply area and extension: $9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$ \bullet Divide: $Y = \frac{450\pi}{9\pi \times 10^{-10}} = \frac{450}{9} \times 10^{10} = 50 \times 10^{10}\ \text{N m}^{-2}$ \bullet 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 .

In an irregular metallic disk, charge density is not uniform due to the shape and proximity of points. Typically, charges distribute more densely at points with sharper curvature or edges. In this case, points A, B, and C are likely to exhibit higher charge densities ($\sigma_1$, $\sigma_2$, and $\sigma_3$) compared to D and E. Therefore, the correct answer is option A.

[CORRECT_OPTION: A]

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)

To find the focal length of the liquid-glass combination, use the lens maker's formula:

where $n = \frac{n_{glass}}{n_{liquid}}$. Here, $n_{glass} = 1.5$, $n_{liquid} = 1.3$, $R_1 = -30$ cm (concave) and $R_2 = 20$ cm (convex). Substituting these values, we calculate:

After simplifying, the focal length $f$ is found to be $\frac{600}{11}$ cm. Therefore, the final answer is [CORRECT_OPTION: D].

A long straight wire carries uniform linear charge density $\lambda = 2\,\text{nC m}^{-1}=2\times10^{-9}\,\text{C m}^{-1}$. We have to find the electric-flux through a Gaussian cube of side length $a=\sqrt{3}\,\text{cm}=0.01\sqrt{3}\,\text{m}$ when the wire passes through the two diagonally opposite corners of the cube. Step 1 : Length of the wire lying inside the cube For a cube of side $a$, the space-diagonal length is $a\sqrt{3}$. Because the straight wire enters at one corner and exits at the diametrically opposite corner, the portion of the wire enclosed by the cube equals the space-diagonal. Hence $\ell_{\text{\in}} = a\sqrt{3}$ Substituting $a=\sqrt{3}\,\text{cm}$, $\ell_{\text{\in}} = (\sqrt{3}\,\text{cm})\sqrt{3}=3\,\text{cm}=0.03\,\text{m}$ Step 2 : Charge enclosed by the Gaussian surface Using $q_{\text{enc}} = \lambda \ell_{\text{\in}}$, $q_{\text{enc}} = (2\times10^{-9})\times (0.03)=6\times10^{-11}\,\text{C}$ Step 3 : Electric-flux through the cube Gauss's law states $\Phi = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ $-(1)$ With $\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$, substitute \in $(1)$: $\Phi = \dfrac{6\times10^{-11}}{8.854\times10^{-12}} = 6.784\times10^{0}\;\text{N m}^2\text{C}^{-1}$ Step 4 : Expressing the result \in terms of $\pi$ Calculate $2.16\pi$: $2.16\pi = 2.16 \times 3.1416 \approx 6.79$ which matches the obtained value (6.784). Hence $\Phi = 2.16\pi\; \text{N m}^2\text{C}^{-1}$ Therefore, $x = 2.16\pi$ and the correct option is Option D .

In this ideal diode circuit, D1 is oriented to conduct when the top node is at +5 V, causing Vout to also be +5 V. However, D2 is reverse biased and does not conduct, maintaining Vout at 0 V. Since D1 cannot provide an output greater than +5 V while D2 blocks current, the output is effectively 0 V. Therefore, the correct answer is [CORRECT_OPTION: B].

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 $V_2 = \frac{V_1}{8}$ \Rightarrow $\frac{V_1}{V_2} = 8$ 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 \bullet $u$ = object distance (negative when the object is on the left of the lens) \bullet $v$ = image distance (positive to the right, negative to the left) \bullet $F$ = 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.

The total current I splits into $i_1$ (path ABC) and $i_2$ (path ADC).
Since paths are in parallel, $i_1 r = i_2 (2r) \Rightarrow i_1 = 2i_2$.
Also, $I = i_1 + i_2 = 2i_2 + i_2 = 3i_2$. So, $i_2 = I/3$ and $i_1 = 2I/3$.

The magnetic field at the center of a square due to one side of length 'a' carrying current 'i' is $B_{side} = \frac{\mu_0 i}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (2 \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\sqrt{2}\pi a}$.

For path ABC (current $i_1$):

• Segments AB and BC both contribute an inward field.
• $B_{ABC} = B_{AB} + B_{BC} = 2 \cdot \frac{\mu_0 i_1}{\sqrt{2}\pi a}$ (inward).

For path ADC (current $i_2$):

• Segments AD and DC both contribute an outward field.
• $B_{ADC} = B_{AD} + B_{DC} = 2 \cdot \frac{\mu_0 i_2}{\sqrt{2}\pi a}$ (outward).

Resultant field $B_{net} = B_{ABC} - B_{ADC}$ (since directions are opposite).
$B_{net} = \frac{2\mu_0}{\sqrt{2}\pi a} (i_1 - i_2) = \frac{\sqrt{2}\mu_0}{\pi a} (\frac{2I}{3} - \frac{I}{3}) = \frac{\sqrt{2}\mu_0}{\pi a} (\frac{I}{3})$.
$B_{net} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$.

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 $(1)$:

$

$ $\Rightarrow\;R=\sqrt{12}=2\sqrt3$ Substitute into $(2)$:

$

$ $\therefore\;\delta=\tfrac{\pi}{6}$ 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 $\lambda_r$. \bullet 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 $A$. Hence, the nuclear mass density is independent of A . Correct option: Option D.

Let the natural (unstretched) length of the spring be $L_0 = 2\,\text{m}$. The free end of the spring is fixed to the wall and the block of mass $m = 2\,\text{kg}$ is attached to the other end. When the spring is compressed to a length $1\,\text{m}$ and released, the compression produced is $y_{\text{max}} = L_0 - 1 = 2 - 1 = 1\,\text{m}$. This maximum compression is the amplitude $A$ of the ensuing simple harmonic motion (SHM): $A = 1\,\text{m}$. The force constant (spring constant) is $k = 200\,\text{N\,m}^{-1}$, so the angular frequency of the SHM is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\,\text{rad\,s}^{-1}$. At any instant, let the actual length of the spring (i.e., the distance of the block from the wall) be $x$ metres. Because $x \lt 2\,\text{m}$, the spring is compressed. Define the instantaneous compression (displacement from the natural length) as $y = L_0 - x = 2 - x.$ For SHM with amplitude $A$, angular frequency $\omega$ and displacement $y$ from equilibrium (natural length here), the speed $v$ is related by the standard formula $v = \omega \sqrt{A^{2} - y^{2}}\;.$ Substituting $\omega = 10\,\text{rad\,s}^{-1}$, $A = 1\,\text{m}$ and $y = 2 - x$: $v = 10 \sqrt{1^{2} - (2 - x)^{2}} = 10\,\$[\,1 - (2 - x)^{2}\,]^{1/2}\,\text{m\,s}^{-1}.$ Hence the speed of the block when it is at a distance $x$ from the wall is given by Option B: $\displaystyle 10\$[1 - (2 - x)^{2}]^{1/2}\,\text{m\,s}^{-1}\;.$

An infinite non-conducting sheet having uniform surface charge density $-\sigma$ produces a uniform electric field of magnitude $E=\frac{\sigma}{2\varepsilon_0} \quad -(1)$ The field lines point toward the sheet because $\sigma$ is negative. Charge on an electron $q=-e$, so the force on the electron is $F = qE = (-e)\,E = e\,\frac{\sigma}{2\varepsilon_0}$ away from the sheet. Its magnitude is $|F| = \frac{e|\sigma|}{2\varepsilon_0} \quad -(2)$ Using Newton's second law, the magnitude of the constant acceleration is $a = \frac{|F|}{m} = \frac{e|\sigma|}{2\varepsilon_0 m} \quad -(3)$ The electron is released from rest, so its speed after time $t$ is obtained from the kinematic relation $v = at$: $v = a t \quad -(4)$ The de-Broglie wavelength is defined as $\lambda = \frac{h}{p}=\frac{h}{mv}$. Substituting $v$ from $(4)$ gives $\lambda = \frac{h}{m (a t)} = \frac{h}{m a}\,\frac{1}{t} \quad -(5)$ Thus $\lambda$ is inversely proportional to time. Differentiate $(5)$ with respect to $t$: $\frac{d\lambda}{dt} = -\frac{h}{m a}\,\frac{1}{t^{2}} \quad -(6)$ Equation $(6)$ shows that the rate of change of the de-Broglie wavelength varies as $t^{-2}$. Therefore, the required power $n=2$. Final answer: $n = 2$.

The bulk modulus $B$ of a liquid is defined as $B = -\,\frac{\Delta P}{\dfrac{\Delta V}{V}}$ where $\Delta P$ is the change \in pressure, $\Delta V$ is the change \in volume (negative for compression), and $V$ is the initial volume. We are given: \bullet Initial pressure $P_1 = 1$ atm, final pressure $P_2 = 5$ atm. Hence the pressure rise is $\Delta P = P_2 - P_1 = 5 - 1 = 4$ atm. \bullet Take $1$ atm $= 10^{5}$ Pa, so $\Delta P = 4 \times 10^{5}\ \text{Pa}$. \bullet Bulk modulus $B = 2$ GPa $= 2 \times 10^{9}$ Pa. \bullet Magnitude of volume decrease $|\Delta V| = 0.8\ \text{cm}^3 = 0.8 \times 10^{-6}\ \text{m}^3$. Using the definition of $B$ and taking magnitudes (since we know the sign corresponds to compression), $B = \frac{\Delta P \, V}{|\Delta V|} \quad\Longrightarrow\quad V = \frac{B\,|\Delta V|}{\Delta P}\,.$ Substituting the numerical values: $V = \frac{2 \times 10^{9}\ \text{Pa} \;\times \; 0.8 \times 10^{-6}\ \text{m}^3}{4 \times 10^{5}\ \text{Pa}}$ Step-wise calculation: $2 \times 0.8 = 1.6$ $10^{9} \times 10^{-6} = 10^{3}$ Numerator $= 1.6 \times 10^{3} = 1600$ Denominator $= 4 \times 10^{5}$ Therefore, $V = \frac{1600}{4 \times 10^{5}} = 0.4 \times 10^{-2}\ \text{m}^3 = 0.004\ \text{m}^3.$ Conversion to litres (as $1\ \text{m}^3 = 1000\ \text{L}$): $V = 0.004\ \text{m}^3 \times 1000\ \frac{\text{L}}{\text{m}^3} = 4\ \text{L}.$ Hence, the initial volume of the liquid was $4$ litres.