JEE Main 2025 7th April Evening Shift Question Paper with Solutions

An air-craft is essentially a hollow metallic body. When lightning (or any external electrostatic discharge) strikes it, charges get redistributed only on the outer metallic surface because of electrostatic induction.

For a conductor in electrostatic equilibrium, two well-known facts hold:
  • The electric field $\mathbf{E}$ anywhere inside the conducting material is zero.
  • Consequently, the electric field in any cavity completely surrounded by the conductor is also zero (principle of electrostatic shielding).

Hence, passengers seated inside the cabin experience no electric field; the metal fuselage acts as a Faraday cage that protects them from lightning. This confirms that Assertion (A) is a correct statement.

Reason (R) restates the principle used: “The electric field inside the cavity enclosed by a conductor is zero.” This statement is true and is exactly the physical explanation for the protection mentioned in (A).

Therefore, both Assertion (A) and Reason (R) are correct, and (R) is the correct explanation of (A).

The appropriate choice is Option A.

The nuclear radius empirical formula is $R = r_0 A^{1/3}$, where
$R$ = radius of the nucleus,
$A$ = mass number,
$r_0$ ≈ $1.2 \times 10^{-15}\,$m (a constant).

Hence the nuclear volume is
$V = \frac{4}{3}\pi R^{3} = \frac{4}{3}\pi\left(r_0 A^{1/3}\right)^{3} = \frac{4}{3}\pi r_0^{3} A$.

The mass of a nucleus is roughly $A m_n$, where $m_n$ is the nucleon mass (proton/neutron mass).
Therefore nuclear density is

$\rho = \frac{\text{mass}}{\text{volume}} = \frac{A m_n}{\frac{4}{3}\pi r_0^{3} A} = \frac{m_n}{\tfrac{4}{3}\pi r_0^{3}}$.

The factor $A$ cancels out, so $\rho$ is the same constant for all nuclei, independent of their mass numbers.

Assertion (A): “The density of the copper $\left(^{64}_{29}Cu\right)$ nucleus is greater than that of the carbon $\left(^{12}_{6}C\right)$ nucleus.”
We just showed that densities are practically identical for all nuclei, so Assertion (A) is incorrect.

Reason (R): “The nucleus of mass number $A$ has a radius proportional to $A^{1/3}$.”
This is exactly the empirical formula stated above, so Reason (R) is correct.

Thus, Assertion (A) is not correct but Reason (R) is correct. The appropriate choice is Option B.

The intensity of a plane electromagnetic wave is related to the peak electric-field amplitude $E_0$ by

$I = \frac{1}{2}\,\epsilon_0\,c\,E_0^{2}$ $-(1)$

Solving $-(1)$ for $E_0$ gives

$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$ $-(2)$

Thus the expression whose unit we must find is simply the unit of the electric field $E_0$.

Unit analysis:

Intensity  $I$ : power per area   $\Rightarrow$  $\text{W m}^{-2} = \left(\text{J s}^{-1}\right)\text{m}^{-2} = \text{kg s}^{-3}$

Permittivity  $\epsilon_0$ : $\text{C}^{2}\,\text{N}^{-1}\,\text{m}^{-2}$ Force  $\text{N}= \text{kg m s}^{-2}$ $\therefore \epsilon_0 = \frac{\text{C}^{2}}{\text{kg m s}^{-2}\,\text{m}^{2}} = \frac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}$

Speed of light  $c$ : $\text{m s}^{-1}$

Compute the unit of $\dfrac{2I}{\epsilon_0 c}$ :

$\dfrac{\text{kg s}^{-3}} {\left(\dfrac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}\right)\;(\text{m s}^{-1})} = \dfrac{\text{kg s}^{-3} \;\text{kg m}^{3}} {\text{C}^{2}\,\text{s}^{2}\;\text{m}} = \dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}}$

Taking the square root (see $-(2)$) gives the unit of $E_0$:

$\sqrt{\dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}}} = \dfrac{\text{kg m s}^{-2}}{\text{C}} = \dfrac{\text{N}}{\text{C}}$

Hence $\sqrt{\dfrac{2I}{\epsilon_0 c}}$ has the SI unit $\mathbf{N\,C^{-1}}$, which is also equal to $\mathbf{V\,m^{-1}}$.

Answer : Option D  $NC^{-1}$

The expression is $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$.
We must find its dimensional formula and compare it with the dimensions of the given physical quantities.

Step 1 : Dimensions of vacuum permeability $\mu_0$
In SI, $\mu_0 = 4\pi \times 10^{-7}\;N\,A^{-2}$, where $N = kg \; m \; s^{-2}$.
Hence, $\mu_0$ has dimensions
$[\,\mu_0\,] = M^{1}\,L^{1}\,T^{-2}\,I^{-2}$.

Step 2 : Dimensions of vacuum permittivity $\epsilon_0$
From Coulomb’s law $F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q^{\,2}}{r^{\,2}}$, we get
$\epsilon_0 = \dfrac{q^{\,2}}{F\,r^{\,2}}$.
Using $q = I\,T$ and $F = M\,L\,T^{-2}$:
$[\,\epsilon_0\,] = \dfrac{(I\,T)^2}{M\,L\,T^{-2}\;L^{2}} = M^{-1}\,L^{-3}\,T^{4}\,I^{2}$.

Step 3 : Dimensions of $\dfrac{\mu_0}{\epsilon_0}$
$\left[\dfrac{\mu_0}{\epsilon_0}\right] = M^{1-(-1)}\,L^{1-(-3)}\,T^{-2-4}\,I^{-2-2}$
$= M^{2}\,L^{4}\,T^{-6}\,I^{-4}$.

Step 4 : Taking the square root
$\left[\sqrt{\dfrac{\mu_0}{\epsilon_0}}\right] = M^{1}\,L^{2}\,T^{-3}\,I^{-2}$.

Step 5 : Dimensions of resistance
Resistance $R = \dfrac{V}{I}$.
Electrical power $P = V\,I$ has dimensions $M\,L^{2}\,T^{-3}$.
Therefore voltage $V = \dfrac{P}{I}$ has dimensions $M\,L^{2}\,T^{-3}\,I^{-1}$.
Thus,
$[\,R\,] = \dfrac{[\,V\,]}{[\,I\,]} = M\,L^{2}\,T^{-3}\,I^{-2}$.

Step 6 : Comparison
The dimension $M\,L^{2}\,T^{-3}\,I^{-2}$ obtained for $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ is exactly the same as that of resistance.

Hence, $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ has the dimensions of resistance.
Correct choice: Option D (Resistance).

According to Kepler’s second law, the line joining the Sun and any planet sweeps out equal areas in equal intervals of time. This statement is the same as saying that the areal velocity $\frac{dA}{dt}$ of the planet is constant.

In mechanics the areal velocity of a particle of mass $m$ moving with linear momentum $\mathbf{p}=m\mathbf{v}$ at an instantaneous position vector $\mathbf{r}$ from the origin is given by

$\frac{dA}{dt}=\frac{1}{2m}\,|\mathbf{r}\times\mathbf{p}|$

The quantity $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ is the angular momentum of the particle about the origin. Hence

$\frac{dA}{dt}=\frac{|\mathbf{L}|}{2m}$

If a particle is subjected to a central force, the force vector $\mathbf{F}$ is always directed along the radius vector $\mathbf{r}$. Therefore the torque about the origin is

$\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}=0$

Zero torque implies $\frac{d\mathbf{L}}{dt}=0$, so the angular momentum $\mathbf{L}$ is conserved (constant in both magnitude and direction).

Because $|\mathbf{L}|$ is constant, the expression $\frac{dA}{dt}=\frac{|\mathbf{L}|}{2m}$ shows that the areal velocity $\frac{dA}{dt}$ must also be constant. This directly yields Kepler’s second law.

Thus:

  • Assertion (A) is correct: the radius vector sweeps out equal areas in equal times, so the areal velocity is constant.
  • Reason (R) is correct: in a central force field the angular momentum is conserved.
  • Reason (R) correctly explains why the areal velocity is constant, since constant angular momentum leads to constant $\frac{dA}{dt}$.

Therefore the appropriate choice is Option A: Both (A) and (R) are correct and (R) is the correct explanation of (A).

For an ideal gas, the average translational kinetic energy of one molecule at absolute temperature $T$ is given by the expression

$\overline{E_k} = \frac{3}{2}\,k\,T$

Here, $k$ is the Boltzmann constant and $T$ is the thermodynamic temperature. Observe that $\overline{E_k}$ depends only on the temperature and the universal constant $k$; it is independent of the nature, molar mass, or molecular mass of the gas.

Both helium and argon are in the same flask at the same temperature $T = 300 \text{ K}$. Therefore, their average kinetic energies per molecule are equal:

$\overline{E_k}(\text{He}) = \frac{3}{2}\,k\,T$
$\overline{E_k}(\text{Ar}) = \frac{3}{2}\,k\,T$

The required ratio is

$\frac{\overline{E_k}(\text{He})}{\overline{E_k}(\text{Ar})} = \frac{\frac{3}{2}\,k\,T}{\frac{3}{2}\,k\,T} = 1$

Hence, the ratio of average kinetic energies (per molecule) of helium to argon is $1 : 1$.

The correct choice is Option D.

The capillary rise formula for a tube kept vertical is  $h = \dfrac{2 T \cos \phi}{\rho g r}$  where

$T$ = surface tension,  $\phi$ = contact angle,  $\rho$ = density of liquid,  $g$ = acceleration due to gravity,  $r$ = radius of the tube.

Given data (in SI units):
Radius, $r = 0.1 \text{ mm} = 0.1 \times 10^{-3}\text{ m} = 1.0 \times 10^{-4}\text{ m}$
Surface tension, $T = 70 \text{ dyn/cm}$. Since $1 \text{ dyn} = 10^{-5}\text{ N}$ and $1 \text{ cm} = 10^{-2}\text{ m}$,
$T = 70 \times 10^{-5}\text{ N} \big/ 10^{-2}\text{ m} = 70 \times 10^{-3}\text{ N/m} = 0.07\text{ N/m}$
Contact angle with glass, $\phi \simeq 0^\circ \;\Rightarrow\; \cos\phi = 1$
Density of water, $\rho = 1000\text{ kg/m}^3$
Acceleration, $g = 9.8\text{ m/s}^2$

Substituting into the formula, the vertical rise is

$h = \dfrac{2 \times 0.07 \times 1}{1000 \times 9.8 \times 1.0 \times 10^{-4}} = \dfrac{0.14}{0.98} = 0.142857\text{ m} = 14.2857\text{ cm}$

The tube is inclined at $30^\circ$ to the vertical. If $L$ is the length of water column along the axis of the tube, its vertical component equals $h$, so

$L \cos 30^\circ = h$ $\Rightarrow\; L = \dfrac{h}{\cos 30^\circ} = \dfrac{14.2857}{\dfrac{\sqrt3}{2}} = \dfrac{200}{7\sqrt3}\text{ cm} \approx 16.5\text{ cm}$

Among the given options, $\dfrac{82}{5}\text{ cm} = 16.4\text{ cm}$ is the closest and matches the calculated value.

Hence the length of water risen in the inclined capillary is approximately $\mathbf{16.4\;cm}$, so the correct choice is Option A.

For a spherical mirror the transverse magnification is defined as
$m \;=\; \frac{h_i}{h_o}\;=\;-\frac{v}{u} \quad -(1)$
where $u$ is the object distance and $v$ is the image distance, both measured from the pole using the Cartesian sign convention.

The given magnitude of magnification is $\frac{1}{4}$.
Because the value is positive and less than $1$, the image is erect and virtual. Hence the mirror must be convex, so we take
$m = +\frac{1}{4} \quad\Longrightarrow\quad \frac{1}{4}= -\frac{v}{u} \quad -(2)$

From $(2)$,
$v = -\frac{u}{4} \quad -(3)$

With the sign convention, the object lies in front of the mirror, so $u$ is negative. Equation $(3)$ therefore makes $v$ positive, placing the virtual image behind the mirror, as expected for a convex mirror.

The distance between the object and its image is given to be $40\ \text{cm}$. Along the principal axis these two points are on opposite sides of the pole, so their separation equals the sum of their magnitudes:
$|u| + |v| = 40 \quad -(4)$

Put $|v| = \frac{|u|}{4}$ from $(3)$ into $(4)$:
$|u| + \frac{|u|}{4} = 40 \;\Longrightarrow\; \frac{5|u|}{4} = 40 \;\Longrightarrow\; |u| = 32\ \text{cm}$

Thus,
$u = -32\ \text{cm}, \qquad v = +8\ \text{cm}$

Apply the mirror formula
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \quad -(5)$

Substituting $v = 8\ \text{cm}$ and $u = -32\ \text{cm}$ into $(5)$ gives
$\frac{1}{f} = \frac{1}{8} - \frac{1}{32} = \frac{4 - 1}{32} = \frac{3}{32}$

Therefore
$f = \frac{32}{3}\ \text{cm} \approx 10.7\ \text{cm}$

The focal length is positive, confirming that the mirror is indeed convex. Hence the correct option is Option C (10.7 cm).

Magnitude of each charge of the dipole: $q = 2\,\mu C = 2 \times 10^{-6}\,C$.

Separation of the charges (length of dipole): $d = 0.5\,\mu m = 0.5 \times 10^{-6}\,m$.

Electric dipole moment is defined as $p = q\,d$ (directed from -ve to +ve charge).

Substituting the values,
$p = (2 \times 10^{-6}) \times (0.5 \times 10^{-6})$
$p = 1 \times 10^{-12}\,C\,m$.

The electric field E between parallel-plate capacitor plates is uniform and given by $E = \frac{V}{\ell}$, where V is the applied potential difference and $\ell$ is plate separation.

Given potential difference: $V = 5\,V$.
Plate separation: $\ell = 0.5\,mm = 0.5 \times 10^{-3}\,m$.

Therefore,
$E = \frac{5}{0.5 \times 10^{-3}} = \frac{5}{5 \times 10^{-4}} = 1 \times 10^{4}\,V\,m^{-1}$.

Torque on a dipole in a uniform field is $\tau = p\,E\,\sin\theta$, where $\theta$ is the angle between $\vec p$ and $\vec E$ after rotation.

The dipole is rotated through $30^{\circ}$, so $\theta = 30^{\circ}$ and $\sin30^{\circ} = 0.5$.

Hence,
$\tau = (1 \times 10^{-12})(1 \times 10^{4})(0.5)$
$\tau = 0.5 \times 10^{-8}\,N\,m$
$\tau = 5 \times 10^{-9}\,N\,m$.

Therefore the torque experienced by the dipole is $5 \times 10^{-9}\,N\,m$.

Option A is correct.

$Y = \overline{(A \cdot B) \cdot (B + \bar{A})}$

Using the distributive law to simplify the expression inside the inversion bar:

$= (A \cdot B \cdot B) + (A \cdot B \cdot \bar{A})$

$= (A \cdot B) + (0) = A \cdot B$

$Y = \overline{A \cdot B}$

$\overline{A \cdot B} = 0 \implies A \cdot B = 1$

$A = 1 \quad \text{and} \quad B = 1$

The dimensional formula of a physical quantity is obtained from the defining equation of that quantity in terms of the fundamental mechanical quantities $M$ (mass), $L$ (length) and $T$ (time).

Case (A) : Mass density

Mass density $\rho$ is mass per unit volume:
$\rho = \frac{\text{mass}}{\text{volume}}$.
Mass has dimension $[M]$ and volume has dimension $[L^3]$, so

$[\rho] = \frac{[M]}{[L^3]} = [M L^{-3} T^{0}]$.
Thus (A) corresponds to (IV).

Case (B) : Impulse

Impulse $J$ is defined as force multiplied by the time interval:
$J = F \, \Delta t$.
Force has dimension $[M L T^{-2}]$ and time has dimension $[T]$, so

$[J] = [M L T^{-2}] \,[T] = [M L T^{-1}]$.
Thus (B) corresponds to (II).

Case (C) : Power

Power $P$ is work done per unit time:
    $P = \frac{W}{t}$.
Work (or energy) has dimension $[M L^{2} T^{-2}]$ and time has dimension $[T]$, hence

$[P] = \frac{[M L^{2} T^{-2}]}{[T]} = [M L^{2} T^{-3}]$.
Thus (C) corresponds to (I).

Case (D) : Moment of inertia

For a point mass $m$ at a distance $r$ from the axis, the moment of inertia $I$ is:
    $I = m r^{2}$.
Mass has dimension $[M]$ and distance squared has dimension $[L^{2}]$, so

$[I] = [M] [L^{2}] = [M L^{2} T^{0}]$.
Thus (D) corresponds to (III).

Collecting the matches:

(A) → (IV), (B) → (II), (C) → ( I ), (D) → (III).

This set of matches is given in Option C. Hence the correct answer is Option C.

The given transverse wave on the string is
$y(x,t)=\sin\!\left(20\pi x+10\pi t\right)$ where $x$ is in metres and $t$ is in seconds.

For any particle of the string the instantaneous transverse (oscillatory) speed is obtained by differentiating $y$ with respect to time:

Formula: $v_y=\frac{\partial y}{\partial t}= \omega \cos\!\left(kx+\omega t\right)$ where $k$ is the wave-number and $\omega$ is the angular frequency.

Here $k=20\pi$ and $\omega=10\pi$, so
$v_y=10\pi \cos\!\left(20\pi x+10\pi t\right)$ $-(1)$

At a fixed instant $t$, two points $x_1$ and $x_2$ will have the same oscillatory speed magnitude when

$\left|v_y(x_1,t)\right|=\left|v_y(x_2,t)\right|$ $\Longrightarrow$ $\left|\cos\!\left(20\pi x_1+10\pi t\right)\right|=\left|\cos\!\left(20\pi x_2+10\pi t\right)\right|$

Let the corresponding phase angles be $\phi_1$ and $\phi_2$: $\phi_1 = 20\pi x_1+10\pi t,\qquad \phi_2 = 20\pi x_2+10\pi t$

Condition for equal absolute value of a cosine:
$\left|\cos\phi_1\right|=\left|\cos\phi_2\right| \; \Longrightarrow \; \phi_2=\phi_1\pm n\pi,\; n\in\mathbb{Z}$ (The plus/minus $\pi$ shift changes the sign of the cosine but keeps its magnitude the same.)

Therefore the phase difference between the two points is
$\Delta\phi = \phi_2-\phi_1 = n\pi$

Since $\Delta\phi = k\,\Delta x$, we get
$k\,\Delta x = n\pi \quad \Longrightarrow \quad \Delta x = \frac{n\pi}{k}$ $-(2)$

With $k = 20\pi$, equation $(2)$ becomes
$\Delta x = \frac{n\pi}{20\pi} = \frac{n}{20}\; \text{metre}$

The minimum non-zero distance corresponds to $n = 1$:

$\Delta x_{\min} = \frac{1}{20}\; \text{m} = 0.05\; \text{m} = 5.0\; \text{cm}$

Hence, the minimum separation between two points on the string having the same oscillating speed is 5.0 cm.

Option A is correct.

The assertion states: “Refractive index of glass is higher than that of air.”
The refractive index $n$ of a medium is defined by $n = \frac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium. For air, $v$ is almost equal to $c$, so $n_{\text{air}}\approx 1.0003$. For ordinary crown glass, $v\approx 2\times 10^{8}\,\text{m s}^{-1}$, so $n_{\text{glass}}\approx 1.5$. Hence $n_{\text{glass}}\gt n_{\text{air}}$, making the assertion true.

The reason claims: “Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index.”
Optical density describes how much a medium slows light, i.e. its refractive index. Mass density is the ratio of mass to volume. There is no fixed proportionality between mass density and refractive index. For example, turpentine has a higher refractive index than water even though its mass density is lower. Therefore the reason is false.

Conclusion: The assertion (A) is correct, but the reason (R) is not correct.
Hence the appropriate choice is Option C.

For a magnetic field $\mathbf{B}$, Gauss’s law for magnetism states $\nabla\!\cdot\!\mathbf{B}=0$ $-(1)$.

Equation $(1)$ means the net magnetic flux through any closed surface is zero. Hence there are no sources or sinks of the magnetic field analogous to electric charges. In other words, isolated north poles or south poles (magnetic monopoles) are absent. This validates Assertion (A).

Because $\nabla\!\cdot\!\mathbf{B}=0$ everywhere, magnetic field lines can neither start nor end; they must be continuous and form closed loops. Therefore Reason (R) is also a correct statement.

Moreover, the continuity of magnetic field lines (closed loops) is the direct physical manifestation of $\nabla\!\cdot\!\mathbf{B}=0$, which is precisely why magnetic monopoles do not exist. Thus Reason (R) correctly explains Assertion (A).

So, both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

Answer: Option C

Potential energy $U$ is defined only for conservative forces. A conservative force satisfies the following statements:

• Work done by the force depends only on the initial and final positions, not on the actual path.
• For a conservative force $\vec F_c$ we can write $\vec F_c = -\,\boldsymbol{\nabla} U$, that is, the force equals the negative gradient of a scalar potential energy function $U(x,y,z)$.

Let us analyse each force in the options.

Case A: Coulomb’s force between two point charges is $\vec F = k\,\frac{q_1 q_2}{r^2}\,\hat r$. It is inverse-square and central, hence conservative. We can define the electrostatic potential energy $U = k\,\dfrac{q_1 q_2}{r}$ such that $\vec F = -\,\dfrac{dU}{dr}\,\hat r$. Therefore it can be expressed in terms of potential energy.

Case B: Gravitational force between two masses is $\vec F = -G\,\dfrac{m_1 m_2}{r^2}\,\hat r$, which is also inverse-square and conservative. The corresponding potential energy is $U = -G\,\dfrac{m_1 m_2}{r}$. Hence gravitational force is expressible through potential energy.

Case C: Frictional force (kinetic or static) depends on the nature of surfaces and usually acts opposite to the direction of motion or impending motion. Work done against friction depends on the path length, not merely on initial and final positions. Therefore friction is a non-conservative force. For a non-conservative force we cannot define a single-valued scalar function $U$ satisfying $\vec F = -\,\boldsymbol{\nabla} U$. So friction cannot be expressed in terms of potential energy.

Case D: A restoring force such as the spring force is $\vec F = -k\,x\,\hat i$. It is conservative; its potential energy is the elastic potential energy $U = \tfrac12 k x^2$ because $\vec F = -\,\dfrac{dU}{dx}\,\hat i$.

Only the frictional force fails to meet the criteria for conservative forces.

Hence the force that cannot be expressed in terms of potential energy is the frictional force → Option C (Option 3).

The first law of thermodynamics gives the relation $\Delta Q = \Delta U + \Delta W$ where:
  • $\Delta Q$ is the heat supplied to the system,
  • $\Delta U$ is the change in internal energy,
  • $\Delta W$ is the work done by the system.

For each thermodynamic process we check which of these three quantities is zero or non-zero.

Case A: Isothermal process

Definition: Temperature remains constant.
For an ideal gas, internal energy depends only on temperature, so when temperature is constant $\Delta U = 0$.
Heat supplied equals work done, but both are generally non-zero.
Hence the correct description is $\Delta U = 0$, i.e. List-II item (IV).

Case B: Adiabatic process

Definition: No heat is exchanged with the surroundings.
Therefore $\Delta Q = 0$.
Neither $\Delta U$ nor $\Delta W$ is automatically zero; they are related by $\Delta U = -\Delta W$.
Thus the matching statement is $\Delta Q = 0$, i.e. List-II item (II).

Case C: Isobaric process

Definition: Pressure remains constant.
At constant pressure, the gas expands or compresses, so work is done: $\Delta W \ne 0$.
Temperature usually changes, so internal energy changes: $\Delta U \ne 0$.
Only $\Delta U \ne 0$ is listed among the four given statements, therefore we match with List-II item (III).

Case D: Isochoric process

Definition: Volume remains constant.
Work done for a volume change is $\Delta W = P\,\Delta V$. With $\Delta V = 0$ we get $\Delta W = 0$.
Heat added goes entirely into changing internal energy: $\Delta Q = \Delta U$, so neither of them is necessarily zero.
Thus the matching statement is $\Delta W = 0$, i.e. List-II item (I).

Collecting the matches:
(A) Isothermal → (IV) $\Delta U = 0$
(B) Adiabatic → (II) $\Delta Q = 0$
(C) Isobaric → (III) $\Delta U \ne 0$
(D) Isochoric → (I) $\Delta W = 0$

This corresponds to Option C.

Final Answer: Option C   [ (A)-(IV), (B)-(II), (C)-(III), (D)-(I) ]

Speed of the helicopter: $v = 360 \text{ km h}^{-1} = 360 \times \frac{1000}{3600}\,\text{m s}^{-1} = 100 \,\text{m s}^{-1}$

Altitude from which the object is released: $h = 2 \text{ km} = 2000 \text{ m}$

Time taken by the object to fall through $h$ (free-fall, neglecting air resistance) is obtained from $h = \frac12 \, g t^{2} \; \Longrightarrow \; t = \sqrt{\frac{2h}{g}}$

Substituting $h = 2000 \text{ m}$ and $g = 10 \text{ m s}^{-2}$, $t = \sqrt{\frac{2 \times 2000}{10}} = \sqrt{400} = 20 \text{ s}$

Thus the object reaches the ground exactly $20 \text{ s}$ after it is dropped.

Horizontal distance travelled in this time by either the helicopter or the object is $x = v t = 100 \,\text{m s}^{-1} \times 20 \text{ s} = 2000 \text{ m} = 2 \text{ km}$

Vertical distance descended by the object is the full altitude, i.e. $y = 2 \text{ km}$ downward.

Hence the displacement vector of the object, measured from the point of release on the helicopter, has components horizontal $2 \text{ km}$ and vertical $2 \text{ km}$. Its magnitude is $\sqrt{(2\text{ km})^{2} + (2\text{ km})^{2}} = 2\sqrt{2}\ \text{km}$

Therefore, the required displacement is $2\sqrt{2}$ km.
Option D.

The speed of the object varies with its position as $v = 4\sqrt{x}$, where $x$ is the coordinate (in metres) measured along the +x-axis.

First convert the mass to SI units: given mass $m = 500\,$g $= 0.5\,$kg.

To find the force we need the acceleration. For one-dimensional motion, acceleration can be expressed in terms of position as
$a = v\,\frac{dv}{dx}$

STEP 1 - Differentiate the speed with respect to $x$:
$v = 4x^{1/2} \; \Rightarrow \; \frac{dv}{dx} = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2}$

STEP 2 - Use $a = v\,\dfrac{dv}{dx}$:
Substitute $v = 4x^{1/2}$ and $\dfrac{dv}{dx} = 2x^{-1/2}$:
$a = \left(4x^{1/2}\right)\left(2x^{-1/2}\right) = 8 \text{ m\,s}^{-2}$

The acceleration turns out to be a constant $8 \text{ m\,s}^{-2}$ (independent of $x$).

STEP 3 - Apply Newton’s second law $F = ma$:
$F = 0.5 \times 8 = 4 \text{ N}$

Hence the force acting on the object is $4 \text{ N}$.

Therefore, Option D is correct.

$\sin \theta_C = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \frac{\mu_1}{\mu_2}$

$\sin \theta_C = \frac{1.0}{1.25} = \frac{4}{5}$

$\cos \theta_C = \sqrt{1 - \sin^2 \theta_C} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$

Using the geometric relation $r = 90^\circ - \theta_C$: $\sin r = \sin(90^\circ - \theta_C) = \cos \theta_C = \frac{3}{5}$

$\mu_1 \sin \theta = \mu_2 \sin r$

$1.0 \times \sin \theta = 1.25 \times \frac{3}{5}$

$\theta = \sin^{-1}\left(\frac{3}{4}\right)$

Let the free charge on the plates of the capacitor be $Q_f = 5 \times 10^{-6}\, \text{C}$.

When a dielectric slab is inserted, the molecules of the slab polarise and create two layers of induced (bound) charge. The magnitude of the induced charge on either face of the slab is given as $Q_{\text{ind}} = 4 \times 10^{-6}\, \text{C}$.

For a parallel-plate capacitor completely (or almost completely) filled with a dielectric of relative permittivity $K$, the relation between the free surface charge density $\sigma_f$ and the bound surface charge density $\sigma_b$ is derived as follows:

• The electric displacement vector $\mathbf{D}$ inside the dielectric is $\mathbf{D} = \sigma_f \hat{n}$ (since $\mathbf{D}$ originates from free charge only).
• Electric field inside the dielectric: $E = \frac{\sigma_f}{K \varepsilon_0}$.
• Polarisation $P$ of the dielectric: $P = \varepsilon_0 \chi_e E = \varepsilon_0 (K-1) E$, where $\chi_e = K-1$.
• Surface bound charge density: $\sigma_b = P = \varepsilon_0 (K-1) \frac{\sigma_f}{K \varepsilon_0} = \frac{K-1}{K} \, \sigma_f$.

Therefore the ratio of the magnitudes of bound to free charge is

$\frac{Q_{\text{ind}}}{Q_f} = \frac{\sigma_b}{\sigma_f} = \frac{K-1}{K}\;.$

Substitute the given values:

$\frac{4 \times 10^{-6}}{5 \times 10^{-6}} = \frac{K-1}{K}$

$\frac{4}{5} = \frac{K-1}{K}$

Cross-multiplying:

$5(K-1) = 4K \quad\Rightarrow\quad 5K - 5 = 4K$

$K = 5$

Hence, the dielectric constant of the slab is $5$.

The given components are connected in series, so the circuit is a series $R\text{-}L\text{-}C$ combination.

Resistance: $R = 50 \,\Omega$
Inductive reactance: $X_L = 100 \,\Omega$
Capacitive reactance: $X_C = 50 \,\Omega$
Applied rms voltage: $V_{\text{rms}} = 10 \text{ V}$

Step 1 Find the net reactance.
For a series circuit, the net reactance is the algebraic difference of the individual reactances:
$X = X_L - X_C = 100 - 50 = 50 \,\Omega$

Step 2 Calculate the impedance.
For a series $R\text{-}L\text{-}C$ circuit, the impedance is
$Z = \sqrt{R^{2} + X^{2}}$
$\Rightarrow Z = \sqrt{50^{2} + 50^{2}}$
$\Rightarrow Z = \sqrt{2500 + 2500}$
$\Rightarrow Z = \sqrt{5000}$
$\Rightarrow Z = 70.71 \,\Omega$

Step 3 Find the rms current.
Ohm’s law for ac gives $I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}$.
$I_{\text{rms}} = \dfrac{10}{70.71} = 0.1414 \text{ A}$

Step 4 Determine the power factor.
Power factor in a series circuit is $\cos\phi = \dfrac{R}{Z}$.
$\cos\phi = \dfrac{50}{70.71} = 0.707$

Step 5 Calculate the average (true) power.
Average power dissipated is
$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$
Substitute the values:
$P = 10 \times 0.1414 \times 0.707$
$P = 1.0 \text{ W}$

Hence, the average power dissipated by the circuit is 1 W.

For steady one-dimensional conduction through a rod, the heat current is
$Q=\frac{K\,A}{L}\,\left(T_{\text{hot}}-T_{\text{cold}}\right)$

When two rods are joined in series, the same heat current $Q$ flows through each of them in the steady state. Let $T$ be the interface temperature (at the junction of rods A and B).

Given ratios
$\frac{L_A}{L_B}=\frac12 \;\Rightarrow\; L_A=\frac{L_B}{2}$
$\frac{r_A}{r_B}=2 \;\Rightarrow\; r_A=2r_B$
$\frac{K_A}{K_B}=\frac12 \;\Rightarrow\; K_A=\frac{K_B}{2}$

Cross-sectional areas: $A=\pi r^{2}$, so
$A_A=\pi r_A^{2}=\pi\,(2r_B)^{2}=4\pi r_B^{2}=4A_B$
thus $\frac{A_A}{A_B}=4$.

Heat current through rod A:
$Q=\frac{K_AA_A}{L_A}\,(400-T)$

Heat current through rod B:
$Q=\frac{K_BA_B}{L_B}\,(T-200)$

Equate the two heat currents:

$\frac{K_AA_A}{L_A}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$

Substitute the ratio values:
$K_A=\frac{K_B}{2},\;A_A=4A_B,\;L_A=\frac{L_B}{2}$

$\frac{\left(\tfrac{K_B}{2}\right)\,(4A_B)}{\tfrac{L_B}{2}}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$

Simplify the fraction on the left:
$\frac{(\tfrac{K_B}{2})\,(4A_B)}{\tfrac{L_B}{2}}=\frac{2K_BA_B}{\tfrac{L_B}{2}}=\frac{2K_BA_B\cdot2}{L_B}=\frac{4K_BA_B}{L_B}$