JEE Main 2025 2nd April Evening Shift Question Paper with Solutions

A polar dielectric molecule (for example $H_2O$) possesses a permanent dipole moment $\vec p_{0}$ even when no external field is present.

However, in a macroscopic sample of such a dielectric, every molecule can point in any direction with equal probability when there is no external electric field. Hence the molecular dipoles are randomly oriented.

When vectors of equal magnitude are randomly oriented, their vector (vectorial) sum averages to zero: $\vec P_{\text{net}} = \sum_i \vec p_{0i} = \vec 0$.

Therefore the bulk or net dipole moment of the whole specimen is zero in the absence of an external electric field.

Now examine the statements:
Assertion (A): Net dipole moment is not zero without the field — this contradicts the conclusion above, so (A) is incorrect.
Reason (R): Permanent dipoles are oriented randomly when there is no field — this is exactly what happens, so (R) is correct.

Because (A) is false while (R) is true, (R) cannot be the correct explanation of (A). Thus the correct choice is Option D: $(A)\ \text{is not correct but}\ (R)\ \text{is correct}$.

Voltage sensitivity $S_V$ of a moving-coil galvanometer is defined as the deflection produced per unit applied voltage.
If the torsional constant of the suspension spring is $k$, then

Current sensitivity $S_I$ is $\displaystyle S_I = \frac{NAB}{k} \quad -(1)$
where
$N$ = number of turns, $A$ = area of each turn, $B$ = magnetic field.

Voltage sensitivity is obtained by dividing current sensitivity by the total resistance $R$ of the coil:

$\displaystyle S_V = \frac{S_I}{R} = \frac{NAB}{kR} \quad -(2)$

Because both coils have the same spring, the torsional constant $k$ is identical for them and cancels in the ratio.

For coil $M_1$:
$N_1 = 15$, $A_1 = 3.6 \times 10^{-3}\,\text{m}^2$, $B_1 = 0.25\,\text{T}$, $R_1 = 5\,\Omega$

For coil $M_2$:
$N_2 = 21$, $A_2 = 1.8 \times 10^{-3}\,\text{m}^2$, $B_2 = 0.50\,\text{T}$, $R_2 = 7\,\Omega$

Using $-(2)$ for each coil and taking the ratio:

$\frac{S_{V1}}{S_{V2}} = \frac{\dfrac{N_1A_1B_1}{kR_1}}{\dfrac{N_2A_2B_2}{kR_2}} = \frac{N_1A_1B_1R_2}{N_2A_2B_2R_1}$

Substituting the given values:

$N_1A_1B_1R_2 = 15 \times 3.6\times10^{-3} \times 0.25 \times 7$
$= 15 \times 0.0036 \times 0.25 \times 7$
$= 0.0945$

$N_2A_2B_2R_1 = 21 \times 1.8\times10^{-3} \times 0.50 \times 5$
$= 21 \times 0.0018 \times 0.50 \times 5$
$= 0.0945$

Therefore

$\frac{S_{V1}}{S_{V2}} = \frac{0.0945}{0.0945} = 1$

Thus, the voltage sensitivities of $M_1$ and $M_2$ are equal, giving the ratio

$S_{V1} : S_{V2} = 1 : 1$

Option A is correct.

The diameter of the ring is given as $r$, hence its radius is
$R = \frac{r}{2}\,.$

Step 1: Moment of inertia about a diameter (axis in the plane, through centre)
Every mass element of a thin ring is at the same distance $R$ from the centre. For an axis along a diameter (say the $x$-axis) lying in the plane of the ring, the perpendicular distance of a point $\bigl(R\cos\theta,\;R\sin\theta\bigr)$ from that axis is $R\lvert\sin\theta\rvert$. Using $I = \int r_\perp^{\,2}\,dm$,

$I_{\text{diameter}} = \int_0^{2\pi} \! \left(R\sin\theta\right)^2 \left(\frac{M}{2\pi}\,d\theta\right) = \frac{MR^2}{2}\,.$

Step 2: Moment of inertia about a tangential axis in the plane
The required axis is parallel to the diameter axis but touches the ring at one point, so it is shifted by a distance $R$ from the centre. Apply the Parallel-Axis Theorem:

$I_{\text{tangent}} \;=\; I_{\text{diameter}} + M R^2 = \frac{MR^2}{2} + M R^2 = \frac{3}{2}\,M R^2 \,.$

Step 3: Express in terms of the given diameter
Substitute $R = \dfrac{r}{2}$:

$I_{\text{tangent}} = \frac{3}{2}\,M\left(\frac{r}{2}\right)^2 = \frac{3}{2}\,M\,\frac{r^{2}}{4} = \frac{3}{8}\,M r^{2}\,.$

Therefore, the moment of inertia of the circular ring about a tangential axis lying in its own plane is $\frac{3}{8}Mr^{2}$.

The correct choice is Option B.

When a liquid drop is formed, its surface possesses energy equal to (surface tension) × (surface area).
Hence the change in surface energy equals the change in surface area multiplied by the surface tension $T$.

Step 1: Initial surface area
Two identical drops, each of radius $r$, have total surface area
$A_i = 2 \times 4\pi r^{2} = 8\pi r^{2}$

Step 2: Radius of the bigger drop after coalescence
Volume is conserved during coalescence.
Initial volume: $V_i = 2 \times \frac{4}{3}\pi r^{3} = \frac{8}{3}\pi r^{3}$
Let $R$ be the radius of the bigger drop.
Final volume: $V_f = \frac{4}{3}\pi R^{3}$

Equating volumes:
$\frac{4}{3}\pi R^{3} = \frac{8}{3}\pi r^{3}$
$\Rightarrow R^{3} = 2\,r^{3}$
$\Rightarrow R = 2^{\frac{1}{3}}\,r$

Step 3: Final surface area
$A_f = 4\pi R^{2} = 4\pi \left(2^{\frac{1}{3}}r\right)^{2} = 4\pi r^{2}\,2^{\frac{2}{3}}$

Step 4: Decrease in surface area
$\Delta A = A_i - A_f = 8\pi r^{2} - 4\pi r^{2}\,2^{\frac{2}{3}}$
Factorising:
$\Delta A = 4\pi r^{2}\left[2 - 2^{\frac{2}{3}}\right]$

Step 5: Surface energy released
Released energy $E = T \times \Delta A$
$E = 4\pi r^{2}\,T \left[2 - 2^{\frac{2}{3}}\right]$

Thus, the surface energy released is
$\boxed{4\pi r^{2}T\left[2 - 2^{\frac{2}{3}}\right]}$

Option A is correct.

The de-Broglie wavelength of a particle is defined as
$\lambda = \frac{h}{p} = \frac{h}{m\,v}$
where $v$ is the magnitude (speed) of the velocity vector.

At $t = 0$ the electron has velocity $\vec v_0 = v_0\hat i$ and de-Broglie wavelength $\lambda_0 = \dfrac{h}{m\,v_0}$.

After entering the magnetic field $\vec B = B_0\hat j$ the electron experiences the Lorentz force
$\vec F = q\,\vec v \times \vec B$
For an electron, $q = -e$ (magnitude $e$). At any instant the velocity has an $\hat i$ component only, so
$\vec F = (-e)\,(v_x\hat i)\times (B_0\hat j) = -e\,v_x B_0\,(\hat i \times \hat j) = -e\,v_x B_0\,\hat k$
The force is always perpendicular to the instantaneous velocity.

Since $\vec F \perp \vec v$, the magnetic force does no work:
$\text{Power} = \vec F\cdot\vec v = 0$

Hence the kinetic energy $\frac12 m v^2$ remains constant, so the speed $v$ of the electron stays equal to its initial value $v_0$ for all time.

Because the speed (and therefore the magnitude of momentum $p = m v$) is constant, the de-Broglie wavelength also stays constant:
$\lambda = \frac{h}{m v} = \frac{h}{m v_0} = \lambda_0$

Thus, after any time $t$, the wavelength is still $\lambda_0$.

Final answer: $\lambda_0$   →   Option D

The standard form of a progressive sinusoidal wave travelling in the +x-direction is
$y = A \cos \bigl(kx - \omega t + \phi\bigr)$ or $y = A \sin \bigl(kx - \omega t + \phi\bigr)$

Here
  • $A$ = amplitude
  • $k = \dfrac{2\pi}{\lambda}$ = wave number (rad cm$^{-1}$ when $\lambda$ is in cm)
  • $\omega = 2\pi f$ = angular frequency (rad s$^{-1}$)
  • Wave speed $v = \dfrac{\omega}{k} = f\lambda$

Given data
  • Wavelength $\lambda = 7.5\,\text{cm}$
  • In $\Delta t = 0.3\,\text{s}$ the disturbance travels $\Delta x = 1.2\,\text{cm}$
  • Amplitude $A = 2\,\text{cm}$

Step 1 - Calculate the wave speed.
$v = \frac{\Delta x}{\Delta t} = \frac{1.2\,\text{cm}}{0.3\,\text{s}} = 4\,\text{cm s}^{-1}$

Step 2 - Calculate the wave number $k$.
$k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5} \,\text{rad cm}^{-1} \approx 0.84\,\text{rad cm}^{-1}$

Step 3 - Calculate the angular frequency $\omega$ using $\omega = vk$.
$\omega = (4\,\text{cm s}^{-1})(0.84\,\text{rad cm}^{-1}) \approx 3.35\,\text{rad s}^{-1}$

Step 4 - Insert $A$, $k$, and $\omega$ into the standard form.
Possible cosine form (with zero phase constant):
$y = 2 \cos\bigl(0.84\,x - 3.35\,t\bigr)\;\text{cm}$

Step 5 - Check the given initial condition.
At $x = 0$ and $t = 0$ the crest is present, so $y(0,0) = +2\,\text{cm}$ (maximum upward displacement).
For the expression above:
$y(0,0) = 2\cos(0) = 2\,\text{cm}$✅ (condition satisfied)

Step 6 - Match with the options.
Option A: $y = 2\cos(0.83x - 3.35t)\;\text{cm}$ - wave number, frequency and amplitude all match the calculated values.
Option B: sine form gives zero displacement at $t = 0$, so it violates the initial condition.
Option C: $k$ and $\omega$ are interchanged, giving an incorrect speed $v = \omega/k \approx 0.25\,\text{cm s}^{-1}$.
Option D: both $k$ and $\omega$ are far from the required values.

Therefore, the correct representation of the wave is given by Option A.

Solution :

Dimensions of momentum :

$[p] = MLT^{-1}$

Given :

Charge,

$[q] = AT$

Current,

$[I] = A$

Permeability of vacuum :

From force per unit length between two current carrying wires,

$\frac{\mu_0 I^2}{r}$

has dimensions of force.

Therefore,

$[\mu_0] = \frac{[F][r]}{[I]^2}$

$= \frac{(MLT^{-2})(L)}{A^2}$

$= MLT^{-2}A^{-2}$

Now consider quantity :

$\mu_0 qI$

Its dimensions are :

$[\mu_0 qI] = (MLT^{-2}A^{-2})(AT)(A)$

$= MLT^{-1}$

which is the dimension of momentum.

Final Answer :

$\mu_0 qI$

The magnetic energy stored in any region is obtained from the energy density formula for a magnetic field inside a linear medium.

Energy density in a magnetic material is given by
$u = \frac{1}{2} B H$

For a linear medium, $B = \mu H$, where $\mu = \mu_0 \mu_r$. Substituting $H = \frac{B}{\mu}$ in the energy density expression:

$u = \frac{1}{2} B \left(\frac{B}{\mu}\right) = \frac{B^2}{2\mu}$ $-(1)$

The solenoid is completely filled with a material of relative permeability $\mu_r = 2$, so
$\mu = \mu_0 \mu_r = 2\mu_0$.

Substituting this value of $\mu$ in equation $(1)$:

$u = \frac{B^2}{2(2\mu_0)} = \frac{B^2}{4\mu_0}$ $-(2)$

The total magnetic energy $U$ stored in the solenoid equals the energy density multiplied by the volume of the field region.
Volume of the solenoid, $V = A l$.

Therefore,
$U = u \, V = \frac{B^2}{4\mu_0} \, (A l) = \frac{B^2 A l}{4\mu_0}$.

Thus, the magnetic energy stored in the solenoid is $\frac{B^2 A l}{4\mu_0}$.

Correct option: Option D

Between parallel plates, electric field is uniform.

If plate separation is 10cm and potential difference is V, field is

$E=\frac{V}{10}​(volts/cm)$

We need potential difference between A and B.

From figure:

AC=3cm

AB=5 cm

Triangle ACB is right angled, so

$CB=\sqrt{\ 5^2-3^2}$

Now electric field is horizontal (between vertical plates), so only horizontal separation matters for potential difference.

Vertical displacement does not affect potential.

Thus only CB=4 cm contributes.

So

$V_{AB}=E(4)$

$=\frac{V}{10}\times4=$

$=\frac{2V}{5}$

For an adiabatic process of an ideal gas, the heat exchanged with the surroundings is zero: $Q = 0$.

First law of thermodynamics: $\Delta U = Q + W$, where $W$ is the work done on the gas.
Because $Q = 0$, we have

$\Delta U = W \quad -(1)$

For a monoatomic ideal gas, the internal energy is $U = \dfrac{3}{2}nRT$. Hence

$\Delta U = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(2)$

Using $(1)$ and $(2)$, the work done on the gas is

$W = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(3)$

Thus, $W$ is directly proportional to $(T_2 - T_1)$.

For an adiabatic change of an ideal gas, the pressure-volume relation is $PV^{\gamma} = \text{constant}$, where $\gamma = \dfrac{C_P}{C_V}$. For a monoatomic gas, $\gamma = \dfrac{5}{3}$. Consequently

$TV^{\gamma - 1} = TV^{2/3} = \text{constant}$,
which is not simply $TV = \text{constant}$, and $PV$ is also not constant.

Now examine each statement:

Statement (A): “Internal energy is constant.”
From $(2)$, $\Delta U \neq 0$ when temperature changes, so (A) is false.

Statement (B): “Work done in the process is equal to the change in internal energy.”
Equation $(1)$ shows $W = \Delta U$ (with the ‘work on the gas’ sign convention). Hence (B) is true.

Statement (C): “The product of temperature and volume is a constant.”
We have $TV^{2/3} = \text{constant}$, not $TV$. Therefore (C) is false.

Statement (D): “The product of pressure and volume is a constant.”
For adiabatic change, $PV^{5/3} = \text{constant}$, so $PV$ is not constant. (D) is false.

Statement (E): “The work done to change the temperature from $T_1$ to $T_2$ is proportional to $(T_2 - T_1)$.”
Equation $(3)$ confirms this proportionality. Hence (E) is true.

Therefore, only statements (B) and (E) are correct.

Option C (B, E only) is the correct choice.

For a hydrogen-like ion (single electron around nucleus of charge $+Ze$), Bohr’s model gives the radius of the $n^{\text{th}}$ orbit as

$r_n = \frac{n^2 a_0}{Z}$

where
$n$ = principal quantum number,
$Z$ = atomic number of the nucleus,
$a_0$ = Bohr radius for the hydrogen ground state.

For $Li^{2+}$ the atomic number is $Z = 3$ because lithium has three protons. The ground state corresponds to $n = 1$.

Substituting $n = 1$ and $Z = 3$ into the formula,

$r_1 = \frac{(1)^2 a_0}{3} = \frac{a_0}{3}$

This can be written as $\frac{1}{X} a_0$ with $X = 3$.

Hence, $X = 3$, which matches Option C.

The energy released in a nuclear reaction equals the increase in total binding energy of the nuclei involved.
$Q = \text{(total BE of products)} - \text{(total BE of reactants)}$

Step 1: Binding energy of the reactants
Each deuteron $_{1}H^{2}$ contains $A = 2$ nucleons.
Given binding energy per nucleon = $1.1$ MeV.
Total binding energy of one deuteron = $1.1 \times 2 = 2.2$ MeV.
Since two deuterons take part, total BE of reactants = $2 \times 2.2 = 4.4$ MeV.

Step 2: Binding energy of the product
Helium-4 nucleus $_{2}He^{4}$ has $A = 4$ nucleons.
Given binding energy per nucleon = $7.0$ MeV.
Total binding energy of product = $7.0 \times 4 = 28.0$ MeV.

Step 3: Energy released
$Q = 28.0 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV}.$

Therefore, the energy liberated when two deuterons fuse into a helium-4 nucleus is $23.6$ MeV.

Option C is correct.

The top-left NAND gate receives inputs $(1, 1)$: $\text{Output} = \overline{1 \cdot 1} = 0$

This $0$ feeds directly into the top AND gate. Since any input of $0$ to an AND gate forces a low output, $P$ is immediately determined: $P = \mathbf{0}$

The two left NOT gates receive inputs $(1, 1)$, turning them into $(0, 0)$.

The first OR gate processes these zeros: $\text{Output} = 0 + 0 = 0$

This $0$ splits and travels to two places:

Directly to the bottom input of the final NOR gate $\rightarrow (0)$.

Through the middle-right NOT gate (which inverts the first NAND's $0$ into a $1$) $\rightarrow (1)$.

Finally, the NOR gate processes inputs $(1, 0)$: $Q = \overline{1 + 0} = \overline{1} = \mathbf{0}$

Radius of curvature of each mirror is $R = 12\ \text{cm}$. For a spherical mirror the focal length is given by $f = \frac{R}{2}$.

Therefore $f = 6\ \text{cm}$ in magnitude.
  • For the concave mirror, the focus lies on the object side, so $f = -6\ \text{cm}$.
  • For the convex mirror, the focus lies behind the mirror, so $f = +6\ \text{cm}$.

The objects are placed at the same distance in front of each mirror: $u = -18\ \text{cm}$ (negative by the new Cartesian sign convention).

Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.

Substituting $f = -6$ and $u = -18$:
$\frac{1}{-6} = \frac{1}{v} + \frac{1}{-18}$

$\frac{1}{v} = -\frac{1}{6} + \frac{1}{18} = -\frac{3}{18} + \frac{1}{18} = -\frac{2}{18} = -\frac{1}{9}$

$v = -9\ \text{cm}$ (image is real and formed in front of the mirror).

Linear magnification: $m_c = -\frac{v}{u} = -\frac{-9}{-18} = -\frac{1}{2}$

Magnitude of image size for concave mirror: $|m_c| = \frac{1}{2}$.

Using $f = +6$ and $u = -18$ in the mirror formula:
$\frac{1}{6} = \frac{1}{v} + \frac{1}{-18}$

$\frac{1}{v} = \frac{1}{6} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}$

$v = \frac{9}{2}\ \text{cm} = 4.5\ \text{cm}$ (image is virtual, behind the mirror).

Linear magnification: $m_v = -\frac{v}{u} = -\frac{4.5}{-18} = \frac{4.5}{18} = \frac{1}{4}$

Magnitude of image size for convex mirror: $|m_v| = \frac{1}{4}$.

Ratio of the sizes (convex : concave) is
$\frac{|m_v|}{|m_c|} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$

Hence the required ratio is $\mathbf{\frac{1}{2}}$, which corresponds to Option A.

Distance travelled :A TO B,B TO A, A TO B

A TO B =\pi\r
B to A =\pi\r
A TO B =\pi\r

Total distance =3*\pi\r
Displacement : A TO B = 2r

Here as the system is in equilibrium 

We have to equate horizontal components of both $T_{1\ }and\ T_2$ as there is no external Force acting in the horizontal direction

Here horizontal component of $T_{1\ }is\ T_1\cos60^{\circ\ }$

Here horizontal component of $T_{2\ }is\ T_2\cos30^{\circ\ }$

Equate these 2

$T_2\cos60^{\circ\ }=\ T_1\cos30^{\circ\ }$

$T_2\times\ \frac{\sqrt{\ 3}}{2}=\ T_1\times\ \frac{1}{2}$

$T_1\ =\ \sqrt{\ 3}T_2$

Now coming to the Vertical direction 

There is gravitational force downwards which is Mg and vertical components of $T_{1\ }and\ T_2$ add up and are opposing the force due to gravitation

so Mg=\ T_1\sin\theta\ \ +\ T_2\sin\ \ \theta\

$Mg=\ T_1\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$

Substituting  $T_1\ =\ \sqrt{\ 3}T_2$ in that equation

$Mg=\ \sqrt{\ 3}T_2\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$

$Mg=\frac{3T_2}{2}+\frac{T_2}{2}$
$Mg=2T_2$

$T_2=\frac{g}{2}\ \ \left(as\ M=1\right)$

$T_2=5N$

as $T_1\ =\ \sqrt{\ 3}T_2$

$T_1=5\sqrt{\ 3}N$

The focal length (and therefore the power) of a thin lens is given by the Lens-Maker’s formula

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

where $\mu$ is the refractive index of the lens material, $R_1$ is the radius of curvature of the first surface (positive for a convex surface facing the incident light) and $R_2$ is the radius of curvature of the second surface (negative for the outgoing convex surface of a bi-convex lens).

Case 1: Given bi-convex lens with equal radii
$R_1 = +\frac16 \text{ cm}, \; R_2 = -\frac16 \text{ cm}$ (sign taken according to the convention).

Curvature term for this lens:
$\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{\frac16} - \frac{1}{-\frac16} = 6 - (-6) = 12 \; \text{cm}^{-1}$

Therefore the power of the original lens is

$P = \frac{1}{f} = (\mu - 1)\times 12 \; \text{cm}^{-1}$ $-(1)$

Case 2: New convex lens with unequal radii $R_1$ and $R_2$ must satisfy the same curvature term so that its power remains unchanged:

$\frac{1}{R_1} - \frac{1}{R_2} = 12 \; \text{cm}^{-1}$ $-(2)$

Now test the four given pairs (remember the second radius must be taken negative for the second convex surface):

Option A: $R_1 = \frac13 \text{ cm}, R_2 = -\frac13 \text{ cm}$
$\frac{1}{R_1} - \frac{1}{R_2} = 3 - (-3) = 6 \neq 12$

Option B: $R_1 = \frac15 \text{ cm}, R_2 = -\frac17 \text{ cm}$
$\frac{1}{R_1} - \frac{1}{R_2} = 5 - (-7) = 12$  ✓  satisfies $(2)$

Option C: $R_1 = \frac13 \text{ cm}, R_2 = -\frac17 \text{ cm}$
$3 - (-7) = 10 \neq 12$

Option D: $R_1 = \frac16 \text{ cm}, R_2 = -\frac19 \text{ cm}$
$6 - (-9) = 15 \neq 12$

Only Option B preserves the value of $\frac{1}{R_1} - \frac{1}{R_2}$ and hence the power.

Therefore the required radii of curvature are $\frac15 \text{ cm}$ and $\frac17 \text{ cm}$, i.e. Option B.

The permeability of free space is denoted by $\mu_0$ and the permittivity of free space by $\varepsilon_0$.

Electromagnetic‐wave theory gives the relation between the speed of light $c$ and these two constants:
$c \;=\;\frac{1}{\sqrt{\mu_0\,\varepsilon_0}} \quad -(1)$

Square both sides of $(1)$ to obtain
$c^{2} \;=\;\frac{1}{\mu_0\,\varepsilon_0} \quad -(2)$

The dimensions of speed are $[c] = L\,T^{-1}$. Squaring this,
$[c^{2}] = (L\,T^{-1})^{2} = L^{2}\,T^{-2} \quad -(3)$

From $(2)$, $\dfrac{1}{\mu_0\,\varepsilon_0}$ has the same dimensions as $c^{2}$. Therefore,
$\left[\frac{1}{\mu_0\,\varepsilon_0}\right] = L^{2}\,T^{-2}$

Hence, the correct option is Option B $\big(L^{2}\,T^{-2}\big)$.

For each physical quantity, first recall its definition, then write the factors on which it depends, and finally write the corresponding SI unit.

Case A - Heat capacity of a body
Definition: Heat required to raise the temperature of the entire body by $1\;{\rm K}$.
Heat $Q$ needed is proportional only to temperature change $\Delta T$: $Q = C\,\Delta T$ where $C$ is heat capacity.
Hence unit of $C$ is Joule per Kelvin, $\rm J\,K^{-1}$. Match: (A) $\rightarrow$ (II).

Case B - Specific heat capacity
Definition: Heat required to raise temperature of unit mass of a substance by $1\;{\rm K}$.
Formula: $Q = m\,c\,\Delta T$. Here $c$ relates heat to both mass and temperature difference, so its unit is Joule per kilogram per Kelvin, $\rm J\,kg^{-1}\,K^{-1}$. Match: (B) $\rightarrow$ (III).

Case C - Latent heat
Definition: Heat required to change the phase of unit mass of a substance without temperature change.
Formula: $Q = m\,L$, where $L$ is latent heat.
Only mass appears in the relation, so the unit is Joule per kilogram, $\rm J\,kg^{-1}$. Match: (C) $\rightarrow$ (I).

Case D - Thermal conductivity
Definition (Fourier’s law for one-dimensional steady conduction):$\displaystyle \frac{dQ}{dt} = k\,A\,\frac{\Delta T}{\Delta x}$.
Rearrange for $k$: $k = \frac{1}{A}\,\frac{dQ}{dt}\,\frac{\Delta x}{\Delta T}$.
Units:
$\frac{dQ}{dt}$ has unit $\rm J\,s^{-1}$ (power), $A$ is $\rm m^{2}$, $\Delta x$ is $\rm m$, $\Delta T$ is $\rm K$.
Thus $k$ has unit $\dfrac{\rm J\,s^{-1}}{\rm m^{2}}\times \rm m \times \rm K^{-1}=J\,m^{-1}\,K^{-1}\,s^{-1}$. Match: (D) $\rightarrow$ (IV).

Combining all matches:
(A)-(II), (B)-(III), (C)-(I), (D)-(IV).

The option with this sequence is Option D.

The ring lies in the $xy$-plane with its centre at the origin and has radius $R = a\sqrt{2}$. The linear charge density is uniform; therefore the axial electric field depends only on the distance $z$ from the centre along the $+z$-axis.

Electric field on the axis of a uniformly charged ring (standard result):
$E(z)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\,z}{\left(z^{2}+R^{2}\right)^{3/2}}$ $-(1)$
where $Q$ is the total charge on the ring.