COMEDK 2026 Afternoon Shift Question Paper with Solutions

Given, $𝐸=5\times {10}^3{\text{N C}}^{-1}$, $𝑞=2\times {10}^{-4}\text{C}$, and displacement $𝑠=5\text{m}$.

The electrostatic force acting on the charge along the positive Y-direction is

$𝐹=𝑞𝐸=(2\times {10}^{-4}\text{C})\times (5\times {10}^3{\text{N C}}^{-1})=1\text{N}$

Since the charge is released from rest, it accelerates from the origin along the direction of the field. By the work-energy theorem, the kinetic energy attained equals the work done by the electric field:

$𝐾=𝑊=𝐹\times 𝑠$

On substituting the values,

$𝐾=1\text{N}\times 5\text{m}=5\text{J}$

Hence, the correct option is (B).

At absolute zero ($𝑇=0\text{ K}$), electrons in a semiconductor possess insufficient thermal energy to overcome the forbidden energy gap between the valence band and the conduction band. Consequently, the valence band remains completely filled and the conduction band remains completely empty, so no free electrons are present in the conduction band.

Hence, the correct option is (D).

According to Newton's law of universal gravitation, the gravitational force between two masses ${𝑚}_1$ and ${𝑚}_2$ separated by a distance $𝑟$ is given by

$𝐹=\frac{𝐺{𝑚}_1{𝑚}_2}{{𝑟}^2}$

The universal gravitational constant $𝐺$ and the gravitational force between two masses are independent of the nature of the intervening medium.

Therefore, the gravitational force between the bodies remains unchanged when transferred from vacuum to water:

${𝐹}_{𝑤}={𝐹}_{𝑣}$

$\therefore \frac{{𝐹}_{𝑣}}{{𝐹}_{𝑤}}=1:1$

Hence, the correct option is (B).

The original resistance of the electric coil is given by

$𝑅=\frac{{𝑉}^2}{{𝑃}_1}$

When the coil is cut into two equal halves, the resistance of each half becomes $𝑅/2$.

Connecting these two halves in parallel across the same source gives an equivalent resistance of

${𝑅}_{\text{eq}}=\frac{𝑅/2}{2}=\frac{𝑅}{4}$

The new energy produced per second is the new power,

${𝑃}_2=\frac{{𝑉}^2}{{𝑅}_{\text{eq}}}=\frac{{𝑉}^2}{𝑅/4}=4\left(\frac{{𝑉}^2}{𝑅}\right)=4{𝑃}_1$

The percentage increase in energy produced per second is

$\frac{{𝑃}_2-{𝑃}_1}{{𝑃}_1}\times 100=\frac{4{𝑃}_1-{𝑃}_1}{{𝑃}_1}\times 100=300\%$

Hence, the correct option is (D).

Given, frequency of the tuning fork $𝑓=800\text{ Hz}$, the first resonance length ${𝑙}_1=9.75\text{ cm}$, and the second resonance length ${𝑙}_2=31.25\text{ cm}$.

For a closed organ pipe (resonance tube), the difference between two successive resonance lengths corresponds to half a wavelength:

$\frac{𝜆}{2}={𝑙}_2-{𝑙}_1=31.25-9.75=21.50\text{ cm}$

$\therefore 𝜆=2\times 21.50\text{ cm}=43.00\text{ cm}=0.43\text{ m}$

The speed of sound in air is:

$𝑣=𝑓𝜆=800\times 0.43=344{\text{ m s}}^{-1}$

The length at which the third resonance occurs is:

${𝑙}_3={𝑙}_2+\frac{𝜆}{2}=31.25+21.50=52.75\text{ cm}$

Hence, the correct option is (A).

Given, the initial horizontal velocity is ${𝑢}_{𝑥}=19.6{\text{m s}}^{-1}$ and the initial vertical velocity is ${𝑢}_{𝑦}=0$.

Since there is no acceleration in the horizontal direction, the horizontal component of velocity after $𝑡=2\text{s}$ remains constant:

${𝑣}_{𝑥}={𝑢}_{𝑥}=19.6{\text{m s}}^{-1}$

Using the first equation of motion with $𝑔=9.8{\text{m s}}^{-2}$, the vertical component of velocity after $𝑡=2\text{s}$ is:

${𝑣}_{𝑦}={𝑢}_{𝑦}+𝑔𝑡=0+\left(9.8\right)\left(2\right)=19.6{\text{m s}}^{-1}$

The ratio of the horizontal component to the vertical component of velocity is:

$\frac{{𝑣}_{𝑥}}{{𝑣}_{𝑦}}=\frac{19.6}{19.6}=\frac{1}{1}$

Hence, the correct option is (C).

Given, ${𝐶}_1=8𝜇\text{F}=8\times {10}^{-6}\text{F}$, ${𝑉}_1=200\text{V}$, ${𝐶}_2=4𝜇\text{F}=4\times {10}^{-6}\text{F}$, and ${𝑉}_2=0\text{V}$.

The loss of electrostatic energy during the sharing of charges between two capacitors is given by:

$\Delta 𝑈=\frac{1}{2}\frac{{𝐶}_1{𝐶}_2}{{𝐶}_1+{𝐶}_2}({𝑉}_1-{𝑉}_2{)}^2$

On substituting the given values,

$\Delta 𝑈=\frac{1}{2}\times \frac{(8\times {10}^{-6})(4\times {10}^{-6})}{8\times {10}^{-6}+4\times {10}^{-6}}\times (200-0{)}^2$

$\Delta 𝑈=\frac{1}{2}\times \frac{32\times {10}^{-12}}{12\times {10}^{-6}}\times 40000=\frac{4}{3}\times {10}^{-6}\times 4\times {10}^4$

$\Delta 𝑈=\frac{16}{3}\times {10}^{-2}\text{J}\approx 5.33\times {10}^{-2}\text{J}$

Hence, the correct option is (B).

The energy of the hydrogen atom in the $𝑛$-th state is

${𝐸}_{𝑛}=-\frac{13.6}{{𝑛}^2}\text{eV}$

For the ground state ($𝑛=1$) and the excited state ($𝑛=3$),

${𝐸}_1=-13.6\text{eV}$

${𝐸}_3=-\frac{13.6}{3^2}=-\frac{13.6}{9}\text{eV}\approx -1.51\text{eV}$

The energy emitted during the transition from $𝑛=3$ to $𝑛=1$ is

$\Delta 𝐸={𝐸}_3-{𝐸}_1=13.6(1-\frac{1}{9})=13.6\times \frac{8}{9}\approx 12.09\text{eV}$

The wavelength of the emitted radiation is given by

$𝜆=\frac{ℎ𝑐}{\Delta 𝐸}$

On substituting $ℎ=6.6\times {10}^{-34}\text{J s}$, $𝑐=3\times {10}^8{\text{m s}}^{-1}$, and $1\text{eV}=1.6\times {10}^{-19}\text{J}$,

$𝜆=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{12.09\times 1.6\times {10}^{-19}}\approx 1.024\times {10}^{-7}\text{m}\approx 1020\text{Å}$

Hence, the correct option is (B).

The magnets $𝐴$ and $𝐵$ are placed along a line perpendicular to the magnetic meridian, with their axes lying along this line. Thus, the magnetic needle lies on the axial line of each short magnet.

The magnetic field produced on the axial line of a short bar magnet at a distance $𝑑$ is given by

$𝐵=\frac{{𝜇}_0}{4𝜋}\frac{2𝑀}{{𝑑}^3}$

Since the south poles of both magnets face the needle from opposite sides, their magnetic fields at the needle oppose each other. For the needle to remain undeflected, these fields must balance:

${𝐵}_{𝐴}={𝐵}_{𝐵}$

$\frac{{𝜇}_0}{4𝜋}\frac{2{𝑀}_{𝐴}}{{𝑑}_3^{𝐴}}=\frac{{𝜇}_0}{4𝜋}\frac{2{𝑀}_{𝐵}}{{𝑑}_3^{𝐵}}$

$\frac{{𝑀}_{𝐴}}{{𝑀}_{𝐵}}={\left(\frac{{𝑑}_{𝐴}}{{𝑑}_{𝐵}}\right)}^3$

Given ${𝑑}_{𝐴}=10\text{cm}$ and ${𝑑}_{𝐵}=20\text{cm}$, on substituting the values,

$\frac{{𝑀}_{𝐴}}{{𝑀}_{𝐵}}={\left(\frac{10}{20}\right)}^3={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

Hence, the correct option is (C).

For a parallel combination of $𝑛$ identical cells, each of electromotive force $𝐸$ and internal resistance $𝑟$, the equivalent electromotive force is

${𝐸}_{\text{eq}}=𝐸$

and the equivalent internal resistance is

${𝑟}_{\text{eq}}=\frac{𝑟}{𝑛}$

The total current $𝐼$ drawn through an external resistance $𝑅$ is given by

$𝐼=\frac{{𝐸}_{\text{eq}}}{𝑅+{𝑟}_{\text{eq}}}=\frac{𝐸}{𝑅+\frac{𝑟}{𝑛}}$

Since the external resistance $𝑅$ is negligibly small, $𝑅\ll \frac{𝑟}{𝑛}$, giving

$𝐼=\frac{𝐸}{\frac{𝑟}{𝑛}}=\frac{𝑛𝐸}{𝑟}$

Hence, the correct option is (C).

The equation of the path is

$𝑦=9{𝑥}^2$

Differentiating with respect to time $𝑡$,

$\frac{d𝑦}{d𝑡}=18𝑥\frac{d𝑥}{d𝑡}=18𝑥{𝑣}_{𝑥}$

Differentiating again with respect to time $𝑡$,

${𝑎}_{𝑦}=\frac{d^2𝑦}{d{𝑡}^2}=18{\left(\frac{d𝑥}{d𝑡}\right)}^2+18𝑥\frac{d^2𝑥}{d{𝑡}^2}$

Since the $𝑥$-component of velocity ${𝑣}_{𝑥}$ is constant, $\frac{d^2𝑥}{d{𝑡}^2}={𝑎}_{𝑥}=0$. Therefore,

${𝑎}_{𝑦}=18{𝑣}_2^{𝑥}$

Given the acceleration $\overrightarrow{𝑎}=2\widehat{𝑗}{\text{m s}}^{-2}$, the component along the $𝑦$-axis is ${𝑎}_{𝑦}=2{\text{m s}}^{-2}$.

$18{𝑣}_2^{𝑥}=2⟹{𝑣}_2^{𝑥}=\frac{1}{9}⟹{𝑣}_{𝑥}=\frac{1}{3}{\text{m s}}^{-1}$

Hence, the correct option is (C).

Given, the alternating emf is

$𝜀={𝜀}_0\sin \left(𝜔𝑡\right)$

The charge $𝑞$ on the capacitor of capacitance $𝐶$ at any instant is

$𝑞=𝐶𝜀=𝐶{𝜀}_0\sin \left(𝜔𝑡\right)$

The instantaneous current $𝐼$ in the circuit is

$𝐼=\frac{𝑑𝑞}{𝑑𝑡}=\frac{𝑑}{𝑑𝑡}[𝐶{𝜀}_0\sin (𝜔𝑡\left)\right]=𝜔𝐶{𝜀}_0\cos \left(𝜔𝑡\right)$

Using $\cos \left(𝜔𝑡\right)=\sin (𝜔𝑡+\frac{𝜋}{2})$ and setting ${𝐼}_0=𝜔𝐶{𝜀}_0$,

$𝐼={𝐼}_0\sin (𝜔𝑡+\frac{𝜋}{2})$

Hence, the correct option is (D).

Let the intensity due to each slit individually be $𝐼$.

At the central maximum, constructive interference occurs with zero phase difference. The resultant intensity ${𝐼}_0$ is given by

${𝐼}_0={(\sqrt{\surd 𝐼}+\sqrt{\surd 𝐼})}^2=4𝐼$

$\therefore 𝐼=\frac{{𝐼}_0}{4}$

When one of the slits is blocked by an opaque object, interference ceases to occur. The intensity at the same region is simply the intensity due to the single open slit:

${𝐼}^{\prime }=𝐼=\frac{{𝐼}_0}{4}$

Hence, the correct option is (C).

Specific resistance (resistivity) $𝜌$ is given by

$𝜌=\frac{𝑅𝐴}{𝑙}$

where $𝑅$ is resistance, $𝐴$ is cross-sectional area, and $𝑙$ is length.

Since resistance $𝑅=\frac{𝑉}{𝐼}=\frac{𝑊}{𝑞𝐼}=\frac{𝑊}{{𝐼}^2𝑡}$, the dimensional formula of resistance is

$\left[𝑅\right]=\frac{\left[\text{M}{\text{L}}^2{\text{T}}^{-2}\right]}{\left[{\text{A}}^2\text{T}\right]}=\left[\text{M}{\text{L}}^2{\text{T}}^{-3}{\text{A}}^{-2}\right]$

Substituting this into the expression for $𝜌$,

$\left[𝜌\right]=\left[𝑅\right]\times \frac{\left[{\text{L}}^2\right]}{\left[\text{L}\right]}=\left[\text{M}{\text{L}}^2{\text{T}}^{-3}{\text{A}}^{-2}\right]\left[\text{L}\right]=\left[\text{M}{\text{L}}^3{\text{T}}^{-3}{\text{A}}^{-2}\right]$

Hence, the correct option is (B).

According to Gauss's law, the total electric flux through a closed surface depends only on the net charge enclosed by that surface:

$\Phi =\frac{{𝑞}_{\text{enclosed}}}{{𝜀}_0}$

For the spherical surface ${𝑆}_1$, the enclosed charge is ${𝑞}_{\text{enclosed}}=5𝜇\text{C}$. Therefore, the flux through ${𝑆}_1$ is:

${\Phi }_1=\frac{5𝜇\text{C}}{{𝜀}_0}$

Since the shell ${𝑆}_2$ encloses the shell ${𝑆}_1$ and carries no additional charge, the net charge enclosed by ${𝑆}_2$ remains the same, namely $5𝜇\text{C}$. The flux through ${𝑆}_2$ is:

${\Phi }_2=\frac{5𝜇\text{C}}{{𝜀}_0}$

The ratio of the flux through ${𝑆}_2$ to ${𝑆}_1$ is:

$\frac{{\Phi }_2}{{\Phi }_1}=\frac{1}{1}=1:1$

Hence, the correct option is (D).

In a nuclear reactor, cadmium is used as a control rod material because cadmium has a very high absorption cross-section for thermal (slow) neutrons.

By absorbing or capturing these slow neutrons, cadmium rods regulate the rate of the nuclear chain reaction or shut it down when required.

Hence, the correct option is (D).

Given, orbital radius $𝑟=0.5\times {10}^{-10}\text{ m}$ and orbital velocity $𝑣=2.2\times {10}^6{\text{ m s}}^{-1}$.

The frequency of revolution of the electron in its orbit is given by

$𝑓=\frac{𝑣}{2𝜋𝑟}$

On substituting the given values,

$𝑓=\frac{2.2\times {10}^6}{2𝜋\times 0.5\times {10}^{-10}}=\frac{2.2\times {10}^{16}}{𝜋}\approx 6.98\times {10}^{15}\text{ Hz}$

Hence, the correct option is (C).

Let the object be dropped from rest at point $𝐴$, so its initial velocity is $𝑢=0$.

The distance covered from $𝐴$ to $𝐵$ in time ${𝑡}_1$ is

$𝐴𝐵=\frac{1}{2}𝑔{𝑡}_2^1$

The total time taken to reach point $𝐶$ from $𝐴$ is ${𝑡}_1+{𝑡}_2$. Thus, the distance $𝐴𝐶$ is

$𝐴𝐶=\frac{1}{2}𝑔({𝑡}_1+{𝑡}_2{)}^2$

Given that $\frac{{𝑡}_1}{{𝑡}_2}=\frac{1}{\sqrt{\surd 2}-1}$, we have ${𝑡}_2=(\sqrt{\surd 2}-1){𝑡}_1$.

Therefore, the total time is

${𝑡}_1+{𝑡}_2={𝑡}_1+(\sqrt{\surd 2}-1){𝑡}_1=\sqrt{\surd 2}{𝑡}_1$

Substituting this into the expression for $𝐴𝐶$ gives

$𝐴𝐶=\frac{1}{2}𝑔(\sqrt{\surd 2}{𝑡}_1{)}^2=2\left(\frac{1}{2}𝑔{𝑡}_2^1\right)=2𝐴𝐵$

The distance $𝐵𝐶$ is

$𝐵𝐶=𝐴𝐶-𝐴𝐵=2𝐴𝐵-𝐴𝐵=𝐴𝐵$

Hence, the ratio of distances is

$\frac{𝐴𝐵}{𝐵𝐶}=\frac{1}{1}$

Hence, the correct option is (A).

Given, slit width $𝑎=1\text{ mm}={10}^{-3}\text{ m}$, distance to the screen $𝐷=2\text{ m}$, and the distance between the first dark fringes on either side of the central maximum (the width of the central maximum) $𝑤=2.2\text{ mm}=2.2\times {10}^{-3}\text{ m}$.

The width of the central maximum in single-slit diffraction is given by

$𝑤=\frac{2𝜆𝐷}{𝑎}$

Rearranging for the wavelength $𝜆$,

$𝜆=\frac{𝑤𝑎}{2𝐷}$

On substituting the given values,

$𝜆=\frac{2.2\times {10}^{-3}\times {10}^{-3}}{2\times 2}=0.55\times {10}^{-6}\text{ m}=5500\text{ Å}$

Hence, the correct option is (D).

The breaking stress of a wire depends only on the material and is independent of its dimensions.

The maximum load a wire can withstand without breaking is given by

$\text{Breaking load}=\text{Breaking stress}\times \text{Area of cross-section}$

Since both the material and the cross-sectional area remain the same, the maximum load is independent of the length of the wire.

Therefore, the maximum load the wire can withstand remains equal to $𝑊$.

Hence, the correct option is (C).

Given, mass of the door $𝑀=12\text{kg}$, width of the door $𝐿=1.0\text{m}$, mass of the bullet $𝑚=10\text{g}=0.01\text{kg}$, and final angular velocity $𝜔=0.625{\text{rad s}}^{-1}$.

The moment of inertia of the door about the vertical hinge is

${𝐼}_{\text{door}}=\frac{1}{3}𝑀{𝐿}^2=\frac{1}{3}\times 12\times (1.0{)}^2=4{\text{kg m}}^2$

The bullet embeds at the centre of the door, at a distance $𝑟=\frac{𝐿}{2}=0.5\text{m}$ from the axis of rotation. Its moment of inertia is

${𝐼}_{\text{bullet}}=𝑚{𝑟}^2=0.01\times (0.5{)}^2=0.0025{\text{kg m}}^2$

Since ${𝐼}_{\text{bullet}}\ll {𝐼}_{\text{door}}$, the total moment of inertia of the system about the hinge is

$𝐼\approx {𝐼}_{\text{door}}=4{\text{kg m}}^2$

By conservation of angular momentum about the hinge axis,

${𝐿}_{𝑖}={𝐿}_{𝑓}$

$𝑚𝑣𝑟=𝐼𝜔$

On substituting the given values,

$0.01\times 𝑣\times 0.5=4\times 0.625$

$0.005𝑣=2.5$

$𝑣=500{\text{m s}}^{-1}$

Hence, the correct option is (D).

The dimensional formulae of the given quantities are:

$\left[𝐺\right]=\left[{\text{M}}^{-1}{\text{L}}^3{\text{T}}^{-2}\right]$

$\left[ℎ\right]=\left[\text{M}{\text{L}}^2{\text{T}}^{-1}\right]$

$\left[𝑐\right]=\left[\text{L}{\text{T}}^{-1}\right]$

Given the relation,

$𝑋={𝐺}^{-1/2}{ℎ}^{1/2}{𝑐}^{5/2}$

Substituting the dimensions:

$\left[𝑋\right]=\left[{\text{M}}^{-1}{\text{L}}^3{\text{T}}^{-2}{]}^{-1/2}\right[\text{M}{\text{L}}^2{\text{T}}^{-1}{]}^{1/2}[\text{L}{\text{T}}^{-1}{]}^{5/2}$

$\left[𝑋\right]={\text{M}}^{1/2+1/2}{\text{L}}^{-3/2+1+5/2}{\text{T}}^{1-1/2-5/2}$

$\left[𝑋\right]=\left[\text{M}{\text{L}}^2{\text{T}}^{-2}\right]$

These are the dimensions of energy.

Hence, the correct option is (B).

The mass of a single atom is given by

$𝑚=\frac{𝑀}{{𝑁}_{\text{A}}}=\frac{108\times {10}^{-3}\text{kg}}{6.023\times {10}^{23}}\approx 1.793\times {10}^{-25}\text{kg}$

The frequency of oscillation in simple harmonic motion is

$𝜈=\frac{1}{2𝜋}\sqrt{\surd \frac{𝑘}{𝑚}}$

On substituting the given values,

$𝜈=\frac{1}{2𝜋}\sqrt{\surd \frac{7.1}{1.793\times {10}^{-25}}}=\frac{1}{2𝜋}\sqrt{\surd 3.96\times {10}^{25}}\approx \frac{6.29\times {10}^{12}}{2𝜋}\approx 1\times {10}^{12}{\text{s}}^{-1}$

Hence, the correct option is (C).

Given,

Radius of the coil, $𝑟=10\text{ cm}=0.1\text{ m}$

Number of turns, $𝑁=300$

Current, $𝐼=2\text{ A}$

Magnetic field, $𝐵=0.7\text{ T}$

Angle made by the plane of the coil with the magnetic field, $𝛼={30}^{\circ }$

The area of the circular coil is

$𝐴=𝜋{𝑟}^2=𝜋(0.1{)}^2=0.01𝜋{\text{ m}}^2$

The angle $𝜃$ between the area vector (normal to the plane of the coil) and the magnetic field is

$𝜃={90}^{\circ }-𝛼={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

The deflecting torque acting on the coil is

$𝜏=𝑁𝐼𝐴𝐵\sin 𝜃=𝑁𝐼𝐴𝐵\cos 𝛼$

On substituting the given values,

$𝜏=300\times 2\times (\frac{22}{7}\times 0.01)\times 0.7\times \cos {30}^{\circ }$

$𝜏=6\times 2.2\times \frac{\sqrt{\surd 3}}{2}=6.6\times 1.732\approx 11.42\text{ N m}$

To prevent the coil from turning, an equal and opposite torque of $11.42\text{ N m}$ must be applied.

Hence, the correct option is (A).

The fringe width in Young's double-slit experiment is given by

$𝛽=\frac{𝜆𝐷}{𝑑}$

When the entire apparatus is immersed in a liquid of refractive index $𝜇$, the wavelength of light becomes ${𝜆}^{\prime }=\frac{𝜆}{𝜇}$, while the slit separation $𝑑$ and screen distance $𝐷$ remain unchanged.

Therefore, the new fringe width ${𝛽}^{\prime }$ is

${𝛽}^{\prime }=\frac{𝛽}{𝜇}$

Given $𝛽=1.82\text{ mm}$ and $𝜇=1.3$,

${𝛽}^{\prime }=\frac{1.82}{1.3}=1.4\text{ mm}$

Hence, the correct option is (A).

The speed of light in free space is given by

$𝑐=\frac{1}{\sqrt{\surd {𝜇}_0{𝜀}_0}}=({𝜇}_0{𝜀}_0{)}^{-1/2}$

Therefore,

$({𝜇}_0{𝜀}_0{)}^{1/2}=\frac{1}{𝑐}$

Since the dimensional formula of speed $𝑐$ is $\left[𝐿{𝑇}^{-1}\right]$,

$\left[\right({𝜇}_0{𝜀}_0{)}^{1/2}]=\frac{1}{\left[𝐿{𝑇}^{-1}\right]}=\left[{𝐿}^{-1}𝑇\right]$

Hence, the correct option is (C).

For the path of the electron beam to remain undeviated, the electric force and the magnetic Lorentz force acting on the electrons must balance:

$𝑞𝐸=𝑞𝑣𝐵⟹𝑣=\frac{𝐸}{𝐵}$

Given, $𝐸=1.8\times {10}^4{\text{V m}}^{-1}$ and $𝐵=6\times {10}^{-3}\text{T}$.

On substituting the given values,

$𝑣=\frac{1.8\times {10}^4}{6\times {10}^{-3}}=3\times {10}^6{\text{m s}}^{-1}$

Hence, the correct option is (C).

Since the two charges are of opposite signs, the null point must lie on the line joining them, outside the charges, and closer to the charge of smaller magnitude ($𝑄$).

Let the distance of the point from $𝑄$ be $𝑥\text{cm}$. The distance of this point from $𝑃$ is $(5+𝑥)\text{cm}$.

Equating the magnitudes of the electric fields due to $𝑃$ and $𝑄$,

$\frac{1}{4𝜋{𝜀}_0}\frac{\left|{𝑞}_{𝑃}\right|}{(5+𝑥{)}^2}=\frac{1}{4𝜋{𝜀}_0}\frac{\left|{𝑞}_{𝑄}\right|}{{𝑥}^2}$

$\frac{25}{(5+𝑥{)}^2}=\frac{16}{{𝑥}^2}$

Taking the square root on both sides,

$\frac{5}{5+𝑥}=\frac{4}{𝑥}$

$5𝑥=20+4𝑥⟹𝑥=20\text{cm}$

Thus, the point is at a distance of $20\text{cm}$ from $𝑄$ and $5+20=25\text{cm}$ from $𝑃$.

Hence, the correct option is (C).

The resolving power of a microscope is inversely proportional to the wavelength of the radiation used. According to de Broglie's hypothesis, moving electrons exhibit wave nature with a wavelength several orders of magnitude smaller than that of visible light. The electron microscope uses this wave nature of electrons to achieve much higher resolving power and magnification than an optical microscope.

Hence, the correct option is (B).

Given, the current in the primary coil ${𝐼}_{𝑝}=2.5\text{A}$, the number of turns in the secondary coil ${𝑁}_{𝑠}=400$, and the magnetic flux linked with each turn of the secondary coil ${𝜙}_{𝑠}=600\times {10}^{-6}{\text{T m}}^2$.

The total magnetic flux linkage of the secondary coil is:

${𝑁}_{𝑠}{𝜙}_{𝑠}=400\times 600\times {10}^{-6}=0.24\text{Wb}$

The mutual inductance $𝑀$ between the coils is given by:

$𝑀=\frac{{𝑁}_{𝑠}{𝜙}_{𝑠}}{{𝐼}_{𝑝}}=\frac{0.24}{2.5}=0.096\text{H}$

The induced emf in the secondary coil when the rate of change of primary current is $\frac{𝑑{𝐼}_{𝑝}}{𝑑𝑡}=0.2{\text{A s}}^{-1}$ is:

$\left|{𝑒}_{𝑠}\right|=𝑀\frac{𝑑{𝐼}_{𝑝}}{𝑑𝑡}=0.096\times 0.2=1.92\times {10}^{-2}\text{V}$

Hence, the correct option is (A).

Let $𝑚$ be the mass of the block.

Since the block is placed on a large block of ice, the final equilibrium temperature of the block is $0^{\circ }C$.

Heat lost by the block as it cools from ${500}^{\circ }C$ to $0^{\circ }C$ equals the heat gained to melt the ice at $0^{\circ }C$:

$𝑚𝑐\Delta 𝑇={𝑚}_{\text{ice}}{𝐿}_{𝑓}$

Given, $𝑐=0.39J{g}^{-1}{}^{\circ }{C}^{-1}=390J{\mathrm{kg}}^{-1}{}^{\circ }{C}^{-1}$, $\Delta 𝑇={500}^{\circ }C$, ${𝑚}_{\text{ice}}=1.455\mathrm{kg}$, and ${𝐿}_{𝑓}=335J{g}^{-1}=3.35\times {10}^{5}J{\mathrm{kg}}^{-1}$.

On substituting the given values,

$𝑚\times 390\times 500=1.455\times 3.35\times {10}^5$

$𝑚\times 195000=487425$

$𝑚=\frac{487425}{195000}\approx 2.5\mathrm{kg}$

Hence, the correct option is (B).

For normal incidence on the first face of the prism, the angle of incidence is ${𝑖}_1=0^{\circ }$, which gives the angle of refraction at this face as ${𝑟}_1=0^{\circ }$.

The angle of incidence at the second face, ${𝑟}_2$, is related to the prism angle $𝐴$ by

${𝑟}_1+{𝑟}_2=𝐴$

$0^{\circ }+{𝑟}_2={45}^{\circ }\Rightarrow {𝑟}_2={45}^{\circ }$

Since the critical angle for the glass-air interface is ${𝜃}_{𝑐}={42}^{\circ }$, the angle of incidence at the second face exceeds the critical angle (${𝑟}_2>{𝜃}_{𝑐}$). Therefore, the light ray undergoes total internal reflection at the second face.

Hence, the correct option is (D).

For two vectors to give a zero resultant, they must be equal in magnitude and opposite in direction. Thus, two vectors of unequal magnitude cannot yield a zero resultant, making statement (C) false.

For three vectors $\overrightarrow{𝐴}$, $\overrightarrow{𝐵}$, and $\overrightarrow{𝐶}$ to have a zero resultant, $\overrightarrow{𝐴}+\overrightarrow{𝐵}+\overrightarrow{𝐶}=\vec{0}$, which implies $\overrightarrow{𝐶}=-(\overrightarrow{𝐴}+\overrightarrow{𝐵})$. Since the vector sum $\overrightarrow{𝐴}+\overrightarrow{𝐵}$ lies in the plane defined by $\overrightarrow{𝐴}$ and $\overrightarrow{𝐵}$, the vector $\overrightarrow{𝐶}$ must also lie in the same plane. Therefore, three non-coplanar vectors can never have a zero resultant, making statement (A) false.

Three coplanar vectors can have a zero resultant if they can be represented by the three sides of a triangle taken in order. Thus, statement (B) is true.

Hence, the correct option is (C).

According to the principle of quantization of electric charge,

$𝑄=𝑛𝑒$

where $𝑛$ must be an integer and $𝑒=1.6\times {10}^{-19}\text{ C}$.

For body $𝑃$:

$𝑛=\frac{0.5\times {10}^{-19}}{1.6\times {10}^{-19}}=\frac{5}{16}\notin \mathbb{Z}$

For body $𝑄$:

$𝑛=\frac{0.7\times {10}^{-19}}{1.6\times {10}^{-19}}=\frac{7}{16}\notin \mathbb{Z}$

For body $𝑅$:

$𝑛=\frac{2.1\times {10}^{-19}}{1.6\times {10}^{-19}}=\frac{21}{16}\notin \mathbb{Z}$

For body $𝑆$:

$𝑛=\frac{4.8\times {10}^{-19}}{1.6\times {10}^{-19}}=3\in \mathbb{Z}$

Since only body $𝑆$ carries an integral multiple of the elementary charge, its charge is physically permissible.

Hence, the correct option is (D).

Given, the work function of the metal is $𝜙=1.4\text{eV}$ and the maximum kinetic energy of the emitted photoelectrons is ${𝐾}_{max}=0.4\text{eV}$.

The stopping potential ${𝑉}_0$ is related to the maximum kinetic energy by:

${𝐾}_{max}=𝑒{𝑉}_0$

On substituting ${𝐾}_{max}=0.4\text{eV}$,

$𝑒{𝑉}_0=0.4\text{eV}\Rightarrow {𝑉}_0=0.4\text{V}$

Hence, the correct option is (C).

In a $𝑃\text{-}𝑉$ diagram starting from a given initial state:

Curve 1 is a vertical line parallel to the pressure axis, representing a constant volume process ($𝑉=\text{constant}$), which is an isochoric process.

Curve 4 is a horizontal line parallel to the volume axis, representing a constant pressure process ($𝑃=\text{constant}$), which is an isobaric process.

Between curves 2 and 3, the magnitude of the slope of an adiabatic expansion is $𝛾\left(\frac{𝑃}{𝑉}\right)$, whereas that of an isothermal expansion is $\frac{𝑃}{𝑉}$. Since $𝛾>1$, curve 2 represents the adiabatic process and curve 3 represents the isothermal process.

Therefore, process 1 is isochoric.

Hence, the correct option is (C).