JEE Advanced 2026 Paper 2 Question Paper with Solutions

For a conductor of uniform cross-sectional area A and length L, the number of conduction electrons per unit

volume (number density) is n.

If the electrons drift with an average velocity ${𝑣}_{𝑑}$, electric current I is defined as the rate of flow of charge:

$𝐼=\frac{\Delta 𝑄}{\Delta 𝑡}=𝑛𝑒𝐴{𝑣}_{𝑑}$

The drift velocity ${𝑣}_{𝑑}$ is,

${𝑣}_{𝑑}=\frac{𝐼}{𝑛𝑒𝐴}$

It is given that each atom contributes exactly one conduction electron. Therefore, the number density of

electrons n is equal to the number density of atoms in the metal.

Let the mass density of the metal be $𝜌$ and its molar mass be $𝑀$.

The volume of one mole of the metal is:

${𝑉}_{\mathrm{mole}}=\frac{𝑀}{𝜌}$

One mole of any substance contains Avogadro's number $\left({𝑁}_{𝐴}\right)$ of atoms.

Since there is 1 conduction electron per atom, one mole contains ${𝑁}_{𝐴}$ conduction electrons. The number of

electrons per unit volume is:

$𝑛=\frac{{𝑁}_{𝐴}}{{𝑉}_{\mathrm{mole}}}=\frac{𝜌{𝑁}_{𝐴}}{𝑀}$

The density is $𝜌=6.35\times {10}^3\mathrm{kg}{m}^{-3}$

The molar mass is $𝑀=63.5\mathrm{gm}/\mathrm{mole}=63.5\times {10}^{-3}\mathrm{kg}/\mathrm{mole}$

Avogadro's number is ${𝑁}_{𝐴}=6\times {10}^{23}\mathrm{atoms}/\mathrm{mole}$

Substituting values into the expression for n:

$𝑛=\frac{(6.35\times {10}^3)\times (6\times {10}^{23})}{63.5\times {10}^{-3}}$

$𝑛=\frac{6.35}{63.5}\times {10}^3\times {10}^3\times 6\times {10}^{23}$

$𝑛=0.1\times 6\times {10}^{29}=6\times {10}^{28}\mathrm{electrons}{m}^{-3}$

The resistance R of a wire depends on its length L, cross-sectional area A, and electrical conductivity $𝜎$. By

definition, resistivity ${𝜌}_{𝑒}$ is the reciprocal of conductivity $({𝜌}_{𝑒}=\frac{1}{𝜎})$.

$R=\frac{{𝜌}_{e}L}{A}=\frac{L}{𝜎A}$

The length of wire is $L=100m$

Electrical conductivity is $𝜎=2\times {10}^8{\mathrm{mhom}}^{-1}$

The cross-sectional area is $A=0.5{\mathrm{mm}}^2=0.5\times {10}^{-6}{m}^2$

Substituting these values:

$R=\frac{100}{(2\times {10}^8)\times (0.5\times {10}^{-6})}$

$R=\frac{100}{2\times 0.5\times {10}^2}=\frac{100}{1\times 100}=1\Omega$

The wire is connected to a battery of electromotive force (e.m.f.) $V=2V$ and internal resistance $r=1\Omega$.

By Ohm's law applied to the total closed loop, the total circuit current I is :

$𝐼=\frac{𝑉}{𝑅+𝑟}$

Substituting the values of $𝑅$ and $𝑟$:

$𝐼=\frac{2}{1+1}=\frac{2}{2}=1A$

Now substituting the values of current $𝐼$, number density $𝑛$, electron charge $𝑒$, and cross-sectional area $𝐴$ into the drift velocity equation:

${𝑣}_{𝑑}=\frac{𝐼}{𝑛𝑒𝐴}$

The current is $𝐼=1A$

The number density is $𝑛=6\times {10}^{28}{m}^{-3}$

The electronic charge is $𝑒=1.6\times {10}^{-19}C$

And area of cross-section is $𝐴=0.5\times {10}^{-6}{m}^2$

${𝑣}_{𝑑}=\frac{1}{(6\times {10}^{28})\times (1.6\times {10}^{-19})\times (0.5\times {10}^{-6})}$

${𝑣}_{𝑑}=\frac{1}{6\times 1.6\times 0.5\times {10}^{28-19-6}}$

${𝑣}_{𝑑}=\frac{1}{4.8\times {10}^3}$

${𝑣}_{𝑑}=\frac{1}{4800}{ms}^{-1}$

${𝑣}_{𝑑}=\frac{{10}^3}{4800}{\mathrm{mm}s}^{-1}=\frac{1000}{4800}{\mathrm{mm}s}^{-1}=\frac{1}{4.8}{\mathrm{mm}s}^{-1}$

${𝑣}_{𝑑}\approx 0.2083{\mathrm{mm}s}^{-1}$

Therefore, the drift velocity of electron in given conductor is $0.2083{\mathrm{mm}s}^{-1}$. Hence, the correct option is (C).

Let $\mathrm{N}(\mathrm{t})$ be the number of active nuclei of the radioactive substance X present in the reactor at any time t.

The net rate of change of the number of nuclei ($\frac{\mathrm{d}N}{\mathrm{d}t}$) is of concurrent processes, the production of nuclei at a constant rate, which is $+\alpha$ and the decay of nuclei which is proportional to the number of nuclei present, given by $-\lambda N$.

$\frac{\mathrm{d}N}{\mathrm{d}t}=\alpha-\lambda N$

$\frac{\mathrm{d}N}{\alpha-\lambda N}=\mathrm{d}t$

At $t=0$, the reactor just starts producing the nuclide, so the initial number of nuclei is $N=0$.

At time $t$, let the number of nuclei be $N$:

$\int\limits_{0}^{N} \frac{\mathrm{d}N}{\alpha-\lambda N}=\int\limits_{0}^{t} \mathrm{d}t$

For integral $\int\limits_{0}^{N} \frac{\mathrm{d}N}{\alpha-\lambda N}$, on substitution $u=\alpha-\lambda N \rightarrow \mathrm{d}u=-\lambda\,\mathrm{d}N \rightarrow \mathrm{d}N=-\frac{\mathrm{d}u}{\lambda}$

$\int\limits_{0}^{N} \frac{\mathrm{d}N}{\alpha-\lambda N}=\int \frac{1}{u}\left(-\frac{\mathrm{d}u}{\lambda}\right)=-\frac{1}{\lambda}\ln|u|=-\frac{1}{\lambda}\ln|\alpha-\lambda N|$

$=\left[-\frac{1}{\lambda}\ln(\alpha-\lambda N)\right]_{0}^{N}=t$

$\ln(\alpha-\lambda N)-\ln(\alpha)=-\lambda t$

Using the logarithmic identity $\ln A-\ln B=\ln\frac{A}{B}$:

$\ln\left(\frac{\alpha-\lambda N}{\alpha}\right)=-\lambda t$

$\frac{\alpha-\lambda N}{\alpha}=e^{-\lambda t}$

$\alpha-\lambda N=\alpha e^{-\lambda t}$

$\lambda N=\alpha-\alpha e^{-\lambda t}$

$\lambda N=\alpha\left(1-e^{-\lambda t}\right)$

The activity $A(t)$ of a radioactive sample is the total number of decays occurring per second.

$A(t)=\lambda N$

$A(t)=\alpha\left(1-e^{-\lambda t}\right)$

Each decay releases a fixed amount of thermal energy $E_0$. Therefore, the total thermal energy released per unit time (the power input to the liquid, $\frac{dQ}{dt}$) is

$\frac{dQ}{dt}=A(t)\cdot E_0=\alpha E_0\left(1-e^{-\lambda t}\right)$

When heat energy $Q$ is supplied to a liquid of mass $m$ and specific heat capacity $s$, the relationship between the heat supplied and the resulting temperature rise $\Delta T$ is given by

$Q=ms\,\Delta T$

For an infinitesimal change over time $dt$,

$dQ=ms\,dT$

$\frac{dQ}{dt}=ms\frac{dT}{dt}$

$\frac{dT}{dt}=\frac{1}{ms}\left(\frac{dQ}{dt}\right)$

$\frac{dT}{dt}=\frac{1}{ms}\left[\alpha E_0\left(1-e^{-\lambda t}\right)\right]$

$\frac{dT}{dt}=\frac{\alpha E_0}{ms}\left(1-e^{-\lambda t}\right)$

Therefore, the rate of increase of temperature of liquid is $\frac{\alpha E_0}{ms}\left(1-e^{-\lambda t}\right)$. Thus, the correct option is (A).

For a thin prism with a small angle $A = 6^\circ$, the deviation $D_m$ is given by:

&#x27; in math mode at position 1:̲D_m = (n - 1)A" style="color:#cc0000">$D_m = (n - 1)A</span>$

To find the wavelength $\lambda_{\text{min}}$ at which $D_m$ is smallest, we minimize the refractive index $n(\lambda)$, where:

$n(\lambda) = \alpha \lambda + \frac{\beta}{\lambda^2}$

Differentiating $n(\lambda)$ with respect to $\lambda$ and setting the derivative to zero:

$\frac{dn}{d\lambda} = \alpha - \frac{2\beta}{\lambda^3} = 0$

$\lambda_{\text{min}}^3 = \frac{2\beta}{\alpha}$

Substituting $\alpha = 3\text{ }\mu\text{m}^{-1}$ and $\beta = 0.096\text{ }\mu\text{m}^2$:

$\lambda_{\text{min}}^3 = \frac{2 \times 0.096}{3} = 0.064\text{ }\mu\text{m}^3$

$\lambda_{\text{min}} = \sqrt[3]{0.064} = 0.4\text{ }\mu\text{m}$

Calculating the refractive index at $\lambda_{\text{min}}$:

$n_{\text{min}} = 3(0.4) + \frac{0.096}{(0.4)^2} = 1.2 + 0.6 = 1.8$

Substituting $n_{\text{min}}$ back into the deviation equation:

$D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$

The correct value of $D_m$ at $\lambda_{\text{min}}$ is $4.8^\circ$.

The particle of mass $m$ moving in a plane under a central force field $\vec{F}(r)=F(r) \hat{r}$.

In polar coordinates $(r, \theta)$, the position vector is $r = r\hat{r}$.

Differentiating the position vector with respect to time gives the velocity vector:

\vec{v} = \left(\frac{dr}{dt}\right) \hat{r} + r\frac{d\theta}{dt} \hat{\theta}

Differentiating once more gives the acceleration vector:

\vec{a} = \left(\frac{d^2 r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right) \hat{r} + \left(r\frac{d^2\theta}{dt^2} + 2\left(\frac{dr}{dt}\right)\right) \hat{\theta}

By Newton's second law, $\vec{F} = m\vec{a}$. Since the force is purely radial, the transverse component of acceleration must be zero, which means the angular momentum $l$ is a conserved quantity.

$l = m r^2 \omega \Rightarrow \frac{d\theta}{dt} = \frac{l}{m\,r^2}$

Using Newton's second law for the radial component:

$m\left(\frac{d^2 r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right) = F(r)$

$m\left(\frac{d^2 r}{dt^2} - r\left(\frac{l}{m\,r^2}\right)^2\right) = F(r)$

$m\frac{d^2 r}{dt^2} - \frac{l^2}{m\,r^3} = F(r)$

$m\frac{d^2 r}{dt^2} = \frac{l^2}{m\,r^3} + F(r)$

The given attractive force is $F(r)=-kr^2$, the equation becomes:

$m\frac{d^2r}{dt^2}=\frac{l^2}{mr^3}-kr^2$

For a perfect circular orbit of constant radius $r_0$, the radial distance does not change over time,

$\frac{dr}{dt}=0 \text{ and } \frac{d^2r}{dt^2}=0.$

From the radial equation of motion:

$\frac{l^2}{mr_0^3}-\frac{k}{r_0^2}=0$

$\frac{l^2}{mr_0^3}=\frac{k}{r_0^2} \Rightarrow r_0=\frac{l^2}{mk}$

When the particle is given a small radial displacement $\delta\, r$ from its equilibrium position $r_0$. Let its instantaneous radius be:

$r=r_0+\delta\, r$

Taking the second time derivative, since $r_0$ is a constant:

$\frac{d^2r}{dt^2}=\frac{d^2}{dt^2}(r_0+\delta\, r)=\frac{d^2\delta r}{dt^2}$

$m\frac{d^2\delta r}{dt^2}=\frac{l^2}{m(r_0+\delta r)^3}-\frac{k}{(r_0+\delta r)^2}$

$m\frac{d^2\delta r}{dt^2}=\frac{l^2}{m r_0^3\left(1+\frac{\delta r}{r_0}\right)^3}-\frac{k}{r_0^2\left(1+\frac{\delta r}{r_0}\right)^2}$

$m\frac{d^2\delta r}{dt^2}=\frac{l^2}{m r_0^3}\left(1+\frac{\delta r}{r_0}\right)^{-3}-\frac{k}{r_0^2}\left(1+\frac{\delta r}{r_0}\right)^{-2}$

$m\frac{d^2\delta r}{dt^2}=\frac{l^2}{m r_0^3}\left(1+\frac{\delta r}{r_0}\right)^{-3}-\frac{k}{r_0^2}\left(1+\frac{\delta r}{r_0}\right)^{-2}$

Since $\delta\, r \ll r_0$, we can use the binomial expansion $(1+x)^n \approx 1+nx$ for $|x| \ll 1$:

$m\frac{d^2\delta r}{dt^2}\approx \frac{l^2}{m r_0^3}\left(1-3\frac{\delta r}{r_0}\right)-\frac{k}{r_0^2}\left(1-2\frac{\delta r}{r_0}\right)$

$m\frac{d^2\delta r}{dt^2}\approx \left(\frac{l^2}{m r_0^3}-\frac{k}{r_0^2}\right)-\left(\frac{3l^2}{m r_0^3}-\frac{2k}{r_0^2}\right)\delta r$

From the equilibrium condition $\frac{l^2}{m r_0^3}-\frac{k}{r_0^2}=0.$

$\mathrm{m} \frac{\mathrm{~d}^2 \delta \mathrm{r}}{\mathrm{dt}^2} \approx=-\left(\frac{3 \mathrm{l}^2}{\mathrm{mr}_0^4}-\frac{2 \mathrm{k}}{\mathrm{r}_0^3}\right) \delta \mathrm{r}$

Substituting $\mathrm{l}^2=\mathrm{mkr}_0$ from the equilibrium condition

$\frac{3\left(\mathrm{mkr}_0\right)}{\mathrm{m} \mathrm{r}_0^4}-\frac{2 \mathrm{k}}{\mathrm{r}_0^3}=\frac{3 \mathrm{k}}{\mathrm{r}_0^3}-\frac{2 \mathrm{k}}{\mathrm{r}_0^3}=\frac{\mathrm{k}}{\mathrm{r}_0^3}$

The standard simple harmonic motion (SHM) form $\frac{\mathrm{d}^2 \delta \mathrm{r}}{\mathrm{dt}^2}=-\omega^2 \delta \mathrm{r}$ :

$\mathrm{m} \frac{\mathrm{~d}^2 \delta \mathrm{r}}{\mathrm{dt}^2}=-\frac{\mathrm{k}}{\mathrm{r}_0^3} \delta \mathrm{r} \Rightarrow \frac{\mathrm{~d}^2 \delta \mathrm{r}}{\mathrm{dt}^2}=-\left(\frac{\mathrm{k}}{\mathrm{mr}_0^3}\right) \delta \mathrm{r}$

Thus, the angular frequency of the radial oscillations is:

$\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{mr}_0^3}}$

The relationship for the time period T of an oscillation is:

$\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{mr}_0^3}{\mathrm{k}}}$

Using $\mathrm{r}_0=\frac{\mathrm{l}^2}{\mathrm{mk}}$ :

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{k}}\left(\frac{\mathrm{l}^2}{\mathrm{mk}}\right)^3}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{k}} \frac{\mathrm{l}^6}{\mathrm{~m}^3 \mathrm{k}^3}}=2 \pi \sqrt{\frac{\mathrm{l}^6}{\mathrm{~m}^2 \mathrm{k}^4}}=\frac{2 \pi \mathrm{l}^3}{\mathrm{mk}^2}$

Therefore, the correct option is (A).

(A) $\sin i_1 = \sin e_1 = n_1 \sin \frac{A_1}{2}$

$\sin i_2 = \sin e_2 = n_2 \sin \frac{A_2}{2}$

$e_1 = i_2 \Rightarrow n_1 \sin \frac{A_1}{2} = n_2 \sin \frac{A_2}{2}$

(B) $i_2 = e_2 = e_1$

$\sin i_2 = n_2 \sin \frac{A_2}{2}$

But $i_1 \neq i_2$

(C) $\delta_1 = (\mu_1 - 1) A_1, \delta_2 = (\mu_2 - 1) A_2$

$\frac{1}{2} \left( \frac{\delta_1}{\mu_1 - 1} \right) + \frac{1}{2} \left( \frac{\delta_2}{\mu_2 - 1} \right) = \frac{1}{2} \frac{A_1}{2} + \frac{A_2}{2} = \theta$

(D) $i_1 = e_1 = i_2$

$\sin i_1 = n_1 \sin \frac{A_1}{2}$

During this first interval $(t=0 \rightarrow t=0.2 \mathrm{~s})$, the particle experiences a uniform electric field $\overrightarrow{\mathrm{E}}=1 \hat{\mathrm{i}} \mathrm{Vm}^{-1}$ and a uniform gravitational field $\vec{g}=-10 \hat{\mathrm{j}} \mathrm{m} \mathrm{s}^{-2}$.

The charge on particle is, $\mathrm{q}=1 \mu \mathrm{C}=10^{-6} \mathrm{C}$

The mass of the particle is, $\mathrm{m}=1 \mathrm{mg}=10^{-6} \mathrm{~kg}$

The net force acting on the particle is the vector sum of the electric force and gravitational force:

$\overrightarrow{\mathrm{F}}_{\mathrm{net}}=\overrightarrow{\mathrm{F}}_{\mathrm{E}}+\overrightarrow{\mathrm{F}}_{\mathrm{g}}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{m} \overrightarrow{\mathrm{~g}}$

Using Newton's second law, $\vec{a}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}$ :

$\vec{a}=\frac{q \overrightarrow{\mathrm{E}}}{m}+\vec{g}$

Substituting $\frac{\mathrm{q}}{\mathrm{m}}=\frac{10^{-6}}{10^{-6}}=1$: $\overrightarrow{\mathrm{a}}=1(1 \hat{\mathrm{i}})-10 \hat{\mathrm{j}}=(1 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}) \mathrm{m} \mathrm{s}^{-2}$

So, the acceleration components are, $\mathrm{a}_{\mathrm{x}}=1 \mathrm{~m} \mathrm{~s}^{-2} ; \mathrm{a}_{\mathrm{y}}=-10 \mathrm{~m} \mathrm{~s}^{-2} ; \mathrm{a}_{\mathrm{z}}=0 \mathrm{~m} \mathrm{~s}^{-2}$

The initial velocity at $\mathrm{t}=0$ is given as $\overrightarrow{\mathrm{v}}_0=1 \hat{\mathrm{i}}+2 \hat{\jmath} \mathrm{~m} \mathrm{~s}^{-1}$.

Using $\vec{v}=\vec{u}+\overrightarrow{\mathrm{at}}$ and $\overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{ut}}+\frac{1}{2} \overrightarrow{\mathrm{at}}^2$ to find the velocity and position at $\mathrm{t}=0.2 \mathrm{~s}$:

$\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}+\mathrm{a}_{\mathrm{x}} \mathrm{t}=1+(1)(0.2)=1.2 \mathrm{~m} \mathrm{~s}^{-1}$

$\mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t}=2+(-10)(0.2)=2-2=0 \mathrm{~m} \mathrm{~s}^{-1}$

$\mathrm{v}_{\mathrm{z}}=\mathrm{u}_{\mathrm{z}}+\mathrm{a}_{\mathrm{z}} \mathrm{t}=0+0=0 \mathrm{~m} \mathrm{~s}^{-1}$

Thus, at the exact moment the electric field is turned off, the velocity vector is $\overrightarrow{\mathrm{v}}(0.2)=1.2 \hat{\mathrm{i}} \mathrm{~m} \mathrm{~s}^{-1}$ (purely along $x$, i.e. no $y$ or $z$ component).

Position at $\mathrm{t}=0.2 \mathrm{~s}$ (particle starts from the $XZ$ plane, so $\mathrm{y}_0=0$):

$y(0.2)=y_0+u_y t+\frac{1}{2} a_y t^2=0+(2)(0.2)+\frac{1}{2}(-10)(0.2)^2 = 0.4-0.2=0.2 \mathrm{~m}=20 \mathrm{~cm}$

At $t=0.2 \mathrm{~s}$, the electric field is switched off and a magnetic field $\vec{B}=6 \hat{\mathrm{j}} \mathrm{T}$ is switched on. From this instant, the only forces are gravity (along $-\hat{\mathrm j}$) and the magnetic Lorentz force $q(\vec v \times \vec B)$. Since $\vec B$ is along $\hat{\mathrm j}$ and the velocity at $t=0.2\,\mathrm s$ has no $y$-component, the magnetic force stays confined to the $XZ$ plane and never affects the $y$-motion. This decouples the problem into two independent parts:

• Along $Y$: pure projectile motion under gravity alone, starting from $y=0.2\,\mathrm m$ with $u_y=0$: $y(t_b) = 0.2 - 5t_b^2$ where $t_b = t - 0.2$ is the time elapsed since the magnetic field was switched on.
• In the $XZ$ plane: uniform circular motion under the magnetic force alone, with speed fixed at $|v|=1.2\,\mathrm{ms^{-1}}$ (magnetic force does no work) and radius $r = \dfrac{mv}{qB} = \dfrac{v}{(q/m)B} = \dfrac{1.2}{1\times 6} = 0.2\,\mathrm m = 20\,\mathrm{cm}$.

Checking option (C): the radius of the trajectory for $t>0.2\,\mathrm s$ is $20\,\mathrm{cm}$, as computed above. Thus, option (C) is correct.

Checking option (A): at $t=0.3\,\mathrm s$, $t_b=0.1\,\mathrm s$:

$y(0.1) = 0.2 - 5(0.1)^2 = 0.2-0.05 = 0.15\,\mathrm m = 15\,\mathrm{cm}$

This matches option (A) exactly. Thus, option (A) is correct.

Checking option (B): at $t=0.4\,\mathrm s$, $t_b=0.2\,\mathrm s$:

$y(0.2) = 0.2 - 5(0.2)^2 = 0.2-0.2 = 0\,\mathrm{cm}$

The vertical distance at $t=0.4\,\mathrm s$ is $0\,\mathrm{cm}$, not $10\,\mathrm{cm}$ as claimed. Thus, option (B) is incorrect.

Checking option (D): at $t=0.35\,\mathrm s$, $t_b=0.15\,\mathrm s$:

$y(0.15) = 0.2 - 5(0.15)^2 = 0.2-0.1125 = 0.0875\,\mathrm m = 8.75\,\mathrm{cm} \ne 0$

Since $y \ne 0$ at $t=0.35\,\mathrm s$, the particle is not in the $XZ$ plane at that instant. Thus, option (D) is incorrect.

Therefore, the correct options are (A) and (C).

Let an arbitrary point in the XY plane be $P(x,y)$.

The electrostatic potential $V$ at a distance $r$ from a point charge $Q$ is derived from the work done per unit charge bringing it from infinity to that point:

$V=\frac{kQ}{r}$

where $k=\frac{1}{4\pi\varepsilon_0}$.

The distance $r_1$ from the point $P_1(a,b)$ to $P(x,y)$ is given by the distance formula:

$r_1=\sqrt{(x-a)^2+(y-b)^2}$

The potential $V_1$ at $P(x,y)$ due to $Q_1=q$ is:

$V_1=\frac{kq}{\sqrt{(x-a)^2+(y-b)^2}}$

The distance $r_2$ from the point $P_2(ma,mb)$ to $P(x,y)$ is:

$r_2=\sqrt{(x-ma)^2+(y-mb)^2}$

The potential $V_2$ at $P(x,y)$ due to $Q_2=mq$ is:

$V_2=\frac{k(mq)}{\sqrt{(x-ma)^2+(y-mb)^2}}$

For the locus of points satisfying the condition $|V_1|=|V_2|$:

$\left|\frac{kq}{\sqrt{(x-a)^2+(y-b)^2}}\right|=\left|\frac{kmq}{\sqrt{(x-ma)^2+(y-mb)^2}}\right|$

$\frac{1}{\sqrt{(x-a)^2+(y-b)^2}}=\frac{m}{\sqrt{(x-ma)^2+(y-mb)^2}}$

Squaring both sides

$\frac{1}{(x-a)^2+(y-b)^2}=\frac{m^2}{(x-ma)^2+(y-mb)^2}$

$(x-ma)^2+(y-mb)^2=m^2\left[(x-a)^2+(y-b)^2\right]$

For $m=-1$

$(x-(-1)a)^2+(y-(-1)b)^2=(-1)^2\left[(x-a)^2+(y-b)^2\right]$

$(x+a)^2+(y+b)^2=(x-a)^2+(y-b)^2$

$x^2+2ax+a^2+y^2+2by+b^2=x^2-2ax+a^2+y^2-2by+b^2$

$2ax+2by=-2ax-2by$

$4ax+4by=0 \rightarrow ax+by=0$

Thus, option (A) is correct.

For $m=2$

$(x-2a)^2+(y-2b)^2=2^2\left[(x-a)^2+(y-b)^2\right]$

$(x-2a)^2+(y-2b)^2=4\left[(x-a)^2+(y-b)^2\right]$

$(x^2-4ax+4a^2)+(y^2-4by+4b^2)=4(x^2-2ax+a^2+y^2-2by+b^2)$

$x^2-4ax+4a^2+y^2-4by+4b^2=4x^2-8ax+4a^2+4y^2-8by+4b^2$

$(4x^2-x^2)+(-8ax+4ax)+(4a^2-4a^2)+(4y^2-y^2)+(-8by+4by)+(4b^2-4b^2)=0$

$3x^2-4ax+3y^2-4by=0$

$x^2-\frac{4}{3}ax+y^2-\frac{4}{3}by=0$

The standard circle form $(x^2+y^2+2gx+2fy+c=0)$:

$\left(x-\frac{2}{3}a\right)^2-\left(\frac{2}{3}a\right)^2+\left(y-\frac{2}{3}b\right)^2-\left(\frac{2}{3}b\right)^2=0$

$\left(x-\frac{2}{3}a\right)^2+\left(y-\frac{2}{3}b\right)^2=\frac{4}{9}a^2+\frac{4}{9}b^2=\frac{4}{9}(a^2+b^2)$

This is a circle equation of the form $(x-x_c)^2+(y-y_c)^2=R^2$:

Where the centre is $(x_c,y_c)=\left(\frac{2}{3}a,\frac{2}{3}b\right)$

And radius is $R=\sqrt{\frac{4}{9}(a^2+b^2)}=\frac{2}{3}\sqrt{a^2+b^2}$

Thus, option (B) is correct.

For $m=-2$

$(x-(-2) a)^2+(y-(-2) b)^2=(-2)^2\left[(x-a)^2+(y-b)^2\right]$

$(x+2 a)^2+(y+2 b)^2=4\left[(x-a)^2+(y-b)^2\right]$

$x^2+4ax+4a^2+y^2+4by+4b^2=4x^2-8ax+4a^2+4y^2-8by+4b^2$

$3x^2 - 12ax + 3y^2 - 12by = 0$

$x^2 - 4ax + y^2 - 4by = 0$

$(x - 2a)^2 - 4a^2 + (y - 2b)^2 - 4b^2 = 0$

$(x - 2a)^2 + (y - 2b)^2 = 4a^2 + 4b^2 = 4(a^2 + b^2)$

This is also a circle with centre $(2a, 2b)$ and radius

$\sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2}$

Thus, option (C) is correct.

For $m = -3$

$(x + 3a)^2 + (y + 3b)^2 = 9[(x - a)^2 + (y - b)^2]$

$x^2 + 6ax + 9a^2 + y^2 + 6by + 9b^2 = 9x^2 - 18ax + 9a^2 + 9y^2 - 18by + 9b^2$

$8x^2 - 24ax + 8y^2 - 24by = 0$

$x^2 - 3ax + y^2 - 3by = 0$

This also represents a circle, not a straight line as given by $3bx + 3ay = 0$. Thus, option (D) is incorrect.

Therefore, the correct options are (A), (B) and (C).