Two particles X and Y having equal charges after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii $r_1$ and $r_2$ respectively. The ratio of the mass of X to that of Y is:
A.$\dfrac{r_1}{r_2}$
B.$\sqrt{\dfrac{r_1}{r_2}}$
C.$\left[\dfrac{r_2}{r_1}\right]^2$
D$\left[\dfrac{r_1}{r_2}\right]^2$

Hint: The particles get kinetic energy for their acceleration from the potential energy stored due to the potential difference in the electric field, and the particles exert a centripetal force in response to the magnetic force acting on them due to which they execute circular motion. Using the above two statements, first determine an expression for the velocity of the particles as they accelerate through a potential difference and when they exert a centripetal force in the magnetic field. Combine the two expressions for velocity to arrive at a relation between the mass of the particles and the radius of their circular path, which is what we need to this end.
Formula Used:
Kinetic energy $KE = \dfrac{1}{2}mv^2$
Electric potential energy $PE_{electrical} = Vq$
Centripetal force $ F_{centripetal} =\dfrac{mv^2}{r}$
Magnetic force $F_{magnetic} = Bqv$

Complete answer:
We are given that two particles X and Y having equal charges, say q, are being accelerated through a potential difference. This means that the motion of the two particles is made possible by the force exerted on the particles by the applied potential difference V. The charges thus get their kinetic energy from the electrical potential energy, i.e., for the charges
$KE = PE_{electrical}$
$\Rightarrow \dfrac{1}{2}mv^2 = Vq \Rightarrow v = \sqrt{\dfrac{2qV}{m}}$
The charges now enter a uniform magnetic field through which they describe circular paths. They do so in order to balance out the magnetic force acting on them. This means that the centripetal force developed by the charge will be equivalent to the magnetic force acting on it, i.e.,
$F_{centripetal} = F_{magnetic}$
$\Rightarrow \dfrac{mv^2}{r} = Bqv \Rightarrow v = \dfrac{Bqr}{m}$
Plugging in the expression for velocity that we got in the energy equation:
$\Rightarrow \sqrt{\dfrac{2qV}{m}} = \dfrac{Bqr}{m}$
We can now solve the above expression to get a correlation between the mass and radius of the charged particle.
$\Rightarrow \sqrt{m} = \dfrac{Bqr}{2qV}$
$\Rightarrow \sqrt{m} \propto r$
Let the masses of the two charged particles be $m_1$ and $m_2$, and let them execute circular paths of radius $r_1$ and $r_2$ respectively.
From the proportionality, we get:
$\sqrt{m_1}\propto r_1$ and $\sqrt{m_2}\propto r_2$
$\Rightarrow \sqrt{\dfrac{m_1}{m_2}}= \dfrac{r_1}{r_2}$
Squaring both sides we get:
$\Rightarrow \dfrac{m_1}{m_2} = \left[\dfrac{r_1}{r_2}\right]^2$

Therefore, the correct choice would be D. $\left[\dfrac{r_1}{r_2}\right]^2$

Note:
Do not get confused between the electrical potential difference and electric potential energy. In general, electric potential energy is the total potential energy of a unit charge at any point in any electric field whereas electric potential difference is the difference in electric potentials between two different points and is given as the amount of work done in moving a unit charge from one point to another, i.e.,
Electric potential energy $E = Vq$
Electric potential difference $E = \dfrac{W}{q}$