1. Kinetic energy gained through potential difference

If a particle of charge $q$ is accelerated through potential difference $V$, then

$qV = \frac{1}{2}mv^2$

So,

$v = \sqrt{\frac{2qV}{m}}$

2. Radius of circular path in magnetic field

When the particle enters normally into a uniform magnetic field $B$, the magnetic force provides centripetal force:

$qvB = \frac{mv^2}{R}$

Hence,

$R = \frac{mv}{qB}$

Substitute $v = \sqrt{\frac{2qV}{m}}$:

$R = \frac{m}{qB}\sqrt{\frac{2qV}{m}}$

$R = \frac{\sqrt{2qV}}{qB}\sqrt{m}$

Thus,

$R \propto \sqrt{m}$

since $q$, $V$, and $B$ are same for both particles.

3. Relating radii and masses

For particles $X$ and $Y$:

$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y}}$

Squaring both sides,

$\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$

4. Matching with options

So the mass ratio of $X$ and $Y$ is

$\left(\frac{R_1}{R_2}\right)^2$

This corresponds to Option D.