Explanation

(a) Let $v$ be velocity acquired by the charged particle when accelerated through the potential difference $V$
$\therefore\frac{1}{2}mv^{2} = qV$
Or $v = \sqrt{\frac{2qV}{m}}$ …(i)
As the charged particle describes a circular path of radius $R$ in the uniform magnetic field
$\therefore\frac{mv^{2}}{R} = qvB$
Or $R = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}}$ [Using (i)]
$= \frac{1}{B}\sqrt{\frac{2Vm}{q}}$
As $q,\ B$ and $V$ remain the same
$\therefore R \propto \sqrt{m}$
$\therefore\frac{R_{A}}{R_{B}} = \sqrt{\frac{m_{A}}{m_{B}}}$
or $\frac{m_{A}}{m_{B}} = \left( \frac{R_{A}}{R_{B}} \right)^{2} = \left( \frac{2}{3} \right)^{2} = \frac{4}{9}$