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Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $\mathrm{X}$ and $\mathrm{Y}$ is :
$\left(\frac{R_2}{R_1}\right)^2$
$\left(\frac{R_1}{R_2}\right)^2$
$\left(\frac{R_1}{R_2}\right)$
$\left(\frac{R_2}{R_1}\right)$
Solution
$ \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} $
$ \mathrm{R} \propto \sqrt{\mathrm{m}}$
$ \mathrm{m} \propto \mathrm{R}^2 $
$ \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2$
