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Two particulars 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 masses of X and Y is

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${(R_{1} / R_{2})}^{1 / 2}$

$(R_{2} / R_{1})$

${(R_{1} / R_{2})}^{2}$

$(R_{1} / R_{2})$

answer is C.

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Detailed Solution

$\frac{1}{2} m v^{2} = q V o r v = \sqrt{\frac{2 q V}{m}}$

Centripetal force, $\frac{m v^{2}}{R} = q v B$

$∴ v = (\frac{q B}{m}) R$

Hence $\sqrt{\frac{2 q V}{m}} = (\frac{q B}{m}) R$

or $R = (\frac{2 m V}{q}) \times \frac{1}{B}$

Here, $V, q$ and B are constant. Hence, $m α R^{2}$
So, $\frac{m_{1}}{m_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$

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