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Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describes circular path of radius $R_{1} and R_{2}$ respectively. The ratio of mass of X to that of Y is

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{(\frac{R_{1}}{R_{2}})}^{^{1 / 2}}

\frac{R_{2}}{R_{2}}

{(\frac{R_{1}}{R_{2}})}^{2}

\frac{R_{1}}{R_{2}}

answer is C.

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Detailed Solution

KE = \frac{1}{2} m_{1} v_{1}^{2} = qV

$y = \frac{1}{2} m_{2} v_{2}^{2} = qv$

Since q , v are same $\bar{F} \propto \bar{F_{c}} \propto q$

$∴ \frac{m_{1} v_{1}^{2}}{R_{1}} = {qv}_{1} B$

$\frac{m_{2} v_{2}^{2}}{R_{2}^{2}} = {qv}_{2} B$

$\frac{m_{1} v_{1}}{R_{1}} = \frac{m_{2} v_{2}}{R_{2}}$

$V_{1} = \frac{m_{2}}{m_{1}} \times \frac{R_{1}}{R_{2}} \times V_{2}$

(or)

$m_{1} {[\frac{m_{2}}{m_{1}} \times \frac{R_{1}}{R_{2}} \times V_{2}]}^{2}$

$= m_{2} v_{2}^{2}$

$m_{2} v_{2}^{2} = \frac{m_{1} m_{2}^{2} R_{1}^{2} r_{2}^{2}}{m_{1}^{2} R_{1}^{2}}$

$m_{1} R_{1}^{2} = m_{1} R_{2}^{2}$

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