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

\sqrt{R_{1} / R_{2}}

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R_{2} / R_{1}

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

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R_{1} / R_{2}

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Solution

The correct option is C $(R_{1} / R_{2})^{2}$
Given that,

The radius of particle X $= R_{1}$

The radius of particle Y $= R_{2}$

We know that,

Work done of each particle = $q V$

Now, suppose the particle starts from rest, and final kinetic energy is

For, X particle,

$\frac{1}{2} m_{1} v_{1}^{2} = q V$

For, Y particle

$\frac{1}{2} m_{2} v_{2}^{2} = q V$

Now,

$m_{1} v_{1}^{2} = m_{2} v_{2}^{2} . . . . . (I)$

Now, from the magnetic force is
$F_{m_{1}} = q v_{1} B$

$F_{m_{2}} = q v_{2} B$

Now, the magnetic force is equal to the centripetal force is

For, X

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

$m_{1} v_{1} = q B R_{1}$

For, Y

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

$m_{2} v_{2} = q B R_{2}$

Now, putting the value in equation (I)

$m_{1} {(\frac{q B R_{1}}{m_{1}})}^{2} = m_{2} {(\frac{q B R_{2}}{m_{2}})}^{2}$

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

$\frac{m_{1}}{m_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$

Hence, the ratio of the mass is ${(\frac{R_{1}}{R_{2}})}^{2}$

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