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(C) When a particle of charge $q$ and mass $M$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2} M v^{2} =

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$\left(\frac{R_{A}}{R_{B}}\right)^{2}$

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(C) When a particle of charge $q$ and mass $M$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2} M v^{2} = qV$. Thus,the velocity $v = \sqrt{\frac{2qV}{M}}$. When this particle enters a uniform magnetic field $B$ perpendicular to its velocity,it describes a circular path of radius $R = \frac{Mv}{qB}$. Substituting the value of $v$,we get $R = \frac{M}{qB} \sqrt{\frac{2qV}{M}} = \frac{1}{B} \sqrt{\frac{2MV}{q}}$. Since $q$,$V$,and $B$ are the same for both particles,$R \propto \sqrt{M}$,which implies $R^{2} \propto M$. Therefore,$\frac{M_{A}}{M_{B}} = \left(\frac{R_{A}}{R_{B}}\right)^{2}$.

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is

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