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(B) The kinetic energy $K$ gained by a particle of charge $Q$ accelerated through a potential difference $V$ is given by $K = QV$.Since $K = \frac{1}{

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$1 : 2$

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(B) The kinetic energy $K$ gained by a particle of charge $Q$ accelerated through a potential difference $V$ is given by $K = QV$.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = QV$.For particles of equal mass $m$ and same potential $V$,$v^2 \propto Q$,which implies $v \propto \sqrt{Q}$.Therefore,the ratio of velocities is $\frac{v_A}{v_B} = \sqrt{\frac{Q_A}{Q_B}}$.Given $Q_A = q$ and $Q_B = 4q$,we get $\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio is $1 : 2$.

Two particles of equal mass $m$ have charges $+q$ and $+4q$. If they are accelerated through the same potential difference $V$,the ratio of their velocities $\frac{v_A}{v_B}$ is:

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