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(A) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.From this,th

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$1: 2$

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(A) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.From this,the speed $v$ is given by $v = \sqrt{\frac{2qV}{m}}$.For particle $A$ with charge $q_A = q$ and mass $m$,the speed is $v_A = \sqrt{\frac{2qV}{m}}$.For particle $B$ with charge $q_B = 4q$ and mass $m$,the speed is $v_B = \sqrt{\frac{2(4q)V}{m}} = \sqrt{\frac{8qV}{m}} = 2\sqrt{\frac{2qV}{m}}$.Taking the ratio of their speeds:$\frac{v_A}{v_B} = \frac{\sqrt{\frac{2qV}{m}}}{2\sqrt{\frac{2qV}{m}}} = \frac{1}{2}$.Thus,the ratio is $1:2$.

$A$ particle $A$ has charge $+q$ and particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference $V$,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

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