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(A) The kinetic energy gained by a particle of charge $q$ and mass $m$ falling through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^

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$1$:$2$

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(A) The kinetic energy gained by a particle of charge $q$ and mass $m$ falling through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.For particle $A$ with charge $q$:$\frac{1}{2}mv_A^2 = qV$  (Equation $1$)For particle $B$ with charge $4q$:$\frac{1}{2}mv_B^2 = 4qV$ (Equation $2$)Dividing Equation $1$ by Equation $2$:$\frac{v_A^2}{v_B^2} = \frac{qV}{4qV} = \frac{1}{4}$Taking the square root on both sides:$\frac{v_A}{v_B} = \sqrt{\frac{1}{4}} = \frac{1}{2}$Therefore,the ratio of their speeds $v_A : v_B$ is $1:2$.

Two particles $A$ and $B$ having the same mass have charges $+q$ and $+4q$ respectively. When they are allowed to fall from rest through the same electric potential difference,the ratio of their speeds $v_A$ to $v_B$ will be:

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