A particle $A$ has charge $ + q$ and a 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, the ratio of their speed $\frac{{{v_A}}}{{{v_B}}}$ will become
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(b) Using $v = \sqrt {\frac{{2QV}}{m}} $ $==>$ $v \propto \sqrt Q $ $==>$ $\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{Q_A}}}{{{Q_B}}}} = \sqrt {\frac{q}{{4q}}} = \frac{1}{2}$
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