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 same electrical potential difference, the ratio of their speed $V_A : V_B$ will be :-
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Acceleration $=\frac{\text { Force }}{\text { Mass }}$
$a=\frac{F}{m}=\frac{q E}{m}$
$a_{A}=\frac{1}{4} a_{B}$
Now $v^{2}=u^{2}+2 a x$
since, $x$ is the same for both the particles.
$\Rightarrow \quad v_{A}^{2}=0+2 a_{A} x$ and $v_{B}^{2}=0+2 a_{B} x$
$\Rightarrow \frac{v_{A}^{2}}{v_{B}^{2}}=\frac{a_{A}}{a_{B}}=\frac{1}{4} \Rightarrow \frac{v_{A}}{v_{B}}=\frac{1}{2}$
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