The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be
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(d)
$\frac{1}{2} m v^2=\frac{k q Q}{r}$
$\frac{1}{2} m(2 v)^2=\frac{k q Q}{r^{\prime}}$
$\frac{1}{4}=\frac{r^{\prime}}{r}$
$r^{\prime}=\frac{r}{4}$
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