Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height as shown in the diagram . The equilibrium separation between the balls now become

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$(\frac{r}{\sqrt[3]{2}})$

$(\frac{2 r}{3})$

${(\frac{r}{\sqrt{2}})}^{2}$

$(\frac{2 r}{\sqrt{3}})$

answer is C.

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Detailed Solution

$\tan θ = \frac{F}{mg} \Rightarrow \frac{r / 2}{y} = \frac{{kq}^{2}}{r^{2} mg} \Rightarrow y \propto r^{3}$
Therefore ${(\frac{r^{/}}{r})}^{3} = \frac{y / 2}{y} \Rightarrow r^{/} = r {(\frac{1}{2})}^{1 / 3}$

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