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. The equilibrium separation between the balls now becomes-
[NEET- 2013]

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Solution

From the FBD of the pith ball,

$T\cos \theta =mg$

$T\sin\theta=\dfrac{Kq^2}{r^2}$

$\tan\theta =\dfrac{Kq^2}{r^2mg}$

$\dfrac{\left(\dfrac r2\right)}{y}=\dfrac{kq^2}{r^2mg}$

$y=\dfrac {r^3 mg}{2 k q^2}$

$\Rightarrow ~y ~\propto~r^3$

$\therefore \left(\dfrac{r'}{r}\right)^3= \dfrac{\left(\dfrac y2\right)}{y}$

$\Rightarrow r'=r\left(\dfrac{1}{2}\right)^{1/3}=\dfrac{r}{\sqrt[3]{2}}$

 Hence, $ (C)$ is the correct answer.

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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. The equilibrium separation between the balls now becomes- [NEET- 2013]

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. The equilibrium separation between the balls now becomes-
[NEET- 2013]