Answer:

Correct Answer: B

Solution:

$ \text{Let the length of the strings be } L \text{ and mass of the ball be } m \text{ and charge be } q. $

$\text{At equilibrium, } \sum F_x = 0 \text{ and } \sum F_y = 0 $

$\therefore T \sin \theta = mg \text{ ……….(1)} $

$\text{Also } T \cos \theta = F_e $

$\Rightarrow T \cos \theta = \frac{Kq^2}{r^2} \text{ …….(2) where } K = \frac{1}{4\pi\epsilon_0} $

$\text{Dividing (1) and (2) gives } r^2 = \frac{mg}{Kq^2} \tan \theta $

$\text{Now as } \frac{mg}{Kq^2} = \text{constant} = C \text{ and } \tan \theta = \frac{y}{r^2} $

$\therefore r^2 = C \times \frac{2y}{r} $

$\Rightarrow r \propto (y)^{\frac{1}{3}} $

$\text{Thus } \frac{r’}{r} = \left(\frac{y’}{y}\right)^{\frac{1}{3}} $

$\text{Now } y’ = y^2 $

$\Rightarrow r’ = r^{\frac{2}{3}} $