Liquid medium of dielectric constant k and of specific gravity 2, two identically charged years are suspended from a fixed point by threats of equal lengths. The angle between them is 90 degrees. In another medium of unknown dielectric, constant k' , and specific gravity 4, the angle between them becomes 120 degrees. If density of material of sphere is 8 grams per cc then k' is

Given:

Density of sphere, $\rho_s=8 \mathrm{~g} / \mathrm{cm}^3$
First medium: specific gravity $=2$, so $\rho_1=2 \mathrm{~g} / \mathrm{cm}^3$
Second medium: specific gravity $=4$, so $\rho_2=4 \mathrm{~g} / \mathrm{cm}^3$
First angle: $90^{\circ}$
Second angle: $120{ }^{\circ}$
Dielectric constants: $k$ and $k^{\prime}$

For two identical charged spheres,

$
\tan \frac{\theta}{2}=\frac{F_e}{W_{\text {effective }}}
$

Since $F_e \propto \frac{1}{k}$ and effective weight is proportional to ( $\rho_s-\rho_m$ ),

$
\begin{aligned}
\frac{\tan 60^{\circ}}{\tan 45^{\circ}} & =\frac{k}{k^{\prime}} \times \frac{8-2}{8-4} \\
\sqrt{3} & =\frac{k}{k^{\prime}} \times \frac{6}{4} \\
\sqrt{3} & =\frac{3 k}{2 k^{\prime}}
\end{aligned}
$

Therefore,

$
k^{\prime}=\frac{\sqrt{3}}{2} k
$