Question

in a liquid medium of dielectric constant K and of specific gravity 2 , two identically charged spheres are suspended from a fixed point by threads of equal lengths .The angle between them 15 ${90}^0$. in another medium of unknown dielectric constant$K^1$ , and specific gravity 4 the angle between them becomes ${120}^0$if density of material of spheres is 8 gm/cc. then $k^1$ is :

• A. $\frac{k}{2}$
• B. $\frac{\sqrt{3}}{k}$
• C. $\frac{\sqrt{3}}{2}k$
• D. $\frac{k}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{k}{\sqrt{3}}$

$\begin{array}{c}\tan {45}^0=\frac{\frac{1}{4\pi E_{0}k}.\frac{q^2}{2l^2}}{\left(8-2\right)vg}..................\left(1\right)\\ \tan {60}^0=\frac{\frac{1}{4\pi E_0{k}^{\prime }}.\frac{q^2}{3l^2}}{\left(8-4\right)vg}..................\left(2\right)\\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{k^{\prime }}{k}\frac{3}{2}.\frac{4}{6}\\ k^{\prime }=\frac{k}{\sqrt{3}}\end{array}$

$\therefore$ Option $D$ is correct.