Two charged small balls each of charge ‘q’ and mass ‘m’ when suspended from a common point by strings of length '$l$'  in air, then these strings from an angle of ${120}^0$  with each other. When the system is immersed in q liquid of dielectric constant '$k$' (ratio of density of liquid and ball’s material$\frac{d_L}{d_M}=\frac{1}{3}$) then the angle between the strings remain same. Mark the correct statements): 

The angle made by the string with vertical will be 60^o.

$\frac{{\mathrm{F}}_{\mathrm{e}}}{\mathrm{mg}}=\sqrt{3}\Rightarrow {\mathrm{F}}_{\mathrm{e}}=\sqrt{3}\mathrm{mg}$........(1)

So the separation between the two charges will be $\sqrt{3}l$ .

$F_e=\frac{k{q}^2}{{\left(\sqrt{3}l\right)}^2}=\frac{k{q}^2}{3l^2}$

Since $\frac{d_L}{d_M}=\frac{1}{3}$, the buoyant force will be $\frac{\mathrm{mg}}{3}$.

As the liquid is dielectric in nature, electrostatic force will be divided by the factor of 'k', So ${{\mathrm{F}}^{\text{'}}}_{\mathrm{e}}=\frac{{\mathrm{F}}_{\mathrm{e}}}{\mathrm{k}}$...(2)

As the angle made by string with vertical is unchanged, $\frac{{{\mathrm{F}}^{\text{'}}}_{\mathrm{e}}}{\mathrm{mg}-\mathrm{B}}=\sqrt{3}$ 

${{\mathrm{F}}^{\text{'}}}_{\mathrm{e}}=\sqrt{3}\times \frac{2}{3}\mathrm{mg}$

From (1),

$\frac{{\mathrm{F}}_{\mathrm{e}}}{\mathrm{k}}=\sqrt{3}\times \frac{2}{3}\mathrm{mg}\Rightarrow \mathrm{k}=\frac{3}{2}$

The electrostatic force from the liquid will be $F_e\left(1-\frac{1}{k}\right)=\frac{k{q}^2}{3l^2}\left(1-\frac{2}{3}\right)=\frac{k{q}^2}{9l^2}$