Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 45° with each other. When the system is immersed in a liquid of relative density 0.8, the angle between the strings remains unchanged. If the density of the material of the sphere is 2.4 $\text{gcm}^{-3}$, what is the dielectric constant of the liquid?

Let the tension in the string be $T$ and the angle made by each string with the vertical be $\alpha$.

In air, the forces acting on each sphere are the electrostatic force $F$, the weight $mg$, and the tension $T$.

For equilibrium,
$T \sin \alpha = F$
$T \cos \alpha = mg$
Dividing the two equations, we get $\tan \alpha = \dfrac{F}{mg}$.

When immersed in the liquid, the electrostatic force becomes $F' = \dfrac{F}{K}$, where $K$ is the dielectric constant of the liquid.

The effective weight of the sphere becomes $W' = mg - F_b = mg - V \rho_l g = mg \left(1 - \dfrac{\rho_l}{\rho_s}\right)$, where $\rho_l$ is the density of the liquid and $\rho_s$ is the density of the sphere.

For equilibrium in the liquid,
$\tan \alpha = \dfrac{F'}{W'} = \dfrac{F}{K mg \left(1 - \dfrac{\rho_l}{\rho_s}\right)}$.

Since the angle remains unchanged,
$\dfrac{F}{mg} = \dfrac{F}{K mg \left(1 - \dfrac{\rho_l}{\rho_s}\right)}$

$K = \dfrac{1}{1 - \dfrac{\rho_l}{\rho_s}} = \dfrac{\rho_s}{\rho_s - \rho_l}$

Given $\rho_s = 2.4 \text{ g cm}^{-3}$ and $\rho_l = 0.8 \text{ g cm}^{-3}$,

$K = \dfrac{2.4}{2.4 - 0.8} = \dfrac{2.4}{1.6} = 1.5$.

Answer: $1.5$