1. For each sphere in air

Since the two identical spheres are suspended symmetrically, each string makes an angle $\theta = \frac{30^\circ}{2}=15^\circ$ with the vertical.

For one sphere in equilibrium:

• वजन (effective downward force in air) $= mg$
• horizontal electrostatic repulsion $= F$
• tension $= T$

So, $T\sin\theta = F, \qquad T\cos\theta = mg$ Hence, $\tan\theta = \frac{F}{mg}$

2. When immersed in liquid

The angle remains the same, so $\theta$ remains unchanged. Therefore, $\frac{F'}{W'} = \frac{F}{mg}$ where:

• $F'$ = electrostatic force in liquid
• $W'$ = apparent weight in liquid

3. Electrostatic force in liquid

If the dielectric constant of the liquid is $K$, then Coulomb force becomes: $F' = \frac{F}{K}$

4. Apparent weight in liquid

Let volume of each sphere be $V$, density of sphere material be $\rho_s = 1.6\,\text{g cm}^{-3}$, and density of liquid be $\rho_l = 0.8\,\text{g cm}^{-3}$.

True weight: $mg = \rho_s Vg$

Buoyant force: $B = \rho_l Vg$

So apparent weight in liquid is: $W' = \rho_s Vg - \rho_l Vg = (\rho_s-\rho_l)Vg$ $W' = (1.6-0.8)Vg = 0.8Vg$

Thus, $\frac{W'}{mg} = \frac{0.8Vg}{1.6Vg} = \frac{1}{2}$ So, $W' = \frac{mg}{2}$

5. Use the condition that angle remains same

Since $\tan\theta$ is unchanged, $\frac{F'}{W'} = \frac{F}{mg}$ Substitute $F' = F/K$ and $W' = mg/2$: $\frac{F/K}{mg/2} = \frac{F}{mg}$

Cancel $F/mg$: $\frac{1/K}{1/2} = 1$ $\frac{2}{K} = 1$ $K=2$

6. Check options

• A: $4$ ❌
• B: $3$ ❌
• C: $2$ ✅
• D: $1$ ❌

Therefore, the dielectric constant of the liquid is $2$.