1. Set up the equilibrium in air

Since the two identical spheres repel each other symmetrically, each string makes an angle $\theta=\frac{37^\circ}{2}$ with the vertical.

For one sphere in air:

• Weight downward: $mg$
• Tension along string: $T$
• Electrostatic repulsion horizontal: $F$

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

2. Set up the equilibrium in the liquid

When immersed in a liquid:

• Electrostatic force reduces by dielectric constant $K$: $F' = \frac{F}{K}$
• Effective weight reduces due to buoyancy: $W' = mg - \rho_l V g$ Now, $\tan\theta = \frac{F'}{W'}$

Given that the angle remains the same, $\theta$ is unchanged. Therefore, $\frac{F}{mg} = \frac{F/K}{mg-\rho_l V g}$

3. Simplify the equation

Cancel $F$ and $g$: $\frac{1}{m} = \frac{1/K}{m-\rho_l V}$ So, $m-\rho_l V = \frac{m}{K}$

Now use $m=\rho_s V$ where $\rho_s$ is density of sphere material.

Thus, $\rho_s V - \rho_l V = \frac{\rho_s V}{K}$ $V(\rho_s-\rho_l)=\frac{\rho_s V}{K}$ $\rho_s-\rho_l=\frac{\rho_s}{K}$ $K=\frac{\rho_s}{\rho_s-\rho_l}$

4. Substitute the given values

Given: $\rho_s=1.4\ \text{g/cm}^3,\qquad \rho_l=0.7\ \text{g/cm}^3$

So, $K=\frac{1.4}{1.4-0.7} = \frac{1.4}{0.7}=2$

5. Final answer

The dielectric constant of the liquid is $\boxed{2}$