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Step-by-Step Solution
Step 1: Identify the Forces Acting on Each Sphere
When two identical charged spheres are suspended in air by strings of equal length, each sphere experiences the following forces:
• Gravitational force (weight), $W = mg$.
• Electrostatic repulsive force between the two charged spheres, let us denote it by $F$.
• Tension in the string, $T$.
Step 2: Condition for Equilibrium in Air
In air, let the angle between the two strings be $30^\circ$. If each string makes an angle of $15^\circ$ with the vertical (because together they make $30^\circ$), then the horizontal component of the tension balances the electrostatic repulsion, and the vertical component balances the weight of the sphere.
However, the key point is that the ratio of the electrostatic force $F$ to the weight $W$ determines the angle in equilibrium:
$$
\frac{F}{W} = \text{a constant depending on } (15^\circ).
$$
The precise geometry leads to the same conclusion: the larger the ratio $F/W$, the larger the angle, and vice versa.
Step 3: Forces Acting when Submerged in Liquid
When the spheres are immersed in a liquid of density $0.8\,\text{g cm}^{-3}$, two additional effects come into play:
1. Buoyant force $B$ acting upward on each sphere due to the displaced liquid. If the volume of each sphere is $V$ and the density of the liquid is $\rho_\text{liquid}$, then
$$
B = V\,\rho_\text{liquid}\,g.
$$
2. The electrostatic force reduces by a factor of the dielectric constant $K$ of the liquid. Hence the electrostatic repulsion in the liquid is
$$
F_\text{liquid} = \frac{F}{K}.
$$
Step 4: Condition for the Angle to Remain the Same
The problem states that the angle between the two strings remains $30^\circ$ even in the liquid. This means the ratio of the effective electrostatic force to the effective weight must remain unchanged:
$$
\frac{F_\text{liquid}}{W - B} = \frac{F}{W}.
$$
Substituting $F_\text{liquid} = \frac{F}{K}$ and $W - B = mg - B = V\,\rho_\text{sphere}\,g - V\,\rho_\text{liquid}\,g = V\,(\rho_\text{sphere} - \rho_\text{liquid})\,g,$ we get
$$
\frac{\frac{F}{K}}{W - B} = \frac{F}{W}.
$$
Canceling $F$ on both sides and rearranging:
$$
\frac{1}{K} \cdot \frac{1}{W - B} = \frac{1}{W}.
$$
Hence
$$
\frac{1}{K} = \frac{W - B}{W}
\quad \Longrightarrow \quad
K = \frac{W}{W - B}.
$$
Step 5: Substitute the Densities to Obtain $K$
The weight $W = V\,\rho_\text{sphere}\,g$ and $W - B = V\,(\rho_\text{sphere} - \rho_\text{liquid})\,g.$ Hence,
$$
K = \frac{\rho_\text{sphere}}{\rho_\text{sphere} - \rho_\text{liquid}}.
$$
Given:
$ \rho_\text{sphere} = 1.6\,\text{g cm}^{-3}, \quad \rho_\text{liquid} = 0.8\,\text{g cm}^{-3}.$
Therefore,
$$
K = \frac{1.6}{1.6 - 0.8}
= \frac{1.6}{0.8}
= 2.
$$
Step 6: Final Answer
Hence, the dielectric constant of the liquid is 2.
