Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of `30^@` with each other. When suspended in a liquid of density `0.8gcm^-3`, the angle remains the same. If density of the material of the sphere is `1.6gcm^-3`, the dielectric constant of the liquid is

Options

A. (a) 4
B. (b) 3
C. (c) 2
D. (d) 1

Explanation

To solve the problem, we need to analyze the forces acting on the charged spheres both in air and in the liquid. Let's break it down step by step.

### Step 1: Analyze the forces acting on the spheres in air

1. **Identify the forces**: Each sphere experiences:
- Gravitational force (weight) \( W = mg \) acting downwards.
- Tension \( T \) in the string acting along the string.
- Electric force \( F \) due to the repulsion between the two identical charged spheres.

2. **Resolve the forces**: Since the strings make an angle of \( 30^\circ \) with each other, we can resolve the forces into components:
- The angle between the tension and the vertical is \( 15^\circ \) (since the total angle is \( 30^\circ \)).
- The vertical component of tension: \( T \cos(15^\circ) \).
- The horizontal component of tension (which balances the electric force): \( T \sin(15^\circ) \).

3. **Set up the equilibrium equations**:
- In the vertical direction:
\[ T \cos(15^\circ) = mg \]
- In the horizontal direction:
\[ T \sin(15^\circ) = F \]

### Step 2: Analyze the forces acting on the spheres in the liquid

1. **Identify the forces in the liquid**: When the spheres are submerged in a liquid, the forces acting on each sphere are:
- Gravitational force \( W = mg \) acting downwards.
- Tension \( T' \) in the string acting along the string.
- Buoyant force \( F_b \) acting upwards.
- Electric force \( F \) remains the same as before.

2. **Buoyant force**: The buoyant force can be calculated using Archimedes' principle:
\[ F_b = \rho_{liquid} V g \]
where \( \rho_{liquid} = 0.8 \, \text{g/cm}^3 \) and \( V \) is the volume of the sphere.

3. **Set up the equilibrium equations in the liquid**:
- In the vertical direction:
\[ T' \cos(15^\circ) + F_b = mg \]
- In the horizontal direction:
\[ T' \sin(15^\circ) = F \]

### Step 3: Relate the two scenarios

1. **From the first scenario**:
\[ \frac{F}{mg} = \tan(15^\circ) \quad \text{(1)} \]

2. **From the second scenario**:
\[ T' \cos(15^\circ) + \rho_{liquid} V g = mg \quad \text{(2)} \]
Substituting \( F_b \):
\[ T' \cos(15^\circ) + \left(0.8 \times V \times g\right) = mg \]

3. **Substituting the buoyant force**:
\[ T' \cos(15^\circ) = mg - 0.8 \times V \times g \]
Rearranging gives:
\[ T' \cos(15^\circ) = mg \left(1 - \frac{0.8 \times V}{m}\right) \]

### Step 4: Solve for the dielectric constant

1. **Using the relationship between forces**:
From equations (1) and (2), we can relate the electric forces:
\[ \frac{T' \sin(15^\circ)}{T \sin(15^\circ)} = \frac{F_b}{F} \]
This leads us to:
\[ \frac{F_b}{F} = \frac{1}{2} \]

2. **Substituting the expressions for forces**:
The electric force in a medium with dielectric constant \( K \) is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{K \cdot r^2} \]
where \( r \) is the distance between the charges.

3. **Finding the dielectric constant**:
From the relationship between the forces, we find that:
\[ K = 2 \]

### Final Answer:
The dielectric constant of the liquid is \( K = 2 \).