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. 4
B. 3
C. 2
D. 1

Answer

To solve the problem, we will analyze the forces acting on the charged spheres both in air and in the liquid. We will use the information given to derive the dielectric constant of the liquid.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical charged spheres suspended by strings making an angle of \(30^\circ\) with each other.
- Each string makes an angle of \(15^\circ\) with the vertical.

2. **Forces in Air**:
- The forces acting on one sphere are:
- Weight (\(mg\)) acting downward.
- Tension (\(T\)) in the string acting along the string.
- Electrostatic force (\(F_e\)) acting horizontally between the two spheres.
- We can resolve the tension into components:
- \(T \cos 15^\circ = mg\) (vertical component)
- \(T \sin 15^\circ = F_e\) (horizontal component)

3. **Electrostatic Force**:
- The electrostatic force between the two charged spheres is given by Coulomb's law:
\[ F_e = \frac{k q^2}{r^2} \]
- Where \(k\) is Coulomb's constant, \(q\) is the charge on each sphere, and \(r\) is the distance between the centers of the spheres.

4. **Setting Up the Equation**:
- From the tension equations, we can write:
\[ \frac{T \sin 15^\circ}{T \cos 15^\circ} = \frac{F_e}{mg} \implies \tan 15^\circ = \frac{k q^2}{mg r^2} \]

5. **Forces in the Liquid**:
- When the spheres are submerged in a liquid, the forces acting on them change slightly:
- The weight is still \(mg\), but we also have a buoyant force (\(F_b\)) acting upward.
- The buoyant force can be calculated as:
\[ F_b = \rho_{liquid} V g \]
- Given the density of the liquid is \(0.8 \, \text{g/cm}^3\) and the density of the sphere is \(1.6 \, \text{g/cm}^3\), the volume submerged is equal to the volume of the sphere.

6. **Weight in Liquid**:
- The effective weight of the sphere in the liquid becomes:
\[ mg' = mg - F_b = mg - \rho_{liquid} V g = mg - 0.8Vg \]
- Since the density of the sphere is \(1.6 \, \text{g/cm}^3\), we have:
\[ V = \frac{m}{1.6} \]
- Therefore, the buoyant force is:
\[ F_b = 0.8 \left(\frac{m}{1.6}\right) g = \frac{mg}{2} \]

7. **Setting Up the New Equations**:
- In the liquid, the tension equations become:
\[ T \cos 15^\circ = \frac{mg}{2} \quad \text{(1)} \]
\[ T \sin 15^\circ = k' \frac{q^2}{r^2} \quad \text{(2)} \]
- Dividing (2) by (1):
\[ \tan 15^\circ = \frac{k' q^2}{\frac{mg}{2} r^2} \]

8. **Equating the Two Expressions**:
- From the air and liquid conditions, we have:
\[ \frac{k q^2}{mg r^2} = \frac{2 k' q^2}{mg r^2} \]
- Canceling out common terms:
\[ k = 2 k' \]

9. **Finding the Dielectric Constant**:
- The relationship between \(k\) and \(k'\) is given by:
\[ k' = \frac{k}{2} \]
- Since \(k = \frac{1}{4 \pi \epsilon_0}\) and \(k' = \frac{1}{4 \pi \epsilon_0 \epsilon_r}\):
\[ \epsilon_r = 2 \]

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