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. 2
B. 4
C. 2.5
D. 3.5

Explanation

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

### Step 1: Analyze the system in air
When the two identical charged spheres are suspended in air, they experience the following forces:
- The gravitational force (weight) acting downwards: \( mg \)
- The tension in the strings, which can be resolved into two components:
- Vertical component: \( T \cos(15^\circ) \)
- Horizontal component: \( T \sin(15^\circ) \)

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

From the equilibrium conditions, we can write:
1. In the horizontal direction:
\[ T \sin(15^\circ) = F_e \]
2. In the vertical direction:
\[ T \cos(15^\circ) = mg \]

### Step 2: Relate the two equations
Dividing the two equations gives:
\[ \tan(15^\circ) = \frac{F_e}{mg} \]
Substituting \( F_e \):
\[ \tan(15^\circ) = \frac{k Q^2 / r^2}{mg} \]

### Step 3: Analyze the system in the liquid
When the spheres are submerged in a liquid, they experience an additional buoyant force \( F_b \) acting upwards. The buoyant force is given by:
\[ F_b = \rho_{\text{liquid}} V g \]
where \( \rho_{\text{liquid}} \) is the density of the liquid and \( V \) is the volume of the sphere.

The weight of the sphere is now effectively reduced by the buoyant force:
\[ T \cos(15^\circ) = mg - F_b \]

In the horizontal direction, the electrostatic force in the liquid \( F_e' \) is given by:
\[ F_e' = \frac{k Q^2}{\varepsilon_r r^2} \]
where \( \varepsilon_r \) is the dielectric constant of the liquid.

### Step 4: Set up the new equilibrium equations
1. In the horizontal direction:
\[ T \sin(15^\circ) = F_e' \]
2. In the vertical direction:
\[ T \cos(15^\circ) = mg - F_b \]

Dividing these two equations gives:
\[ \tan(15^\circ) = \frac{F_e'}{mg - F_b} \]

### Step 5: Substitute the expressions for forces
Substituting for \( F_e' \) and \( F_b \):
\[ \tan(15^\circ) = \frac{k Q^2 / \varepsilon_r r^2}{mg - \rho_{\text{liquid}} V g} \]

### Step 6: Relate the two scenarios
Since the angle remains the same, we can equate the two expressions for \( \tan(15^\circ) \):
\[ \frac{k Q^2 / r^2}{mg} = \frac{k Q^2 / \varepsilon_r r^2}{mg - \rho_{\text{liquid}} V g} \]

### Step 7: Simplify and solve for \( \varepsilon_r \)
Cross-multiplying gives:
\[ (mg - \rho_{\text{liquid}} V g) = \varepsilon_r mg \]
Rearranging leads to:
\[ mg - \rho_{\text{liquid}} V g = \varepsilon_r mg \]
\[ mg(1 - \varepsilon_r) = \rho_{\text{liquid}} V g \]
Dividing through by \( g \) and substituting \( V = \frac{m}{\rho_{\text{solid}}} \):
\[ m(1 - \varepsilon_r) = \rho_{\text{liquid}} \frac{m}{\rho_{\text{solid}}} \]
Cancelling \( m \) from both sides:
\[ 1 - \varepsilon_r = \frac{\rho_{\text{liquid}}}{\rho_{\text{solid}}} \]
Substituting the given densities:
\[ 1 - \varepsilon_r = \frac{0.8}{1.6} = \frac{1}{2} \]
Thus:
\[ \varepsilon_r = 1 - \frac{1}{2} = \frac{1}{2} \]

### Conclusion
The dielectric constant of the liquid is:
\[ \varepsilon_r = 2 \]