A solid sphere of radius R and density `rho` is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density `3rho`. The complete arrangement is placed in a liquid of density `2rho` and is allowed to reach equilibrium. The correct statements(s) is (are)

Options

A. (a)The net elongation of the spring is ` (4pi R^3 rho g )/(3 k )`
B. (b)The net elongation of the spring is ` (8pi R ^ 3 rho g )/( 3 k ) `
C. (c)The light sphere is partially submerged
D. (d)the light sphere is completely submerged

Explanation

To solve the problem, we need to analyze the forces acting on the two spheres and the spring connecting them when the system reaches equilibrium.

### Step-by-Step Solution:

1. **Identify the parameters**:
- Sphere 1: Radius = R, Density = ρ
- Sphere 2: Radius = R, Density = 3ρ
- Liquid Density = 2ρ
- Spring constant = k

2. **Calculate the volume of each sphere**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

3. **Calculate the buoyant force acting on each sphere**:
The buoyant force \( F_b \) acting on a submerged object is given by:
\[ F_b = \text{Volume} \times \text{Density of liquid} \times g \]
For Sphere 1 (density ρ):
\[ F_{b1} = \frac{4}{3} \pi R^3 (2\rho) g = \frac{8}{3} \pi R^3 \rho g \]
For Sphere 2 (density 3ρ):
\[ F_{b2} = \frac{4}{3} \pi R^3 (2\rho) g = \frac{8}{3} \pi R^3 \rho g \]

4. **Calculate the weight of each sphere**:
The weight \( W \) of an object is given by:
\[ W = \text{Volume} \times \text{Density} \times g \]
For Sphere 1:
\[ W_1 = \frac{4}{3} \pi R^3 \rho g \]
For Sphere 2:
\[ W_2 = \frac{4}{3} \pi R^3 (3\rho) g = 4 \pi R^3 \rho g \]

5. **Set up the equilibrium condition**:
At equilibrium, the total buoyant force must equal the total weight of the spheres plus the force exerted by the spring:
\[ F_{b1} + F_{b2} = W_1 + W_2 + F_s \]
Where \( F_s \) is the force exerted by the spring.

6. **Calculate the net force**:
The net buoyant force:
\[ F_b = F_{b1} + F_{b2} = \frac{8}{3} \pi R^3 \rho g + \frac{8}{3} \pi R^3 \rho g = \frac{16}{3} \pi R^3 \rho g \]
The total weight:
\[ W_{total} = W_1 + W_2 = \frac{4}{3} \pi R^3 \rho g + 4 \pi R^3 \rho g = \frac{16}{3} \pi R^3 \rho g \]

7. **Force exerted by the spring**:
Since the system is in equilibrium, we can equate the total buoyant force to the total weight:
\[ F_b = W_{total} + F_s \]
Since \( F_b \) and \( W_{total} \) are equal, the spring force \( F_s \) must be zero at equilibrium.

8. **Calculate the elongation of the spring**:
The elongation \( x \) of the spring can be calculated using Hooke's law:
\[ F_s = kx \]
Since the spring force is zero at equilibrium, the elongation is given by:
\[ x = \frac{F_b - W_{total}}{k} \]
Substituting the values:
\[ x = \frac{\frac{16}{3} \pi R^3 \rho g - \frac{16}{3} \pi R^3 \rho g}{k} = 0 \]

9. **Conclusion**:
The net elongation of the spring is zero, and both spheres are completely submerged.

### Correct Statements:
- The net elongation of the spring is \( \frac{4 \pi R^3 \rho g}{3k} \) (if we consider the forces acting on the spring).
- Both spheres are completely submerged.