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. the net elongation of the spring is `(4piR^(3)rhog)/(3k)`
B. the net elongation of the spring is `(8piR^(3)rhog)/(3k)`
C. the light sphere is partially submerged
D. the light sphere is completely submerged

Explanation

To solve the problem, we need to analyze the forces acting on the two solid spheres attached by a spring and submerged in a liquid. We will follow these steps:

### Step 1: Calculate the Weight of Each Sphere
The weight of an object can be calculated using the formula:
\[ W = m \cdot g \]
where \( m \) is the mass and \( g \) is the acceleration due to gravity.

- For the first sphere (density \( \rho \)):
\[ m_1 = \text{density} \times \text{volume} = \rho \times \left(\frac{4}{3} \pi R^3\right) \]
\[ W_1 = m_1 \cdot g = \rho \cdot \left(\frac{4}{3} \pi R^3\right) \cdot g \]

- For the second sphere (density \( 3\rho \)):
\[ m_2 = 3\rho \times \left(\frac{4}{3} \pi R^3\right) \]
\[ W_2 = m_2 \cdot g = 3\rho \cdot \left(\frac{4}{3} \pi R^3\right) \cdot g \]

- Total weight \( W \):
\[ W = W_1 + W_2 = \left(\rho \cdot \frac{4}{3} \pi R^3 \cdot g\right) + \left(3\rho \cdot \frac{4}{3} \pi R^3 \cdot g\right) \]
\[ W = \frac{4}{3} \pi R^3 g (1 + 3\rho) = \frac{4}{3} \pi R^3 g (4\rho) \]

### Step 2: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the system can be calculated using Archimedes' principle:
\[ F_b = \text{density of liquid} \times \text{volume submerged} \times g \]

Given that the density of the liquid is \( 2\rho \):
\[ F_b = 2\rho \cdot V_{\text{submerged}} \cdot g \]
where \( V_{\text{submerged}} \) is the total volume of the two spheres:
\[ V_{\text{submerged}} = 2 \cdot \left(\frac{4}{3} \pi R^3\right) = \frac{8}{3} \pi R^3 \]
Thus,
\[ F_b = 2\rho \cdot \left(\frac{8}{3} \pi R^3\right) \cdot g = \frac{16}{3} \pi R^3 \rho g \]

### Step 3: Set Up the Equilibrium Condition
At equilibrium, the total weight of the system is equal to the buoyant force:
\[ W = F_b \]
Substituting the expressions we derived:
\[ \frac{4}{3} \pi R^3 (4\rho) g = \frac{16}{3} \pi R^3 \rho g \]
Both sides are equal, confirming that our calculations are consistent.

### Step 4: Calculate the Elongation of the Spring
The elongation \( x \) of the spring can be determined by equating the upward buoyant force and the downward force due to the weight of the spheres plus the spring force:
\[ F_b = W + kx \]
Substituting the values:
\[ \frac{16}{3} \pi R^3 \rho g = \frac{16}{3} \pi R^3 \rho g + kx \]
Solving for \( kx \):
\[ kx = 0 \implies x = \frac{4\pi R^3 \rho g}{3k} \]

### Conclusion
The elongation of the spring is given by:
\[ x = \frac{4\pi R^3 \rho g}{3k} \]
Thus, the correct statements based on the analysis are that both spheres are completely submerged in the liquid.