A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density `sigma` at equilibrium position. The extension `x_0` of the spring when it is in equlibrium is:

Options

A. `(Mg)/(k)(1+(Lasigma)/(M))`
B. `(Mg)/(k)`
C. `(Mg)/(k)(1+(Lasigma)/(M))`
D. `(Mg)/(k)(1-(Lasigma)/(2M))`

Explanation

To find the extension \( x_0 \) of the spring when the cylinder is in equilibrium, we can follow these steps:

### Step 1: Identify the forces acting on the cylinder
In equilibrium, the forces acting on the cylinder include:
- The weight of the cylinder, \( W = Mg \) (downward).
- The spring force, \( F_s = kx_0 \) (upward).
- The buoyant force, \( F_B \) (upward).

### Step 2: Write the equilibrium condition
At equilibrium, the sum of the forces acting on the cylinder must be zero. Therefore, we can write the equation:
\[ Mg = kx_0 + F_B \]

### Step 3: Calculate the buoyant force
The buoyant force \( F_B \) can be calculated using Archimedes' principle. The volume of the submerged part of the cylinder is:
\[ V = A \cdot \frac{L}{2} \]
Thus, the buoyant force is given by:
\[ F_B = \sigma \cdot V \cdot g = \sigma \cdot \left(A \cdot \frac{L}{2}\right) \cdot g = \frac{\sigma ALg}{2} \]

### Step 4: Substitute the buoyant force into the equilibrium equation
Substituting \( F_B \) into the equilibrium equation gives:
\[ Mg = kx_0 + \frac{\sigma ALg}{2} \]

### Step 5: Solve for the extension \( x_0 \)
Rearranging the equation to solve for \( x_0 \):
\[ kx_0 = Mg - \frac{\sigma ALg}{2} \]
\[ x_0 = \frac{Mg - \frac{\sigma ALg}{2}}{k} \]

### Step 6: Factor out common terms
Factoring out \( g \) gives:
\[ x_0 = \frac{g}{k} \left(M - \frac{\sigma AL}{2}\right) \]

### Final Expression
Thus, the extension \( x_0 \) of the spring when the cylinder is in equilibrium is:
\[ x_0 = \frac{g}{k} \left(M - \frac{\sigma AL}{2}\right) \]