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)(1-(LAsigma)/(2M))`
C. `(Mg)/(k)(1+(LAsigma)/(2M))`
D. `(Mg)/(k)`

Answer

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

### Step 1: Understand the forces acting on the cylinder
The forces acting on the cylinder when it is in equilibrium are:
- The weight of the cylinder \( W = Mg \) acting downwards.
- The buoyant force \( F_b \) acting upwards due to the liquid.
- The restoring force of the spring \( F_s = kx_0 \) acting upwards.

### Step 2: Write the expression for the buoyant force
The buoyant force \( F_b \) can be calculated using Archimedes' principle:
\[ F_b = \sigma V_{submerged} g \]
Since half of the cylinder is submerged, the volume of the submerged part is:
\[ V_{submerged} = A \left(\frac{L}{2}\right) = \frac{AL}{2} \]
Thus, the buoyant force becomes:
\[ F_b = \sigma \left(\frac{AL}{2}\right) g \]

### Step 3: Set up the equilibrium condition
At equilibrium, the sum of the forces acting on the cylinder is zero. Therefore, we can write:
\[ F_s + F_b = W \]
Substituting the expressions for these forces:
\[ kx_0 + \sigma \left(\frac{AL}{2}\right) g = Mg \]

### Step 4: Rearranging the equation
Rearranging the equation to solve for \( x_0 \):
\[ kx_0 = Mg - \sigma \left(\frac{AL}{2}\right) g \]
\[ x_0 = \frac{Mg - \sigma \left(\frac{AL}{2}\right) g}{k} \]

### Step 5: Factor out \( g \)
Factoring \( g \) out of the numerator 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:
\[ x_0 = \frac{g}{k} \left(M - \frac{\sigma AL}{2}\right) \]