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 equilibrium position is:

Step 1: Identify the forces acting on the cylinder at equilibrium.

The forces acting on the cylinder are the downward weight, the upward spring force, and the upward buoyant force.

$$W = Mg$$ $$F_s = kx_0$$ $$F_B = \sigma g V_{\text{sub}}$$

Step 2: Determine the submerged volume and buoyant force.

Since the cylinder is half submerged, the submerged volume is half of the total volume.

$$V_{\text{sub}} = A \cdot \frac{L}{2}$$

Substitute this volume into the buoyant force formula.

$$F_B = \frac{\sigma g A L}{2}$$

Step 3: Set up the equilibrium equation.

At equilibrium, the net force on the cylinder is zero.

$$Mg = kx_0 + F_B$$

Substitute the expression for $F_B$ into the equation.

$$Mg = kx_0 + \frac{\sigma g A L}{2}$$

Step 4: Solve for the spring extension $x_0$.

Isolate the spring force term.

$$kx_0 = Mg - \frac{\sigma g A L}{2}$$

Factor out $Mg$ from the right-hand side.

$$kx_0 = Mg \left(1 - \frac{LA\sigma}{2M}\right)$$

Divide by the spring constant $k$.

$$x_0 = \frac{Mg}{k} \left(1 - \frac{LA\sigma}{2M}\right)$$

Answer: $\dfrac{Mg}{k}\left(1 - \dfrac{LA\sigma}{2M}\right)$