91Ó°ÊÓ

The cylinder is in equilibrium, which means the total force acting on it is zero. There are three forces acting on the cylinder: the weight acting downwards, the buoyancy force acting upwards (due to the submerged part of the cylinder) and the force exerted by the spring acting upwards. Let's express each of these forces: 1. The weight of the cylinder: \(F_{\text{weight}} = M \cdot g\) 2. The buoyancy force: since only half of the cylinder is submerged, the volume of the cylinder submerged in fluid is \(\frac{1}{2} \cdot A \cdot \ell\). Therefore, \(F_{\text{buoyancy}} = \rho \cdot \frac{1}{2} \cdot A \cdot \ell \cdot g\), where \(\rho\) is the density of the fluid. 3. The spring force: according to Hooke's law, \(F_{\text{spring}} = k \cdot x\), where \(x\) is the extension of the spring.

Since the cylinder is in equilibrium, the sum of forces acting on it is zero. Therefore, we can write the following equation: \(F_{\text{spring}} = F_{\text{weight}} - F_{\text{buoyancy}}\) Substituting the expressions for the forces from Step 1: \(k \cdot x = M \cdot g - \rho \cdot \frac{1}{2} \cdot A \cdot \ell \cdot g\)

First, it is necessary to rearrange the equation to solve for \(x\): \(k \cdot x = g \cdot M(1 - \frac{\rho \cdot A \cdot l}{2 \cdot M})\) Then, divide by the spring constant \(k\) to find the extension \(x\): \(x = \frac{M \cdot g}{k}(1 - \frac{\rho \cdot A \cdot l}{2 \cdot M})\) So, the spring extension when the system is in equilibrium is given by \(\frac{M \cdot g}{k}(1 - \frac{A \cdot \ell \rho}{2 \cdot M})\). Comparing with the given options, it matches with (B), hence it's the correct option.