A uniform cylinder of length L and mass M hav | vedclass

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(D) At equilibrium,the forces acting on the cylinder are the gravitational force $(Mg)$ acting downwards,the spring force $(kx_0)$ acting upwards,and

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) At equilibrium,the forces acting on the cylinder are the gravitational force $(Mg)$ acting downwards,the spring force $(kx_0)$ acting upwards,and the buoyant force $(F_B)$ acting upwards.The buoyant force is given by $F_B = V_{submerged} \cdot \sigma \cdot g$,where $V_{submerged} = A \cdot \frac{L}{2}$.Thus,$F_B = \frac{LA\sigma g}{2}$.For the cylinder to be in equilibrium,the net force must be zero:$kx_0 + F_B = Mg$Substituting the value of $F_B$:$kx_0 + \frac{LA\sigma g}{2} = Mg$$kx_0 = Mg - \frac{LA\sigma g}{2}$$x_0 = \frac{Mg - \frac{LA\sigma g}{2}}{k}$$x_0 = \frac{Mg}{k} \left( 1 - \frac{LA\sigma}{2M} \right)$

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