Question

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 of constant $K$ such that it is half-submerged in the liquid of density $\sigma $ in the equilibrium position. The extension of the spring $x_{0}$ in equilibrium position is

• A. $Mg/K$
• B. $\left(\right.Mg/K\left.\right)$ $\left(1 \text{-} \textit{LA} \sigma / \textit{M}\right)$
• C. $\left(\textit{Mg} / \textit{K}\right)\left[1 - \left(\textit{LA} \sigma / 2 \textit{M}\right)\right]$
• D. $\textit{MgLA}/\textit{K}\sigma $

Answer 1

Answer: (C)

Let $x_{0}$ = extension in equilibrium position
$F$ = Restoring force in the spring
$F=Kx_{0}$
The net downward force acting on the cylinder
= Mg - upward thrust
$=\textit{Mg}-\left(\textit{L} / 2\right)\times \textit{A}\times \sigma \times \textit{g}$
For equilibrium $\left(\textit{Kx}\right)_{\text{o}}=\textit{Mg}-\left(\textit{LA } \sigma \textit{g} / 2\right)$
$\left(\textit{Kx}\right)_{\text{o}}=\textit{Mg}\left[1 - \left(\textit{LA} \, \sigma / 2 \textit{M}\right)\right]$
$x_{o}=\left(\frac{M g}{K}\right)\left[1 - \left(\frac{L A \sigma }{2 M}\right)\right]$