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

Answer 1

Answer: (C)

Let $x_0$ = extension in equilibrium position
$F$ = Restoring force in the spring
$F=K{x}_0$
The net downward force acting on the cylinder
= Mg - upward thrust
$=\text{Mg}-\left(\text{L}/2\right)\times \text{A}\times \sigma \times \text{g}$
For equilibrium ${\left(\text{Kx}\right)}_{\text{o}}=\text{Mg}-\left(\text{LA }\sigma \text{g}/2\right)$
${\left(\text{Kx}\right)}_{\text{o}}=\text{Mg}\left[1-\left(\text{LA}\sigma /2\text{M}\right)\right]$
$x_o=\left(\frac{Mg}{K}\right)\left[1-\left(\frac{LA\sigma }{2M}\right)\right]$