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 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 is :

• A. $\frac{Mg}{k}(1+\frac{LA\sigma }{M})$
• B. $\frac{Mg}{k}$
• C. $\frac{Mg}{k}(1-\frac{LA\sigma }{2M})$
• D. $\frac{Mg}{k}(1-\frac{LA\sigma }{M})$

Answer 1

The correct option is C $\frac{Mg}{k}(1-\frac{LA\sigma }{2M})$

Let $k$ be the spring constant of spring and it gets extended by length $x_0$ in the equilibrium position.

In equilibrium,


Using force balance along vertical direction,

$k{x}_0+F_B=Mg$

$\Rightarrow k{x}_0+\sigma \frac{L}{2}Ag=Mg$

$\Rightarrow x_0=\frac{Mg-\frac{\sigma LAg}{2}}{k}$

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

Hence, option $\left(C\right)$ is correct.