Question 105 of 210

A uniform cylinder of length $\text{L}$ and mass $\text{M}$ having cross-sectional area $\text{A}$ 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 ${\text{x}}_{0}$ of the spring when it is in equilibrium is :

A$\dfrac{\text{Mg}}{\text{k}}\left(1-\dfrac{\text{LA}\sigma }{2\text{M}}\right)$B$\dfrac{\text{Mg}}{\text{k}}\left(1+\dfrac{\text{LA}\sigma }{\text{M}}\right)$C$\dfrac{\text{Mg}}{\text{k}}$D$\dfrac{\text{Mg}}{\text{k}}\left(1-\dfrac{\text{LA}\sigma }{\text{M}}\right)$

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Answer: (A) $\dfrac{\text{Mg}}{\text{k}}\left(1-\dfrac{\text{LA}\sigma }{2\text{M}}\right)$

${\text{kx}}_{0}+{\text{F}}_{\text{B}}=\text{mg}$

${\text{kx}}_{0}+\sigma \dfrac{\text{L}}{2}\text{Ag}=\text{Mg}$


${\text{x}}_{0}=\dfrac{\text{Mg}-\dfrac{\sigma \text{LAg}}{2}}{\text{k}}$

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