A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length

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9-1.Fluid Mechanics

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A uniform cylinder of length $L$ and mass $M$ having crosssectional 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

$\frac{{Mg}}{k}$

B

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

C

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

D

$\;\frac{{Mg}}{k}\left( {1 + \frac{{LA\sigma }}{M}} \right)$

(JEE MAIN-2013)

Solution

From figure, $k{x_0} + {F_B} = Mg$

$k{x_0} + \sigma \frac{L}{2}Ag = Mg$

$\left[ {mass = density \times volume} \right]$

$ \Rightarrow k{x_0} = Mg – \sigma \frac{L}{2}Ag$

$ \Rightarrow {x_0} = \frac{{Mg – \frac{{\sigma LAg}}{2}}}{k} = \frac{{Mg}}{k}\left( {1 – \frac{{LA\sigma }}{{2M}}} \right)$

Standard 11

Physics

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