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A uniform cylinder of length L and mass M having cross sectional area A is suspended, with its length vertical, from fixed point by a mass less spring such that it is half sub merged in a liquid of density $σ$ at equilibrium position. The extension x₀ of the spring when it is in equilibrium is

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$\frac{Mg}{k}$

$\frac{Mg}{k} (1 - \frac{LAσ}{M})$

$\frac{Mg}{k} (1 - \frac{L G Aσ}{2 M})$

$\frac{Mg}{k} (1 + \frac{LAσ}{M})$

answer is C.

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Detailed Solution

In the equilibrium position, the spring is expanded by length $x_{0}$ and has a spring constant of k,

thus

${Kx}_{0} + F_{B} = Mg$

${Kx}_{0} + σ \frac{L}{2} AG = Mg$

${Kx}_{0} = Mg - σ \frac{L}{2} AG$

$x_{0} = \frac{Mg - σ \frac{L}{2} AG}{K}$

$x_{0} = \frac{Mg}{K} (1 - \frac{σLAG}{2 M})$

Hence the correct answer is $\frac{Mg}{K} (1 - \frac{σLAG}{2 M})$

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