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A vertical thin uniform rod is held in equilibrium after immersing completely by applying a vertically downward force at its top end as shown. Net compression in the rod is (length of the rod is L, its density is $ρ$ and Young’s modulus of rod’s material is Y and density of fluid is $2 ρ$ )

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$\frac{ρ L^{2} g}{2 Y}$

$\frac{ρ L^{2} g}{4 Y}$

$\frac{3 ρ L^{2} g}{2 Y}$

$\frac{2 ρ L^{2} g}{Y}$

answer is C.

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Detailed Solution

$F + A l ρ g = A l 2 ρ g$

$\Rightarrow F = A l ρ g$

At a point which is at a distance of x from bottom , $T_{x} = m g + \frac{m g}{l} x$

Strain in dx $= \frac{Δ x}{d x}$ $= \frac{T_{x}}{A Y} \Rightarrow Δ x = \frac{(m g + \frac{m g}{l} x) d x}{A Y}$

Net Compression $= {[\frac{mgx + \frac{mg}{l} \frac{x^{2}}{2}}{AY}]}_{0}^{l}$ $= \frac{3}{2} \frac{m g l}{A Y} = \frac{3}{2} ρ g \frac{l^{2}}{Y}$

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