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A thin uniform elastic rod of natural length ‘l’, area of cross-section A, density $ρ$ and Young’s modulus Y is just completely immersed in vertical position in fluid of density $2 ρ$ by applying a vertically downward force (F) at its top end. The rod is in equilibrium. Elastic deformation energy stored in the rod is (Neglect lateral strain)

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$\frac{7 ρ^{2} A l^{3} g^{2}}{6 Y}$

$\frac{9 ρ^{2} A l^{3} g^{2}}{8 Y}$

$\frac{3 ρ^{2} A l^{3} g^{2}}{4 Y}$

$\frac{4 ρ^{2} A l^{3} g^{2}}{3 Y}$

answer is A.

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Detailed Solution

$F_{t o p} = m g ↓, F_{b o t t o m} = 2 m g ↑ U = \int \frac{σ^{2}}{2 Y} A d y$

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