JEE Main · 2013 · 07 Apr 2013 · Offline · Fluid Mechanics

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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 a fixed point by a massless spring, such that it is half submerged in a liquid of density

σ

at equilibrium position. The extension

x_{0}

of the spring when it is in equilibrium is:
(Here k is spring constant)

(A) $\frac{Mg}{k} (1 - \frac{LA σ}{M})$
(B) $\frac{Mg}{k} (1 - \frac{LA σ}{2 M})$ ✓ Correct
(C) $\frac{Mg}{k} (1 + \frac{LA σ}{M})$
(D) $\frac{Mg}{k}$

Answer: (B)

Solution

At equilibrium the spring tension together with the upward buoyant thrust must hold up the full weight of the cylinder. Only half the cylinder of length

L

and area

A

dips into the liquid, so the buoyant force is

F_{b} = \frac{A L}{2} σ g

. Balancing forces,

k x_{0} + \frac{A L σ g}{2} = M g

, and solving for the extension gives

x_{0} = \frac{M g}{k} (1 - \frac{L A σ}{2 M})

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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JEE Main · 2013 · 07 Apr 2013 · Offline · Fluid Mechanics

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