A wooden cube (density of wood ‘d’) of side ‘ $l$ ’ floats in a liquid of density ‘ $ρ$ ’ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period ‘T’

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$2 π \sqrt{\frac{l d}{ρ g}}$

$2 π \sqrt{\frac{l ρ}{d g}}$

$2 π \sqrt{\frac{l d}{(ρ - d) g}}$

$2 π \sqrt{\frac{l ρ}{(ρ - d) g}}$

answer is A.

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Detailed Solution

At equilibrium $F_{b} = m g$
$ρ A l_{0} g = d A \lg$
Restoring force,
F=mg- ${F_{b}}^{'}$
F=mg- $ρ A (l_{0} + x) g$
$\begin{array}{l} d A l a = d A \lg - ρ A l_{0} g - ρ g A x \\ a = - \frac{ρ g}{d l} x \end{array}$
Therefore, wooden cube performs S.H.M
$∴ ω = \sqrt{\frac{ρ g}{d l}} \Rightarrow T = 2 π \sqrt{\frac{l d}{ρ g}}$