A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then (a) T ∝ 1/√(m) (b) T ∝ √(ρ) (c) T ∝ 1/√(A) (d) T ∝ 1/ρ

Answer: (c). Extra depth x gives restoring force Aρ g x, so keff = Aρ g. T = 2π√((m)/(Aρ g)): T ∝ √(m), T ∝ (1)/(√(ρ)), T ∝ 1/√(A). So the answer is (c).

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A rectangular block of mass m and area of cross-section A floats in a liquid of density $\rho$ . If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then

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Answer: (c)

Extra depth $x$ gives restoring force $A\rho g x$ , so $k_{eff} = A\rho g$ .
$T = 2\pi\sqrt{\frac{m}{A\rho g}}$ : $T \propto \sqrt{m}$ , $T \propto \frac{1}{\sqrt{\rho}}$ , $T \propto \frac{1}{\sqrt{A}}$ .
So the answer is (c).

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