Rankers Physics

miscellaneous: Practice Problem & Solution

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 (2006)
$T \propto \frac{1}{\sqrt{m}}$
$T \propto \sqrt{\rho}$
$T \propto \frac{1}{\sqrt{A}}$
$T \propto \frac{1}{\rho}$

Solution Explained:

To solve this problem, we apply the core principles of miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:

Restoring force on the block is $F = -(\rho A g)y$. Acceleration $a = F/m = -(\frac{\rho A g}{m})y$. This is SHM with $\omega^2 = \frac{\rho A g}{m}$. Time period $T = 2\pi\sqrt{\frac{m}{\rho A g}}$. Therefore, $T \propto \frac{1}{\sqrt{A}}$.

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