Question

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 oscillation with a time period $T$, then :

• A. $T \propto \sqrt{ m }$
• B. $ T \propto \sqrt \rho $
• C. $T \propto \frac{1}{\sqrt{ A }}$
• D. $ T \propto \frac{1}{\rho} $

Answer 1

The Correct Option is D

To understand the oscillation time period $T$ of a floating block in a liquid, we need to consider the principles of buoyancy and simple harmonic motion.

When a block of cross-sectional area $A$ and mass $m$ floats on a liquid of density $\rho$, it displaces some volume of the liquid equal to its own weight at equilibrium.

Let's briefly analyze why the time period $T \propto \frac{1}{\rho}$ is the correct answer:

1. When the block is slightly displaced vertically, it will experience a restoring force due to the change in the buoyant force. This leads to a simple harmonic motion.
2. The buoyant force can be expressed as: $$ F_b = \rho g V $$ where $V$ is the submerged volume and $g$ is the acceleration due to gravity.
3. For small displacements, the change in buoyant force provides the restoring force: $$ F = -k \cdot \Delta x $$ where $k$ is the effective spring constant related to the buoyant force changes, and $\Delta x$ is the displacement.
4. The period of oscillation $T$ is related to the effective mass and the effective spring constant: $$ T = 2\pi \sqrt{\frac{m}{k}} $$
5. For buoyancy-driven systems, $k$ is directly proportional to $\rho A g$ because: $$ k = \rho A g $$ Thus, we can modify the time period expression to: $$ T = 2\pi \sqrt{\frac{m}{\rho A g}} $$ Simplifying further, it becomes: $$ T \propto \frac{1}{\rho} $$
6. Thus, this shows that the oscillation time period is inversely proportional to the density of the liquid, confirming $T \propto \frac{1}{\rho}$ is indeed correct.

Therefore, among the given options, the correct one is $T \propto \frac{1}{\rho}$.