Question

A rectangular block of mass 'm' and cross-sectional area 'A' floats on a liquid of density '\(ρ\)'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency (g=acceleration due to gravity)

• A. \(2π\sqrt{\frac{\text{m}}{\text{A}ρ\text{g}}}\)
• B. \(2π\sqrt{\frac{\text{A}ρ\text{g}}{\text{m}}}\)
• C. \(\frac{1}{2π}\sqrt{\frac{\text{A}ρ\text{g}}{\text{m}}}\)
• D. \(\frac{1}{2π}\sqrt{\frac{\text{m}}{\text{A}ρ\text{g}}}\)

Hint:

The extra buoyant force is A rho g times the displacement, like a spring of constant A rho g.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
At equilibrium, weight equals buoyant force. If the block is pushed down a small distance \(y\), it displaces extra liquid volume \(Ay\).

Step 2: Restoring force:
\[ F = -A\rho g\,y \]
This has the form \(F = -ky\) with \(k = A\rho g\), so the motion is simple harmonic.

Step 3: Frequency:
\[ \omega = \sqrt{\frac km} = \sqrt{\frac{A\rho g}{m}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{A\rho g}{m}} \]

Step 4: Check the options:
Option (C) matches. Options (A) and (B) are time periods with \(2\pi\) in the numerator, and (D) has the ratio upside down.

Final Answer:
The extra buoyant force acts like a spring of constant A rho g. \[ \boxed{\text{(C) }\dfrac{1}{2\pi}\sqrt{\dfrac{A\rho g}{m}}} \]