Question

A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi \times 10^{-2}$ s, where the value of y is (Acceleration due to gravity, g = 10 m/s², density of water = $10^3$ kg/m³)

• A. 2
• B. 6
• C. 4
• D. 1

Answer 1

Answer: (A)

2

Answer 2

For a floating body oscillating vertically, a small displacement x changes the displaced volume by A·x (where A is the cross‐sectional area), so the additional buoyant force is

$\Delta F = \rho g A x$.

This gives a restoring force $F = -\rho g A\, x$. Thus, the effective spring constant is

$k_{\text{eff}} = \rho g A$.

The equation of motion is:

$m\ddot{x} + \rho g A\, x = 0 \quad \Longrightarrow \quad \omega = \sqrt{\frac{\rho g A}{m}}$.

The time period is:

$T = 2\pi\sqrt{\frac{m}{\rho g A}}$.

Given:

• Mass $m = 10\,\text{g} = 0.01\,\text{kg}$
• Side length = 10 cm $\implies A = (0.1)^2 = 0.01\,\text{m}^2$
• $\rho = 10^3\, \text{kg/m}^3$
• $g = 10\, \text{m/s}^2$

Substitute:

$T = 2\pi\sqrt{\frac{0.01}{10^3 \times 10 \times 0.01}} = 2\pi\sqrt{\frac{0.01}{100}} = 2\pi\sqrt{10^{-4}} = 2\pi\times10^{-2}$.

The problem states $T = y\pi \times 10^{-2}$ s so comparing gives $y=2$.