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, \( g = 10 \, \text{m/s}^2 \), density of water = \( 10^3 \, \text{kg/m}^3 \))

• A. 2
• B. 6
• C. 4
• D. 1

Hint:

For oscillations of floating objects, use the formula \( T = 2\pi \sqrt{\frac{m}{L^2 \rho g}} \) to calculate the time period of oscillation. Remember to convert all units into SI units.

Answer 1

The Correct Option is A

The time period \( T \) of oscillations for a floating object undergoing simple harmonic motion is given by: \[ T = 2 \pi \sqrt{\frac{m}{L^2 \rho g}}, \] where: - \( m \) is the mass of the cube (10g = 0.01 kg), - \( L \) is the length of the cube's side (10 cm = 0.1 m), - \( \rho \) is the density of water (1000 kg/m\(^3\)), - \( g \) is the acceleration due to gravity (10 m/s\(^2\)). Now, we calculate the time period: \[ T = 2 \pi \sqrt{\frac{0.01}{(0.1)^2 \times 1000 \times 10}} = 2 \pi \sqrt{\frac{0.01}{0.1^2 \times 10000}} = 2 \pi \sqrt{\frac{0.01}{10}} = 2 \pi \sqrt{10^{-3}}. \] Thus, the time period of oscillation is \( y \pi \times 10^{-2} \), and solving for \( y \), we get: \[ y = 2. \] Thus, the correct answer is: \[ \boxed{2}. \]