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

To determine the time period of oscillations for a cube undergoing simple harmonic motion in water, principles from fluid mechanics and simple harmonic motion are applied.

The cube floats, indicating its weight is balanced by its buoyant force, consistent with Archimedes' principle.

The time period \( T \) for simple harmonic motion is defined by:

\(T = 2\pi \sqrt{\frac{m}{k}}\)

For this floating object:

• m represents the mass of the cube.
• k signifies the effective spring constant, derived from the buoyant force.

The buoyant force on the cube, when displaced by a small distance \( x \), is given by:

\(F_b = - \rho g A x\)

In this equation:

• \(\rho = 1000 \, \text{kg/m}^3\) is the density of water.
• \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity.
• \(A = 0.1 \, \text{m} \times 0.1 \, \text{m} = 0.01 \, \text{m}^2\) is the cross-sectional area of the cube.

This force corresponds to an effective spring constant \(k\):

\(k = \rho g A\)

Substituting the values yields:

\(k = 1000 \times 10 \times 0.01 = 100 \, \text{N/m}\)

The mass of the cube \(m\) is 10 g, equivalent to 0.01 kg.

Substituting \( m \) and \( k \) into the time period formula results in:

\(T = 2\pi \sqrt{\frac{0.01}{100}}\)

\(T = 2\pi \sqrt{0.0001} = 2\pi \times 0.01 = 0.02\pi \, \text{s}\)

This result aligns with the expression \( y \pi \times 10^{-2} \, \text{s}\), with \( y \) being 2.

Therefore, the value of \( y \) is 2.