A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is y π × 10⁻² s, where the value of y is (Acceleration due to gravity, g=10 m / s², density of water =10³ kg / m³ )

Solution

Physics sir · 4 steps

Push the cube down by x and the extra buoyancy is your spring force, so find k from water, not a spring.

1. Restoring force. Push the cube down by x from its floating position. It displaces an extra volume Ax of water, so the extra upthrust is ρ A g x, pointing back up. This force is proportional to x and opposite to it, so the motion is SHM with k = ρ g A.
2. Find k. The face area is A = (0.1 m)² = 10⁻² m². So k = 10³ × 10 × 10⁻² = 100 N m⁻¹, and the mass is m = 10 g = 10⁻² kg.

   T = 2π√(m/k) = 2π√(10⁻²/100) = 2π × 10⁻² s

3. Quick check. At equilibrium the cube sinks to depth h = m/(ρ A) = 10⁻³ m. Then T = 2π√(h/g) = 2π√10⁻⁴ s = 2π × 10⁻² s, the same result.
4. So. T = 2π × 10⁻² s, so y = 2, option (a).

The idea to keep. For any floating body bobbing up and down, the effective spring constant is ρ g A, where A is the cross-section at the waterline. Then use T = 2π√(m/k).

Answer from the NTA final answer key. Solution by PrepWiser.