If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant force?

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

Part B

The buoyant force is approximately 40.93 N

Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg

The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.

see earlier answer