The buoy, weighing 30 kg, floats on the water. The volume of the above-water part is 12 dm3.

The buoy, weighing 30 kg, floats on the water. The volume of the above-water part is 12 dm3. Determine the volume of the entire buoy.

The volume of the whole buoy is equal to:
V = V1 + V2, where V1 is the volume of the above-water part of the buoy (V1 = 12 dm ^ 3 = 0.012 m ^ 3), V2 is the volume of the underwater part of the buoy.
Since the system is in equilibrium: Fa = Fт.
Fa – the strength of Archimedes; Fa = ρ * g * V2, where ρ is the density of water (ρ = 1000 kg / m ^ 3), g is the acceleration of gravity (we take g = 10 m / s ^ 2).
Ft is the force of gravity; Fт = m * g, where m is the mass of the buoy (m = 30 kg).
m * g = ρ * g * V2.
V2 = m * g / (ρ * g) = 30 * 10 / (1000 * 10) = 0.03 m ^ 3.
V = V1 + V2 = 0.012 + 0.03 = 0.042 m ^ 3.
Answer: The volume of the buoy is 0.042 m ^ 3.