Hollow aluminum ball with a volume of 120 m3. Meters floats in the water half submerged.

Hollow aluminum ball with a volume of 120 m3. Meters floats in the water half submerged. Find the volume of the cavity of the ball

Given:

V = 120 m ^ 3 – the volume of the ball floating in the water;

ro1 = 2700 kg / m ^ 3 is the density of aluminum;

ro2 = 1000 kg / m ^ 3 is the density of water.

It is required to find the volume of the cavity in the ball V1 (m ^ 3).

The weight of aluminum in the ball will be equal to:

P1 = ro1 * Vа * g = ro1 * g * (V – V1).

Since the ball floats in water, having sunk into it by half, then the weight of the displaced water will be equal to:

P2 = ro2 * V * g / 2.

According to Archimedes’ law:

P1 = P2

ro1 * g * (V – V1) = ro2 * V * g / 2

V – V1 = ro2 * V / (2 * ro1)

V1 = V – ro2 * V / (2 * ro1) = 120 – 1000 * 120 / (2 * 2700) = 120 – 120000/5400 =

= 120 – 22.2 = 97.8 m ^ 3.

Answer: the volume of the cavity in the ball is 97.8 m ^ 3.