A piece of ice with a volume of V = 400 m3 floats in a vessel filled with water.

A piece of ice with a volume of V = 400 m3 floats in a vessel filled with water. What is the volume of the part of the ice above the surface of the water?

The “piece” of ice given in our example cannot take up a volume of 400 m³ – it will not fit into any vessel with such a volume. Let’s consider its volume equal to 400 cm³ (or 0.0004 m³, since in 1 m³ – 1000 dm³ (liters) or 1,000,000 cm³).

The specific gravity of ice is 0.9 of the specific gravity of water (900 kg / m³). This means that a buoyant force will act on a piece of ice of any size (thickness, area) greater than the weight of the piece itself.

That is, the ice will not only float flush with the surface of the water, but will protrude from it by about a tenth of its thickness (volume) (10%). Because this is how a balance (balance) will be achieved between the buoyancy force and the weight of the ice.

Pushing out the ice floe by one tenth of its thickness (volume), F pushing (F pushing = Vρg, where V is the volume of the body, ρ is the density of the liquid (in our case, water, ρ of water – 1000 kg / m³), ​​g is the acceleration of the free fall) will “reduce” the volume of the underwater (submerged) part of the piece also by one tenth (10%) and, therefore, will itself decrease by the same 10%, and thereby achieve a balance of forces – between itself and the weight of the ice.

In our case, the answer will be: the sought surface part of our piece of ice will be equal to 40 cm³ or 0.00004 m³ (V surface = 0.1 V ice (where V ice is the entire volume of the ice block), V = 0.1 * 400 cm³ = 40 cm³).