High School

A piece of gold (\( r = 19.3 \, \text{g/cm}^3 \)) has a cavity in it. It weighs 38.2 g in air and 36.2 g in water. What is the volume of the cavity in the gold?

Answer :

Final answer:

The volume of the cavity in the gold is determined to be 2.0 cm³ using the principle of buoyancy and Archimedes' principle, by calculating the weight difference in air and water which equals the weight and thus the volume of the displaced water.

Explanation:

To find the volume of the cavity in the gold, we use the concept of buoyancy and Archimedes' principle. This states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces. Therefore, when the gold with cavity is weighed in water, it appears lighter because of the buoyant force.

First, we calculate the apparent loss of weight when the object is submerged in water, which equals the weight of displaced water. This loss of weight is 38.2 g (in air) - 36.2 g (in water) = 2.0 g. Since the density of water is 1 g/cm³, the volume of water displaced, which also equals the volume of the cavity, is 2.0 cm³ (as 2.0 g of water corresponds to 2.0 cm³).