Question

A piece of pure gold of density (9.3G/cc) is suspected to be hollow inside.? It weighs 38.250 g in air and 33.865 g in water. Calculate the volume of the hollow portion in the gold if any

Answer 1

mass = 38.250
density= 9.3
volume of gold in air= mass/density
= 38.250/9.3
= 4.112

loss of mass= 38.250- 33.865
= 4.385g

according to laws of flotation: mass of water = volume of object
so volume of gold in water = 4.385

volume of hollow portion = volume of gold in water - volume of old in air
= 4.385 - 4.112
= 0.273 cm3