Question

A solid, homogeneous sphere with a mass of m₀, a radius of r₀ and a density of ρ₀ is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same.

r f s  The new sphere has a radius of r < r₀ and a density of ρ = ρ₀.
r f s  The new sphere has a density of ρ = ρ₀ and a mass of m > m₀.
r f s  The new sphere has a radius of r > r₀ and a mass of m = m₀.

Answer 1

1) s (stays the same) because density is same for new sphere. Since radius is less volume will also be less. To balance this the mass will also decrease.

2) s (stays the same) because density is same for new sphere. Mass is greater which means volume has to be greater or radius has to be greater. So increase in mass is balanced by increase in radius.

3) f (falls) here the mass remains same and radius increases(volume increases). So density decreases(density=

mass/volume) when density decreases sphere will float higher. Part of sphere under water will be less or water displaced by it will be less. Therefore level of water in container falls.