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. Combination answers like 'f or s' are possible answers in some of the cases.

The new sphere has a mass of m = m₀ and a radius of r < r₀. (choose from r, f, s, r or s, f or s )
The new sphere has a radius of r = r₀ and a density of ρ > ρ₀. (choose from r, f, s, r or s, f or s )
The new sphere has a radius of r = r₀ and a mass of m < m₀. (choose from r, f, s, r or s, f or s )

Answer 1

From Archimedes' principle, the buoyant force on the solid is equal to the weight of the water displaced by the object. If the object is floating (at rest), this upward buoyant force will be equal to the weight of the object (net force on the object being zero).

where V_w is the volume of the water displaced.

1] This equation is independent on the radius or the density of the mass. So, as long as the mass of the new sphere remains the same, the volume of the water displaced will also be the same. Therefore, the water level will remain the same [s]

2] Mass of the object is given by:

so, increasing the density increases the mass of the object and so by the equation above, the volume of the displaced water will increase. The water will therefore rise [r].

3] Here, the mass decreases which implies that the volume of the water displaced is less and so the water level will fall [f].