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

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

Answer 1

The amount of water displaced equals the weight of the sphere (as long as it floats), due to buoyant force.

So weight of ball =force of buoyancy Fb =vg

Where rho=density of fluid, v=volume of fluid displaced and g=acceleration due to gravity.

So here rho and g is constant so whenever volume changes the weight changes.

So when m<mo and ρ > ρ0 then as we know volume depends on mass so here mass is decreasing so volume will also decrease .

So in first case the level will fall that is f

Case2– when radius of r < r0 and a mass of m > m0,

Here clearly mass is increasing due to which weight of ball is more so the volume displaced by the ball will also increase and so the water level will rise that is r

case3– when density of ρ < ρ0 and a radius of r > r0

here as the radius increases and density decreases which means mass will either constant or decreasing that is mmo

Now since volume is directly proportional to mass so water level may stay same or fall that is so Or f