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. (So for each, pick either r, s, or f- or a combo.. thank you)

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

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

7. The new sphere has a radius of r < r₀ and a mass of m > m₀.
8. The new sphere has a radius of r < r₀ and a density of ρ > ρ₀.
9. The new sphere has a mass of m > m₀ and a density of ρ < ρ₀.

Answer 1

Hello,

Given: solid homogeneous sphere

mass = m₀

Radius =r₀

Density =

1) Solution: For new sphere,

In this case, and m<m₀ , where m and are the mass and density of new sphere respectively.

The expression of the density for first sphere is

where, V₀ is the volume of the first solid sphere.

For new sphere, the density is

Where, V is the volume of the new sphere.

But, , hence we equate both of these equations

Simply this for V as

But the condition is m<m₀ then new volume is less than first one. The densities are similar hence this sphere will float but V>V₀ It indicates the water level falls in the container. (f)

According to the Newton's third law, force applied by new sphere is small as compared to first one hence the force exerted by the water level on it also small.

2) Solution:

In this case the densities for both spheres are same but r<r₀ ,

The expression of the volume of the sphere is

The volumes for both both spheres are different but the densities are same hence both will float on the same level. Water level stay same (s)

3) Solution:

In this case, and .

Equating both equations as

But

Volume of new sphere is greater than the first one hence . The masses are same but density of new sphere is small as compare to the first hence force applied on the water level is less hence force exerted by water level on it small. hence water level falls (f).

4) Solution:

In this case, and .

The density of the new sphere is less than the first one. But first sphere is floating on water hence second one will be float. The radius for both spheres are same hence volumes are same.

After simplifying above equations of density we get,

But

Simply density is directly proportional to mass, hence density of new sphere is less then its mass is also less. The gravitational force of the new sphere is small and force exerted on it by water is small as compare to first sphere. The water level falls (f).

Thanking you!