A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of 0

Question

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of 0 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.

Answers can be f, r, s, f or s, r or s .

1. The new sphere has a mass of m > m0 and a radius of r = r0.
2. The new sphere has a density of > 0 and a mass of m = m0.
3. The new sphere has a density of > 0 and a radius of r = r0.

Explanation / Answer

water level is decided by the buoyany force by water on sphere

buoyant force = rho*g*V

rho is the density of water

g is the accelaration due to gravity

and V is the volume of the displaced water or volume of the sphere = m/rho_sphere

downward force= weight

so in equilibrium

weight = buoyant force

m*g = rho*g*V

g cancels

m = rho*V

in case 1)

m>mo and r = ro

since m > mo ,then water level rises in the container

2) if rho > rho_o, and m = mo then

initially the sphere is floating then ,now the high density sphere ,the buoyany force decreases hence the water level falls

so the answer is f

3) for rho > rho_o and r = ro

Now the water level raises

so the answer is r

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