Here we make the point that sometimes geomertry is as easy as pie ( π ) and then sometimes knowledge of geometry is needed. Seemingly simple things can get nasty but the geometry must be solved first!
Sphere, Tank and Water The diameter of the sphere is one foot (D = 1 ft). When it is lowered into the tank the water rises to a final depth of two feet. No water spills over.
Calculate the mass of water.
♦ Since the final depth of water is said to be 2 feet, the sphere will be completely submerged.
Thus the mass is: m = ρ V:
Suppose the initial depth of water were 0.5 feet (that is, h = 0.5 ft). Were the sphere lowered into the tank, what final depth of the water would occur?
♦ Hoping this calculation is simple, we assume the sphere becomes completely submerged. That way the sum of the spherical and rectangular volumes will become a rectangular total. We calculate the second depth of water (rectangular) - as a sum of the water and sphere volumes (subject to our assumption).
A problem has arisen. The sphere will not be completely submerged. Therefore our calculation must be recast to include a "cap" of the sphere extending above the water level. Do you know the volume of a spherical cap? Some tedious algebra will obtain an answer. We leave it here.