A spherical shell of iron (?iron = 7800 kg/m3 ) has an inner diameter of 2.50 m
ID: 1458812 • Letter: A
Question
A spherical shell of iron (?iron = 7800 kg/m3 ) has an inner diameter of 2.50 m and an outer diameter of 2.60 m. The interior of the shell is empty. The shell is placed on water.
(a) What is the mass of the shell?
(b) Show that the shell will float.
(c) What is the volume of the portion of the shell under water?
Suppose the interior of the shell is now completely filled with a material of density ? and the filled shell floats fully submerged in the water.
(d) What is the mass of the liquid filling the interior?
(e) Determine ?.
Explanation / Answer
given,
density of iron = 7800 kg/m^3
inner diameter = 2.5 m
outer diameter = 2.6 m
volume of sphere = (4/3) * pi * r^3
volume of the the iron = (4/3) * pi * outer radius^3 - (4/3) * pi * inner radius^3
volume of the the iron = (4/3) * pi * (2.6/2)^3 - (4/3) * pi * (2.5/2)^3
volume = 1.0215 m^3
density = mass / volume
7800 = mass / 1.0215
mass = 7800 * 1.0215
mass of the shell = 7967.7 kg
density of the shell = 7967.7 / ((4/3) * pi * (2.6/2)^3)
density of shell = 865.7934 kg/m^3
since the density of shell is less than density of water that is 1000 kg/m^3 so it'll float
volume of portion under water = 865.7934 / 1000
volume of portion under water = 0.8657934
volume = ((4/3) * pi * (2.6/2)^3)
volume = 9.20277 kg/m^3
mass of water needs to displaced = 1000 * 9.20277
mass of water needs to displaced = 9202.77 kg
mass of liquid needed = 9202.77 - 7967.7
mass of liquid needed = 1235.07 kg
density of liquid = 1235.07 / ((4/3) * pi * (2.5/2)^3)
density of liquid = 150.9638 kg/m^3
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