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Answer to Question #1396 in Mechanics | Relativity for Mike
Question #1396
A hollow brass tube has outer diameter D =
3.1 cm. The tube is sealed at one end and
loaded with lead shot to give it a total mass
of M = 88 g. The tube floats in water (of
density 1 g/cm3) in vertical position, loaded
end down.
What is the depth of the bottom end of the
tube?
Answer in units of cm.
3.1 cm. The tube is sealed at one end and
loaded with lead shot to give it a total mass
of M = 88 g. The tube floats in water (of
density 1 g/cm3) in vertical position, loaded
end down.
What is the depth of the bottom end of the
tube?
Answer in units of cm.
Expert's answer
The weight of the tube with the lead shot should be equal to the Archimedes force.
Denote the depth of the bottom end as H, the volume of the tube under the surface would be V = H πd2/4.
ρ g V = M g
V = M / ρ = 88 [g] / 1 [g/cm3] = 88 cm3.
H πd2/4 = 88.
H = 4*88 / 3.14* 3.12 = 11.7 cm.
Denote the depth of the bottom end as H, the volume of the tube under the surface would be V = H πd2/4.
ρ g V = M g
V = M / ρ = 88 [g] / 1 [g/cm3] = 88 cm3.
H πd2/4 = 88.
H = 4*88 / 3.14* 3.12 = 11.7 cm.
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#340153 on Dec 2023
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