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.
1
Expert's answer
2011-01-27T08:23:59-0500
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.

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