Fluid Mechanics Viscosity Question 258
Question: A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is T and its mass M. It is suspended by a string in a liquid of density p where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of thecylinder by the liquid is
Options:
A) $ Mg $
B) $ Mg-V\rho g $
C) $ Mg+\pi R^{2}h\rho g $
D) $ \rho g(V+\pi R^{2}h) $
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Answer:
Correct Answer: D
Solution:
[d] According to Archimedes principle Upthrust = Wt. of fluid displaced $ F_{bottom}-F_{top}=V\rho g $ $ \therefore ,,,F_{bottom}=F_{top}+V\rho g $ $ =P_{1}\times A+V\rho g $ $ =(h\rho g)\times (\pi R^{2})+V\rho g $ $ =\rho g[\pi R^{2}h+V] $
BETA
