Fluid Mechanics Viscosity Question 146
Question: A hollow cylinder of mass m made heavy at its bottom is floating vertically in water. It is tilted from its vertical position through an angle $ \theta $ and is left. The respecting force acting on it is
Options:
A) $ mg,\cos \theta $
B) $ \frac{mg,}{\cos \theta } $
C) $ mg\left[ \frac{1}{\cos \theta }-1 \right] $
D) $ mg\left[ \frac{1}{\cos \theta }+1 \right] $
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Answer:
Correct Answer: C
Solution:
[c] Let l be the length of the cylinder, when vertical, in water. Let A be the cross-sectional area of the cylinder. Equating weight of the cylinder with the up thrust, we get $ Mg=Al\rho g,,orm=Al\rho $ When the cylinder is tilted through an angle $ \theta $ , length of Cylinder in water $ =\frac{1}{\cos \theta } $ Weight of water displaced $ =\frac{l}{\cos \theta }A\rho g $ Restoring force $ =\frac{lA\rho g}{\cos \theta }-lA\rho g $ $ =lA\rho g\left[ \frac{1}{\cos \theta }-1 \right]=mg\left[ \frac{1}{\cos \theta }-1 \right] $
BETA
