Fluid Mechanics Viscosity Question 304
Question: An air bubble of radius 1 cm rises with terminal velocity 0.21 cm/s in liquid column. If the density of liquid is$ 1.47\times 10^{3}kg/m^{3} $ . Then the value of coefficient of viscosity of liquid ignoring the density of air, will be
Options:
A) $ 1.71\times 10^{4}poise $
B) $ 1.82\times 10^{4}poise $
C) $ 1.78\times 10^{4}poise $
D) $ 1.52\times 10^{4}poise $
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Answer:
Correct Answer: D
Solution:
[d] Using the formula of the terminal velocity of a body falling through a viscous medium, $ V=\frac{2r^{2}(\rho -\sigma )g}{9\eta }\Rightarrow \eta =\frac{2r^{2}(\rho -\sigma )g}{9v} $ Where $ \rho $ is the density of material of body and $ \sigma $ is the density of medium. In case of the air bubble$ \rho =1 $ and$ \sigma =1.47\times 10^{3}kg/ms $ the air bubble rises up. $ \eta =\frac{2r^{2}\sigma g}{9v}=\frac{2\times {{({{10}^{-2}})}^{2}}\times 1.47\times 10^{3}\times 9.8}{9\times 0.21\times {{10}^{-2}}} $ $ =1.52\times 10^{3} $ decapoise $ =1.52\times 10^{4} $ Poise
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