Fluid Mechanics Viscosity Question 77
Question: A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is r and angular velocity of rotation is $ \omega $ , then the difference in the heights of the liquid at the centre of the vessel and the edge is
Options:
A) $ \frac{r\omega }{2g} $
B) $ \frac{r^{2}{{\omega }^{2}}}{2g} $
C) $ \sqrt{2gr\omega } $
D)$ \frac{{{\omega }^{2}}}{2gr^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
From Bernoulli’s theorem, $ P_{A}+\frac{1}{2}dv_{A}^{2}+dgh_{A}=P_{B}+\frac{1}{2}dv_{B}^{2}+dgh_{B} $ Here,$ h_{A}=h_{B} $ $ \therefore \ P_{A}+\frac{1}{2}dv_{A}^{2}=P_{B}+\frac{1}{2}dv_{B}^{2} $ Þ $ P_{A}-P_{B}=\frac{1}{2}d[v_{B}^{2}-v_{A}^{2}] $ Now,$ v_{A}=0,\ v_{B}=r\omega $ and $ P_{A}-P_{B}=hdg $ $ \therefore \ \ hdg=\frac{1}{2}dr^{2}{{\omega }^{2}} $ or $ h=\frac{r^{2}{{\omega }^{2}}}{2g} $
BETA
