Properties Of Solids And Liquids Question 195
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}} $
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Answer:
Correct Answer: B
Solution:
From Bernoulli’s principle,
$ P _{A}+\frac{1}{2}\rho v _{A}^{2}+\rho gh _{A}=P _{B}+\frac{1}{2}\rho v _{B}^{2}+\rho gh _{B} $
Here,$ h _{A}=h _{B} $
$ \therefore \ P _{A}+\frac{1}{2}\rho v _{A}^{2}=P _{B}+\frac{1}{2}\rho v _{B}^{2} $
therefore $ P _{A}-P _{B}=\frac{1}{2}\rho[v _{B}^{2}-v _{A}^{2}] $
Now,$ v _{A}=0,\ v _{B}=r\omega $
and $ P _{A}-P _{B}=hdg $
$ \therefore \ \ h=\frac{1}{2}dr^{2}{{\omega }^{2}} $ or $ h=\frac{r^{2}{{\omega }^{2}}}{2g} $
BETA
