Question

A vertical $U-$tube has a liquid up to a height $h$. If the tube is slowly rotated to an angular speed $\omega =\sqrt{glh}$, find the
(i) heights of the liquid column,
(ii) pressure at the points $1,2$ and $3$ in the limbs in steady state.

Answer 1

We have $\frac{dP}{dx}=\rho {\omega }^{2}x\Rightarrow dP=\rho {\omega }^{2}xdx\to \left(a\right)$

On integerating ${\int }_{P_1}^{P_3}dP={\int }_{-h}^{2h}\rho {\omega }^{2}xdx$

$P_3-P_1=\frac{\rho {\omega }^2}{2}\left[\right(2h{)}^2-h^2]\Rightarrow \rho g(h_3-h_1)=\frac{3\rho {\omega }^2{h}^2}{2}\Rightarrow \rho g(h_3-h_1)=\frac{3\rho }{2}{h}^2(\frac{g}{h})\Rightarrow (h_3-h_1)=\frac{3h}{2}\to (1)$

From figure, $h_1+h_3=2h\to \left(2\right)$

From $\left(1\right)and\left(2\right)h_1=\frac{h}{4}\&h_3=\frac{7h}{4}$

$\left(ii\right)\therefore P_3=P_0+\rho g{h}_3=P_0+\rho g\left(\frac{7h}{4}\right)\to \left(3\right)P_1=P_0+\rho g{h}_1=P_0+\rho g\left(\frac{h}{4}\right)\to \left(4\right)$

Integerating $\left(a\right)$ from $P_{2}to{P}_3,$

$P_3-P_2=\frac{\rho {\omega }^2}{2}\left(4h^2\right)=\frac{\rho }{2}\left(\frac{g}{h}\right)\left(4h^2\right)\to \left(5\right)\left(3\right)in\left(5\right)\Rightarrow P_2=P_0-\frac{\rho gh}{4}\to \left(6\right)$