Question

A vertical $U$-tube with two limbs 0.75 m apart is filled with water and rotated about a vertical axis 0.5 m from the left limb, as shown in the figure.



If the speed of rotation is 60 rpm, difference in elevation of water levels in the two limbs is: (Upto 2 decimal digit)

Answer 1

For a small element of length dr at a distance r from the axis of rotation the pressure difference can be given by


$(p+dp)-p=\rho {\omega }^{2}rdr$ or $dp=\rho {\omega }^{2}rdr$
On integration we get

$p_1-p_2=\rho {\omega }^2{\int }_{r_1}^{r_2}rdr$

$p_1-p_2=\frac{\rho {\omega }^2}{2}[{r_1}^2-{r_2}^2]$

⇒$p_1-p_2=\frac{\rho {\omega }^2}{2}(r_1^2-r_2^2)$

or $h_1-h_2=\frac{{\omega }^2}{2g}(r_1^2-r_2^2)$ = $\frac{(2\pi {)}^2}{2\times 10}({0.5}^2-{0.25}^2)$
$△h=0.37m$ (approx.)