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 ism. (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


$(p+dp)-p=\rho {\omega }^{2}rdrordp=\rho {\omega }^{2}rdr$
On integration we get$p_1-p_2=\rho {\omega }^2{\int }_{-r_2}^{r_1}rdr\Rightarrow p_1-p_2=\frac{\rho {\omega }^2}{2}({r_1}^2-{r_2}^2)or{h}_1-h_2=\frac{{\omega }^2}{2g}({r_1}^2-{r_2}^2)=\frac{{\left(2\pi \right)}^2}{2\times 10}\left(0.5^2-0.{25}^2\right]$
= 0.37 m (approx.)