A U-tube filled with liquid is being rotated about an axis with angular speed ω as shown in the figure. The difference in heights of liquid in two arms is

Let us take a small element of length dr at a distance r from axis of rotation.
Mass of the element dm = (Adr.ρ) where A = cross-sectional area of the tube. p = pressure at a radial distance r.
p + dp = pressure at a radial distance (r + dr)
Net radially inward thrust on the element = (p + dp)A - p.A
$\therefore$ (p + dp)A - p.A = dm. ω².r
(p + dp) - p = ρω²rdr
$\Rightarrow {\mathrm{p}}_1-{\mathrm{p}}_2={\mathrm{\rho \omega }}^2{\int }_{-{\mathrm{r}}_2}^{{\mathrm{r}}_1}\mathrm{rdr}\Rightarrow {\mathrm{p}}_1-{\mathrm{p}}_2=\frac{{\mathrm{\rho \omega }}^2}{2}\left({\mathrm{r}}_1^2-{\mathrm{r}}_2^2\right)$
$\Rightarrow {\mathrm{h}}_1-{\mathrm{h}}_2=\frac{{\mathrm{\omega }}^2}{2\mathrm{g}}\left({\mathrm{r}}_1^2-{\mathrm{r}}_2^2\right)$