When the tube is rotated, the liquid starts to flow radially outwards and air in sealed arm is compressed. Let the shift of the liquid be y as shown in figure.

Let the cross-sectional area of the tube be S. Here element of length dx is rotating with the tube. From the free-body diagram of the element, we can write

$\begin{array}{l}(\mathrm{P}+\mathrm{dP})\mathrm{S}-\mathrm{PS}={\mathrm{dm\omega }}^2\mathrm{x}=\left(\mathrm{\rho Sdx}\right){\mathrm{\omega }}^2\mathrm{x}\\ \mathrm{dPS}={\mathrm{\rho \omega }}^2\mathrm{Sxdx}\end{array}$

Here the pressure difference between points A and B can be given by integrating the pressure difference across an element of width dx, which is given as

$\mathrm{dP}={\mathrm{\rho \omega }}^2\mathrm{xdx}$

Now integrating from A to B, we get

$\begin{array}{l}{\mathrm{P}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}}={\int }_{\mathrm{v}}^{\mathrm{L}} {\mathrm{\rho \omega }}^2\mathrm{xdx}=\frac{{\mathrm{\rho \omega }}^2}{2}\left({\mathrm{L}}^2-{\mathrm{y}}^2\right)\\ \Rightarrow {\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_0+\frac{{\mathrm{\rho \omega }}^2}{2}\left({\mathrm{L}}^2-{\mathrm{y}}^2\right)\end{array}$

The pressure at point C can be given as

${\mathrm{P}}_{\mathrm{C}}={\mathrm{P}}_{\mathrm{a}}-{\mathrm{y}}_{\mathrm{\rho }}\mathrm{g}$

At point A, pressure is atmospheric . Thus, we have

${\mathrm{P}}_{\mathrm{C}}=\frac{{\mathrm{\rho \omega }}^2}{2}\left({\mathrm{L}}^2-{\mathrm{y}}^2\right)+{\mathrm{P}}_0-{\mathrm{y}}_{\mathrm{\rho }}\mathrm{g}={10}^5\mathrm{N}/\mathrm{m}$