Hi Ravija and welcome to PF.
Perhaps the easiest way to see this is in the rotating frame of the tube. Assuming that the rotation is in a horizontal plane, so that we don't have to worry about gravity, the centripetal acceleration points towards the center and is ##\omega^2 r## where ##r## is the distance from the center. In the reference frame of the rotating tube, this is seen as a centrifugal acceleration that has the same magnitude but points away from the center. The effective acceleration of gravity then is is ##g_{eff}=\omega^2 r## and becomes larger the more one moves away from the center. At the center the pressure is zero. At a point away from the center, the pressure is ##p = \rho g_{eff} r = \rho \omega^2 r^2##.
On edit: Here, ##\omega## is the angular speed of the tube and ##\rho## is the density of water.