A cylindrical vessel of radius 40 cm and large height contains water to a depth of 10 cm when at rest. It is rotated about its vertical axis with a…

In steady rotation the free surface is a paraboloid $z=\dfrac{\omega^2r^2}{2g}$ measured from its lowest point. When the vertex just touches the bottom, the water forms the region between the paraboloid and the vessel wall. The volume of a paraboloid 'bowl' of height $H$ inside a cylinder of radius $R$ is $\tfrac12\pi R^2H$, so the water volume is $\pi R^2H-\tfrac12\pi R^2H=\tfrac12\pi R^2H$. Equating to the initial volume $\pi R^2h_0$ gives $H=2h_0=20\ \mathrm{cm}$. Now the height of the surface at the wall is $H=\dfrac{\omega^2R^2}{2g}$, so $\omega^2=\dfrac{2gH}{R^2}=\dfrac{2\times10\times0.2}{0.16}=25$, hence $\omega=5\ \mathrm{rad\,s^{-1}}$.