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(A) When a cylindrical vessel containing liquid is rotated with an angular velocity $\omega$ about its vertical axis, the surface of the liquid takes

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$125 	imes 10^{-4} \,m$

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(A) When a cylindrical vessel containing liquid is rotated with an angular velocity $\omega$ about its vertical axis, the surface of the liquid takes the shape of a paraboloid.Let $r$ be the radius of the vessel and $h$ be the height difference between the liquid level at the center and at the sides.For a particle of liquid at the surface at a distance $r$ from the axis, the forces acting in the rotating frame are the centrifugal force $(m r \omega^2)$ and the gravitational force $(mg)$.The slope of the liquid surface is given by $\frac{dh}{dr} = \frac{r \omega^2}{g}$.Integrating this from $r=0$ to $r=R$:$\int_0^h dh = \int_0^R \frac{\omega^2}{g} r dr$$h = \frac{\omega^2 R^2}{2g}$Given $R = 0.05 \,m$, $\omega = 10 \,rad \,s^{-1}$, and $g = 10 \,m \,s^{-2}$:$h = \frac{(10)^2 	imes (0.05)^2}{2 	imes 10}$$h = \frac{100 	imes 0.0025}{20} = \frac{0.25}{20} = 0.0125 \,m$$h = 125 	imes 10^{-4} \,m$.

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