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(B) Consider a point $A$ at the center of the liquid surface and a point $B$ at the side of the container at the same horizontal level as $A$ in the r

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$\frac{r^2\omega^2}{2g}$

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(B) Consider a point $A$ at the center of the liquid surface and a point $B$ at the side of the container at the same horizontal level as $A$ in the rotating frame.Using Bernoulli's principle in the rotating frame,the pressure difference between the center and the side is given by the hydrostatic pressure difference due to the height $h$:$P_B - P_A = h \rho g$In the rotating frame,the effective pressure difference due to centrifugal force is:$P_B - P_A = \int_0^r \rho \omega^2 x \, dx = \frac{1}{2} \rho \omega^2 r^2$Equating the two expressions for pressure difference:$h \rho g = \frac{1}{2} \rho \omega^2 r^2$$h = \frac{r^2 \omega^2}{2g}$

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