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(B) Consider two points $A$ (at the centre) and $B$ (at the edge) on the surface of the liquid.Let $P_A$ and $P_B$ be the pressures at points $A$ and

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

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(B) Consider two points $A$ (at the centre) and $B$ (at the edge) on the surface of the liquid.Let $P_A$ and $P_B$ be the pressures at points $A$ and $B$ respectively.Since both points are on the free surface of the liquid,$P_A = P_B = P_{atm}$.In the rotating frame of the vessel,the effective force on a fluid element is the combination of gravity and centrifugal force.The pressure variation in the rotating liquid is given by $dP = \rho \omega^2 x dx$,where $x$ is the distance from the axis.Integrating from $x = 0$ to $x = r$:$\int_{P_A}^{P_B} dP = \int_{0}^{r} \rho \omega^2 x dx$$P_B - P_A = \frac{1}{2} \rho \omega^2 r^2$Since $P_A = P_B$,this pressure difference is balanced by the hydrostatic pressure difference due to the height difference $h$:$P_B - P_A = \rho g h$Equating the two expressions:$\rho g h = \frac{1}{2} \rho \omega^2 r^2$$h = \frac{r^2 \omega^2}{2g}$

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