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(A) Consider a point $1$ at the centre of the rotating liquid surface and a point $2$ at the edge of the liquid surface.Let $h$ be the difference in h

Vedclass · Accepted Answer

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

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(A) Consider a point $1$ at the centre of the rotating liquid surface and a point $2$ at the edge of the liquid surface.Let $h$ be the difference in height between point $2$ and point $1$.In the rotating frame of reference,the pressure at any point in the liquid is given by the hydrostatic pressure variation in a non-inertial frame.The pressure difference between the edge (point $2$) and the centre (point $1$) is due to the centrifugal force acting on the liquid elements.Using the equation of motion for a fluid element in a rotating frame,the pressure gradient is $\frac{dP}{dr} = \rho \omega^2 r$.Integrating this from $r=0$ to $r=R$,we get $P_2 - P_1 = \int_0^r \rho \omega^2 r dr = \frac{1}{2} \rho \omega^2 r^2$.Also,the pressure difference due to the height difference $h$ is $P_2 - P_1 = \rho g h$.Equating the two expressions for the pressure difference:$\rho g h = \frac{1}{2} \rho \omega^2 r^2$$h = \frac{r^2 \omega^2}{2g}$

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