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(D) When a cylindrical vessel containing liquid is rotated at an angular speed $\omega$ about its vertical axis,the liquid particles at a distance $r$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) When a cylindrical vessel containing liquid is rotated at an angular speed $\omega$ about its vertical axis,the liquid particles at a distance $r$ from the axis rotate with a linear velocity $v = r\omega$.Applying Bernoulli's principle in the rotating frame of reference,the effective pressure at a point at distance $r$ from the axis is given by $P(r) = P_0 + \frac{1}{2}\rho r^2 \omega^2$,where $P_0$ is the pressure at the center $(r=0)$.At the edge of the vessel,$r = R$,so the pressure is $P_R = P_0 + \frac{1}{2}\rho R^2 \omega^2$.The difference in pressure between the edge and the center is $\Delta P = P_R - P_0 = \frac{1}{2}\rho R^2 \omega^2$.This pressure difference is balanced by the hydrostatic pressure difference due to the height difference $h$ of the liquid column,given by $\Delta P = \rho g h$.Equating the two expressions: $\rho g h = \frac{1}{2}\rho R^2 \omega^2$.Solving for $h$,we get $h = \frac{R^2 \omega^2}{2g}$.

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