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(C) Let $P_B$ be the pressure at point $B$ at the bottom of the vertical column of liquid of height $h$. Since the top of the column is open to the at

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$P_0 + \rho gh - \frac{\rho R^2 \omega^2}{2}$

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(C) Let $P_B$ be the pressure at point $B$ at the bottom of the vertical column of liquid of height $h$. Since the top of the column is open to the atmosphere,$P_B = P_0 + \rho gh$,where $P_0$ is the atmospheric pressure.Now,consider the horizontal segment of the tube rotating with angular velocity $\omega$ about the vertical axis passing through $A$. Point $A$ is at a distance $r = 0$ from the axis,and point $B$ is at a distance $r = R$ from the axis.The pressure variation in a rotating fluid is given by $dP = \rho \omega^2 r dr$. Integrating from $A$ to $B$:$P_B - P_A = \int_0^R \rho \omega^2 r dr = \frac{\rho \omega^2 R^2}{2}$.Therefore,$P_A = P_B - \frac{\rho \omega^2 R^2}{2}$.Substituting the value of $P_B$,we get $P_A = P_0 + \rho gh - \frac{\rho \omega^2 R^2}{2}$.

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