Question

A horizontally oriented tube AB of length 'l' rotates with a constant angular velocity ${}^{\prime }{\omega }^{\prime }$ about a stationary vertical axis $O_1{O}_2$ passing through the end A as shown in the figure. The tube is filled with an ideal fluid. The end A of the tube is open. [Consider $P_0=P_{atm}$ ]

If end B is closed, the pressure just inside the tube at B as a function of the length of liquid column ‘h’ is

• A. $P_0+\rho {\omega }^2[l^2-(l-h{)}^2]$
• B. $P_0+\frac{\rho {\omega }^2}{2}[l^2-(l-h{)}^2]$
• C. $P_0+\frac{\rho {\omega }^2}{4}[l^2-(l-h{)}^2]$
• D. $\frac{1}{2}[P_0+\rho {\omega }^2[l^2-(l-h{)}^2]]$

Answer 1

The correct option is B $P_0+\frac{\rho {\omega }^2}{2}[l^2-(l-h{)}^2]$

Consider an element of length ‘dx’ at a distance ‘x’ from the axis.

Pressure force $=\left(\rho Adx\right){\omega }^{2}x$

${\int }_{P_0}^{P_B}dP\left(A\right)={\int }_{l-h}^l\rho A{\omega }^{2}xdx$
or, $P_B=P_0+\frac{\rho {\omega }^2}{2}[l^2-(l-h{)}^2]$