Question

A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is -

• A. $\frac{M{\omega }^{2}L}{2}$
• B. $M{\omega }^{2}L$
• C. $\frac{M{\omega }^{2}L}{4}$
• D. $\frac{M{\omega }^2{L}^2}{2}$

Answer 1

The correct option is A $\frac{M{\omega }^{2}L}{2}$


When the fluid is rotated in horizontal plane the gradient of pressure at any location $x$ from the axis of rotation can be written as
$\frac{dP}{dx}=\rho {\omega }^{2}x$
Hence pressure at the other end of the tube is
$P=\frac{\rho {\omega }^2{l}^2}{2}$
$F=PA=\rho Al\frac{{\omega }^{2}l}{2}=\frac{M{\omega }^{2}L}{2}$ (Since $\rho Al$ is the mass of the liquid column)