Question

A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both 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. $\frac{1}{2}M{\omega }^{2}L$
• B. $M{\omega }^{2}L$
• C. $\frac{1}{4}M{\omega }^{2}L$
• D. $\frac{1}{2}M{\omega }^2{L}^2$

Answer 1

Answer: (A)

$\frac{1}{2}M{\omega }^{2}L$

The mass of the element $=dm=\left(M/L\right)dx$
The force on the element towards the axis $=(F+dF)-F=dF$
$\therefore dF=\left(dm\right){\omega }^{2}x=\left(\frac{M}{L}dx\right){\omega }^{2}x$

Integrating w.r.t to $x$
$F=\frac{1}{2}\frac{M{\omega }^2}{L}{x}^2+constant$

For $x=0$,$F=0$. $\therefore constant=0$

For $x=L$,$F=\frac{M{\omega }^{2}L}{2}$