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. $\frac{1}{2}M{\omega }^{2}L$
• B. $\frac{M{L}^2\omega }{2}$
• C. $ML{\omega }^2$
• D. $\frac{M{L}^2{\omega }^2}{2}$

Answer 1

Answer: (A)

Let the length of a small element of tube be $dx$. Mass of this element
$dm=\frac{M}{L}dx$
where $M$ is mass of filled liquid and $L$ is length of tube.
Force on this element,
$dF=dm\times x{\omega }^2$
$\underset{0}{\overset{F}{\int }}dF=\frac{M}{L}{\omega }^2\underset{0}{\overset{L}{\int }}xdx$
or $F=\frac{M}{L}{\omega }^2\left[\frac{L^2}{2}\right]=\frac{ML{\omega }^2}{2}$
or $F=\frac{1}{2}ML{\omega }^2$