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. $\frac {ML^2 \omega}{2}$
• C. $ML \omega ^2$
• D. $\frac {ML^2 \omega^2}{2}$

Answer 1

Answer: (A)

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

Answer 2

Let the length of a small element of tube be dx. Mass of this element \hspace15mm dm= \frac {M}{L}dx where Mis mass of filled liquid and Lis length of tube. Force on this element, \hspace15mm dF=dm \times x \omega^2 \hspace15mm \int _0^FdF= \frac {M}{L}\omega^2 \int _0^Lxdx or \hspace15mm F= \frac {M}{L}\omega^2 \bigg [\frac {L^2}{2}\bigg ]=\frac {ML\omega^2}{2} or \hspace15mm F=\frac {1}{2}ML\omega^2