Q. 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) $\frac{ML\omega^{2}}{2}$
B) $ML\omega^{2}$
C) $\frac{ML\omega^{2}}{4}$
D) $\frac{ML^{2}\omega^{2}}{2}$
Explanation
(a) $dM = \left( \frac{M}{L} \right)dx$
Force on $'dM'$ mass is
$dF = (dM)\omega^{2}x$
By integration we can get the force exerted by whole liquid
$\Rightarrow F = \int_{0}^{L}\frac{M}{L}\omega^{2}x\ dx = \frac{1}{2}M\omega^{2}L$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.