question_answer

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) \[\frac{M{{L}^{2}}\omega }{2}\]

C) \[ML{{\omega }^{2}}\]

D) \[\frac{M{{L}^{2}}{{\omega }^{2}}}{2}\]

Correct Answer: D

Solution :

 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 the length of tube. Face on this element \[dF=(dm){{\omega }^{2}}x=\left( \frac{M}{L} \right)dx.{{\omega }^{2}}x\] Integrating \[\int_{0}^{F}{dF}=\frac{M}{L}{{\omega }^{2}}\int_{0}^{L}{x\,dx}\] or \[F=\frac{M}{L}{{\omega }^{2}}\left[ \frac{{{L}^{2}}}{2} \right]=\frac{ML{{\omega }^{2}}}{2}\] or \[F=\frac{1}{2}ML{{\omega }^{2}}\]