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: A

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