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 co. The force exerted by the liquid at the other end is: [AIPMT (S) 2005]

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

Where M is mass of filled liquid and I is length of tube.

Force on this element

\[dF=dm\times x{{\omega }^{2}}\]

\[\int\limits_{0}^{F}{dF=\frac{M}{L}{{\omega }^{2}}\int\limits_{0}^{L}{x\,dx}}\]

or         \[F=\frac{M}{L}{{\omega }^{2}}\left[ \frac{{{L}^{2}}}{2} \right]=\frac{ML{{\omega }^{2}}}{2}\]

or         \[F=\frac{1}{2}ML{{\omega }^{2}}\]