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                                 where M is mass of filled liquid and I is length of tube.                 Force on this element                 \[dF=dm\times x{{\omega }^{2}}\]                 \[\int\limits_{0}^{F}{dF=\frac{M}{L}{{\omega }^{2}}\int\limits_{0}^{L}{x\,dx}}\] \[orF=\frac{M}{L}{{\omega }^{2}}\left[ \frac{{{L}^{2}}}{2} \right]=\frac{ML{{\omega }^{2}}}{2}\] \[orF=\frac{1}{2}ML{{\omega }^{2}}\]