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

Correct Answer: A

Solution :

[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 \[\Rightarrow F=\int_{0}^{L}{\frac{M}{L}{{\omega }^{2}}xdx=\frac{1}{2}M{{\omega }^{2}}L}\]