Kinematics Question 483
Question: A tube of length $ L $ it’s filled completely with an incompressible liquid of mass $ M $ and closed at both the ends. The tube it’s then rotated in a horizontal plane about one of it’s ends with a uniform angular velocity $ \omega $ . The force exerted by the liquid at the other end it’s [IIT 1992]
Options:
A) $ \frac{ML{{\omega }^{2}}}{2} $
B) $ ML{{\omega }^{2}} $
C) $ \frac{ML{{\omega }^{2}}}{4} $
D) $ \frac{ML^{2}{{\omega }^{2}}}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ dM=( \frac{M}{L} )dx $
force on dM mass is $ dF=(dM){{\omega }^{2}}x $
By integration we can get the force exerted by whole liquid therefore
$ F=\int _{0}^{L}{\frac{M}{L}}{{\omega }^{2}}xdx $
$ =\frac{1}{2}M{{\omega }^{2}}L $
