Question

A tube of length $L$ is filled with an incompressible liquid of mass $M$ and closed at both 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 will be

• A. $\frac{1}{2}{M}^2{\omega }^{2}L$
• B. $\frac{1}{2}M{\omega }^{2}L$
• C. $\frac{1}{2}M{\omega }^2{L}^2$
• D. None of these

Answer 1

Answer: (B)

A tube of length $L$ is filled with an incompressible liquid of mass $M$. When the tube is rotated, the liquid will experience a centrifugal force

Now we take an element of length $dl$ and its distance from the axis $OO$ʹ is $l$
Then the mass per unit length of the tube = $\frac{M}{L}$
Hence, the mass of the element, $dm=\frac{M}{L}\cdot dl$
So, the centrifugal force experienced by the element
$dF=dm\cdot r{\omega }^2$
Here, $dm=\frac{M}{L}dl,r=l,\omega$ =angular velocity
$\therefore dF=\frac{M}{l}dl\cdot l{\omega }^2$
$=\frac{M}{L}{\omega }^{2}ldl\dots \left(i\right)$
So, the total force experienced by the liquid[on integrating Eq. $(i)$ with limit 0 to $L$]
$\Rightarrow F={\int }_0^L\frac{M}{L}{\omega }^{2}ldl$
or $F=\frac{M{\omega }^2}{L}{\int }_0^{L}ldl$
or $F=\frac{M{\omega }^2}{L}{\left[\frac{l^2}{2}\right]}_0^L$
or $F=\frac{M{\omega }^2}{2L}\left[L^2-0\right]$
or $F=\frac{1}{2}M{\omega }^{2}L$