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 $L$]
$\Rightarrow F=\int_{0}^{L} \frac{M}{L} \omega^{2}l \,dl$
or $ F=\frac{M\omega^{2}}{L} \int_{0}^{L}l\,dl$
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 $F=\frac{1}{2}M\omega^{2}L$