Question 202 of 210

A tube of length $50cm$ is filled completely with an incompressible liquid of mass $250g$ and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $x\sqrt{F}rad{s}^{-1}$. If $F$ be the force exerted by the liquid at the other end then the value of $x$ will be _____ .

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Answer: 4

Considering an element of tube of length $dx$ having mass $dm$ at a distance of $x$ from right end as shown in figure below.

Now, force acting of that element is $F=\int \left(dm\right){\omega }^{2}x$
Integrating the above relation for whole length of tube, we have
$F={\int }_{0}^{L}\left(\dfrac{m}{L}dx\right){\omega }^{2}x$
$=\dfrac{m}{L}{\omega }^{2}\dfrac{{L}^{2}}{2}$ $=\dfrac{m{\omega }^{2}L}{2}$
$\Rightarrow \omega =\sqrt{\dfrac{2}{mL}}\sqrt{F}$ $=\sqrt{\dfrac{2}{0.25\times 0.5}}\sqrt{F}$
$=\sqrt{16}\sqrt{F}$$=4\sqrt{F}rad{s}^{-1}$
Hence, the value of $x=4$.