A tube of length 50cm is filled completely with an incompressible...

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

Now, force acting of that element is $F = ∫ \left(d m\right){\omega}^{2}x$

Integrating the above relation for whole length of tube, we have

$F = {∫}_{0}^{L}\left(\dfrac{m}{L} d x\right){\omega}^{2}x$

$= \dfrac{m}{L}{\omega}^{2}\dfrac{{L}^{2}}{2}$ $= \dfrac{m{\omega}^{2}L}{2}$

$⇒ \omega = \sqrt{\dfrac{2}{mL}}\sqrt{F}$ $= \sqrt{\dfrac{2}{0 . 25 \times 0 . 5}}\sqrt{F}$

$= \sqrt{16}\sqrt{F}$$= 4\sqrt{F}rad{s}^{- 1}$

Hence, the value of $x = 4$.