1. Given data

• Length of tube: $L = 50\text{ cm} = 0.5\text{ m}$
• Mass of liquid: $m = 250\text{ g} = 0.25\text{ kg}$
• Tube is completely filled with an incompressible liquid.
• Tube is rotated in a horizontal plane about one end with angular speed $\omega = x\sqrt{F}$ where $F$ is the force exerted by the liquid at the other end.

We need to find $x$.

2. Mass per unit length of liquid

Since the liquid is uniformly distributed along the tube, $\lambda = \frac{m}{L} = \frac{0.25}{0.5} = 0.5\ \text{kg m}^{-1}$

3. Force at the outer end

Consider a small element of liquid at distance $r$ from the axis and thickness $dr$.

Its mass is $dm = \lambda\,dr$

Required centripetal force for this element is $dF = dm\,\omega^2 r = \lambda \omega^2 r\,dr$

The force transmitted to the outer end equals the total force needed for all liquid elements from $r=0$ to $r=L$: $F = \int_0^L \lambda \omega^2 r\,dr$

$F = \lambda \omega^2 \int_0^L r\,dr = \lambda \omega^2 \cdot \frac{L^2}{2}$

Now substitute $\lambda = m/L$: $F = \frac{m}{L}\omega^2\frac{L^2}{2} = \frac{1}{2}m\omega^2 L$

4. Substitute numerical values

$F = \frac{1}{2}(0.25)\omega^2(0.5)$

$F = 0.0625\,\omega^2$

So, $\omega^2 = 16F$

Taking square root, $\omega = 4\sqrt{F}$

Comparing with $\omega = x\sqrt{F}$ we get $x=4$

5. Comparison with stored answer

Stored correct answer: $4$

Our derived answer is also $4$, so it agrees.