A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is \(\sqrt{\frac{\mathrm{F}}{\alpha \mathrm{M}}}\) in SI unit. The value of \(\alpha\) is __________.

Explanation:
Let us calculate the angular velocity by determining the total centrifugal force exerted by the continuous liquid column:

1. Setting up the Integration Element:
Let \( L = 1\text{ m} \) be the total length of the tube and \( m = 2M \) be the total mass of the liquid. Consider a tiny element of the liquid column having a width \( dx \) located at a distance \( x \) from the axis of rotation.
The mass of this small element is:
\( dm = \frac{m}{L} dx \)

2. Integrating the Centrifugal Force:
The outward centrifugal force \( dF \) exerted by this small element during rotation is:
\( dF = (dm) \omega^2 x = \left(\frac{m}{L} dx\right) \omega^2 x \)
To find the total force \( F \) acting on the outer wall, we integrate this expression from the pivot \( x = 0 \) to the outer end \( x = L \):
\( F = \int_{0}^{L} \frac{m}{L} \omega^2 x dx = \frac{m}{L} \omega^2 \left[ \frac{x^2}{2} \right]_{0}^{L} \)
\( F = \frac{m}{L} \omega^2 \frac{L^2}{2} = \frac{1}{2} m \omega^2 L \)
Notice that this matches the center of mass approach: \( F = m \omega^2 r_{cm} \), where \( r_{cm} = \frac{L}{2} = 0.5\text{ m} \).

3. Substituting the Given Constants:
Now, substitute the values \( L = 1\text{ m} \) and \( m = 2M \) into the derived equation:
\( F = \frac{1}{2} (2M) \omega^2 (1) = M \omega^2 \)
Solving explicitly for the angular velocity \( \omega \):
\( \omega^2 = \frac{F}{M} \implies \omega = \sqrt{\frac{F}{M}} \)

Comparing this result with the expression given in the problem \( \omega = \sqrt{\frac{F}{\alpha M}} \), we find that \( \alpha = 1 \).

Hence, Option D is the correct choice.