A tube of length 1m is filled completely with an ideal liquid of mass 2M, 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 ______.

Step 1: Consider the Force at a Distance $x$

When the tube rotates with angular velocity $\omega$, each element of the liquid experiences a centrifugal force.

The force at a distance $x$ from the pivot can be found by considering the differential force due to an element of liquid.

$dF=\left(\rho Adx\right){\omega }^{2}x$

where $\rho$ is the volumetric mass density of the liquid.

Step 2: Find the Volumetric Mass Density

Since the mass of the liquid is $2M$ , the length of the tube is $1$ m, and let the area of cross-section is $A$, the volumetric mass density is:

$\rho =\frac{2M}{AL}=\frac{2M}{A}$

Step 3: Calculate the Total Force at the Other End

The force at $x = L$ is obtained by integrating:

$F={\int }_0^L\rho x{\omega }^{2}dx$

 $F=2M{\omega }^2{\int }_0^{1}xdx$ 

 $F=\frac{2M{\omega }^2}{2}=M{\omega }^2$

Step 4: Solve for $\omega$

$$\omega^2 = \frac{F}{M}$$

 $$\omega = \sqrt{\frac{F}{M}}$$