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 $$\sqrt{\frac{F}{\alpha M}}$$ in SI unit. The value of $$\alpha$$ is __________.

We need to find $$\alpha$$ where the angular velocity is $$\sqrt{\frac{F}{\alpha M}}$$ for a tube of length 1 m filled with liquid of mass 2M, rotated about one end.

Set up the problem

A tube of length $$L = 1$$ m is filled with liquid of mass $$2M$$. It rotates about one end with angular velocity $$\omega$$. We need the force at the other end (the far end).

Find the centripetal force

Linear mass density: $$\lambda = \frac{2M}{L} = 2M$$ (since $$L = 1$$).

Consider an element at distance $$r$$ from the axis: $$dm = \lambda \, dr = 2M \, dr$$.

The centrifugal force on this element: $$dF = dm \cdot \omega^2 r = 2M\omega^2 r \, dr$$.

The force at the far end ($$r = L = 1$$) is due to the liquid beyond some point — but since it's the end, the force is exerted by the liquid on the closed end.

Actually, the force at the far end is the integral of centripetal acceleration of all the liquid:

$$F = \int_0^L \lambda \omega^2 r \, dr = 2M\omega^2 \int_0^1 r \, dr = 2M\omega^2 \times \frac{1}{2} = M\omega^2$$

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

So $$\alpha = 1$$.

Conclusion

$$\alpha = 1$$.

Therefore, the answer is 1.