A tube of length L = 1 m is filled completel | vedclass

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(D) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dm = \frac{M_{tota

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$1$

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(D) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dm = \frac{M_{total}}{L} dx = \frac{2M}{1} dx = 2M dx$.The centrifugal force $dF$ acting on this element is $dF = (dm) \omega^2 x = (2M dx) \omega^2 x$.The total force $F$ exerted by the liquid at the outer end is the integral of these forces from $x = 0$ to $x = L = 1 \ m$:$F = \int_{0}^{L} 2M \omega^2 x dx = 2M \omega^2 \left[ \frac{x^2}{2} \right]_{0}^{1} = 2M \omega^2 \left( \frac{1}{2} \right) = M \omega^2$.Given that the angular velocity is $\omega = \sqrt{\frac{F}{\alpha M}}$,we have $\omega^2 = \frac{F}{\alpha M}$.Comparing $F = M \omega^2$ with $\omega^2 = \frac{F}{M}$,we get $\alpha = 1$.

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