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

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$\frac{M\omega^2 L}{2}$

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(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation.The mass of this small element is $dm = \frac{M}{L} dx$.The centripetal force required for this element is $dF = (dm) x \omega^2 = \left(\frac{M}{L} dx\right) x \omega^2 = \frac{M\omega^2}{L} x dx$.To find the total force $F$ exerted by the liquid at the outer end (at $x = L$),we integrate the force from $x = 0$ to $x = L$:$F = \int_{0}^{L} \frac{M\omega^2}{L} x dx = \frac{M\omega^2}{L} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{M\omega^2}{L} \cdot \frac{L^2}{2} = \frac{M\omega^2 L}{2}$.

$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is

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