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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 $dm = (m/L) dx$.The

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$4$

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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 $dm = (m/L) dx$.The centrifugal force $dF$ acting on this element is $dF = (dm) \omega^2 x = (m/L) \omega^2 x dx$.To find the total force $F$ exerted at the outer end,we integrate from $x = 0$ to $x = L$:$F = \int_{0}^{L} \frac{m}{L} \omega^2 x dx = \frac{m \omega^2}{L} \left[ \frac{x^2}{2} ight]_{0}^{L} = \frac{m \omega^2 L}{2}$.Given $m = 250\,g = 0.25\,kg$ and $L = 50\,cm = 0.5\,m$,we have:$F = \frac{0.25 	imes \omega^2 	imes 0.5}{2} = \frac{0.125}{2} \omega^2 = 0.0625 \omega^2$.Thus,$\omega^2 = F / 0.0625 = 16F$,which gives $\omega = 4 \sqrt{F}$.Comparing this with $\omega = x \sqrt{F}$,we get $x = 4$.

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