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A uniform U – Tube containing a liquid of uniform density $ρ,$ rotates about a vertical axis with constant angular velocity $ω,$ as shown in the figure
Length of the horizontal section is L. Difference in height of the liquid columns in the two vertical sections of the tube is $Δ h = \frac{ω^{2} L^{2}}{x g} .$ Find $x .$

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answer is 6.

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Detailed Solution

On $' d x'$ element shown,

$\begin{array}{l} - d T = d m x ω^{2} \\ T = \frac{m ω^{2}}{2 L} (L^{2} - x^{2}) \end{array}$

The required force is difference of value of $T$ at $x = 0$ and $x = \frac{L}{2}$

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