A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the left dotted line as shown in the figure. Find the difference in the levels of the liquid column.

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$\frac{2 \sqrt{2} ω^{2} L^{2}}{g}$

$\frac{ω^{2} L^{2}}{2 \sqrt{2} g}$

$\frac{2 ω^{2} L^{2}}{g}$

$\frac{ω^{2} L^{2}}{2 g}$

answer is B.

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Detailed Solution

$dF = dm x ω^{2} and (dp) A = (ρ Adx) \times ω^{2}$

$\int_{p_{1}}^{p_{2}} d p = ρ ω^{2} \int_{0}^{L} x d x =$

$or P_{2 -} P_{1} = \frac{ρ ω^{2} L^{2}}{2}$

$\Rightarrow ρ g H = \frac{ρ ω^{2} L^{2}}{2}; H = \frac{ω^{2} L^{2}}{2 g}$

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