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A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the ${OO}^{1}$ . The manometers A and B fixed in the tube at distances $r_{1}$ and $r_{2}$ from rotational axis indicate pressures $p_{1}$ and $p_{2}$ respectively. Determine the angular velocity $ω$ of rotation of the tube.

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$ω = \sqrt{\frac{2 (p_{2} - p_{1})}{ρ (r_{2}^{2} - r_{1}^{2})}}$

$ω = \sqrt{\frac{2 (p_{2} - p_{1})}{ρ (r_{2}^{2} + r_{1}^{2})}}$

$ω = \sqrt{\frac{2 (r_{2}^{2} - r_{1}^{2})}{p_{1} - p_{2}}}$

$ω = \sqrt{\frac{2 (p_{2} - p_{1})}{ρ r_{1} r_{2}}}$

answer is A.

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Detailed Solution

Consider the condition of equilibrium for the mass of water contained between cross section separated by x and x + dx from rotation axis.

$d p = ρ ω^{2} x d x \Rightarrow \int_{p_{1}}^{p_{2}} d p = \int_{r_{1}}^{r_{2}} ρ ω^{2} x d x$

$p_{2 -} p_{1 =} ρ ω^{2} {[\frac{x^{2}}{2}]}_{r_{1}}^{r_{2}} = ρ ω^{2} (\frac{r_{2}^{2} - r_{1}^{2}}{2})$

$ω = \sqrt{\frac{2}{ρ} \frac{(p_{2 -} p_{1})}{(r_{2}^{2} - r_{1}^{2})}}$

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