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Variation of G Due to Height and Depth

A tube filled...

Question

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

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

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

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

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

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Solution

The correct option is B $ω = \sqrt{\frac{2 (p_{2} - p_{1})}{ρ (r_{1}^{2} - r_{2}^{2})}}$

Applying Bernoulli's principle, we have
$p_{1} + \frac{1}{2} ρ {v_{1}}^{2} = p_{2} + \frac{1}{2} ρ {v_{2}}^{2} \Rightarrow p_{2} - p_{1} = \frac{1}{2} ρ {v_{1}}^{2} - \frac{1}{2} ρ {v_{2}}^{2} \Rightarrow p_{2} - p_{1} = \frac{1}{2} ρ ω^{2} r_{1}^{2} - \frac{1}{2} ρ ω^{2} r_{2}^{2} \Rightarrow \frac{2 (p_{2} - p_{1})}{ρ (r_{1}^{2} - r_{2}^{2})} = ω^{2} \Rightarrow ω = \sqrt{\frac{2 (p_{2} - p_{1})}{ρ (r_{1}^{2} - r_{2}^{2})}}$

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A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the OO' axis. The manometers A and B fixed in the tube at distances r1 and r2 from rotational axis indicate pressure p1 and p2, respectively. Determine the angular velocity ω of rotation of the tube:

The correct option is B ω=√2(p2−p1)ρ(r21−r22) Applying Bernoulli's principle, we have p1+12ρv12=p2+12ρv22⇒p2−p1=12ρv12−12ρv22⇒p2−p1=12ρω2r21−12ρω2r22⇒2(p2−p1)ρ(r21−r22)=ω2⇒ω=√2(p2−p1)ρ(r21−r22)