Length of a horizontal arm of a U-tube is ℓ = 21 cm and ends of both of the vertical arms are open to surroundings of pressure 10500 Nm − 2 . A liquid of density ρ = 1 0 3 kg m − 3 is poured into the

Question

Length of a horizontal arm of a U-tube is ℓ = 21 cm and ends of both of the vertical arms are open to surroundings of pressure 10500 Nm − 2 . A liquid of density ρ = 1 0 3 kg m − 3 is poured into the tube such that liquid just fills horizontal part of the tube. Now one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity ω 0 ​ = 10 radians/sec. If length of each vertical arm be a = 6 cm , calculate the length of air column in the sealed arm.

TestHub · Accepted Answer

Let the liquid shift by x; air compressed isothermally: $P = \frac{P _{0} a}{a - x}$

Pressure at the bend B: $P_{B} = P + x ρ g$

Excess pressure supplies centripetal force for the horizontal column: $(P_{B} - P_{0}) S = Sρ (ℓ - x) ω_{0}^{2} \frac{ℓ + x}{2}$

Substituting $ℓ = 21$ cm, $a = 6$ cm, $P_{0} = 10500$ N/m $^{2}$ , $ω_{0} = 10$ rad/s gives $x = 1$ cm

Length of air column $= a - x = 5$ cm

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