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Physics - Fluid Question with Solution | TestHub

PhysicsFluidPressureMedium3 minQB

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Passage / Comprehension

A U-tube with liquids of densities $ρ$ and $2 ρ$ stands vertically on a rotating table.
Its rotation axis passes through the table centre. Initially the interface is at A, the liquid column above A is $8 a /3$ , and $A B = 2 a$ .
The tube has cross-sectional area S. At angular speed $ω$ , the interface shifts $2 a /3$ toward B. Use the shown geometry for Q16 to Q18.

Find the value of $ω$ :-

Options:

Answer:

Solution:

Across the light-liquid horizontal segment, from radius a to 5a/3:
$S (P_{x} - P_{A}) = \int_{a}^{5 a /3} ρS ω^{2} r d r$
$S (P_{x} - P_{A}) = \frac{8}{9} ρS ω^{2} a^{2}$
Across the heavy-liquid segment, from 5a/3 to 3a:
$S (P_{B} - P_{x}) = \int_{5 a /3}^{3 a} 2 ρS ω^{2} r d r$
$S (P_{B} - P_{x}) = \frac{56}{9} ρS ω^{2} a^{2}$
Adding and using Q16, P_B - P_A = 2aρg:
$2 ρ g a = \frac{64}{9} ρ ω^{2} a^{2}$
$ω^{2} = \frac{9 g}{32 a}$
$ω = \frac{9 g}{32 a}$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Pressure

⏱ 3mℹ️ Source: QB

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