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PhysicsFluidPressureMedium2 minQB

PhysicsMediuminteger

The interface of two liquids of densities $ρ$ and $2 ρ$ respectively lies at point A in a uniform $U$ tube at rest. The height of liquid column above $A$ is $8 a /3$ where $A B = a$ . The cross sectional area of the tube is whirled about a vertical axis at distance 'a' as shown in diagram with angular velocity $ω_{1}$ , the interface of the liquids shifts towards B by $2 a /3$ . Find the value of $19 \frac{ω _{1} ^{2}}{g}$

Answer:

Solution:

Unbalanced pressure forces on the liquid column in horizontal arm of u-tube is equal tocentripetal force. Solving we get $ω_{1} = \frac{18}{19} \frac{g}{a}$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Pressure

⏱ 2mℹ️ Source: QB

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