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The interface of two liquids of densities & 2 respectively li

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The interface of two liquids of densities $ρ$ & $2 ρ$ respectively lies at the point A in a light U tube at rest. The height of liquid column above A is $\frac{8}{3} a$ , where $A B = a$ . The cross-sectional area of the tube is S. With what angular velocity it must be whirled about a vertical axis at a distance 'a' such that the interface of the liquid shifts towards B by $\frac{2}{3} a$ .

Name: Video solution 1: The interface of two liquids of densities ρ & 2ρ respectively lies at the point A in a light U tube at rest. The height of liquid column above A is 38a, where AB=a. The cross-sectional area of the tube is S. With what angular velocity it must be whirled about a vertical axis at a distance 'a' such that the interface of the liquid shifts towards B by 32a.
Uploaded: 2023-10-12T03:43:09.000Z
Duration: 9 min
Description: Video solution 1: The interface of two liquids of densities ρ & 2ρ respectively lies at the point A in a light U tube at rest. The height of liquid column above A is 38a, where AB=a. The cross-sectional area of the tube is S. With what angular velocity it must be whirled about a vertical axis at a distance 'a' such that the interface of the liquid shifts towards B by 32a.

2 \frac{6 g}{a}

\frac{20 g}{9 a}

\frac{g}{2 a}

\frac{9 g}{20 a}

Views: 5,783 students

Updated on: Oct 12, 2023

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Lucas Discussed

The interface of two liquids of densities

ρ

2 ρ

respectively lies at the point A in a light U tube at rest. The height of liquid column above A is

\frac{8}{3} a

, where

A B = a

. The cross-sectional area of the tube is S. With what angular velocity it must be whirled about a vertical axis at a distance 'a' such that the interface of the liquid shifts towards B by

\frac{2}{3} a

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

The densities of two liquids are

ρ

and

2 ρ

Height,

h = \frac{8 a}{3}

The velocity at point A

v = \frac{2 g h ρ}{ρ}

= 2 g \times \frac{8 a}{3} \times \frac{2 ρ}{ρ}

= \frac{32 a g}{3}

\Rightarrow v = ω \times \frac{2 a}{3}

\Rightarrow ω = \frac{3}{2 a} \times v = \frac{3}{2 a} \times \frac{32 a g}{3}

= 2 \frac{6 g}{a}

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Question Text	The interface of two liquids of densities $ρ$ & $2 ρ$ respectively lies at the point A in a light U tube at rest. The height of liquid column above A is $\frac{8}{3} a$ , where $A B = a$ . The cross-sectional area of the tube is S. With what angular velocity it must be whirled about a vertical axis at a distance 'a' such that the interface of the liquid shifts towards B by $\frac{2}{3} a$ .
Updated On	Oct 12, 2023
Topic	Mechanical Properties of Fluids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	434
Avg. Video Duration	10 min

Text solutionVerified

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