An air bubble of radius 10−2 m is rising up at a steady rate of 2×10−3 ms−1 through a liquid of density 1.5×103 kg m−3, the coefficient of viscosity neglecting the density of air, will be (g=10 ms−2)

Name: Video solution 1: An air bubble of radius 10−2 m is rising up at a steady rate of 2×10−3 ms−1 through a liquid of density 1.5×103 kg m−3, the coefficient of viscosity neglecting the density of air, will be (g=10 ms−2)
Uploaded: 2023-11-13T12:09:45.000Z
Duration: 6 min
Description: Video solution 1: An air bubble of radius 10−2 m is rising up at a steady rate of 2×10−3 ms−1 through a liquid of density 1.5×103 kg m−3, the coefficient of viscosity neglecting the density of air, will be (g=10 ms−2)

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Mechanical Properties of Fluids

Viscosity

An air bubble of radius 10^{-2} {~m} is rising up at a steady

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Solving time: 2 mins

An air bubble of radius $1 0^{- 2} m$ is rising up at a steady rate of $2 \times 1 0^{- 3} ms^{- 1}$ through a liquid of density $1.5 \times 1 0^{3} kg m^{- 3}$ , the coefficient of viscosity neglecting the density of air, will be $(g = 10 ms^{- 2})$

23.2

units

83.5

units

334 units

167 units

Views: 6,088 students

Updated on: Nov 13, 2023

Found 5 tutors discussing this question

Theodore Discussed

An air bubble of radius

1 0^{- 2} m

is rising up at a steady rate of

2 \times 1 0^{- 3} ms^{- 1}

through a liquid of density

1.5 \times 1 0^{3} kg m^{- 3}

, the coefficient of viscosity neglecting the density of air, will be

(g = 10 ms^{- 2})

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Let a bubble of radius

r

and densityp is rising up in a liquid whose density is

σ

and coefficient of viscosity

η

.

Then effective force acting downward

= V (ρ - σ) g = \frac{4}{3} π r^{3} (ρ - σ) g

Viscous force acting upward

= 6 π n rv

.
Since bubble is moving up with constant velocity

v

, there is no acceleration in it, the net force
acting on it must be zero.

∴ 6 π ηr v = \frac{4}{3} π r^{3} (ρ - σ) g

\Rightarrow η = \frac{2}{9} \frac{r ^{2} ( ρ - σ )}{v} g

Given,

v = - 2 \times 1 0^{- 3} ms^{- 1}, r = 1 0^{- 2} m

ρ = 0, σ = 1.5 \times 1 0^{3} kgm^{- 3}, g = 1 0^{- 2}

∴ η = \frac{2 \times ( 1 0 ^{- 2} ) ^{2} \times ( - 1.5 \times 1 0 ^{3} ) \times 10}{9 \times ( - 2 \times 1 0 ^{- 3} )}

= \frac{3}{18 \times 1 0 ^{- 3}} = \frac{1}{6} \times 1 0^{3}

= 0.167 \times 1 0^{3} = 167

units.

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Question Text	An air bubble of radius $1 0^{- 2} m$ is rising up at a steady rate of $2 \times 1 0^{- 3} ms^{- 1}$ through a liquid of density $1.5 \times 1 0^{3} kg m^{- 3}$ , the coefficient of viscosity neglecting the density of air, will be $(g = 10 ms^{- 2})$
Updated On	Nov 13, 2023
Topic	Mechanical Properties of Fluids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	116
Avg. Video Duration	6 min

An air bubble of radius $1 0^{- 2} m$ is rising up at a steady rate of $2 \times 1 0^{- 3} ms^{- 1}$ through a liquid of density $1.5 \times 1 0^{3} kg m^{- 3}$ , the coefficient of viscosity neglecting the density of air, will be $(g = 10 ms^{- 2})$

Text solutionVerified

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An air bubble of radius 10−2 m is rising up at a steady rate of 2×10−3 ms−1 through a liquid of density 1.5×103 kg m−3, the coefficient of viscosity neglecting the density of air, will be (g=10 ms−2)

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An air bubble of radius $1 0^{- 2} m$ is rising up at a steady rate of $2 \times 1 0^{- 3} ms^{- 1}$ through a liquid of density $1.5 \times 1 0^{3} kg m^{- 3}$ , the coefficient of viscosity neglecting the density of air, will be $(g = 10 ms^{- 2})$