JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Fluid Mechanics

The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density

1750

kg m

^{- 3}

at the rate of

0.35

cms

^{- 1}

. The coefficient of viscosity of the solution is _____ poise (in nearest integer). (the density of air is negligible).

Answer: 11

Solution

When the bubble rises steadily (terminal velocity), the net force is zero. The buoyant force equals the viscous drag force (weight is negligible).

Buoyant force:

B = \frac{4}{3} π R^{3} ρ g

Viscous drag (Stokes' law):

F = 6 π η R v

At equilibrium:

\frac{4}{3} π R^{3} ρ g = 6 π η R v

Solving for viscosity:

η = \frac{2 R ^{2} ρ g}{9 v}

Given:

$R = 1$ mm $= 1 0^{- 3}$ m
$ρ = 1750$ kg/m³
$g = 10$ m/s²
$v = 0.35$ cm/s $= 3.5 \times 1 0^{- 3}$ m/s

η = \frac{2 \times ( 1 0 ^{- 3} ) ^{2} \times 1750 \times 10}{9 \times 3.5 \times 1 0 ^{- 3}} = \frac{2 \times 1 0 ^{- 6} \times 17500}{31.5 \times 1 0 ^{- 3}} = \frac{0.035}{0.0315} \approx 1.11 Pa\cdots

Converting to poise (1 Pa·s = 10 poise):

η \approx 11

poise.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Fluid Mechanics

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