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With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m2 and specific gravity 0.9? Density of air is 1.29 kg/m3.

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Answer= - 0.782 × 10-3 m/s  (upwards)

Given: d=0.4 mm, r=d/2 = 0.2mm= 2×10-4 m, , h = 0.1 Pa-s, pL = 0.9 × 103 kg/m= 900 kg/m3 

pair = 1.29 kg/m3 g=9.8 m/s2

Solution:

Since the density of air is less than that of oil, the air bubble will rise up through the liquid. Hence, the viscous force is downward, At terminal velocity, this downward viscous force is equal in magnitude to the net upward force.

Viscous force = buoyant force - gravitational force

6πηrvt = \(\frac{4}{3}πr^3(p_L - p_{air})g\) 

∴ Terminal velocity,

vt =\(\frac{2r^2g(p_{L} - p_{air})}{9η}\) 

=\(\frac{2×10^{-4}9.8(900 - 1.29)}{9×0.1}\) 

= - 7.829 × 10-4 m/s  (upwards)

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