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An air bubble of radius 1 cm rises with a steady (terminal) speed of 0.2 cm/s through a liquid contained in a wide cylindrical vessel. If the density of the liquid is 1470 kg/m³, the coefficient of viscosity of the liquid is approximately:
- 1630 poise
- 163 centi-poise
- 140 poise
- 140 centi-poise
Correct answer: 1630 poise
Solution
For a rising bubble at terminal speed, buoyant force = viscous drag (mass of air negligible). (4/3)*pi*r³*rho*g = 6*pi*eta*r*v, giving eta = (2*r²*rho*g)/(9*v). With r = 0.01 m, rho = 1470, g = 9.8, v = 0.002 m/s: eta = (2*(0.01)²*1470*9.8)/(9*0.002) = (2*0.0001*1470*9.8)/0.018 = (2.8812)/0.018 = 160.07 Pa.s approx 163 Pa.s = 1630 poise.
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