With what terminal velocity will an air bubble `0.8 mm` in diameter rise in a liquid of viscosity `0.15 N-s//m^(2)` and specific gravity `0.9`? Density of air is `1.293 kg//m^(3)`.
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The terminal velocity of the bubble is given by, `v_(T)=(2)/(9)(r^(2)(rho-sigma)g)/(eta)` Here, `r=0.4xx10^(-3) m`, `sigma=0.9xx10^(3) kgm^(-3) rho=1.293 kgm^(-3)` `eta=0.15, Nsm^(-2) " and " g=9.8 ms^(-2)` Substituting the values, we have, `v_(T)=(2)/(9)xx((0.4xx10^(-3))^(2)(1.293-0.9xx10^(3))xx9.8)/(0.15)` `= -0.0021 ms^(-1)` or `v_(T)= -0.21 cms^(-1)`
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