Saved Bookmarks
| 1. |
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)`. |
|
Answer» Correct Answer - A::B::C 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)kg//m^(3), rho=1.293 kg//m^(3), eta=0.15 N-s//m^(2)` and `g=9.8m//s^(2)` Substituting the values, we have `v_(T)=(2)/(9)xx ((0.4 xx 10^(-3))^(2)(1.293-0.9 xx 10^(3))xx9.8)/(0.15)` `= - 0.0021 m//s` or `v_(T) =- 0.21 cm` |
|