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A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is `1.8xx10^(-8)kgm^(-1)s^(-1)` what will be the terminal velocity of the drop. Density of air can be neglected. |
Answer» `V_(T)=(2)/(9)(r^(2)(rho-sigma)g)/(eta)=(2xx[(15xx10^(-4))/(1000)]^(2)xx10^(3)xx9.8)/(9xx1.8xx10^(-5))=2.72xx10^(-4)m//s` | |