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An oil drop of density 950 kg. m^-3 and ...

An oil drop of density 950 kg. `m^-3` and radius `10^-6` m is falling through air. The density of air is `1.3 kg. m^-3` and its coefficient of viscosity is `181xx10^-7` SI unit. Determine the terminal velocity of the oil drop. `[g=9.8m. S^-2]`

Text Solution

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Terminal velocity,
`v=(2)/(9)(r^2(rho-sigma)g)/(eta)`
[Here, `rho=950 kg. m^-3, r= 10^-6`,
`sigma=1.3kg. M^-3, eta=181xx10^-7` SI]
`=2/9((10^-6)^(2)(950-1.3)xx9.8)/(181xx10^-7)`
`=1.14xx10^-4m. S^-1`.
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