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A charged oil dop falls 4.0 mm in 16.0 s...

A charged oil dop falls 4.0 mm in 16.0 s at constant speed in air in the absence of an electric field. The relative density of oil is 0.80, that of air is `1.3 xx 10^(-3)` and the viscosityf of air is `1.8 xx 10^(-5)` Ns `m^(-2)`. Find (i) the radius of the drop (ii) the mass of the drop. (iii) If the drop carries one electronic unit of charge and is an electric field of 2000 V `cm^(-1)`, what is the ratio of the force of the electric field on the drop to its weight ?

Text Solution

Verified by Experts

The correct Answer is:
`(i) 1.61 xx 10^(-6) m " "(ii)1.41 xx 10^(-14) kg " "(iii) 0.23`
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