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An oil drop of charge of 2 electrons fall freel with as terminal speed. Calculate the mas of oil drop so, it can move upward with same terminal speed, if electric field of `2xx10^(3) V//m` is applied.
A. `3.0xx10^(-17)kg`
B. `3.2xx10^(-17)kg`
C. `2.5xx10^(-17)kg`
D. `3.3xx10^(-17)kg`

1 Answer

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Best answer
(b) `mg=F=6pietarv` ………..(i)
`qE-mg=6pi eta rv=F` …………..(ii)
From eqs (i) and (ii) we get
`qE=6 pi eta rv+mg=2mg`
`E=(2mg)/q=(2mg)/(2e)=(mg)/e`
`m=(2xx10^(3)xx1.6xx10^(-19))/10=3.2xx10^(-17)kg`

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