1.

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.

Answer»

`3.0xx10^(-17)kg`
`3.2xx10^(-17)kg`
`2.5xx10^(-17)kg`
`3.3xx10^(-17)kg`

Solution :(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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