An oil drop of 12 excess electrons is held stationary under a constant electric field of `2.55xx10^(4)NC^(-1)` in Millikan's oil drop experiment. The density of the oil is `1.26xx10^(3)kgm^(-3)`. Estimate the radius of the drop `(g=9.81ms^(-2),e=1.60xx10^(-19)C)`.

Options

A. `4 . 3 xx 10^(-7) m`
B. `7 . 8 xx 10^(-7) m`
C. `0 . 078 xx 10^(-7) m `
D. `3 . 4 xx 10^(-7) m`

Answer

B

Explanation

` F = qE = mg (q = 6 e = 6 xx 1 . 6 xx 10^(-19))`
Density `(d) = ("mass")/( "volume") = (m) /((4)/(3) pi r ^(3)) or r^(3) = (m)/((4)/(3) pi d)`
Putting the value of d and m `(- (qE)/(g))` and solving we get ` = 7 . 8 xx 10^(-7) m`