A steel ball of density `8.0 xx 10^(3) kg//m^(3)` and radius 2 mm is observed to fall with a terminal velocity `1.0 xx 10^(-2) m//s` in a liquid of density `1.8 xx 10^(3) kg//m^(3)` . Calculate the viscosity of the liquid.

Explanation

We have,
`eta = (2r^(2)(d-d_(0))g)/(9u)`
`=(2(2 xx 10^(-3))^(2) (8.0 xx 10^(3) - 1.8 xx 10^(3)) xx 9.8)/(9 xx 1.0 xx 10^(-2))`
`=5.4 Nm^(-2) s`