A liquid drop of mass `0.0129 g` drips from a capillary. When the drop breaks away, the diameter of the neck of capliiary is 1mm. Find surface tension of liquid.

Explanation

Here, `m=0.0129 g`
`=0.0129 xx 10^(-3) kg = 1.29 xx 10^(-5)kg`,
`D=1 mm = 10^(-3)m`
When drop breaks away from the capillary,
weight of drop = force of S.T acing on capillary
`mg=(pi D) xx S`
or,` S=(mg)/(pi D) = ((1.29xx10^(-5))xx9.8)/(3.14 xx (10^(-3))`
`=4.3 xx 10^(-2) Nm^(-1)`.