The potential energy of the free surface of a liquid drop is ` 2pixx 10^(-5)` times the surface tension of the liquid. What is the radius of the drop ? (Assume all quantities are in SI units )

Explanation

Data : Surface energy = `(2pi xx 10^(-5)) `T
Surface tension =, ` T = (" Surface energy")/(" Surface area")`
Surface energy = ` T xx ` surface area
`(2pi xx 10^(-5)) T =T (pi r^(2))`
The radius of the drop r = ` sqrt(2xx 10^(-5)) = sqrt(20 xx 10^(-6))`
` = 4.472 xx 10^(-3) ` m or 4.472 mm