The total energy of free surface of a liquid drop is `2pi` times the surface tension of the liquid. What is the diameter of the drop ?
[Assume all terms in SI unit].

Explanation

Given, `E=2pi T`
`E=T Delta A`
`therefore 2pi T=T(4pi r^(2)) " " [because Delta A=4pi r^(2)]`
`2r^(2)=1`
`r^(2)=(1)/(2)`
`therefore r=(1)/(sqrt(2))`
Now, Diameter (d) `=2r=2xx(1)/(sqrt(2))`
`=sqrt(2)`
`=1.414 m.`