Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is `rho` and L is its latent heat of vaporization.

Options

A. `sqrt(T//pL)`
B. `T//pL`
C. `2T//pL`
D. `pL/T`

Answer

C

Explanation

Let initial radius of drop is r and on evaporation it decreases by dr
Change in surface area `=triangeA=8pi rdr`
Decreases in surface energy =Latent heat required for vaporisation
`T triangleA=triangleQ=mL`
`T xx (8pi rdr)=p L (4pir^(2)dr)`
`rArr r=(2T)/(pL)`