JEE Main · 2013 · 07 Apr 2013 · Offline · Fluid Mechanics

If the drop is to evaporate at constant temperature, the surface energy released as it shrinks must exactly supply the latent heat for the vapour. Peel off a thin shell of thickness $\Delta R$: the mass lost is $\rho 4\pi R^2\Delta R$, needing heat $\rho 4\pi R^2\Delta RL$, while the surface energy given up is $T\cdot 8\pi R\Delta R$. Setting the two equal, $\rho R^{2}L=2TR$, so the smallest radius for which evaporation can sustain itself this way is $R=\frac{2T}{\rho L}$.