Assume that a drop of a liquid evaporates by a decrease in its surface energy so that its temperature remains unchanged. The minimum radius of the drop for this to be possible is. (The surface tension is T, the density of the liquid is $$\rho$$ and L is its latent heat of vaporisation.)

We consider a spherical drop of liquid of radius $$r$$. Its surface energy is equal to the product of its surface tension and its surface area, while the energy required to vaporise a small mass of the liquid is obtained from the definition of latent heat. We wish to see whether the surface energy released when the drop shrinks by an infinitesimal amount can supply the latent heat needed for the corresponding infinitesimal mass to evaporate, all at the same temperature.

The surface area of a sphere is $$A = 4\pi r^{2}$$. If the radius decreases by an infinitesimal amount $$dr$$ (with $$dr<0$$ as the drop is getting smaller), the change in surface area is obtained by differentiating:

$$dA = \frac{d}{dr}(4\pi r^{2})\,dr = 8\pi r\,dr.$$

Surface energy equals $$T\times A$$. Therefore the change in surface energy is

$$dE_{\text{s}} = T\,dA = T\,(8\pi r\,dr)=8\pi rT\,dr.$$

(Because $$dr<0$$, $$dE_{\text{s}}$$ is negative, signifying that energy is released. We will shortly compare magnitudes, so the sign will not matter.)

Next, the volume of the drop is $$V = \dfrac{4}{3}\pi r^{3}$$. Its infinitesimal change is

$$dV = \frac{d}{dr}\!\left(\frac{4}{3}\pi r^{3}\right)\!dr = 4\pi r^{2}\,dr.$$

The mass corresponding to this change of volume is, using density $$\rho$$,

$$dm = \rho\,dV = \rho\,(4\pi r^{2}\,dr)=4\pi r^{2}\rho\,dr.$$

Latent heat of vaporisation $$L$$ is defined as the heat required per unit mass to convert liquid to vapour at constant temperature. Hence the heat needed to vaporise the mass $$dm$$ is

$$dQ = L\,dm = L\,(4\pi r^{2}\rho\,dr)=4\pi r^{2}\rho L\,dr.$$

For the evaporation to proceed solely at the expense of the drop’s surface energy, the magnitude of the surface energy released must at least equal the latent heat required, so we set

$$|dE_{\text{s}}| = dQ.$$

Using the expressions obtained, this becomes

$$8\pi rT\,|dr| = 4\pi r^{2}\rho L\,|dr|.$$

The common factors $$4\pi|dr|$$ cancel immediately, leaving

$$2T = r\rho L.$$

Solving for the radius, we get

$$r = \frac{2T}{\rho L}.$$

This value represents the minimum radius. For a drop of exactly this radius, the surface energy released during an infinitesimal contraction is just sufficient to supply the latent heat of the infinitesimal mass that evaporates. If the drop were any smaller, the released surface energy would be insufficient.

Hence, the correct answer is Option B.