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Step-by-Step Solution
Step 1: Understand the Energy Balance
When the drop evaporates, the energy required for vaporization (latent heat) must come from the decrease in surface energy. This ensures that the temperature of the drop remains unchanged.
Step 2: Express Change in Surface Energy
The surface area of a spherical drop of radius $R$ is $4\pi R^2$. A small change in radius by $\Delta R$ changes its surface area approximately by
$4\pi R^2 - 4\pi (R - \Delta R)^2.$
For very small $\Delta R$, this area change (denoted $\Delta A$) is approximately:
$4\pi R^2 - 4\pi (R^2 - 2R\,\Delta R) = 8\pi R\,\Delta R.$
So, the decrease in surface energy is $T \times \Delta A = T \times 8\pi R\,\Delta R,$
where $T$ is the surface tension.
Step 3: Express Latent Heat Energy Required
Let the mass of liquid that evaporates due to the radius change $\Delta R$ be $\Delta m$. If $\rho$ is the density of the liquid, the volume change of the drop is about:
$\ 4\pi R^2 \Delta R,$
so
$\ \Delta m = \rho \times (4\pi R^2 \Delta R).$
The latent heat required for this mass to evaporate is
$ \Delta E_{\text{latent}} = \Delta m \times L = \rho \,L \,(4\pi R^2 \Delta R).$
Step 4: Equate Decrease in Surface Energy to Latent Heat
For no temperature change, the energy lost from the surface must entirely supply the latent heat:
$ T \times 8\pi R \,\Delta R = \rho \,L \,(4\pi R^2 \,\Delta R).$
Cancelling common factors like $4\pi \,\Delta R$ on both sides, we get:
$ 2T = \rho\,L\, R.$
Step 5: Find the Minimum Radius
From the above relation,
$ R = \frac{2T}{\rho \, L}.$
This is the minimum radius of the drop for the evaporation to occur without any temperature change.
Answer
The correct answer is $ \displaystyle \frac{2T}{\rho \, L}. $
