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(D) The energy required for the evaporation of a mass $dm$ of the liquid is given by $dQ = L \cdot dm$. Since the mass of the drop is $m = \rho \cdot

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$\frac{2T}{\rho L}$

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(D) The energy required for the evaporation of a mass $dm$ of the liquid is given by $dQ = L \cdot dm$. Since the mass of the drop is $m = \rho \cdot V = \rho \cdot (\frac{4}{3} \pi r^3)$,the change in mass for a change in radius $dr$ is $dm = \rho \cdot (4 \pi r^2) dr$. Thus,the energy required is $dQ = L \cdot \rho \cdot (4 \pi r^2) dr$. This energy is supplied by the decrease in surface energy. The surface area of a drop is $A = 4 \pi r^2$,so the surface energy is $U = T \cdot A = T \cdot (4 \pi r^2)$. The change in surface energy for a change in radius $dr$ is $dU = T \cdot d(4 \pi r^2) = T \cdot (8 \pi r) dr$. Equating the energy required for evaporation to the decrease in surface energy: $L \cdot \rho \cdot (4 \pi r^2) dr = T \cdot (8 \pi r) dr$. Solving for $r$: $4 \pi \rho L r^2 = 8 \pi T r$ $r = \frac{8 \pi T}{4 \pi \rho L} = \frac{2T}{\rho L}$.

Assume that a drop of liquid evaporates by decrease 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.

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