The surface tension and vapour pressure of a liquid at $25^{\circ} \mathrm{C}$ are $8 \times 10^{-2}…
The surface tension and vapour pressure of a liquid at $25^{\circ} \mathrm{C}$ are $8 \times 10^{-2} \mathrm{Nm}^{-1}$ and $2.5 \times 10^3 \mathrm{~Pa}$ respectively. The radius of the smallest spherical water droplet which can from without evaporating at $25^{\circ} \mathrm{C}$ is
$64 \mu \mathrm{m}$
$30 \mu \mathrm{m}$
$60 \mu \mathrm{m}$
$32 \mu \mathrm{m}$
Solution
The drop will evaporate if the water pressure is greater than the vapour pressure $\mathrm{P}$.
As $\mathrm{P}=\frac{2 \mathrm{~S}}{\mathrm{R}}$, where $\mathrm{R}=$ radius of drop
$\begin{aligned} & \mathrm{R}=\frac{2 \mathrm{~S}}{\mathrm{P}}=\frac{2 \times 8 \times 10^{-2}}{2.5 \times 10^3}=6.4 \times 10^{-5} \mathrm{~m} \\ & =64 \mu \mathrm{m}\end{aligned}$