Question

The surface tension and vapour pressure of water at $20^\circ C$ are $7.28\times 10^{- 2} \, Nm^{- 1}$ and $2.33\times 10^{3}Pa$ respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

• A. $1.25\times 10^{-5} \, m$
• B. $6.25\times 10^{-5} \, m$
• C. $4.3\times 10^{-5} \, m$
• D. $3.4\times 10^{-5} \, m$

Answer 1

Answer: (B)

Here, $S=7.28\times 10^{- 2}Nm^{- 1},P=2.33\times 10^{3}Pa$
The drop will evaporate if the water pressure is greater than the vapour pressure $P$
As $P=\frac{2 S}{R}$ where $R$ is the radius of drop
$\therefore R=\frac{2 S}{P}=\frac{2 \times 7.28 \times 10^{- 2}}{2.33 \times 10^{3}}=6.25\times 10^{- 5}$