The surface tension and vapour pressure of water at $20^{\circ} C$ are $7.28 \times 10^{-2} \, N/m$ and $2.33 \times 10^{3} \, Pa$ respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{\circ} C$?

(D) Given:
Surface tension of water $S = 7.28 \times 10^{-2} \, N/m$
Vapour pressure $P = 2.33 \times 10^{3} \, Pa$
The drop will remain stable without evaporating if the excess pressure inside the drop is equal to the vapour pressure.
The excess pressure inside a spherical droplet is given by $\Delta P = \frac{2S}{R}$.
Setting the excess pressure equal to the vapour pressure:
$P = \frac{2S}{R}$
Rearranging for the radius $R$:
$R = \frac{2S}{P}$
Substituting the values:
$R = \frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^{3}}$
$R = \frac{14.56 \times 10^{-2}}{2.33 \times 10^{3}}$
$R \approx 6.25 \times 10^{-5} \, m$