Explanation:
The excess of pressure due to surface tension on a spherical liquid drop is given by
\(p = \frac{{2T}}{R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Where, \(T = \) surface tension and
\(R = \) radius of liquid drop So, from eq.(1), we get
\(p \propto \dfrac{1}{R} \text { or } p \propto R^{-1}\)