Understanding the Concept:
For a spherical liquid droplet to form or stay stable, the pressure inside the droplet must exceed the pressure outside. This difference is known as excess pressure ($\Delta P$). For a spherical droplet with a single surface boundary layer, the excess pressure is calculated as:
\[
\Delta P = \frac{2T}{r}
\]
where $T$ represents the surface tension and $r$ is the radius of the droplet.
Step 1: Substitute values to calculate excess pressure ($\Delta P$).
Given values:
Surface tension, $T = 7.28 \times 10^{-2}\text{ N m}^{-1}$
Radius, $r = 6.25 \times 10^{-5}\text{ m}$
Let's compute the value of $\Delta P$:
\[
\Delta P = \frac{2 \times 7.28 \times 10^{-2}}{6.25 \times 10^{-5}}
\]
\[
\Delta P = \frac{14.56 \times 10^{-2}}{6.25 \times 10^{-5}} = \frac{14.56}{6.25} \times 10^3
\]
\[
\Delta P = 2.3296 \times 10^3\text{ Pa} \approx 2.33 \times 10^3\text{ Pa}
\]