Surface tension — excess pressure inside a soap bubble and liquid drop

A liquid drop has one surface (the outer surface in contact with air). Surface tension acts along this surface, trying to minimise the area. This inward pull creates an excess pressure inside.

Consider a spherical drop of radius $R$ with surface tension $T$. Imagine cutting it in half. The surface tension force along the rim of the cut acts inward: $F_{surface} = T \times 2\pi R$.

The pressure force pushing outward on the cross-section: $F_{pressure} = \Delta P \times \pi R^2$.

At equilibrium:

$\Delta P \times \pi R^2 = T \times 2\pi R$

$\boxed{\Delta P_{drop} = \frac{2T}{R}}$

A soap bubble has two surfaces — an inner surface and an outer surface (it's a thin film). Both surfaces contribute to the inward force.

The surface tension force is now doubled: $F_{surface} = 2 \times T \times 2\pi R = 4\pi RT$.

Equating with pressure force:

$\Delta P \times \pi R^2 = 4\pi RT$

$\boxed{\Delta P_{bubble} = \frac{4T}{R}}$

The factor of 4 (instead of 2) is because of the two free surfaces of the soap film.

Given: $R = 2$ cm $= 0.02$ m, $T = 0.03$ N/m

$\Delta P = \frac{4T}{R} = \frac{4 \times 0.03}{0.02} = \frac{0.12}{0.02}$

$\boxed{\Delta P = 6 \text{ Pa}}$