Q.A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.

Why Does a Drop Have Extra Pressure Inside?

Imagine you're blowing a tiny soap bubble on a wand. You have to push air into it — that push is extra pressure. Now think of a raindrop hanging from a leaf. It's round, not flat. Something is squeezing it into that shape. That something is surface tension.

Surface tension acts like a stretched elastic skin on the liquid surface. It always tries to shrink the surface area. For a spherical drop, the surface is curved outward. The tension pulls tangentially along the surface, and because the surface is curved, that pull has a net inward component — like a rubber band squeezing a ball from all sides. That inward squeeze compresses the liquid inside, raising its pressure above the outside.

So the excess pressure is the extra pressure inside the drop (or bubble) compared to the atmosphere outside. It's what keeps the drop from collapsing.

The Precise Physics

For a spherical liquid drop (like a water droplet in air), there is only one liquid-air interface. Surface tension acts along that single surface. The excess pressure $\Delta P$ inside is given by:

where $T$ is the surface tension of the liquid and $R$ is the radius of the drop.

For a soap bubble, there are two liquid-air interfaces — an inner surface and an outer surface, each with its own surface tension. Both surfaces are curved and both contribute to the inward squeeze. So the excess pressure inside a soap bubble is double that of a single-surface drop:

Why the Factor of 2?

Here's the intuition without heavy math. For a single surface, the net inward force from surface tension on a hemisphere is $T \times (2\pi R)$ (tension times circumference). This force is balanced by the excess pressure acting on the cross-sectional area $\pi R^2$. Equating:

For a bubble, the same argument applies twice — once for the outer surface and once for the inner surface. Both pull inward, so the total inward force is doubled, giving $\Delta P = 4T/R$. …