Q.What is the pressure inside the drop of mercury of radius $3.00\ \text{mm}$ at room temperature? Surface tension of mercury at that temperature ($20\,^{\circ}\text{C}$) is $4.65 \times 10^{-1}\ \text{N m}^{-1}$. The atmospheric pressure is $1.01 \times 10^{5}\ \text{Pa}$. Also give the excess pressure inside the drop.

Concept: Capillary Action — For a liquid drop, the curved surface creates an excess pressure due to surface tension.

Step 1: Excess pressure formula

For a spherical drop, the excess pressure inside is given by

where $T$ is surface tension and $r$ is radius.

Step 2: Substitute values

$T = 4.65 \times 10^{-1}\ \text{N/m}$, $r = 3.00\ \text{mm} = 3.00 \times 10^{-3}\ \text{m}$

Step 3: Total pressure inside …

Why This Works — The Physics of Capillary Action

A liquid drop is not just a blob of fluid — it's held together by surface tension. The molecules at the surface experience a net inward pull because they have fewer neighbours on the outside. This creates a curved interface that is always under tension, like a stretched rubber membrane.

For a spherical drop, this tension tries to shrink the surface area. To keep the drop from collapsing, the pressure inside must be greater than the pressure outside. That difference — the excess pressure — is what we calculate.

The key formula comes from balancing forces on a hemisphere of the drop: the surface tension force around the circumference ($T \times 2\pi r$) must equal the net pressure force on the flat circular face ($\Delta P \times \pi r^2$). Solving gives:

This is the Laplace pressure for a spherical interface with one free surface (a liquid drop in air). Notice: smaller radius → larger excess pressure. That's why tiny droplets are much harder to deform than large ones.

Step-by-Step Calculation

1. Identify the given data

• Radius of mercury drop: $r = 3.00\ \text{mm} = 3.00 \times 10^{-3}\ \text{m}$
• Surface tension of mercury at $20\,^{\circ}\text{C}$: $T = 4.65 \times 10^{-1}\ \text{N/m}$
• Atmospheric pressure: $P_{\text{atm}} = 1.01 \times 10^{5}\ \text{Pa}$

2. Compute the excess pressure

Using $\Delta P = \frac{2T}{r}$:

First, multiply numerator: $2 \times 4.65 \times 10^{-1} = 9.30 \times 10^{-1} = 0.930$

Then divide: $\frac{0.930}{3.00 \times 10^{-3}} = \frac{0.930}{0.00300} = 310\ \text{Pa}$

So the excess pressure is $310\ \text{Pa}$. …

Drop has one interface: deltaP=2T/r=2*(4.65e-1)/(3.00e-3)=310 Pa. P_inside=P_ …