Q.The lower end of a capillary tube of diameter $2.00\ \text{mm}$ is dipped $8.00\ \text{cm}$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is $7.30 \times 10^{-2}\ \text{N m}^{-1}$. $1$ atmospheric pressure $= 1.01 \times 10^{5}\ \text{Pa}$, density of water $= 1000\ \text{kg/m}^{3}$, $g = 9.80\ \text{m s}^{-2}$. Also calculate the excess pressure.

Concept: Surface Tension — Excess Pressure in a Bubble

The bubble is hemispherical and formed at the tube's end inside water. The pressure inside the bubble must overcome both the hydrostatic pressure at that depth and the excess pressure due to surface tension.

Step 1: Hydrostatic pressure at depth

Depth $h = 8.00\ \text{cm} = 0.0800\ \text{m}$.

Step 2: Excess pressure for a hemispherical bubble

For a bubble in liquid (one surface), excess pressure is:

Radius $r = 1.00\ \text{mm} = 1.00 \times 10^{-3}\ \text{m}$.

$\Delta P = \frac{2 \times 7.30 \times 10^{-2}}{1.00 \times 10^{-3}} = 146\ \text{Pa}$ …

Concept and Intuition

When you blow a bubble at the end of a capillary tube submerged in water, you're fighting two things. First, the water itself pushes inward because of its weight — that's hydrostatic pressure, which increases with depth. Second, the bubble's curved surface creates an additional inward squeeze called excess pressure (or Laplace pressure). For a hemispherical bubble, the excess pressure is given by $P_{\text{excess}} = \frac{2T}{r}$, where $T$ is surface tension and $r$ is the bubble's radius.

The pressure you need to supply inside the tube must equal the sum of these two: the hydrostatic pressure at the depth of the bubble, plus the excess pressure from curvature. The atmospheric pressure is already acting on the water surface, so we need to account for that too.

Step-by-step Solution

1. Identify the given data

• Tube diameter = $2.00\ \text{mm}$, so radius $r = 1.00\ \text{mm} = 1.00 \times 10^{-3}\ \text{m}$
• Depth of tube end below water surface: $h = 8.00\ \text{cm} = 0.0800\ \text{m}$
• Surface tension: $T = 7.30 \times 10^{-2}\ \text{N/m}$
• Atmospheric pressure: $P_{\text{atm}} = 1.01 \times 10^{5}\ \text{Pa}$
• Density of water: $\rho = 1000\ \text{kg/m}^3$
• $g = 9.80\ \text{m/s}^2$

2. Calculate the hydrostatic pressure at depth

The pressure due to the water column at depth $h$ is:

This is the additional pressure from the water's weight, over and above atmospheric pressure.

3. Calculate the excess pressure due to surface tension

For a hemispherical bubble of radius $r$ in a liquid, the excess pressure inside the bubble relative to the surrounding liquid is:

Substitute the values:

Step 1: three additive contributions -- atmospheric, hydrostatic, excess pressure. Step 2: P_hydro=rhogh=784 Pa. Step 3: bubble inside water (one surface only) uses deltaP=2T/r, not 4T/r: =146 Pa. Step 4: total=1.01e5+784+146~=1.02e5 Pa. Com …