Find the work done in expanding a soap bubble from radius \(1\,\text{cm}\) to \(2\,\text{cm}\). Surface tension is \( \gamma = 7.2 \times 10^{-2}\,\text{N/m} \).

Question

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 ms^–1and 65 ms^–1 respectively. If the wing area is 2 m2 the lift of the wing is _______ N.
(Given density of air = 1.2 kg m^-3)

Answer 1

The work done in expanding a soap bubble can be calculated using the formula for the work done against the surface tension:

W = 2 \times \gamma \times (\Delta A)

where:

\gamma = 7.2 \times 10^{-2} \, \text{N/m} (surface tension)
\Delta A is the change in surface area.

Since a soap bubble has two surfaces (inner and outer), the factor of 2 is included in the formula.

The surface area A of a sphere is given by:

A = 4 \pi r^2

Initially, the radius r_1 = 1\,\text{cm} = 0.01\,\text{m} and finally r_2 = 2\,\text{cm} = 0.02\,\text{m}.

The change in surface area is:

\Delta A = 4 \pi (r_2^2 - r_1^2)

Substituting the values:

\Delta A = 4 \pi ((0.02)^2 - (0.01)^2)

= 4 \pi (0.0004 - 0.0001)

= 4 \pi \times 0.0003

= 0.0012 \pi \, \text{m}^2

Now, calculate the work done:

W = 2 \times 7.2 \times 10^{-2} \times 0.0012 \pi

= 0.0144 \times 0.0012 \times \pi

= 0.0000542 \times \pi \, \text{J}

= 0.0001702 \, \text{J}

Convert to microjoules = 170.2 \, \mu\text{J}

Therefore, the work done is 542.6 \times 10^{-6}\,\text{J}.

Answer 2