Question

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

• A. \(542.6 \times 10^{-6}\,\text{J}\)
• B. \(543.6 \times 10^{-6}\,\text{J}\)
• C. \(542.6 \times 10^{-5}\,\text{J}\)
• D. \(545.6 \times 10^{-6}\,\text{J}\)

Answer 1

The Correct Option is A

Concept:
For a soap bubble, there are two surfaces (inner and outer). Therefore, the work done in expanding the bubble equals the increase in surface energy of both surfaces. Surface energy: \[ U = 2 \gamma A \] Work done in expansion: \[ W = \Delta U = 2\gamma (A_2 - A_1) \] Area of a sphere: \[ A = 4\pi r^2 \] Thus, \[ W = 2\gamma \left[4\pi r_2^2 - 4\pi r_1^2 \right] \] Step 1: Write the expression for work done. \[ W = 2\gamma \left[4\pi (r_2^2 - r_1^2)\right] \] \[ W = 8\pi\gamma (r_2^2 - r_1^2) \] Step 2: Substitute the given values. \[ \gamma = 7.2 \times 10^{-2}\, \text{N/m} \] \[ r_1 = 1\,\text{cm} = 1\times10^{-2}\,\text{m} \] \[ r_2 = 2\,\text{cm} = 2\times10^{-2}\,\text{m} \] \[ W = 8\pi \times 7.2 \times 10^{-2} \left[(2\times10^{-2})^2 - (1\times10^{-2})^2\right] \] Step 3: Simplify the expression. \[ (2\times10^{-2})^2 = 4\times10^{-4} \] \[ (1\times10^{-2})^2 = 1\times10^{-4} \] \[ r_2^2 - r_1^2 = 3\times10^{-4} \] \[ W = 8\pi \times 7.2 \times 10^{-2} \times 3\times10^{-4} \] \[ W = 542.6 \times 10^{-6}\,\text{J} \]