Question

The surface tension of a soap bubble is \(2 \times 10^{-2} \, \text{N/m}\). Work done to increase the radius of the bubble from 3.5 cm to 7 cm will be:

• A. \(4.072 \times 10^{-3} \, \text{J}\)
• B. \(5.76 \times 10^{-3} \, \text{J}\)
• C. \(9.24 \times 10^{-3} \, \text{J}\)
• D. \(1.848 \times 10^{-3} \, \text{J}\)

Hint:

For work done on bubbles:
• Account for both inner and outer surfaces when calculating surface area.
• Use the formula W = T ×∆A consistently with units.

Answer 1

The Correct Option is D

1. Work Done: - Work done is equal to the change in surface energy:

\[W = \text{Surface tension} \times \Delta A.\]

2. Surface Area of a Bubble: - Surface area of a sphere:

\[A = 4\pi R^2.\]

- Change in surface area for a bubble (two surfaces):

\[\Delta A = 2 \times 4\pi (R_2^2 - R_1^2).\]

3. Substitute Values: - \(R_1 = 3.5 \, \text{cm}\), \(R_2 = 7 \, \text{cm}\), \(T = 2 \times 10^{-2} \, \text{N/m}\)

\[W = 2 \times 10^{-2} \times 2 \times 4\pi ((0.07)^2 - (0.035)^2).\]

- Simplify:

\[W = 2 \times 10^{-2} \times 2 \times 4\pi \times (0.0049 - 0.001225) \approx 1.848 \times 10^{-3}.\]

Final Answer: \(\boxed{1.848 \times 10^{-3} \, \text{J}}\)