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:
Show 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.
\(4.072 \times 10^{-3} \, \text{J}\)
\(5.76 \times 10^{-3} \, \text{J}\)
\(9.24 \times 10^{-3} \, \text{J}\)
\(1.848 \times 10^{-3} \, \text{J}\)
The Correct Option is D
Solution and Explanation
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: \(\boxed{1.848 \times 10^{-3} \, \text{J}}\)
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