Question

What is the work done to increase the radius of a soap bubble from 1 cm to 1.1 cm, if the surface tension of the soap solution is 0.025 N/m?

• A. $ 1.32 \times 10^{-5} \, \text{J} $
• B. $ 2.64 \times 10^{-5} \, \text{J} $
• C. $ 1.32 \times 10^{-6} \, \text{J} $
• D. $ 2.64 \times 10^{-6} \, \text{J} $

Hint:

For a soap bubble, always use the factor $ 8\pi \sigma $ because there are two free surfaces — the inner and outer surfaces of the bubble.

Answer 1

The Correct Option is A

Step 1: Recall the formula for work done against surface tension
The work done to increase the radius of a soap bubble is given by: \[ W = 8 \pi \sigma (R_2^2 - R_1^2) \] where: \( \sigma \) is the surface tension,
\( R_1 \) is the initial radius,
\( R_2 \) is the final radius.
This formula accounts for both the inner and outer surfaces of the soap bubble. Step 2: Use the given values
Given:
\( R_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
\( R_2 = 1.1 \, \text{cm} = 0.011 \, \text{m} \)
\( \sigma = 0.025 \, \text{N/m} \)
Step 3: Calculate the difference in squared radii
\[ R_1^2 = (0.01)^2 = 0.0001 \, \text{m}^2 \] \[ R_2^2 = (0.011)^2 = 0.000121 \, \text{m}^2 \] \[ R_2^2 - R_1^2 = 0.000121 - 0.0001 = 0.000021 \, \text{m}^2 \] Step 4: Substitute into the formula
\[ W = 8 \pi \times 0.025 \times 0.000021 \] First compute: \[ 8 \pi \approx 25.1327 \] Now: \[ W = 25.1327 \times 0.025 \times 0.000021 = 0.6283175 \times 0.000021 = 1.319 \times 10^{-5} \, \text{J} \] Step 5: Conclusion
The work done is approximately: \[ {1.32 \times 10^{-5} \, \text{J}} \]