Question

Surface tension of soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is \( \alpha \pi \times 10^{-4} \) J. Find the value of \( \alpha \).

Hint:

To calculate the work done by surface tension, use the formula \( W = 4 \pi \gamma (r_2^2 - r_1^2) \), where \( \gamma \) is surface tension and \( r_1, r_2 \) are the initial and final radii.

Answer 1

Correct Answer: 3.84

Step 1: Work done in expanding the soap bubble.
The work done in increasing the size of a soap bubble is equal to the increase in its surface energy. Since a soap bubble has two surfaces, the work done is given by: \[ W = 4 \pi \gamma \left( r_2^2 - r_1^2 \right) \] where \( \gamma \) is the surface tension, \( r_1 \) is the initial radius, and \( r_2 \) is the final radius.
Step 2: Substituting the given values.
Given: \[ \gamma = 0.03 \, \text{N/m}, \quad r_1 = 2 \, \text{cm} = 0.02 \, \text{m}, \quad r_2 = 6 \, \text{cm} = 0.06 \, \text{m} \] Substituting these values into the formula: \[ W = 4 \pi \times 0.03 \times \left[ (0.06)^2 - (0.02)^2 \right] \]
Step 3: Simplifying the calculation.
\[ W = 4 \pi \times 0.03 \times (0.0036 - 0.0004) \] \[ W = 4 \pi \times 0.03 \times 0.0032 \] \[ W = 4 \pi \times 0.000096 \] \[ W = 0.000384 \pi \, \text{J} \]
Step 4: Comparing with the given form.
The work done is expressed as \( \alpha \pi \times 10^{-4} \) J. Comparing: \[ 0.000384 \pi = 3.84 \pi \times 10^{-4} \] Hence, \[ \alpha = 3.84 \]
Final Answer: \( \alpha = 3.84 \).