Calculate Surface Tension Work for Soap Bubble Expansion

To find the work done in increasing the diameter of the soap bubble, we need to calculate the change in total surface area of the bubble.

A soap bubble has two free surfaces (an inner surface and an outer surface) in contact with air. Therefore, the total surface area of a soap bubble of radius $r$ is: $A = 2 \times (4\pi r^2) = 8\pi r^2$

1. Given Data:

• Surface tension of soap solution, $T = 0.03\text{ N/m}$
• Initial diameter, $D_1 = 2\text{ cm} \implies$ Initial radius, $r_1 = 1\text{ cm} = 1 \times 10^{-2}\text{ m}$
• Final diameter, $D_2 = 6\text{ cm} \implies$ Final radius, $r_2 = 3\text{ cm} = 3 \times 10^{-2}\text{ m}$

2. Increase in Total Surface Area ($\Delta A$): $\Delta A = 8\pi (r_2^2 - r_1^2)$ $\Delta A = 8\pi \left[(3 \times 10^{-2})^2 - (1 \times 10^{-2})^2\right]$ $\Delta A = 8\pi \left[9 \times 10^{-4} - 1 \times 10^{-4}\right]$ $\Delta A = 8\pi \times 8 \times 10^{-4} = 64\pi \times 10^{-4}\text{ m}^2$

3. Work Done ($W$): The work done in increasing the surface area against surface tension is given by: $W = T \times \Delta A$ $W = 0.03 \times 64\pi \times 10^{-4}\text{ J}$ $W = 1.92\pi \times 10^{-4}\text{ J}$

4. Finding the Value of $\alpha$: Comparing the calculated work done with the given expression $W = \alpha\pi \times 10^{-4}\text{ J}$: $\alpha = 1.92$

Correct Answer: C (1.92)