Question

The energy required to increase the radius of a soap bubble from 3 cm to 4 cm is (Surface tension = $3\times10^{-2}$ N/m):

• A. $528\times10^{-6}$ J
• B. $264\times10^{-6}$ J
• C. $1056\times10^{-6}$ J
• D. $478\times10^{-6}$ J

Hint:

Soap bubbles have two surfaces contributing to surface energy. Always convert cm to meters before substitution. Work done equals surface tension × increase in total area. For liquid films, both inner and outer surfaces must be included.

Answer 1

The Correct Option is A

• For a soap bubble, work done = $4\pi T (r_2^2 - r_1^2)$ (two surfaces).
Substitute: \[ W = 4\pi (3\times10^{-2})[(0.04)^2 - (0.03)^2] \] \[ W = 0.12\pi(0.0016 - 0.0009) = 0.12\pi(0.0007) = 2.64\times10^{-4}\pi \] \[ W = 8.28\times10^{-4} \text{ J } = 528\times10^{-6} \text{ J} \] Hence, energy required is $528\times10^{-6$ J}.