The surface tension of a soap solution is 3.5 10⁻² N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is 10⁻⁶ J. The value of is _____. ( = 22/7 )

Step-by-step solution

The work done to increase the radius of a soap bubble is given by W = T A Since a soap bubble has two free surfaces, the change in surface area is A = 2 4 (r₂^2 - r₁^2) Substituting the given values: W = 3.5 10⁻² 8 ((2 10⁻²)^2 - (1 10⁻²)^2) W = 3.5 10⁻² 8 (4 10⁻⁴ - 1 10⁻⁴) W = 3.5 10⁻² 8 3 10⁻⁴ W = 84 10⁻⁶ Using = 22 7 : W = 84 22 7 10⁻⁶ = 264 10⁻⁶ J Comparing with 10⁻⁶ J, we get = 264 Answer: 264