The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is 10⁻⁴ J. The value of is _______. (Take = 3.14 )

Step-by-step solution

Initial radius r₁ = 2 2 = 1 cm = 10⁻² m Final radius r₂ = 6 2 = 3 cm = 3 10⁻² m Surface tension T = 0.03 N/m A soap bubble has two free surfaces. The change in total surface area is given by: A = 2 4 (r₂^2 - r₁^2) A = 8 ((3 10⁻²)^2 - (10⁻²)^2) A = 8 (9 10⁻⁴ - 10⁻⁴) = 64 10⁻⁴ m ^2 Work done W = T A W = 0.03 64 10⁻⁴ W = 1.92 10⁻⁴ J Comparing with 10⁻⁴ J, we get = 1.92 . Answer: 1.92