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Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m⁻¹)

1. 0.2π mJ
2. 2π mJ
3. 0.4π mJ
4. 4π mJ

Correct answer: 0.4π mJ

Solution

A soap bubble has two surfaces, so W = T*2*4*pi*(r2^2 - r1^2) = 0.03*8*pi*(0.05^2 - 0.03^2) = 0.03*8*pi*0.0016 = 3.84e-4 * pi J ~ 0.4*pi mJ.

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