The work done in blowing a soap bubble of surface tension 0.06 Nm ⁻¹ from 2 cm radius to 5 cm radius is

Step-by-step solution

Here, S=0.06 Nm ⁻¹ , r ₁=2 ~cm =0.02 ~m , r ₂=5 ~cm =0.05 ~m Since, bubble has two surfaces, initial surface area of the bubble aligned & =2 4 r₁^2=2 4 (0.02)^2 & =32 10⁻⁴ ~m ^2 aligned Final surface area of the bubble aligned & =2 4 r₂^2=2 4 (0.05)^2 & =200 10⁻⁴ ~m ^2 aligned Final surface area of the bubble aligned & =2 4 r ₂^2=2 4 (0.05)^2 & =200 10⁻⁴ ~m ^2 aligned Increase in surface area aligned & =200 10⁻⁴-32 10⁻⁴ & =168 10⁻⁴ ~m ^2 aligned Work done = S Increase in surface area =0.06 168 10⁻⁴=0.003168 ~J