MHT CET202523 Apr 2025Evening ShiftPhysicsMechanical Properties of FluidsActual
The amount of work done in blowing a soap bubble such that its diameter increases from d₁ to d₂ is ( T= surface tension of soap solution)
Options
- A4 ( ~d ₂^2- d ₁^2 ) T
- B8 ( ~d ₂^2- d ₁^2 ) T
- C( d ₂^2- d ₁^2 ) T
- D2 ( ~d ₂^2- d ₁^2 ) T
Correct answer
D. 2 ( ~d ₂^2- d ₁^2 ) T
Step-by-step solution
The work done equals the change in surface energy, given by W = T A , where T is the surface tension and A the area difference. A soap bubble has two surfaces, so its total area is 8 R^2 . Given initial diameter d₁ and final d₂ , radii are R₁ = d₁ / 2 and R₂ = d₂ / 2 . Initial area is A₁ = 8 (d₁ / 2)^2 = 2 d₁^2 , final area A₂ = 8 (d₂ / 2)^2 = 2 d₂^2 , so A = 2 (d₂^2 - d₁^2) . Therefore, work done is W = T 2 (d₂^2 - d₁^2) = 2 T (d₂^2 - d₁^2) . This corresponds to option D .