MHT CET20239 May 2023Evening ShiftPhysicsMechanical Properties of FluidsActual
A film of soap solution is formed between two straight parallel wires of length 10 ~cm each separated by 0.5 ~cm . If their separation is increased by 1 ~mm while still maintaining their parallelism. How much work will have to be done? (surface tension of solution =65 10⁻² ~N / m )
Options
- A7.22 10⁻⁶ ~J
- B13.0 10⁻⁵ ~J
- C2.88 10⁻⁵ ~J
- D5.76 10⁻⁵ ~J
Correct answer
B. 13.0 10⁻⁵ ~J
Step-by-step solution
The increase in surface area of the film is, A = A ₂- A ₁ A ₁=2 l b =2 10 10⁻² 0.5 10⁻² A ₂=2 l ( b +1)=2 10 10⁻² (0.5+0.1) 10⁻²A₂-A₁= [2 10 10⁻² (0.5+0.1) 10⁻² ]- [2 10 10⁻² 0.5 10⁻² ]=2 10⁻⁴ ~m ^2 Work done = Increase in surface energy = TdA = (65 10⁻² ) (2 10⁻⁴ )=1.3 10⁻⁴=13 10⁻⁵ ~J