MHT CET20244 May 2024Evening ShiftPhysicsMechanical Properties of FluidsActual
A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires is increased by 1 mm . The work done in the process is (surface tension of water =72 ~N / m ).
Options
- A2.88 10⁻² ~J
- B7.2 10⁻² ~J
- C1.44 10⁻² ~J
- D3.6 10⁻² ~J
Correct answer
C. 1.44 10⁻² ~J
Step-by-step solution
Surface area of film, A ₁=l d =10 0.5=5 ~cm ^2 The distance between wires ' d ' is increased by 1 ~mm =0.1 ~cm Surface area of film becomes, A ₂=10 (0.5+0.1)=6 ~cm ^2 Increase in the surface area of the film is, aligned & A = A ₂- A ₁ & ~A =6-5=1 ~cm ^2=10⁻⁴ ~m ^2 aligned Work done, aligned & W =2 ~T . A ( where T is surface tension) & W =2 72 10⁻⁴ ~J & ~W =1.44 10⁻² ~J aligned