A rectangular film of liquid is extended from (4cm × 2cm) to (5cm × 4cm). If the work done is 3 × 10⁻⁸ J, what is the value of the surface tension of the liquid?

Initial area of the film Aᵢ = 4 × 2 = 8 cm² = 8 × 10⁻⁴ m²
Final area of the film Aƒ = 5 × 4 = 20 cm² = 20 × 10⁻⁴ m²
Thus change in area ΔA = (20 − 8) × 10⁻⁴ = 12 × 10⁻⁴ m²
Work done W = 3 × 10⁻⁸ J
Using for soap films W = S(2ΔA) where S is the surface tension
∴ 3 × 10⁻⁸ = S × 2 × 12 × 10⁻⁴
⇒ S = 0.125 Nm⁻¹