A rectangular film of liquid is extended from (4 cm×2 cm) to (5 cm×4 cm). If the work done is 3×10−4 J, the value of the surface tension of the liquid is
Initial area of the film
Ai=4×2=8 cm2=8×10−4 m2
Final area of the film
Af=5×4=20 cm2=20×10−4 m2
Thus change in area
ΔA=(20−8)×10−4=12×10−4 m2
Work done W=3×10−4 J
Using for soap films W=S(2ΔA)
2ΔA because both side area will be taken (top and bottom)
where S is the surface tension
∴ 3×10−4=S×2×12×10−4
⟹S=0.125 N/m