Consider a soap film on a rectangular frame of wire of area 4 xx 4 cm^(2). If the area of the soap film is increased to 4 xx 5 cm^(2), the work done in the process will be (The surface tension of the soap film is 3 xx 10^(-2)N//m)

Consider a soap film on a rectangular frame of wire of area 4 xx 4 cm^

1.	Consider a soap film on a rectangular frame of wire of area 4 xx 4 cm^(2). If the area of the soap film is increased to 4 xx 5 cm^(2), the work done in the process will be (The surface tension of the soap film is 3 xx 10^(-2)N//m)
Answer» `12 xx 10^(-6)J` `24 xx 10^(-6)J` `60 xx 10^(-6)J` `96 xx 10^(-6)J` Solution :Since the soap film has two surfaces, `:.` Initial surface area of the film `=2 xx 4 xx 4 CM^(2) = 32 cm^(2) = 32 xx 10^(-4) m^(2)` Final surface area of the film `=2 xx 4 xx 5 cm^(2) = 40CM^(2) = 40 xx 10^(-4) m^(2)` Increase in surface area `= (40 xx 10^(-4) -32 xx 10^(-4)) m^(2) = 8 xx 10^(-4) m^(2)` Work done W= surface tension `xx` increase in surface area `=(3 xx 10^(-2) N//m) xx (8 xx 10^(-4) m^(2)) = 24 xx 10^(-6)J`