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`)

Options

A. `12 xx 10^(-6)J`
B. `24 xx 10^(-6)J`
C. `60 xx 10^(-6)J`
D. `96 xx 10^(-6)J`

Answer

B

Explanation

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`