The work done in increasing the size of a soap film from `10cmxx6cm` to `10cmxx11cm` is `3xx10^-4` Joule. The surface tension of the film is

Options

A. `1.5xx10^(2) N//m`
B. `3xx10^(-2) N//m`
C. `2.5xx10^(-2) N//m`
D. `1.2xx10^(-2) N//m`

Answer

B

Explanation

Here, `W=3.0xx10^(-4)j`
Initial area, `A_(i)=10 cmxx6 cmxx10^(4) m^(2)`
Final area, `A_(f)=10 cmxx11 cm=110xx10^(4)m^(2)`
Change in area `=2[110-60]xx10^(4)` [because soap film has 2 free surfaces]
Surface tension `=("Work")/("Increase in area")`
`=(3xx10^(4))/(2(110-60)xx10^(4))=(3)/(2xx50) =3xx10^(-2) N//m`