Area of liquid film is `6xx10cm^(2)` and surface tension is T = 20 dyne/cm, what is the work done to change area up to `12xx10cm^(2)`

The surface tension of a liquid is 10^(8) dyne/cm .It is equivalent to

The surface tension of a liquid is 10^(8) dyne cm^(-1) . It is equivalent to

The surface tension of a liquid is 10^(8) dyne cm^(-1) . It is equivalent to

The surface tension soap solution is 3 xx 10^(-2) N/m then then the work done in forming a soap film of 20 cm xx 5 cm will be

A rectengular wire frame of size 2 cm xx 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm xx 3 cm, calculate the work done in the process. The surface tension of soap film is 3 xx 10^(-2) N/m .

A rectangular wire frame of size 2 cm xx 2 cm, is dipped in a soap solution and taken out. A soap film is formed, it the size of the film is changed to 3 cm x 3 cm, Calculate the work done in the process. The surface tension of soap film is 3 xx 10^(-2) N/m.

Assertion : if work done in increasing th size of a soap film from 10 cm xx 4 cm to 10cm xx8cm is 2xx10^(-4) J, then the surface tension is 5xx10^(-4)N m^(-1) Reason : work done = surface tension xx increase in area .