Question

a rectangular wire frame of size 2cm×2cm,is dipped in a soap solution and taken out. A soap film is Formed, if the size of the film is changed to 3cm ×3cm, calculate the work done ❣in the process. the surface tension of soap film is 3×10^-²N/m.
[Ans.3×10^-5]​

Answer 1

Given that,

Surface tension $T=3\times10^{-2}\ N/m$

A rectangular wire frame of size is

side a = 2 cm

A soap film is Formed,

If the size of the film is

side a '= 3 cm

We need to calculate the initial area

Using formula of area

$A_{1}=a^2$

Put the value into the formula

$A_{1}=2^2\ cm^2$

$A_{1}=4\ cm^2$

We need to calculate the final area

Using formula of area

$A_{2}=a'^2$

Put the value into the formula

$A_{2}=3^2\ cm^2$

$A_{2}=9\ cm^2$

In case of rectangular frame the water wets it from two sides,

We need to change in area of rectangular wire frame

Using formula of change area

$dA=2(A_{2}-A_{1})$

Put the value into the formula

$dA=2(9-4)$

$dA=10\ cm^2$

$dA=10\times10^{-4}\ m^2$

We need to calculate the work done in the process

Using formula of work done

$W=TdA$

Put the value into the formula

$W=3\times10^{-2}\times10\times10^{-4}$

$W=0.00003\ J$

$W=3.0\times10^{-5}\ J$

Hence, The work done in the process is $3.0\times10^{-5}\ J$