Question

There is a soap film on a rectangular frame of wire of area 4 cm x 4 cm. If the area of the frame is increased to 4 cm x 5 cm, find the work done in the process.(Surface tension of soap film = 3 x 10⁻² N/m)
(Ans : 2.4 x 10⁻⁵ J)

Answer 1

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Answer 2

Answer:

The work done in the process is $2.4\times10^{-5}\ J$

Explanation:

Given that,

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

Initial area $A_{i}=16\times10^{-4}\ m$

Final area $A_{f}=20\times10^{-4}\ m$

Change in area is

$\Delta A=2(A_{f}-A_{i})$

$\Delta A=2(20-16)\times10^{-4}$

$\Delta A=8\times10^{-4}$

The work done is

$W= T\Delta A$

$W= 3\times10^{-2}\ N/m\times8\times10^{-4}$

$W=2.4\times10^{-5}\ J$

Hence, The work done in the process is $2.4\times10^{-5}\ J$