The surface tension of a soap solution is 30 | vedclass

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Question

(A) soap film has two surfaces,so the change in area is doubled. Initial area $A_1 = 5 \, cm 	imes 5 \, cm = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$.

Vedclass · Accepted Answer

$4.5 	imes 10^{-4} \, J$

Vedclass · Answer

(A) soap film has two surfaces,so the change in area is doubled. Initial area $A_1 = 5 \, cm 	imes 5 \, cm = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$. Final area $A_2 = 10 \, cm 	imes 10 \, cm = 100 \, cm^2 = 100 	imes 10^{-4} \, m^2$. Change in area $\Delta A = A_2 - A_1 = (100 - 25) 	imes 10^{-4} \, m^2 = 75 	imes 10^{-4} \, m^2$. Work done $W = 2 	imes S 	imes \Delta A$,where $S$ is surface tension. $W = 2 	imes (30 	imes 10^{-3} \, N/m) 	imes (75 	imes 10^{-4} \, m^2) = 4500 	imes 10^{-7} \, J = 4.5 	imes 10^{-4} \, J$.

The surface tension of a soap solution is $30 \times 10^{-3} \, N/m$. The work done to increase the surface area of a soap film from $5 \, cm \times 5 \, cm$ to $10 \, cm \times 10 \, cm$ is:

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