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A frame made of metallic wire enclosing a surface area $A$ is covered with a soap film. If the area of the frame of metallic wire is reduced by $50 \%$, then the energy of the soap film will be changed by
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The correct answer is:
$50 \%$
Given, initial surface area, $A_{1}=A$
Final surface area, $A_{2}=A_{1}-50 \%$ of $A$
$A_{2}=A-50 \%$ of $A=A-\frac{A}{2}=\frac{A}{2}$ Initial surface energy, $E_{1}=T \cdot 2 A=2 T A$ Final surface energy, $E_{2}=$ Surface tension $\times$ Surface $\Rightarrow \quad E_{2}=T \times 2\left(A_{1}-A_{2}\right)$ $=T \times 2\left(A-\frac{A}{2}\right)$ $\Rightarrow \quad \quad E_{2}=T A$
$\therefore$ Percentage change in the energy of soap film
$$
\begin{aligned}
&=\frac{E_{1}-E_{2}}{E_{1}} \times 100 \\
&=\frac{2 T A-T A}{2 T A} \times 100 \\
&=50 \%
\end{aligned}
$$
Final surface area, $A_{2}=A_{1}-50 \%$ of $A$
$A_{2}=A-50 \%$ of $A=A-\frac{A}{2}=\frac{A}{2}$ Initial surface energy, $E_{1}=T \cdot 2 A=2 T A$ Final surface energy, $E_{2}=$ Surface tension $\times$ Surface $\Rightarrow \quad E_{2}=T \times 2\left(A_{1}-A_{2}\right)$ $=T \times 2\left(A-\frac{A}{2}\right)$ $\Rightarrow \quad \quad E_{2}=T A$
$\therefore$ Percentage change in the energy of soap film
$$
\begin{aligned}
&=\frac{E_{1}-E_{2}}{E_{1}} \times 100 \\
&=\frac{2 T A-T A}{2 T A} \times 100 \\
&=50 \%
\end{aligned}
$$
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