Q132 / 180 · COMEDK 2020 · Physics

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

  • A
    $100 \%$
  • B
    $75 \%$
  • Correct answer
    C
    $50 \%$
  • D
    $25 \%$

Solution

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 $$ \begin{aligned} &=\frac{E_{1}-E_{2}}{E_{1}} \times 100 \\ &=\frac{2 T A-T A}{2 T A} \times 100 \\ &=50 \% \end{aligned} $$

Hence, the correct answer is option: 'C'.

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