Consider a soap film on a rectangular frame of wire of area $3 \times 3 \mathrm{~cm}^2$. If the area of the…

Consider a soap film on a rectangular frame of wire of area $3 \times 3 \mathrm{~cm}^2$. If the area of the soap film is increased to $5 \times 5 \mathrm{~cm}^2$, the work done in the process will be (surface tension of soap solution is $2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ )
  1. $9 \times 10^{-6} \mathrm{~J}$
  2. $16 \times 10^{-6} \mathrm{~J}$
  3. $40 \times 10^{-6} \mathrm{~J}$
  4. $80 \times 10^{-6} \mathrm{~J}$

Solution

$\begin{aligned} & A_1=9 \times 10^{-4} \mathrm{~m}^2, A_2=25 \times 10^{-4} \mathrm{~m}^2 \\ & T=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m} \end{aligned}$ Work done, $\begin{aligned} & W=2 T \Delta A=2 \times 2.5 \times 10^{-2} \times(25-9) \times 10^{-4} \\ & W=80 \times 10^{-6} \mathrm{~J} \end{aligned}$ .

Asked in: MHT CET 2023 (12 May Shift 1)

Practice more Mechanical Properties of Fluids questions on Aicharya