MHT CET · Physics · Mechanical Properties of Fluids
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 \(2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) )
- A \(9 \times 10^{-6} \mathrm{~J}\)
- B \(16 \times 10^{-6} \mathrm{~J}\)
- C \(40 \times 10^{-6} \mathrm{~J}\)
- D \(80 \times 10^{-6} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(80 \times 10^{-6} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& A_1=9 \times 10^{-4} \mathrm{~m}^2, A_2=25 \times 10^{-4} \mathrm{~m}^2 \\
& T=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
Work done,
\(
\begin{aligned}
& W=2 T \Delta A=2 \times 2.5 \times 10^{-2} \times(25-9) \times 10^{-4} \\
& W=80 \times 10^{-6} \mathrm{~J}
\end{aligned}
\)
\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,
\(
\begin{aligned}
& W=2 T \Delta A=2 \times 2.5 \times 10^{-2} \times(25-9) \times 10^{-4} \\
& W=80 \times 10^{-6} \mathrm{~J}
\end{aligned}
\)
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