MHT CET · Physics · Mechanical Properties of Fluids
A water film is formed between the two straight parallel wires, each of length 10
\(\mathrm{cm}\), kept at a separation of \(0 \cdot 5 \mathrm{~cm}\). Now, the separation between them is
increased by \(1 \mathrm{~mm}\) without breaking the water film.
The work done for this is
(surface tension of water \(=7 \cdot 2 \times 10^{-2} \mathrm{Nm}^{-1}\) )
- A \(7 \cdot 22 \times 10^{-6} \mathrm{~J}\)
- B \(5 \cdot 76 \times 10^{-5} \mathrm{~J}\)
- C \(1 \cdot 44 \times 10^{-5} \mathrm{~J}\)
- D \(2 \cdot 88 \times 10^{-5} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(1 \cdot 44 \times 10^{-5} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
(A)
Increase in the area of the film is
\(\Delta \mathrm{A}=10 \mathrm{~cm} \times 0.1 \mathrm{~cm}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}\)
Work done \(W=2 T . \Delta A\)
\(\begin{array}{l}
=2 \times 7.2 \times 10^{-2} \times 10^{-4} \mathrm{~J} \\
=1.44 \times 10^{-5} \mathrm{~J}
\end{array}\)
Increase in the area of the film is
\(\Delta \mathrm{A}=10 \mathrm{~cm} \times 0.1 \mathrm{~cm}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}\)
Work done \(W=2 T . \Delta A\)
\(\begin{array}{l}
=2 \times 7.2 \times 10^{-2} \times 10^{-4} \mathrm{~J} \\
=1.44 \times 10^{-5} \mathrm{~J}
\end{array}\)
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