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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 $=7 \cdot 2 \times 10^{-2} \mathrm{Nm}^{-1}$ )
Options:
$\mathrm{cm}$, 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 $=7 \cdot 2 \times 10^{-2} \mathrm{Nm}^{-1}$ )
Solution:
1066 Upvotes
Verified Answer
The correct answer is:
$1 \cdot 44 \times 10^{-5} \mathrm{~J}$
(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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