A water film is formed between the two straight parallel wires, each of length 10 $\mathrm{cm}$, kept at a…

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 $=7 \cdot 2 \times 10^{-2} \mathrm{Nm}^{-1}$ )
  1. $7 \cdot 22 \times 10^{-6} \mathrm{~J}$
  2. $5 \cdot 76 \times 10^{-5} \mathrm{~J}$
  3. $1 \cdot 44 \times 10^{-5} \mathrm{~J}$
  4. $2 \cdot 88 \times 10^{-5} \mathrm{~J}$

Solution

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 $W=2 T . \Delta A$ $\begin{array}{l} =2 \times 7.2 \times 10^{-2} \times 10^{-4} \mathrm{~J} \\ =1.44 \times 10^{-5} \mathrm{~J} \end{array}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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