A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires…

A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires is increased by 1 mm . The work done in the process is (surface tension of water $=72 \mathrm{~N} / \mathrm{m}$ ).
  1. $2.88 \times 10^{-2} \mathrm{~J}$
  2. $7.2 \times 10^{-2} \mathrm{~J}$
  3. $1.44 \times 10^{-2} \mathrm{~J}$
  4. $3.6 \times 10^{-2} \mathrm{~J}$

Solution

Surface area of film, $\mathrm{A}_1=l \times \mathrm{d}=10 \times 0.5=5 \mathrm{~cm}^2$
The distance between wires ' $d$ ' is increased by $1 \mathrm{~mm}=0.1 \mathrm{~cm}$
Surface area of film becomes, $\mathrm{A}_2=10 \times(0.5+0.1)=6 \mathrm{~cm}^2$
Increase in the surface area of the film is, $\begin{aligned} & \Delta \mathrm{A}=\mathrm{A}_2-\mathrm{A}_1 \\ & \Delta \mathrm{~A}=6-5=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2 \end{aligned}$
Work done, $\begin{aligned} & \mathrm{W}=2 \mathrm{~T} . \Delta \mathrm{A} \quad \ldots(\text { where } \mathrm{T} \text { is surface tension) } \\ & \mathrm{W}=2 \times 72 \times 10^{-4} \mathrm{~J} \\ & \mathrm{~W}=1.44 \times 10^{-2} \mathrm{~J} \end{aligned}$ ^

Asked in: MHT CET 2024 (04 May Shift 2)

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