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MHT CET 2024 (4 May, Shift 2) · Surface Energy & Work Done

Surface Energy & Work Done question from MHT CET 2024 (4 May, Shift 2)

Q15MHT CET 2024 · 4 MAY · SHIFT 2Medium

Surface Energy & Work Done

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 N/m= 72\text{ N/m})

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Answer: (c)

l = 10 cm0.5 cm+1 mmfilm: 2 surfaces, ΔA = l·Δx
  1. Film has two surfaces: W=2TΔAW = 2T\Delta A, with ΔA=lΔx\Delta A = l \Delta x.
  2. ΔA=0.1 m×10−3 m=10−4 m2\Delta A = 0.1\text{ m} \times 10^{-3}\text{ m} = 10^{-4}\text{ m}^2
  3. W=2×72×10−4=1.44×10−2 JW = 2 \times 72 \times 10^{-4} = 1.44\times10^{-2}\text{ J}
  4. So the answer is (c) 1.44×10−2 J1.44\times10^{-2}\text{ J}.
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