Consider a soap film on a rectangular frame of wire of area 3 × 3 cm². If the area of the soap film is increased to 5 × 5 cm², the work done in the process will be (surface tension of soap is 2.5 × 10⁻² N/m) (a) 9 × 10⁻⁶ J (b) 16 × 10⁻⁶ J (c) 40 × 10⁻⁶ J (d) 80 × 10⁻⁶ J

Answer: (d). A film has two surfaces: W = 2T Δ A. Δ A = (25 - 9) cm² = 16×10⁻⁴ m². W = 2 × 2.5×10⁻² × 16×10⁻⁴ = 80×10⁻⁶ J. So the answer is (d).

Consider a soap film on a rectangular frame of wire of area 3 × 3… | MHT CET 2023 (12 May, Shift 1)

Q16MHT CET 2023 · 12 MAY · SHIFT 1Easy

Surface Energy & Work Done

Consider a soap film on a rectangular frame of wire of area $3 \times 3\text{ cm}^{2}$ . If the area of the soap film is increased to $5 \times 5\text{ cm}^{2}$ , the work done in the process will be (surface tension of soap is $2.5 \times 10^{-2}\text{ N/m})$

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Answer: (d)

A film has two surfaces: $W = 2T\,\Delta A$ .
$\Delta A = (25 - 9)\text{ cm}^2 = 16\times10^{-4}\text{ m}^2$ .
$W = 2 \times 2.5\times10^{-2} \times 16\times10^{-4} = 80\times10^{-6}\text{ J}$ . So the answer is (d).

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Surface Energy & Work Done question from MHT CET 2023 (12 May, Shift 1)

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