1. 0.125 N m ⁻¹ 2. 0.2 N m ⁻¹ 3. 8.0 N m ⁻¹ 4. 0.250 N m ⁻¹

Answer: 1. 0.125 N m ⁻¹ Work done in stretching a liquid film is W=T Δ A , where the film has two free surfaces, so Δ A=2×(change in area) . Initial area: 4×2=8\ cm² ; final area: 5×4=20\ cm² . Change in one surface: 12\ cm²=12×10⁻⁴\ m² ; both surfaces: 24×10⁻⁴\ m² . T=W/Δ A=frac3×10⁻⁴24×10⁻⁴=0.125\ N m⁻¹ Option (1).

Topic: Surface energy, surface tension, angle of contact, excess pressure, drops, bubbles, capillary rise· Properties of Solids and Liquids

A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3×10^-4 J, the value of the surface tension of the liquid is

1. 0.125 N m$^{-1}$

2. 0.2 N m$^{-1}$

3. 8.0 N m$^{-1}$

4. 0.250 N m$^{-1}$

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Answer: 1. 0.125 N m$^{-1}$

Work done in stretching a liquid film is $W=T\,\Delta A$, where the film has two free surfaces, so $\Delta A=2\times(\text{change in area})$.

Initial area: $4\times2=8\ \text{cm}^2$; final area: $5\times4=20\ \text{cm}^2$.
Change in one surface: $12\ \text{cm}^2=12\times10^{-4}\ \text{m}^2$; both surfaces: $24\times10^{-4}\ \text{m}^2$.

$$T=\frac{W}{\Delta A}=\frac{3\times10^{-4}}{24\times10^{-4}}=0.125\ \text{N m}^{-1}$$

Option (1).

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A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3×10^-4 J, the value of the surface tension of the liquid is

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