A capillary tube with an inside radius of 638 𝑛𝑚 can support a 100 𝑚𝑚 column of
liquid that has a density of 820 𝑘𝑔𝑚-3. The observed contact angle is 12.5°. Find
the surface tension of the liquid for ℎ = 20 𝑐𝑚
According to the Jurin's law
h=2σ(ρ−ρ0)gr⋅cosθ→σ=h(ρ−ρ0)gr/(2cosθ)=h=\frac{2\sigma}{(\rho-\rho_0)gr}\cdot\cos\theta\to \sigma=h(\rho-\rho_0)gr/(2\cos\theta)=h=(ρ−ρ0)gr2σ⋅cosθ→σ=h(ρ−ρ0)gr/(2cosθ)=
=0.2⋅(820−1.29)⋅9.8⋅638⋅10−9/(2⋅cos12.5°)=5.24⋅10−4 (N/m)=0.2\cdot(820-1.29)\cdot9.8\cdot638\cdot10^{-9}/(2\cdot\cos12.5°)=5.24\cdot10^{-4} \ (N/m)=0.2⋅(820−1.29)⋅9.8⋅638⋅10−9/(2⋅cos12.5°)=5.24⋅10−4 (N/m) . Answer
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