Question
In a capillary with diameter $d=0.5 \mathrm{~mm}$ water will rise to a height$$\begin{gathered}\frac{2 \alpha}{\rho g r}=\frac{4 \alpha}{\rho g d} \\=\frac{4 \times 73 \times 10^{-3}}{10^{3} \times 9.8 \times 0.5 \times 10^{-3}}=59.6 \mathrm{~mm}\end{gathered}$$Since this is greater than the height $(=25 \mathrm{~mm})$ of the tube, a meniscus of radius $R$ will be formed at the top of the tube, where $R=\frac{2 \alpha}{\rho g h}=\frac{2 \times 73 \times 10^{-3}}{10^{3} \times 9.8 \times 25 \times 10^{-3}} \sim 0.6 \mathrm{~mm}$
Step 1
This is given by the formula: $$ h=\frac{4 \alpha}{\rho g d} $$ where $\alpha$ is the surface tension of the water, $\rho$ is the density of the water, $g$ is the acceleration due to gravity, and $d$ is the diameter of the capillary tube. Show more…
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