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Question 1.2: Determine the distance h that mercury will be depressed with...

Determine the distance h that mercury will be depressed with a 4-mm-diameter glass tube inserted into a pool of mercury at 20ºC (Figure 1.11).

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Equation (1-20) applies, so we have

h=\frac{2 \sigma  \cos \theta}{\rho g r}

Recall that, for mercury and glass, θ = 130º.

For mercury at 20°C r = 13,580 kg/m³, and for mercury in air σ= 0.44 N/m (Table 1.2) giving

Table 1.2      Surface tensions of some fluids in air at latm and 20 °C
Fluid σ(N/m)
Ammonia 0.021
Ethyl alcohol 0.028
Gasoline 0.022
Glycerin 0.063
Kerosene 0.028
Mercury 0.440
Soap solution 0.02S
SAE 30 oil 0.035
h=\frac{2(0.44 \mathrm{~N} / \mathrm{m})\left(\cos 130^{\circ}\right)}{\left(13580 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(2 \times 10^{-3} \mathrm{~m}\right)}

 

=2.12 \times 10^{-3} \mathrm{~m} \quad(2.12 \mathrm{~mm})

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