Answer:
Explanation:
From the given information, the rise in a capillary tube h = [tex]\dfrac{2T cos \theta }{r \rho g}[/tex]
where:
For the height of water:
Surface Tension T = 66.2 mN/m = 66.2 N/m
θ = 10⁰
Cos θ = cos 10 = 0.985
radius r = 0,257 mm = 0.275 × 10⁻³ m
density of water [tex]\rho[/tex] = 1000 kg/m³
g = 9.8 m/s²
∴
replacing our values:
h = [tex]\dfrac{2T cos \theta }{r \rho g}[/tex]
h = [tex]\dfrac{2\times 66.2 \times 10^{-3} \times 0.985}{0.275 \times 10^{-3} \times 1000\times 9.8}[/tex]
h = 0.0484 meteres
h = 48.3 mm
Since the height h = [tex]\dfrac{2T cos \theta }{r \rho g}[/tex]
r = 0.275 mm = 0.275 × 10⁻³ m
the density of mecury now [tex]\rho_H{g}}[/tex]
= 13593 kg/m³
the surface tension of the mercury [tex]T_{Hg} =470 \times 10^{-3} \ N/m[/tex]
θ = 130⁰
Cos θ = cos 130 = -0.6428
Using the same above equation:
h = [tex]\dfrac{2\times 470 \times 10^{-3} \times (-0.6428) }{0.275 \times 10^{-3} \times 13593 \times 9.8}[/tex]
h = -0.016494 m
h = 16.49 mm
