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A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is 25.0 cm.
(a) What is the gauge pressure at the water-mercury interface?
(b) Calculate the vertical distance h from the top of the mercury in the right-hand arm of the tube to the top of the water in the left-hand arm.

Respuesta :

Answer:

a) [tex]P=2450\ Pa[/tex]

b) [tex]\delta h=23.162\ cm[/tex]

Explanation:

Given:

height of water in one arm of the u-tube, [tex]h_w=25\ cm=0.25\ m[/tex]

a)

Gauge pressure at the water-mercury interface,:

[tex]P=\rho_w.g.h_w[/tex]

we've the density of the water [tex]=1000\ kg.m^{-3}[/tex]

[tex]P=1000\times 9.8\times 0.25[/tex]

[tex]P=2450\ Pa[/tex]

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

[tex]\rho_w.g.h_w=\rho_m.g.h_m[/tex]

[tex]1000\times 9.8\times 0.25=13600\times 9.8\times h_m[/tex]

[tex]h_m=0.01838\ m=1.838\ cm[/tex]

Now the difference in the column is :

[tex]\delta h=h_w-h_m[/tex]

[tex]\delta h=25-1.838[/tex]

[tex]\delta h=23.162\ cm[/tex]

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