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 Ushaped tube until the vertical height of the water column is 15.0 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. (The density of mercury is 13.6 ⇥ 103 kg/m3).

Let due of weight of water column , mercury level in left column goes down by distance h . The level of mercury in right column will rise by the same distance ie by distance h .

So mercury column of 2h height is balancing the water column of height 15 cm

2h x 13.6 x 10³ x g = .15 x 10³ x g

h = .15 / (2 x 13.6)

= .55 x 10⁻² m

= . 55 cm

Difference of height of water column and mercury column

= 15 - 2 x .55 cm

= 13.9 cm .

boffeemadrid boffeemadrid · Answer 2 · 2021-11-22T05:40:37+01:00

The gauge pressure and height is required.

The gauge pressure at the water–mercury interface is [tex]1471.5\ \text{N}[/tex]

The vertical height [tex]h[/tex] is [tex]13.89\ \text{cm}[/tex]

[tex]\rho[/tex] = Density = [tex]13.6\times 10^3\ \text{kg/m}^3[/tex]

[tex]\rho_w[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]

[tex]h_w[/tex] = Height of water column = 15 cm

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

For water-mercury interface the gauge pressure is

[tex]P=\rho_wgh_w\\\Rightarrow P=1000\times 9.81\times 0.15\\\Rightarrow P=1471.5\ \text{N}[/tex]

Pressure balance for mercury is given by

[tex]\rho g\Delta h=P\\\Rightarrow \rho g(h_w-h)=P\\\Rightarrow h=h_w-\dfrac{P}{\rho g}\\\Rightarrow h=0.15-\dfrac{1471.5}{13.6\times 10^3\times 9.81}\\\Rightarrow h=0.1389=13.89\ \text{cm}[/tex]

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