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Amiraneli [1.4K]
2 years ago
9

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.
Physics
1 answer:
Gekata [30.6K]2 years ago
7 0

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

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

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

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

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

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

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

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I hope it helps you!      

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