Question

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.

Answer 1

a)
`P=2450\ Pa`

b)
`\delta h=23.162\ cm`

Step-by-step 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* 9.8* 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* 9.8* 0.25=13600* 9.8* h_m`

`h_m=0.01838\ m=1.838\ cm`

Now the difference in the column is :

`\delta h=h_w-h_m`

`\delta h=25-1.838`

`\delta h=23.162\ cm`