Suppose water is used in a barometer instead of mercury. If the barometric pressure is `760 mm Hg`, what is the height of the water column in the barometer at `0^(@) C`. The densities of water and mercury at `0^(@) C` are `0.99987 g cm^(-3)` and `13.596 g cm^(-3)` , respectively.
Strategy : The prtessure exerted by a column of liquid `h` whose density is `d` is `hdg`. Because the pressure are equal , we can equate the expressions for water `(W)` and mercury (M):
`h_(W)d_(W)g = h_(M) d_(M)g`
or `h_(W)d_(W) = h_(M)d_(M)`
Rearranging gives
`(h_(W))/(h_(M)) = (d_(M))/(d_(W))`
This implies that the height of the liquid column in inversely proportional to its density . Solve the equation to find the height of the water column, `h_(W)`.

Explanation

After substituting the data into the equation , we obtain
`(h_(W))/(760 mm) = (13.596 g cm^(-3))/(0.99987 g cm^(-3))`
`:. H_(W) = 760 mm xx (13.596)/(0.99987)`
`= 10334 mm`
`= 1.0334 xx 10^(4) mm`
`= 10.334 m` `(1 mm = 10^(-3) m)`