Question

assuming the density of air to be 1.295 kg/m^3 find the fall in barometric height in mm of hg at a height of 107m above sea leavel.

Answer 1

Let us take a cuboid of height 107 meters and a square base of area 1 m².
Volume of air in this cuboid = 107 m³
mass of air in this cuboid = 107 * 1.295 = 138.565 kg

Pressure on the base of the cuboid due to the weight of air in the cuboid
    = ρ g h
    = 1.295 * g * 107   Pa
    = 138.565*g   Pa

This is fall in pressure.
Fall in the barometric height of Mercury  column
    = 138.565 /(density of mercury)
    = 138.565 / 13,600
    = 0.01019 meters
    = 10.19 mm
==============================

Height of mercury column in barometer, which is equivalent the height of air of 107 m high = 107 * 1.295 / 13,600 = 10.19 mm

Answer 2

Answer:

ρ = 1.295 kg/m³

decrease in the pressure of air = ρ g h = 1.295 * g * 107  Pa

ρ g h = 13.6 * 10³ * g * H

H = decrease in barometric height of the mercury

   = ρ g h / (13.6*10³ * g) = 10.188 mm

Explanation: