Assuming the density of air to be 1.295 kgm-3, find the fall in barometric height in mm of Hg at a height of 107 m above the sea level . Take density of mercury = 13.6 × 10pawer3 kg m-3.

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Solution

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
= 138.565g 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
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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

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Assuming the density of air to be $1.295 k g / m^{3}$ , find the fall in the barometric height in mm of Hg at a height of 107 m above sea level. Take density of mercury = $13.6 \times 10^{3} k g / m^{3}$ .

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