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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Solution

First, let us assume that the height above the sea level is h = 107 m.
$∴ P_{h} - P_{s e a} = ρ_{a i r} g h$
$ρ_{H g} g h_{f} - ρ_{H g} g h_{i} = ρ_{a i r} g h$
$△ h = \frac{ρ_{a i r} g h}{ρ_{H g} g} = \frac{1.295 \times 107}{13.6 \times 10^{3}} = 0.010 m$ .
$∴ △ h = 0.010 m$ of Hg.

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Assuming the density of air to be 1.295kg/m3, find the fall in the barometric height in mm of Hg at a height of 107 m above sea level. Take density of mercury = 13.6×103kg/m3.

First, let us assume that the height above the sea level is h = 107 m.∴Ph−Psea=ρairghρHgghf−ρHgghi=ρairgh△h=ρairghρHgg=1.295×10713.6×103=0.010 m.∴△h=0.010 m of Hg.

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 = $13.6 \times 10^{3} k g / m^{3}$ .

First, let us assume that the height above the sea level is h = 107 m.
$∴ P_{h} - P_{s e a} = ρ_{a i r} g h$
$ρ_{H g} g h_{f} - ρ_{H g} g h_{i} = ρ_{a i r} g h$
$△ h = \frac{ρ_{a i r} g h}{ρ_{H g} g} = \frac{1.295 \times 107}{13.6 \times 10^{3}} = 0.010 m$ .
$∴ △ h = 0.010 m$ of Hg.