# Question #2e732

Mar 11, 2017

The building's height $h = 200 m$.
So the barometer reading at lower floor is higher than the upper floor due to the pressure exerted by 200m air column.

Pressure exerted by the air column
$= h \times d \times g$
Where $d = \text{density of air and g = acceleration due to gravity}$

The barometer reading difference gives the pressure difference of $76 - 74.15 = 1.85 c m = 1.85 \times {10}^{-} 2 m$ of Hg column.

Here the barometric pressure difference is equal to the pressure of 200m air column.

Let the density of air be $d \text{kg/} {m}^{3}$

So

$200 \times d \times g = \left(76 - 74.15\right) \times {10}^{-} 2 \times 13600 \times g$

$d = \frac{1.85 \times {10}^{-} 2 \times 13600}{200} \text{kg/} {m}^{3}$

$= 1.285 \text{ kg/} {m}^{3}$