Question #2e732

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Question #2e732

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Answer 1 · 2017-03-11T16:38:52

The building's height $h = 200 m$ $h=200m$ .
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
$= h \times d \times g$ $=h xxd xxg$
Where $d = density of air and g = acceleration due to gravity$ $d= "density of air and g = acceleration due to gravity"$

The barometer reading difference gives the pressure difference of $76 - 74.15 = 1.85 c m = 1.85 \times 10^{- 2} m$ $76-74.15=1.85cm=1.85 xx10^-2m$ of Hg column.

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

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

So

$200 \times d \times g = (76 - 74.15) \times 10^{- 2} \times 13600 \times g$ $200xxd xxg=(76 -74.15)xx10^-2xx13600xxg$

$d = \frac{1.85 \times 10^{- 2} \times 13600}{200} kg/ m^{3}$ $d=(1.85xx10^-2xx13600)/200"kg/"m^3$

$= 1.285 kg/ m^{3}$ $=1.285" kg/"m^3$

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Question #2e732

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