# The atmospheric pressures at the top and the bottom of a building are

The atmospheric pressures at the top and the bottom of a building are read b a barometer to be 96.0 and 98.0 kPa. If the density of air is $1.0k\frac{g}{{m}^{3}}$, the height of the building is (a) 17 m (b) 20 m (c) 170 m (d) 204 m (e) 252 m
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Jordan Mitchell
Given:
${P}_{\mathrm{\top }}=96kPa$
${P}_{\mathrm{\perp }\to m}=98kPa$
$\rho =1k\frac{g}{{m}^{3}}$
Solution:
We can easily show that,
$\mathrm{△}P={\rho }_{a}g\mathrm{△}h=1×9.81×\mathrm{△}h=2×{10}^{3}Pa⇒\mathrm{△}h=\frac{2000}{9.81}=203.874=\approx 204m$
$\mathrm{△}h=204m$

Ronnie Schechter

203.9 m
Step-by-step explanation:
Pressure at top of building $=98kPa$
Pressure at bottom of building $=96kPa$
difference in pressure $=98.0kPa-96kPa=2kPa=2,000Pa$
$\text{Pressure difference}=\text{density of air x acceleration due to gravity}×\text{height}$
$2,000Pa=1.0kg/{m}^{3}×9.81m/{s}^{2}×\text{height}$
$2,000=9.81×\text{height}$
$height=\frac{2,000}{9.81}$
$height=203.9m$