Question

The density of air in atmosphere decreases with height and can be expressed by the relation $\rho ={\rho }_0{e}^{-\alpha h}$ , where ${\rho }_0$ is the density at sea level, $\alpha$ is a constant and $h$ is the height. The atmospheric pressure at the sea level is

• A. $\frac{{\rho }_{0}g}{\alpha }$
• B. $\frac{{\rho }_0\alpha h}{g}$
• C. $\frac{\alpha h}{{\rho }_{0}g}$
• D. $\frac{h}{{\rho }_0\alpha }$

Answer 1

Answer: (A)

Because variation of density with respect to height
is given by the relation,
$\rho ={\rho }_0{e}^{-\alpha h}$
where, ${\rho }_0=$ density at sea level
$\alpha =$ a constant
$h=$ height
Hence, the pressure due to a small air column of
length $dh$ at height $h$ is given by
$dp=(dh)\cdot \rho g$
$dp=(dh){\rho }_0{e}^{-\alpha h}\cdot g$
$={\rho }_{0}g\cdot e^{-\alpha h}dh$
$p=\underset{0}{\overset{\infty }{\int }}\left({\rho }_{0}g\right)e^{-\alpha h}dh$
$={\rho }_{0}g\underset{0}{\overset{\infty }{\int }}{e}^{-\alpha h}dh$
$={\rho }_{0}g\left(\frac{e^{-\alpha h}}{-\alpha }\right)$
Atmospheric pressure at the sea level
$p=\frac{{\rho }_{0}g}{\alpha }$