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
$d p = (d h )\cdot \rho g$
$d p = (dh) \rho_{0} e^{-\alpha h} \cdot g$
$ = \rho_{0} g \cdot e^{ -\alpha h} dh$
$p = \int\limits_{0}^{\infty} \left(\rho_{0} g\right)e^{-\alpha h} dh $
$ = \rho_{0} g \int\limits_{0}^{\infty} e^{-\alpha h} dh $
$= \rho_{0} g\left(\frac{e^{-\alpha h}}{-\alpha}\right)$
Atmospheric pressure at the sea level
$p = \frac{\rho_{0}g}{\alpha}$