91Ó°ÊÓ

1. Density of air,$p=1.3kg/m^3$
2. Air pressure at sea level, $p=1.0atm$
3. Acceleration due to gravity,$g=9.8m/s^2$

The atmospheric pressure depends on the density of the air. Therefore, if the density is constant, we can find the atmospheric pressure using the formula for pressure in terms of density, acceleration due to gravity, and height. From this, we can calculate the height. If the density is changing linearly with height, then we have to define a function, which would give us this change in density with height, and at a certain height, the density will be zero. Using this function, we can find the height for the second case.

Formula:

Pressure applied on a body, $p=pgh$ (i)

Using equation (i) and the given values, the height can be found as:

$$\begin{gathered} 1.01×{10}^5=1.3×9.8×h \\ h=7927.79m \\ =7.9km \end{gathered}$$

Hence, the value of height at uniform density is 7.9 km.

Now, density changes uniformly with atmosphere, so we can write the density as

$p=p_0-p_0\frac{y}{h}$

$p_0$is the density of the air at the earth’s surface, ‘y’ is the height at which the density is measured, and h is the height at which the density is zero.

We can simplify this as:

$p=p_0\left(1-\frac{y}{h}\right)..........................................................\left(1\right)$

Now, pressure as a function of height can be written as:

$p={∫}_0^{h}pgdy....................................................\left(2\right)$

Putting value of equation (1) in equation (2), we get

$$\begin{gathered} p={∫}_o^h{p}_{0}g\left(1-\frac{y}{h}\right)dy \\ p={{\left[p_{0}gy\right]}_o}^h-{{\left[\frac{p_{o}gh}{2h}\right]}_0}^h \\ p=p_{0}gh-\frac{p_{0}gh}{2} \\ p=\frac{p_{0}gh}{2} \\ h=\frac{2p}{p_{0}g}(equationforheight) \end{gathered}$$

$$\begin{gathered} h=\frac{2×1.01×{10}^5}{1.3×9.8} \\ =15855.573m \\ =16km \end{gathered}$$

Hence, the value of height when the atmospheric pressure changes linearly to zero is 16 km.