Question

Assuming that the atmosphere has the same density anywhere as at sea level $(\rho =1.3{\text{kg/m}}^3)$ and $g$ to be constant $(g=10{\text{m/s}}^2)$, what should be the approximate height of the atmosphere so that the atmospheric pressure at sea level is $p_0=1.01\times {10}^5{\text{N/m}}^2$ ?

• A. $6\text{km}$
• B. $8\text{km}$
• C. $12\text{km}$
• D. $18\text{km}$

Answer 1

The correct option is B $8\text{km}$

Let $h$ be the height of the atmosphere and assume pressure at height $h$ to be zero
Then, pressure at sea level $P=\rho gh$
i.e $1\text{atmosphere}=1.01\times {10}^5{\text{N/m}}^2=\rho gh$
Given, $\rho =1.3{\text{kg/m}}^3,g=10{\text{m/s}}^2$

$\therefore h=\frac{1.01\times {10}^5}{1.3\times 10}$
$=7769\text{m}\approx 8\text{km}$