Question

The density of atmosphere varies with height H above the ground as per relation , $\rho = {\rho _0}{e^{ - aH}}$ , where a is a constant. The pressure at a height $h$.
(A) $\dfrac{{\rho_0}{g}{e^{aH}}}{a}$
(B) $\dfrac{{\rho_0}{g}{e^{aH}}}{h}$
(C) $\dfrac{{\rho_0}{g}{e^{-aH}}}{h}$
(D) $\dfrac{{\rho_0}{g}{e^{-aH}}}{a}$

Answer 1

Hint
Pressure is the force per unit area. This means that the pressure a solid object exerts on another solid surface is its weight in newtons divided by its area in square metres.According to formula pressure (P) is equal to force (F) by area (A).

Complete step by step answer
As we know “$\rho{gh}$” is the pressure difference between 2 points at a vertical height difference $h$.
Applying it for multiple layers of dH width
We know pressure at point $H= \infty $is 0,
∴ $P_{\infty} = 0$.
Here dP is the magnitude of change in pressure
We also know that increase in height is quals to decrease of pressure,
Now from the given equation,
$\Rightarrow d\rho = {\rho _0}{e^{ - aH}}.g(dH)$
Now integrating both sides
$\Rightarrow \int_H^\infty {dP = {\rho _0}g\int\limits_H^\infty {{e^{ - aH}}} .dH} $
$\Rightarrow \int_H^\infty {dP = {\rho _0}g(\dfrac{1}{{ - a}})\left[ {{e^{ - aH}}} \right]_H^\infty } $
$\Rightarrow \int_H^\infty {dP = {\rho _0}g(\dfrac{1}{{ - a}})(0 - {e^{ - aH}})} $
$\therefore {P_H} = \dfrac{{{\rho _0}g{e^{ - aH}}}}{a} $.

Note
That pressure is called atmospheric pressure, or air pressure. It is the force exerted on a surface by the air above it as gravity pulls it to Earth. Atmospheric pressure is commonly measured with a barometer.