First slide

Fluid statics

Question

Considering the pressure ‘P’ to be proportional to the density of air ‘d’, the pressure at height ‘h’ if the pressure on surface of earth is ‘P₀’ and density ‘d₀’, under isothermal conditions

Difficult

A

$\large {P_0}{e^{ - \frac{{{P_0}}}{{{d_0}}}gh}}$
B

$\large {P_0}{e^{ - \frac{{{d_0}}}{{{P_0}}}gh}}$
C

$\large {P_0}{e^{\frac{{{d_0}}}{{{P_0}}}gh}}$
D

$\large {P_0}{e^{\frac{{{P_0}}}{{{d_0}}}gh}}$

Solution

$\large p\propto d;\;\frac pd=constant=K(let)$
$\large \therefore d=\frac pk \;and\;\frac {p_0}{d_0}=k$
$\large \frac {d(p)}{dy}=-(d)(g)\Rightarrow(dp)=-(d)(g)(dy)=\frac {-P}{k}gdy$
$\large \frac {dp}{p}=\left ( \frac {-g}{k} \right )dy$
$\large \int_{p_0}^{p}dp=\int _{0}^{h}\left ( \frac {-g}{k} \right )dy\Rightarrow log_e\frac {p}{p_0}=\frac {-gh}{k}=\frac {-gh}{\left ( p_0/d_0 \right )}$
$\large \therefore log_e\frac {p}{p_0}=\frac {-gd_0h}{p_0}\Rightarrow p=p_0e^{\frac {-gd_0h}{p_0}}$

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