Question

Assuming the density of air to be $\sf 1.295kg \: {m}^{-3}$ , find the fall of barometric height in mm of Hg at a height of 107m above the sea level.

Take density of mercury as $\sf 13.6 \times 10^3 kg \: {m}^{-3}$

Answer should come

10mm of Hg

I want explaination.​

Answer 1

The pressure difference at a height $\sf{h}$ from sea level is,

$\longrightarrow\sf{\Delta P=\rho gh}$

From this height is given by,

$\longrightarrow\sf{h=\dfrac{\Delta P}{\rho g}}$

Since $\sf{\Delta P}$ and $\sf{g}$ are constants,

$\longrightarrow\sf{h\propto\dfrac{1}{\rho}}$

Therefore,

$\longrightarrow\sf{\dfrac{h_2}{h_1}=\dfrac{\rho_1}{\rho_2}}$

Here,

• $\sf{\rho_1=1.295\ kg\,m^{-3}}$

• $\sf{h_1=107\ m}$

• $\sf{\rho_2=13.6\times10^3\ kg\,m^{-3}}$

Then fall in barometric height is,

$\longrightarrow\sf{h_2=h_1\times\dfrac{\rho_1}{\rho_2}}$

$\longrightarrow\sf{h_2=107\times\dfrac{1.295}{13.6\times10^3}}$

$\longrightarrow\sf{h_2=0.01\ m\ of\ Hg}$

$\longrightarrow\underline{\underline{\sf{h_2=10\ mm\ of\ Hg}}}$

Hence 10 mm of Hg is the answer.