Question

A barometer reads 75.5 cm of Hg at the foot of a mountain and 65.5 cm of Hg at

the top of the mountain. Estimate the height of the mountain in metre, assuming the
density of mercury as 13.6 x 103 kg m-3 and the average density of air as 1.25 kg m-3
​

Answer 1

Explanation:

Pressure exerted by a liquid column is given by hρg

Where, h is the height of column, ρ is density and g is the acceleration due to gravity.

Thus, difference of the pressure at foot and top point of the mountain = (75.5 – 65.5)ρmg.

Here, density ρm = density of mercury.

This difference is caused by the column of air between foot of the mountain and height of the mountain.

Let height of mountain be H, then pressure of this air column = Hρag

Where, ρa = density of air.

Now, Hρag = (75.5 – 65.5)ρmg

Solving this equation, we get H = 1088 m

Thus, height of the mountain = 1088 m

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Answer 2

The height of the mountain = 112.064 m.

Explanation:

According to the given information, a barometer reads 75.5 cm of Hg at the foot of a mountain and 65.5 cm of Hg at the top of the mountain. Therefore, their pressure differences has to be found out before proceeding further in the problem.

Firstly 75.5 cm can be written as 75.5 * $10^{-2}$ m. and in a similar manner, 65.5 cm. can be written as 65.5 *  $10^{-2}$ m.

Now, the pressure differences when the barometer is at the top of the mountain and when the barometer is at the foot of the mountain

= (75.5 - 65.5) * $10^{-2}$ * (density of mercury) * g,

where g is the acceleration due to gravity.

Then, we get,

10* $10^{-2}$ * (density of mercury) * g

= $10^{-1}$ * (density of mercury) * g...(1)

Equating the pressure difference when the height of the mountain is h1 and (1), we get,

$10^{-1}$ * (density of mercury) * g = h1 * (density of air) * g

Or,  $\frac{10^{-1}}{h1}$ = (density of air)/ (density of mercury)

Now, the values are given as density of mercury as 13.6 x 103 kg m-3 and the average density of air as 1.25 kg m-3.

Thus, we get,

$\frac{10^{-1}}{h1}$ = (density of air)/ (density of mercury)

Or,  $\frac{h1}{10^{-1} }$ = (density of mercury)/ (density of air)

Or, h1 = 112.064 m.

Thus, the height of the mountain = 112.064 m.

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