Question

If at the top of a hill $2000$m above sea level, the atmospheric pressure is $50$ cm of Hg and at sea level the atmospheric pressure is $74.5$cm of Hg, and you need as much oxygen to breathe at sea level as on the top of the hill, how much faster need you breathe on the hill top?
A.$2.44$ times
B.$1.49$ times
C.$5$ times
D.$7$ times

Answer 1

Hint: First convert the unit of atmospheric pressure from cm into m as height is also in m. Then use the formula of atmospheric pressure which is given as-
$ \Rightarrow $ P=egh
Where P is atmospheric pressure, g is gravity and h is height and e is average density.
Find the average density for sea level and for hill top. Then divide the average density at sea level by the density at hill top to get the answer.

Complete step by step answer:
Given, the height of the hill above the sea level=$2000$m
The atmospheric pressure at hill top =$50$ cm of Hg
The atmospheric pressure at sea level =$74.5$cm of Hg
We know that $1m = 100cm$
Then $1cm = {10^{ - 2}}m$
So the atmospheric pressure at hill top=$50 \times {10^{ - 2}} = 0.5m$
And the atmospheric pressure at sea level=$74.5 \times {10^{ - 2}} = 0.745m$
We have to find how much faster we need to breathe on the hill top.
We will use the formula of atmospheric pressure which gives relation between the height and average density –
$ \Rightarrow $ P=egh
Where P is atmospheric pressure, g is gravity and h is height and e is average density.
Then we can write-
$ \Rightarrow $ e=$\dfrac{P}{{g \times h}}$ -- (i)
Now above sea level or at the hilltop, on putting the given values in the eq. (i0, we get-
$ \Rightarrow $ e=$\dfrac{{0.5}}{{9.8 \times 2000}}$
On solving the equation, we get-
$ \Rightarrow $ e=$\dfrac{{0.5}}{{19600}} = 2.55 \times {10^{ - 5}}$-- (ii)
Now, at sea level, on putting the given values in eq. (i), we get-
$ \Rightarrow $ e=$\dfrac{{0.745}}{{9.8 \times 2000}}$
On solving, we get-
$ \Rightarrow $ e=$\dfrac{{0.745}}{{19600}} = 3.8 \times {10^{ - 5}}$--- (iii)
On dividing, eq. (iii) by eq. (ii), we get-
The need to breathe oxygen=$\dfrac{{3.8 \times {{10}^{ - 5}}}}{{2.55 \times {{10}^{ - 5}}}}$
On solving, we get-
The need to breathe oxygen=$\dfrac{{3.8}}{{2.55}} = 1.49$ times

The correct answer is option B.

Note:
When we climb up an altitude the atmospheric pressure starts decreasing but when we dive under water go towards the depth, the pressure exerted on us is increased. This is because water is about $775$ times denser than air.