The heights of mercury in a barometer at the | vedclass

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Question

(B) The pressure difference between the base and the top of the mountain is equal to the weight of the air column of height $h$ and density $\rho_{air

Vedclass · Accepted Answer

$2.5 \, km$

Vedclass · Answer

(B) The pressure difference between the base and the top of the mountain is equal to the weight of the air column of height $h$ and density $ho_{air}$.This pressure difference is also equal to the difference in pressure exerted by the mercury columns.Let $h$ be the height of the mountain,$ho_{Hg}$ be the density of mercury,and $ho_{air}$ be the density of air.The pressure difference is given by: $\Delta P = (h_1 - h_2) ho_{Hg} g = h ho_{air} g$.Substituting the given values: $(75 \, cm - 50 \, cm) 	imes ho_{Hg} = h 	imes ho_{air}$.$25 	imes 10^{-2} \, m 	imes ho_{Hg} = h 	imes ho_{air}$.$h = 25 	imes 10^{-2} 	imes \left( \frac{ho_{Hg}}{ho_{air}} ight)$.Given $\frac{ho_{Hg}}{ho_{air}} = 10^4$.$h = 25 	imes 10^{-2} 	imes 10^4 = 2500 \, m$.Therefore,the height of the mountain is $2.5 \, km$.

The heights of mercury in a barometer at the base and at the top of a mountain are $75 \, cm$ and $50 \, cm$,respectively. If the ratio of the density of mercury to the density of air is $10^4$,what is the height of the mountain?

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