To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.
For mercury, density, gravity and height are defined as
[tex]\rho_m = 846lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_1 = 1in = \frac{1}{12} ft[/tex]
For the air the defined properties would be
[tex]\rho_a = 0.0075lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_2 = ?[/tex]
We have for equilibrium that
[tex]\text{Pressure change in Air}=\text{Pressure change in Mercury}[/tex]
[tex]\rho_m g h_1 = \rho_a g h_2[/tex]
Replacing,
[tex](846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)[/tex]
Rearranging to find [tex]h_2[/tex]
[tex]h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}[/tex]
[tex]h = 9400ft[/tex]
Therefore the elevation of the mountain top is 9400ft
