Question

On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument?

Answer 1

Answer:

The barometer reading will be 29.43 in

Explanation:

Using the formula of pressure variation

p2 - p1 = -yair * H

= 7.65 * $10^{-2} \frac{lb}{ft^{3} } * 500 ft$

= 38.5 $\frac{lb}{ft^{2} }$

According to the relationship between the pressure and the height of the mercury column

p = yHg * h --> where yHg and h is the barometer reading

yHg $(\frac{29.97}{12} ft)$ - yHg * h1 = 38.5 $\frac{lb}{ft^{2} }$

h1 = ($\frac{29.97}{12} ft$) - $\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }$

     $[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in$  

Answer 2

The barometer reading be "29.43 in".

According to the question,

Barometer reading,

• $H' = 29.97 \ in$

Pressure variation for incompressible and static fluid will be:

→ $P_2-P_1 = V_{air} H$

                $= 7.65\times 10^{-2}\times 500$

                $= 38.5 \ lb/ft^2$

Pressure and height relationship for mercury will be:

→ $P = V_{Hg} H'$

→ $V.Hg\times \frac{29.97}{12} - V_{Hg} h_1 = 38.5$

→ $h_1 = \frac{29.97}{12} - \frac{38.5}{847}$

       $= [\frac{29.97}{12} - 0.0455]$

       $= 29.42 \ in$

Thus the above answer is right.

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