At the base of a mountain, a mercury barometer reads 700 mm whereas at the top it reads 500 mm of Mercury. The density of air is 1.23 kg/m³, is assumed constant. Determine the height of the mountain. (Take specific gravity of mercury as 13.6, and the density of water as 1000 kg/m³)
Try it first: pick an option, then read the explanation.
Asked in SSC JE 11-12-2020 (Evening)
Detailed Explanation
As altitude increases, atmospheric pressure decreases because the weight of the overlying air column decreases. Using the hydrostatic law with constant air density, the mountain height can be determined from the difference in barometric readings.
1. Given parameters:
- Barometer reading at mountain base \(h_1 = 700\text{ mm of Hg} = 0.700\text{ m}\)
- Barometer reading at mountain top \(h_2 = 500\text{ mm of Hg} = 0.500\text{ m}\)
- Mercury column difference \(\Delta h_{\text{Hg}} = 700 - 500 = 200\text{ mm} = 0.200\text{ m}\)
- Density of mercury \(\rho_{\text{Hg}} = 13.6 \times 1000 = 13,600\text{ kg/m}^3\)
- Density of air \(\rho_{\text{air}} = 1.23\text{ kg/m}^3\) (assumed uniform)
2. Calculate the atmospheric pressure drop (\(\Delta p\)) from base to top:
\[ \Delta p = \rho_{\text{Hg}} \cdot g \cdot \Delta h_{\text{Hg}} \]
3. By the hydrostatic equation for the air column of height \(H_{\text{mountain}}\):
\[ \Delta p = \rho_{\text{air}} \cdot g \cdot H_{\text{mountain}} \]
4. Equate the two expressions and solve for \(H_{\text{mountain}}\):
\[ \rho_{\text{air}} \cdot g \cdot H_{\text{mountain}} = \rho_{\text{Hg}} \cdot g \cdot \Delta h_{\text{Hg}} \]
\[ H_{\text{mountain}} = \frac{\rho_{\text{Hg}} \cdot \Delta h_{\text{Hg}}}{\rho_{\text{air}}} = \frac{13,600 \times 0.200}{1.23} = \frac{2720}{1.23} \approx 2211.38\text{ m} \approx 2211\text{ m} \]
Why “2211 m” Is Correct
The \(200\text{ mm}\) drop in the mercury column corresponds to an atmospheric pressure difference of \(13,600 \times 9.81 \times 0.2 = 26,683\text{ Pa}\). Balancing this against an air column of density \(1.23\text{ kg/m}^3\) gives a mountain altitude of \(2211\text{ m}\).
Why the Other Options Are Wrong
- A. 3423 m: 3423 m results from mistakenly dividing 2720 by 0.794 or inverted ratio calculations.
- C. 1584 m: 1584 m results from omitting the specific gravity factor of 13.6 correctly.
- D. 1122 m: 1122 m results from dividing the result by 2 (using half the pressure difference).
Key Point
Barometric altimetry: H_mountain = (ρ_Hg · Δh_Hg) / ρ_air.
Key Concepts for Students
- Because air is compressible, true air density decreases with altitude, making H_mountain slightly higher in reality.
- Under constant density assumption (isothermal low altitude), Δp = ρ_air · g · Δh.
- A barometric pressure drop of 1 mm Hg corresponds roughly to an ascent of about 10 to 11 metres near sea level.
Practice more: this is question 8 in Measurement of Pressure MCQs – Part 1. Try the whole set and check every answer instantly.