JAMB Physics 2019 Paper

Concept:The atmospheric pressure decreases with height, and the drop in pressure is equal to the hydrostatic pressure exerted by the 20 km air column.Explanation:Let the pressure at 20 km be $p$.The decrease in pressure is $760 - p$ mmHg.Use the hydrostatic relation: $\rho_m h_m = \rho_{\text{air}} h_{\text{air}}$.Convert 20 km to cm: $20\text{ km} = 2 \times 10^6\text{ cm}$.Calculate the mercury height equivalent to this air column:$$h_m = \frac{0.00013 \times 2 \times 10^6}{13.6} = 19.12\text{ cm}$$Convert to mmHg: $19.12\text{ cm} = 191.2\text{ mmHg}$.Thus,$$760 - p = 191.2$$$$p = 760 - 191.2 = 568.8 \approx 570\text{ mmHg}$$Answer:Option B: $570\text{ mmHg}$.