JAMB 2019 Physics · Question 42

The height at which the atmosphere ceases to exist is about 80km. If the atmospheric pressure on the ground…

Question

The height at which the atmosphere ceases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is (ρm = 13.6g/cm $^3$ ρ = 0.00013g/cm $^3$ )

A.380mmHg
B.570mmHg
C.190mmHg
D.480mmHg

Answer: B. 570mmHg

Explanation written with AI and checked against the answer key

Analyze the question: The atmosphere is taken to extend 80 km above the ground. The pressure at ground level is 760 mmHg. We need the pressure 20 km above the ground. Concept: The question is built on the idea that pressure comes from the weight of the air above you. If we treat the air as having uniform density, the pressure falls linearly with height and reaches zero at the top of the atmosphere (80 km). Check with the densities given: P = ρgh, so the height of air equivalent to 760 mmHg is h = (ρm × 76 cm) / ρ = (13.6 × 76) / 0.00013 ≈ 7.95 × 10⁶ cm ≈ 79.5 km ≈ 80 km This confirms the uniform-density assumption: 80 km of air produces 760 mmHg. Solution: The pressure drop from climbing 20 km is the weight of the 20 km layer of air that is now below you: ΔP = (20 / 80) × 760 = 190 mmHg Pressure at 20 km: P = 760 − 190 = 570 mmHg (Equivalently, the air above you is 60 km thick: (60/80) × 760 = 570 mmHg.) Answer: B. 570 mmHg. Tip: When a question says the atmosphere ends at a fixed height and gives constant density, use the linear method. Pressure at height h = P₀ × (1 − h/H). Option C (190 mmHg) is the trap: it is only the pressure drop, not the pressure at that height. Always check whether you are asked for the drop or for the remaining pressure.

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