a pilot records the atmospheric pressure outside his plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg with his barometer. Assuming that the density of air is constant, calculate the height of the plane above the ground. (take the relative densities of air and mecury as 0.00136 and 13.6 respectively)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
pressure change for air = pressure change for mercury
hpg = Hpg
h x 0.00136 x g = (75 - 63) x 13.6 x g
h = \(\frac{12 \times 13.6}{0.00136}\)
= 120,000cm
= 1,200m
pressure change for air = pressure change for mercury
hpg = Hpg
h x 0.00136 x g = (75 - 63) x 13.6 x g
h = \(\frac{12 \times 13.6}{0.00136}\)
= 120,000cm
= 1,200m