WAEC - Physics (1991 - No. 54)
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
120m
138m
1200m
274m
120,000m
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 \(\frac{12 \times 13.6}{0.00136}\)
= 120,000
hpg = Hpg
h x 0.00136 x g = (75 - 63) x 13.6 x g
h \(\frac{12 \times 13.6}{0.00136}\)
= 120,000
Comments (0)
