When at cruising altitude, a typical airplane cabin will have an air pressure eq

Question

When at cruising altitude, a typical airplane cabin will have an air pressure equivalent to an altitude of about 2190 m. During the flight, ears often equilibrate, so that the air pressure inside the Inner ear equalizes with the air pressure outside the plane. The Eustachian tubes allow for this equalization, but can become clogged. If an Eustachian tube is clogged, pressure equalization may not occur on descent and the air pressure inside an inner ear may remain equal to the pressure at 2190 m. In that case, by the time the plane lands and the cabin is repressurized to sea-level air pressure, what is the net force on one ear drum due to this pressure difference, assuming the ear drum has an area of 0.53 cm^2?

Explanation / Answer

At 2190m above sea level, air pressure will be 79kPa
whereas at sea level, air pressure is 101kPa

This pressure difference creates force in ear drum.
F = 0.53 * 10^-4 * (101 - 79) * 10^3 = 1.166 N

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