Question

A person drives to the top of a mountain. On the way up, the person’s ears fail to “pop,” or equalize the pressure of the inner ear with the outside atmosphere. The pressure of the atmosphere drops from 1.010 x 10^5 Pa at the bottom of the mountain to 0.998 x 10^5 Pa at the top. Each eardrum has a radius of 0.40 cm. What is the pressure difference between the inner and outer ear at the top of the mountain?

a.

1.2 x 10^3 Pa

b.

1.1 x 10^5 Pa


c.

1.0 x 10^2 Pa


d.

1.2 x 10^2 Pa

Answer 1

The pressure on the inner ear is calculated by subtracting the pressure of the atmosphere of the bottom from the top, so calculating this will give us: 1.010x10^5 - .998x10^5 = 1200Pa outward which would be letter A.
And so the net force would be now calculated as P*A = 1200Pa*π*(0.40x10^-2m)^2 = 0.0603N