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 105 Pa at the bottom of the mountain to 0.998 x 105 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?

Question

anonymous · Answer

It seems like to solve this all you need to do is subtract the pressures. The atmosphere and inner ear are at 1.010 x 10^5 Pa at the bottom of the mountain. Then at the top, the atmosphere is now 0.998 x 10^5 Pa and the inner ear is still 1.010 x 10^5 Pa, then I would imagine the pressure difference is just the difference between the two. I don't see how the radius of the ear has anything to do with this.