Question

Under normal circumstances, the vitreous humor, a jelly-like substance in the main part of the eye, exerts pressure of up to 24 mm of mercury that maintains the shape of the eye. If blockage of the drainage duct for aqueous humor causes this pressure to increase to about 44 mm of mercury, the conditions is called glaucoma. What is the increase in the total force (in newtons) on the walls of the eye if the pressure increases from 24 mm to 44 mm of mercury, and we can accurately model the eye as a sphere 2.5 cm in diameter? (1 mm mercury = 133.3 Pa)

Answer 1

Answer:

F = 5.235 N

Explanation:

This is a pressure problem in which we have got all the data to solve it.

They are asking us about the increase in the total force on the walls of the eye because of a pressure increase.

First let's calculate the pressure increase.

Δp = final pressure - initial pressure = 44 mm of mercury - 24 mm of mercury

Δp = 20 mm of mercury = 20 mm Hg

Using this 1 mm Hg = 133.3 Pa

20 mm Hg x (133.3 Pa) / (1 mm Hg) = 2666 Pa

Where Pa = N / (m^2)

Our area is the sphere area = π x (d^2)

d = 2.5 cm = 0.025 m

Area = π x [(0.025 m)^2]

Area = [1.9634 x (10^ -3)] m^2

p = F/A

2666 Pa = F / [1.9634 x (10^ -3)] m^2

2666  N / (m^2) = F / [1.9634 x (10^ -3)] m^2

F = 5.235 N