Question

Under normal circumstances, the vitreous humor, a jelly-like substance in the main part of the eye, exerts a pressure of up to 24.0mm 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 50.0mmof mercury, the condition is called glaucoma. What is the increase in the total force (in newtons) on the walls of the eye if the pressure increases from 24.0mm to 50.0mm of mercury? We can quite accurately model the eye as a sphere 2.10cm in diameter.

Answer 1

Change in pressure=50-24=26 mm Hg=3466.4 Pa

Area of eye=4*pi*d²/4=13.85 cm²=13.85*10^-4 m²

Force=Pressure*area=13.85*10^-4*3466.4=4.80 N

Answer 2

change in pressure:
?p = 50.0 - 24.0 mm Hg
?p = 26.0 mm Hg

change in force:
?F = A?p
?F = (?d