The spring of the pressure gauge has a force constant of 2000N/m and the piston has a diameter of 3.00 cm .As the gauge is lowered into water, what change in depth causes the piston to move in by 0.55 cm?
The pressure needs to be enough to exert a force on the piston of
F = kX = 2000*0.0055 = 11 N
P = 11 N/(pi D^2/4) N/m^2
(Make sure D in in meters)
Set that equal to (depth)(density)*g and solve for depth