Question

4. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compared to what it was at the lake surface? The area of the eardrum is 0.60 cm2.
Can someone do this problem in the most simplistic way? Thank you

Answer 1

Pressure at depth h in water given by P = P_0 + gh Increase In Pressure delta P = P - P_0 = gh Increase in Pressure of eardrum given by delat P middot a = gh middot a =)pi r^2) gh = 0.60 times 10^-4 times 1000 times 9.8 times 3.2 delta P middot a = 1.88p increase in pressure of eardrum at h = 3.2 m w.r.t surface P_0 = pressure surface = Density of water g = gravitational acc h = depth in water delta P = increase in pressure a = area of eardrum = pi r^2 = 0.60 cm^2 r = radius of drum a = 0.60 times 10^-4 m^2 g = 9.8 m/s^2 = 1000 kg/m^3