A force of 3.00 N is enough to rupture an eardrum of area 1.00 x 10^-4 m^2. Taking fresh water as 1.00 x 10^3 kg/m^3, g as 9.80 m/s^2 and the gauge pressure in the middle ear as zero, at what depth would the eardrum rupture?

1. First find the pressure the eardrum can stand, using pressure = force / area. 2. Pressure = (3.00 N) / (1.00 x 10^-4 m^2) = 3.00 x 10^4 N/m^2. 3. This is a gauge pressure, because the middle ear behind the drum is at zero gauge pressure. 4. Water depth gives gauge pressure = h rho g, so h = gauge pressure / (rho g). 5. Denominator = (1.00 x 10^3 kg/m^3)(9.80 m/s^2) = 9.80 x 10^3 kg/(m^2 s^2). 6. h = (3.00 x 10^4 N/m^2) / (9.80 x 10^3 kg/(m^2 s^2)) = 3.06 m. 7. The values 0.306 m, 30.6 m and 0.031 m all carry the right digits with the power of ten mishandled, which is what happens if the area is entered in square centimetres. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.9 Pressures in the Body_