Question

In scuba diving, a regulator is used so that the pressure of the air the diver breathes is close to that of the ambient water. A reckless swimmer decides to use a hose sticking out of the surface to breathe underwater while diving in a lake. When the air pressure in the lungs is at a pressure of around 0.145atm below the ambient pressure, lung injury may occur. Find the depth at which the swimmer would experience such a pressure differential.

Thank you for your help

Answer 1

Let the depth be d,

Ambient pressure at the given depth = Patm +rho g h

Pressure in his lungs = Patm

Pressure difference = rho g h

0.145 atm = 1000*9.8*h

0.145*101325 = 1000*9.8*h

h = 0.145*101325/[1000*9.8]

= 1.5 m answer