Avinash practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth 'L' and swimming to the surface and failing to exhale during his ascent. At the surface ∆P between the external pressure on him and the air pressure in his lungs is 9.3 kpa. From what depth does he start?
0.93 m
Here, when the driver fills his lungs at depth L, the external pressure on him and thus the air pressure within his lungs is greater than the normal and is given by,
P=P0+pgL
As he ascends, the external pressure on him decreases, until it is atmospheric pressure P0. His blood pressure also decreases until it is normal. However because he does not exhale, the air pressure in his lungs remains at the value it had at depth L. at the surface the pressure difference Δ P is
ΔP=P−P0=pgL
L=ΔPpg=93001000×10=0.93 m (approx)