Question #04317

1 Answer
Dec 29, 2014

The absolute pressure increases by a factor of 1.97.

The absolute pressure exerted on a submerged body is given by:

#P=rhogh +A#

#rho# is the density of the liquid
#g# is the acceleration due to gravity = #9.8m.s^(-2)#
#h#= depth
#A# = atmospheric pressure.

The pressure felt by the swimmer is effectively decided by the total weight of the water and air above them. Since we are not given the air pressure in the question we cannot assess the factor by which the pressure on the swimmer will change. However I will assume standard atmospheric pressure to be #1.01xx10^5N.m^-2# and we can see that there will be a significant change in absolute pressure on the swimmer.

So initial absolute pressure:

#P_1=(1000xx9.8xx1)+1.01xx10^5Pa#

#P_1=1.108xx10^5Pa#

#P_2=(1000xx9.8xx12)+1.01xx10^5Pa#

#P_2=2.186xx10^5Pa#

#(P_2)/(P_1)=(2.186xx10^5)/(1.108xx10^5)=1.97#

So we can see that the absolute pressure is almost doubled.