Answer
a) $h=1.90m$
b) $P=118KPa$
Work Step by Step
(a) We can find the required depth as
$h=\frac{P-P_{at}}{\rho g}$
We plug in the known values to obtain:
$h=\frac{116\times 10^3-1.01\times 10^5}{(806)(9.81)}$
$h=1.90m$
(b) We can find the pressure as
$h_2=\frac{2.05\times 10^{-3}}{65.2\times 10^{-4}}=0.314m$
$h_{total}=h+h_2=1.897+0.314=2.211m$
Now $P=P_{at}+\rho gh$
We plug in the known values to obtain:
$P=1.01\times 10^5+(806)(9.81)(2.211)$
$P=118KPa$
