Answer
(a) The water pressure at a depth of 10.915 km is $1.10\times 10^8~Pa$
(b) The force exerted on the top of the hull due to the water pressure is $1.10\times 10^8~N$
Work Step by Step
(a) We can find the water pressure at a depth of 10.915 km:
$P_{abs} = P_{atm}+P_g$
$P_{abs} = P_{atm}+\rho~g~h$
$P_{abs} = 101~kPa+(1025~kg/m^3)(9.80~m/s^2)(10,915~m)$
$P_{abs} = 1.10\times 10^8~Pa$
The water pressure at a depth of 10.915 km is $1.10\times 10^8~Pa$
(b) We can assume that the pressure inside the vessel is 1 atmosphere. Therefore, the pressure difference between the outside and inside of the hull is $1.10\times 10^8~Pa$. We can find the force on the top of the hull:
$F = P~A = (1.10\times 10^8~N/m^2)(1.0~m^2) = 1.10\times 10^8~N$
The force exerted on the top of the hull due to the water pressure is $1.10\times 10^8~N$.
