At a depth of $1000 \; m$ in an ocean:
$(a)$ What is the absolute pressure?
$(b)$ What is the gauge pressure?
$(c)$ Find the force acting on the window of area $20 \; cm \times 20 \; cm$ of a submarine at this depth,the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is $1.03 \times 10^3 \; kg \; m^{-3}$,$g = 10 \; m \; s^{-2}$.)

$(a)$ The absolute pressure $P$ is given by $P = P_a + \rho g h$.
Given $P_a = 1.01 \times 10^5 \; Pa$,$\rho = 1.03 \times 10^3 \; kg \; m^{-3}$,$g = 10 \; m \; s^{-2}$,and $h = 1000 \; m$.
$P = 1.01 \times 10^5 + (1.03 \times 10^3 \times 10 \times 1000) = 1.01 \times 10^5 + 103 \times 10^5 = 104.01 \times 10^5 \; Pa \approx 104 \; atm$.
$(b)$ The gauge pressure $P_g$ is the difference between absolute pressure and atmospheric pressure: $P_g = P - P_a = \rho g h$.
$P_g = 1.03 \times 10^3 \times 10 \times 1000 = 103 \times 10^5 \; Pa \approx 103 \; atm$.
$(c)$ The net pressure acting on the window is the gauge pressure $P_g = \rho g h$. The area of the window $A = 20 \; cm \times 20 \; cm = 0.2 \; m \times 0.2 \; m = 0.04 \; m^2$.
The force $F$ acting on the window is $F = P_g \times A$.
$F = 103 \times 10^5 \; Pa \times 0.04 \; m^2 = 4.12 \times 10^5 \; N$.