The force acting on a window of area 50 cm x 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = ${10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$, g = l0 ${\mathrm{ms}}^{-2}$)

Here, h = 2000 m, $\mathrm{\rho } = {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}, \mathrm{g} = 10 {\mathrm{ms}}^{-2}$

The pressure outside the submarine is $\mathrm{P} = {\mathrm{P}}_{\mathrm{a}}+\mathrm{\rho gh}$

where ${\mathrm{P}}_{\mathrm{a}} \mathrm{is} \mathrm{the} \mathrm{atmospheric} \mathrm{pressure}$

Pressure inside the submarine is ${\mathrm{P}}_{\mathrm{a}}$.

Hence, net pressure acting on the window is guage pressure.

Gauge pressure, ${\mathrm{P}}_{\mathrm{g}} = \mathrm{P}-{\mathrm{P}}_{\mathrm{a}} = \mathrm{\rho gh}$

   = ${\text{10}}^3 \mathrm{kg} {\mathrm{m}}^{-3} \times 10 {\mathrm{ms}}^{-2}\times 2000 \mathrm{m} = 2 \times {10}^7 \mathrm{Pa}$

Area of a window is

A = 50 cm $\times 50 \mathrm{cm} = 2500 \times {10}^{-4} {\mathrm{m}}^2$

Force acting on the window is

F = ${\mathrm{P}}_{\mathrm{g}}\mathrm{A} = 2\times {10}^7 \mathrm{Pa} \times 2500 \times {10}^{-4} {\mathrm{m}}^2$

   = 5 $\times {10}^{6 }\mathrm{N}$