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 xx 20cm` 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 xx 10^(3) kg m^(-3)`, `g=10ms^(-2)`. Atmospheric pressure = `1.01 xx 10^(5) Pa`.

Explanation

Here `h=100m` and `rho=1.03xx10^(3)kgm^(-3)` `P=P_(a)+rhogh` ltb rgt `=1.01xx10^(5)Pa`
`+1.03xx10^(3)kgm^(-3)xx10ms^(-2)xx1000m`
`=104.01xx10^(5)Pa`
`~~104atm`
(b). Gauge pressure is `P-P=rhogh=P`
`P_(g)=1.03xx10^(3)kgm^(-3)xx10ms^(2)xx1000m`
`=103xx10^(5)Pa`
`~~103atm`
(c). The pressure outside the submarine is `P=P_(a)+rhogh` and the pressure inside it is `P_(a)`. Hence, the net pressure acting on the window is gauge pressure `P_(g)=rhogh`. Since the area of the window is `A=0.04m^(2)`, the force acting on it is
`F=P_(g)A=103xx10^(5)Paxx0.04m^(2)=4.12xx10^(5)N`