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

Given, `h=1000 m, p=1.03 xx 10^(3) kg//m^(3)`
The absolute pressure at depth h is
`P=P_(a)+hpg`
where `P_(a)=1 atm =1.01 xx 10^(5) Pa`
`P=1.01 xx 10^(5)+1000 xx 1.03 xx 10^3 xx 10`
`=104.04 xx 10^(5)Pa=(104.0 xx 10^(5))/(1.01 xx 10^(5)) atm`
=104 atm
Gauge pressure is
`P_(g)=P-P_(a)=hpg`
`=1000 xx 1.03 xx 10^(3) xx 10`
`=103 xx 10^(5) Pa`
=103 atm
Are aof window
`A=20 cm xx 20 cm =400 cm^(2)-0.04 m^(2)`
Pressure outside the submarine is
`P=P_(a)+hpg`
Pressure inside the submarine `=P_(a)`
Net pressure on window is equal to gauge pressure `P_(g)=hpg`
Force acting on the window,
`F=P_(g) A=103 xx 10^(5) Pa xx 0.04 m^(2)`
`=4.12 xx 10^(5)N`