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=1000m and `p=1.03 times 10^3 kg m^-3`
(a) From Eq. 10.6 absolute pressure
`P=P_a+pgh`
`=1.01 times 10^5 Pa`
`+1.03 times 10^3 kg m^-3 times 10 ms^-2 times 1000m`
`=104.01 times 10^5 Pa`
=104 atm
(b) Gauge pressure is `P-P_a=pgh=P_s`
`P_l=1.03 times 10^3 kg m^-3 times 10 ms^2 times 1000m`
`=103 atm`
(c ) the pressure outside the submarine is `P=P_a+pgh` and the pressure inside it is `P_a`. hence the net pressure acting on the window is gauge pressure `P_g=pgh` Since the area of the window is `A=0.04 m^2` the force acting on it is
`F=P_gA=103 times 10^6Pa times 0.04 m^2=4.12 times 10^5 N`