At a depth of 1000m 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 20cm 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 ms^-2

At a depth of 1000m in an ocean (a) What is the absolute pressure? (b)

1.	At a depth of 1000m 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 20cm 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 ms^-2
Answer» Solution :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`