InterviewSolution
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InterviewSolution
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At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) what is the auge pressure? (c ) Find the force acting on the windown 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`. |
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Answer» Here `P_(a) = 1.01 xx 10^(5)Pa`, `rho = 1.03 xx 10^(4)=3 kg m^(-3), h=1000m` `A= 0.20 xx 0.20 =0.04 m^(2)` (a) Absolute pressure `P=P_(a) + rho g h` `= 1.01 xx 10^(5) + 1.03 xx 10^(3) xx 10 xx 1000` `=104.01 xx 10^(5) Pa` (b) Gauge pressure, `P=P-P_(a) = rho g h` `=(1.03 xx 10^(3)) xx 10 xx 1000 = 103 xx 10^(5)Pa` (c ) The pressure outside the submarine is `P=P_(a)+rho gh` and the pressure inside the submarine is `P_(a)`. Pressure difference on the window of the submarine =`P-P+(a) = rho gh` Force on the window = pressure difference `xx` area of window `= rho gh xx A = 1.3 xx 10^(5) xx (0.04)` `=4.12 xx 10^(5) N`. |
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