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Question

A barrel contains a 0.120 m layer of oil floating on water that is 0.250 m deep. The density of the oil is 600kg/m3. (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

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Solution

Let P be the atmospheric pressure P, be the pressure at oil-water interface and P2 be the pressure at the bottom of the barrel
Let, ρo be the density of oil and ρw be the density of water
ρo=600 kg/m3,ρw=1000 kg/m3
(a) We know that
P1=P0+ρogh - (1) (h-depth of oil layer)
P1P0=600×9.8×0.12
=72×9.8 Pa
=705.6 Pa
Gauge pressure at the oil-water
interface =705.6 Pa
( Gauge pressure at interface =P1P0)

(b) P2=P1ρwgh(h-depth of water layer)
Now, P1=P0+ρogh (from (1))
P2=P0+ρogh+ρwgh
(P2P0)=ρ0gh+ρwgh
=(600×9.8×0.12)+(1000×9.8×0.25)
=705.6+2450 Pa
=3155.6 Pa
Gauge pressure at the bottom of the barrel
=3155.6 pa
( Gauge pressure at bottom =P2P0)









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