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)
⇒P1−P0=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 =P1−P0)
(b) P2=P1ρwgh′(h-depth of water layer)
Now, P1=P0+ρogh (from (1))
⇒P2=P0+ρogh+ρwgh
⇒(P2−P0)=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 =P2−P0)