Answer
(a) The gauge pressure at the oil-water interface is $706~N/m^2$.
(b) The gauge pressure at the bottom of the barrel is $3160~N/m^2$.
Work Step by Step
(a) We can find the gauge pressure $P_1$ at the bottom of the oil layer.
$P_1 = \rho_o~g~h_1$
$P_1 = (600~kg/m^3)(9.80~m/s^2)(0.120~m)$
$P_1 = 706~N/m^2$
The gauge pressure at the oil-water interface is $706~N/m^2$.
(b) We can find the gauge pressure $P_2$ at the bottom of the water layer.
$P_2 = P_1 + \rho_w~g~h_2$
$P_2 = (706~N/m^2) + (1000~kg/m^3)(9.80~m/s^2)(0.250~m)$
$P_2 = 3160~N/m^2$
The gauge pressure at the bottom of the barrel is $3160~N/m^2$.
