Answer
a) $P=116~Pa$
b) $P=921~Pa$
c) $m=0.82~kg$
$~~~~\rho=822~kg/m^3$
Work Step by Step
a)Pressure acting on the top surface of the cube:
$P=\rho g h=(790~kg/m^3)(9.8~m/s^2)({1.5~cm \over 100~m})=116~Pa$
b)Pressure acting on the bottom surface of the cube:
$P_{oil+water}=(1000~kg/m^3)(9.8~m/s^2)({10~cm \over 100~m})+(1000~kg/m^3)(9.8~m/s^2)({1.5~cm \over 100~m})=921~Pa$
c) Find total force keeping the cube in equilibrium by finding the difference between the force acting on top and bottom surfaces of the cube:
$P={F \over A}$
$F=PA$
$\Delta F=P_2A-P_1B=(P_2-P_1)A$
$\Delta F=(921~Pa-116~Pa)(0.1)^2=8.05~N$
$m={8.05~N \over 9.8~m/s^2}=0.82~kg$
$\rho={m \over V}={0.82~kg \over (0.1~m)^3}=822~kg/m^3$
