## Physics: Principles with Applications (7th Edition)

The density of the oil is $0.683\times 10^3~kg/m^3$
The gauge pressure at points $a$ and $b$ are equal because they are at the same height and only water is below these two points. We can equate the gauge pressure of oil and the gauge pressure of water. $\rho_o~g~h_o = \rho_w~g~h_w$ $\rho_o = \frac{\rho_w~h_w}{h_o}$ $\rho_o = \frac{(1.00\times 10^3~kg/m^3)(0.1858~m)}{0.272~m}$ $\rho_o = 0.683\times 10^3~kg/m^3$ The density of the oil is $0.683\times 10^3~kg/m^3$.