Question

The indicating liquid in a manometer is water (sg = 0.997) and the other liquid is benzene (sg = 0.879). The two liquids are essentially insoluble in each other.

If the manometer reading is 13.8 cm of water, what is the pressure difference in Pa?

Answer 1

$\begin{aligned} ∆P &= \rho_wh_wg_w - \rho_mh_mg_m\\ &= g(\rho_wh_w-\rho_mh_m)\\ &= g\rho_{H_2O}(sg_w-sg_b)∆h \\&= 9.8× 1000×(0.997-0.879)(13.8×10^{-2})\\ &= 159.6\ Pa \end{aligned}\\$