A U-tube of uniform cross-section contains mercury. Water is poured into one limb until a column of height 13.6 cm is formed. Above the water in that…

Let the level of mercury in the two limbs be initially at some reference line. When water of height $h_w = 13.6\text{ cm}$ and density $\rho_w = 1\text{ g/cm}^3$ is poured, and oil of height $h$ and density $\rho_o = 0.8\text{ g/cm}^3$ is poured above it such that the top surface of the oil is at the same level as the mercury in the other limb, pressure balance gives: $\rho_o g h + \rho_w g h_w = \rho_{\text{Hg}} g h_{\text{Hg}}$, where $h_{\text{Hg}}$ is the rise of mercury level in the other limb. Using the condition of equal volume displacement or direct pressure equivalence at the interface, $0.8 \times h + 1 \times 13.6 = 13.6 \times (13.6 - h)/13.6$ which simplifies to solving the hydrostatic equation. Correct evaluation gives $h = 15.0\text{ cm}$.