Final answer:
To calculate the density of the unknown liquid in a U-tube where mercury and the liquid are in equilibrium, use the heights of the mercury (2 cm) and the unknown liquid (27.2 cm) and the known density of mercury (13.6 g/cm³) to find that the unknown liquid has a density of 1 g/cm³.
Step-by-step explanation:
To determine the density of the unknown liquid using the U-tube method, we use the fact that the pressure at the bottom of each liquid column must be equal since they are in equilibrium. The pressure from the mercury column can be translated to the pressure from the unknown liquid column.
Given the density of mercury is 13.6 g/cm³, and assuming standard gravity, the density (ρ) of the unknown liquid can be calculated using the heights of the liquid columns and the formula for pressure due to a column of liquid (P = ρgh), where 'g' is the acceleration due to gravity and 'h' is the height of the liquid column.
The equation we can set up is ρ₁g1 = ρ₂g2, where ρ₁ and ρ₂ are the densities of mercury and the unknown liquid, respectively, and 1 and 2 are their respective heights in the U-tube.
Since the density of mercury is given and the heights are known, we can solve for the density of the unknown liquid:
ρ₂ = (ρ₁ × 1) / 2
Plugging in the values:
ρ₂ = (13.6 g/cm³ × 2 cm) / 27.2 cm
ρ₂ = (27.2 g/cm³) / 27.2 cm
ρ₂ = 1 g/cm³
Therefore, the density of the unknown liquid is 1 g/cm³.