Final answer:
The density of the liquid is calculated using the pressure-depth relationship and is found to be 500 kg/m³. The specific gravity, which is the ratio of its density to that of water is 0.5.
Step-by-step explanation:
To determine the density and specific gravity of a liquid when the pressure 3m below the free surface is 15 kN/m² in excess of atmospheric pressure, we can use the pressure-depth relationship in fluids, which is given by the equation p = ρgh, where p is the pressure, ρ (rho) is the density of the liquid, g is the acceleration due to gravity, and h is the depth.
Since the excess pressure is 15 kN/m² (which is 15,000 N/m²) at a depth of 3 meters, and with a given gravitational acceleration of 10 m/s², the density of the liquid can be calculated by rearranging the formula to ρ = p / (g × h).
Substituting the given values, ρ = 15,000 N/m² / (10 m/s² × 3m) = 500 kg/m³.
The specific gravity of the liquid is the ratio of its density to the density of water (1000 kg/m³ at 4°C), which is 500 kg/m³ / 1000 kg/m³ = 0.5.