Fluid properties: A liquid has mass 3000 kg and volume 4 m^3. The numerical value 0.75 corresponds to which property?

Introduction / Context:

Identifying fluid properties from basic mass–volume data is fundamental in hydraulics and process engineering. Density, specific weight, and specific gravity are related but distinct. This question tests your ability to compute and interpret these values.

Given Data / Assumptions:

• Mass m = 3000 kg.
• Volume V = 4 m^3.
• Reference for specific gravity: water at about 4°C with density 1000 kg/m^3.

Concept / Approach:

Compute density rho = m / V. Then determine specific gravity SG = rho / rho_w. Specific weight gamma = rho * g; its numerical value would be in N/m^3, not a pure number.

Step-by-Step Solution:

Verification / Alternative check:

Unit check: specific gravity is dimensionless; the calculated 0.75 fits. Density has units kg/m^3; the number 0.75 without units cannot represent density or specific weight.

Why Other Options Are Wrong:

• Specific weight needs N/m^3 units and a value near 7.36 × 10^3.
• Specific mass (density) equals 750 kg/m^3 (not 0.75).
• “None of these” is false because 0.75 correctly matches specific gravity.

Common Pitfalls:

• Confusing density (kg/m^3) with specific gravity (ratio to water).
• Forgetting that specific gravity is unitless.

Final Answer:

Specific gravity