Specific Weight from Weight and Volume — Oil in a Vessel A closed vessel contains V = 4 m^3 of oil with total weight W = 30 kN. Determine the specific weight (unit weight) gamma of the oil in kN/m^3.

Introduction:
Specific weight (unit weight) gamma is the weight per unit volume of a fluid. It is central to hydrostatics because pressure head p/w translates pressure into an equivalent height of fluid.

Given Data / Assumptions:

• Total weight W = 30 kN.
• Volume V = 4 m^3.
• Fluid is homogeneous; gravity standard.

Concept / Approach:
The definition is gamma = W / V. Units: kN divided by m^3 produces kN/m^3, which is the desired unit.

Step-by-Step Solution:
Write gamma = W / V.Substitute: gamma = 30 kN / 4 m^3.Compute: gamma = 7.5 kN/m^3.

Verification / Alternative check:
Compare with water's specific weight ≈ 9.81 kN/m^3; the obtained value 7.5 kN/m^3 is lower, which is reasonable for many oils that are lighter than water.

Why Other Options Are Wrong:
4.5 and 6 kN/m^3 would correspond to lower W or higher V; 10 kN/m^3 would require heavier fluid or smaller volume than given.

Common Pitfalls:
Confusing specific weight with specific gravity; using mass density instead of weight density; unit mistakes between N and kN.

Final Answer:
7.5 kN/m^3