The Value of mass density in kg·sec²/m⁴ for water at 0°C is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 4 MARCH 2017 Morning Shift
Detailed Explanation
Before the universal adoption of the SI system, gravitational metric units (MKS technical system) defined force in kgf (or kg) and derived mass as force divided by acceleration.
In the technical (gravitational) metric system, the unit of weight/force is \(\text{kgf}\):
Specific weight of water \(\gamma = 1000\text{ kgf/m}^3\).
Mass is derived from Newton's second law: \(m = W/g\), so:
\[ [m] = \frac{\text{kgf}}{\text{m/s}^2} = \text{kgf}\cdot\text{s}^2/\text{m} \]
Mass density \(\rho = \frac{\gamma}{g}\):
\[ \rho = \frac{1000\text{ kgf/m}^3}{9.81\text{ m/s}^2} \approx 101.94\text{ kgf}\cdot\text{s}^2/\text{m}^4 \approx 101.9\text{ kg}\cdot\text{sec}^2/\text{m}^4 \]
Why “101.9” Is Correct
Dividing the gravitational specific weight of water (1000 kgf/m³) by g = 9.81 m/s² gives 101.9 kg·s²/m⁴.
| Property | Formula | SI unit | Standard value (water, 4°C) |
|---|---|---|---|
| Density ρ | m/V | kg/m³ | 1000 kg/m³ |
| Specific weight γ | ρg | N/m³ | 9810 N/m³ |
| Specific gravity S | ρ/ρ_water | — | 1.0 |
| Specific volume v | 1/ρ | m³/kg | 0.001 m³/kg |
| Bulk modulus K | −dp/(dV/V) | Pa | 2.06 GPa |
| Compressibility β | 1/K | Pa⁻¹ | 4.85 × 10⁻¹⁰ Pa⁻¹ |
Why the Other Options Are Wrong
- A. 1: 1 is the density of water in g/cm³ in CGS units.
- B. 1000: 1000 is the density of water in kg/m³ in absolute SI units.
- C. 100: 100 is an approximation error assuming g = 10 m/s².
Key Point
Mass density of water in gravitational MKS units: ρ = 1000 / 9.81 ≈ 101.9 kg·s²/m⁴.
Key Concepts for Students
- In gravitational units, mass unit is the hyl (or metric slug) = 9.81 kg.
- 1 hyl/m³ = 1 kgf·s²/m⁴.
- 101.9 hyl/m³ × 9.81 kg/hyl = 1000 kg/m³ (SI density).
Practice more: this is question 60 in Properties of Fluid MCQs – Part 4. Try the whole set and check every answer instantly.