Two litres of an oil weight 16 N. Calculate the specific gravity of the oil. Take density of water as 998 kg/m³.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 15.11.2022 Shift-III
Detailed Explanation
Specific gravity expresses the density of a fluid relative to pure water. Using the provided volume and weight, oil density is calculated via mass = W/g and divided by the reference water density.
\[ V = 2\text{ litres} = 0.002\text{ m}^3, \quad W = 16\text{ N}, \quad \rho_w = 998\text{ kg/m}^3 \]
\[ \rho_{\text{oil}} = \frac{W}{g \cdot V} = \frac{16}{9.81 \times 0.002} \approx 815.49\text{ kg/m}^3 \]
\[ S = \frac{\rho_{\text{oil}}}{\rho_w} = \frac{815.49}{998} \approx 0.816 \]
Why “0.816” Is Correct
Dividing the mass density of the oil (815.49 kg/m³) by water's density (998 kg/m³) gives 0.816, corresponding to option (c).
| 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. 0.846: 0.846 does not match the computed density ratio.
- B. 0.856: 0.856 is an incorrect value from improper arithmetic.
- D. 0.786: 0.786 results from using g = 10 m/s² and rounding incorrectly.
Key Point
Specific gravity S = ρ_fluid / ρ_water; 815.49 / 998 ≈ 0.816.
Key Concepts for Students
- Specific gravity is dimensionless and identical in all unit systems.
- Weight divided by g yields mass: m = 16 / 9.81 ≈ 1.631 kg.
- Water density varies slightly with temperature (1000 kg/m³ at 4°C, 998 kg/m³ at 20°C).
Practice more: this is question 21 in Properties of Fluid MCQs – Part 2. Try the whole set and check every answer instantly.