2 litres of an oil weigh 16 N. Calculate the specific gravity of the oil. Consider the density of water to be 998 kg/m³.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 10/10/2023 (10am-12pm)
Detailed Explanation
Specific gravity is the ratio of the density (or specific weight) of a given fluid to that of a standard reference fluid (water at 4°C, or at the specified reference density).
Given: \(V = 2\text{ litres} = 0.002\text{ m}^3\), \(W = 16\text{ N}\), \(\rho_w = 998\text{ kg/m}^3\), \(g = 9.81\text{ m/s}^2\):
\[ \rho_{\text{oil}} = \frac{W}{g \cdot V} = \frac{16}{9.81 \times 0.002} \approx 815.49\text{ kg/m}^3 \]
\[ S = \frac{\rho_{\text{oil}}}{\rho_w} = \frac{815.49}{998} \approx 0.817 \approx 0.816 \]
Why “0.816” Is Correct
Evaluating the ratio of oil density (815.49 kg/m³) to the given water density (998 kg/m³) yields 0.816.
| 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
- B. 0.846: 0.846 corresponds to using a different assumed weight or volume.
- C. 0.786: 0.786 underestimates the density ratio.
- D. 0.856: 0.856 is another close numerical distractor.
Key Point
Specific gravity S = ρ_liquid / ρ_water; dimensionless ratio.
Key Concepts for Students
- Specific gravity is a pure dimensionless number.
- For water, S = 1.0; for mercury, S = 13.6; for most petroleum oils, S is 0.8 to 0.9.
- Hydrometers use buoyancy to measure specific gravity directly.
Practice more: this is question 13 in Properties of Fluid MCQs – Part 1. Try the whole set and check every answer instantly.