Two liters of a petrol having specific gravity 0.805 weigh 15.8 N. Calculate its density in kg/m³.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 16.11.2022
Detailed Explanation
Specific gravity is defined as the ratio of the density of a fluid to the density of pure water at standard temperature (1000 kg/m³). Therefore, multiplying specific gravity by 1000 kg/m³ gives the mass density directly.
Given specific gravity \(S = 0.805\) and standard water density \(\rho_{\text{water}} = 1000\text{ kg/m}^3\):
\[ \rho = S \times \rho_{\text{water}} = 0.805 \times 1000\text{ kg/m}^3 = 805\text{ kg/m}^3 \]
Verification with weight and volume:
\[ m = \frac{W}{g} = \frac{15.8}{9.81} \approx 1.6106\text{ kg} \]
\[ \rho = \frac{m}{V} = \frac{1.6106}{0.002} \approx 805.3\text{ kg/m}^3 \]
Why “805” Is Correct
Specific gravity 0.805 corresponds directly to a mass density of 805 kg/m³ when water density is taken as 1000 kg/m³.
| 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. 830: 830 kg/m³ is an arbitrary distractor.
- B. 840: 840 kg/m³ does not match specific gravity 0.805.
- D. 780: 780 kg/m³ is typical of pure gasoline/heptane blends but not this given value.
Key Point
Mass density = Specific gravity × 1000 kg/m³; 0.805 × 1000 = 805 kg/m³.
Key Concepts for Students
- Specific gravity is the ratio of liquid density to standard water density (1000 kg/m³).
- The additional data (volume = 2 L, weight = 15.8 N) is consistent with g ≈ 9.81 m/s².
- Density has units of kg/m³ in SI and g/cm³ in CGS.
Practice more: this is question 18 in Properties of Fluid MCQs – Part 2. Try the whole set and check every answer instantly.