It is found that the weight of liquid is 15 N which measures 6 litres. Calculate its specific weight.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 04/12/2023 (09am-11am)
Detailed Explanation
Specific weight (or weight density) of a fluid is defined as its total weight per unit volume. Converting liters to cubic meters allows direct evaluation in SI units (N/m³).
Given weight \(W = 15\text{ N}\) and volume \(V = 6\text{ litres}\):
\[ V = 6 \times 10^{-3}\text{ m}^3 \]
\[ \gamma = \frac{W}{V} = \frac{15\text{ N}}{6 \times 10^{-3}\text{ m}^3} = \frac{15{,}000}{6} = 2500\text{ N/m}^3 \]
Why “2500 N/m³” Is Correct
Dividing 15 N by 0.006 m³ gives exactly 2500 N/m³, which corresponds to option (a).
| 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. 4000 N/m³: 4000 N/m³ corresponds to an inverted ratio (24/6) or arithmetic mistake.
- C. 3200 N/m³: 3200 N/m³ is an arbitrary incorrect distractor.
- D. 1500 N/m³: 1500 N/m³ results from dividing by 10 litres instead of 6 litres.
Key Point
Specific weight γ = Weight / Volume; 1 litre = 10⁻³ m³.
Key Concepts for Students
- Specific weight γ has SI units of N/m³ and dimensional formula [M L⁻² T⁻²].
- Relation with mass density: γ = ρg.
- Water has a standard specific weight of 9810 N/m³ at 4°C.
Practice more: this is question 10 in Properties of Fluid MCQs – Part 1. Try the whole set and check every answer instantly.