The head produced by a liquid of relative density 1.62 is equal to the pressure of 250 kPa. What is the head (m) produced by the liquid?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 22.1.2018 (Evening Shift)
Detailed Explanation
The pressure head of a liquid represents the vertical column height of that fluid needed to create a specified hydrostatic pressure intensity at its base.
1. Identify the given parameters:
- Pressure intensity \(p = 250\text{ kPa} = 250,000\text{ N/m}^2\)
- Relative density (specific gravity) \(S = 1.62\)
- Liquid density \(\rho = S \times \rho_{\text{water}} = 1.62 \times 1000 = 1620\text{ kg/m}^3\)
- Gravitational acceleration \(g = 9.81\text{ m/s}^2\)
2. Calculate the specific weight of the liquid (\(\gamma\)):
\[ \gamma = \rho g = 1620\text{ kg/m}^3 \times 9.81\text{ m/s}^2 = 15,892.2\text{ N/m}^3 \]
3. Calculate the required pressure head (\(h\)):
\[ h = \frac{p}{\gamma} = \frac{250,000\text{ N/m}^2}{15,892.2\text{ N/m}^3} \approx 15.7309\text{ m} \approx 15.73\text{ m} \]
Why “15.73” Is Correct
Dividing the pressure of \(250,000\text{ N/m}^2\) by the liquid's specific weight of \(15,892.2\text{ N/m}^3\) yields a column height of \(15.73\text{ m}\).
Why the Other Options Are Wrong
- A. 5.04: 5.04 m results from multiplying by 1.62 instead of dividing or incorrect power-of-ten operations.
- B. 7.53: 7.53 m results from dividing 15.73 by 2 or arbitrary scaling.
- D. 25.21: 25.21 m is the equivalent head of standard water: 250,000 / (1000 × 9.81) ≈ 25.48 m.
Key Point
Pressure head: h = p / (ρ·g) = p / (S · 9810) = 250,000 / (1.62 × 9810) = 15.73 m.
Key Concepts for Students
- Heavier liquids (S > 1) require shorter columns than water for the same pressure.
- Water would require h_w = 25.48 m; the dense liquid requires only 25.48 / 1.62 = 15.73 m.
- Mercury (S = 13.6) would require only 25.48 / 13.6 = 1.87 m.
Practice more: this is question 22 in Measurement of Pressure MCQs – Part 2. Try the whole set and check every answer instantly.