Pressure of 200 kPa is equivalent to the head of z metre of liquid having relative density 1.59. The value of z (m) is ..........
Try it first: pick an option, then read the explanation.
Asked in SSC JE 22.1.2018 (Morning Shift)
Detailed Explanation
The hydrostatic pressure intensity developed at the base of a liquid column is directly proportional to the column height, liquid density, and acceleration due to gravity.
1. State the given data:
- Pressure \(p = 200\text{ kPa} = 200,000\text{ N/m}^2\)
- Relative density (specific gravity) \(S = 1.59\)
- Liquid density \(\rho = S \times \rho_w = 1.59 \times 1000 = 1590\text{ kg/m}^3\)
- Gravitational acceleration \(g = 9.81\text{ m/s}^2\)
2. Apply the hydrostatic column equation:
\[ p = \rho g z = (S \times 1000) \cdot g \cdot z \]
3. Solve for column height \(z\):
\[ z = \frac{p}{\rho g} = \frac{200,000\text{ N/m}^2}{1590\text{ kg/m}^3 \times 9.81\text{ m/s}^2} = \frac{200,000}{15,597.9} \approx 12.822\text{ m} \approx 12.82\text{ m} \]
Why “12.82” Is Correct
Dividing the pressure of \(200,000\text{ N/m}^2\) by the specific weight of the liquid (\(1590 \times 9.81 = 15,598\text{ N/m}^3\)) gives \(z = 12.82\text{ m}\).
Why the Other Options Are Wrong
- A. 11.6: 11.6 m results from using g = 10.8 m/s² or arithmetic errors.
- B. 11.82: 11.82 m results from subtracting 1 m from the true value.
- D. 13.14: 13.14 m results from taking S = 1.55 or using an incorrect denominator.
Key Point
Equivalent head: z = p / (S · ρ_w · g) = 200,000 / (1.59 × 9810) = 12.82 m.
Key Concepts for Students
- Carbon tetrachloride (CCl₄) has a specific gravity of approximately 1.59.
- Equivalent water head would be 200,000 / 9810 = 20.39 m.
- Higher density fluids develop the same pressure within smaller vertical heights.
Practice more: this is question 24 in Measurement of Pressure MCQs – Part 2. Try the whole set and check every answer instantly.