An oil of mass density 800 kg/m³ is contained in a vessel. Calculate the height of water required to develop an equivalent hydrostatic pressure as that developed by oil of height 30 m. Take acceleration due to gravity as 9.81 m/sec²
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (09am-11am)
Detailed Explanation
Hydrostatic pressure developed by a static liquid column depends on the product of its mass density and vertical column height. An equivalent pressure head represents the height of an alternative reference liquid (typically water) producing the identical hydrostatic pressure.
1. Given data:
- Density of oil \(\rho_{\text{oil}} = 800\text{ kg/m}^3\)
- Height of oil column \(h_{\text{oil}} = 30\text{ m}\)
- Density of standard water \(\rho_{\text{water}} = 1000\text{ kg/m}^3\)
- Gravitational acceleration \(g = 9.81\text{ m/s}^2\)
2. Calculate the hydrostatic pressure produced by the oil column:
\[ p = \rho_{\text{oil}} g h_{\text{oil}} = 800 \times 9.81 \times 30 = 235,440\text{ N/m}^2 = 235.44\text{ kPa} \]
3. Equate to the pressure produced by a water column of height \(h_{\text{water}}\):
\[ \rho_{\text{water}} g h_{\text{water}} = \rho_{\text{oil}} g h_{\text{oil}} \]
\[ h_{\text{water}} = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} \times h_{\text{oil}} = \frac{800}{1000} \times 30 = 0.80 \times 30 = 24\text{ m} \]
Why “24 m” Is Correct
Because the specific gravity of the oil is \(S = \frac{800}{1000} = 0.80\), oil is \(20\%\) lighter than water. Therefore, a water column needs only \(0.80 \times 30\text{ m} = 24\text{ m}\) to develop the identical base pressure.
Why the Other Options Are Wrong
- A. 32 m: 32 m results from inverted density ratio: 30 / 0.80 = 37.5 m with calculation mistakes.
- C. 42 m: 42 m is an unrealistic value larger than the original oil column height.
- D. 23 m: 23 m is an incorrect round-off.
Key Point
Equivalent pressure head relation: h₁·S₁ = h₂·S₂ ⇒ h_water = S_oil · h_oil.
Key Concepts for Students
- Specific gravity S is the ratio of fluid density to standard water density (ρ/1000).
- Denser liquids require a smaller vertical column height to generate the same hydrostatic pressure.
- Pressure head conversion is fundamental in piezometers and differential manometers.
Practice more: this is question 4 in Hydrostatic Pressure MCQs – Part 1. Try the whole set and check every answer instantly.