Pressure in terms of metres of oil (specific gravity = 0.9) equivalent to 4.5 m of water is
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2014, Morning
Detailed Explanation
Hydrostatic pressure can be expressed as an equivalent column height of any reference liquid. The required column height is inversely proportional to the specific gravity of the liquid.
1. Given data:
- Water pressure head \(h_w = 4.5\text{ m}\)
- Specific gravity of water \(S_w = 1.0\)
- Specific gravity of oil \(S_o = 0.9\)
2. Equate the hydrostatic pressure developed by both columns:
\[ p = \rho_w g h_w = \rho_o g h_o \]
3. Simplify using specific gravities (\(S = \rho / \rho_w\)):
\[ h_w \cdot S_w = h_o \cdot S_o \]
\[ h_o = \frac{h_w \cdot S_w}{S_o} = \frac{4.5 \times 1.0}{0.9} = \frac{45}{9} = 5.0\text{ m} \]
Why “5.0” Is Correct
Because oil is \(10\%\) lighter than water (\(S = 0.9\)), a taller column of oil is needed to match the pressure of \(4.5\text{ m}\) of water. Dividing \(4.5\) by \(0.9\) gives exactly \(5.0\text{ m}\).
Why the Other Options Are Wrong
- A. 4.05: 4.05 results from mistakenly multiplying by 0.9: 4.5 × 0.9 = 4.05 m.
- C. 3.6: 3.6 results from dividing 4.5 by 1.25 or arbitrary arithmetic.
- D. 0.298: 0.298 is an erroneous fraction with no relation to the head formula.
Key Point
Equivalent head formula: h_oil = h_water / S_oil = 4.5 / 0.9 = 5.0 m.
Key Concepts for Students
- Pressure head is inversely proportional to fluid density: h ∝ 1/ρ.
- Multiplying head by specific gravity gives equivalent water head (h_w = h · S).
- Dividing water head by specific gravity gives equivalent fluid head (h = h_w / S).
Practice more: this is question 9 in Measurement of Pressure MCQs – Part 1. Try the whole set and check every answer instantly.