The pressure in meters of oil (specific gravity 0.85) equivalent to 42.5 m of water is____:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 1 MARCH 2017 Morning Shift
Detailed Explanation
A given pressure intensity can be expressed as an equivalent column height of any liquid. The relationship between equivalent heads of different fluids is inversely proportional to their specific gravities.
1. State the given data:
- Water pressure head \(h_w = 42.5\text{ m}\)
- Specific gravity of water \(S_w = 1.0\)
- Specific gravity of oil \(S_o = 0.85\)
2. Equate the hydrostatic pressure developed by both liquid columns:
\[ p = \rho_w g h_w = \rho_o g h_o \]
3. Divide both sides by \((\rho_w g)\):
\[ h_w = \left(\frac{\rho_o}{\rho_w}\right) h_o = S_o \cdot h_o \]
4. Solve for the oil column height \(h_o\):
\[ h_o = \frac{h_w}{S_o} = \frac{42.5\text{ m}}{0.85} = 50\text{ m} \]
Why “50 m” Is Correct
Because oil is less dense than water (\(S = 0.85\)), a taller column of oil is required to exert the same base pressure. Dividing \(42.5\text{ m}\) by \(0.85\) yields exactly \(50\text{ m}\) of oil.
Why the Other Options Are Wrong
- A. 12.5 m: 12.5 m is an erroneous number with no physical connection to the problem.
- C. 52.5 m: 52.5 m results from mistakenly adding 10 to 42.5 m.
- D. 85 m: 85 m results from dividing 42.5 by 0.50 or mistakenly multiplying by 2.
Key Point
Equivalent head formula: h₁ · S₁ = h₂ · S₂ ⇒ h_oil = h_water / S_oil.
Key Concepts for Students
- Lighter fluids (S < 1) require greater vertical column heights for the same pressure.
- Heavier fluids (like mercury, S = 13.6) require much smaller heights: h_Hg = 42.5 / 13.6 = 3.125 m.
- Pressure head represents potential energy per unit weight of fluid (dimensions of length [L]).
Practice more: this is question 14 in Hydrostatic Pressure MCQs – Part 1. Try the whole set and check every answer instantly.