The difference in pressure head, measured by a mercury water differential manometer for a 20 cm difference of mercury head will be
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2009
Detailed Explanation
A differential U-tube manometer containing mercury connected between two points in a water pipeline converts a small mercury deflection into an equivalent head of water.
1. Identify the given parameters:
- Manometer deflection (mercury level difference) \(x = 20\text{ cm} = 0.20\text{ m}\)
- Specific gravity of mercury \(S_m = 13.6\)
- Specific gravity of flowing liquid (water) \(S_w = 1.0\)
2. Apply the differential head formula for a heavier manometric liquid:
\[ h = x \left(\frac{S_m}{S_w} - 1\right) \]
3. Substitute the values:
\[ h = 0.20\text{ m} \left(\frac{13.6}{1.0} - 1\right) = 0.20 \times 12.6 = 2.52\text{ m of water} \]
Why “2.52 m” Is Correct
The mercury column of \(0.20\text{ m}\) displaces water, giving an effective specific gravity difference of \(13.6 - 1.0 = 12.6\). Multiplying \(0.20\text{ m}\) by \(12.6\) yields exactly \(2.52\text{ m}\) of water head.
Why the Other Options Are Wrong
- A. 2.72 m: 2.72 m results from omitting the displaced water subtraction: 0.20 × 13.6 = 2.72 m.
- C. 2.0 m: 2.0 m results from arbitrary approximation without accounting for mercury's specific gravity accurately.
- D. 0.2 m: 0.2 m is simply the mercury deflection in metres, not the equivalent water head.
Key Point
Differential head for heavy manometric fluid: h = x · (S_m/S_w − 1) = 0.20 × 12.6 = 2.52 m.
Key Concepts for Students
- Always subtract 1 (representing the displaced flowing liquid column) for upright U-tube manometers.
- For an inverted U-tube with lighter manometric liquid: h = x · (1 − S_m/S_w).
- 2.52 m of water head equals 2.52 × 9.81 = 24.72 kPa of differential pressure.
Practice more: this is question 31 in Measurement of Pressure MCQs – Part 3. Try the whole set and check every answer instantly.