Water Resources·MCQ explained

The total pressure on the surface of a vertical sluice gate 2 m × 1 m with its top 2 m surface being 0.5 m below the water level will be ........

Try it first: pick an option, then read the explanation.

Correct answer: D. 2000 kg

Asked in SSC JE 2 MARCH 2017 Evening Shift

Detailed Explanation

In older engineering examinations and MKS units, force is frequently expressed in kilogram-force (abbreviated as kg or kgf), where \(1\text{ kgf} = 9.81\text{ N}\) (the gravitational force acting on a 1 kg mass).

1. Identify the gate geometry:

  • Width of gate \(W = 2\text{ m}\)
  • Height of gate \(H = 1\text{ m}\)
  • Area of gate \(A = W \times H = 2\text{ m} \times 1\text{ m} = 2\text{ m}^2\)
  • Depth of top edge below water surface \(= 0.5\text{ m}\)

2. Calculate the depth of the centroid (\(\bar{h}\)):

\[ \bar{h} = 0.5\text{ m} + \frac{H}{2} = 0.5 + \frac{1}{2} = 1.0\text{ m} \]

3. Calculate total hydrostatic force in SI units:

\[ P = \rho g A \bar{h} = 1000\text{ kg/m}^3 \times 9.81\text{ m/s}^2 \times 2\text{ m}^2 \times 1.0\text{ m} = 19,620\text{ N} \]

4. Convert force to kilogram-force (\(\text{kgf}\) or \(\text{kg}\)):

\[ P = \frac{19,620\text{ N}}{9.81\text{ N/kgf}} = 2000\text{ kgf} = 2000\text{ kg} \]

(Or directly in MKS: \(P = w \cdot A \cdot \bar{h} = 1000\text{ kgf/m}^3 \times 2\text{ m}^2 \times 1.0\text{ m} = 2000\text{ kgf}\)).

Why “2000 kg” Is Correct

The centroid of the sluice gate is at a depth of \(0.5 + 0.5 = 1.0\text{ m}\). With a wetted area of \(2\text{ m}^2\) and water weight density of \(1000\text{ kgf/m}^3\), the total thrust is \(1000 \times 2 \times 1.0 = 2000\text{ kgf}\).

Water LevelSluice Gate 2m × 1m0.5 mH = 1.0 mG (h̄ = 1.0 m)P = 2000 kgfP = ρgAh̄ = 1000 × 10 × (2×1) × 1.0 = 20,000 N = 2000 kgf
Vertical sluice gate (2 m wide × 1 m high) submerged with top edge 0.5 m below water: P = 2000 kgf.

Why the Other Options Are Wrong

  • A. 500 kg: 500 kg results from taking centroid depth as 0.25 m: 1000 × 2 × 0.25 = 500 kg.
  • B. 1000 kg: 1000 kg results from taking area as 1 m² or centroid depth as 0.5 m.
  • C. 1500 kg: 1500 kg results from taking centroid depth as 0.75 m: 1000 × 2 × 0.75 = 1500 kg.

Key Point

Total thrust in MKS: P = w · A · h̄ = 1000 kgf/m³ × A × h̄; 1 kgf = 9.81 N.

Key Concepts for Students

  • Top edge is at 0.5 m; bottom edge is at 0.5 + 1 = 1.5 m; centroid is at 1.0 m.
  • In MKS, unit weight of water w = 1000 kgf/m³; in SI, γ = 9.81 kN/m³ = 9810 N/m³.
  • When older questions say 'kg' for force, they mean kilogram-force (kgf).

Practice more: this is question 28 in Hydrostatic Pressure MCQs – Part 2. Try the whole set and check every answer instantly.