Water Treatment
JKSSB Junior Engineer (Civil) · Water Supply and Waste Water Engineering · 23 practice questions with answers and explanations: 15 Easy, 7 Medium, 1 Hard.
Sample questions
Sample Easy question 1
The common coagulant used in water treatment is:
- Alum
- Quicklime only
- Bleaching powder
- Sodium chloride
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Answer: A. Aluminium sulphate makes fine particles clump together into flocs.
Sample Easy question 2
The main purpose of sedimentation in water treatment is to remove:
- Bacteria completely
- Suspended solids that settle under gravity
- Hardness
- Dissolved salts
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Answer: B. Settling lowers the load on the filters.
Sample Easy question 3
The overflow rate of a settling tank is:
- Discharge divided by surface area
- Discharge divided by volume
- Volume divided by discharge
- Surface area divided by discharge
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Answer: A. Overflow rate = Q ÷ A, in m³/m²/day.
Sample Medium question 4
Five rapid sand filters with a total area of 50 m² treat 6,000 m³ of water a day. The filtration rate is:
- 5 m³/m²/day
- 120 m³/m²/day
- 8.3 m³/m²/day
- 240 m³/m²/day
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Answer: B. Rate = Q ÷ A = 6,000 ÷ 50 = 120 m/day, which is 5 m/h.
Sample Medium question 5
A settling tank of volume 300 m³ receives 1,200 m³/day. The detention time is:
- 6 hours
- 24 hours
- 12 hours
- 4 hours
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Answer: A. 300 ÷ 1,200 = 0.25 day = 6 hours.
Sample Medium question 6
The overflow rate of a tank handling 1,200 m³/day with a surface area of 30 m² is:
- 36 m³/m²/day
- 400 m³/m²/day
- 4 m³/m²/day
- 40 m³/m²/day
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Answer: D. 1,200 ÷ 30 = 40.
Sample Hard question 7
By Stokes' law, a particle of 0.02 mm diameter and density 2,650 kg/m³ settling in water at 20°C (μ = 0.001 Pa·s) has a settling velocity of about:
- 3.6 mm/s
- 0.04 mm/s
- 6.5 mm/s
- 0.36 mm/s
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Answer: D. v = g (ρs − ρ) d² ÷ (18 μ) = 9.81 × 1,650 × (2 × 10⁻⁵)² ÷ (18 × 0.001) = 0.36 mm/s.