GATE CE 2019 Set 1 — Question 44

MCQ+2 / -0.67MediumSedimentationDrinking Water TreatmentEnvironmental Engineering

Environmental Engineering → Drinking Water Treatment → Sedimentation

Last updated 5 September 2026

Question

0.2\text{ m}^3/\text{s}

\text{SOR} = \frac{Q}{L \times W} = \frac{0.2\text{ m}^3/\text{s}}{32\text{ m} \times 8\text{ m}} = 7.8125 \times 10^{-4}\text{ m/s} = 0.078125\text{ cm/s}

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\text{SOR} = \frac{Q}{L \times W} = \frac{0.2\text{ m}^3/\text{s}}{32\text{ m} \times 8\text{ m}} = 7.8125 \times 10^{-4}\text{ m/s} = 0.078125\text{ cm/s}

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