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ICAR PG Water Science And Technology 29 Jun 2024Shift 1
MCQ (Single Correct) +4 −1
A watershed has an area of 600 ha. Due to a 10 cm rainfall event over the watershed a streamflow is generated and at the outlet of the watershed it lasts for 10 hours. Assuming a runoff/rainfall ratio of 0.2 for this event, the average stream flow rate at the outlet in this period of 10 hours is:
एक जलग्रहण क्षेत्र (watershed) का क्षेत्रफल 600 ha है। इस पर 10 cm वर्षा होने के कारण आउटलेट पर 10 घंटे तक जलधारा प्रवाह (streamflow) बना रहता है। इस घटना के लिए अपवाह/वर्षा अनुपात (runoff/rainfall ratio) 0.2 मानते हुए, 10 घंटे की इस अवधि में आउटलेट पर औसत प्रवाह दर क्या है?

Calculation of average stream flow rate:
1. Watershed Area \(A = 600\text{ ha} = 600 \times 10,000\text{ m}^2 = 6 \times 10^6\text{ m}^2\). 2. Rainfall depth \(P = 10\text{ cm} = 0.10\text{ m}\). 3. Runoff/Rainfall ratio \(C = 0.2\). Runoff depth \(R = C \times P = 0.2 \times 0.10\text{ m} = 0.02\text{ m}\). 4. Total runoff volume \(V = A \times R = (6 \times 10^6\text{ m}^2) \times 0.02\text{ m} = 120,000\text{ m}^3\). 5. Flow duration \(t = 10\text{ hours} = 600\text{ minutes}\). 6. Average stream flow rate: - In \(\text{m}^3/\text{h}\): \(\frac{120,000\text{ m}^3}{10\text{ h}} = 12,000\text{ m}^3/\text{h}\).

  • In \(\text{m}^3/\text{minute}\): \(\frac{120,000\text{ m}^3}{600\text{ min}} = 200\text{ m}^3/\text{minute}\). - In \(\text{m}^3/\text{s}\): \(\frac{120,000\text{ m}^3}{36,000\text{ s}} = 3.33\text{ m}^3/\text{s}\).
Therefore, the correct matching option is \(200\text{ m}^3/\text{minute}\).

गणना:

  • क्षेत्रफल \(A = 600\text{ ha} = 6,000,000\text{ m}^2\)
  • वर्षा \(P = 10\text{ cm} = 0.1\text{ m}\)
  • अपवाह गहराई \(R = 0.2 \times 0.1\text{ m} = 0.02\text{ m}\) 4. कुल अपवाह आयतन \(V = 6,000,000 \times 0.02 = 120,000\text{ m}^3\) 5. अवधि \(t = 10\) घंटे \(= 600\) मिनट 6. औसत प्रवाह दर \(Q = \frac{120,000\text{ m}^3}{600\text{ min}} = 200\text{ m}^3/\text{मिनट}\)।
अतः सही उत्तर \(200\text{ m}^3/\text{minute}\) है।
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