Question:
How much area can be irrigated with a flow rate of 30 liters per second for 10 hours a day when irrigation requirement of the 100 days duration crop is 50 cm?
How much area can be irrigated with a flow rate of 30 liters per second for 10 hours a day when irrigation requirement of the 100 days duration crop is 50 cm?
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Volume = Flow rate \(\times\) Time.
Area = Volume Depth.
1 L/s = 3600 L/h = 3.6 m\(^3\)/h.
Area = Volume Depth.
1 L/s = 3600 L/h = 3.6 m\(^3\)/h.
Updated On: Aug 18, 2026
- 60.0 ha
- 0 ha
- 6 ha
- 16.6 ha
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
We need to calculate the area that can be irrigated.
We have the flow rate, time, and irrigation requirement.
Step 2: Key Formula or Approach:
Volume of water available = Flow rate \(\times\) Time.
Area = Volume of water Irrigation requirement.
Step 3: Detailed Explanation:
Flow rate = 30 L/s = 30 \(\times\) 3600 = 108,000 L/h.
Time = 10 hours/day.
Volume per day = 108,000 \(\times\) 10 = 1,080,000 L = 1080 m\(^3\).
Irrigation requirement = 50 cm = 0.5 m.
Area = Volume Depth = 1080 0.5 = 2160 m\(^2\) per day.
For 100 days, total area = 2160 \(\times\) 100 = 216,000 m\(^2\) = 6 ha.
Step 4: Final Answer:
The area that can be irrigated is 6 ha.
Hence, the correct option is (C).
We need to calculate the area that can be irrigated.
We have the flow rate, time, and irrigation requirement.
Step 2: Key Formula or Approach:
Volume of water available = Flow rate \(\times\) Time.
Area = Volume of water Irrigation requirement.
Step 3: Detailed Explanation:
Flow rate = 30 L/s = 30 \(\times\) 3600 = 108,000 L/h.
Time = 10 hours/day.
Volume per day = 108,000 \(\times\) 10 = 1,080,000 L = 1080 m\(^3\).
Irrigation requirement = 50 cm = 0.5 m.
Area = Volume Depth = 1080 0.5 = 2160 m\(^2\) per day.
For 100 days, total area = 2160 \(\times\) 100 = 216,000 m\(^2\) = 6 ha.
Step 4: Final Answer:
The area that can be irrigated is 6 ha.
Hence, the correct option is (C).
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