Q80: A mustard crop is irrigated in $30\text{ hours}$, with a... - Water Management - ICAR AIEEA PG Agronomy 14 July 2023 (Shift 1) Paper | Agrigyan

Computation of irrigation depth:

• Discharge rate: $Q = 200\text{ L/min} = \frac{200}{1000} = 0.2\text{ m}^3/\text{min}$
• Total time: $t = 30\text{ hours} = 30 \times 60 = 1800\text{ minutes}$
• Total volume: $V = 0.2 \times 1800 = 360\text{ m}^3$
• For a standard hectare field ($A = 10,000\text{ m}^2$):
  $\text{Depth} = \frac{V}{A} = \frac{360\text{ m}^3}{10,000\text{ m}^2} = 0.036\text{ m} = 36\text{ mm}$.