GATE CE · Irrigation Engineering

Crop Water Requirements Practice Questions

Generate GATE-level questions on Crop Water Requirements in Irrigation Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

26 questions · 20 PYQs · 0 AI practice · GATE CE 2027

🎯 These are sample questions

Get the full Crop Water Requirements experience — 100% free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Q1GATE 2023NAT

A canal supplies water to an area growing wheat over 100 hectares. The duration between the first and last watering is 120 days, and the total depth of water required by the crop is $35 \mathrm{~cm}$ . The most intense watering is required over a period of 30 days and requires a total depth of water equal to $12 \mathrm{~cm}$ . Assuming precipitation to be negligible and neglecting all losses, the minimum discharge (in $\mathrm{m}^{3} / \mathrm{s}$ , rounded off to three decimal places) in the canal to satisfy the crop requirement is _____

🚀 Solve in Practice Mode

📖 Explanation

${Area}(A)=100$ ha Total duration of irrigation $=120$ days Total depth of water required $=35 \mathrm{~cm}$ Intense irrigation $(A)$ Time $\left(B_{A}\right)=30$ days Depth of water $\left(\Delta_{A}\right)=12 \mathrm{~cm}$ ${Duty}\left(D_{A}\right)=\frac{8.64 B_{A}}{\Delta_{A}}$ $\mathrm{D}_{\mathrm{A}}=\frac{8.64 \times 30}{0.12}$ $D_{A}=2160 \mathrm{ha} / \text { cumec }$ Discharge required,

\begin{aligned} & Q_{A}=\frac{A}{D_{A}} \\ & Q_{A}=\frac{100}{2160}=0.0463 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Remaining irrigation (B) Duration $\left(B_{B}\right)=90$ days Depth of water $\left(\Delta_{\mathrm{B}}\right)=23 \mathrm{~cm}$

\begin{aligned} \text { Duty } \left(D_{B}\right) & =\frac{8.64 \times 90}{0.23} \\ D_{B} & =3380.87 \mathrm{ha} / \text { cumec } \end{aligned}

Discharge required $\left(Q_{B}\right)=\frac{A}{D_{B}}$ $\Rightarrow \quad \mathrm{Q}_{\mathrm{B}}=0.0296 \mathrm{~m}^{3} / \mathrm{s}$ Hence, minimum capacity required $=0.0463 \mathrm{~m}^{3} / \mathrm{s}$ .

Q19GATE 2006MCQ

In a cultivated area, the soil has porosity of 45% and field capacity of 38%. For a particular crop, the root zone depth is 1.0m, the permanent wilting point is 10% and the consumptive use is 15 mm/d. If the irrigation efficiency is 60%, what should be the frequency of irrigation such that the moisture content does not fall below 50% of the maximum available moisture ?

🚀 Solve in Practice Mode

📖 Explanation

Depth of maximum available moisture, $d_{w}=\frac{\gamma d}{\gamma_{w}}(F-\phi)$ Field capacity

\begin{aligned} &= \frac{\text { weight of water retained by soil }}{\text { weight of same area of soil }} \\ &=\frac{\gamma_{w} \times V_{v}}{\gamma \times V} \\ \text { But } \frac{V_{v}}{V}&=n \\ \therefore \text { Field capacity } &=\frac{\gamma_{w}}{\gamma} \times n\\ \Rightarrow \frac{\gamma_{\mathrm{w}}}{\gamma} &=\frac{n}{\text{Field capicity}} \\ \Rightarrow \frac{\gamma_{\mathrm{w}}}{\gamma} &=\frac{0.45}{0.38} \\ \therefore \quad \quad d_{w} &=\frac{0.45}{0.38} \times 1 \times [0.38-0.1] \\ &=0.331 \mathrm{m} \end{aligned}

But the moisture content should not fall below 50 % $\therefore$ Depth of water required $=0.5 \times 0.331=0.165 \mathrm{m}$ Actual depth stored $=\frac{0.165 \times 60}{100}=0.099 \mathrm{m}=99 \mathrm{mm}$ Frequency of irrigation

\begin{array}{l} =\frac{\text { Actual depth stored }}{\text { Consumptive use }}=\frac{99}{15} \\ =6.63 \mathrm{d} \simeq 6 \mathrm{d} \end{array}