ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersHow much water a crop needs, when to give it, how to measure and convey it, and how to design surface irrigation.
In the exams: GATE AG Section 5 (irrigation depth, interval and command area in 5 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subject 43; NCEES PE irrigation; PRC ABE; IA CAIS
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit VII)
Single-answer MCQs on soil-water constants and their suctions, ET methods and pan coefficients, weirs and flumes, duty and delta, Lacey and Kennedy regime relations, lining types and surface-method definitions, with short calculations.
How our questions for ICAR AIEEA-PG spread across the units of Irrigation Engineering (Soil-Water-Plant, Conveyance, Surface Methods). Difficulty: 29 easy, 34 medium, 9 hard.
| Evapotranspiration and FAO-56 Penman-Monteith | 8 | |
| Design of irrigation channels | 7 | |
| Soil-water-plant relations: field capacity, wilting point and MAD | 6 | |
| Crop coefficients and crop water requirement | 6 | |
| Water measurement: weirs, Parshall flume and orifices | 6 | |
| Effective rainfall | 5 | |
| Irrigation scheduling | 5 | |
| Furrow irrigation | 5 | |
| Irrigation efficiencies and distribution uniformity | 5 | |
| Canal lining | 4 | |
| Underground pipeline systems | 4 | |
| Border irrigation | 4 | |
| Command area development | 4 | |
| Basin irrigation | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In the gravimetric method, soil samples are dried to constant mass at what temperature?
A. 105 °C
Soil is dried at 105 °C (usually about 24 h) to remove water without burning organic matter. 60 °C is used for plant tissue and leaves water in soil; higher temperatures begin to oxidise organic matter and remove structural water.
Further reading: Michael, Irrigation: Theory and Practice
Using the USDA SCS formula as simplified in FAO CROPWAT, what is the effective rainfall for a month with 200 mm of rain?
C. 136 mm
For P ≤ 250 mm: Pe = P (125 − 0.2P)/125 = 200 × (125 − 40)/125 = 136 mm. 145 mm uses the P > 250 mm branch (125 + 0.1P); 160 mm is a fixed 80 %; 200 mm treats all rain as effective.
Further reading: FAO CROPWAT documentation
A wheat scheduling recommendation is an IW/CPE ratio of 0.8 with an irrigation depth of 50 mm. At what cumulative pan evaporation since the last irrigation should the next irrigation be given?
D. 62.5 mm
CPE = IW / ratio = 50 / 0.8 = 62.5 mm. 40 mm multiplies instead of dividing; 50 mm ignores the ratio; 80 mm confuses the ratio with the CPE.
Further reading: Majumdar, Irrigation Water Management
A 90° V-notch weir operates with a head of 0.20 m. Using Q = 1.38 H^2.5 (SI), what is the discharge?
D. 24.7 L/s
Q = 1.38 × 0.20^2.5 = 1.38 × 0.01789 = 0.0247 m³/s = 24.7 L/s. 123 L/s uses the rectangular exponent 1.5; 42.3 L/s omits the discharge coefficient (Cd ≈ 0.58); 2.47 L/s is a unit slip.
Further reading: USBR Water Measurement Manual
The side slopes of a Cipolletti weir are:
B. 1 horizontal to 4 vertical
Cipolletti chose 1H:4V so that the added flow through the sloping sides offsets the end contractions, giving Q = 1.86 L H^1.5 (SI) with no contraction correction. 1:1 sides make a 90° notch-like trapezoid, and 4H:1V is far too flat.
Further reading: USBR Water Measurement Manual
By Kennedy's theory, what is the critical (non-silting, non-scouring) velocity in a channel 2.0 m deep with critical velocity ratio m = 1.0?
D. 0.86 m/s
V₀ = 0.55 m D^0.64 = 0.55 × 1.0 × 2^0.64 = 0.55 × 1.558 = 0.86 m/s. 1.10 treats the exponent as 1; 0.55 ignores depth; 0.71 uses the wrong exponent.
Further reading: Garg, Irrigation Engineering and Hydraulic Structures
A buried pipeline must carry 40 L/s with a velocity not exceeding 1.5 m/s. What is the minimum internal diameter?
C. 184 mm
A = Q/V = 0.040/1.5 = 0.0267 m²; D = √(4A/π) = 184 mm. 163 mm takes A = D² (forgets π/4); 92 mm uses the radius formula as diameter; 583 mm takes 40 L/s as 0.4 m³/s.
Further reading: Continuity equation
Under Criddle's quarter-time rule of thumb, if the required infiltration opportunity time for a furrow is 8 h, the advance time to the end of the furrow should be about:
C. 2 h
The rule sets advance time at about one quarter of the required opportunity time: 8/4 = 2 h. 4 h is half; 8 h would leave the tail badly under-irrigated; 0.25 h confuses the fraction with the time.
Further reading: Walker and Skogerboe, Surface Irrigation: Theory and Practice
A wheat crop yields 6,000 kg/ha of grain with a seasonal ET of 500 mm. What is its crop water use efficiency?
A. 1.2 kg/m³
500 mm on 1 ha = 0.5 m × 10,000 m² = 5,000 m³. WUE = 6,000 / 5,000 = 1.2 kg/m³. The others are factor-of-ten or hundred slips in converting mm·ha to m³. 12 and 0.12 kg/m³ are factor-of-ten slips, and 120 kg/m³ a factor-of-hundred slip.
Further reading: Majumdar, Irrigation Water Management
A distributary runs for 12 days in each 20-day rotation period. What is its time factor?
C. 0.6
Time factor = days the channel runs / days in the period = 12/20 = 0.6. 1.67 inverts it; 0.4 is the idle fraction; 12 is the number of days, not a ratio.
Further reading: Garg, Irrigation Engineering and Hydraulic Structures
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