Irrigation Association Certified Agricultural Irrigation Specialist (CAIS) · USA · Irrigation Association
Original practice questions written to the published syllabus · not past papersIrrigation Association Certified Agricultural Irrigation Specialist (CAIS) is a professional certification run by Irrigation Association, for candidates in USA. 100 MCQs, 3 hours.
Source: irrigation.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
The soil an engineer tills, irrigates, drains and builds on: its physical, water and chemical properties.
Our 25 Irrigation Association Certified Agricultural Irrigation Specialist questions in this subject: 9 easy, 12 medium, 4 hard, across 6 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to Irrigation Association Certified Agricultural Irrigation Specialist and are not shown on any other exam's page.
A drip designer has a clay loam field and a loamy sand field. For the same emitter discharge, in which field should emitter spacing along the lateral generally be closer to give a continuous wetted strip?
B. The loamy sand, because water spreads less laterally and moves more vertically
In coarse soils gravity dominates over capillarity, so the wetted bulb is narrow and deep and emitters must be closer. Fine soils spread water further sideways. Lateral spread depends on soil as well as discharge, and the reason given in the bulk-density option is wrong.
A moist soil sample weighs 150 g and 125 g after oven drying at 105 °C. What is its gravimetric water content?
D. 20.0%
w = mass of water/mass of dry soil = (150 − 125)/125 = 25/125 = 20.0%. Dividing by the wet mass gives 16.7%, the classic slip; 25.0% confuses grams of water with percent; 83.3% is the dry-to-wet mass ratio.
A soil has field capacity 28% and permanent wilting point 12% (both by mass) and bulk density 1.40 g/cm³. How much available water does a 1.0 m root zone hold?
A. 224 mm
Available water = (0.28 − 0.12) × 1.40 × 1,000 mm = 224 mm. 160 mm forgets the bulk density (treats gravimetric as volumetric); 114 mm divides instead of multiplying; 392 mm is the total water at field capacity, not the available part.
The permanent wilting point is conventionally estimated as the soil water content at a matric potential of:
C. −1,500 kPa (−15 bar)
PWP is conventionally taken at −1,500 kPa. −33 kPa is the usual laboratory field capacity for medium and fine soils, −10 kPa is used for field capacity of sands, and about −3,100 kPa is the hygroscopic coefficient.
Time-domain reflectometry (TDR) estimates soil water content mainly because:
C. Water's dielectric permittivity (about 80) far exceeds that of solids and air
TDR measures the travel time of an electromagnetic pulse, which depends on the bulk dielectric permittivity; water (about 80) dominates over solids (3–5) and air (1). Neutron thermalisation is the neutron probe, electrical resistance is the gypsum block, and gamma attenuation is the gamma densitometer.
One dS/m of electrical conductivity is equal to:
B. 1 mmho/cm
1 dS/m = 0.1 S/m = 1 mS/cm = 1 mmho/cm. 1 µmho/cm is a thousandth of that; 1 mho/m is 10 dS/m.
An irrigation water contains Na⁺ 20, Ca²⁺ 4 and Mg²⁺ 2 meq/L. What is its sodium adsorption ratio?
C. 11.55
SAR = Na/√((Ca + Mg)/2) = 20/√(6/2) = 20/√3 = 11.55. Omitting the division by 2 gives 8.16; forgetting the square root gives 6.67; 14.14 leaves magnesium out and uses calcium alone.
A tubewell water has CO₃²⁻ 0.5, HCO₃⁻ 5.5, Ca²⁺ 1.8 and Mg²⁺ 1.2 meq/L. Under the US Salinity Laboratory guidance, how is it rated for residual sodium carbonate?
A. RSC 3.0 meq/L, unsuitable (above 2.5)
RSC = (0.5 + 5.5) − (1.8 + 1.2) = 3.0 meq/L, above 2.5, so unsuitable without treatment (for example gypsum or blending). The same value rated marginal misreads the class limits; −3.0 reverses the subtraction; 2.0 is a ratio, not RSC.
A soil at field capacity contains 0.32 volumetric water content and at wilting point 0.14. Root depth is 0.9 m. With p = 0.4 (allowable depletion fraction), what is the readily available water?
D. 64.8 mm
TAW = (0.32 − 0.14) × 900 mm = 162 mm; RAW = p × TAW = 0.4 × 162 = 64.8 mm. 162 mm is total available water; 115.2 uses field capacity instead of the available range; 97.2 uses 1 − p.
A drip-irrigated field is monitored with a 1:5 soil:water EC meter, and a consultant converts the readings to ECe with a single fixed factor taken from another region. What is the main problem?
B. The EC1:5-to-ECe conversion depends on texture and salt type and must be calibrated locally
Because the 1:5 ratio dilutes the solution far more in a sand than in a clay relative to its saturation water content, and dissolves sparingly soluble salts such as gypsum, the factor varies by soil and must be calibrated against saturated paste extracts. EC1:5 is easily measured, ECe is generally higher, and there is no single universal factor.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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