GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersGATE — Agricultural Engineering (AG) is a postgraduate admission exam run by IITs and IISc on rotation (2026: IIT Guwahati), for candidates in India. Computer-based, 3 hours, 100 marks (15 General Aptitude + 85 technical: 25 one-mark and 30 two-mark questions). MCQ, MSQ and NAT; negative marking on MCQ only (−1/3 for one-mark, −2/3 for two-mark); no partial credit on MSQ. On-screen virtual calculator.
Source: gate2026.iitg.ac.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Hydrology section numericals that overlap this subject: flood frequency (return period, risk, Gumbel), rational method, unit hydrograph application and Muskingum routing; reservoir capacity calculations.
Managing water at the scale of reservoirs, canals and river basins, including floods and farmer participation.
Our 16 GATE questions in this subject: 3 easy, 7 medium, 6 hard, across 4 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 GATE and are not shown on any other exam's page.
Monthly inflows to a site over a typical year are 20, 35, 60, 80, 45, 25, 12, 8, 5, 4, 6 and 10 Mm³. Using the sequent peak method, what live storage is needed to supply a uniform 20 Mm³ per month?
C. 75 Mm³
K_t = max(0, K_(t−1) + 20 − Q_t). Deficits build in the six months of 12, 8, 5, 4, 6 and 10: 8 + 12 + 15 + 16 + 14 + 10 = 75 Mm³ (the record is run twice; the maximum is 75). 16 is the worst single month; 70 is the annual surplus 310 − 240; 110 is the storage for a 25 Mm³ draft.
A weir floor 30 m long has a single downstream cutoff 5 m deep and holds a head of 4 m. By Khosla's exit gradient formula, what is the exit gradient?
B. About 1/7.4
α = b/d = 6; λ = [1 + √37]/2 = 3.54; G_E = (H/d)/(π√λ) = 0.8 ÷ (π × 1.882) = 0.135 ≈ 1/7.4. 1/1.25 is H/d alone (0.8); 1/4.4 divides by λ instead of π√λ; 1/3.9 omits √λ (0.8 ÷ π).
Using the rational method, what is the peak runoff from a 2.5 km² catchment with runoff coefficient 0.35 and design rainfall intensity 60 mm/h?
A. 14.6 m³/s
Q = CiA/3.6 = 0.35 × 60 × 2.5 ÷ 3.6 = 14.6 m³/s. 52.5 forgets the 3.6; 1.46 slips a decimal; 41.7 omits C.
Using Dickens' formula Q = C·A^(3/4) with C = 11.4 (Q in m³/s, A in km²), what is the peak flood from a 400 km² catchment?
C. About 1,020 m³/s
Q = 11.4 × 400^0.75 = 11.4 × 89.4 = 1,020 m³/s. 4,560 uses A to the power 1; 620 uses A^(2/3) (Ryves' exponent); 230 uses A^0.5.
A 6-h unit hydrograph has a peak of 50 m³/s per cm. A storm gives 3 cm of effective rainfall in 6 h and baseflow is 20 m³/s. What is the peak of the flood hydrograph?
A. 170 m³/s
Direct runoff peak = 3 × 50 = 150 m³/s; add baseflow 20: 170 m³/s. 150 omits baseflow; 70 adds instead of multiplying; 210 multiplies the baseflow by the rainfall depth as well, 3 × (50 + 20).
The ordinates of a 6-h unit hydrograph for 1 cm of effective rainfall, read at 6-h intervals, sum to 300 m³/s. What is the catchment area?
D. 648 km²
Volume = 300 × 6 × 3,600 = 6.48 × 10⁶ m³ = 1 cm over A, so A = 6.48 × 10⁶ ÷ 0.01 = 648 × 10⁶ m² = 648 km². 108 uses a 1-h interval; 64.8 and 6,480 are decimal slips.
A structure has a design life of 25 years and the accepted risk of failure in that time is 10%. What design return period is needed?
C. About 238 years
T = 1/[1 − (1 − 0.10)^(1/25)] = 1 ÷ 0.0042 = 237.8 years. 250 comes from the approximation T ≈ n/R; 25 sets T equal to the life; 2.5 is n × R.
Annual flood peaks for 40 years are ranked in descending order. Using the Weibull plotting position, what return period is assigned to the second-largest flood?
D. 20.5 years
T = (n + 1)/m = 41 ÷ 2 = 20.5 years. 20.0 uses n/m; 41 is the top-ranked flood; 13.7 uses rank 3.
Annual flood peaks at a site have mean 1,200 m³/s and standard deviation 400 m³/s. Using Gumbel's method with K_T = −(√6/π)[0.5772 + ln ln(T/(T−1))] (very long record), what is the 100-year flood?
B. About 2,455 m³/s
K100 = −0.7797 × [0.5772 + ln ln(100/99)] = −0.7797 × [0.5772 − 4.6002] = 3.137; Q = 1,200 + 3.137 × 400 = 2,455 m³/s. 2,237 is the 50-year flood (K = 2.592); 1,722 is the 10-year flood; 2,018 is the 25-year flood (K = 2.044).
For Muskingum routing with K = 10 h and X = 0.25, which time step keeps all routing coefficients non-negative?
A. Δt = 6 h
Coefficients stay non-negative when 2KX ≤ Δt ≤ 2K(1 − X), i.e. 5 h ≤ Δt ≤ 15 h. Only 6 h lies in the range; 2 h makes C0 negative; 16 h and 20 h make C2 negative.
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.
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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