GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersAquifers, well hydraulics, well design and the pumps that lift the water — asked in some form in every GATE AG paper.
In the exams: GATE AG Section 5 (well hydraulics in 7 of 8, pumps in 5 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subjects 40 and 43; NCEES PE pumps; IA CID pumps 15–20%; RPSC AAE
GATE — Agricultural Engineering (AG)
Numerical-answer and MCQ items on Thiem and Dupuit discharge, transmissivity from test data and well interference appear in most papers; pump power, specific speed and affinity-law scaling are frequent. Theis unsteady flow is in the syllabus but has been asked rarely; learn Cooper-Jacob straight-line analysis so it is quick. Watch natural versus common logarithms and unit conversions (L/s, m³/day, minutes to days).
How our questions for GATE spread across the units of Groundwater, Wells & Pumps. Difficulty: 21 easy, 38 medium, 11 hard.
| Aquifer types and properties: T, S and Sy | 6 | |
| Centrifugal pump characteristic curves | 6 | |
| Darcy's law in aquifers | 5 | |
| Steady flow to wells: Thiem and Dupuit | 5 | |
| Unsteady flow: Theis and Cooper-Jacob | 5 | |
| Pumping tests | 5 | |
| Well design: screen and gravel pack | 5 | |
| Well drilling and development | 5 | |
| Well interference | 4 | |
| Groundwater exploration: resistivity | 4 | |
| Classification of pumps | 4 | |
| Specific speed and affinity laws | 4 | |
| Recharge estimation and artificial recharge | 3 | |
| NPSH and cavitation | 3 | |
| Pump selection and installation | 3 | |
| Solar pumping | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
An unconfined aquifer underlying 3 km² has a specific yield of 0.15. If the water table falls by 2 m over a dry season, the volume of water released from storage is:
C. 0.90 million m³
ΔV = Sy × A × Δh = 0.15 × 3,000,000 m² × 2 m = 900,000 m³ = 0.90 million m³. Forgetting Sy gives 6 million m³ (the whole rock volume); writing 900 m³ is a factor-of-1000 unit slip. 0.0018 million m³ uses a confined storage coefficient (3 × 10⁻⁴) in place of Sy.
A fully penetrating well pumps a confined aquifer with T = 500 m²/day. At steady state, observation wells at 10 m and 100 m show drawdowns of 3.2 m and 1.4 m. The pumping rate is:
A. 2,456 m³/day
Thiem: Q = 2πT(s₁ − s₂)/ln(r₂/r₁) = 2π × 500 × 1.8 / ln 10 = 5,654.9/2.3026 = 2,456 m³/day. Using log₁₀ 10 = 1 gives 5,655; using π instead of 2π halves the answer; adding the drawdowns (4.6 m) is wrong because only the difference drives flow between the two wells.
A well pumps 1,500 m³/day from a confined aquifer with T = 600 m²/day. At a certain observation well and time, W(u) = 4.04. The drawdown is:
D. 0.80 m
s = QW(u)/(4πT) = 1500 × 4.04/(4π × 600) = 6060/7539.8 = 0.80 m. Using 2π doubles it; dividing by W(u) inverts the relation. 10.10 m leaves out the 4π (1500 × 4.04 ÷ 600).
From a Cooper-Jacob plot, T = 704 m²/day and the straight line intercepts zero drawdown at t₀ = 1.2 min. The observation well is 60 m from the pumped well. The storativity is:
D. 3.67 × 10⁻⁴
S = 2.25Tt₀/r² with t₀ in days: t₀ = 1.2/1440 = 8.33 × 10⁻⁴ day; S = 2.25 × 704 × 8.33 × 10⁻⁴/3600 = 3.67 × 10⁻⁴. Leaving t₀ in minutes gives 0.528 (impossible for a confined aquifer); using r instead of r² or converting to hours are the other slips.
In a Theis recovery analysis, residual drawdown s′ is plotted on an arithmetic scale against:
C. log(t/t′), with t from pump start and t′ from pump stop
Theis recovery: s′ = (2.3Q/4πT) log₁₀(t/t′), so the plot against log(t/t′) is straight and T = 2.3Q/(4πΔs′). Recovery data are free of pumping-rate fluctuations.
The primary purpose of a cement or bentonite grout seal in the upper annulus of a well is to:
C. Stop surface water and contaminants moving down the outside of the casing
The sanitary seal closes the annular pathway for surface contamination. It has no role in yield.
In mechanical surging, well development is achieved by:
D. Moving a close-fitting surge block up and down in the casing to force water through the screen
The surge block creates alternating inflow and outflow through the screen. Overpumping moves water in one direction only and is less effective.
In a Wenner survey with electrode spacing a = 20 m, a current of 150 mA produces a potential difference of 45 mV. The apparent resistivity is:
A. 37.7 Ω·m
R = ΔV/I = 0.045/0.150 = 0.3 Ω; ρₐ = 2πaR = 2π × 20 × 0.3 = 37.7 Ω·m. π·a·R and a·R omit parts of the geometric factor; I/ΔV inverts the ratio.
The most common operational problem in surface recharge basins, and its usual management, is:
B. Clogging of the infiltration surface, managed by wet-dry cycling and scraping
Suspended sediment, algae and bacteria seal the basin floor. Drying cracks and oxidises the layer; scraping removes it. Injection wells need filtered water and backflushing for the same reason.
A pump delivers 25 L/s against a total head of 35 m. Pump efficiency is 70% and motor efficiency 88%. The electrical input power is:
D. 13.93 kW
Water power = ρgQH = 1000 × 9.81 × 0.025 × 35 = 8.58 kW. Shaft = 8.58/0.70 = 12.26 kW. Input = 12.26/0.88 = 13.93 kW. Multiplying by efficiencies instead of dividing is a classic slip.
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