GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersFluids at rest and in motion, pipe flow and open channels: the base of irrigation, drainage, pumping and conservation structures.
In the exams: GATE AG Section 4 (hydraulics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit III; ICAR AICE subject 43 Unit 2; NCEES FE 12–18 questions (largest FE area)
GATE — Agricultural Engineering (AG)
Numerical-answer and MCQ items in Section 4: manometer readings, force and centre of pressure on gates, venturimeter discharge, Darcy-Weisbach head loss and laminar flow, pipes in series and parallel, V-notch and rectangular weir discharge, Manning uniform flow, the best hydraulic section, critical depth and specific energy, hydraulic jump sequent depth and energy loss, and Froude model scaling. Most items need a calculator and careful units.
How our questions for GATE spread across the units of Fluid Mechanics & Open Channel Hydraulics. Difficulty: 11 easy, 53 medium, 23 hard.
| Specific energy and critical depth | 8 | |
| Darcy-Weisbach equation and Moody diagram | 7 | |
| Hydraulic jump | 7 | |
| Pipes in series and parallel | 6 | |
| Most economical channel section | 6 | |
| Forces on submerged surfaces and buoyancy | 5 | |
| Momentum equation | 5 | |
| Orifices, mouthpieces and tank emptying | 5 | |
| Weirs and notches | 5 | |
| Dimensional analysis and similitude | 5 | |
| Properties of fluids | 4 | |
| Hydrostatics and manometry | 4 | |
| Bernoulli's equation: venturimeter, orifice meter and pitot tube | 4 | |
| Laminar and turbulent flow in pipes | 4 | |
| Uniform flow in open channels: Manning and Chezy | 4 | |
| Continuity equation | 3 | |
| Minor losses | 3 | |
| Hazen-Williams equation | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A U-tube mercury manometer (relative density 13.6) is connected to a pipe carrying water; the other limb is open to air. The pipe centre is 0.2 m above the mercury surface in the pipe-side limb, and the mercury in the open limb stands 0.3 m above that surface. What is the gauge pressure at the pipe centre?
C. 38.1 kPa
Start at the pipe centre and move to the open surface: p_A + ρ_w g(0.2) − ρ_Hg g(0.3) = 0, so p_A = 13 600 × 9.81 × 0.3 − 1000 × 9.81 × 0.2 = 40025 − 1962 = 38063 Pa ≈ 38.1 kPa. Adding the water column gives 42.0 kPa; ignoring it gives 40.0 kPa. 35.1 kPa counts the whole 0.5 m (0.2 + 0.3 m) as a water column on the pipe side.
Further reading: Munson et al., Fundamentals of Fluid Mechanics
A vertical rectangular gate 1.5 m wide and 2 m high has its top edge 1 m below the water surface. At what depth below the water surface does the resultant hydrostatic force act?
A. 2.17 m
h_cp = h̄ + I_G/(A h̄). h̄ = 2 m, A = 3 m², I_G = bd³/12 = 1.5 × 8/12 = 1.00 m⁴. h_cp = 2 + 1/(3 × 2) = 2.17 m. The centre of pressure is always below the centroid (2.00 m); forgetting to divide by h̄ gives 2.33 m. 1.83 m subtracts I_G/(A h̄) and so puts the centre of pressure above the centroid.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Oil (density 900 kg/m³, dynamic viscosity 0.2 Pa·s) flows at a mean velocity of 1 m/s through 100 m of 100 mm pipe. What is the friction head loss?
D. 7.25 m of oil
Re = ρVD/μ = 900 × 1 × 0.1/0.2 = 450, laminar. Hagen-Poiseuille: h_f = 32μVL/(ρgD²) = 32 × 0.2 × 1 × 100/(900 × 9.81 × 0.01) = 7.25 m. The same comes from f = 64/Re = 0.142 in Darcy-Weisbach. Assuming a turbulent f = 0.02 gives 1.02 m, far too low. 3.62 m uses 16 instead of 32 in Hagen-Poiseuille; 14.5 m uses 64.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
An Indian textbook writes the pipe friction loss as h_f = 4fLV²/(2gD) and quotes f = 0.005. What is the equivalent Darcy friction factor used in h_f = fLV²/(2gD)?
B. 0.020
The 4f form uses the Fanning coefficient, which is one quarter of the Darcy factor. 4 × 0.005 = 0.020. Mixing the two forms is a common source of answers four times too large or too small. 0.00125 divides by 4 instead of multiplying; 0.010 doubles the Fanning value.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Use the Swamee-Jain equation to estimate the Darcy friction factor for turbulent flow at Re = 2 × 10⁵ in a pipe of relative roughness ε/D = 0.001.
A. 0.0212
f = 0.25/[log₁₀(ε/3.7D + 5.74/Re^0.9)]² = 0.25/[log₁₀(0.000270 + 0.0000973)]² = 0.0212. The Blasius smooth-pipe value, 0.316/Re^0.25 = 0.0149, ignores roughness; 64/Re applies only to laminar flow. 0.000320 is 64/Re, valid only for laminar flow; 0.00530 is the Fanning value (one quarter of Darcy).
Further reading: White, Fluid Mechanics
A single 300 mm pipe is to be replaced by two identical pipes laid in parallel, of the same length, carrying the same total discharge with the same head loss and friction factor. What diameter should each new pipe be?
D. 227 mm
Each new pipe carries Q/2 at the same head loss: (Q/2)²/d⁵ = Q²/D⁵, so d⁵ = D⁵/4 and d = D/4^0.2 = D/2^0.4 = 0.758D = 227 mm. Keeping the same total area (D/√2 = 212 mm) is not enough, because smaller pipes lose more head per unit flow. 261 mm takes the fifth root of 2 instead of 4; 150 mm simply halves the diameter.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A 90° V-notch with C_d = 0.6 is used to measure a tubewell's discharge. The head over the vertex is 0.2 m. What is the discharge?
D. 25.4 L/s
Q = (8/15) C_d √(2g) tan(θ/2) H^2.5 = (8/15) × 0.6 × 4.429 × tan 45° × 0.2^2.5 = 1.417 × 0.0179 = 0.0254 m³/s = 25.4 L/s. Using H^1.5 gives 127 L/s, a large overestimate. 56.7 L/s uses H² instead of H^2.5; 31.7 L/s uses the rectangular 2/3 instead of 8/15.
Further reading: USBR Water Measurement Manual; Modi and Seth
The head over a triangular notch is measured with an error of 2 per cent. What is the resulting percentage error in the computed discharge?
D. 5 %
Q ∝ H^2.5, so dQ/Q = 2.5 dH/H = 2.5 × 2 = 5 %. For a rectangular weir (Q ∝ H^1.5) it would be 3 %, which is why head must be measured more precisely on a V-notch. 4 % uses an exponent of 2.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
If both the bed width and the side slope of a trapezoidal channel are free to be chosen, what shape is the most economical section?
B. Half a regular hexagon, with sides at 60° to the horizontal
Optimising side slope as well as width gives z = 1/√3 (sides at 60°), with the bed and the sloping sides of equal length: half a regular hexagon. The b = 2y rectangle is the best only among rectangles.
Further reading: Chow, Open-Channel Hydraulics
For a given discharge in an open channel, what is true of the flow at critical depth?
B. The specific energy is a minimum
On the specific-energy curve for a fixed q, the critical depth is the point of minimum E; equivalently, for a fixed E, discharge is maximum there. In a rectangular channel the velocity head at critical depth is y_c/2, not y_c, and Fr = 1.
Further reading: Subramanya, Flow in Open Channels
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