Rajasthan PSC — Assistant Agriculture Engineer (Agriculture Engineering syllabus) · India · Rajasthan Public Service Commission
Original practice questions written to the published syllabus · not past papersRajasthan PSC — Assistant Agriculture Engineer (Agriculture Engineering syllabus) is a government recruitment exam run by Rajasthan Public Service Commission, for candidates in India. Objective paper; the paper structure was not captured.
Source: rpsc.rajasthan.gov.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Fluids at rest and in motion, pipe flow and open channels: the base of irrigation, drainage, pumping and conservation structures.
Our 17 Rajasthan PSC questions in this subject: 15 easy, 2 medium, 0 hard, across 14 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 Rajasthan PSC and are not shown on any other exam's page.
As the temperature rises, how do the viscosities of water and of air change?
B. Water's viscosity falls; air's viscosity rises
In liquids viscosity comes mainly from intermolecular cohesion, which weakens as temperature rises. In gases it comes from molecular momentum exchange, which increases with temperature. So the two trends are opposite.
What is the gauge pressure at a depth of 5 m in a fresh-water tank open to the atmosphere (take ρ = 1000 kg/m³, g = 9.81 m/s²)?
D. 49.05 kPa
p = ρgh = 1000 × 9.81 × 5 = 49050 Pa = 49.05 kPa gauge. Adding 101.3 kPa gives the absolute pressure (150.3 kPa), which was not asked. 490.5 kPa is a factor-of-ten slip in converting Pa to kPa.
For a floating body to be in stable equilibrium, which condition must hold?
C. The metacentre lies above the centre of gravity
A floating body is stable if the metacentre M is above G (positive GM), because the shifted buoyancy then produces a restoring couple. B above G is the condition for a fully submerged body, and is sufficient but not necessary for a floating one; ships and pontoons are commonly stable with G above B.
Water flows through a pipe that reduces from 300 mm to 150 mm diameter. The velocity in the 300 mm section is 1.5 m/s. What is the velocity in the 150 mm section?
A. 6.00 m/s
A₁V₁ = A₂V₂, and area varies as D², so V₂ = V₁(D₁/D₂)² = 1.5 × (300/150)² = 1.5 × 4 = 6.00 m/s. Using the diameter ratio without squaring gives 3.00 m/s. 0.375 m/s inverts the area ratio; 12.0 m/s cubes the diameter ratio.
Which of the following is NOT an assumption made in deriving Bernoulli's equation p/ρg + V²/2g + z = constant?
A. The flow is horizontal
Bernoulli's equation assumes steady, incompressible, frictionless flow along a streamline with no shaft work. Elevation is one of its terms (z), so the flow need not be horizontal.
In fully developed laminar flow in a circular pipe, what is the ratio of the maximum (centreline) velocity to the mean velocity?
B. 2.0
The laminar profile is parabolic, u = u_max(1 − r²/R²); integrating over the area gives a mean of u_max/2. The ratio 1.5 applies to laminar flow between parallel plates; turbulent flow has a much flatter profile with a ratio nearer 1.2. 1.33 is 4/3, the laminar momentum correction factor, not the velocity ratio.
What is the loss coefficient K for the exit of a pipe discharging into a large reservoir?
A. 1.0
At the exit the jet's entire velocity head is dissipated in the reservoir, so h = V²/2g and K = 1.0. K ≈ 0.5 is a sharp-edged entrance; about 0.05 a bell-mouthed entrance; 0.375 a sudden contraction with C_c = 0.62.
An orifice has a coefficient of contraction of 0.64 and a coefficient of velocity of 0.97. What is its coefficient of discharge?
D. 0.621
C_d = C_c × C_v = 0.64 × 0.97 = 0.621. Averaging the two (0.805) or dividing has no physical meaning. 0.660 divides C_c by C_v and 1.52 divides C_v by C_c.
A sharp-crested rectangular weir with a crest 1.5 m long operates under a head of 0.3 m. With C_d = 0.62 and no end contractions or approach velocity, what is the discharge?
A. 451 L/s
Q = (2/3) C_d L √(2g) H^1.5 = (2/3) × 0.62 × 1.5 × 4.429 × 0.3^1.5 = 0.451 m³/s = 451 L/s. Using H^2.5 (the V-notch exponent) gives 135 L/s; dropping the 2/3 gives 677 L/s. 1504 L/s uses H^0.5 instead of H^1.5.
Which dimensionless number must be matched between a model and prototype of a spillway or a hydraulic jump stilling basin?
C. Froude number
Free-surface flows over spillways and in stilling basins are gravity dominated, so Froude similarity is used. Reynolds governs viscous-dominated closed-conduit flow; Mach governs compressibility; Weber matters where surface tension is important, such as droplets.
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.
Do these early in revision so they are fast on the day.
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