LAN-based computer test, 2 hours, 120 MCQs: 20 Part A (general knowledge and reasoning), 50 Part B (core group), 50 Part C (specialisation); four options.
Sections
38 Dairy Engineering (under group 9)
39 Farm Machinery and Power Engineering
40 Soil and Water Conservation Engineering
41 Processing and Food Engineering
42 Renewable Energy Engineering
43 Irrigation and Drainage Engineering
Over the years
PhD admission and JRF/SRF fellowship. Each specialisation syllabus has 7–10 units at PG depth (e.g. FMPE: design of tillage and planting machinery, testing, tractor design, ergonomics, soil dynamics, manufacturing, instrumentation, energy in agriculture). Question papers are not published. 2026 bulletin: https://cdnbbsr.s3waas.gov.in/s388a839f2f6f1427879fc33ee4acf4f66/uploads/2026/05/20260508269763347.pdf
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. What distinguishes hot working from cold working of a metal?
Deformation above the metal's recrystallisation temperature, so new grains form and it does not work-harden
Deformation above 500 °C for any metal, whatever its composition, melting point or grain size before working
Deformation once the metal is heated until it glows visibly red when viewed in a darkened workshop
Deformation with power hammers or presses rather than hand tools, whatever temperature the metal reaches
Answer and working
A. Deformation above the metal's recrystallisation temperature, so new grains form and it does not work-harden
The dividing line is the recrystallisation temperature, which differs by metal: lead recrystallises near room temperature, so working it there is hot working, while steel must be far hotter. A fixed 500 °C or the onset of red heat is not a general criterion, and the type of machine does not define it.
Workshop Technology & Manufacturing · Fitting, carpentry and smithy
2. Reservoir areas enclosed by contours at 100, 102, 104, 106 and 108 m are 1200, 2600, 4100, 5900 and 8000 m². What is the storage between 100 m and 108 m by the prismoidal formula?
68533 m³
28400 m³
34400 m³
34267 m³
Answer and working
D. 34267 m³
Prismoidal: V = (d/3)[(A1 + A5) + 4(A2 + A4) + 2A3] = (2/3)[9200 + 4 × 8500 + 2 × 4100] = 34267 m³. The trapezoidal formula gives 34400 m³; swapping the 4 and 2 weights is a common slip; using d = 4 m double-counts. Using d = 4 m gives 68533 m³; swapping the weights gives 28400 m³.
Surveying & Levelling · Contouring
3. A soil scientist describing a profile notes that clay has moved from the surface horizon and accumulated as coatings in the subsoil. Which pair of processes is being described?
Illuviation from the upper horizon and eluviation in the lower one
Eluviation from the upper horizon and illuviation in the lower one
Calcification above and salinisation of the lower subsoil
Gleisation above and podzolisation of the lower subsoil
Answer and working
B. Eluviation from the upper horizon and illuviation in the lower one
Eluviation is removal from a horizon; illuviation is deposition in a lower one. The reversed pair is the usual slip. Calcification and salinisation concern carbonate and soluble salts, and gleisation is reduction under waterlogging, none of which is clay movement.
Soil Science for Engineers · Soil formation and profile
4. Clean water (surface tension 0.073 N/m, contact angle 0°) stands in a glass tube of 2 mm bore dipped into a water bath. How high does the water rise by capillarity?
29.8 mm
7.44 mm
14.9 mm
1.49 mm
Answer and working
C. 14.9 mm
h = 4σ cos θ/(ρ g d) = 4 × 0.073 × 1/(1000 × 9.81 × 0.002) = 0.0149 m = 14.9 mm. Using 2σ/(ρgd) halves it (7.44 mm); putting the radius (1 mm) into the diameter formula doubles it (29.8 mm). 1.49 mm is a decimal slip that reads the 2 mm bore as 20 mm.
Fluid Mechanics & Open Channel Hydraulics · Properties of fluids
5. A soil has a bulk unit weight of 19.0 kN/m³, a water content of 12% and G = 2.65. Taking γw = 9.81 kN/m³, what is its degree of saturation?
0.532
0.597
0.318
0.864
Answer and working
B. 0.597
γd = γ/(1 + w) = 19.0/1.12 = 16.96 kN/m³. e = Gγw/γd − 1 = 2.65 × 9.81/16.96 − 1 = 0.532. S = wG/e = 0.12 × 2.65/0.532 = 0.597. Using the bulk instead of the dry unit weight to find e gives e = 0.368 and S = 0.864; 0.318 is just wG with no division by e.
Soil Mechanics · Phase relations
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