UPHESC Assistant Professor Agricultural Engineering PYQs

Concept: Heat moves in a greenhouse by all three modes.

1. Radiation: solar gain in, long-wave loss out.
2. Conduction: through the glazing and the soil.
3. Convection: air movement, ventilation and infiltration.

Short trick: R + C + C – all three.

Concept: A lean mix of 1 : 4 : 8 (cement : sand : coarse aggregate) is the usual bed concrete under the foundation of farm buildings.

1. 1 : 2 : 4 (M15) is used for RCC work such as lintels and slabs.
2. 1 : 3 : 6 is used for floors and PCC work; 1 : 4 : 8 is the leanest and goes under foundations.

Short trick: Foundation bed = leanest mix 1 : 4 : 8.

Concept: Width of a strip foundation is B = load per metre run ÷ safe bearing capacity of soil.

1. B = 450 kg/m ÷ 5000 kg/m² = 0.09 m.
2. 0.09 m (9 cm) is not among the options. 0.90 m would need a load of 4500 kg/m.
3. As printed, no option is correct.

Short trick: B = W/q → check the units and the order of magnitude.

Concept: Macadam road construction was developed by John Loudon McAdam (1756–1836), a Scottish engineer.

1. It uses layers of compacted broken stone; with tar binder it becomes tar-macadam (tarmac).
2. Trésaguet (a French engineer) and Telford (also Scottish) gave other methods.

Short trick: Mc = Scottish name → McAdam the Scot.

Concept: A gully passes through four stages (Ireland et al.).

1. Formation – channel erosion and down-cutting of the topsoil.
2. Development – upstream movement of the gully head and widening.
3. Healing – vegetation begins to grow in the channel ('heating stage' in the question means this healing stage).
4. Stabilisation – bed and banks reach stable slope; no further erosion.

Short trick: F-D-H-S: Form, Develop, Heal, Stabilise.

Concept: Earth from each borrow pit forms a length of bund. The pit length plus the berm (gap between pits) equals that bund length.

1. Bund length per pit = pit earthwork ÷ bund section = 6.00 ÷ 1.55 = 3.87 m.
2. Berm = 3.87 − 3.50 = 0.37 m.

Short trick: Berm = (pit quantity ÷ bund section) − pit length.

Concept: Runoff coefficient C = runoff ÷ rainfall. In practice it ranges from about 0.1 (sandy, wooded land) to about 0.8 and above (paved or rocky surfaces).

1. Cultivated lands: about 0.3–0.6.
2. Impervious or rocky catchments: about 0.7–0.9.
3. Of the options given, the largest practical value is about 0.80.

Short trick: Concrete or rock ≈ 0.8 → top of the usual range.

Concept: Woodruff–Zingg formula: $d=17H\left(\frac{V_m}{V}\right)\cos \theta$.

1. H = 18 m, V_m = 18 kmph (threshold), V = 14 kmph, θ = 22°.
2. $d=17\times 18\times \frac{18}{14}\times \cos {22}^{\circ }=306\times 1.2857\times 0.9272$.
3. d ≈ 364.78 m.

Short trick: Protection ≈ 17 H, adjusted by velocity ratio and cos θ.

Concept: Rhizosphere (Greek rhiza = root) is the soil zone occupied and influenced by plant roots.

1. It is the zone of intense microbial activity and water and nutrient uptake.
2. Hydrosphere is all the water on Earth; 'hygrosphere' and 'capillary sphere' are not terms for the root zone.

Short trick: Rhizo = root.

Concept: USLE (A = R·K·LS·C·P) was developed by Wischmeier and Smith of the USDA (1965, revised 1978).

1. It is based on more than 10,000 plot-years of runoff data from the USA.
2. Revised as RUSLE in the 1990s.

Short trick: USLE → USDA → USA.

Concept: FAO design formula for micro-catchment water harvesting: $\frac{C}{CA}=\frac{CWR-DR}{DR\times RC\times EF}$, where C = catchment area and CA = cultivated area.

1. CWR − DR = 575 − 380 = 195 mm.
2. DR × RC × EF = 380 × 0.13 × 0.5 = 24.7 mm.
3. C : CA = 195 ÷ 24.7 ≈ 7.9, i.e. about 8 : 1 — the catchment must be about 8 times the cultivated area.
4. The question asks for catchment : cultivated, but the options give 1 : 5, 1 : 8, 1 : 7, 1 : 3 — each makes the catchment smaller than the cultivated area. '1 : 8' is the reverse of the correct 8 : 1, so as printed no option is correct.

Short trick: Rainfall short of crop need → catchment must be LARGER than the field, so C : CA is always more than 1.

Concept: Area of a trapezoidal section is $A=(b+zd)d$.

1. b = 1 m, z = 1, d = 1 m.
2. A = (1 + 1×1) × 1 = 2 m².

Short trick: Top width = b + 2zd = 3 m; mean width 2 m × depth 1 m = 2 m².

Concept: Surface drainage layouts are chosen according to land slope.

1. Bedding system – flat lands (< 1.5%).
2. Random drains – flat land with scattered depressions.
3. Cross-slope (interception) drain system – sloping lands up to about 4%; drains run across the slope to intercept runoff.

Short trick: Cross-slope drains → gently sloping land up to ~4%.

Concept: Brake power = $\frac{\rho gQH}{\eta }$, or in hp = $\frac{Q(\text{l/s})\times H(\text{m})}{75\eta }$.

1. Water hp = 7.6 × 50 ÷ 75 = 5.07 hp.
2. Brake hp = 5.07 ÷ 0.70 ≈ 7.24 hp.
3. The next standard motor size is 7.5 hp.

Short trick: hp = QH/75η (Q in l/s).

Concept: Water distribution efficiency: $E_d=100(1-\frac{y}{d})$.

1. d = average depth of water stored along the run; y = average numerical deviation from d.
2. Storage efficiency = water stored in root zone ÷ water needed in root zone.

Short trick: Deviation (y) in the formula → uniformity → distribution.

Concept: Excess sodium disperses soil clay, reduces permeability and creates sodic (alkali) soils.

1. It is measured by SAR = $\frac{Na}{\sqrt{(Ca+Mg)/2}}$.
2. Ca and Mg improve soil structure; K is a nutrient.

Short trick: SAR has Na on top → sodium is the villain.

Concept: Volume needed in field = area × depth; supply needed = field volume ÷ conveyance efficiency.

1. 5 cm over 1 ha = 0.05 × 10,000 = 500 m³ = 5,00,000 L.
2. With 10% loss, efficiency 90%: 5,00,000 ÷ 0.9 = 5,55,556 L.
3. Time = 5,55,556 ÷ 16,000 ≈ 34.7 h.

Short trick: 1 ha-cm = 100 m³ = 1 lakh litres.

Concept: Motor power for a pump = $\frac{QH}{75{\eta }_p{\eta }_m}$ hp, with Q in litres (kg) per second and H in metres.

1. The question gives the discharge as '500 metres per minute' — metres is a length, not a volume, so no discharge Q can be taken from the printed data.
2. Only if the figure were 500 litres per minute: Q = 8.33 l/s, water hp = 8.33 × 50 ÷ 75 = 5.56 hp, input = 5.56 ÷ (0.7 × 0.9) ≈ 8.8 hp → a 10 hp motor.
3. That needs changing the printed unit, so as printed no option can be marked correct.

Short trick: Check units first: hp = (litres/s × m) ÷ (75 × overall efficiency).

Concept: Hammer mills are impact grinders used for feed and grain grinding.

1. Agricultural-processing texts list commercial hammer mills in capacities of about 400–1500 kg/h, driven by roughly 5–25 hp motors.
2. Capacity depends on screen size, hammer tip speed (about 50–110 m/s) and the moisture of the material.

Short trick: Hammer mill: impact grinder, roughly 0.4–1.5 t/h.

Concept: Henderson's equation (1952): $1-RH=e^{-cT{M}_e^n}$.

1. It relates equilibrium moisture content $M_e$ to relative humidity and temperature.
2. Chung–Pfost and Chen–Clayton are other EMC models.

Short trick: EMC ↔ Henderson.