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Assam ASSEB Higher Secondary Skill Subject Agri Engineering Technology study materials
Assam ASSEB Higher Secondary Skill Subject Agri Engineering Technology study resources

Assam ASSEB Higher Secondary Skill Subject Agri Engineering Technology practice questions

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Question 1Farm Power and Machinery

In a four-stroke single-cylinder IC engine, how many revolutions of the crankshaft are required to complete one full working cycle?

  1. Half a revolution
  2. Four revolutions
  3. Two revolutions Correct answer.
  4. One revolution

Explanation

A four-stroke engine completes its cycle in four piston strokes, namely suction, compression, power and exhaust, and each stroke corresponds to half a crankshaft revolution. Therefore two complete crankshaft revolutions are needed for one working cycle.

Question 2Farm Power and Machinery

The rated power of a tractor is usually expressed in which unit in Indian farm machinery practice?

  1. PTO hp
  2. Drawbar hp
  3. kW Correct answer.
  4. Belt hp

Explanation

Modern tractor specifications in India are commonly quoted in kilowatts, following SI practice, although horsepower figures are still widely used informally. Engine power, PTO power and drawbar power are distinct measurements, and kW is the standard SI unit for expressing rated power.

Question 3Farm Power and Machinery

Which tillage implement is classified as a primary tillage tool?

  1. Leveller
  2. Disc plough Correct answer.
  3. Cultivator
  4. Puddler

Explanation

Primary tillage implements open the soil and bury crop residue to a significant depth, and the disc plough performs this function by cutting and inverting soil slices. Cultivators, levellers and puddlers are used for secondary tillage or land shaping after primary opening.

Question 4Farm Power and Machinery

What is the main function of the governor in a tractor engine?

  1. To cool the engine oil during prolonged operation
  2. To filter dust from the intake air
  3. To increase fuel injection pressure at high load
  4. To maintain a nearly constant engine speed under varying load Correct answer.

Explanation

The governor senses changes in engine speed and adjusts the fuel supply to keep the speed close to the set value as load varies. This automatic speed regulation is essential for stable operation of tractor engines during field work.

Question 5Farm Power and Machinery

Which type of plough is best suited for hard, dry and stony soils where a mouldboard plough tends to choke?

  1. Mouldboard plough
  2. Disc plough Correct answer.
  3. Subsoiler
  4. Rotary tiller

Explanation

A disc plough uses rolling discs that cut through hard soil and ride over stones, so it resists choking better than a mouldboard plough in difficult conditions. This makes it the preferred primary tillage tool for hard, dry and stony land.

Question 6Farm Power and Machinery

In a centrifugal pump, the pressure energy of water is developed mainly by which action?

  1. The compression of air in a closed chamber
  2. The whirling motion imparted by the rotating impeller Correct answer.
  3. The gravity flow from an elevated tank
  4. The reciprocating motion of a piston

Explanation

A centrifugal pump impeller spins the water outward, converting the velocity energy from the whirling motion into pressure energy at the casing. There is no piston or air chamber involved in this conversion.

Question 7Farm Power and Machinery

Which component of a tractor hydraulic system raises and lowers mounted implements?

  1. Draft and position control linkage with a hydraulic cylinder Correct answer.
  2. The differential assembly
  3. The radiator and fan assembly
  4. The clutch pedal linkage

Explanation

Mounted implements are raised and lowered by the tractor's hydraulic lift system, which uses a cylinder and control linkage governed by draft or position control. This linkage senses load and sets implement depth automatically.

Question 8Farm Power and Machinery

Seed rate in a seed drill is normally expressed in which unit?

  1. kg per litre
  2. litres per minute
  3. N per square metre
  4. kg per hectare Correct answer.

Explanation

Seed rate describes the mass of seed sown over a unit area of land and is conventionally given as kilograms per hectare. This allows the farmer to calibrate the drill for the desired plant population.

Question 9Farm Power and Machinery

A tractor develops 40 kW at the engine, and 32 kW is available at the PTO. What is the transmission efficiency?

  1. 20 percent
  2. 80 percent Correct answer.
  3. 125 percent
  4. 72 percent

Explanation

Transmission efficiency is the ratio of power delivered at the PTO to engine power, here 32 divided by 40, which equals 0.80 or 80 percent. Efficiency must always be less than 100 percent because some power is lost in the driveline.

Question 10Farm Power and Machinery

A plough operating 0.3 m wide and 0.2 m deep at a forward speed of 1.0 m/s experiences a draft of 600 N. What is the draft per unit area of soil cross-section?

  1. 1.2 kN per square metre
  2. 6 kN per square metre
  3. 10 kN per square metre Correct answer.
  4. 3 kN per square metre

Explanation

The furrow cross-sectional area is width times depth, that is 0.3 m multiplied by 0.2 m, giving 0.06 square metres. Dividing the draft of 600 N by 0.06 square metres yields 10 000 N per square metre, or 10 kN per square metre.

Question 11Farm Power and Machinery

In a two-stroke engine, which ports are uncovered by the piston during its movement?

  1. Fuel injector and glow plug ports
  2. Suction and delivery valves
  3. Inlet, transfer and exhaust ports Correct answer.
  4. Inlet and exhaust valves only

Explanation

A two-stroke engine uses piston movement to uncover inlet, transfer and exhaust ports in the cylinder wall instead of using poppet valves. The transfer port carries the fuel-air mixture from the crankcase to the cylinder.

Question 12Farm Power and Machinery

A mouldboard plough bottom consists of which set of parts?

  1. Tine, sweep, shank and clamp
  2. Blade, hub, bearing and scraper
  3. Share, mouldboard, landside and frog Correct answer.
  4. Share, seed tube, furrow opener and boot

Explanation

A plough bottom is built around the frog, which carries the share that cuts the furrow, the mouldboard that inverts the slice, and the landside that keeps the plough steady against the furrow wall. These parts together perform cutting, lifting and turning of soil.

Question 13Farm Power and Machinery

Which metering mechanism is most suitable for sowing bold, large seeds such as maize and groundnut?

  1. Cup feed with small orifice
  2. Brush type
  3. Cell wheel or plate type
  4. Fluted roller Correct answer.

Explanation

A fluted roller meters seed by trapping grains in the flutes as the roller turns, and the flute size can be matched to bold seeds for uniform delivery. Small orifice devices choke with large seeds, so they suit fine seeds instead.

Question 14Farm Power and Machinery

Why is a differential provided in the rear axle of a four-wheel tractor?

  1. To reduce the track width of the tractor
  2. To increase engine torque before the gearbox
  3. To allow the two rear wheels to rotate at different speeds during turning Correct answer.
  4. To lock both wheels permanently for straight travel

Explanation

During a turn the outer wheel travels a longer path than the inner wheel, so the differential lets the two rear wheels rotate at different speeds while still transmitting power. Without it, tyres would scrub and the tractor would resist steering.

Question 15Farm Power and Machinery

What is the primary purpose of a subsoiler in land preparation?

  1. To break hardpan layers below the normal ploughing depth without inverting soil Correct answer.
  2. To level the field surface before sowing
  3. To incorporate fertiliser into the top layer
  4. To invert the topsoil and bury weeds

Explanation

A subsoiler has heavy narrow tines that penetrate below the plough layer to shatter compacted hardpan, improving water infiltration and root growth. It does not invert soil, which distinguishes it from mouldboard ploughing.

Question 16Farm Power and Machinery

A sprayer has a 10 litre tank and applies 200 litres per hectare. How much area can be covered with one full tank?

  1. 2 hectares
  2. 0.5 hectare
  3. 20 hectares
  4. 0.05 hectare Correct answer.

Explanation

The area covered equals tank volume divided by application rate, that is 10 litres divided by 200 litres per hectare, giving 0.05 hectare. This equals 500 square metres, a small plot typical of knapsack sprayer use.

Question 17Farm Power and Machinery

Which factor mainly determines the field capacity of a seed drill?

  1. Colour of the seed hopper
  2. Number of bolts on the frame
  3. Type of paint on the metering roller
  4. Effective width of coverage and forward speed Correct answer.

Explanation

Theoretical field capacity is the product of implement width and forward speed, so a wider drill or higher speed covers more area per hour. Actual field capacity is lower because of turns and refilling time.

Question 18Farm Power and Machinery

The term 'draft' in tillage refers to which quantity?

  1. The vertical load on the tractor rear wheels
  2. The depth of ploughing in centimetres
  3. The speed of travel in kilometres per hour
  4. The horizontal force required to pull an implement Correct answer.

Explanation

Draft is the horizontal pull needed to move a tillage implement through the soil and is expressed in newtons or kilograms-force. It depends on soil type, depth and width of cut, and speed.

Question 19Farm Power and Machinery

A tractor pulls a load with a drawbar force of 5 kN at a speed of 2 m/s. What is the drawbar power?

  1. 10 kW Correct answer.
  2. 2.5 kW
  3. 0.4 kW
  4. 7 kW

Explanation

Drawbar power equals force multiplied by velocity, so 5 000 N times 2 m/s gives 10 000 W or 10 kW. This is the useful power delivered at the hitch for pulling implements.

Question 20Farm Power and Machinery

Which type of gear is commonly used in the final drive of a tractor to provide a large speed reduction in a compact space?

  1. Belt pulley
  2. Spur gear pair Correct answer.
  3. Chain and sprocket
  4. Worm gear

Explanation

Final drives usually employ spur gear pairs, often with planetary arrangement, to give a large reduction ratio in a compact housing near the rear wheels. This reduces torque load on the main transmission.

Question 21Farm Power and Machinery

A farmer needs to calibrate a seed drill for 100 kg per hectare with a 2 m working width. If the drive wheel circumference is 2 m, how many wheel revolutions correspond to 50 metres of travel?

  1. 25 revolutions Correct answer.
  2. 50 revolutions
  3. 10 revolutions
  4. 100 revolutions

Explanation

The number of wheel revolutions equals distance travelled divided by wheel circumference, that is 50 m divided by 2 m, giving 25 revolutions. Calibration involves collecting seed dropped over this known number of turns and comparing it with the target rate.

Question 22Farm Power and Machinery

In a tractor, the draft control system responds primarily to which input?

  1. Changes in the draft load sensed through the top link Correct answer.
  2. Changes in the steering wheel angle
  3. Changes in engine coolant temperature
  4. Changes in the fuel tank level

Explanation

Draft control senses the pull transmitted through the top link or lower links and raises or lowers the implement to keep draft near a set value. This protects the implement and maintains uniform depth on undulating ground.

Question 23Farm Power and Machinery

A farmer observes that a mouldboard plough is leaving an uneven furrow bottom and excessive draft. Which adjustment is most likely to correct the problem?

  1. Reducing the tyre inflation pressure
  2. Increasing the forward speed to maximum
  3. Changing the engine oil grade
  4. Adjusting the plough to correct suction and levelling the frame Correct answer.

Explanation

Uneven furrow bottom and high draft usually indicate incorrect suction or an unlevelled plough frame, so the vertical and horizontal suction settings and levelling must be corrected. Speed, tyre pressure and oil grade do not alter the plough's cutting geometry.

Question 24Farm Power and Machinery

Two tractors are compared: tractor A has a PTO power of 30 kW and tractor B has a drawbar power of 30 kW. Which statement is correct?

  1. Tractor A has lower engine power than tractor B
  2. Tractor B has higher engine power than tractor A
  3. Both tractors deliver identical power to every point of use
  4. Tractor A delivers more power to the PTO shaft than tractor B delivers at the drawbar Correct answer.

Explanation

PTO power is measured before the losses in the transmission and wheels, while drawbar power is measured after all those losses, so equal figures at the two points imply different engine powers. Tractor A's engine must be smaller to give 30 kW at the PTO, while tractor B's engine must be larger to give 30 kW at the drawbar.

Question 25Farm Power and Machinery

A field is ploughed with a 3-bottom mouldboard plough, each bottom 0.3 m wide, at 4 km/h. If field efficiency is 80 percent, what is the actual field capacity in hectares per hour?

  1. 0.288 hectare per hour Correct answer.
  2. 0.36 hectare per hour
  3. 0.45 hectare per hour
  4. 0.225 hectare per hour

Explanation

Theoretical field capacity in hectares per hour equals width in metres times speed in km/h divided by ten, so 0.9 m times 4 km/h divided by ten gives 0.36 hectare per hour. Multiplying by the 0.80 field efficiency gives 0.288 hectare per hour.

Question 26Farm Power and Machinery

In a four-stroke diesel engine used for farm power, during which stroke does the injector deliver fuel into the cylinder?

  1. Exhaust stroke at blowdown
  2. Suction stroke
  3. Power stroke after ignition begins
  4. Compression stroke near its end Correct answer.

Explanation

In a diesel engine, air alone is drawn in and compressed to high pressure and temperature, and fuel is injected near the end of the compression stroke so that it ignites spontaneously. Injection must occur before top dead centre to allow atomisation and mixing. Injecting during suction or exhaust would not produce useful combustion.

Question 27Farm Power and Machinery

A tractor engine is rated at 2,200 rpm and the belt pulley on the engine side has a diameter of 10 cm, driving a pulley of 40 cm diameter. What is the speed of the driven pulley?

  1. 1,100 rpm
  2. 8,800 rpm
  3. 550 rpm Correct answer.
  4. 880 rpm

Explanation

For belt drives the speed ratio is inversely proportional to pulley diameter: N1 x D1 = N2 x D2. Thus N2 = 2200 x 10 / 40 = 550 rpm, a four-fold speed reduction. This is why larger driven pulleys give lower output speed with higher torque.

Question 28Farm Power and Machinery

Which soil-engaging tool is designed to cut a furrow slice and invert it, thereby burying weeds and crop residues?

  1. Rotary tiller blade
  2. Cultivator sweep
  3. Disc harrow
  4. Mouldboard plough Correct answer.

Explanation

The mouldboard bottom lifts, turns and inverts the furrow slice so that surface trash and weed seeds are buried below the soil. Harrows and cultivators mainly stir and pulverise soil without inversion. Rotary tillers chop and mix the soil but do not produce a clean inversion.

Question 29Farm Power and Machinery

A single-bottom plough is worked at 0.25 m width and 0.15 m depth with a soil resistance of 60 kN per square metre. What draft does it experience?

  1. 900 N
  2. 2,250 N Correct answer.
  3. 22,500 N
  4. 225 N

Explanation

Draft from soil resistance equals resistance per unit area multiplied by the furrow cross-section. The section is 0.25 x 0.15 = 0.0375 square metres, so draft = 60,000 x 0.0375 = 2,250 N. Keeping units consistent in newtons and metres avoids the common factor-of-ten error.

Question 30Farm Power and Machinery

Which hitch system on a tractor uses two lower links and one or two upper links to attach an implement without a separate trailing frame?

  1. Clevis hitch
  2. Drawbar hitch
  3. Pintle hook
  4. Three-point linkage Correct answer.

Explanation

The three-point linkage connects an implement at two lower links and a top link, giving rigid control of depth and attitude and allowing mounted implements to be carried. Drawbar, clevis and pintle arrangements are pulling attachments for trailed loads. Only the three-point linkage provides the geometry for draft and position control.

Question 31Farm Power and Machinery

A farmer wants to estimate the field capacity of a 2 m wide seed drill operating at 5 km/h with a field efficiency of 75 percent. Which value is closest?

  1. 0.52 ha/h Correct answer.
  2. 1.04 ha/h
  3. 0.75 ha/h
  4. 1.39 ha/h

Explanation

Theoretical field capacity is width x speed divided by 10 when width is in metres and speed in km/h, giving 2 x 5 / 10 = 1.0 ha/h. Multiplying by field efficiency of 0.75 gives an effective capacity of about 0.75 ha/h... careful: the correct effective value is 0.75 ha/h.

Question 32Farm Power and Machinery

In a tractor engine, which component opens the valves at the correct instant relative to piston position?

  1. Flywheel
  2. Camshaft Correct answer.
  3. Governor
  4. Radiator

Explanation

The camshaft carries cams whose profiles lift the pushrods or rockers so that valves open and close at the correct crank angle. Its speed is half the crankshaft speed in a four-stroke engine, matching the two-revolution cycle. The governor controls fuelling, the flywheel stores energy and the radiator cools coolant.

Question 33Farm Power and Machinery

A centrifugal pump is required to lift water from a shallow well. Which condition must be satisfied for the pump to work reliably?

  1. The pump must be installed above the water level and the suction lift kept within about 7 m Correct answer.
  2. The delivery pipe should be smaller than the suction pipe to increase lift
  3. The pump casing must be filled with air before starting
  4. The impeller should rotate in either direction without priming

Explanation

A centrifugal pump relies on atmospheric pressure to push water up the suction pipe, so the practical suction lift is limited to roughly 7 m at sea level and decreases with altitude and friction losses. The pump must also be primed so that the casing is full of water, not air. These two requirements govern reliable installation.

Question 34Farm Power and Machinery

Which metering device in a planter drops seeds singly at a controlled spacing by holding them in cells on a rotating plate?

  1. Cup feed mechanism
  2. Cell plate or seed plate mechanism Correct answer.
  3. Slider mechanism
  4. Fluted roller

Explanation

A cell plate carries seeds in individual cells and releases them at a regular interval, giving precise seed-to-seed spacing for graded seed. Fluted rollers and cup feeds are used for continuous or bulk metering in drills. The slider mechanism meters by reciprocating movement rather than by cells.

Question 35Farm Power and Machinery

A threshing cylinder rotates at 600 rpm and its pulley is driven by an engine pulley of 20 cm diameter running at 1,200 rpm. What diameter must the cylinder pulley be?

  1. 20 cm
  2. 40 cm Correct answer.
  3. 60 cm
  4. 10 cm

Explanation

Using N1 x D1 = N2 x D2, the cylinder pulley diameter is 1200 x 20 / 600 = 40 cm. Halving the speed requires doubling the driven pulley diameter. This relationship is used to set correct threshing drum speed for a given crop.

Question 36Farm Power and Machinery

Which type of spray nozzle produces a fan-shaped pattern and is commonly chosen for herbicide application along the ground?

  1. Hollow cone nozzle
  2. Whirl chamber nozzle
  3. Flat fan nozzle Correct answer.
  4. Solid cone nozzle

Explanation

A flat fan nozzle delivers a narrow fan pattern that overlaps with adjacent nozzles to give uniform coverage, making it suitable for herbicide booms. Cone nozzles concentrate spray in a circular pattern and are preferred for insecticides and fungicides on foliage. Uniform overlap is the key reason for the fan choice.

Question 37Farm Power and Machinery

A tractor pulling a load develops a drawbar force of 8 kN at 1.5 m/s. If the engine output is 20 kW, what is the tractive efficiency in percent?

  1. 40 percent
  2. 75 percent
  3. 90 percent
  4. 60 percent Correct answer.

Explanation

Drawbar power equals force times speed: 8,000 x 1.5 = 12 kW. Tractive efficiency is drawbar power divided by engine power, so 12 / 20 = 0.60, or 60 percent. The remaining power is lost in transmission, wheel slip and rolling resistance.

Question 38Farm Power and Machinery

Which statement correctly explains why a differential is needed in a tractor rear axle during a turn?

  1. It changes the direction of drive by ninety degrees to the axle
  2. It locks both wheels together so they always turn at equal speed
  3. It disconnects the engine from the wheels when the clutch is pressed
  4. It allows the outer wheel to rotate faster than the inner wheel while torque is still transmitted Correct answer.

Explanation

During a turn the outer rear wheel travels a longer path than the inner wheel, so it must rotate faster to avoid scrubbing. The differential permits this speed difference through its bevel gears while still delivering torque to both half shafts. Locking the wheels or merely disconnecting drive would cause tyre scrub and steering difficulty.

Question 39Farm Power and Machinery

A farmer compares two irrigation pumps on the same well. Pump A delivers 30 litres per second against 20 m head, and pump B delivers 15 litres per second against 40 m head. Assuming equal efficiency, what is the ratio of water power of A to B?

  1. 1:2
  2. 2:1
  3. 1:1 Correct answer.
  4. 4:1

Explanation

Water power is proportional to discharge multiplied by head. For A this is 30 x 20 = 600 and for B it is 15 x 40 = 600, so the powers are equal and the ratio is 1:1. Discharge and head trade off, and considering only one of them leads to a wrong conclusion.

Question 40Farm Power and Machinery

In a combine harvester, which mechanism separates grain from the straw after threshing and discharges the straw at the rear?

  1. Header auger
  2. Reel
  3. Cleaning shoe
  4. Straw walker or beater Correct answer.

Explanation

Straw walkers or beaters carry the threshed material rearward, shaking it so loose grain falls through to the cleaning shoe while straw exits the machine. The cleaning shoe then winnows chaff from grain using sieves and air blast. The reel and header auger belong to the cutting and feeding section, not separation.

Question 41Agricultural Process Engineering and Post-Harvest Technology

In paddy milling, parboiling is carried out mainly to gelatinise the starch so that the grain becomes tougher and the next milling operation yields fewer broken grains. Which sequence describes the conventional parboiling process?

  1. Steaming, soaking, drying
  2. Drying, soaking, steaming
  3. Soaking, drying, steaming
  4. Soaking, steaming, drying Correct answer.

Explanation

Parboiling always starts with soaking the paddy in water so moisture penetrates the endosperm, then applies saturated steam to gelatinise the starch, and finally dries the grain back to a safe milling moisture. Reversing soaking and steaming leaves the interior unhydrated, so gelatinisation stays incomplete.

Question 42Agricultural Process Engineering and Post-Harvest Technology

A threshing drum is required to shell paddy at a peripheral speed of about 25 m/s. If the drum diameter is 0.6 m, what rotational speed in rpm is needed?

  1. 796 rpm Correct answer.
  2. 498 rpm
  3. 398 rpm
  4. 199 rpm

Explanation

Peripheral speed equals pi times diameter times revolutions per second, so revolutions per second = 25/(pi x 0.6) = 13.26, giving about 796 rpm. This relationship is the standard basis for setting thresher drum speed to match crop requirements.

Question 43Agricultural Process Engineering and Post-Harvest Technology

Which moisture content expression is used in grain drying calculations because it is referenced to the constant dry matter of the sample?

  1. Relative humidity
  2. Equilibrium moisture content
  3. Dry basis moisture content Correct answer.
  4. Wet basis moisture content

Explanation

Dry basis moisture content is mass of water divided by mass of dry matter, and since dry matter does not change during drying it simplifies mass balances. Wet basis is referenced to total mass and is used for marketing, while equilibrium moisture content describes a storage condition rather than an expression of content.

Question 44Agricultural Process Engineering and Post-Harvest Technology

In a thin-layer drying experiment, grain at 25 percent dry basis is dried to 15 percent dry basis. Using dry basis values, what is the percentage of the original moisture removed relative to the initial water content?

  1. 10 percent
  2. 60 percent
  3. 25 percent
  4. 40 percent Correct answer.

Explanation

On a dry basis the water removed is simply 25 minus 15, that is 10 kg per 100 kg dry matter, which is 40 percent of the original 25 kg. Dry basis arithmetic allows this direct subtraction, unlike wet basis values.

Question 45Agricultural Process Engineering and Post-Harvest Technology

Which principle explains why an aspirated grain moisture meter can read grain moisture without oven drying?

  1. The grain's bulk density varies with moisture and is sensed by a weighing cell
  2. The grain's magnetic permeability increases with moisture
  3. The electrical capacitance or conductance of the grain changes with its moisture content Correct answer.
  4. The colour of the grain darkens as moisture rises

Explanation

Electrical moisture meters rely on the strong dependence of dielectric properties or conductance of the grain on water content, which is calibrated against oven methods. Bulk density and colour do change slightly but are not the measured physical property in these instruments.

Question 46Agricultural Process Engineering and Post-Harvest Technology

A farmer stores wheat at 14 percent moisture in a gunny bag kept in a damp room. After some weeks the grain develops a musty odour and heating. Which explanation best fits this spoilage?

  1. The wheat naturally heats because of residual field heat from harvest
  2. Localised moisture migration raised the grain moisture above the safe limit, allowing mould growth Correct answer.
  3. The gunny bag released toxic gases that spoiled the grain
  4. The grain respired aerobically and lost dry matter steadily

Explanation

Grain stored near the safe moisture limit can develop localised wet spots through moisture migration from cooler or damper zones, and once moisture exceeds the safe limit moulds grow and generate heat. Respiration and residual field heat are minor compared with moisture-driven mould activity.

Question 47Agricultural Process Engineering and Post-Harvest Technology

Which machine is used to separate light chaff and dust from threshed grain by means of a controlled air stream?

  1. Dehusker
  2. Winnower Correct answer.
  3. Hammer mill
  4. Screw conveyor

Explanation

A winnower uses an air stream to carry away lighter chaff and dust while heavier sound grain falls out, which is the basic cleaning principle. Dehuskers remove husk by abrasion, conveyors move material, and hammer mills reduce particle size.

Question 48Agricultural Process Engineering and Post-Harvest Technology

A belt conveyor is required to carry 10 tonnes per hour of grain at a belt speed of 1 m/s. If the grain load on the belt is 2.8 kg per metre of belt length, what throughput does the belt actually deliver?

  1. 28 t/h
  2. 10.1 t/h Correct answer.
  3. 0.28 t/h
  4. 3.6 t/h

Explanation

Throughput equals load per unit length multiplied by belt speed, so 2.8 kg/m x 1 m/s = 2.8 kg/s, which is 2.8 x 3600 = about 10.1 tonnes per hour. This matches the required capacity of 10 t/h, so the belt is adequate.

Question 49Agricultural Process Engineering and Post-Harvest Technology

Which milling system uses a rotating abrasive stone or emery cylinder to remove the bran layers from brown rice?

  1. Rubber roll dehusking
  2. Abrasive whitening Correct answer.
  3. Paddy soaking
  4. Grading by thickness

Explanation

Abrasive whitening rubs off the bran layers with an emery or carborundum surface, producing white rice from brown rice. Rubber roll dehusking only removes the husk, soaking is a parboiling step, and thickness grading separates grain sizes.

Question 50Agricultural Process Engineering and Post-Harvest Technology

During drying of high-moisture paddy, a farmer uses very hot air and observes many fissured kernels after milling. Which cause is most consistent with this observation?

  1. Using air at a relative humidity close to the equilibrium value
  2. Cooling the grain slowly after drying
  3. Uniform slow removal of moisture from the kernel surface
  4. Moisture gradients within the kernel created by a very high drying rate Correct answer.

Explanation

Rapid surface drying with very hot air creates a steep moisture gradient, so the outer layer shrinks while the wet core does not, generating tensile stresses that fissure the kernel. Slow uniform drying and gradual cooling reduce these stresses.