BCS Agriculture cadre — Agricultural Engineering (subject code 821) · Bangladesh · Bangladesh Public Service Commission
Original practice questions written to the published syllabus · not past papersBCS Agriculture cadre — Agricultural Engineering (subject code 821) is a government recruitment exam run by Bangladesh Public Service Commission, for candidates in Bangladesh. Preliminary 200-mark MCQ (2 h); written 900 marks; viva 200 marks. Code 821 is a 200-mark post-related paper.
Source: bpsc.gov.bd.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Sizing the parts of farm machines so they do not fail: shafts, keys, joints, springs, bearings, drives and overload devices.
Our 27 BCS Agriculture cadre questions in this subject: 15 easy, 12 medium, 0 hard, across 12 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 BCS Agriculture cadre and are not shown on any other exam's page.
Why is grey cast iron a common material for gearbox housings and machine beds but not for highly stressed tension members?
B. It casts easily, damps vibration and is strong in compression, but it is brittle and weak in tension
Grey cast iron flows well into complex moulds, its graphite flakes damp vibration and give good machinability, and its compressive strength is several times its tensile strength. The same flakes make it brittle and weak in tension. Its modulus is lower than that of steel, not higher.
The rods of a cotter joint carry an axial tensile load of 50 kN. The allowable tensile stress is 80 MPa. What is the minimum rod diameter?
D. 28.2 mm
P = (pi/4) d^2 sigma t, so d = sqrt(4P/(pi sigma)) = sqrt(4 x 50000/(pi x 80)) = 28.2 mm. Omitting the 4 gives 14.1 mm; using 2 gives 19.9 mm, treating it like a double-shear pin; sqrt(P/sigma) = 25.0 mm treats the rod as a square bar.
A bracket plate is welded with two parallel fillet welds of 8 mm leg size and carries 50 kN along the welds. The allowable shear stress on the throat is 60 MPa. What length is needed for each weld?
D. 73.7 mm
Parallel fillets are in shear on the throat: P = 2 x 0.707 h l tau, so l = 50000/(1.414 x 8 x 60) = 73.7 mm each. Using leg size instead of throat gives 52.1 mm; ignoring the second weld gives 147.3 mm. 36.8 mm halves the answer again, counting the two welds twice.
Two 12 mm plates are joined by a full-penetration butt weld 150 mm long. The allowable tensile stress in the weld is 100 MPa. What tensile load can the joint carry?
B. 180 kN
For a butt weld the throat equals the plate thickness: P = sigma t l = 100 x 12 x 150 = 180000 N = 180 kN. Applying the fillet-weld factor 0.707 gives 127 kN; adding both plate thicknesses (24 mm) gives 360 kN; 180000 kN forgets to convert N to kN.
At the moment a heavily loaded journal bearing starts from rest, which lubrication regime exists?
C. Boundary lubrication
A hydrodynamic film needs relative sliding speed to pump oil into the wedge. At start-up the speed is near zero, so the surfaces touch through thin adsorbed films: boundary lubrication, with higher friction and wear. Hydrostatic lubrication needs an external pressure supply.
What is the minimum number of teeth on a 20-degree full-depth involute pinion to avoid interference when it is generated by a rack cutter?
A. 18
The theoretical limit is 2/sin^2(20 degrees) = 17.1, so at least 18 teeth. For 14.5-degree teeth it is 32, which is why the 20-degree system allows smaller pinions. 13 is the minimum when the pinion meshes with an equal gear rather than a rack; 14 is the figure for 20-degree stub teeth.
What does the property class 8.8 marked on the head of a metric bolt indicate?
D. Ultimate tensile strength about 800 MPa and yield strength about 80 per cent of that (about 640 MPa)
In ISO 898-1 the first number times 100 gives the nominal ultimate strength in MPa, and the second number times 10 gives the yield-to-ultimate ratio in per cent, so 8.8 means about 800 MPa ultimate and 640 MPa yield. Size is given separately, for example M16.
Which bearing is best suited to a long implement shaft supported on a light frame where some misalignment between bearings is unavoidable?
B. A self-aligning bearing, such as an insert bearing with a spherical outer ring, or a spherical roller bearing
Self-aligning designs let the inner ring tilt with the shaft, so frame deflection and mounting errors do not overload the rolling elements. Cylindrical, needle and preloaded angular-contact bearings are intolerant of misalignment.
What is the difference between a shaft and an axle?
D. A shaft rotates and transmits torque; an axle supports rotating parts and carries bending but no significant torque
A shaft is a rotating member that transmits power, so it carries torque and usually bending. An axle, which may be fixed or rotating, carries bending (for example a trailer wheel axle) but does not transmit torque. Either may be solid or hollow.
A pinion with 20 teeth drives a gear with 60 teeth. The pinion runs at 900 rpm. What is the gear speed, and what happens to the torque (ignoring losses)?
C. 300 rpm; torque is multiplied by 3
Speed ratio = zg/zp = 60/20 = 3, so gear speed = 900/3 = 300 rpm. With power constant, torque rises in the same ratio, three times the pinion torque.
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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