ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersHow mechanisms move and transmit power: linkages, cams, gears, belts, flywheels, governors, balancing and vibration.
In the exams: GATE AG Sections 2 and 3 (power transmission, belts and gears recur); ICAR AIEEA-PG Unit III
ICAR AIEEA-PG — Agricultural Engineering & Technology
Unit III names theory of machines: definitions (pairs, inversions, isochronism, hunting, interference, creep), standard formulas and short calculations across all topics. Speed and accuracy on formula recall pay.
How our questions for ICAR AIEEA-PG spread across the units of Theory of Machines. Difficulty: 30 easy, 40 medium, 13 hard.
| Gears and gear trains: simple, compound and epicyclic | 13 | |
| Flywheels and turning-moment diagrams | 10 | |
| Free and forced vibrations and critical speed | 10 | |
| Mechanisms, links, pairs and Grashof's law | 8 | |
| Belt, rope and chain drives: tension ratio and power | 8 | |
| Velocity and acceleration analysis | 7 | |
| Cams and followers | 6 | |
| Governors | 6 | |
| Balancing of rotating and reciprocating masses | 5 | |
| Clutches | 5 | |
| Brakes and dynamometers | 3 | |
| Differential gear | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
The Scotch yoke, which gives a slider exact simple harmonic motion, is an inversion of which kinematic chain?
C. The double slider-crank chain
The double slider-crank chain (two turning and two sliding pairs) gives the elliptical trammel, the Scotch yoke and the Oldham coupling. The single slider-crank chain has three turning pairs and one sliding pair and its slider motion is not exactly harmonic because of connecting-rod obliquity.
Further reading: Rattan, Theory of Machines
A slider moves outward along a slotted link at 2 m/s relative to the link while the link rotates at a constant 5 rad/s. What is the magnitude of the Coriolis component of the slider's acceleration?
A. 20 m/s²
Coriolis acceleration = 2ωv = 2 x 5 x 2 = 20 m/s², directed perpendicular to the slot (the sliding velocity rotated 90° in the sense of ω). Dropping the factor 2 gives 10. Constant link speed removes tangential acceleration, not the Coriolis term, which exists whenever a point slides on a rotating link.
Further reading: Rattan, Theory of Machines
A flat belt drive carries 1000 N on the tight side and 400 N on the slack side. Assuming the belt obeys Hooke's law and neglecting centrifugal tension, what was the initial tension?
D. 700 N
T0 = (T1 + T2)/2 = (1000 + 400)/2 = 700 N. 600 N is the effective pull T1 - T2; 1400 N is the sum. 1000 N is simply the tight-side tension.
Further reading: Rattan, Theory of Machines
A chain sprocket has 20 teeth and the chain pitch is 12.7 mm. What is the pitch-circle diameter of the sprocket?
B. 81.18 mm
d = p/sin(180°/T) = 12.7/sin 9° = 12.7/0.1564 = 81.18 mm. pT/π = 80.85 mm treats the pitch as an arc on a circle, but the chain wraps as a polygon, so the chord formula is correct. Using 360°/T gives 41.10 mm. 80.18 mm uses tan 9° instead of sin 9°.
Further reading: Rattan, Theory of Machines
What is the hoop stress in the thin rim of a cast-iron flywheel of density 7200 kg/m³ when the rim speed is 30 m/s?
D. 6.48 MPa
For a thin rotating rim σ = ρv² = 7200 x 900 = 6480000 Pa = 6.48 MPa, independent of the rim section. Using v instead of v² gives 0.216 MPa; multiplying by g treats density as a weight density. 3.24 MPa is half the correct value.
Further reading: Rattan, Theory of Machines
Which statement correctly contrasts a flywheel with a governor?
C. A flywheel limits speed fluctuation within each cycle; a governor controls mean speed as load changes
A flywheel is an energy reservoir: it absorbs and releases energy during each cycle but has no effect on mean speed if the load changes permanently. A governor senses a sustained change in mean speed and adjusts the fuel supply.
Further reading: Rattan, Theory of Machines
What condition must a centrifugal governor satisfy to be stable?
D. Radius must rise with speed
Stability means one equilibrium radius for each speed, which requires the controlling force to rise faster than proportionally with radius so that radius increases with speed. Equal speed at all radii is isochronism, the limit of stability; a central load (Porter) does not affect whether the governor is stable in this sense.
Further reading: Rattan, Theory of Machines
Two masses rotate in one plane: m1r1 = 0.6 kg m at 0° and m2r2 = 0.8 kg m at 90°. A single balancing mass is to be placed at 0.25 m radius. What mass is needed and at what angle?
B. 4.0 kg at 233.1°
Resultant mr = √(0.6² + 0.8²) = 1.0 kg m at tan⁻¹(0.8/0.6) = 53.1°. The balancing mass must supply equal and opposite mr: m = 1.0/0.25 = 4.0 kg at 53.1° + 180° = 233.1°. Placing it at 53.1° doubles the unbalance; adding mr arithmetically gives 5.6 kg.
Further reading: Rattan, Theory of Machines
A machine is mounted on springs to reduce the force transmitted to its foundation. For the transmitted force to be less than the exciting force, the ratio of forcing frequency to natural frequency must be:
C. greater than √2
Transmissibility falls below 1 only when ω/ωn > √2, whatever the damping. Below √2, and especially near r = 1 (resonance), the mounts amplify the force. Mounts therefore need to be soft enough to put the natural frequency well below the running frequency.
Further reading: Rao, Mechanical Vibrations
A multi-plate clutch has 5 plates on the driving shaft and 4 plates on the driven shaft. How many pairs of friction surfaces transmit torque?
C. 8
Number of pairs of contact surfaces n = n1 + n2 - 1 = 5 + 4 - 1 = 8. Adding the plates without subtracting one gives 9; counting both faces of every plate gives 18 or 20.
Further reading: Rattan, Theory of Machines
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