ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 is a PhD, fellowship and lectureship exam run by National Testing Agency for ICAR, for candidates in India. 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.
Source: icar.org.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
How mechanisms move and transmit power: linkages, cams, gears, belts, flywheels, governors, balancing and vibration.
Our 16 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 0 medium, 16 hard, across 9 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 ICAR AICE-JRF/SRF and are not shown on any other exam's page.
In a crank-and-slotted-lever quick-return mechanism, the distance between the fixed centres is 300 mm and the crank is 150 mm long. What is the ratio of the time of the cutting stroke to the time of the return stroke?
B. 2.0
At the ends of the stroke the crank is perpendicular to the slotted lever, so cos(β/2) = crank/centre distance = 150/300 = 0.5, β/2 = 60°, and the return stroke occupies β = 120° of crank rotation. The cutting stroke occupies 360 - 120 = 240°. At constant crank speed, time ratio = 240/120 = 2.0. 0.5 is the ratio inverted (return to cutting). 1.5 divides the whole revolution by the cutting angle (360/240); 3.0 divides it by the return angle (360/120).
A lossless planetary reduction with the ring gear fixed gives a speed ratio of 5:1 from sun (input) to carrier (output). The input torque on the sun is 100 N m. What torque must the fixed ring gear's mounting resist?
B. 400 N m
Power balance: T_out = 100 x 5 = 500 N m at the carrier. The three external torques must sum to zero: T_sun + T_carrier + T_ring = 0. The carrier torque acts against the input, so the ring (holding) torque = 500 - 100 = 400 N m. 500 N m is the output torque; 600 N m adds instead of subtracting.
A tractor's bevel pinion (8 teeth) drives a 40-tooth crown wheel, and each half-shaft then drives a planetary final reduction with a 20-tooth sun as input, a fixed 70-tooth ring and the carrier driving the wheel. With the propeller shaft at 1500 rpm in straight running, what is the wheel speed?
C. 66.7 rpm
Crown-wheel stage: 40/8 = 5, so the half-shafts turn at 1500/5 = 300 rpm when running straight. Planetary with ring fixed: ratio = 1 + T_ring/T_sun = 1 + 70/20 = 4.5, so wheel speed = 300/4.5 = 66.7 rpm. Using 70/20 = 3.5 as the planetary ratio gives 85.7 rpm; 300 rpm stops before the final reduction.
A tractor's differential cage turns at 120 rpm while the midpoint of the rear axle follows a circle of 8 m radius; the rear track width is 1.8 m. What is the speed of the outer rear half-shaft?
B. 133.5 rpm
The half-shaft speeds must average 120 rpm (N_o + N_i = 240) and be in the ratio of path radii (8.9/7.1). So N_o = 240 x 8.9/(8.9 + 7.1) = 240 x 8.9/16 = 133.5 rpm and N_i = 106.5 rpm. 106.5 rpm is the inner wheel; using the full track width on one side gives 147.0.
A Hartnell governor has balls of 2 kg each, ball arms of 100 mm and sleeve arms of 80 mm. The ball radius is 0.12 m at 300 rpm and 0.14 m at 320 rpm. Neglecting sleeve mass, the balls' weight and arm obliquity, what is the spring stiffness?
B. 12.1 N/mm
F1 = 2 x 31.416² x 0.12 = 236.9 N; F2 = 2 x 33.510² x 0.14 = 314.4 N. Spring forces S = 2Fx/y: S1 = 592.2 N, S2 = 786.1 N. Sleeve lift = (0.14 - 0.12) x 80/100 = 16 mm. Stiffness = (786.1 - 592.2)/16 = 12.1 N/mm. Taking the lift as the 20 mm change in ball radius gives 9.7 N/mm. 6.1 N/mm forgets that two levers share the spring; 7.8 N/mm inverts the arm ratio for the lift (20 × 100/80 = 25 mm).
A single-cylinder engine has 3 kg of reciprocating mass, a 60 mm crank, and runs at 600 rpm. Two-thirds of the reciprocating mass is balanced by a rotating mass opposite the crank. Considering primary forces only, what is the resultant unbalanced force when the crank is at 45° from inner dead centre?
C. 375 N
mω²r = 3 x 62.83² x 0.06 = 710.6 N. Along the line of stroke: (1 - c)mω²r cos θ = (1/3)(710.6)(0.7071) = 167.5 N. Perpendicular: c mω²r sin θ = (2/3)(710.6)(0.7071) = 335.0 N. Resultant = √(167.5² + 335.0²) = 375 N. Giving only the in-line component (167 N) forgets the new perpendicular force that the balance mass creates. 502 N is the unbalanced primary force with no balance mass (710.6 × 0.7071); 335 N is the perpendicular component alone.
An in-line four-cylinder engine with cranks at 0°, 180°, 180° and 0° has 1.5 kg of reciprocating mass per cylinder, a crank radius of 50 mm and a connecting-rod-to-crank ratio of 4. What is the maximum unbalanced secondary force at 2400 rpm?
B. 4737 N
Secondary forces from all four cylinders add: maximum = 4 x mω²r/n = 4 x 1.5 x 251.3² x 0.05/4 = 4737 N, at twice engine frequency. One cylinder alone gives 1184 N; 0 N would be true of the primary forces. 18950 N adds four primary forces (4 × mω²r).
A band brake has an angle of wrap of 270° on a drum of 500 mm diameter, with a coefficient of friction of 0.25. The slack-side tension is 300 N. What braking torque does it produce?
A. 168.6 N m
T1 = T2 e^(μθ) = 300 x e^(0.25 x 4.7124) = 300 x 3.2482 = 974.5 N. Torque = (T1 - T2) r = 674.5 x 0.25 = 168.6 N m. Using the diameter as radius doubles it; T1 x r ignores the slack-side tension. 89.5 N m uses 180° of wrap instead of 270°.
A pivoted long shoe brake has a contact angle of 90° on the drum and a lining with μ = 0.3. What equivalent coefficient of friction should be used in place of μ?
C. 0.330
μ' = 4μ sin θ/(2θ + sin 2θ), where 2θ is the contact angle: θ = 45°, so μ' = 4 x 0.3 x 0.7071/(1.5708 + 1) = 0.330. The equivalent value exceeds μ because pressure on a long shoe is not concentrated at the centre. Using 2θ = 180° in the formula gives 0.382; dropping the sin 2θ term gives 0.540.
A centrifugal clutch has 4 shoes, each of 0.5 kg with its centre of mass at 100 mm radius when engaged. Each shoe is held back by a spring force of 150 N. The drum's inner radius is 120 mm and μ = 0.3. What torque does it transmit at 1500 rpm?
D. 156.1 N m
Centrifugal force per shoe = mω²r = 0.5 x 157.08² x 0.1 = 1233.7 N. Net force on the drum = 1233.7 - 150 = 1083.7 N. Torque = μ x net force x drum radius x shoes = 0.3 x 1083.7 x 0.12 x 4 = 156.1 N m. Ignoring the spring gives 177.7; using the shoe radius instead of the drum radius gives 130.0. 39.0 N m counts one shoe only.
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.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
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