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 papersICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) is a postgraduate admission exam run by National Testing Agency for ICAR, for candidates in India. Computer-based, 120 minutes, 120 MCQs, +4 / −1, English only (2026 bulletin). One paper for the whole subject group.
Source: icar.org.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Rapid MCQs on electric motors, instruments and power distribution: instrument types and range extension, two-wattmeter method, star–delta relations, transformer EMF equation, losses and efficiency, rectifier formulas, and earthing and protective devices. One-step numericals at about one minute each.
Circuits, power, transformers, electronic devices and safe wiring: what a farm engineer needs before motors and sensors.
Our 28 ICAR AIEEA-PG questions in this subject: 5 easy, 17 medium, 6 hard, across 10 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 AIEEA-PG and are not shown on any other exam's page.
Three equal 30 Ω resistors connected in delta are to be replaced by an equivalent star. What is the value of each star resistor?
B. 10 Ω
For a balanced network R_Y = R_Δ/3 = 30/3 = 10 Ω (from R_Y = R_ab·R_ac/(R_ab + R_bc + R_ca) = 900/90). 90 Ω is the reverse (star-to-delta) conversion; 30√3 confuses resistance conversion with the line-to-phase voltage ratio.
Why does an induction motor driving a pump operate at a lagging power factor?
A. It draws magnetising current that lags the voltage by nearly 90°
The motor needs a magnetising current to set up the air-gap flux; this component is almost purely inductive and lags the voltage, so the total current lags. Rotor bars are not capacitors, rotor frequency is slip times supply frequency (lower, not higher), and resistive losses keep current in phase rather than shifting it ahead.
A pump delivers 20 L/s against a total head of 30 m. Pump efficiency is 70 % and motor efficiency 88 %. What electrical input power does the motor take?
D. 9.56 kW
Water power = ρgQH = 1000 × 9.81 × 0.020 × 30 = 5.89 kW. Shaft power = 5.89/0.70 = 8.41 kW; electrical input = 8.41/0.88 = 9.56 kW. 5.89 kW is water power only; 8.41 kW stops at the shaft; multiplying by the efficiencies gives a number smaller than the water power, which is impossible.
Three 50 Ω heating elements of a grain dryer are connected to a 400 V three-phase supply. What total power do they take in star and in delta?
A. 3200 W in star; 9600 W in delta
Star: each element sees 400/√3 = 231 V, P = 3 × 231²/50 = 400²/50 = 3200 W. Delta: each sees 400 V, P = 3 × 400²/50 = 9600 W. Delta takes three times the star power. Reversing them, or assuming equal power, are the usual slips; 1067 W treats the star total as one element's share.
A transformer has an iron loss of 0.3 kW and a full-load copper loss of 1.8 kW. At what fraction of full load is its efficiency a maximum?
A. 40.8 %
Maximum efficiency occurs where copper loss equals iron loss: x² × 1.8 = 0.3, so x = √(0.3/1.8) = 0.408 = 40.8 % of full load. 16.7 % forgets that copper loss varies with the square of load; 245 % inverts the ratio; efficiency is not necessarily greatest at full load.
In the two-wattmeter method on a balanced three-phase load, one wattmeter reads zero. What is the power factor?
C. 0.5
tan φ = √3(W₁ − W₂)/(W₁ + W₂); with W₂ = 0, tan φ = √3, φ = 60°, pf = 0.5. At unity pf the two readings are equal; at zero pf they are equal and opposite.
Two 100 kΩ resistors form a divider across 10 V. A voltmeter of 1 MΩ input resistance is connected across the lower resistor. What does it read?
A. 4.76 V
The meter in parallel with the lower 100 kΩ gives 90.9 kΩ, so V = 10 × 90.9/(100 + 90.9) = 4.76 V. 5.00 V is the true unloaded value; 9.09 V and 0.909 V wrongly treat the meter and one resistor as the whole divider. This loading effect is why sensor circuits are measured with high-input-resistance meters.
A non-inverting amplifier conditioning a soil-moisture sensor uses R_f = 47 kΩ and R_i = 10 kΩ. What is its voltage gain?
B. 5.7
Non-inverting gain = 1 + R_f/R_i = 1 + 47/10 = 5.7. 4.7 omits the 1; a negative value is the inverting amplifier; 0.21 inverts the ratio.
Which flip-flop transfers its single data input to its output at the active clock edge, making it the basic storage cell of registers?
C. The D flip-flop
The D (data) flip-flop copies D to Q at the clock edge and holds it until the next edge. A T flip-flop toggles; an SR latch has a forbidden input state; a JK flip-flop with J = K = 1 toggles rather than storing a data bit.
Before servicing a grain auger, which procedure is correct?
B. Isolate, lock off, tag, then prove dead with a tested voltage detector
Lockout–tagout means isolating at the isolator, applying a personal lock and tag, and proving dead with a detector checked on a known live source before and after. A stopped motor is not proof of isolation; a neutral fuse leaves the circuit live; a colleague is not a lock.
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.
Do these early in revision so they are fast on the day.
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