Nepal Engineering Council · Mechanical Engineering · Chapter 8
Industrial Engineering and Automation
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187 questions in 6 syllabus topics.
8.1 Metrology and measurement
31 questions · AMeE0801
1. Which of the following best defines metrology?
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Think of the word's Greek root meaning 'measure'.
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Answer: A. The science of measurement, including its theory and practice
Metrology is the science of measurement and covers the units, standards, methods and instruments used to measure.
2. The ability of an instrument to give readings close to the true value of the quantity measured is called its
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Closeness to the true value, not to each other.
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Answer: D. accuracy
Accuracy is closeness of the measured value to the true value; precision is closeness of repeated readings to one another.
3. A rifle shoots five shots tightly grouped but far from the bull's eye. The shooting is
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Look at the spread and at the offset separately.
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Answer: C. precise but not accurate
A tight group means good repeatability (precision); a large offset from the target means a systematic error, so poor accuracy.
4. The least count of a vernier caliper in which 50 vernier scale divisions coincide with 49 main scale divisions of 1 mm each is
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LC = value of 1 MSD minus value of 1 VSD.
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Answer: B. 0.02 mm
LC = 1 MSD − 1 VSD = 1 − 49/50 = 0.02 mm.
5. A micrometer has a screw of pitch 0.5 mm and 50 divisions on the thimble. Its least count is
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Divide pitch by thimble divisions.
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Answer: B. 0.01 mm
LC = pitch / number of thimble divisions = 0.5/50 = 0.01 mm.
6. In a micrometer of least count 0.01 mm the sleeve shows 7.5 mm and the 28th thimble division coincides with the datum line. The reading is
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Add the thimble part to the sleeve reading.
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Answer: A. 7.78 mm
Reading = 7.5 + 28 × 0.01 = 7.78 mm.
7. Slip gauges are generally assembled into a stack by
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It needs no tool or fastener.
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Answer: D. wringing, i.e. sliding the faces together so they adhere
Wringing makes very flat, clean faces adhere by molecular attraction and a thin oil film, with negligible gap between blocks.
8. A dimension of 48.795 mm is to be built from slip gauges. Which combination of blocks gives it?
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Eliminate the last decimal places first.
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Answer: C. 1.005 + 1.29 + 6.5 + 40
1.005 + 1.29 + 6.5 + 40 = 48.795 mm. The first block removes the last digit (5) and the next removes the remaining hundredths.
9. A sine bar of 200 mm centre distance is to be set at 30°. The height of the slip gauge stack under one roller must be
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h = L sin θ.
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Answer: C. 100 mm
h = L sinθ = 200 × sin30° = 100 mm.
10. A 250 mm sine bar is to be set to 20°. The required gauge-block height is nearest to
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Use the sine, not the cosine or tangent.
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Answer: B. 85.5 mm
h = 250 × sin20° = 250 × 0.3420 = 85.5 mm.
11. A sine bar is least accurate for angles
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Think about where the sine curve is flattest.
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Answer: A. above about 45°
Near 90° the sine changes slowly with angle and small errors in height or centre distance cause large angular errors; sine bars are preferred below 45°.
12. The least count of a standard vernier bevel protractor is
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It reads in minutes of arc.
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Answer: D. 5 minutes
A vernier bevel protractor has 12 vernier divisions spanning 23 degrees, giving a least count of 5 minutes.
13. An autocollimator is mainly used to measure
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Uses a reflected beam of light.
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Answer: D. very small angular tilts and flatness or straightness errors
An autocollimator uses a reflected collimated beam to detect small angular deviations, which are converted into straightness or flatness errors.
14. In taper measurement with two equal rollers placed in the tapered hole, the difference in readings over the rollers at two heights 50 mm apart is 10 mm. The included angle of the taper is nearest to
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The measured difference is a total across both sides.
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Answer: C. 11.4°
tan(α/2) = (M2 − M1)/(2H) = 10/100 = 0.1, so α/2 = 5.71° and α = 11.42°.
15. In the measurement of the length of a workpiece, the measured value is 25.12 mm while the true value is 25.00 mm. The percentage error is
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Divide by the true value.
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Answer: B. 0.48 %
Error = 0.12 mm; % error = 0.12/25.00 × 100 = 0.48 %.
16. Zero error of a micrometer that has not been adjusted is an example of
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It is repeatable and correctable.
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Answer: A. systematic error
A constant offset that is repeated in every reading and can be corrected is a systematic error.
17. Errors caused by the operator misreading an instrument or recording a wrong value are called
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Operator blunders, not instrument faults.
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Answer: A. gross errors
Gross errors are human blunders such as misreading, wrong recording or improper use of the instrument.
18. Random errors in measurement are best reduced by
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They change from reading to reading.
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Answer: D. taking many readings and using their mean
Random errors vary unpredictably in both sign and size, so averaging a large number of readings reduces their effect.
19. Parallax error occurs when
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It depends on the observer's position.
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Answer: C. the observer's line of sight is not perpendicular to the scale
Viewing the pointer and scale from an oblique angle gives an apparent shift in the reading.
20. Five readings of a shaft diameter are 10.02, 10.04, 10.03, 10.01 and 10.05 mm. The sample standard deviation is nearest to
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Use n − 1 in the denominator.
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Answer: B. 0.016 mm
Mean = 10.03; squared deviations sum to 0.001; s = √(0.001/4) = 0.0158 mm.
21. A steel part 500 mm long (α = 11.5 × 10⁻⁶ /°C) is measured at 30 °C while the standard reference temperature is 20 °C. The error due to temperature is nearest to
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Use ΔL = LαΔT.
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Answer: B. 0.0575 mm
ΔL = L α ΔT = 500 × 11.5 × 10⁻⁶ × 10 = 0.0575 mm.
22. Two blocks of lengths each known to ±0.02 mm are measured end to end. The maximum (limiting) error in the total length is
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Worst case means adding the two errors.
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Answer: A. ±0.04 mm
Limiting errors add algebraically for a sum: 0.02 + 0.02 = ±0.04 mm.
23. The standard reference temperature for dimensional measurement is
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International standard value, close to room temperature.
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Answer: D. 20 °C
Gauges and components are standardised at 20 °C so that thermal expansion does not distort comparisons.
24. The SI base unit of length, the metre, is now defined in terms of
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Modern definition uses a fixed physical constant.
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Answer: C. the speed of light in vacuum
Since 1983 the metre is the distance light travels in vacuum in 1/299 792 458 s.
25. Abbe's principle states that, for minimum error, the measuring axis
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Alignment between the part and the scale axes.
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Answer: C. should be in line with the axis of the scale or standard
Alignment of the measured line with the scale line removes first-order errors due to tilt; micrometers follow it while vernier calipers do not.
26. Calibration of an instrument means
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Comparison with a reference.
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Answer: B. comparing its readings with a standard of higher accuracy
Calibration establishes the relation between instrument indication and a reference standard, ideally traceable to national standards.
27. The tolerance on a dimension is 0.05 mm. By the common 'rule of ten', the least count of the instrument used to inspect it should be at most
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One tenth of the tolerance.
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Answer: A. 0.005 mm
The instrument discrimination should be about one tenth of the tolerance: 0.05/10 = 0.005 mm.
28. The main purpose of inspection in a manufacturing plant is to
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What does a tolerance allow?
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Answer: D. ensure parts conform to specification and are interchangeable
Inspection separates conforming from non-conforming parts and supports interchangeability and quality control.
29. In Taylor's principle of limit gauging, the GO gauge should check
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The GO gauge must assemble with the part.
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Answer: D. all dimensions of the feature at once, at maximum material limit
The GO gauge has full-form contact and checks the maximum-material limit; the NO-GO gauge checks only one dimension at the minimum-material limit.
30. Flatness of a surface is checked with an optical flat and monochromatic light of wavelength 0.589 µm. Five fringes are seen across the part. The departure from flatness is nearest to
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Each fringe equals half a wavelength.
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Answer: C. 1.47 µm
Each fringe corresponds to a gap change of λ/2: 5 × 0.589/2 = 1.47 µm.
31. In acceptance tests on a machine tool, the Schlesinger tests are intended to check
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They are for verifying accuracy of the machine.
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Answer: B. the geometrical accuracy and the accuracy of test pieces machined
Acceptance tests include geometrical tests (straightness, flatness, parallelism, squareness, run-out) and practical tests on a machined test piece.
8.2 Sensors and actuators
32 questions · AMeE0802
32. A transducer is a device that
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Think of energy conversion.
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Answer: A. converts one form of energy or signal into another, usually electrical
A sensor/transducer converts the measured physical quantity (pressure, temperature, force etc.) into a convenient, usually electrical, signal.
33. An active transducer differs from a passive transducer in that it
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Self-generating versus externally excited.
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Answer: D. generates its own output voltage or current without an external supply
Thermocouples and piezoelectric crystals are active (self-generating); strain gauges, RTDs and LVDTs are passive and need excitation.
34. Pressure above local atmospheric pressure, as read by a Bourdon gauge, is called
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Relative to the atmosphere.
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Answer: C. gauge pressure
Gauge pressure = absolute pressure − atmospheric pressure; a Bourdon gauge reads relative to atmosphere.
35. The sensing element of a Bourdon tube pressure gauge is a
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It uncoils when pressurised.
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Answer: B. curved elliptical tube that tends to straighten under internal pressure
Pressure inside the flattened, curved tube causes it to uncurl; the tip movement is geared to a pointer.
36. A U-tube mercury manometer (density 13 600 kg/m³) shows a height difference of 300 mm. The gauge pressure is nearest to
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p = ρgh with h in metres.
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Answer: B. 40.0 kPa
p = ρgh = 13 600 × 9.81 × 0.3 = 40 024 Pa ≈ 40.0 kPa.
37. A piezoelectric pressure sensor is best suited to measuring
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Charge leaks away with time.
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Answer: A. rapidly changing (dynamic) pressures
The charge leaks away through the circuit, so piezoelectric sensors respond to changes but cannot hold a true static pressure.
38. The principle of an orifice meter or venturimeter for flow measurement is
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Bernoulli's equation.
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Answer: D. the pressure drop created by a constriction in the flow
A restriction raises velocity and lowers pressure (Bernoulli); the flow rate is proportional to the square root of the pressure difference.
39. Water flows through a 50 mm diameter orifice with discharge coefficient 0.62 under a pressure difference of 20 kPa (ρ = 1000 kg/m³). Neglecting approach velocity, the flow rate is nearest to
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Q = Cd·A·√(2Δp/ρ).
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Answer: C. 7.7 L/s
Q = Cd A √(2Δp/ρ) = 0.62 × 1.9635×10⁻³ × √40 = 7.70×10⁻³ m³/s = 7.7 L/s.
40. The volumetric flow rate through an orifice meter is proportional to
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Velocity head relation.
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Answer: C. the square root of the pressure difference
From Bernoulli's equation v = √(2Δp/ρ), so Q ∝ √Δp.
41. A Pitot-static tube in an air duct (ρ = 1.2 kg/m³) records a dynamic pressure of 500 Pa. The air velocity is nearest to
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Dynamic pressure = ½ρv².
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Answer: B. 28.9 m/s
v = √(2 Δp/ρ) = √(2×500/1.2) = 28.9 m/s.
42. A rotameter indicates flow rate by
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A variable-area meter.
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Answer: A. the height of a float in a tapered vertical tube
A float rises until the annular area between float and tube balances the drag and weight; its height is read as flow.
43. An electromagnetic flowmeter can be used only for fluids that
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Faraday's law needs a conductor.
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Answer: D. are electrically conductive
It works on Faraday's law; the voltage induced across a conducting fluid moving through a magnetic field is proportional to velocity.
44. A turbine flowmeter gives an output which is a
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The rotor speed is counted.
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Answer: D. frequency of pulses proportional to the flow rate
Rotor speed is proportional to fluid velocity; a pickup counts blade passages giving a pulse frequency.
45. The Seebeck effect is the basis of the
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Two dissimilar metals joined.
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Answer: C. thermocouple
A voltage is produced at the junction of two dissimilar metals when it is at a different temperature from the reference junction.
46. A Pt100 RTD has R = 100 Ω at 0 °C and temperature coefficient 0.00385 /°C. Its resistance at 100 °C is nearest to
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R = R0(1+αΔT).
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Answer: B. 138.5 Ω
R = R0(1 + αΔT) = 100(1 + 0.00385×100) = 138.5 Ω.
47. A negative temperature coefficient (NTC) thermistor
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Semiconductor behaviour.
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Answer: A. decreases in resistance as temperature rises
NTC thermistors are semiconductor devices with strongly nonlinear resistance that falls as temperature increases.
48. Which temperature sensor is preferred for measuring very high temperatures of a furnace without contact?
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No contact with the hot body.
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Answer: A. Radiation pyrometer
A pyrometer senses the thermal radiation emitted, so it does not touch the hot body and can read above the limit of contact sensors.
49. Which of the following gives the most linear and accurate response over a moderate temperature range?
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Platinum is very stable.
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Answer: D. Platinum RTD
Platinum RTDs are stable, repeatable and nearly linear, whereas thermistors are highly nonlinear and thermocouples have a nonlinear output.
50. The commonly used 'Type K' thermocouple is made of
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Nickel alloys.
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Answer: C. chromel and alumel
Type K = Chromel (Ni-Cr) positive leg and Alumel (Ni-Al) negative leg.
51. A strain gauge of resistance 120 Ω and gauge factor 2.0 is subjected to a strain of 500 microstrain. The change in resistance is
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GF = (ΔR/R)/ε.
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Answer: B. 0.12 Ω
ΔR = GF × ε × R = 2.0 × 500×10⁻⁶ × 120 = 0.12 Ω.
52. A single strain gauge in a quarter Wheatstone bridge (excitation 10 V, GF = 2) senses 1000 microstrain. The approximate bridge output is
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The quarter-bridge factor is 1/4.
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Answer: B. 5 mV
V0 ≈ (Vex/4) × GF × ε = 2.5 × 2 × 10⁻³ = 5 mV.
53. A load cell used to measure mass or force generally uses
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Elastic member plus gauges.
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Answer: A. strain gauges bonded to an elastic member
Force deforms the elastic member; the bonded gauges convert the strain into a resistance change measured in a bridge.
54. A motor transmits 15 kW at 1500 rpm. The torque measured by a torque transducer on its shaft is nearest to
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P = Tω with ω in rad/s.
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Answer: D. 95.5 N·m
T = P/ω = 15 000 × 60/(2π × 1500) = 95.5 N·m.
55. A proving ring is a device used to measure
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Used to calibrate testing machines.
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Answer: C. force
The deflection of an elastic ring is measured with a dial gauge or transducer and calibrated to give force.
56. A linear variable differential transformer (LVDT) measures
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Movable core in a transformer.
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Answer: C. linear displacement
A movable core changes the mutual inductance between primary and two secondaries; the differential output indicates displacement.
57. A hydraulic cylinder with a 50 mm bore operates at 10 MPa. Neglecting losses, its extending force is nearest to
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F = p × piston area.
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Answer: B. 19.6 kN
F = pA = 10×10⁶ × π/4 × 0.05² = 19 635 N.
58. A pneumatic cylinder of 80 mm bore works at 6 bar gauge. The theoretical force on the piston is nearest to
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Convert bar to Pa first.
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Answer: A. 3.0 kN
F = pA = 6×10⁵ × π/4 × 0.08² = 3016 N.
59. Oil at 30 L/min is supplied to the cap end of a cylinder of 50 mm bore. The piston speed is nearest to
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Speed = flow ÷ area.
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Answer: D. 0.25 m/s
v = Q/A = (30×10⁻³/60)/1.9635×10⁻³ = 0.255 m/s.
60. A hydraulic pump delivers 20 L/min at 100 bar. The hydraulic power output is nearest to
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P = p × Q in SI units.
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Answer: D. 3.33 kW
P = pQ = 100×10⁵ × (20×10⁻³/60) = 3333 W.
61. Compared with pneumatic actuators, hydraulic actuators are preferred where
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Compressibility of the working fluid.
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Answer: C. large forces and stiff, precise positioning are required
Oil is nearly incompressible and can be used at high pressure, giving large forces and stiffness; air is compressible, cheaper and cleaner.
62. A stepper motor with a step angle of 1.8° requires how many full steps per revolution?
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Divide a full turn by the step angle.
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Answer: B. 200
Steps per revolution = 360/1.8 = 200.
63. The main advantage of a servo motor over a stepper motor in a positioning application is
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Feedback.
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Answer: A. closed-loop feedback control giving high speed and torque at precise positions
A servo motor uses an encoder or resolver for feedback so that it corrects position errors under varying load.
8.3 Measurement of physical quantity
31 questions · AMeE0803
64. The SI unit of thermodynamic temperature, the kelvin, is a
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Count the seven fundamental units.
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Answer: A. base unit
The seven SI base units are metre, kilogram, second, ampere, kelvin, mole and candela.
65. A standard that is kept at a national laboratory and is used to calibrate secondary standards is called the
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Top of the calibration chain.
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Answer: D. primary standard
The primary standard is the highest-accuracy standard; secondary and working standards are calibrated from it in a chain of traceability.
66. Traceability of a measurement means that its result can be related to
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A chain back to a standard.
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Answer: C. national or international standards through an unbroken chain of calibrations
Traceability links a measurement to recognised standards through documented calibrations, each with stated uncertainty.
67. The angle of 1 degree is equal to
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60 × 60.
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Answer: B. 3600 arc seconds
1° = 60 arc minutes = 3600 arc seconds.
68. The span of an instrument whose range is from −20 °C to 100 °C is
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Subtract the lower value.
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Answer: B. 120 °C
Span = upper range value − lower range value = 100 − (−20) = 120 °C.
69. The ratio of the change in output of an instrument to the change in the input that caused it is called
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The slope of the calibration curve.
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Answer: A. sensitivity
Sensitivity = Δoutput/Δinput, i.e. the slope of the calibration curve.
70. A temperature sensor gives 20 mV output when the temperature changes by 4 °C. Its sensitivity is
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Output change divided by input change.
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Answer: D. 5 mV/°C
S = 20 mV / 4 °C = 5 mV/°C.
71. The smallest change in input that produces a detectable change in the output of an instrument is its
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Smallest detectable step.
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Answer: C. resolution
Resolution is the smallest increment of input that can be detected; threshold is the same idea starting from zero input.
72. The maximum difference in output for the same input when the input is approached from above and from below is called
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Up-scale versus down-scale.
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Answer: C. hysteresis
Hysteresis is the difference between upscale and downscale readings, caused by friction, backlash or elastic effects.
73. A 0 to 10 unit instrument shows a maximum up-scale versus down-scale difference of 0.4 units. The hysteresis as a percentage of full scale is
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Divide by the full-scale span.
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Answer: B. 4 %
Hysteresis = 0.4/10 × 100 = 4 % of full scale.
74. An instrument with accuracy ±0.5 % of full scale, range 0–200 °C, reads 50 °C. The maximum possible error is
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The percentage is applied to full scale.
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Answer: A. ±1 °C
Error = 0.5 % × 200 = ±1 °C, which is 2 % of the reading of 50 °C.
75. The error of an instrument specified as a percentage of full scale is relatively
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Constant absolute error.
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Answer: D. largest when reading near the bottom of the scale
Since the absolute error is constant, it forms a larger percentage of a small reading, so use the upper part of the scale.
76. Closeness of agreement between repeated measurements of the same quantity under the same conditions is called
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Same conditions.
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Answer: D. repeatability
Repeatability refers to same operator, instrument and conditions in a short time; reproducibility refers to changed conditions or operators.
77. Calibration of a measuring instrument involves
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Known input, recorded output.
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Answer: C. applying known inputs and recording the instrument outputs
Static calibration applies known standard inputs over the range and records the output to determine the calibration curve and errors.
78. A potentiometer with an ideal output proportional to the shaft position is a good example of a
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No time delay or dynamics.
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Answer: B. zero-order instrument
A zero-order instrument has output proportional to input with no lag; its response is y = Kx.
79. The step response of a first-order instrument reaches what fraction of its final value after one time constant?
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Evaluate 1 − e⁻¹.
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Answer: A. 63.2 %
y = 1 − e^(−t/τ); at t = τ this is 1 − e⁻¹ = 0.632.
80. A thermometer behaves as a first-order system with time constant 5 s. It is plunged into a bath 100 °C above its initial reading. Its reading rise after 10 s is nearest to
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t/τ = 2.
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Answer: A. 86.5 °C
ΔT = 100 (1 − e^(−10/5)) = 100 (1 − 0.1353) = 86.5 °C.
81. A first-order thermometer with a time constant of 5 s reaches 95 % of its final value after nearly
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About three time constants.
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Answer: D. 15 s
t = −τ ln(0.05) = 5 × 3.0 ≈ 15 s (about 3τ).
82. For a first-order system, the 10 % to 90 % rise time is approximately
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A bit more than twice the time constant.
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Answer: C. 2.2 τ
t90 = 2.303 τ and t10 = 0.105 τ, so the rise time is about 2.2 τ.
83. For a first-order instrument with time constant τ, the amplitude ratio at the input frequency ω = 1/τ is
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Evaluate at ωτ = 1.
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Answer: B. 0.707
M = 1/√(1 + (ωτ)²) = 1/√2 = 0.707, with a phase lag of 45°.
84. Which of the following describes a second-order instrument such as a spring–mass accelerometer?
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Two parameters describe the dynamics.
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Answer: B. Its response is governed by natural frequency and damping ratio
The equation of motion m x″ + c x′ + k x = F gives two parameters: ωn = √(k/m) and ζ = c/(2√(km)).
85. A spring–mass system has m = 10 kg and k = 1000 N/m. The critical damping coefficient is
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cc = 2√(km).
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Answer: A. 200 N·s/m
cc = 2√(km) = 2√(1000×10) = 200 N·s/m.
86. A spring–mass system (m = 10 kg, k = 1000 N/m) has a damper with c = 100 N·s/m. Its damping ratio is
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ζ = c/cc.
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Answer: D. 0.5
ζ = c/cc = 100/200 = 0.5; the system is underdamped.
87. A second-order instrument has damping ratio 0.5. The percent maximum overshoot of its step response is nearest to
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Use the overshoot formula.
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Answer: C. 16 %
Mp = exp(−πζ/√(1−ζ²)) = exp(−1.814) = 0.163, i.e. 16 %.
88. For a second-order instrument with ωn = 100 rad/s and ζ = 0.5, the damped natural frequency is nearest to
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ωd < ωn.
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Answer: C. 86.6 rad/s
ωd = ωn √(1 − ζ²) = 100 × 0.866 = 86.6 rad/s.
89. Most measuring instruments of the second-order type are designed with a damping ratio near
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Slightly underdamped.
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Answer: B. 0.6 to 0.7
ζ ≈ 0.65–0.7 gives fast response with small overshoot and a nearly flat amplitude response over a wide frequency range.
90. A system with damping ratio greater than 1 is called
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Real distinct roots.
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Answer: A. overdamped
For ζ > 1 the roots are real and distinct and the response is slow and non-oscillatory.
91. The undamped natural frequency of a measuring element with stiffness 4×10⁴ N/m and mass 1 kg is nearest to
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Convert rad/s to Hz.
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Answer: D. 31.8 Hz
ωn = √(k/m) = 200 rad/s; fn = ωn/2π = 31.8 Hz.
92. Drift in an instrument refers to
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It changes slowly with time.
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Answer: D. a gradual change in output with time for a constant input
Drift is a slow, undesired change in reading caused by ageing, temperature or supply variations.
93. Dead zone (dead band) of an instrument is
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No response over a band.
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Answer: C. the range of input values over which there is no change in output
Within the dead band the input can change without any detectable response because of friction or backlash.
94. Linearity of an instrument expresses
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Straight-line fit.
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Answer: B. how closely the calibration curve follows a straight line
Non-linearity is the maximum deviation of the calibration curve from the best-fit straight line, usually as a percentage of full scale.
8.4 Design of production systems
30 questions · AMeE0804
95. Plant layout refers to
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It is about arrangement inside the plant.
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Answer: A. the physical arrangement of machines, equipment and facilities within a plant
Layout is the arrangement of departments, machines, workstations and material-handling facilities to give efficient flow of men and material.
96. Which layout is best suited to mass production of a standardised product?
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Machines follow the sequence of operations.
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Answer: D. Product (line) layout
Machines are arranged in the sequence of operations, giving smooth flow, low handling and high output for standard items.
97. Which layout is best suited to job-shop production with a wide variety of products in small batches?
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Similar machines grouped by function.
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Answer: C. Process (functional) layout
Similar machines are grouped together, giving flexibility for variety at the cost of more handling and work-in-process.
98. A fixed-position layout is used for
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Product stays where it is.
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Answer: B. ship building and large bridge construction
The product is too large or heavy to move, so workers, tools and materials are brought to it.
99. Group technology (cellular) layout is based on
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Part families.
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Answer: B. grouping parts with similar shape or process into families made in a cell
Parts with similar design or manufacturing features are made in cells of dissimilar machines, combining benefits of product and process layouts.
100. Which of the following is an advantage of a process layout over a product layout?
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Flexibility versus efficiency.
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Answer: A. Greater flexibility to handle changes in product mix
General-purpose machines of the same type are grouped, so different products can be routed freely; the price is more handling and inventory.
101. Which of the following is NOT a main factor in the choice of plant location?
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Think of economic factors.
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Answer: D. Colour scheme of the office buildings
Location depends on raw materials, labour, power, transport, market, climate, laws and costs, not on decorative features.
102. A plant making a product whose raw material loses a lot of weight in processing is best placed
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Avoid hauling waste.
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Answer: C. near the source of raw materials
Weight-losing raw materials are cheaper to process at the source and transport as finished product (Weber's principle).
103. The factor rating method of plant location involves
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Weighted scoring.
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Answer: C. assigning weights and scores to location factors and comparing total weighted scores
Each factor gets an importance weight; each site is scored and the site with the highest weighted total is selected.
104. Plant A has fixed costs of Rs 100 000 per year and variable cost Rs 50 per unit; the product sells for Rs 90 per unit. The break-even volume is
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Divide fixed cost by contribution per unit.
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Answer: B. 2500 units
BEP = fixed cost/(price − variable cost) = 100 000/(90 − 50) = 2500 units.
105. Three warehouses A(10, 4), B(20, 1) and C(30, 6) receive weekly volumes of 10, 20 and 30 units respectively. The centre of gravity x-coordinate for a new distribution centre is
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Weight each x-coordinate by volume.
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Answer: A. 23.3
x = Σwx/Σw = (10×10 + 20×20 + 30×30)/60 = 1400/60 = 23.3.
106. Material handling cost is generally considered
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Does it change the form of the product?
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Answer: D. a non-value-adding cost that should be minimised
Handling adds cost but does not change the product, so good layout aims to reduce distance, number of moves and storage.
107. Which of the following is a continuous material-handling device?
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Fixed path, continuous flow.
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Answer: D. Belt conveyor
Conveyors move material continuously along a fixed path; cranes, trucks and trolleys are intermittent or variable-path equipment.
108. A belt conveyor moves loads spaced 2 m apart at a belt speed of 0.5 m/s. The throughput is
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Speed ÷ spacing, then convert to hours.
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Answer: C. 900 loads per hour
Rate = 0.5/2 = 0.25 loads/s = 900 loads/h.
109. The 'unit load' principle of material handling states that
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Handling many items as one.
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Answer: B. items should be grouped into a single unit (pallet or container) so they are handled together
A unit load (e.g. palletised goods) reduces the number of trips and handling effort.
110. Automated guided vehicles (AGVs) are mainly used for
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They transport loads without a driver.
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Answer: A. moving materials along programmed paths between workstations
AGVs are driverless vehicles that follow wires, markers or laser guidance to carry loads in factories and warehouses.
111. A forklift truck is classified as which type of equipment?
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It can go anywhere on the floor.
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Answer: A. Industrial truck (variable-path equipment)
Industrial trucks move loads over variable paths and can lift and stack them.
112. Production planning and control (PPC) mainly aims at
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Delivery on time with good utilisation.
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Answer: D. meeting delivery schedules with optimal use of resources
PPC plans what, when and how much to produce and controls execution to meet due dates economically.
113. Which of the following is the correct sequence of functions in production control?
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Path first, monitoring last.
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Answer: C. Routing, scheduling, dispatching, follow-up
Routing decides the path and operations, scheduling decides when, dispatching issues orders, follow-up monitors and expedites.
114. Routing in production control determines
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Sequence of operations and machines.
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Answer: B. the sequence of operations and the path of work through the plant
A route sheet lists operations, machines and tooling required to convert raw material to a finished part.
115. Scheduling in production control determines
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Time dimension.
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Answer: B. when each operation is to start and finish
Scheduling fixes the time sequence of operations on each machine to meet due dates.
116. A Gantt chart is used to
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Bars versus time.
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Answer: A. show planned and actual progress of jobs against time
A Gantt chart plots jobs or machines against a time scale as bars, comparing schedule with actual progress.
117. In an assembly line, the cycle time is the
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Available time ÷ demand.
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Answer: D. maximum time allowed at each workstation, = available time / required output
Cycle time = available production time per day/required units per day.
118. An assembly line must produce 240 units in an 8-hour shift (480 min). The cycle time is
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Time ÷ units.
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Answer: C. 2 min
CT = 480/240 = 2 min per unit.
119. Total task time for a product is 7.5 min and cycle time is 2 min. The theoretical minimum number of workstations is
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Always round up.
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Answer: C. 4
N = ΣT/CT = 7.5/2 = 3.75, rounded up to 4.
120. For 5 workstations, cycle time of 2 min and total task time of 7.5 min, the line balance efficiency is
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ΣT divided by N×CT.
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Answer: B. 75 %
Efficiency = ΣT/(N × CT) = 7.5/(5×2) = 0.75.
121. A part requires 3 min per unit on a machine. The annual demand is 60 000 units. Each machine is available 2000 h per year with 75 % utilisation. The number of machines required is
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Compare required hours with effective hours.
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Answer: A. 2
Machine hours needed = 60 000×3/60 = 3000 h. Effective hours per machine = 2000×0.75 = 1500 h. Machines = 2.
122. Three jobs with processing times 3, 1 and 2 hours are sequenced on one machine in shortest processing time (SPT) order. The mean flow time is
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Completion times accumulate.
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Answer: D. 3.33 h
SPT order 1, 2, 3: completion times 1, 3, 6; mean = 10/3 = 3.33 h.
123. Which sequencing rule minimises the average flow time on a single machine?
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Shortest first.
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Answer: D. Shortest processing time (SPT)
SPT places short jobs first so that fewer jobs wait for long ones, minimising mean completion time.
124. Johnson's rule is applied to sequence jobs on
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Two-machine flow shop.
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Answer: C. two machines in series to minimise total completion time (makespan)
Jobs with the shortest time on machine 1 go first; jobs with the shortest time on machine 2 go last.
8.5 Inventory control
31 questions · AMeE0805
125. Economic order quantity (EOQ) is the order size that
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A balance of two cost components.
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Answer: A. minimises the sum of annual ordering cost and annual holding cost
EOQ balances ordering cost (falling with larger Q) against holding cost (rising with Q) to give minimum total relevant cost.
126. The basic EOQ formula is
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Annual demand and ordering cost go in the numerator.
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Answer: D. √(2DS/H)
With D = annual demand, S = ordering cost per order, H = holding cost per unit per year: Q* = √(2DS/H).
127. An item has annual demand 12 000 units, ordering cost Rs 300 per order and holding cost Rs 24 per unit per year. The EOQ is nearest to
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Q = √(2DS/H).
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Answer: C. 548 units
Q* = √(2×12 000×300/24) = √300 000 = 547.7 ≈ 548 units.
128. For the item above (D = 12 000, S = Rs 300, H = Rs 24), the minimum annual ordering plus holding cost is nearest to
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At EOQ the two costs are equal.
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Answer: B. Rs 13 150
TC = √(2DSH) = √(2×12 000×300×24) = Rs 13 145, which is twice the ordering (or holding) cost of Rs 6 573.
129. In the basic EOQ model, at the optimum order quantity
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The total cost curve is flat at the minimum.
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Answer: B. annual ordering cost equals annual holding cost
Setting d(TC)/dQ = 0 gives DS/Q = QH/2, i.e. the two components are equal.
130. For the item with D = 12 000 per year, the EOQ of about 548 units means the time between orders is nearest to
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Cycle length = Q/D.
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Answer: A. 17 days
Orders per year = 12 000/548 = 21.9; interval = 365/21.9 = 16.7 days.
131. An item costs Rs 200 and the annual holding cost is 20 % of its price. Annual demand is 8000 units and the ordering cost is Rs 500 per order. The EOQ is nearest to
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Compute H first.
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Answer: D. 447 units
H = 0.20×200 = Rs 40; Q* = √(2×8000×500/40) = √200 000 = 447 units.
132. If the annual demand doubles while S and H remain unchanged, the EOQ
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Q* is proportional to the square root of D.
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Answer: C. increases by a factor of √2
Q* ∝ √D, so doubling D multiplies Q* by √2 ≈ 1.41.
133. If the holding cost per unit is increased four times, the EOQ
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Q* ∝ 1/√H.
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Answer: C. becomes half
Q* ∝ 1/√H; with H × 4, Q* is divided by 2.
134. The reorder point in a continuous review system is
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A stock level, not a quantity.
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Answer: B. the stock level at which a new order is placed
When on-hand stock falls to the reorder point, an order is placed so that it arrives after the lead time.
135. Daily demand is 40 units, lead time is 7 days and safety stock is 50 units. The reorder point is
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ROP = dL + SS.
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Answer: A. 330 units
ROP = demand during lead time + safety stock = 40×7 + 50 = 330 units.
136. Lead time in inventory control is the
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From placing the order to receipt.
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Answer: D. time between placing an order and receiving the goods
Lead time includes order preparation, supplier processing, transport and inspection.
137. The main purpose of safety stock is to
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Uncertainty protection.
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Answer: D. protect against uncertainty in demand or lead time
Safety (buffer) stock absorbs variations in demand during lead time and in supplier delays to avoid stock-outs.
138. Daily demand has a standard deviation of 10 units, lead time is 9 days, and a service level of 95 % (z = 1.65) is required. The safety stock is nearest to
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Variation grows with √L.
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Answer: C. 50 units
SS = z σd √L = 1.65 × 10 × √9 = 49.5 ≈ 50 units.
139. If the service level is raised, the safety stock
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Higher service needs more buffer.
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Answer: B. increases, and so does the holding cost
A higher z-value requires a larger buffer, raising average inventory and holding cost.
140. In ABC analysis, class A items are
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Pareto principle.
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Answer: A. few in number but account for a large share of annual consumption value
Typically about 10–20 % of items make up 70–80 % of annual usage value and need the tightest control.
141. ABC analysis classifies items on the basis of
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Value of usage.
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Answer: A. annual usage value (consumption × unit cost)
Items are ranked by annual usage value and divided into A, B and C classes.
142. Which control is most suitable for class C items?
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Low value, low effort.
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Answer: D. Simple control such as two-bin system with large safety stock
Low-value, high-count items warrant minimal control effort; the cost of control exceeds possible savings.
143. In a two-bin system, an order is placed when
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The trigger is an empty bin.
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Answer: C. the first bin is empty and stock is drawn from the second bin
The second bin holds the quantity for lead-time demand plus safety stock; emptying the first triggers a reorder.
144. Which of the following is a holding (carrying) cost?
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Cost of keeping stock.
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Answer: B. Insurance and storage costs of inventory
Holding costs include capital cost, storage, insurance, obsolescence and handling; the others are ordering/setup or shortage costs.
145. The economic production quantity (EPQ) is larger than the EOQ because
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Inventory is not delivered all at once.
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Answer: B. inventory builds up gradually while production is on
Items are used while being produced so average inventory is lower, allowing larger lots: EPQ = EOQ/√(1 − d/p).
146. For D = 12 000 per year and H and S as before (EOQ = 548), the production rate is 24 000 per year. The EPQ is nearest to
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Divide EOQ by √(1 − d/p).
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Answer: A. 775 units
EPQ = EOQ/√(1 − d/p) = 548/√0.5 = 775 units.
147. Forecasting by exponential smoothing uses
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Recent data count more.
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Answer: D. a weighted average giving more weight to recent data
Forecast F(t+1) = αA(t) + (1 − α)F(t), where weights decline geometrically with age of data.
148. A previous forecast was 100 units and actual demand was 120. With α = 0.2, the next exponentially smoothed forecast is
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F(new) = F + α(A − F).
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Answer: C. 104 units
F = 100 + 0.2(120 − 100) = 104.
149. Demands in the last four months were 100, 120, 110 and 130. The three-month simple moving average forecast for the fifth month is
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Use only the most recent three values.
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Answer: C. 120
Average of the last three: (120+110+130)/3 = 120.
150. A weighted moving average uses weights 0.5, 0.3, 0.2 for the latest, previous and earlier months (130, 110, 120). The forecast is
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Multiply each demand by its weight.
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Answer: B. 122
0.5×130 + 0.3×110 + 0.2×120 = 65 + 33 + 24 = 122.
151. A larger value of the smoothing constant α in exponential smoothing
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Response versus smoothness.
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Answer: A. makes the forecast respond faster to demand changes but be less smooth
A high α puts more weight on the latest actual and less on the history.
152. The linear trend demand over four periods is 10, 12, 14, 16. Using the line of best fit, the forecast for period 5 is
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The data are exactly linear.
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Answer: D. 18
Demand = 8 + 2t; for t = 5, forecast = 18.
153. Mean absolute deviation (MAD) of forecast is the
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Ignore the signs of errors.
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Answer: D. average of the absolute forecast errors
MAD = Σ|A − F|/n, a measure of forecast accuracy.
154. Forecast errors for five periods are +5, −5, +2, +4, −5. The MAD is
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Average of the absolute values.
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Answer: C. 4.2
MAD = (5+5+2+4+5)/5 = 21/5 = 4.2.
155. Which of the following is a qualitative forecasting technique?
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It relies on expert judgement.
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Answer: B. Delphi method
The Delphi method collects and iterates expert opinions; the others are quantitative time-series or causal methods.
8.6 Engine performance and testing
32 questions · AMeE0806
156. Indicated power of an engine is the
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It is measured at the piston.
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Answer: A. power developed inside the cylinder by the gases acting on the piston
IP is obtained from the indicator diagram (mean effective pressure); brake power = IP − friction power.
157. The brake power of an engine is measured by
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It is measured at the shaft.
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Answer: D. a dynamometer coupled to the output shaft
A dynamometer absorbs the engine output and measures torque and speed; BP = 2πNT/60.
158. A dynamometer measures 150 N·m torque at 3000 rpm. The brake power is nearest to
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BP = 2πNT/60.
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Answer: C. 47.1 kW
BP = 2πNT/60 = 2π × 3000 × 150/60 = 47 124 W.
159. A rope brake dynamometer has brake drum diameter 1.0 m and rope diameter 20 mm; load 300 N, spring balance reading 40 N, speed 600 rpm. The brake power is nearest to
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Effective radius includes the rope diameter.
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Answer: B. 8.3 kW
BP = π(D + d) N (W − S)/60 000 = π × 1.02 × 600 × 260/60 000 = 8.33 kW.
160. A four-cylinder, four-stroke engine has bore 80 mm, stroke 90 mm, imep 8 bar and speed 2400 rpm. The indicated power is nearest to
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A four-stroke fires once every two revolutions.
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Answer: B. 29 kW
IP = pm L A (N/2) n/60 = 8×10⁵ × 0.09 × 5.027×10⁻³ × 1200/60 × 4 = 28.9 kW.
161. Mechanical efficiency of an engine is the ratio of
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Ratio of shaft output to cylinder output.
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Answer: A. brake power to indicated power
ηmech = BP/IP = 1 − (friction power/IP).
162. An engine has an indicated power of 60 kW and a brake power of 48 kW. Its friction power and mechanical efficiency are
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Subtract for FP, divide for efficiency.
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Answer: D. 12 kW and 80 %
FP = IP − BP = 12 kW; ηmech = 48/60 = 0.80.
163. Brake thermal efficiency is defined as
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Output over fuel energy input.
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Answer: C. brake power divided by the heat energy supplied by fuel per unit time
ηbth = BP/(ṁf × CV).
164. An engine develops 30 kW brake power while consuming 8 kg/h of fuel of calorific value 44 000 kJ/kg. The brake thermal efficiency is nearest to
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Convert kg/h to kg/s.
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Answer: C. 30.7 %
Fuel energy = 8/3600 × 44 000 = 97.8 kW; ηbth = 30/97.8 = 0.307.
165. For the same engine (30 kW at 8 kg/h of fuel) the brake specific fuel consumption is nearest to
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Fuel per hour per kW.
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Answer: B. 0.267 kg/kWh
bsfc = fuel flow/BP = 8/30 = 0.267 kg/kWh.
166. Volumetric efficiency of a four-stroke engine is the ratio of
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Ability to breathe.
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Answer: A. actual volume of air drawn in to the swept volume
ηv = actual intake volume (at intake conditions)/swept volume per unit time.
167. A 1.6 litre four-stroke engine at 3000 rpm inducts 0.034 m³/s of air measured at intake conditions. The volumetric efficiency is
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Each cylinder takes in air only once every two revolutions.
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Answer: D. 85 %
Swept volume rate = 1.6×10⁻³ × 3000/(2×60) = 0.04 m³/s; ηv = 0.034/0.04 = 0.85.
168. The air-standard efficiency of an Otto cycle with compression ratio 8 (γ = 1.4) is nearest to
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η = 1 − r^(1−γ).
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Answer: D. 56.5 %
η = 1 − 1/r^(γ−1) = 1 − 8^(−0.4) = 0.565.
169. Relative efficiency (efficiency ratio) of an engine is the ratio of
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Actual versus ideal cycle.
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Answer: C. indicated thermal efficiency to air-standard efficiency
It compares the actual indicated performance with the ideal air-standard cycle.
170. The Morse test is used to find the
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Cut out one cylinder at a time.
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Answer: B. indicated power of a multi-cylinder engine by cutting out cylinders in turn
BP is measured with all cylinders running and with each cylinder cut off in turn at constant speed; the drop in BP gives the IP of that cylinder.
171. In a Morse test the brake power with all cylinders firing is 40 kW. When cylinder 1 is cut out it falls to 28 kW at the same speed. The indicated power of cylinder 1 is
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Subtract the two BP values.
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Answer: A. 12 kW
The difference 40 − 28 = 12 kW is the IP of cylinder 1 (friction is unchanged).
172. The Willans line method of finding friction power is applicable to
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Fuel consumption against brake power.
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Answer: A. diesel engines at constant speed
Fuel consumption vs BP is almost linear for CI engines; extrapolating to zero fuel gives the negative BP, i.e. the friction power.
173. The coolant of an engine receives 25 kW and may rise by 8 K across the engine. With cp = 4.18 kJ/kg·K, the coolant mass flow rate is nearest to
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Q = ṁ cp ΔT.
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Answer: D. 0.75 kg/s
ṁ = Q/(cp ΔT) = 25/(4.18×8) = 0.748 kg/s.
174. In a water-cooled engine the thermostat is used to
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It is a temperature-actuated valve.
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Answer: C. keep the coolant at a steady operating temperature by regulating flow to the radiator
The thermostat valve stays closed until the coolant reaches its set temperature, then opens to allow circulation through the radiator.
175. Fins on the cylinders of a motorcycle engine are provided to
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More surface area.
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Answer: B. increase the surface area for heat dissipation to air
Air cooling relies on extended surfaces to enhance convection from the cylinder walls.
176. Knocking in a spark-ignition engine is caused by
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The last portion of the charge burns abnormally.
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Answer: B. auto-ignition of the end gas before the flame front reaches it
The unburnt end gas, compressed and heated by the advancing flame, ignites spontaneously, causing a sudden pressure rise and a metallic knock.
177. Which of the following would reduce knocking in a spark-ignition engine?
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Octane number measures knock resistance.
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Answer: A. Using fuel of higher octane number
Higher octane fuel resists auto-ignition; lowering compression ratio, retarding the spark and cooler intake charge also help.
178. The octane number of a fuel is the percentage by volume of
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Iso-octane is the 100 reference.
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Answer: D. iso-octane in a blend with n-heptane that matches the knock behaviour of the fuel
Iso-octane is assigned an octane number of 100 and n-heptane 0; a fuel's rating is the iso-octane percentage in the equivalent reference blend.
179. Knocking in a diesel engine is mainly the result of
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Opposite to what causes SI knock.
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Answer: C. a long ignition delay causing a large amount of fuel to burn suddenly
A long delay allows accumulated fuel to burn in one rapid combustion with a high rate of pressure rise; high-cetane fuels shorten the delay.
180. Pre-ignition in an SI engine means
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Ignition before the spark occurs.
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Answer: C. ignition of the charge before the spark due to a hot spot
Hot surfaces such as an overheated spark plug or carbon deposits ignite the mixture before the normal spark.
181. The main working principle of a simple carburettor is that
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Bernoulli's principle.
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Answer: B. air flowing through a venturi creates a pressure drop that draws in fuel from the jet
The suction at the venturi throat, lower than the float-chamber pressure (atmospheric), pushes fuel through the jet.
182. The stoichiometric air–fuel ratio of petrol by mass is approximately
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Chemically correct mixture.
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Answer: A. 14.7 : 1
Complete combustion of petrol needs about 14.7 kg of air per kg of fuel.
183. A carburettor supplies a rich mixture mainly during
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Cruise is the lean condition.
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Answer: D. cold starting, idling and full-load operation
Starting and idling need rich mixtures to compensate for poor vaporisation, and full power needs a rich mixture for maximum output.
184. The choke in a carburettor is used to
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Cold engine needs more fuel.
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Answer: D. enrich the mixture for starting a cold engine
The choke valve restricts the air inlet, increasing suction at the jet and giving a rich mixture.
185. The engine sucks 120 kg/h of air at an air–fuel ratio of 15 : 1. The fuel consumption is
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Divide air by the ratio.
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Answer: C. 8 kg/h
Fuel = air/AFR = 120/15 = 8 kg/h.
186. Multi-point fuel injection (MPFI) offers which advantage over a carburettor?
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One injector per cylinder.
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Answer: B. Better distribution of fuel to every cylinder and precise control of mixture
Individual injectors with electronic control give precise metering and equal cylinder-to-cylinder distribution, improving efficiency and emissions.
187. In a diesel engine, the fuel injection system must
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Needs high pressure.
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Answer: A. deliver an accurately metered fuel quantity at high pressure at the right time
High-pressure injection atomises the fuel into a fine spray in the hot compressed air for rapid mixing and combustion.