Nepal Engineering Council · Mechanical Engineering · Chapter 5
Manufacturing Technology
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181 questions in 6 syllabus topics.
5.1 Foundry
31 questions · AMeE0501
1. A casting is to be 500 mm long in cast iron with a shrinkage allowance of 1%. The pattern length should be nearly:
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Pattern must compensate for contraction on cooling.
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Answer: B. 505 mm
Pattern is made larger than the casting: 500 × 1.01 = 505 mm.
2. Which allowance is provided on a pattern so that it can be withdrawn from the mould without damaging the mould cavity?
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Think of the taper on vertical faces.
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Answer: D. Draft (taper) allowance
Draft is a slight taper on vertical faces of the pattern that eases its withdrawal from the sand.
3. In a sand casting, a cube of side 100 mm and another of side 200 mm are cast from the same metal in the same sand. By Chvorinov's rule the larger cube solidifies in a time that is how many times that of the smaller?
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Solidification time depends on the square of the volume-to-area ratio.
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Answer: A. 4
t ∝ (V/A)²; for a cube V/A = a/6, so doubling the side doubles V/A and gives t × 4.
4. Which property of moulding sand allows steam and gases formed during pouring to escape through the mould?
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Gas escape through the sand grains.
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Answer: C. Permeability
Permeability is the ability of packed sand to let gases pass; poor permeability causes blowholes.
5. A casting has a modulus (V/A) of 1 cm and solidifies in 4 minutes. Using t = B·M², how long will a casting of the same metal and mould with modulus 2.5 cm take to solidify?
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Time varies with the square of the modulus.
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Answer: D. 25 min
B = 4 min/cm². t = 4 × 2.5² = 25 min.
6. The main purpose of a riser in a casting mould is to:
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It must freeze after the casting does.
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Answer: A. Feed molten metal to compensate for solidification shrinkage
A riser is a reservoir of liquid metal that solidifies last and feeds the casting as it shrinks.
7. Molten steel falls freely through a sprue from a height of 0.5 m above the sprue exit. Neglecting losses, the velocity at the sprue exit is nearly (g = 9.81 m/s²):
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Use Bernoulli's / free-fall relation.
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Answer: C. 3.13 m/s
v = √(2gh) = √(2 × 9.81 × 0.5) = 3.13 m/s.
8. The vertical passage through which molten metal is first poured into the mould is called the:
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It is the vertical channel.
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Answer: B. Sprue
The sprue carries metal down from the pouring basin to the runner system.
9. A casting cavity of volume 6000 cm³ is filled through a gating system with an effective flow rate of 1.2 × 10⁻³ m³/s. The pouring time is:
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Time = volume ÷ flow rate, mind the units.
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Answer: C. 5 s
V = 6000 cm³ = 6 × 10⁻³ m³; t = V/Q = 6 × 10⁻³ / 1.2 × 10⁻³ = 5 s.
10. Which furnace is most widely used for melting cast iron in foundries in large quantities?
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It is a shaft furnace fired with coke.
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Answer: A. Cupola
The cupola is a vertical shaft furnace charged with pig iron, scrap, coke and flux, and is the standard cast-iron melter.
11. The mass of a steel casting of volume 2000 cm³ (density 7.85 g/cm³) is:
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Mass = density × volume, then convert g to kg.
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Answer: B. 15.7 kg
m = 2000 × 7.85 = 15 700 g = 15.7 kg.
12. In die casting, the hot-chamber process is generally restricted to low-melting-point alloys such as:
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The metal pot is part of the machine.
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Answer: D. Zinc and tin alloys
In the hot-chamber machine the injection plunger is immersed in the molten metal, so high-melting alloys would attack it; zinc, tin and lead alloys are used.
13. A cylindrical sand core of volume 1000 cm³ (density 1.6 g/cm³) is surrounded by molten cast iron of density 7.2 g/cm³. The net upward buoyancy force on the core is nearly (g = 9.81 m/s²):
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Weight of displaced metal minus weight of the core.
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Answer: A. 54.9 N
F = V(ρmetal − ρcore)g = 1000 × 10⁻⁶ × (7200 − 1600) × 9.81 = 54.9 N.
14. Investment (lost-wax) casting is preferred for turbine blades mainly because it gives:
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Think precision, not speed or cost.
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Answer: C. Excellent dimensional accuracy and surface finish for complex shapes
The wax pattern is coated with refractory slurry and melted out, so the mould has no parting line and reproduces fine detail.
15. A melting practice uses a coke-to-iron ratio of 1 : 10 by mass in a cupola. How much coke is needed to melt 1000 kg of iron?
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Ratio is coke : iron.
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Answer: D. 100 kg
Coke = 1000/10 = 100 kg.
16. Which casting defect is caused by two streams of molten metal meeting without fusing properly?
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Look at the name: metal that is too cold.
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Answer: B. Cold shut
A cold shut is a seam where metal fronts have cooled too much to fuse, often due to low pouring temperature or slow filling.
17. Gating ratio sprue : runner : ingate = 1 : 3 : 3 is used. If the sprue choke area is 150 mm², the total ingate area is:
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Multiply the choke area by the ratio.
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Answer: B. 450 mm²
Ingate area = 3 × 150 = 450 mm².
18. In shell moulding the mould is made from sand coated with:
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The sand is pre-coated with a binder that hardens on heat.
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Answer: D. Thermosetting resin
Resin-coated sand falls onto a heated pattern, cures to a thin hard shell, and the shells are clamped together.
19. Centrifugal casting is best suited to producing:
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Rotation pushes metal outward.
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Answer: A. Hollow cylindrical parts such as pipes and bushes
Rotation forces the metal outward against the mould wall, giving hollow cylinders without a core and with dense outer metal.
20. The process of removing sprues, gates, risers, fins and adhering sand from a casting is called:
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It is the foundry word for finishing the rough casting.
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Answer: C. Fettling
Fettling is the cleaning of castings, involving cutting off gates and risers, grinding, and shot or sand blasting.
21. A casting weighs 80 kg and the gating and risers weigh 40 kg. The casting yield is:
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Yield = useful casting ÷ total metal poured.
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Answer: D. 66.7%
Yield = 80 / (80 + 40) = 0.667 = 66.7%.
22. Which sand binder is most commonly used in green sand moulds?
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'Green' means damp, not coloured.
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Answer: A. Bentonite clay
Green sand is silica sand with about 5–10% clay (bentonite) and 2–5% water.
23. Chills are placed in a mould to:
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They are metal inserts that remove heat fast.
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Answer: C. Promote directional solidification
A metal chill extracts heat quickly from a thick section so that solidification proceeds towards the riser.
24. How much heat is needed only to melt (latent heat 397 kJ/kg) 100 kg of aluminium already at its melting point?
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Latent heat × mass.
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Answer: B. 39.7 MJ
Q = m L = 100 × 397 = 39 700 kJ = 39.7 MJ.
25. Misrun in a casting means that:
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Incomplete casting.
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Answer: C. The mould cavity was not completely filled
A misrun occurs when metal solidifies before filling the cavity, due to low fluidity or low pouring temperature.
26. Core prints on a pattern are provided to:
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They are extensions on the pattern.
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Answer: A. Locate and support the core in the mould
Core prints form recesses in the mould where the core ends rest.
27. Which of the following is a typical application of sand casting?
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Large, moderately accurate parts.
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Answer: B. Engine blocks and machine bed frames
Sand casting makes large, heavy parts like engine blocks and machine tool beds.
28. Hot tears in a casting are mainly caused by:
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Cracks during cooling.
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Answer: D. Restraint to contraction during cooling
When a casting is restrained by the mould or by its own geometry, tensile stress in the weak semi-solid state causes cracks.
29. In a permanent mould (gravity die) casting compared with sand casting, the product generally has:
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Metal moulds cool faster than sand.
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Answer: A. Finer grain and better surface finish
Metal moulds chill the metal faster and give a fine structure and good finish, but have a limited mould life.
30. The temperature to which molten metal is heated above its melting point before pouring is called:
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The extra heat above melting.
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Answer: C. Superheat
Superheat helps fluidity and mould filling.
31. A good moulding sand must be refractory. This means it:
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Think of fire-resistant.
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Answer: D. Withstands high temperature without fusing
Refractoriness is the ability of sand to resist the high temperature of molten metal without softening or fusing.
5.2 Heat treatment
30 questions · AMeE0502
32. In full annealing of a hypoeutectoid steel, the steel is heated to about 30–50 °C above the upper critical temperature and then:
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The main aim is maximum softness.
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Answer: B. Cooled very slowly in the furnace
Slow furnace cooling gives coarse pearlite and ferrite, which is soft and ductile with relieved stresses.
33. Normalizing differs from full annealing in that normalizing uses:
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Look at the cooling medium.
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Answer: D. Cooling in still air
Air cooling is faster, giving a finer pearlite and somewhat higher strength than annealing.
34. Which structure is formed when austenite of a plain carbon steel is cooled so rapidly that diffusion cannot occur?
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It is the hardest structure in plain steel.
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Answer: A. Martensite
Fast quenching traps carbon in a supersaturated body-centred tetragonal lattice, called martensite, which is hard and brittle.
35. A 0.4% carbon steel is slowly cooled to just below the eutectoid temperature. Taking ferrite at 0.02% C and pearlite at 0.77% C, the percentage of pearlite is nearly:
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Apply the lever rule with pearlite at the end of the tie line.
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Answer: C. 50.7%
Pearlite fraction = (0.40 − 0.02)/(0.77 − 0.02) = 0.507, i.e. 50.7%.
36. Tempering of a hardened steel is carried out by:
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It is done after quenching.
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Answer: D. Reheating below the lower critical temperature and cooling
Tempering at about 150–650 °C reduces brittleness and internal stress at some cost in hardness.
37. Which quenching medium gives the fastest cooling rate (most severe quench)?
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Salt breaks up the vapour film.
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Answer: A. Brine (salt water)
Brine breaks the vapour blanket quickly and cools faster than plain water, oil or air.
38. Plain carbon steel of 0.77% carbon contains pearlite. The percentage of cementite in this pearlite, taking cementite at 6.67% C and ferrite at 0.02% C, is:
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Lever rule from the ferrite end.
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Answer: C. 11.3%
Wcementite = (0.77 − 0.02)/(6.67 − 0.02) = 0.113.
39. Carburizing is the case-hardening process in which:
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The name contains the element added.
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Answer: B. Carbon is diffused into the surface of low carbon steel
Low carbon steel (0.1–0.25% C) is heated in a carbon-rich atmosphere so that the case gains carbon; it is then quenched.
40. Nitriding of steel is usually carried out at about:
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It is much lower than carburizing temperature.
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Answer: C. 500–550 °C in an ammonia atmosphere
At the low nitriding temperature no phase change occurs, so little distortion and no quenching are needed.
41. Cyaniding is a case-hardening process that introduces into the surface of steel:
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Cyanide contains C and N.
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Answer: A. Both carbon and nitrogen
A molten sodium cyanide bath at 800–870 °C supplies both carbon and nitrogen, followed by quenching.
42. A steel part is cooled from 850 °C to 350 °C in 5 s during quenching. The mean cooling rate is:
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Temperature drop ÷ time.
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Answer: B. 100 °C/s
Rate = (850 − 350)/5 = 100 °C/s.
43. Which heat treatment is applied mainly to improve the machinability of high-carbon steel by producing globular carbides?
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Cementite takes a rounded shape.
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Answer: D. Spheroidizing
Prolonged heating just below A1 makes the cementite coalesce into spheroids in a ferrite matrix, giving a soft, machinable structure.
44. 5 kg of steel (specific heat 0.5 kJ/kg·K) is heated from 20 °C to 850 °C for hardening. The heat absorbed by the steel is:
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ΔT is 830 K.
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Answer: A. 2075 kJ
Q = m c ΔT = 5 × 0.5 × 830 = 2075 kJ.
45. Induction hardening is mainly suited to steels containing about:
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Medium carbon content is needed.
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Answer: C. 0.4–0.6% carbon
Medium carbon steel can form enough martensite on rapid surface heating and quenching, giving a hard case over a tough core.
46. The main steps of powder metallurgy in correct order are:
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Press before you heat.
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Answer: D. Blending, compacting, sintering
Powders are mixed, pressed into a green compact in a die, then heated below the melting point to bond the particles.
47. A green compact has density 6.0 g/cm³ while the fully dense solid metal has 7.8 g/cm³. The porosity of the compact is nearly:
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Porosity = 1 − relative density.
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Answer: B. 23.1%
Porosity = 1 − 6.0/7.8 = 0.231.
48. Sintering is carried out at a temperature:
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The powder is not melted.
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Answer: B. Below the melting point of the major constituent
Atoms diffuse across particle contacts and bond them without full melting.
49. Self-lubricating bearings are made by powder metallurgy because the process can produce:
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Think of what powder processing keeps that casting cannot.
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Answer: D. Controlled porosity that is impregnated with oil
Interconnected pores in the sintered part are filled with lubricant by vacuum impregnation.
50. What mass of powder is needed to compact a part of volume 50 cm³ to a density of 6.5 g/cm³ (ignoring losses)?
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Mass = density × volume.
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Answer: A. 325 g
m = ρV = 6.5 × 50 = 325 g.
51. During the solidification of a pure metal, the temperature during freezing:
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Alloys differ because they freeze over a range.
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Answer: C. Remains constant at the freezing point
A pure metal freezes at a single temperature, with latent heat released at a flat part of the cooling curve.
52. Compared with slow cooling, rapid cooling during solidification generally produces:
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Many nuclei, little growth time.
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Answer: D. A finer grain structure
A high cooling rate gives more nuclei per unit volume and less time for growth, so grains are finer.
53. An ingot undergoes 5% volumetric shrinkage on solidification. A liquid volume of 2000 cm³ will, after solidification, occupy:
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Shrinkage reduces volume.
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Answer: A. 1900 cm³
2000 × (1 − 0.05) = 1900 cm³.
54. The Jominy end-quench test is used to determine the steel's:
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It measures depth of hardening.
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Answer: C. Hardenability
A heated bar is quenched at one end only and hardness is measured along its length, showing how deep hardening extends.
55. The case depth in carburizing varies as the square root of time. If the case depth is 0.5 mm after 4 h, then after 16 h it will be about:
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Four times the time gives twice the depth.
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Answer: B. 1.0 mm
x ∝ √t, so x = 0.5 × √(16/4) = 1.0 mm.
56. The eutectoid temperature of the iron–carbon system is about:
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It is below 800 °C.
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Answer: C. 727 °C
Austenite transforms into pearlite at 727 °C and 0.77% C (the A1 temperature).
57. In powder compaction, a force is applied by a punch of diameter 20 mm at a pressure of 400 MPa. The compacting force is nearly:
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Area of a circle × pressure.
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Answer: A. 125.7 kN
F = pA = 400 × (π/4 × 20²) = 400 × 314.16 = 125 664 N ≈ 125.7 kN.
58. A 1.2% C steel is slowly cooled below the eutectoid temperature. The proeutectoid phase (taking cementite at 6.67% C and the eutectoid at 0.77% C) is about:
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Lever rule between the eutectoid composition and cementite.
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Answer: B. 7.3% cementite
Wcementite = (1.2 − 0.77)/(6.67 − 0.77) = 0.0729, forming along grain boundaries.
59. Austempering of steel produces which microstructure?
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Isothermal hold gives this intermediate structure.
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Answer: D. Bainite
The steel is quenched to a temperature above Ms, held until isothermal transformation to bainite, then cooled; it is tough with little distortion.
60. Which of the following steels is most suitable for nitriding?
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Nitride-forming alloying elements.
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Answer: A. Steels containing aluminium, chromium and molybdenum
Elements such as Al, Cr and Mo form hard stable nitrides on the surface.
61. Case hardening gives a component:
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Hard outside, tough inside.
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Answer: C. A hard wear-resistant surface with a tough core
Only the surface layer is hardened, so the core stays ductile to resist shock.
5.3 Metal working
29 questions · AMeE0503
62. Hot working of a metal is carried out:
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Compare with the recrystallization temperature.
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Answer: B. Above the recrystallization temperature
In hot working, recrystallization occurs simultaneously with deformation, so there is no strain hardening.
63. The recrystallization temperature of a pure metal is approximately:
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It is less than half the melting point in kelvin.
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Answer: D. 0.4 Tm (absolute melting temperature)
As a rule of thumb recrystallization begins near 0.3–0.5 of the absolute melting temperature, taken as 0.4 Tm.
64. Which of the following is an advantage of hot working over cold working?
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Metal is softer when hot.
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Answer: A. Lower force required for large deformation
Metal is soft and ductile at high temperature, so large reductions need smaller forces and power.
65. A plate is rolled from 25 mm to 20 mm thickness. The draft in this pass is:
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Draft is a length.
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Answer: C. 5 mm
Draft = h0 − hf = 25 − 20 = 5 mm; the percentage reduction is 20% but the draft is expressed in length.
66. A strip is reduced from 25 mm to 20 mm in one pass. The percentage reduction in thickness is:
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Reduction ÷ original thickness.
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Answer: D. 20%
(25 − 20)/25 × 100 = 20%.
67. In rolling, the maximum possible draft is Δh = μ²R. For a coefficient of friction 0.1 and roll radius 250 mm the maximum draft is:
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Square the friction coefficient first.
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Answer: A. 2.5 mm
Δh = 0.1² × 250 = 2.5 mm.
68. Cold working of metals produces:
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Work hardening.
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Answer: C. Higher strength and hardness but reduced ductility
Dislocation density increases due to strain hardening, raising strength and hardness at the cost of ductility.
69. A rolling pass with roll radius 200 mm and draft 2 mm has a projected contact length of nearly:
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Use the chord geometry L = √(RΔh).
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Answer: B. 20 mm
L = √(RΔh) = √(200 × 2) = 20 mm.
70. The process in which a heated billet is forced through a die opening to give a long product of constant cross-section is called:
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The material is pushed rather than pulled.
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Answer: C. Extrusion
A ram pushes the billet through the die; the process is used for aluminium sections and tubes.
71. Open-die forging is characterized by:
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Think of a blacksmith's hammer and anvil.
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Answer: A. Metal being free to flow laterally between flat dies
The workpiece is compressed between simple flat or shaped dies without full confinement, used for large simple shapes.
72. A cylindrical workpiece is upset in a cold forging operation. The flow stress is 300 MPa and diameter 40 mm. Neglecting friction, the force required is nearly:
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Force = stress × area.
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Answer: B. 377 kN
F = σA = 300 × π/4 × 40² = 300 × 1256.6 = 377 kN.
73. In which extrusion process does the billet remain stationary relative to the container, resulting in less friction and power?
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Friction against the container wall is the key.
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Answer: D. Indirect extrusion
In indirect (backward) extrusion the die is moved against the billet so no billet–container sliding occurs.
74. A billet of diameter 100 mm is extruded into a round bar of diameter 25 mm. The extrusion ratio is:
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Ratio of cross-sectional areas.
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Answer: A. 16
R = A0/Af = (100/25)² = 16.
75. Seamless tubes are most commonly produced by hot:
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No welded joint exists.
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Answer: C. Piercing and rolling (Mannesmann process)
A heated billet is pierced by a mandrel over rolls and then rolled to the desired wall thickness.
76. In wire drawing, a wire of diameter 20 mm is reduced to 18 mm. The reduction in area is nearly:
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Area goes as the square of the diameter.
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Answer: D. 19%
1 − (18/20)² = 0.19.
77. The 'four-high' rolling mill is used because:
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Backup rolls support the smaller ones.
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Answer: B. Small working rolls reduce roll force and are supported by large backup rolls
Smaller work rolls need lower roll force and power while backup rolls prevent bending.
78. Cold rolling compared with hot rolling gives:
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No scale forms on the metal.
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Answer: B. Better surface finish and closer tolerance
The absence of scale and the strain hardening result in good finish and accuracy, but the work needs more force.
79. A wire is drawn so that its cross-sectional area is halved. The true strain in drawing is:
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Use the natural logarithm of the area ratio.
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Answer: D. 0.693
ε = ln(A0/Af) = ln 2 = 0.693.
80. Shot peening of a metal surface is used to:
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It improves fatigue resistance.
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Answer: A. Induce compressive residual stresses and improve fatigue life
Impacts of small shots plastically deform the surface layer, leaving compressive residual stress which delays crack initiation.
81. A circular hole of diameter 20 mm is punched in a 2 mm sheet with shear strength 300 MPa. The punching force is nearly:
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Shear area is the perimeter times thickness.
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Answer: C. 37.7 kN
F = π d t τ = π × 20 × 2 × 300 = 37 699 N ≈ 37.7 kN.
82. In sheet metal operations, blanking differs from piercing in that in blanking:
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Which part do you keep?
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Answer: D. The punched-out piece is the required product
Blanking cuts out the workpiece from the sheet; piercing makes a hole and the slug is scrap.
83. A 90° bend is made in a 2 mm sheet with inner radius 5 mm and k-factor 0.33. The bend allowance BA = θ(R + kt) is nearly:
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Use radians for the angle.
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Answer: A. 8.9 mm
θ = π/2 rad; BA = 1.5708 × (5 + 0.33 × 2) = 8.89 mm.
84. The tendency of a bent sheet to partly return to its original shape after load removal is called:
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An elastic effect.
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Answer: C. Springback
Elastic recovery makes the bend angle less than the die angle, so overbending is used.
85. In forging, grain flow lines that follow the shape of the component give:
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Compare with casting and machining.
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Answer: B. Better strength and fatigue resistance
Forging aligns the grain flow with part contour, unlike machining which cuts across it.
86. A cylindrical billet of diameter 40 mm and height 100 mm is forged to a height of 50 mm. Assuming constant volume, the new diameter is nearly:
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Volume stays constant.
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Answer: C. 56.6 mm
d² h = const; d2 = 40 × √(100/50) = 56.6 mm.
87. Cold hobbing in metalworking is a process in which:
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Used for making dies, not gears.
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Answer: A. A hardened steel hob is pressed into a soft blank to form a cavity
A hard master (hob) is pressed into a soft steel block to form a die or mould cavity with fine detail.
88. The process of squeezing a metal blank between two dies to reproduce fine surface details such as on coins is called:
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Think of coins.
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Answer: B. Coining
In coining the metal is confined and squeezed so that it flows to fill the die impression.
89. A rolling mill roll of radius 0.25 m rotates at 100 rpm. The surface speed of the roll is nearly:
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Convert rpm to rad/s.
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Answer: D. 2.62 m/s
v = 2πRN/60 = 2π × 0.25 × 100/60 = 2.62 m/s.
90. Which disadvantage applies to hot working?
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Oxygen reacts with hot metal.
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Answer: A. Surface oxidation and scale formation
At high temperature the surface oxidizes, so finish and tolerance are poorer.
5.4 Machine tools
31 questions · AMeE0504
91. The size of a centre lathe is specified by:
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Think of the largest workpiece it can hold.
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Answer: B. Swing over the bed and distance between centres
A lathe is designated by the largest diameter it can swing over the bed together with the maximum length between centres.
92. A workpiece of diameter 50 mm is turned at 400 rpm. The cutting speed is nearly:
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Surface speed = πDN.
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Answer: D. 62.8 m/min
V = πDN/1000 = π × 50 × 400/1000 = 62.8 m/min.
93. The operation of producing a flat surface perpendicular to the lathe axis is called:
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The tool feed is perpendicular to the axis.
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Answer: A. Facing
Facing moves the tool across the end of the work, perpendicular to the rotational axis.
94. The spindle speed required for a cutting speed of 30 m/min on a 60 mm diameter job is nearly:
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Rearrange V = πDN/1000.
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Answer: C. 159 rpm
N = 1000 V/(πD) = 30 000/(π × 60) = 159 rpm.
95. Knurling on a lathe is performed to:
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It improves hand gripping.
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Answer: D. Produce a rough diamond or straight pattern for grip
Knurling tools press patterns onto the surface by plastic deformation; no chip is removed.
96. A shaft 200 mm long is turned in one pass at feed 0.2 mm/rev and speed 500 rpm (neglecting approach and overrun). The machining time is:
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Feed per minute = fN.
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Answer: A. 2 min
T = L/(fN) = 200/(0.2 × 500) = 2 min.
97. Which lathe is best suited to repetitive batch production of small parts with several tools mounted on a hexagonal head?
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Think of a head that indexes through tools.
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Answer: C. Turret lathe
The turret holds several tools in sequence and speeds up operations on many identical parts.
98. Metal removal rate in turning at V = 60 m/min, feed 0.25 mm/rev and depth of cut 2 mm is:
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MRR = V × f × d, with V in mm/min.
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Answer: B. 30 cm³/min
MRR = V f d = 60 000 mm/min × 0.25 × 2 = 30 000 mm³/min = 30 cm³/min.
99. A taper from 50 mm to 40 mm is to be turned over a length of 200 mm on a 400 mm long job between centres by the tailstock set-over method. The set-over is:
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Total job length enters the formula.
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Answer: C. 10 mm
Set-over = L(D − d)/(2l) = 400 × 10/(2 × 200) = 10 mm.
100. In up (conventional) milling:
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Think relative directions at the contact.
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Answer: A. The cutter rotates opposite to the direction of table feed
The chip thickness starts at zero and increases to a maximum; the cutter rotation opposes feed.
101. A milling cutter with 8 teeth rotates at 300 rpm with a feed of 0.1 mm per tooth. The table feed rate is:
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Multiply feed per tooth, teeth and rpm.
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Answer: B. 240 mm/min
Feed = ft z N = 0.1 × 8 × 300 = 240 mm/min.
102. Plain milling of flat surfaces with a cylindrical cutter on a horizontal arbor is called:
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The cutter axis is parallel to the surface.
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Answer: D. Slab milling
The axis of the cutter is parallel to the surface being machined.
103. To mill a gear blank with 24 divisions using simple indexing (40:1 ratio of worm to wheel), the index crank must be turned:
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Divide 40 by the number of divisions.
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Answer: A. 1 full turn plus 16 holes in a 24-hole circle
Turns = 40/N = 40/24 = 1 16/24, i.e. one full turn and 16 holes on a 24-hole circle.
104. The standard point angle of a general purpose twist drill is:
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It is a bit more than a right angle.
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Answer: C. 118°
A 118° point angle is the standard for mild steel and general work.
105. Enlarging an existing hole accurately to a specified size and good finish using a multi-flute tool is called:
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It follows drilling.
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Answer: D. Reaming
Reaming removes a small stock from a drilled hole giving precise diameter and finish.
106. The time to drill a hole 30 mm deep at feed 0.2 mm/rev and 500 rpm is:
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Depth ÷ (feed × rpm).
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Answer: B. 0.3 min
T = 30/(0.2 × 500) = 0.3 min.
107. A flat-bottomed enlargement of a drilled hole to accommodate bolt heads is called:
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Compare with conical recess.
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Answer: B. Counterboring
A counterbore leaves a cylindrical recess with a flat bottom; a countersink is conical.
108. Which operation is performed on a drilling machine to cut internal threads?
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Tool is a tap.
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Answer: D. Tapping
A tap is rotated in a drilled hole to cut an internal thread.
109. In grinding wheel specification, the 'grade' refers to:
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It is a bond property.
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Answer: A. Hardness of the bond holding the abrasive grains
Grade denotes the strength of the bond: the grain-holding strength, from soft to hard.
110. A grinding wheel of 250 mm diameter rotates at 2800 rpm. Its peripheral speed is nearly:
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Convert rpm to rev/s.
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Answer: C. 36.7 m/s
V = πDN/60 = π × 0.25 × 2800/60 = 36.7 m/s.
111. Glazing of a grinding wheel occurs when:
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Worn-out cutting points.
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Answer: D. Abrasive grains become dull and the wheel face becomes smooth
Dull grains do not fracture, so the wheel does not self-sharpen; dressing restores the cutting surface.
112. Centreless grinding is used to:
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The workpiece is not held on centres.
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Answer: A. Grind cylindrical parts without holding them between centres
The work rests on a blade between a grinding wheel and a regulating wheel; it suits long, thin bars and mass production.
113. The main purpose of a boring operation is to:
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The hole already exists.
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Answer: C. Enlarge and finish an existing hole accurately
Boring uses a single-point tool to enlarge a hole already drilled or cast, correcting its position and size.
114. In a shaper, the cutting stroke takes 0.6 s and the return stroke 0.4 s. The quick return ratio is:
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Cutting time over return time.
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Answer: B. 1.5
Ratio = cutting time/return time = 0.6/0.4 = 1.5.
115. Which machine tool is used to machine a flat surface by a reciprocating motion of the tool while the work moves in steps across the stroke?
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The ram moves, the work only feeds.
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Answer: C. Shaper
In a shaper the ram carries the tool and the table is indexed at the end of each stroke; in a planer the work reciprocates.
116. A job 120 mm wide is machined on a shaper with a feed of 0.5 mm per stroke at 60 strokes per minute. The time for one pass is:
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Number of strokes ÷ strokes per minute.
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Answer: A. 4 min
Strokes = 120/0.5 = 240; time = 240/60 = 4 min.
117. Broaching is best suited to:
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One pass with a multi-tooth tool.
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Answer: B. Cutting keyways and shaping holes in a single stroke
A broach with progressively larger teeth removes all the material in one pass, giving high production rates.
118. A hacksaw blade cuts on:
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Teeth point away from the handle.
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Answer: D. The forward stroke only
Teeth point forward, so the blade cuts on the forward stroke and is lifted on the return.
119. The thread of pitch 1.5 mm is cut on a lathe with lead screw pitch 6 mm. The ratio of lead screw speed to spindle speed is:
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The carriage moves one thread pitch per spindle revolution.
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Answer: A. 0.25
Lead screw turns/spindle turns = thread pitch/lead screw pitch = 1.5/6 = 0.25.
120. A press runs at 60 strokes per minute producing one component per stroke. In an 8-hour shift with 85% efficiency, the number of components produced is:
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Strokes per shift times efficiency.
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Answer: C. 24 480
60 × 480 × 0.85 = 24 480.
121. The press which stores energy in a flywheel and delivers it through a crank is a:
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Flywheel and crank.
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Answer: D. Mechanical press
The crank press uses a flywheel and crank slider to give a fixed stroke; hydraulic presses give a constant force through the stroke.
5.5 Welding
30 questions · AMeE0505
122. In fusion welding the joint is formed by:
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The base metal itself melts.
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Answer: B. Melting the edges of the parts, usually with filler metal
Fusion welding melts the base metal at the joint, which then solidifies to form a continuous bond.
123. Which of the following is NOT a basic type of welded joint?
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Think of carpentry.
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Answer: D. Dovetail joint
The basic weld joints are butt, lap, T, corner and edge; a dovetail is a woodwork joint.
124. An arc welding operation is performed at 24 V and 200 A, travel speed 4 mm/s and heat transfer efficiency 0.8. The heat input per unit length is:
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Power ÷ speed, with efficiency.
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Answer: A. 960 J/mm
H = ηVI/v = 0.8 × 24 × 200/4 = 960 J/mm.
125. In shielded metal arc welding (SMAW) the flux coating on the electrode is used to:
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Think of protection from air.
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Answer: C. Shield the weld pool from the atmosphere and stabilize the arc
The coating burns to give a gas shield and slag that protect the metal from oxygen and nitrogen.
126. Tungsten inert gas (TIG) welding uses:
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The electrode is not consumed.
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Answer: D. A non-consumable tungsten electrode with inert gas shielding
The tungsten electrode does not melt; filler is added separately and argon or helium protects the weld.
127. Submerged arc welding is characterized by:
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The arc is hidden.
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Answer: A. A granular flux blanket covering the arc, giving high deposition rates
The arc is buried under flux, so there is no spatter and high currents can be used; it suits long straight joints in thick plate.
128. An arc welding power source delivers 25 V at 150 A. The arc power is:
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Power = voltage × current.
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Answer: C. 3.75 kW
P = VI = 25 × 150 = 3750 W = 3.75 kW.
129. The temperature of a neutral oxy-acetylene flame is nearly:
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It is hotter than the melting point of steel.
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Answer: B. 3200 °C
The inner cone of a neutral oxy-acetylene flame reaches about 3100–3200 °C.
130. Which flame in gas welding has excess acetylene?
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More fuel than oxygen.
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Answer: C. Carburizing (reducing) flame
Excess acetylene leaves a feather beyond the inner cone; it adds carbon and is used for hard-facing.
131. In resistance spot welding the current is 8000 A, resistance 100 µΩ and weld time 0.2 s. The heat generated is:
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Joule heating.
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Answer: A. 1280 J
Q = I²Rt = 8000² × 100 × 10⁻⁶ × 0.2 = 1280 J.
132. Spot welding is widely used in:
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Thin overlapping sheets.
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Answer: B. Sheet metal assemblies such as car bodies
Overlapping sheets are clamped between electrodes and joined with a local nugget; it is easily automated.
133. In a fillet weld with leg length 10 mm, the effective throat thickness (for equal legs) is nearly:
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The throat is the shortest distance across a 45° triangle.
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Answer: D. 7.07 mm
Throat = 0.707 × leg = 7.07 mm.
134. The safe load of a butt weld in a 10 mm plate, weld length 150 mm, allowable stress 100 N/mm² is:
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Load = stress × throat × length.
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Answer: A. 150 kN
P = σ t l = 100 × 10 × 150 = 150 000 N.
135. A fillet weld has leg 8 mm, length 100 mm and allowable shear stress 80 N/mm². The load capacity is nearly:
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Use the throat as the weld section.
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Answer: C. 45.2 kN
P = 0.707 s l τ = 0.707 × 8 × 100 × 80 = 45.2 kN.
136. Brazing differs from soldering in that brazing uses a filler metal that melts:
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The dividing temperature is 450 °C.
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Answer: D. Above 450 °C but below the base metal melting point
Brazing fillers (e.g. brass, silver alloys) melt above 450 °C; solders melt below 450 °C. Neither melts the base metal.
137. What mass of acetylene is obtained from 3.2 kg of calcium carbide (CaC₂ = 64 g/mol) by CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂? (C₂H₂ = 26 g/mol)
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One mole of carbide gives one mole of acetylene.
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Answer: B. 1.3 kg
3200/64 = 50 mol of CaC₂ gives 50 mol of C₂H₂ = 50 × 26 = 1300 g.
138. Which defect is a groove melted into the base metal at the toe of a weld and left unfilled?
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It is cut under the plate surface.
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Answer: B. Undercut
Excessive current or arc length melts away base metal along the weld edge, leaving a notch that concentrates stress.
139. Porosity in a weld is mainly caused by:
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Gas dissolved in the molten pool.
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Answer: D. Gas trapped in the solidifying weld metal
Gases from moisture, contamination or poor shielding are trapped as bubbles when the metal freezes.
140. A weld 600 mm long is made in 2 minutes. The welding speed is:
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Convert minutes to seconds.
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Answer: A. 5 mm/s
v = 600 mm/(120 s) = 5 mm/s.
141. The heat-affected zone (HAZ) in a weld is the region:
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The metal did not melt.
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Answer: C. Of base metal not melted but changed in microstructure by heat
HAZ is next to the fusion line and shows grain growth or hardening, which may weaken the joint.
142. The volume of weld metal for an equal-leg fillet weld of 8 mm leg and 1 m length, density 7.85 g/cm³, gives a mass of nearly:
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Cross-sectional area is a right triangle.
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Answer: D. 251 g
Area = ½ × 8 × 8 = 32 mm²; volume = 32 000 mm³ = 32 cm³; m = 32 × 7.85 = 251 g.
143. A welding machine is rated 300 A at 60% duty cycle. The allowable current at 100% duty is nearly:
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Heating goes with I² × duty.
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Answer: A. 232 A
I100 = I × √(duty) = 300 × √0.6 = 232 A.
144. A power source draws 6.25 kW and delivers 5 kW to the arc circuit. Its efficiency is:
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Output over input.
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Answer: C. 80%
η = 5/6.25 = 0.8.
145. Which welding method uses a continuously fed consumable wire electrode with shielding gas?
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Metal inert gas.
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Answer: B. MIG/MAG (GMAW)
In gas metal arc welding the wire is both the electrode and filler metal and is fed automatically.
146. For steel with C = 0.2, Mn = 0.9, Cr = 0.1, Mo = 0.05, V = 0.05, Ni = 0.3, Cu = 0.15 (wt%), the carbon equivalent CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 is:
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Add the contributions.
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Answer: C. 0.42
CE = 0.2 + 0.15 + 0.04 + 0.03 = 0.42.
147. The purpose of flux in brazing is to:
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Oxides stop wetting.
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Answer: A. Clean oxides from surfaces and allow filler to wet them
Flux dissolves oxide films and prevents new oxidation so that the molten filler flows by capillary action.
148. Which brazing method heats the assembly in a molten salt bath?
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Parts are dipped.
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Answer: B. Dip brazing
Parts with pre-placed filler are immersed in a molten salt bath, which also acts as flux.
149. Soldering is typically done using an alloy of:
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Low-melting alloy of tin.
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Answer: D. Tin and lead (or tin-silver-copper)
Soft solders are tin-based and melt well below 450 °C; lead-free tin–silver–copper solders replace tin–lead for electronics.
150. Preheating before welding a medium carbon steel plate mainly helps to:
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Cooling too fast is harmful.
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Answer: A. Reduce cooling rate and the risk of cracking
Slower cooling reduces martensite formation in the HAZ and lowers residual stress.
151. Which of the following is a solid-state welding process?
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No melting occurs.
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Answer: C. Friction welding
In friction welding heat from rubbing under pressure joins parts without melting.
5.6 CAD/CAM (Computer-aided Design & manufacturing)
30 questions · AMeE0506
152. In additive manufacturing, a part is built by:
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The opposite of subtractive.
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Answer: B. Adding material layer by layer from a 3D model
AM builds objects from successive layers sliced from a CAD model, unlike subtractive machining.
153. The file format most commonly exported from CAD to rapid prototyping machines is:
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It describes the surface by triangles.
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Answer: D. STL
STL represents the surface as triangular facets and is the standard input for slicing software.
154. A part 60 mm high is built with a layer thickness of 0.2 mm. The number of layers is:
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Height ÷ layer thickness.
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Answer: A. 300
60/0.2 = 300 layers.
155. Stereolithography (SLA) works by:
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Light-sensitive resin.
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Answer: C. Curing liquid photopolymer with a UV laser
A UV laser traces each layer on the surface of a photosensitive resin, which solidifies; the platform lowers for the next layer.
156. Fused deposition modelling (FDM) builds parts by:
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Filament and nozzle.
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Answer: D. Extruding molten thermoplastic filament through a nozzle
A heated nozzle deposits filament layer by layer along the tool path.
157. Each layer in an additive process takes 30 s, and 300 layers are needed. The total build time is:
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Convert seconds to hours.
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Answer: A. 2.5 h
300 × 30 s = 9000 s = 2.5 h.
158. Which G-code produces a linear interpolation movement at the programmed feed rate in CNC machining?
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The 'cutting' straight-line move.
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Answer: C. G01
G00 is rapid positioning, G01 linear interpolation, G02 clockwise and G03 counter-clockwise circular interpolation.
159. A CNC tool moves from (10, 20) to (50, 60) in the XY plane in a straight line. The length of the move is:
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Use Pythagoras on the increments.
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Answer: B. 56.6 mm
Δx = 40, Δy = 40; length = √(40² + 40²) = 56.6 mm.
160. A CNC axis traverses 100 mm at a feed of 250 mm/min. The time taken is:
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Distance ÷ feed, then convert to seconds.
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Answer: C. 24 s
t = 100/250 = 0.4 min = 24 s.
161. In CNC programming G90 means:
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Compare G91.
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Answer: A. Absolute positioning
G90 sets coordinates relative to the program zero; G91 gives incremental positioning.
162. A circular arc of radius 25 mm and angle 90° is cut with G02/G03. The arc length is nearly:
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Arc length = radius × angle in radians.
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Answer: B. 39.3 mm
L = rθ = 25 × π/2 = 39.27 mm.
163. The spindle speed required for cutting speed 120 m/min with a 40 mm diameter cutter is nearly:
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Use N = 1000V/πD.
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Answer: D. 955 rpm
N = 1000V/(πD) = 120 000/(π × 40) = 955 rpm.
164. A stepper motor with 200 steps per revolution drives a lead screw of 5 mm pitch directly. The movement per step is:
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Pitch ÷ steps per revolution.
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Answer: A. 0.025 mm
5/200 = 0.025 mm per step.
165. An encoder of 1000 pulses/rev is coupled 1:1 to a 4 mm pitch lead screw. The position resolution is:
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Pitch ÷ pulses per revolution.
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Answer: C. 0.004 mm
Resolution = 4/1000 = 0.004 mm.
166. To move 25 mm with a basic length unit (BLU) of 0.01 mm, the controller must send:
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Distance ÷ BLU.
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Answer: D. 2500 pulses
25/0.01 = 2500 pulses.
167. A machine with an automatic tool changer that can perform milling, drilling and boring in one set-up is called a:
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It changes tools automatically.
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Answer: B. Machining centre
The ATC and multi-axis control let one machine perform many operations without re-fixturing.
168. In a closed-loop CNC system, compared with open-loop, the machine has:
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Feedback is the key.
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Answer: B. Feedback of actual position to the controller
Position sensors (encoders, resolvers) compare actual and commanded position and correct errors.
169. Computer numerical control (CNC) differs from conventional NC in that CNC:
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Computer inside the controller.
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Answer: D. Has a dedicated computer in the controller that stores and edits programs
The microcomputer allows program storage, editing and diagnostics at the machine.
170. Which of the following is a benefit of CAD/CAM?
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Think of integrated design and production.
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Answer: A. Reduced design and production lead time
A common geometric database moves directly from design to tool paths, saving time and reducing errors.
171. Which solid-modelling approach combines primitives using Boolean operations (union, intersection, difference)?
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Boolean operations.
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Answer: C. Constructive solid geometry (CSG)
CSG builds a solid as a tree of primitives combined by Boolean operations.
172. The geometric modelling type that contains the most complete information about volume and mass properties is:
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Interior is defined.
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Answer: D. Solid modelling
Only a solid model defines the inside and outside, allowing calculation of volume, mass and interference.
173. A point (4, 0) is rotated 90° counter-clockwise about the origin. Its new coordinates are:
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Apply the 2D rotation matrix.
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Answer: A. (0, 4)
x′ = x cosθ − y sinθ = 0; y′ = x sinθ + y cosθ = 4.
174. Rapid prototyping reduces the lead time of a product from 12 weeks to 3 weeks. The saving in lead time is:
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Saving is relative to the original time.
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Answer: C. 75%
(12 − 3)/12 = 75%.
175. A part of volume 40 cm³ needs support material of 10 cm³ in an ABS process (density 1.04 g/cm³). The total material mass consumed is:
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Include supports.
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Answer: B. 52 g
m = (40 + 10) × 1.04 = 52 g.
176. Selective laser sintering (SLS) differs from FDM in that SLS:
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The powder bed supports the part.
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Answer: C. Fuses powder particles with a laser and the surrounding powder supports the part
Unsintered powder acts as support, so complex shapes can be built without separate supports.
177. A flexible manufacturing system (FMS) typically consists of:
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Automated integrated cell.
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Answer: A. CNC machines, automated material handling and central computer control
FMS integrates workstations, AGVs or conveyors and a supervisory computer to produce a variety of parts.
178. An FMS of 2 machines each with a cycle time of 3 min per part works 20 h/day. The daily production is:
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Machines work in parallel.
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Answer: B. 800 parts
Each machine makes 20 × 60/3 = 400; two machines give 800.
179. The main advantage of FMS over a dedicated transfer line is:
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Flexibility.
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Answer: D. Ability to handle a variety of part types
FMS can switch among part types with little changeover time, while transfer lines suit a single high-volume part.
180. In CNC programming, the M03 code is used to:
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The 'M' means miscellaneous.
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Answer: A. Start spindle clockwise
M03 starts clockwise spindle rotation; M05 stops it; M30 ends the program.
181. According to the usual CNC axis convention, the Z axis is:
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Defined by the spindle.
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Answer: C. Along the spindle axis
Z is defined parallel to the spindle axis, with the positive direction away from the workpiece.