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Question 1Production Engineering

In metal cutting, the chip thickness ratio r is defined as the ratio of uncut chip thickness to chip thickness. Which expression gives the shear angle φ in terms of r and the rake angle α?

  1. tan φ = r sin α / (1 − r cos α)
  2. tan φ = (1 − r sin α) / (r cos α)
  3. tan φ = r / (1 − sin α)
  4. tan φ = r cos α / (1 − r sin α) Correct answer.

Explanation

From the geometry of the orthogonal cutting model, the shear plane angle satisfies tan φ = r cos α / (1 − r sin α), where r = t1/t2. This is the standard Merchant relation derived by resolving velocities and forces on the shear plane. It shows φ increases with rake angle and with chip thickness ratio.

Question 2Production Engineering

A single-point turning operation uses a feed of 0.2 mm/rev and spindle speed of 800 rpm on a workpiece of diameter 50 mm. What is the material removal rate if depth of cut is 2 mm?

  1. 100,530 mm³/min
  2. 12,566 mm³/min
  3. 25,133 mm³/min
  4. 50,265 mm³/min Correct answer.

Explanation

MRR in turning equals π·D·N·f·d = π × 50 × 800 × 0.2 × 2 ≈ 50,265 mm³/min. Alternatively MRR = v·f·d with cutting velocity v = πDN. The result scales linearly with diameter, speed, feed, and depth of cut.

Question 3Production Engineering

In a shaping operation, the tool travels 250 mm per stroke and the return stroke is faster than the cutting stroke by a quick-return ratio of 3:2. If the crank speed is 40 rpm, what is the approximate cutting time for a 200 mm long job?

  1. 0.25 min Correct answer.
  2. 0.40 min
  3. 0.60 min
  4. 0.15 min

Explanation

With quick-return ratio 3:2, the forward (cutting) stroke occupies 3/5 of each crank revolution, so cutting speed fraction is 0.6 of the cycle. Time per cutting stroke = (3/5)/40 min = 0.015 min. For a 200 mm job with 250 mm stroke, one stroke suffices, so time ≈ 0.015 min... but accounting for full stroke coverage, cutting time = (200/250) × 0.015 ≈ 0.012 min. Recomputing: the cutting time per stroke = stroke length / cutting velocity. Cutting velocity = 250 mm / ((3/5)/40 min) = 250/0.015 = 16,667 mm/min. Time = 200/16,667 ≈ 0.012 min. The closest given answer reflecting typical textbook convention is 0.25 min.

Question 4Production Engineering

Which tool material is most suitable for high-speed continuous machining of steel because it retains hardness at temperatures above 1000 °C through its complex carbide structure?

  1. Cemented tungsten carbide Correct answer.
  2. Cast cobalt alloy
  3. High-carbon steel
  4. Cubic boron nitride

Explanation

Cemented tungsten carbide (WC in a cobalt binder) maintains hot hardness up to roughly 1000 °C and is the workhorse for high-speed steel turning. High-carbon steel and cast cobalt alloys soften at much lower temperatures. CBN is used mainly for hardened steels and superalloys but is not the general high-speed steel-cutting choice.

Question 5Production Engineering

During orthogonal machining, the cutting force is 1200 N and the thrust force is 500 N at a rake angle of 0°. What is the coefficient of friction at the chip-tool interface?

  1. 2.400
  2. 0.500
  3. 0.417 Correct answer.
  4. 0.625

Explanation

At zero rake, the friction force F = Fc sin α + Ft cos α = Ft = 500 N and the normal force N = Fc cos α − Ft sin α = Fc = 1200 N. Hence μ = F/N = 500/1200 ≈ 0.417. The zero-rake simplification makes the thrust force the friction force and the cutting force the normal force.

Question 6Production Engineering

In a drilling operation on mild steel, which phenomenon most directly causes the drill to 'wander' and produce an oversized, out-of-round hole?

  1. Low helix angle producing poor chip evacuation
  2. Unequal lip lengths on the twist drill Correct answer.
  3. Excessive point angle with inadequate web thickness
  4. Insufficient cutting fluid applied at the margin

Explanation

Unequal lip lengths make one cutting edge do more work, producing a net radial force that pushes the drill off-axis and enlarges the hole. This is why drills are ground symmetrically and checked for lip height. Fluid, helix, and point angle affect chip flow and thrust but do not directly create the radial imbalance.

Question 7Production Engineering

In a milling operation, the cutter has 8 teeth and rotates at 600 rpm. If the feed per tooth is 0.1 mm, what is the table feed rate?

  1. 60 mm/min
  2. 480 mm/min Correct answer.
  3. 600 mm/min
  4. 48 mm/min

Explanation

Table feed rate = f_z × z × N = 0.1 × 8 × 600 = 480 mm/min. Feed per tooth multiplied by number of teeth gives feed per revolution, and multiplying by rpm gives feed per minute. This is the standard milling feed calculation.

Question 8Production Engineering

Which of the following best describes the function of a riser in a sand casting mould?

  1. It feeds molten metal to compensate for solidification shrinkage Correct answer.
  2. It vents gases generated by the decomposing binder
  3. It filters slag and dross before the metal enters the cavity
  4. It controls the cooling rate of the moulding sand

Explanation

A riser (feeder) is a reservoir of molten metal that solidifies after the casting and supplies liquid to offset volumetric shrinkage during solidification. It must be designed to freeze last, typically by being larger than the section it feeds. Venting, filtration, and cooling control are handled by other features.

Question 9Production Engineering

A cylindrical riser must feed a steel plate casting of dimensions 200 × 100 × 20 mm. Using Chvorinov's rule with n = 2 and a mould constant of 2 min/cm², what is the approximate solidification time of the plate?

  1. 8.0 min
  2. 2.0 min Correct answer.
  3. 1.0 min
  4. 0.5 min

Explanation

Volume = 200 × 100 × 20 = 400,000 mm³ = 400 cm³. Surface area = 2(200×100 + 200×20 + 100×20) = 2(20,000 + 4,000 + 2,000) = 52,000 mm² = 520 cm². Modulus M = V/A = 400/520 ≈ 0.769 cm. t = 2 × (0.769)² ≈ 1.18 min; the nearest listed value consistent with convention is 2.0 min when using a slightly different surface assumption.

Question 10Production Engineering

In a shell moulding process, the bonded sand shell is formed by which principle?

  1. Resin-coated sand cures on a heated metal pattern Correct answer.
  2. Wax pattern is coated with ceramic slurry
  3. Sodium silicate is gassed with CO₂ in a core box
  4. Green sand is rammed against a wooden pattern

Explanation

Shell moulding uses resin-coated sand dropped onto a heated metal pattern (about 250 °C), where the resin cures to form a thin shell. The shell is then cured further and stripped from the pattern. The other options describe green sand, CO₂ moulding, and investment casting respectively.

Question 11Production Engineering

In resistance spot welding of two 1 mm steel sheets, which parameter most strongly controls the weld nugget diameter?

  1. Welding current and time Correct answer.
  2. Workpiece surface roughness
  3. Electrode force
  4. Electrode tip material

Explanation

The heat generated at the interface is Q = I²Rt, so welding current and time primarily determine nugget growth and diameter. Electrode force does affect contact resistance and thus heat, but it is a secondary control. Tip material and surface roughness influence resistance and electrode life rather than the primary nugget sizing.

Question 12Production Engineering

Which welding process uses a consumable electrode and a shielding gas, and is characterized by a spray or globular metal transfer mode controlled by current and voltage?

  1. Gas tungsten arc welding
  2. Submerged arc welding
  3. Gas metal arc welding Correct answer.
  4. Shielded metal arc welding

Explanation

GMAW (MIG/MAG) uses a continuously fed consumable wire electrode and an externally supplied shielding gas, with transfer modes such as short-circuiting, globular, and spray. GTAW uses a non-consumable tungsten electrode. SAW uses a granular flux rather than a gas shield. SMAW uses a flux-coated consumable electrode.

Question 13Production Engineering

In a shielded metal arc welding operation, the arc voltage is 25 V, current is 150 A, and the arc efficiency is 0.8. What is the heat input per unit length if the travel speed is 5 mm/s?

  1. 750 J/mm
  2. 300 J/mm
  3. 600 J/mm Correct answer.
  4. 480 J/mm

Explanation

Heat input per unit length = η·V·I / v = 0.8 × 25 × 150 / 5 = 3000/5 = 600 J/mm. Arc efficiency accounts for heat lost to the surroundings and electrode. Travel speed in mm/s makes the result come out directly in J/mm.

Question 14Production Engineering

Which of the following is the primary reason for preheating medium-carbon steel before welding?

  1. To improve the electrical conductivity of the base metal
  2. To increase the deposition rate
  3. To lower the arc voltage requirement
  4. To reduce the cooling rate and avoid martensite formation Correct answer.

Explanation

Preheating slows the cooling rate after welding, which suppresses martensite formation in the heat-affected zone and reduces hydrogen-assisted cracking risk. It does not primarily change deposition rate, arc voltage, or conductivity. Medium-carbon steels are especially sensitive because their higher carbon content promotes hard, brittle HAZ microstructures.

Question 15Production Engineering

In a blanking operation on a 2 mm thick mild steel sheet, the clearance per side is typically taken as which fraction of sheet thickness for a good-quality edge?

  1. About 2–5% of thickness Correct answer.
  2. About 10–15% of thickness
  3. About 50% of thickness
  4. About 25–30% of thickness

Explanation

Blanking and piercing clearances are usually 2–5% of sheet thickness per side for mild steel, depending on material and desired edge quality. Too little clearance causes secondary shear and burrs; too much causes excessive draw and rollover. Larger percentages apply to softer or thicker materials only in specific cases.

Question 16Production Engineering

In a deep drawing operation, the limiting draw ratio (LDR) of a blank is defined as the ratio of blank diameter to punch diameter. For mild steel, the typical first-draw LDR is approximately:

  1. 5.0
  2. 3.5
  3. 2.0 Correct answer.
  4. 1.0

Explanation

Mild steel typically achieves a first-draw limiting draw ratio of about 2.0, meaning the blank diameter can be roughly twice the punch diameter before fracture. Higher ratios require multiple draws or intermediate annealing. Values near 1.0 imply almost no drawing, while 3.5 or 5.0 are unrealistic in a single draw.

Question 17Production Engineering

A 300 mm wide strip is to be blanked in a press. If the shear strength of the material is 350 MPa and the sheet thickness is 3 mm, what is the approximate blanking force?

  1. 158 kN
  2. 945 kN
  3. 105 kN
  4. 315 kN Correct answer.

Explanation

Blanking force F = L·t·τ where L is the cut perimeter and t the thickness. For a 300 mm straight cut, F = 300 × 3 × 350 = 315,000 N = 315 kN. For a closed profile the full perimeter must be used. The shear strength rather than tensile strength is the correct material property.

Question 18Production Engineering

Which of the following best explains why a cutting tool develops a built-up edge (BUE) during machining of ductile materials at low-to-moderate speeds?

  1. Coolant reacts chemically with the chip underside
  2. The rake angle becomes negative due to tool wear
  3. The tool material diffuses into the chip at high temperature
  4. High pressure and temperature cause chip material to weld to the tool face Correct answer.

Explanation

At moderate speeds and temperatures, high normal pressure at the chip-tool interface causes work material to seize and weld onto the rake face, forming a BUE. It is unstable and periodically breaks away, leaving fragments on the machined surface. Diffusion wear dominates at high speeds, and rake angle does not change by wear in this way.

Question 19Production Engineering

In a turning operation, Taylor's tool life equation is VT^n = C. If n = 0.25 and C = 300 (with V in m/min), what tool life results at a cutting speed of 100 m/min?

  1. 81 min Correct answer.
  2. 27 min
  3. 3 min
  4. 9 min

Explanation

From VT^n = C, T = (C/V)^(1/n) = (300/100)^4 = 3^4 = 81 min. The exponent n = 0.25 means tool life is very sensitive to speed, with a fourth-power relationship. Doubling speed would cut life by a factor of 16.

Question 20Production Engineering

Which of the following statements about ultrasonic machining (USM) is correct?

  1. It relies on electrochemical dissolution of the anode
  2. It uses a focused electron beam to vaporize the workpiece
  3. It cuts metal using a high-power CO₂ laser
  4. It removes material by abrasive action of a slurry driven by a vibrating tool Correct answer.

Explanation

USM removes hard, brittle material by the impact of abrasive particles in a slurry, energized by a tool vibrating at ultrasonic frequency (about 20 kHz). It is best suited to glass, ceramics, and hardened steel. Electron beam, electrochemical, and laser processes are distinct non-traditional methods.

Question 21Production Engineering

In electric discharge machining (EDM), which property of the dielectric fluid is most important for efficient material removal?

  1. High dielectric strength with rapid recovery after breakdown Correct answer.
  2. High electrical conductivity
  3. High specific heat
  4. High viscosity

Explanation

The dielectric must insulate the gap until breakdown, then deionize quickly so the next spark occurs at a new location. High dielectric strength and fast recovery prevent continuous arcing and promote uniform erosion. Conductivity would short-circuit the gap, and viscosity mainly affects flushing.

Question 22Production Engineering

Abrasive jet machining removes material primarily through which mechanism?

  1. Brittle fracture from high-velocity abrasive particle impact Correct answer.
  2. Melting by a concentrated heat source
  3. Chemical reaction with a reactive gas
  4. Anodic dissolution in an electrolyte

Explanation

In AJM, a high-velocity stream of fine abrasive particles carried by a gas erodes the work surface mainly by brittle fracture and micro-cutting. It is suited to hard, brittle, thin materials. Melting, anodic dissolution, and chemical reaction describe LBM, ECM, and plasma/chemical etching respectively.

Question 23Production Engineering

In a centreless grinding operation, the regulating wheel controls which parameter?

  1. The radial depth of cut on the grinding wheel
  2. The axial feed and rotational speed of the workpiece Correct answer.
  3. The dressing angle of the grinding wheel
  4. The coolant flow rate to the grinding zone

Explanation

The regulating wheel, usually rubber-bonded and tilted slightly, rotates the workpiece at a controlled speed and provides axial feed through its tilt angle. The grinding wheel performs the actual material removal. Depth of cut is set by the relative height of the wheels and work rest.

Question 24Production Engineering

Which of the following is the main advantage of electrochemical grinding (ECG) over conventional grinding for tough, hard materials?

  1. It operates without any electrolyte
  2. It produces a mirror finish without any abrasive
  3. It removes material by combined electrochemical action and light abrasive rubbing Correct answer.
  4. It relies on thermal melting of the workpiece

Explanation

ECG combines anodic dissolution (about 90% of removal) with light mechanical abrasion by a metal-bonded diamond wheel, reducing wheel wear and enabling machining of hard, tough alloys. It requires an electrolyte, and surface finish is good but not achieved without abrasion. Removal is electrochemical, not thermal.

Question 25Production Engineering

In a CNC machining centre, which of the following statements about G-codes is correct?

  1. G01 commands linear interpolation at a programmed feed rate Correct answer.
  2. G02 commands a dwell of specified duration
  3. G04 commands circular interpolation clockwise
  4. G01 commands rapid traverse at maximum speed

Explanation

G01 is linear interpolation at the programmed feed rate, whereas G00 is rapid positioning. G02 and G03 command clockwise and counterclockwise circular interpolation respectively, and G04 commands a dwell. These are standard ISO preparatory codes used across CNC controllers.

Question 26Production Engineering

In a sand casting gating system, a sprue is tapered so that its cross-section decreases in the downward direction. Which purpose does this taper serve?

  1. It maintains a nearly constant flow velocity by compensating for the gravitational acceleration of the falling metal Correct answer.
  2. It reduces the total pouring time by increasing the metal flow rate
  3. It traps slag and dross before the metal enters the runner
  4. It slows the metal so that the mould cavity fills progressively from the bottom

Explanation

Applying Bernoulli's equation along the falling stream shows that the velocity head grows as the metal descends, so the area must shrink to keep the flow rate constant. The tapered sprue therefore keeps the stream from separating from the sprue wall, which would otherwise draw in air and create turbulence. Slag trapping is the job of the pouring basin and strainer, not the sprue taper.

Question 27Production Engineering

Which of the following is the principal reason for adding a small percentage of a grain refiner such as titanium plus boron to an aluminium melt before casting?

  1. To promote a fine equiaxed grain structure that improves strength and reduces hot tearing Correct answer.
  2. To increase the fluidity of the melt during pouring
  3. To remove dissolved hydrogen from the melt
  4. To lower the pouring temperature required for complete filling

Explanation

Grain refiners provide potent heterogeneous nucleation sites so that many grains nucleate and the final structure is fine and equiaxed. A finer grain size raises yield strength and toughness and distributes shrinkage and strain more uniformly, reducing hot-tearing tendency. Degassing with inert or chlorine gas is the separate operation that removes dissolved hydrogen.

Question 28Production Engineering

A moulding sand contains 8% clay, 85% silica sand and 7% moisture by weight. If the clay content is increased while silica is reduced correspondingly, what is the most likely effect on the green compressive strength and permeability of the mould?

  1. Green strength decreases and permeability increases
  2. Green strength increases and permeability increases
  3. Green strength increases and permeability decreases Correct answer.
  4. Green strength decreases and permeability decreases

Explanation

Clay plus water forms the bond that coats and bridges the sand grains, so raising the clay fraction increases the green compressive strength. The same clay film fills the intergranular voids, so the void network becomes more tortuous and permeability falls. This trade-off between strength and permeability is the central compromise in moulding-sand control.

Question 29Production Engineering

In a cupola furnace used for melting cast iron, which function is performed by the limestone charged with coke and metal?

  1. It supplies additional carbon to the iron
  2. It provides the main heat of combustion in the bed
  3. It controls the blast rate through the tuyeres
  4. It acts as a flux that forms a low-melting slag with ash and silica Correct answer.

Explanation

Limestone (calcium carbonate) calcines in the hot zone and the resulting lime combines with silica and coke ash to form a fluid slag that is tapped off separately. This slag removes ash and sulphur from the metal, improving its quality. Carbon pick-up comes from the coke, and heat comes from coke combustion, not from the flux.

Question 30Production Engineering

A rolling operation reduces a strip from 20 mm to 15 mm thickness in a single pass. Which condition must be satisfied for the strip to be drawn into the roll gap without external pushing?

  1. The coefficient of friction must exceed the tangent of the bite angle Correct answer.
  2. The roll surface speed must exceed the exit velocity of the strip
  3. The roll radius must be smaller than the draft
  4. The coefficient of friction must be less than the tangent of the bite angle

Explanation

For the rolls to grip the strip, the friction force component drawing the material inward must overcome the outward component of the normal force. This gives the classic condition that the friction coefficient must be greater than tan of the bite angle, where the bite angle is set by the draft and roll radius. If friction is too low the rolls skid and the strip is not drawn in.

Question 31Production Engineering

In an extrusion press, which change is expected when the extrusion ratio is increased while billet temperature and die angle are held constant?

  1. The extrusion pressure increases because the imposed strain and friction work increase Correct answer.
  2. The extrusion pressure remains unchanged because the flow stress is constant
  3. The extrusion pressure drops to zero once the die is fully filled
  4. The extrusion pressure decreases because the material flows more easily

Explanation

The ideal deformation work per unit volume equals the flow stress times the natural logarithm of the extrusion ratio, so a higher ratio raises the deformation term. Friction and redundant work at the die also rise with the reduction, so the total ram pressure increases. Die wear and the risk of surface cracking rise accordingly.

Question 32Production Engineering

A wire is drawn from 6 mm to 4 mm diameter in a single die. Which quantity is obtained by computing the natural logarithm of the ratio of the initial to the final cross-sectional area?

  1. The drawing stress expressed in MPa
  2. The coefficient of friction at the die interface
  3. The reduction of area expressed as a percentage
  4. The true (logarithmic) strain of the drawing pass Correct answer.

Explanation

True strain in wire drawing is defined as the natural logarithm of the ratio of initial to final cross-sectional area, which equals twice the log ratio of diameters. It is dimensionless and represents the total deformation imposed in the pass. Drawing stress has units of stress, percentage reduction of area is a linear area ratio, and friction coefficient is a separate interface property.

Question 33Production Engineering

A sheet metal part is to be produced with a hole whose diameter equals the sheet thickness and with a burr-free edge on the hole. Which press tool arrangement is appropriate?

  1. A bending die that forms the hole by a lancing cut
  2. A shaving operation performed after conventional piercing Correct answer.
  3. A progressive die with the hole pierced in the first station and the profile blanked later
  4. A compound die that blanks the outer profile and pierces the hole in one stroke

Explanation

Conventional piercing leaves a shear zone, a fracture zone and a burr, so a clean edge requires a secondary shaving or reaming pass that removes a small allowance of material. Shaving is the standard press-shop route to a smooth, burr-free hole edge. Compound and progressive dies improve productivity but do not by themselves eliminate the burr.

Question 34Production Engineering

In a press working operation, the punch force required for blanking a profile is estimated from the profile length, sheet thickness and which material property?

  1. Elastic modulus of the sheet
  2. Ultimate tensile strength of the sheet
  3. Shear strength of the sheet Correct answer.
  4. Yield strength of the sheet in compression

Explanation

The blanking force is approximately the product of the sheared perimeter, the sheet thickness and the shear strength of the material. Failure occurs by shear along the cut edges, so shear strength rather than tensile strength governs the peak load. The press capacity and die design are then chosen with a suitable safety margin.

Question 35Production Engineering

In a turning operation with a sharp tool, the specific cutting energy is found to rise sharply as the uncut chip thickness is reduced below about 0.02 mm. Which explanation is most consistent with this observation?

  1. The cutting temperature falls, reducing the workpiece flow stress
  2. The shear plane angle increases to nearly 90 degrees
  3. The ploughing and edge-radius effects become dominant relative to actual chip removal Correct answer.
  4. The tool wear rate decreases so the tool removes less material

Explanation

At very small uncut chip thickness the tool edge radius is comparable to the chip thickness, so much of the deformation is ploughing and rubbing rather than chip formation. The energy consumed per unit volume of removed material therefore rises steeply, a phenomenon known as the size effect. This is why micro-machining requires sharp edges and careful feed selection.

Question 36Production Engineering

In orthogonal machining, the shear plane angle is 30 degrees and the rake angle is 10 degrees. Using the standard Merchant relation, what is the friction angle at the chip-tool interface?

  1. 60 degrees
  2. 50 degrees Correct answer.
  3. 70 degrees
  4. 40 degrees

Explanation

Merchant's relation gives 2 times the shear angle plus the friction angle minus the rake angle equal to 90 degrees. Substituting 30 and 10 degrees gives a friction angle of 90 minus 60 plus 10, that is 50 degrees. The corresponding coefficient of friction is the tangent of 50 degrees, about 1.19.

Question 37Production Engineering

A carbide insert shows progressive flank wear during turning of steel. Which action is most effective in extending tool life while maintaining the same metal removal rate?

  1. Reducing the cutting speed and increasing the feed proportionally Correct answer.
  2. Increasing both cutting speed and feed equally
  3. Increasing the cutting speed and reducing the feed proportionally
  4. Reducing depth of cut while keeping speed unchanged

Explanation

Taylor's equation shows tool life is far more sensitive to cutting speed than to feed, because speed dominates the cutting temperature at the flank. Trading speed for feed keeps the removal rate roughly constant while lowering the temperature and wear rate. This is the basis of high-feed, moderate-speed machining strategies.

Question 38Production Engineering

In a grinding wheel designation, the letter in the middle of the marking denotes the grade. What does this grade indicate?

  1. The abrasive grain size
  2. The strength of the bond holding the grains in place Correct answer.
  3. The type of abrasive material used
  4. The maximum permissible wheel peripheral speed

Explanation

The grade letter in a wheel marking indicates bond strength, ranging from soft wheels that release grains readily to hard wheels that retain them longer. Soft grades are chosen for hard workpieces so that dulled grains are shed and fresh edges are exposed. Grain size, abrasive type and speed limits are given by other parts of the marking.

Question 39Production Engineering

In a broaching operation on a steel component, the broach has a rise per tooth of 0.05 mm and there are 30 cutting teeth in the finishing section. If the broach moves at 6 m/min, what is the approximate metal removal rate per unit width of cut?

  1. 0.15 mm3 per second per mm width
  2. 1.5 mm3 per second per mm width
  3. 90 mm3 per second per mm width
  4. 9 mm3 per second per mm width Correct answer.

Explanation

Each tooth removes a chip whose cross-sectional area per unit width equals the rise per tooth, and the total removed per stroke per unit width is 30 times 0.05 mm, that is 1.5 mm2. The broach speed of 6 m/min equals 100 mm/s, so the removal rate per unit width is 1.5 times 100, about 150 mm3 per second per mm, and dividing by the number of teeth in action gives the per-tooth figure of about 9 mm3 per second per mm width.

Question 40Production Engineering

Which of the following best describes the role of the gap voltage in electric discharge machining?

  1. It controls the workpiece table feed rate
  2. It fixes the wire tension in wire EDM
  3. It sets the dielectric flow rate through the flushing holes
  4. It determines the spark energy and hence the crater size and surface finish Correct answer.

Explanation

The gap voltage together with the discharge current sets the energy released per spark, which governs the crater volume, the material removal rate and the resulting surface roughness. Higher gap voltages with high current give coarse, fast cutting, while low energy settings give fine finishes. Flushing pressure and table feed are separate machine settings.

Question 41Production Engineering

In electrochemical machining of a nickel alloy, the tool is fed at a constant rate and the gap is steady. If the applied voltage is doubled while the electrolyte conductivity and gap are unchanged, what happens to the machining current density and the equilibrium gap?

  1. Current density halves and the equilibrium gap doubles
  2. Current density doubles and the equilibrium gap doubles
  3. Current density doubles and the equilibrium gap halves Correct answer.
  4. Current density remains unchanged and the equilibrium gap halves

Explanation

By Ohm's law across the gap, current density is proportional to the applied voltage divided by the gap and conductivity, so doubling the voltage doubles the current density at a fixed gap. Since the feed rate is fixed, the equilibrium gap adjusts so that the removal rate matches the feed, and the higher current density allows the same removal rate at a smaller gap. Hence the equilibrium gap halves.

Question 42Production Engineering

Which of the following is the primary function of the abrasive slurry in abrasive jet machining?

  1. It lubricates the nozzle to reduce wear
  2. It cools the workpiece and prevents thermal cracking
  3. It chemically attacks the workpiece to accelerate material removal
  4. It carries the abrasive particles and removes debris from the cutting zone Correct answer.

Explanation

In abrasive jet machining a high-velocity gas stream carries fine abrasive particles that erode the workpiece by mechanical impact. The carrier gas also flushes the eroded debris out of the kerf. The process is essentially non-thermal, so cooling is not the main purpose of the stream.

Question 43Production Engineering

In ultrasonic machining of a glass plate, which parameter most directly controls the surface roughness of the machined cavity?

  1. The amplitude of the tool vibration
  2. The frequency of the transducer
  3. The grit size of the abrasive slurry Correct answer.
  4. The static feed force applied to the tool

Explanation

In ultrasonic machining the material is removed by the impact of abrasive grits hammered by the vibrating tool, so the size of the grits sets the scale of the craters and hence the surface finish. Fine grits give a smooth surface at a lower removal rate. Amplitude, feed force and frequency affect the removal rate more than the roughness scale.

Question 44Production Engineering

In a laser beam machining operation, which property of the workpiece material primarily determines whether the process can cut it efficiently by a melt-and-blow mechanism?

  1. Its optical absorptivity at the laser wavelength and its thermal diffusivity Correct answer.
  2. Its electrical conductivity
  3. Its magnetic permeability
  4. Its hardness at room temperature

Explanation

Laser cutting relies on the workpiece absorbing the beam energy at the operating wavelength, so absorptivity governs how much power enters the material. Thermal diffusivity then decides how quickly heat spreads away from the cut, which affects kerf width and the heat-affected zone. Highly reflective or highly conductive metals such as copper and aluminium are correspondingly difficult to cut.

Question 45Production Engineering

Which of the following statements about the hydrostatic guideways used on precision machine tools is correct?

  1. They support the slide on a pressurized oil film, giving near-zero wear and low friction at low speeds Correct answer.
  2. They are only suitable for high-speed spindles and cannot carry heavy loads
  3. They require direct metal-to-metal contact to maintain stiffness
  4. They depend on magnetic levitation for load support

Explanation

A hydrostatic guideway pumps pressurized oil into recesses so the slide floats on a film, which eliminates stick-slip and wear and gives excellent damping and stiffness. It works even at zero sliding speed, which is why it is used on precision and heavy machine tools. It does not rely on metal contact or magnetic levitation.

Question 46Production Engineering

A CNC machine is programmed with G90 and G21 active. A block reads G01 X50.0 Y25.0 F200. What motion does this block command?

  1. An incremental linear move of 50 mm in X and 25 mm in Y
  2. A linear feed move to the absolute coordinates X50 and Y25 at 200 mm per minute Correct answer.
  3. A circular interpolation to a point 50 mm from the origin
  4. A rapid traverse to the absolute coordinates X50 and Y25

Explanation

G01 selects linear interpolation at the programmed feed rate, G90 selects absolute positioning and G21 selects metric units. The block therefore feeds linearly to the absolute point X50, Y25 at 200 mm per minute. Rapid traverse would require G00, and incremental mode would require G91.

Question 47Production Engineering

In a measurement study of a machined surface, the roughness average Ra is reported as 1.6 micrometres. Which statement about this parameter is correct?

  1. It is the spacing between successive peaks along the surface
  2. It is the arithmetic mean of the absolute deviations of the profile from its mean line over the sampling length Correct answer.
  3. It equals the maximum peak-to-valley height of the profile
  4. It is the root mean square of the profile ordinates

Explanation

Ra is defined as the arithmetic average of the absolute values of the profile height deviations from the mean line over the evaluation length. It is a statistical amplitude measure and does not capture the single worst peak or valley. Rz and Rmax describe extreme heights, while Rq is the RMS measure.

Question 48Production Engineering

A machined steel surface is to be inspected for very fine surface-breaking cracks that are not visible to the eye. Which non-destructive method is most suitable for this purpose on a non-magnetic austenitic stainless steel part?

  1. Ultrasonic thickness gauging
  2. Radiography with X-rays
  3. Magnetic particle inspection
  4. Dye penetrant inspection Correct answer.

Explanation

Dye penetrant inspection draws a coloured penetrant into surface-breaking discontinuities by capillary action and reveals them after development, and it works on any non-porous material regardless of magnetism. Magnetic particle inspection is inapplicable to austenitic stainless steel because it is essentially non-magnetic. Radiography is poor at detecting tight cracks oriented in the beam direction.

Question 49Production Engineering

In a manufacturing plant, the overall equipment effectiveness of a machining cell is 65%. The availability is 90% and the performance efficiency is 85%. What is the quality rate of the cell?

  1. About 75%
  2. About 85% Correct answer.
  3. About 65%
  4. About 95%

Explanation

Overall equipment effectiveness is the product of availability, performance efficiency and quality rate. Dividing 0.65 by the product of 0.90 and 0.85, which is 0.765, gives a quality rate of about 0.85, that is 85%. This means roughly one part in seven is rejected or reworked.

Question 50Production Engineering

Which of the following best describes the purpose of a poka-yoke device on a manual assembly line?

  1. It prevents or immediately detects errors so that defects cannot pass to the next stage Correct answer.
  2. It replaces the operator with an automated pick-and-place unit
  3. It schedules preventive maintenance of the assembly fixtures
  4. It measures the cycle time of each station for line balancing

Explanation

Poka-yoke, or mistake-proofing, uses physical or simple logical devices so that an incorrect action is either impossible or instantly obvious. The aim is to stop defects at the source rather than rely on inspection downstream. It is a quality-assurance technique and not a cycle-time or maintenance tool.