Category: Manufacturing Engineering Objective Questions

Manufacturing Processes Multiple Choice Questions on “Laser Welding – 6”.

1. Weld bead quality and the penetration are functions of _____
a) shroud gas velocity
b) material of the component
c) wavelength of the laser beam
d) power
Answer: d
Clarification: Penetration is a function of power and likewise the weld bead quality.
– Too much power causes vaporization and material ejection as in drilling
– Thus for welding the pulse is usually longer than for drilling and shaped to have a smaller initial peak.

2. Joining efficiency is dependent on _____
a) spot size
b) mode
c) polarisation
d) surface conditions
Answer: b
Clarification: In the laser welding process:
• The joining efficiency is greatly affected by the mode of the laser beam
• True TEM00 modes provide highest joining efficiencies.

3. Which of the following has no effect on laser welding?
a) Plane of polarisation
b) Joint geometries
c) Shroud gas composition
d) Wavelength
Answer: a
Clarification: Polarization should have no effect on laser welding since the beam is absorbed inside a keyhole and hence it should be absorbed regardless of the plane of polarization unlike cutting.

4. In some cases, polarisation can affect _____
a) weld fusion zones
b) power
c) focal position
d) spot size
Answer: a
Clarification: Some second order events have been noted by Beyer et al. which show the effect of polarization;
– variation in penetration
– the weld fusion zones are also wider for the case of s-polarisation (perpendicular to the plane of incidence).

5. At slow speeds, the beam is absorbed by _____ effects.
a) the george clooney
b) inverse Bremsstrahlung
c) acousto-optic
d) bridgman
Answer: b
Clarification: At slow speeds the plasma absorption dominates and the beam is absorbed by inverse Bremsstrahlung effects in the keyhole generating a plasma. Acousto-optic effects can be seen in the nonlinear optics field. Bridgman effect is used in electricity.

6. _____ is advantageous for CO₂ lasers.
a) High reflectivity
b) High absorptivity
c) Low reflectivity with short wave length
d) Low absorptivity with high wave length
Answer: c
Clarification: In welding with a conduction limited weld then the surface reflectivity becomes extremely important. The lower reflectivity with the shorter wavelengths gives a distinct advantage to Excimer, YAG or CO lasers over the CO₂ laser.

7. The effect of welding speed can be seen on ____
a) weld bead
b) penetration
c) power consumption
d) polarisation
Answer: a
Clarification: The effect of speed on the welding process is principally described by the
• Overall heat balance equation due to time for conduction
• Weld bead
• Shrouding high speed welds.

8. Which of the following is not affected by the speed of welding?
a) Pool size
b) Pool flow pattern
c) Weld bead
d) Focal position
Answer: d
Clarification: As the speed increases the pool flow pattern and size changes. At higher speeds, if the power is high enough and the pool large enough then undercut is produced which proceeds and edge freezing occurs leaving a slight undercut but the thread of the pool in the centre.

9. At slow speeds, smaller pool is formed.
a) True
b) False
Answer: b
Clarification: At slow speeds:
– the pool is large and wide and may result in drop out
– the ferrostatic head is too large for the surface tension to keep the pool in place and so it drops out of the weld leaving a hole or depression.

10. At high speeds, undercut occurs.
a) True
b) False
Answer: a
Clarification: At higher speeds:
– the strong flow towards the centre of the weld in the wake of the keyhole has no time to redistribute and is hence frozen as an undercut at the sides of the weld.

11. Which of the following lasers is suitable for welding of components made of aluminium?
a) Nd-YAG laser
b) Dye laser
c) Combination of CO and Excimer lasers
d) Combination of ruby and CO₂ lasers
Answer: c
Clarification: An Excimer and CO₂ laser beam combination showed improved coupling for the welding of high reflectivity materials, such as aluminium or copper could be obtained. Using two electron beams, the keyhole could be stabilized causing fewer waves on the weld pool and giving a better penetration and bead shape.

12. The improved coupling in case of welding of high reflectivity materials using twin beam lasers is due to _____
a) surface rippling
b) surface tension
c) surface smoothness
d) surface area
Answer: a
Clarification: The enhanced coupling was considered principally due to,
• altering the reflectivity by surface rippling caused by the Excimer
• a secondary effect coining from coupling through the Excimer generated plasma.

13. Which of the following is the problem on welding aluminium by laser welding?
a) Porosity
b) Cracking
c) Less grain growth
d) Brittle weld
Answer: a
Clarification: Following problems are faced while laser welding aluminium:
– High reflectivity
– Porosity
– Excessive fluidity.

14. Which of the following is the problem on welding heat resistant alloys by laser welding?
a) Brittle welds
b) Hot Cracking
c) Fluidity
d) High power consumption
Answer: a
Clarification: Following problems are faced while laser welding heat resistant alloys:
– Brittle welds
– Segregation problems
– Cracking.

15. Which of the following materials faces the problem of hot cracking while laser welding?
a) Inco 718
b) Steels
c) Hastelloy
d) Iridium alloys
Answer: d
Clarification: Problems faced during laser welding of materials like Inco718, Jetehet, M152, Hastelloy are-
– Weld is more brittle
– Segregation problems
– Cracking.
While laser welding Iridium alloys, there are chances of hot cracking. Steels can be laser welded very easily without any problems.

Manufacturing Processes Questions & Answers for Exams on “Hardening – 2”.

1. How many types of hardening processes are commonly used?
a) 3
b) 5
c) 6
d) 7
Answer: b
Clarification: There are several other ways the strength or the hardness of the surface can be increased without adversely affecting the toughness of the core. Some of the most common techniques are as follows:
• Induction hardening
• Case carburizing + case hardening
• Nitriding
• Shot peening
• Hard facing, coating or surface alloying.

2. Which of the following is not the purpose of the surface hardening?
a) To improve wear resistance
b) To increase fatigue life
c) Prevention from cracking
d) To improve ductility
Answer: d
Clarification: The purpose of surface hardening is to develop a hard surface with compressive residual stress, to improve its wear resistance, to increase its fatigue life and to avoid susceptibility to distortion and cracking.

3. How many surface hardening methods are there which are commonly used?
a) 2
b) 3
c) 4
d) 5
Answer: c
Clarification: The most commonly used methods of surface hardening are as follows:
• Shot peening: general applicable to all metals
• Coating / hard facing
• Surface (local) heating & cooling: steel
• Surface diffusion & subsequent treatment.

4. Shot peening technique is applicable to all metals and alloys that are prone to _____
a) plastic deformation
b) brittle failure
c) fatigue failure
d) ductile deformation
Answer: a
Clarification: Shot peening technique is applicable to all metals and alloys that are amenable to plastic deformation. The part to be hardened is placed in a chamber where extremely fine hard particles moving at a high speed keep striking at its surface. The energy of the moving particles is high enough to cause local plastic deformation at its surface.

5. Which of the following is not subjected to shot peening?
a) Landing gears of an aircraft
b) Automotive gears
c) Shafts
d) Coil springs
Answer: c
Clarification: BLanding gears of aircraft are subjected to shot peening to develop residual compressive stress on its surface. Even automotive gears, following carburizing, are subjected to hot peening to raise the value of compressive residual stress (to as high as 1000 – 1200 MPa particularly at depths of 30 – 40 microns. This help resist crack propagation during service as result of fatigue loading.

6. Which of the following is not the outcome of hard facing?
a) Improved resistance to particle erosion
b) Improved resistance to abrasion
c) Improved resistance to fretting
d) Improved resistance to plastic deformation
Answer: d
Clarification: Engineering components that are required to resist solid particle erosion, abrasion, fretting orcavitation are usually given a hard surface coating. This consists of a fine dispersion of hard metal carbides in a compatible metal matrix. Thermal spray is the most commonly used technique to apply such coatings on the component.

7. Which of the following is the most commonly used coating materials for hard facing?
a) Mixture of chromium carbide and cobalt
b) Mixture of calcium nitrate and iron carbide
c) Mixture of copper oxide and zinc oxide
d) Mixture of boron carbide and vanadium pentoxide
Answer: a
Clarification: The most commonly used coating materials are mixtures of chromium or tungsten carbides in either cobalt or nickel-chromium alloy matrix. Hard facing is also a commonly used technique to salvage worn out parts so that they could be reused.

8. Induction hardening is used for _____
a) steels
b) aluminium alloys
c) copper alloys
d) zinc alloys
Answer: a
Clarification: This is applicable only for steel. An induction coil is used to heat the component to be hardened. Only the surface gets heated. Its microstructure transforms into austenite from a mixture of ferrite and cementite, but the structure of the core remains intact as it remains cold all through the process.

9. During induction hardening, the microstructure of the surface gets transformed into _____
a) austenite
b) perlite
c) bainite
d) martensite
Answer: d
Clarification: Once the process is complete the microstructure of the surface gets transformed into martensite while that at its core remains unaltered. Hardness of induction hardened steel component may often be higher than that in through hardened steel having identical composition. One of the main advantages of induction hardening is good surface finish and little distortion.

10. Which of following is not the feature of induction hardening?
a) Heat the surface to a temperature above an austenitic region
b) Good surface finish
c) Fast heating & short hold time
d) Applicable to carbon steels having (0.8– 1% C)
Answer: d
Clarification: The salient features of induction hardening are as follows:
• Heat the surface to a temperature above austenitic region
• Core does not get heated: the structure remains unaltered
• Surface converts to martensite on quenching.
• Fast heating & short hold time: needs higher austenization temperature
• Martensite forms in fine inhomogeneous grains of austenite
• Applicable to carbon steels (0.35 – 0.7C)
• Little distortion & good surface finish.

11. The hardness of steel depends only on the ______
a) carbon content
b) temperature
c) yield strength
d) tensile strength
Answer: a
Clarification: The hardness of steel depends only on the concentration of carbon in steel. Therefore it may be enough to have high carbon only at the surface. This can be achieved by increasing the concentration of carbon in a component made of low carbon steel by allowing carbon to diffuse into it.

12. A major limitation of pack carburizing is poor control over_____
a) temperature
b) ductility
c) hardness
d) strength
Answer: a
Clarification: A major limitation of pack carburizing is poor control over temperature & carburization depth. On completion of the process, the steel parts are cooled slowly. Direct quenching is not possible as the job is surrounded by carburizing mixture packed in a sealed box having high thermal mass. This can be overcome by using a gaseous or liquid carburizing medium.

13. Which of the following is most commonly used as carburizing gas?
a) CH₄
b) CO
c) N₂
d) C₂H₂
Answer: a
Clarification: CH₄ and CO are the most commonly used carburizing gas. It is usually mixed with decarburizing (H₂ and CO₂) and neutral gases (N₂). This helps maintain close control over carbon potential. Gas carburization is done by keeping the samples at the carburizing temperature for a specified time in a furnace having a mixture of carburizing and neutral gas.

14. The main purpose of this stage is to harden the case consisting of austenite and globules of un‐dissolved carbide.
a) True
b) False
Answer: a
Clarification: The main purpose of case hardening is to harden the case. Therefore the component after case refining is heated to 30⁰C-40⁰C above the lower critical temperature. At this temperature, the case consists of austenite and globules of undissolved carbide. The structure of the core during this stage of heat treatment should have ferrite and austenite.

15. The reasons for case hardening is to develop compressive residual stress.
a) True
b) False
Answer: a
Clarification: One of the main reasons for case hardening is to develop compressive residual stress at the surface of the components that are subjected to fatigue loading. A general thumb rule is that the region that transforms last has a compressive stress. In the case of a carburized steel there is a large difference in the concentration of carbon at the surface and that at the centre. The difference is of large that although the surfaced on quenching cools faster it transforms to martensite later than the core. Since it transforms last it should be under compression. This is the reason why case hardened components have compressive residual stress.

 

To practice all exam questions on Manufacturing Processes,

Manufacturing Engineering Multiple Choice Questions on “Magnetic Properties of Material”.

1. Which of the following parameter is used to assess the magnetic ability of a material?
a) Magnetic flux density
b) Magnetization
c) Magnetic dipole moment
d) Susceptibility
Answer: d
Clarification: Magnetic susceptibility is a measure to quantify the ability of a material to undergo magnetization in an applied magnetic field. It is the ratio of magnetization (M) to the applied magnetic field intensity (H).

2. For a diamagnetic material, which of the following statement is correct?
a) Magnetic susceptibility < 0
b) Magnetic susceptibility > 0
c) Magnetic susceptibility = 0
d) Magnetic susceptibility = 1
Answer: a
Clarification: Diamagnetic materials are those which repel magnetic field and hence their magnetic susceptibility (χ) is negative.

3. For a diamagnetic material, which of the following statement is correct (μ_r = relative permeability)?
a) μ_r > 2
b) μ_r < 1
c) μ_r > 1
d) μ_r = 1
Answer: b
Clarification: A diamagnetic material has a constant relative permeability (μ_r) slightly less than 1.

4. For a paramagnetic material, which of the following statement is correct?
a) Magnetic susceptibility < 0
b) Magnetic susceptibility > 0
c) Magnetic susceptibility = 0
d) Magnetic susceptibility = -1
Answer: b
Clarification: Magnetic susceptibility (χ) is very small positive quantity for a paramagnetic material.

5. For a paramagnetic material, which of the following statement is correct (μ_r = relative permeability)?
a) μ_r = 0
b) μ_r < 1
c) μ_r > 1
d) μ_r < 0
Answer: c
Clarification: A paramagnetic material has a constant relative permeability (μr) slightly greater than 1.

6. What is the curie temperature of iron (in kelvin scale)?
a) 2195 K
b) 495 K
c) 895 K
d) 1095 K
Answer: d
Clarification: The curie temperature of iron is about 1095K. It changes its magnetic behaviour from ferromagnetic to paramagnetic.

7. With an increase in temperature, magnetic susceptibility of a ferromagnetic material ____________
a) Increases
b) Decreases
c) Remains constant
d) First increases and then decreases
Answer: a
Clarification: Magnetic susceptibility of a ferromagnetic material decreases with increase in temeprature.

8. With an increase in temperature, magnetic susceptibility of an anti-ferromagnetic material ____________
a) Increases
b) Decreases
c) First decreases and then increases
d) First increases and then decreases
Answer: d
Clarification: Susceptibility of an anti-ferromagnetic material is first increases and then decreases with increase in temperature.

9. With an increase in the area of hysteresis curve, power loss will ___________
a) Increases
b) Decreases
c) First decreases and then increases
d) First increases and then decreases
Answer: a
Clarification: Power loss is directly proportional to the area of hysteresis curve.

10. Magnetic Bubbles are used as __________
a) Storage device
b) Strain gauge
c) Thermostat
d) Potentiometer
Answer: a
Clarification: Magnetic bubbles are small magnetized areas used as storage devices (data bites). One good thing about magnetic bubbles are they do not disappear when power is turned off.

Manufacturing Engineering Multiple Choice Questions on “Machinability”.

1. The specific cutting energy used for establishing the machinability of the metal depends upon its
a) coefficient of friction
b) micro-structure
c) work hardening characteristics
d) all of the mentioned
Answer: d
Clarification: None

2. For machining a mild steel workpiece using carbide tool, the maximum material will be removed at a temperature of
a) 50⁰
b) 100⁰
c) 175⁰
d) 275⁰
Answer: b
Clarification: None

3. For machining a mild steel workpiece by a high speed steel tool, the average cutting speed is
a) 5 m/min
b) 10 m/ min
c) 15 m/min
d) 30 m/min
Answer: d
Clarification: For machining a cast iron workpiece by a high speed steel tool, the average cutting speed is 22 m/min and for a mild steel is 30 m/min.

4. For machining a cast iron workpiece by a high speed steel tool, the average cutting speed is
a) 10 m/min
b) 15 m/min
c) 22 m/min
d) 30 m/min
Answer: c
Clarification: For machining a cast iron workpiece by a high speed steel tool, the average cutting speed is 22 m/min and for a mild steel is 30 m/min.

5. The machining of titanium is difficult due to
a) high thermal conductivity of titanium
b) chemical reaction between tool and work
c) low tool-chip contact area
d) none of the mentioned
Answer: c
Clarification: None

6. The factor considered for evaluation of maintainability is
a) cutting forces and power consumption
b) tool life
c) type of chips and shear angle
d) all of the mentioned
Answer: d
Clarification: None

7. In machining metals, chips break due to _____________ of work material.
a) toughness
b) ductility
c) elasticity
d) work hardening
Answer: d
Clarification: None

8. In machining metals, surface roughness is due to
a) feed marks or ridges left by the cutting tool
b) fragment of built up edge on the machined surface
c) cutting tool vibrations
d) all of the mentioned
Answer: d
Clarification: None

9. In machining soft materials, a tool with a negative relief angle is used
a) True
b) False
Answer: a
Clarification: None

10. The tool material, for faster machining, should have
a) wear resistance
b) red hardness
c) toughness
d) all of the mentioned
Answer: d
Clarification: None

Manufacturing Engineering Question Bank on “Milling Operations”.

1. Diameter of milling cutter is 100 mm, running at 210 rpm. Cutting speed in m/min is equal to
a) 26
b) 23
c) 66
d) 78
Answer: c
Clarification: V= (3.14*D*N)/1000. V is the cutting speed, D is diameter and N is the revolution per minute.

2. Distance moved by table in mm in one minute in any direction is known as
a) Feed per minute
b) Feed per tooth
c) Feed per revolution
d) None of the mentioned
Answer: a
Clarification: Distance moved by table in mm in one minute in any direction is known as feed per minute.

3. Distance moved by a table in mm during time when cutter revolve through angle corresponding to distance between two cutting edges of two adjacent teeth is known as
a) Feed per minute
b) Feed per tooth
c) Feed per revolution
d) None of the mentioned
Answer: b
Clarification: Distance moved by a table in mm during time when cutter revolves through angle corresponding to distance between two cutting edges of two adjacent teeth is known as feed per tooth.

4. In a milling operation, feed per tooth is 0.020 mm and the total number of teeth on milling cutter is 50. Feed per revolution in mm is equal to
a) 0.2
b) 1.4
c) 1.0
d) 0.7
Answer: c
Clarification: Feed per revolution= feed per tooth * number of teeth.

5. In a milling operation, feed per revolution is 5 mm and the total number of teeth on milling cutter is 50. Feed per tooth in mm is equal to
a) 0.1
b) 0.2
c) 0.5
d) 0.05
Answer: a
Clarification: Feed per revolution= feed per tooth * number of teeth.

6. In a milling operation, feed per revolution is 0.05 mm and speed of 400 rpm. Feed per min in mm/min is equal to
a) 1
b) 2
c) 0.5
d) 0.05
Answer: b
Clarification: Feed per minute= feed per revolution* speed.

7. In a milling operation, feed per min is 10 mm and speed of 500 rpm. Feed per min in mm/min is equal to
a) 1
b) 2
c) 0.5
d) 0.04
Answer: d
Clarification: Feed per minute= feed per revolution* speed.

8. In a milling operation feed per tooth is .002 mm and number of teeth is 50 rotating with 60 rpm. Feed per min in mm/min is equal to
a) 3
b) 4
c) 6
d) None of the mentioned
Answer: c
Clarification: Feed per minute= feed per tooth* speed*number of teeth.

9. In a milling operation feed per tooth is .002 mm and number of teeth is 30 rotating with 40 rpm. Feed per min in mm/min is equal to
a) 3
b) 4.3
c) 2.4
d) None of the mentioned
Answer: c
Clarification: Feed per minute= feed per tooth* speed*number of teeth.

10. In a milling operation feed per revolution is 10 mm/rev and number of teeth is 50 rotating with 10 rpm. Feed per tooth in mm is equal to
a) .02
b) .04
c) .06
d) None of the mentioned
Answer: a
Clarification: Feed per minute= feed per tooth* speed*number of teeth.

To practice Manufacturing Engineering Question Bank,

Manufacturing Engineering Multiple Choice Questions on “Electrochemical Machining”.

1. Which of the following is un-conventional machining process?
a) Grinding
b) Milling
c) Turning
d) Electro chemical machining
Answer: d
Clarification: Electro chemical machining is an un-conventional machining process used for large material removal from the surface using electricity generated due to chemical reactions.

2. Which of the following is conventional machining process?
a) Electro chemical machining
b) Milling
c) Electron discharge machining
d) None of the mentioned
Answer: b
Clarification: Milling is a conventional machining process used for material removal from the surface.

3. In ECM, heavy electrical sparks are created.
a) True
b) False
Answer: b
Clarification: Sparks are generated in EDM, by using high voltage current, for proper erosion of material.

4. In ECM, tool does not touch the work piece.
a) True
b) False
Answer: a
Clarification: Tool is very near to work piece but does not touch it in ECM.

5. Which of the following is correct about ECM?
a) Erosion of metal takes place as a reverse process of electroplating
b) Thermal stresses are induced
c) Mechanical stresses are induced
d) None of the mentioned
Answer: a
Clarification: In ECM, erosion of metal takes place as reverse process of electroplating.

6. Which of the following material cannot be machined using electro chemical machining?
a) Iron
b) Aluminum
c) Copper
d) Wood
Answer: d
Clarification: Wood is a bad conductor of electricity and hence cannot be machined using electro chemical machining.

7. Which of the following material can be machined using electro chemical machining?
a) Iron
b) Rubber
c) Plastic
d) Wood
Answer: a
Clarification: Iron is a good conductor of electricity and hence can be machined using electro chemical machining.

8. Electrode gap in electro chemical machining is generally ranged from
a) 0.5 mm to 0.9 mm
b) 1.1 mm to 1.2 mm
c) 0.1 mm to 0.2 mm
d) 3.1 mm to 4.2 mm
Answer: c
Clarification: Electrode gap in electro chemical machining is generally ranged from 0.1 mm to 0.2 mm.

9. Electrolytes used in ECM must posses
a) Low electrical conductivity
b) Low chemical stability
c) High electrical conductivity
d) None of the mentioned
Answer: c
Clarification: Electrolytes used in ECM must have high electrical conductivity.

10. Which of the following is not a function of electrolyte in ECM?
a) It completes the circuit
b) It helps in electrochemical reaction
c) It carries away heat and waste product
d) It provide non reactive environment
Answer: d
Clarification: Electrolyte provides reactive environment for chemical reactions to takes place.