250+ TOP MCQs on Micromachining – 2 and Answers

Manufacturing Processes Multiple Choice Questions & Answers on “Micromachining – 2”.

1. There is no thermal damage to the workpiece in IBM.
a) True
b) False
Answer: a
Clarification: In case of IBM. An ion hits an atom at the top surface of the workpiece and removes the material atom by atom or in the groups of atoms. Therefore, there is no thermal damage to the workpiece in IBM.

2. Thermal micromachining processes can be used to produce reduced hole diameter, lower hole pitch.
a) True
b) False
Answer: a
Clarification: Reduced hole diameter, lower hole pitch and longer head can be manufactured by thermal micromachining processes. These processes are used to manufacture computer hard disc drive heads, inkjet printer heads, sensors, infrared images.

3. Chemical micromachining is used for engraving the metal.
a) True
b) False
Answer: a
Clarification: It is an ancient process being used for engraving the metal for making ornaments and other products. It removes material in a controlled manner by the application of maskant and etchant.

4. It is required to remove material in the form of atoms for finishing of the surface.
a) True
b) False
Answer: a
Clarification: To finish surfaces to nano level, it is required to remove material in the form of atoms or molecules individually or in groups. Most of the nano finishing processes are using abrasive particles either suspended inliquid or held by the viscoelastic material, carbonyl iron particles, or by the magnetorheological fluid as a carrier.

5. Magneto rheological abrasive finishing (MRF) is a magnetic field assisted process.
a) True
b) False
Answer: a
Clarification: MRF is a deterministic and magnetic field assisted precision finishing process. MRF uses MRP fluid which is invented by Rabinow in late 1940s consist of CIP (Magnetic), abrasive (non-magnetic), carrierliquid (Oilorwater), additives (glycerol, grease).

6. MRF is used for finishing of brittle materials.
a) True
b) False
Answer: a
Clarification: MRF has been used for finishing a large variety of brittle material ranging from optical glasses to hard crystals. The major limitation of this process is that Internal and especially complex surfaces can’t be finished.

7. Abrasive flow machining (AFM) is used for _____
a) de-burring
b) etching
c) drilling
d) cutting
Answer: a
Clarification: AFM was developed by Extude Home Corporation USA in 1960 as a method to deburr, polish and radius difficult to reach surface like intricate geometries and edges by flowing an abrasive laden viscoelastic medium over them.

8. In chemo-mechanical polishing (CMP) process, material is removed due to abrading.
a) True
b) False
Answer: a
Clarification: CMP uses both chemical and mechanical type for material removal mechanism. Chemical reaction is used to soften material and then mechanically polish off this layer. Mechanical removal takes place due to abrading.

9. CMP is used for flat surfaces only.
a) True
b) False
Answer: a
Clarification: CMP is not deterministic in nature. End point of CMP is difficult to control for a desired thickness. This process is used for only flat surfaces. Basically, CMP is used to polish the silicon wafer.

10. MAF was developed to produce efficiently and economically good quality finish material.
a) True
b) False
Answer: a
Clarification: MAF was developed to produce efficiently and economically good quality finish on the internal and external surface of tubes as well as flat surface made of magnetic or non-magnetic material.

11. Magnetic Float Polishing is a technique based on_____
a) magneto-dynamic behaviour
b) magneto-hydrodynamic behaviour
c) kinematic behaviour
d) viscosity
Answer: b
Clarification: Magnetic Float Polishing is a technique based on the Magnet on hydrodynamic behaviour of the magnetic fluid which in the presence of magnetic field can levitate a non-magnetic float and abrasive particles suspended in it.

12. Ferromagnetic particles are attracted towards the area of a higher magnetic field.
a) True
b) False
Answer: a
Clarification: When the magnetic field is applied the ferromagnetic particles in the ferrofluid are attracted downward to the area of higher magnetic field and upward buoyant force is exerted on all non-magnetic materials to push them to the area of lower magnetic field.

13. MRAFF is the hybrid finishing process of MRF and AFF.
a) True
b) False
Answer: a
Clarification: MRAFF is the hybrid finishing process to take advantage of both the finishing processes MRF (magneto-rheological finishing) & AFF (abrasive flow finishing). Any complex geometry can be finished by this process.

14. In magnetic flow polishing process, very small force is applied by the abrasives.
a) True
b) False
Answer: a
Clarification: The forces applied by abrasives are extremely small and controllable. The balls are polished by the abrasive particles mainly due to the action of the magnetic buoyancy force when the spindle rotates.

15. For replication of micro parts moulding is preferred
a) True
b) False
Answer: a
Clarification: Nowadays, focus is on miniaturization through the development of novel production concepts (especially micro & nano) for the processing of non-ceramic materials. The replication of microparts through moulding is one of the preferred routes for micromanufacture because of its mass-production capability and relatively low cost.

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250+ TOP MCQs on Ultrasonic Machining – 2 and Answers

Manufacturing Processes Multiple Choice Questions on “Ultrasonic Machining – 2”.

1. Which of the following materials is not used as abrasive in ultrasonic machining?
a) SiC
b) Boronsilicarbide
c) Diamond
d) TiC
Answer: d
Clarification: Abrasive materials used as abrasive in USM are as follows;
• Al2O3
• SiC
• B4C
• Boronsilicarbide
• Diamond.

2. Pick the odd one out.
a) Frequency of vibration
b) Abrasive size
c) Coolant flow rate
d) Flow strength of work material
Answer: c
Clarification: The process parameters which govern the ultrasonic machining process are listed below;
• Amplitude of vibration – 15 – 50 μm
• Frequency of vibration – 19 – 25 kHz
• Feed force – related to tool dimensions
• Feed pressure
• Abrasive size – 15 μm – 150 μm
• Flow strength of work material
• Flow strength of the tool material
• Contact area of the tool
• Volume concentration of abrasive in water slurry
• Abrasive material.

3. Which of the following is not the component of USM machine?
a) Slurry delivery
b) Transducer
c) Concentrator
d) Lead screw
Answer: d
Clarification: The typical elements of a USM are as follows;
• Slurry delivery and return system
• Feed mechanism to provide a downward feed force on the tool during machining
• The transducer, which generates the ultrasonic vibration
• The horn or concentrator.

4. In USM, horn is used for amplifying the vibrations.
a) True
b) False
Answer: a
Clarification: The horn or concentrator in USM mechanically amplifies the vibration to the required amplitude of 15 – 50 μm and accommodates the tool at its tip.

5. The transducer for USM works on how many principles?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: The ultrasonic vibrations are produced by the transducer. The transducer is driven by suitable signal generator followed by a power amplifier. The transducer for USM works on the following principle
• Piezoelectric effect
• Magnetostrictive effect
• Electrostrictive effect.

6. The horn or concentrator can have how many different shapes?
a) 2
b) 5
c) 3
d) It can have any shape
Answer: c
Clarification: The horn or concentrator is a wave-guide, which amplifies and concentrates the vibration to the tool from the transducer. The horn or concentrator can be of different shape like
• Tapered or conical
• Exponential
• Stepped.

7. Which of the following is not machined by USM?
a) Brittle metals
b) Glass
c) Ceramics
d) Ductile materials
Answer: d
Clarification: USM is used for machining;
• hard metallic alloys
• brittle metallic alloys
• semiconductors
• glass
• ceramics
• carbides.

8. For which of the following operations USM is not used?
a) Round shaped hole
b) Square shaped hole
c) Irregular shaped hole
d) Boring
Answer: d
Clarification: USM is used for machining;
• round shaped holes
• square shaped holes
• irregular shaped holes
• surface impressions.

9. USM has low material removal rate.
a) True
b) False
Answer: a
Clarification: Limitations of USM are as follows;
• Low MRR
• Rather high tool wear
• Low depth of hole.

10. USM machine uses _____ axis table.
a) single
b) 2
c) 3
d) 5
Answer: b
Clarification: The basic mechanical structure of a USM is very similar to a drill press. However, it has additional features to carry out USM of brittle work material. The workpiece is mounted on a vice, which can be located at the desired position under the tool using a 2 axis table. The table can further be lowered or raised to accommodate work of different thickness.

250+ TOP MCQs on Wire-cut EDM – 3 and Answers

Manufacturing Processes Multiple Choice Questions on “Wire-cut EDM – 3”.

1. Wire electric discharge (ED) machining is based on the same principle as that of _____
a) hydro-dynamic EDM
b) die-sink EDM
c) polar EDM
d) non-conventional EDM
Answer: b
Clarification: Wire electric discharge (ED) machining is based on the same principle as die-sink ED machining. The basic elements in all EDM methods are dielectric fluid, a workpiece and an electrode.

2. The only difference between die-sink EDM and wire cut EDM is the _____
a) way of material removal
b) electrode used for the machining
c) type of materials machined
d) processing time
Answer: b
Clarification: In the die-sink EDM method the electrode has the same shapes as the wished machining results. In the wire cut EDM method the electrode is a moving wire made from some electrically conducting material. The workpiece is cut with the electrode wire.

3. During wire cut EDM, the size of the cavity produced by the wire while machining depends upon _____
a) material of the workpiece
b) di-electric fluid used
c) wire material
d) electric current
Answer: d
Clarification: While machining a high electric current passes through the dielectric fluid and heats the workpiece surface from a very small area. The corresponding workpiece area melts and what is left, is a small round cavity. The cavity size depends on the electric current and potential.

4. Sparking gap is the distance between _____
a) the workpiece and the CNC table
b) the workpiece and the electrode wire
c) the electrode wire and the di-electric fluid
d) the workpiece and the spark plug
Answer: b
Clarification: The distance between the electrode wire and the workpiece is called a sparking gap. The electrode produces shapes that are a sparking gap dimension larger than the programmed shape through which the electrode wire passes.

5. The absolute minimum inner corner radius is _____
a) the wire radius minus the sparking gap
b) the sparking gap minus the wire radius
c) the wire radius plus the sparking gap
d) double of the wire radius
Answer: c
Clarification: The absolute minimum inner corner radius is the wire radius added with the sparking gap . Therefore during machining, the shapes produced have larger dimensions as compared to the shapes mentioned in the program.

6. The wire ED machines have _____ programmable axes.
a) 2
b) 2-5
c) 6
d) 3-9
Answer: b
Clarification: The wire cut EDM machines have 2 – 5 programmable axes. The machines that are used in mould making applications typically have 5 programmable axes. These axes are Wire guide, wire tilting in x and y –directions and workpiece or wire system movements in x and y –directions.

7. Which of the following component of the wire cut EDM machine does not get heated?
a) Workpiece
b) Electrode wire
c) Di-electric fluid
d) Coils
Answer: b
Clarification: The electrode wire moves between two coils with a moderate speed. The part of the wire that actually machines the workpiece is constantly changing. There is no time for the wire to heat up.

8. Which of the following material properties sets restrictions to use wire cut EDM?
a) Material type
b) Melting point
c) Material hardness
d) Electrical conductivity
Answer: d
Clarification: As the electrode wire does not get heated during machining, the problems with electrode wear are not an issue like in the case of die-sink ED machining and it is possible to use wire cut EDM also for materials with high melting ranges. The material hardness sets no restrictions. The only restriction is that the material needs to be electrically conductive.

9. Wires used in wire cut EDM are usually disposed after one usage.
a) True
b) False
Answer: a
Clarification: Despite the minimum wear, wires are usually disposed after one usage. Sparking and high temperature during the machining reduces the wire tensile strength and the wire could easily break if re-used.

10. The electrode wires are usually made form _____
a) graphite
b) iron
c) nickel
d) brass
Answer: d
Clarification: Wire used in wire cut EDM are usually made of brass – either zinc-coated or uncoated. Brass wire can be purchased in different hardnesses and different diameters. Zinc coated wire is used in machining high melting point workpiece materials.

250+ TOP MCQs on Annealing – 4 and Answers

Tricky Manufacturing Processes Questions and Answers on “Annealing – 4”.

1. The yield point can be completely recovered in a zinc crystal at _____
a) room temperature
b) 32⁰C
c) 50⁰C
d) 88⁰C
Answer: a
Clarification: Recovery of the yield point thus begins very rapidly. The yield point can be completely recovered in a zinc crystal at room temperature in a period of a day. The stress-strain diagrams indicate a well-known fact: the rate at which a property recovers isothermally is a decreasing function of the time.

2. Polygonization is associated with_____
a) grain size
b) plastically bent crystals
c) conductivity
d) strength of the material
Answer: b
Clarification: Another recovery process is called polygonization. In its simplest form it is associated with crystals that have been plastically bent. Many workers have shown that the Laue spots of bent crystals assume a fine structure after a recovery anneal (anneal that does not recrystallize the specimen).

3. Polygonization is a process of _____
a) annealing a bent crystal
b) annealing a pure metal
c) heating the metal above recrystallization temperature
d) quenching
Answer: a
Clarification: When a bent crystal is annealed, the curved crystal breaks up into a number of closely related small perfect crystal segments. This process has been given the name polygonization.

4. The regrouping of edge dislocations into low-angle boundaries helps in _____
a) lowering the strain energy
b) increasing the stored energy
c) lowering the recrystallization temperature
d) increasing the strength of the metal
Answer: a
Clarification: In addition to lowering the strain energy, the regrouping of edge dislocations into low-angle boundaries has a second important effect. This is the removal of general lattice curvature. As a result of polygonization, crystal segments lying between a pair of low-angle boundaries approach the state of strain-free crystals with flat uncurved planes.

5. An edge dislocation is capable of moving by a climb in a direction _____ to the slip plane.
a) 30⁰
b) 45⁰
c) 90⁰
d) 180⁰
Answer: c
Clarification: An edge dislocation is capable of moving either by slip on its slip plane or by a climb in a direction perpendicular to its slip plane. The driving force for these movements comes from the strain energy of the dislocations, which decreases as a result of polygonization.

6. The rate of polygonization increases rapidly with _____
a) density
b) temperature
c) humidity
d) grain size
Answer: b
Clarification: The strain field of dislocations grouped on slip planes produces an effect force that makes them move into sub-boundaries. This force exists at all temperatures, but at low temperatures edge dislocations cannot climb. Since dislocation climb depends on the movement of vacancies (an activated process), the rate of polygonization increases rapidly with temperature.

7. As the polygon walls become more widely separated, the rate of ______becomes a decreasing function of time and temperature.
a) polygonization
b) decay
c) coalescence
d) grain growth
Answer: c
Clarification: As the polygon walls become more widely separated, the rate of coalescence becomes a decreasing function of time and temperature so that the polygonization process approaches a more or less stable state with widely spaced, approximately parallel (in a single crystal deformed by simple bending) sub-boundaries.

8. The rate at which a recovery process occurs always decreases with _____
a) temperature
b) time
c) yield strength
d) ductility
Answer: b
Clarification: In an isothermal anneal, the rate at which a recovery process occurs always decreases with time; that is, it starts rapidly and proceeds at a slower and slower rate as the driving force for the reaction is expended.

9. In deformed polycrystalline metals, high-temperature recovery is considered to be essentially a matter of polygonization.
a) True
b) False
Answer: a
Clarification: In deformed polycrystalline metals, high-temperature recovery is considered to be essentially a matter of polygonization and annihilation of dislocations. At lower temperatures other processes such as occur in dynamic recovery are of greater importance, current theories picture the recovery process as primarily a matter of reducing the number of point defects to their equilibrium value.

10. Recrystallization during an isothermal anneal begins very slowly.
a) True
b) False
Answer: a
Clarification: Recrystallization during an isothermal anneal begins very slowly, and builds up to a maximum reaction rate, after which it finishes slowly. Recovery processes start at the beginning of the annealing cycle and account for the initial energy release, while recrystallization starts later and accounts for the second (larger) energy release.

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250+ TOP MCQs on Miller Indices and Answers

Manufacturing Engineering Multiple Choice Questions on “Miller Indices”.

1. Stacking sequence in hexagonal close packed (HCP) structure is?
a) AAAAA
b) ABABAB
c) ABCABC
d) AABBAA
Answer: b
Clarification: The geometry of close packed hexagonal unit cell can be understood from the below figure, which indicates a plane view of atoms.

Figure: Stacking sequences of close packed layers of atoms. A-first layer (with outlines of atoms shown in gold colour); B-second layer.

A is the first layer (with the circular outlines of the atoms drawn in gold colour) and B is the second layer (outlines of the atoms not shown for clarity). In hexagonal close packed unit cell, the third layer of atoms go directly above the A layer, the fourth layer over the B layer, and so on; the sequence becomes ABABAB.

2. Stacking sequence in face centered cubic (FCC) close packed structure is?
a) AAAAA
b) ABABAB
c) ABCABC
d) AABBAA
Answer: c
Clarification: The stacking sequence in FCC can be best understood from the below figure.

Figure: Stacking sequences of close packed layers of atoms. A-first layer (with outlines of atoms shown in green colour); B-second layer; C-third layer.

A is the first layer (with the circular outlines of the atoms drawn in green colour) and B is the second layer (outlines of the atoms not shown for clarity). The third layer of atoms goes above the interstices marked C and the sequence only repeats at the fourth layer, which goes directly above the first layer. Thus, the stacking sequence is now ABCABC.

3. For pane (1 1 1) of FCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 2.31/a2
b) 2.31/a3
c) 1.31/a2
d) 1.31/a3
Answer: a
Clarification: Upon visualizing (1 1 1) plane of FCC, one can identify that there the equilibrium triangle (1 1 1) of FCC consists 2 atoms (16×3 + 12×3=2). Also, the area of the equilibrium triangle (1 1 1) is 0.886 a2.
Therefore, the number of atoms per square inch which is nothing but its planar density = (frac{2}{0.866 a^2} = frac{2.31}{a^2}).

4. For pane (1 1 1) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 1.07/a2
b) 0.58/a2
c) 2.07/a2
d) 0.78/a2
Answer: b
Clarification: From the geometry of the triangle (1 1 1), it is clear that it has 0.5 atoms in it (1/6×3 = 0.5) and the area of the triangle (1 1 1) is 0.866 a2.
Therefore, the number of atoms per square inch which is nothing but its planar density = (frac{0.5}{0.866 a^2} = frac{0.58}{a^2}).

5. For pane (1 0 0) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 1/a3
b) 2/a2
c) 3/a4
d) 1/a2
Answer: d
Clarification: The atomic arrangement on square (1 0 0) of BCC indicates that it has 1 atom and area of this square (1 0 0) is a2. Thus, its planar density = 1/a2.

6. For pane (1 1 0) of BCC having a lattice parameter ‘a’, planar atomic density is given by?
a) 3.690/a2
b) 2.312/a2
c) 1.414/a2
d) 0.580/a2
Answer: c
Clarification: From the geometry of the rectangle (1 1 0), it is clear that it has 2 atoms in it (14×4+1=2) and the area of the rectangle (1 1 0) is 0.866 a2.
Therefore, the number of atoms per square inch which is nothing but its planar density = (frac{2}{1.414 a^2} = frac{1.414}{a^2}).

7. For pane (1 1 0) of SC having a lattice parameter ‘a’, planar atomic density is given by?
a) 0.508/a2
b) 0.707/a2
c) 0.707/a3
d) 0.508/a3
Answer: b
Clarification: The atomic arrangement on (1 1 0) plane in simple cubic (SC) unit cell indicates that it has 1 atom (14×4=1) in this plane having an area = (sqrt{2} a^2).
Therefore, the number of atoms per square inch which is nothing but its planar density = (frac{1}{1.414 a^2} = frac{0.707}{a^2}).

8. For pane (1 1 1) of SC having a lattice parameter ‘a’, planar atomic density is given by?
a) 0.58/a2
b) 0.78/a3
c) 0.68/a2
d) 0.88/a2
Answer: a
Clarification: The atomic arrangement on triangle (1 1 1) in simple cubic (SC) unit cell indicates that it has 0.5 atoms (16×3=0.5) in this plane having an area = 0.866 a2.
Therefore, the number of atoms per square inch which is nothing but its planar density = (frac{0.5}{0.866 a^2} = frac{0.577}{a^2}).

9. Which of the following equation describes Bragg’s law of diffraction? (Assume that all symbols have their usual meaning.)
a) 2d sinθ = λ
b) 2d = nλ
c) 2d = nλ sinθ
d) 2d sinθ = nλ
Answer: d
Clarification: Bragg’s law, which describes the constructive interference conditions for θ to be at its strongest: nλ = 2d sinθ, where λ is the wavelength of the incident wave and and n is the order of diffraction which is always a positive integer and d is the interplanar spacing and θ = angle of diffraction. This equation can be easily derived upon constructing a basic wave reflection diagram taking incident angle as θ.

10. Powder X-ray technique can be used to determine crystal structures.
a) True
b) False
Answer: a
Clarification: The powder method is used to determine the values of lattice parameters accurately. As we know, lattice parameters defines the axis system, which in turns can predict the crystal structure of the powder sampled.

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250+ TOP MCQs on Tool Geometery and Nomenclature-1 and Answers

Manufacturing Engineering Multiple Choice Questions on “Tool Geometery and Nomenclature-1”.

1. Angle between the rake face and plane perpendicular to rake face is known as:
a) Side rake angle
b) Side relief angle
c) End relief angle
d) Back rake angle

Answer: a
Clarification: Side rake angle is the angle between the rake face and plane perpendicular to rake face.

2. Angle between the rake face flank of tool and perpendicular line drawn from cutting point to base of tool is known as:
a) Side rake angle
b) Side relief angle
c) End relief angle
d) Back rake angle

Answer: b
Clarification: Side relief angle is the angle between the flank of tool and perpendicular line drawn from cutting point to base of tool.

3. Angle between side cutting edge and axis of tool is known as:
a) Side rake angle
b) Side relief angle
c) Side cutting edge angle
d) Back rake angle

Answer: c
Clarification: Side cutting angle is the angle between side cutting edge and axis of tool.

4. Angle between end cutting edge and axis of tool is known as:
a) Side rake angle
b) Side relief angle
c) End cutting edge angle
d) Back rake angle

Answer: c
Clarification: End cutting angle is the angle between end cutting edge and axis of tool.

5. Angle between side cutting edge and end cutting edge in the top surface plane of tool.
a) Side rake angle
b) Side relief angle
c) Side cutting edge angle
d) Nose angle

Answer: d
Clarification: Nose angle is the angle between side cutting edge and end cutting edge.

6. With an increase in lip angle keeping side rake angle constant, strength of tool.
a) Increases
b) Decreases
c) Remains constant
d) None of the mentioned

Answer: a
Clarification: Thickness of tool tip increase with an increase in lip angle, hence the strength of tool increase.

7. For large positive back rake angle, tool will be
a) Weaker
b) Stronger
c) Smoother
d) Harder

Answer: a
Clarification: With the increase in positive back rake angle, lip angle decreases and tool tip become thin.

8. For large negative back rake angle, tool will be
a) Weaker
b) Stronger
c) Smoother
d) Harder

Answer: b
Clarification: With an increase in negative back rake angle, lip angle increases and tool tip become thick and hence the strength of tool will increase.

9. Which of the following will give better chip flow?
a) Positive back rake angle tool
b) Negative back rake angle tool
c) Zero back rake angle tool
d) None of the mentioned

Answer: a
Clarification: With positive back rake angle, lip angle decreases and space for chip flow increases and hence it will give better chip flow.

10. Which of the following will give large friction during chip flow?
a) Positive back rake angle tool
b) Negative back rake angle tool
c) Zero back rake angle tool
d) Small lip angle tool

Answer: b
Clarification: With negative back rake angle, lip angle increases and space for chip flow decreases due to which it offers large resistance to chip flow.