1. In ultrasonic machining the tool ____ at very high frequency with the help of ____ transducers.
(1 Mark, 1987)

Ans: Vibrates, Piezoelecric/ Magnetostrictive
2. ECM ___ (Can/ Cannot) be used for all such materials for which ultrasonic machining is possible, while EDM ____ (Can/ Cannot) be used for all such materials for which ECM is possible.
(1 Mark, 1987)

Ans: Cannot/ can
3. In ECM, the material removal rate will be higher for metal with
(a) Large density
(b) Large valency
(c) Large chemical absorption tendency
(d) Large chemical weight
(2 Mark, 1989)

Ans: d
We know, In ECM
MRR = \frac { AI }{ \rho ZF }
=> MRR \propto A
So, MRR is proportional to chemical weight
4. The two main criteria for selecting the electrolyte in Electrochemical Machining (ECM) is that the electrolyte should
(a) Be chemically stable
(b) Not allow dissolution of cathode material
(c) Not allow dissolution of anode material
(d) Have high electrical conductivity
(1 Mark, 1992)

Ans: a, b, d
5. In Ultrasonic Machining (USM) the material removal rate would
(a) Increase
(b) Decrease
(c) Increase and then decrease
(d) Decrease and then increase
with increasing mean grain diameter of the abrasive material.
(1 Mark, 1992)
6. In ultrasonic machining process, the material removal rate will be higher for materials with
(a) Higher toughness
(b) Higher ductility
(c) Lower toughness
(d) Higher fracture strain
(1 Mark, 1993)

Ans: c
7. Electric discharge machining is more efficient process than Electrochemical machining for producing large noncircular holes. (T/F)
(1 Mark, 1994)
8. EDM imposes larger forces on tool than ECM. (T/F)
(1 Mark, 1994)

Ans: False
As there is no contact between tool and workpiece in both of the machining EDM and ECM, the force on the tool is negligible.
9. Ultrasonic machining is about the best process for making holes in glass which are comparable in size with the thickness of the sheet. (T/F)
(1 Mark, 1994)

Ans: True
10. Selection electrolyte for ECM is as follows:
(a) Nonpassivating electrolyte for stock removal and passivating electrolyte for finish control.
(b) Passivating electrolyte for stock removal and nonpassivating electrolyte for finish control.
(c) Selection of electrolyte is dependent on current density.
(d) Electrolyte selection is based on toolwork electrodes
(1 Mark, 1997)

Ans: a
ECM electrolytes are classified into two categories: Passivating electrolytes containing oxidizing anions i.e. sodium nitrate, sodium chlorate and nonpassivating electrolytes containing relatively aggressive anions such as sodium chloride. Passivating electrolytes are known to give better machining precision.
Source: A journal paper “RECENT DEVELOPMENTS IN ELECTROCHEMICAL MICRO MACHINING” by Prof. B. Bhattacharyya Department of Production Engg., Jadavpur University, Kolkata, India
11. Inter electrode gap in ECG is controlled by
(a) Controlling the pressure of electrolyte flow
(b) Controlling the applied static load
(c) Controlling the size of diamond particle in the wheel
(d) Controlling the texture of the work piece
(1 Mark, 1997)

Ans: c

Ans: A – 5, B – 4, C – 2, D – 4
14. In ElectroDischarge Machining (EDM), the tool is made of:
(a) Copper
(b) High Speed Steel
(c) Cast Iron
(d) Plain Carbon Steel
(1 Mark, 1999)

Ans: a
Pure copper is extensively used as an electrode material in EDM.
15. Choose the correct statement:
(a) A fixture is used to guide the tool as well as to locate and clamp the work piece.
(b) A jig is used to guide the tool as well as to locate and clamp the work piece.
(c) Jigs are used on CNC machines to locate and clamp the work piece and also to guide the tool.
(d) No arrangements to guide the tool is provided in a jig
(1 Mark, 1999)

Ans: b, c
16. Deep hole drilling of small diameter, say 0.2 mm is done with EDM by selecting the tool material as
(a) Copper wire
(b) Tungsten wire
(c) Brass wire
(d) Tungsten carbide
(1 Mark, 2000)

Ans: b
Tungsten wire is used for making small diameter hole in EDM.
For more information:
PN Rao (2) Book —> Unconventional machining —> EDM —–> Electrodes
17. In ECM, the material removal is due to
(a) Corrosion
(b) Erosion
(c) Fusion
(d) Ion displacement
(1 Mark, 2001)

Ans: d
18. 3 – 2 – 1 method of location in a jig or fixture would collectively restrict the workpiece in n degrees of freedom, where the value of n is
(a) 6
(b) 8
(c) 9
(d) 12
(2 Mark, 2001)

Ans: c
19. Estimate the metal removal rate (in cc/hr) of an alloy containing 18% Cobalt, 62% Nickel and 20% Chromium during ElectroChemical Machining (ECM) with a current of 500 Amperes. The density of the alloy is 8.28 gm/cc. The following data is available:
Assume Faraday’s constant as 96,500 Coulombs/mole.
(2 Mark, 2002)

Ans: 44.77
For a compound, The MRR in ECM:
MRR = \frac { I }{ \rho F\sum { \frac { { \alpha }_{ i }{ Z }_{ i } }{ { A }_{ i } } } }
Where, { \alpha }_{ i } = Weight percentage of different element
In this question,
MRR = \frac { I }{ \rho F\left( \frac { { \alpha }_{ co }{ Z }_{ co } }{ { A }_{ co } } +\frac { { \alpha }_{ n }{ Z }_{ n } }{ { A }_{ n } } +\frac { { \alpha }_{ ch }{ Z }_{ ch } }{ { A }_{ ch } } \right) }
=> MRR = \frac { 500 }{ 8.28\times 96500\times \left( \frac { 0.18\times 2 }{ 58.93 } +\frac { 0.62\times 2 }{ 58.71 } +\frac { 0.20\times 6 }{ 51.99 } \right) }
=> MRR = 0.0124 cc/hr
=> MRR = 44.77 cc/hr
20. As tool and work are not in contact in EDM process
(a) No relative motion occurs between them
(b) No wear of tool occurs
(c) No power is consumed during metal cutting
(d) No force between tool and work occurs
(1 Mark, 2003)

Ans: d
21. The mechanism of material removal in EDM process is
(a) Melting and Evaporation
(b) Melting and Corrosion
(c) Erosion and Cavitation
(d) Cavitation and Evaporation
(1 Mark, 2004)

Ans: a
22. Typical machining operations are to be performed on handtomachine materials by using the processes listed below. Choose the best set of Operationprocess combinations.
(a) P – 1, Q – 5, R – 3, S – 4
(b) P – 1, Q – 4, R – 1, S – 2
(c) P – 5, Q – 1, R – 2, S – 6
(d) P – 2, Q – 3, R – 5, S – 6
(2 Mark, 2004)

Ans: d
23. A zigzag cavity in a block of high strength alloy is to be finish machined. This can be carried out by using
(a) Electric discharge machining
(b) Electrochemical machining
(c) Laser beam machining
(d) Abrasive flow machining
(1 Mark, 2005)

Ans: c
Laser beam machining may be more suitable.
24. Arrange the processes in the increasing order of their maximum material removal rate.
Electrochemical Machining (ECM),
Ultrasonic Machining (USM)
Electron Beam Machining (EBM),
Laser Beam Machining (LBM)
Electric Discharge Machining (EDM)
(a) USM, LBM, EBM, EDM, ECM
(b) EBM, LBM, USM, ECM, EDM
(c) LBM, EBM, USM, ECM, EDM
(d) LBM, EBM, USM, EDM, ECM
(2 Mark, 2006)
25. In electrodischarge machining (EDM), if the thermal conductivity of tool is high and the specific heat of work piece is low, then the tool wear rate and material removal rate are expected to be respectively.
(a) High and high
(b) Low and low
(c) High and low
(d) Low and high
(2 Mark, 2007)

Ans: d
High thermal conductivity of tool:
For the the same heat load, the local temperature rise would be less due to faster heat conducted to the bulk of the tool and thus less tool wear.Low specific heat of workpiece:
For the same heat load, the temperature rise of the workpiece would be more. So, high material removal rate.
27. A researcher conducts electrochemical machining (ECM) on a binary allow (density 6000 kg/mm^{3}) of iron (atomic weight 56, valency 2) and metal P (atomic weight 24, valency 4). Faraday’s constant = 96500 coulomb/mole. Volumetric material removal rate of the alloy is 50 mm^{3}/s at a current of 2000 A. The percentage of the metal P in the alloy is closest to
(a) 40
(b) 25
(c) 15
(d) 79
(2 Mark, 2008)

Ans: b
For a compound, The MRR in ECM:
MRR = \frac { I }{ \rho F\sum { \frac { { \alpha }_{ i }{ Z }_{ i } }{ { A }_{ i } } } }
Where, { \alpha }_{ i } = Weight percentage of different element
In this question,
MRR = \frac { I }{ \rho F\left( \frac { { \alpha }_{ I }{ Z }_{ I } }{ { A }_{ I } } +\frac { { \alpha }_{ P }{ Z }_{ P } }{ { A }_{ P } } \right) }
Where, \rho = 600 kg/m^{3 }= 6 g/cm^{3}
MRR = 50 mm^{3}/s = 0.050 cm^{3}/s
Now, 0.050 = \frac { I }{ 6\times 96500\left( \frac { { \alpha }_{ I }{ Z }_{ I } }{ { A }_{ I } } +\frac { { \alpha }_{ P }{ Z }_{ P } }{ { A }_{ P } } \right) }
=> \frac { 1 }{ 14.475 } =\frac { 100x }{ 100 } \times \frac { 2 }{ 56 } +\frac { x }{ 100 } \times \frac { 4 }{ 24 }
=> x \simeq 25%
28. Electrochemical machining is performed to remove material from an iron surface of 20mm × 20mm under the following conditions:
Inter electrode gap = 0.2mm,
Faraday ‘s constant = 96540 Coulombs,
Supply voltage (DC) = 12V,
Specific resistance of electrolyte = 2 Ω cm,
Atomic weight of Iron = 55.85,
Valency of Iron = 2
The material removal rate (in g/s) is
(a) 0.3471
(b) 3.471
(c) 34.71
(d) 347.1
(2 Mark, 2009)

Ans: a
Resistance (R) = \rho \frac { L }{ A }
Where, Specific resistance (\rho) = 2\Omega cm
L = 0.02 cm
Area (A) = 2 * 2 = 4 cm^{2}
Now, R = 2\times \frac { 0.02 }{ 2\times 2 } = 0.01 \Omega
Current (I) = \frac { V }{ R } =\frac { 12 }{ 0.01 }
=> I = 1200 Amp
Material removal rate (MRR) = \frac { AI }{ ZF }
=> MRR = \frac { 55.85\times 1200 }{ 2\times 96540 }
=> MRR = 0.3471 g/s
29. Match the following nontraditional machining processes with the corresponding material removal mechanisms:
(a) P2, Q3, R4, S1
(b) P2, Q4, R3, S1
(c) P3, Q2, R4, S1
(d) P2, Q3, R1, S4
(2 Mark, 2011)

Ans: a
30. The operation in which oil is permeated into the pores of a powder metallurgy product is known as
(a) Mixing
(b) Sintering
(c) Impregnation
(d) Infiltration
(1 Mark, 2011)

Ans: c
30. In abrasive jet machining, as the distance between the nozzle tip and the work surface increases, the material removal rate
(a) Increases continuously.
(b) Decreases continuously.
(c) Decreases, becomes stable and then increases.
(d) Increases, becomes stable and then decreases.
(1 Mark, 2012)
31. During the electrochemical machining (ECM) of iron (atomic weight = 56, valency = 2) at current of 1000 A with 90% current efficiency, the material removal rate was observed to be 0.26 gm/s. If Titanium (atomic weight = 48, valency = 3) is machined by the ECM process at the current of 2000 A with 90% current efficiency, the expected material removal rate in gm/s will be
(a) 0.11
(b) 0.23
(c) 0.30
(d) 0.52
(2 Mark, 2013)

Ans: c
In ECM, MRR = \frac { AI }{ ZF } gm/s
Current efficiency (\eta ) = 90%
MRR for Titanium = \frac { 48\times (0.9\times 2000) }{ 3\times 96500 }
=> MRR = 0.3 gm/s
32. The following four unconventional machining processes are available in a shop floor. The most appropriate one to drill a hole of square cross section of 6 mm × 6 mm and 25 mm deep is
(a) Abrasive Jet Machining
(b) Plasma Arc Machining
(c) Laser Beam Machining
(d) Electro Discharge Machining
(1 Mark, 2014[2])

Ans: d
33. The process utilizing mainly thermal energy for removing material is
(a) Ultrasonic Machining
(b) Electrochemical Machining
(c) Abrasive Jet Machining
(d) Laser Beam Machining
(1 Mark, 2014[3])

Ans: d
34. The principle of material removal in Electrochemical machining is
(a) Fick’s law
(b) Faraday’s laws
(c) Kirchhoff’’s laws
(d) Ohm’s law
(1 Mark, 2014[4])

Ans: b
35. The primary mechanism of material removal in electrochemical machining (ECM) is
(a) Chemical corrosion
(b) Etching
(c) Ionic dissolution
(d) Spark erosion
(1 Mark, 2015[2])

Ans: c
Option ‘c’ is correct.
36. A resistancecapacitance relaxation circuit is used in an electrical discharge machining process. The discharge voltage is 100 V. At a spark cycle time of 25 μs, the average power input required is 1 kW. The capacitance (in μF) in the circuit is
(a) 2.5
(b) 5.0
(c) 7.5
(d) 10.0
(2 Mark, 2015[2])

Ans: b
Power = \frac { Work }{ time }
=> P = \frac { C{ V }^{ 2 } }{ 2\times t }
=> 1000 = \frac { C\times { 100 }^{ 2 } }{ 2\times \left( 25\times { 10 }^{ 6 } \right) }
=> C = 5 \mu F
37. The nontraditional machining process that essentially requires vacuum is
(a) Electron beam machining
(b) Electro chemical machining
(c) Electro chemical discharge machining
(d) Electro discharge machining
(1 Mark, 2016[1])

Ans: a
Option ‘a’ is correct.
38. In an ultrasonic machining (USM) process, the material removal rate (MRR) is plotted as a function of the feed force of the USM tool. With increasing feed force, the MRR exhibits the following behavior:
(a) Increases linearly
(b) Decreases linearly
(c) Does not change
(d) First increases and then decreases
(1 Mark, 2016[2])
39. The surface irregularities of electrodes used in an electrochemical machining (ECM) process are 3 μm and 6 μm as shown in the figure. If the workpiece is of pure iron and 12V DC is applied between the electrodes, the largest feed rate is ______mm/min.
Assume the iron to be dissolved as Fe^{+2} and the Faraday constant to be 96500 Coulomb.
(2 Mark, 2016[2])

Ans: 51.5
Resistance (R) = \rho \frac { L }{ A }
Where, Resistivity (\rho) =\frac { 1 }{ Conuctivity }
=> \rho =\frac { 1 }{ 0.02 } = 50 ohm.mm
L = 3 + 6 = 9 \mu m = 0.009 mm
Area (A) = Unknown
Now, R = 50\times \frac { 0.009 }{ A } =\frac { 0.45 }{ A }
Current (I) = \frac { V }{ R } =\frac { \left( 121.5 \right) A }{ 0.45 }
=> I = 23.33A Amp
Material removal rate (MRR) = \frac { AI }{ \rho ZF }
Where, A = Atomic weight = 55.85 gm
\rho = Density = 7860 \frac { kg }{ { m }^{ 3 } }
=> \rho =7860\times { 10 }^{ 6 }\frac { gm }{ { mm }^{ 3 } }
Now, MRR = \frac { 55.85\times 23.33A }{ \left( 7860\times { 10 }^{ 6 } \right) \times 2\times 96500 }
=> \frac { MRR }{ A } = 0.859 mm/s
=> \frac { MRR }{ A } = 51.5 mm/min
40. In a wirecut EDM process the necessary conditions that have to be met for making a successful cut are that
(a) Wire and sample are electrically nonconducting
(b) Wire and sample are electrically conducting
(c) Wire is electrically conducting and sample is electrically nonconducting
(d) Sample is electrically conducting and wire is electrically nonconducting
(1 Mark, 2016[3])

Ans: b
Option ‘b’ is correct.
41. Match the processes with their characterstics.
(a) P – 2, Q – 3, R – 1, S – 4
(b) P – 3, Q – 2, R – 1, S – 4
(c) P – 3, Q – 2, R – 4, S – 1
(d) P – 2, Q – 4, R – 3, S – 1
(1 Mark, 2017[1])

Ans: a
Option ‘a’ is correct.
42. Which one of the following statements is TRUE for the ultrasonic machining (USM) process?
(a) In USM, the tool vibrates at subsonic frequency.
(b) USM does not employ magnetostrictive transducer.
(c) USM is an excellent process for machining ductile materials.
(d) USM often uses a slurry comprising abrasive particles and water.
(1 Mark, 2017[2])

Ans: d
Option ‘d’ is correct.
43. An electrochemical machining (ECM) is to be used to cut a through hole into a 12 mm thick aluminum plate. The hole has a rectangular crosssection, 10 mm × 30 mm. The ECM operation will be accomplished in 2 minutes, with efficiency of 90%. Assuming specific removal rate for aluminum as 3.44 × 10^{2} mm^{3}/(A s), the current (in A) required is ______ (correct to two decimal places).
(2 Mark, 2018[1])

Ans: 968.99
Volume of metal to be removed (V) = 10\times 30\times 12 = 3600 mm^{3}
Given, time (t) = 2 min. = 120 s
Now, Specific removal rate = \frac { V }{ I\times t\times \eta }
=> 3.44\times { 10 }^{ 2 }=\frac { 3600 }{ I\times 120\times 0.9 }
=> I = 968.992 A
44. Metal removal in electric discharge machining takes place through
(a) Ion displacement
(b) Melting and vaporization
(c) Corrosive reaction
(d) Plastic shear
(1 Mark, 2018[2])

Ans: b
Option ‘b’ is correct.
45. A circular hole of 25 mm diameter and depth of 20 mm is machined by EDM process. The material removal rate (in mm^{3}/min) is expressed as 4\times { 10 }^{ 4 }I{ T }^{ 1.23 }. Where, I = 300 A and the melting point of the material, T = 1600^{0} C. The time (in minutes) for machining this hole is ________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 7.143
Given, MRR = 4\times { 10 }^{ 4 }I{ T }^{ 1.23 }
=> MRR = 4\times { 10 }^{ 4 }\times 300\times 1600^{ 1.23 }
=> MRR = 1374.4 \frac { { mm }^{ 3 } }{ min }
Volume of hole (V) = \frac { \pi }{ 4 } { d }^{ 2 }l
=> V = \frac { \pi }{ 4 } \times 25^{ 2 }\times 20 = 9817.77 mm^{3}
Time for machining:
t = \frac { V }{ MRR } =\frac { 9817.77 }{ 1374.4 } = 7.143 min
46. In an electrical discharge machining process, the breakdown voltage across inter electrode gap (IEG) is 200V and the capacitance of the RC circuit is 50 μF. The energy (in J) released per spark across the IEG is ____.
(1 Mark, 2019[2])

Ans: 1
Energy (E) = \frac { 1 }{ 2 } C{ V }^{ 2 }
=> E = \frac { 1 }{ 2 } \times \left( 50\times { 10 }^{ 6 } \right) \times 200^{ 2 } = 1J