1. The standard specification of a grinding wheel is A46M6V21. It means a wheel of
(a) Aluminum oxide of mesh size 6
(b) Boron carbide of mesh size 46
(c) Aluminum oxide of mesh size 46
(d) Silicon carbide of mesh size 6
(1 Mark, 1987)

Ans: c
A46M6V21:
‘A’ denotes that the type of abrasive. Here, it is aluminium oxide.
The number ‘46’ specifies the average grit size in inch mesh.
The letter ‘M’ denotes the hardness of the wheel.
The number ‘6’ denotes the structure or porosity of the wheel.
The letter code ‘V’ means that the bond material used is vitrified.
The number ‘21’ is a wheel manufacturer’s identifier.
2. The ideal cutting fluid for low speed machining of metals should be one which
(a) Removes the heat faster from the cutting zone
(b) Forms a coating on the cutting tools by chemical reaction
(c) Forms a low shear strength film of work material at the tool chip interface
(d) Serves as a dielectric, minimizing thereby reactions due to EMF at the interface
(1 Mark, 1988)

Ans: a
3. In twist drills _____ (small/ large) point angle and _____ (small/ large) helix angle are provided for drilling soft, low strength steel.
(1 Mark, 1988)
4. Cutting speed in grinding is set to high value to
(a) Reduce the cutting time
(b) Increase the bond strength
(c) Improve cooling of job and wheel
(c) Reduce the wheel wear
(1 Mark, 1988)

Ans: a
5. Gear hobbing produces more accurate gears than milling because in hobbing
(a) There is a continuous indexing operation
(b) Pressure angle is larger than in milling
(c) Hob and work piece both are rotating
(d) A special multitooth cutter (hob) is used
(1 Mark, 1989)

Ans: a
6. Teeth of internal spur gears can be accurately cut in a
(a) Milling machine
(b) Gear shaping machine
(c) Slotting machine
(d) Hobbing machine
(1 Mark, 1989)

Ans: b
Gear planning and gear hobbing are limited to external gears only whereas gear shaping with pinion cutter is for both external and internal gears.
7. Minimum dimensional and form accuracy can be obtained in the cylinder bores of automobile engines if the bores are finished by
(a) Lapping
(b) Reaming
(c) Internal grinding
(d) Honing
(1 Mark, 1989)

Ans: d
Honing operation is performed as the final operation and mostly used for internal surfaces.
8. In turning operation the feed rate could be doubled to increase the metal removal rate. To keep the same level of surface finish, the nose radius of the tool has to be
(a) Doubled
(b) Halved
(c) Multiplied by 4 times
(d) Kept unchanged
(2 Mark, 1989)

Ans: c
Surface roughness (h) =\frac { { f }^{ 2 } }{ 8r }
To keep the same level of surface finish,
\frac { { f }_{ 1 }^{ 2 } }{ { r }_{ 1 } } =\frac { { f }_{ 2 }^{ 2 } }{ { r }_{ 2 } }
=> \frac { { f }^{ 2 } }{ { r } } =\frac { (2f)^{ 2 } }{ { r }_{ 2 } }
=> { r }_{ 2 }=4r
9. If the longitudinal feed in centreless grinding is expressed by V_{f} = πDN.sinα, D stands for
(a) Diameter of blank
(b) Diameter of finished workpiece
(c) Diameter of control wheel
(d) Diameter of grinding wheel
(1 Mark, 1990)

Ans: c
10. In small lot production for machining T slots on machine tables, it is expected to use
(a) Shaping machine
(b) Broaching machine
(c) Vertical milling machine
(d) Horizontal milling machine
(1 Mark, 1990)

Ans: c
11. For cutting double start screw threads of pitch 1.0 mm on a lathe, the thread cutting toll should have a feed rate of
(a) 0.5 mm/rev
(b) 1.0 mm/rev
(c) 2.0 mm/rev
(d) 4.0 mm/rev
(2 Mark, 1991)

Ans: c
For double start screw threads,
Feed rate = Pitch of the screw thread \times 2
= 1 \times 2 = 2 mm/rev
12. A milling cutter having 10 teeth is rotating at 100 rpm. The table feed is set at 50 mm per minute. The feed per tooth in mm is
(a) 5
(b) 0.5
(c) 0.2
(d) 0.05
(2 Mark, 1991)

Ans: d
Table feed (f_{tab}) = N\times Z\times f
Where, N = Rotational speed of cutter
Z = Number of teeth
f = feed per tooth
=> 50 = 100\times 10\times f
=> f = 0.05 mm per tooth

Ans: A – T, B – U, C – Q, D – R
14. In horizontal milling process ____ (up/down) milling provides better surface finish and ____ (up/down) milling provides longer tool life.
(1 Mark, 1992)

Ans: Down, Down
15. A milling cutter having 8 teeth is rotating at 150 rpm. If the feed per tooth is 0.1, the table speed in mm per minute is
(a) 120
(b) 187
(c) 125
(d) 70
(2 Mark, 1993)

Ans: a
Table speed in mm per minute (f_{tab}) = N\times Z\times f
Where, N = Rotational speed of cutter
Z = Number of teeth
f = feed per tooth
f_{tab }= 150\times 8\times 0.1 = 120 mm/min.
16. To get good surface finish on a turned job, one should use a sharp tool with a ___ feed rate and ___ speed of rotation of the job.
(1 Mark, 1994)

Ans: Small, High
17. Among the conventional machining processes, maximum specific energy is consumed in
(a) Turning
(b) Drilling
(c) Planing
(d) Grinding
(1 Mark, 1995)

Ans: d
Specific power consumption = \frac { Power }{ MRR }
In grinding, the MRR is low. So, it has maximum specific energy consumption.
18. Diamond wheels should not be used for grinding steel components. (T/F)
(1 Mark, 1995)

Ans: T
Diamond being a form of carbon, tends to dissolve in iron when exposed to higher temperature and eventually loose its cutting efficiency.
19. Helix angle of fast helix drill is normally
(a) 35°
(b) 60°
(c) 90°
(d) 5°
(1 Mark, 1997)

Ans: b
Fasthelix drills have been developed especially for drilling deep holes in materials of low tensile strength, such as aluminum, magnesium, copper, die casting materials, wood, some plastic materials, and stacked aluminum sheets.
Helix angle of the flutes:
Usual – 20^{0} to 35^{0} – most common
Large helix : 45^{0} to 60^{0} – suitable for deep holes and softer work materials
Small helix : for harder / stronger materials
20. A cylinder of 155 mm is to be reduced 150 mm diameter in one turning cut with a feed of 0.15 mm/revolution and a cutting speed of 150 m/min on a NC lathe. What are the programmed feed rate, and the material removal rate?
(2 Mark, 1998)

Ans: 46.95 mm/min., 56250 mm^{3}/min.
Average diameter = \frac { 155+150 }{ 2 } = 152.5 mm
Cutting speed (V) = \pi dn
=> 150\times 1000 = \pi \times 152.5\times n
=> n = 313 rev/min.
Feed rate = feed\times n
= 0.15 \times 313 = 46.95 mm/min.
Depth of cut (d) = \frac { 155150 }{ 2 } = 2.5 mm
MRR = fdV = 0.15\times 2.5\times (150\times 1000)
= 56250 mm^{3}/min.
21. Abrasive material used in grinding wheel selected for grinding ferrous alloys is:
(a) Silicon carbide
(b) Diamond
(c) Aluminium oxide
(d) Boron carbide
(1 Mark, 2000)

Ans: c
22. A conventional lathe and a CNC lathe are under consideration for machining a given part. The relevant data are shown below.
The machine preferred for producing 100 pieces is
(a) Conventional lathe
(b) CNC lathe
(c) Any of the Above
(d) None of these
(2 Mark, 2000)

Ans: b
Conventional lathe:
production time per part = 30 min.
Time required to produce 100 parts
= 100 \times 30 = 3000 min. = 50 hr.
Cost for producing 100 pieces
= 30 + 75 \times 50 = Rs.3780
CNC lathe:
production time per part = 15 min.
Time required to produce 100 parts
= 100 \times 15 = 1500 min. = 25 hr.
Cost for producing 100 pieces
= 150 + 120 \times 25 = Rs.3150
As cost for producing in CNC lathe is less than the conventional lathe, so CNC lathe is preferred.
23. A leadscrew with half nuts in a lathe, free to rotate in both directions has
(a) Vthreads
(b) Whitworth threads
(c) Buttress threads
(d) Acme threads
(1 Mark, 2002)

Ans: d
24. The hardness of a grinding wheel is determined by the
(a) Hardness of abrasive grains
(b) Ability of the bond to retain abrasives
(c) Hardness of the bond
(d) Ability of the grinding wheel to penetrate the work piece
(1 Mark, 2002)

Ans: b
The bond strength of the grinding wheel, commonly known as wheel hardness.
25. The time taken to drill a hole through a 25 mm thick plate with the drill rotating at 300 rpm and moving at a feed rate of 0.25 mm/revolution is
(a) 10 sec
(b) 20 sec
(c) 60 sec
(d) 100 sec
(2 Mark, 2002)

Ans: b
Machining time to drill the hole (t)= \frac { L }{ fN }
=> t = \frac { 25 }{ 0.25\times 300 } =\frac { 1 }{ 3 } min = 20 sec.
26. Trepanning is performed for
(a) Finishing a drilled hole
(b) Producing a large hole without drilling
(c) Truing a hole for alignment
(d) Enlarging a drilled hole
(1 Mark, 2003)

Ans: b
Trepanning is the operation of producing a hole by removing metal along the circumference of a hollow cutting tool. Trepanning is performed for producing large holes.
27. Quality screw threads are produced by
(a) Thread milling
(b) Thread chasing
(c) Thread cutting with single point tool
(d) Thread casting
(1 Mark, 2003)

Ans: c
Tread milling: This process gives quite fast production by using suitable thread milling cutters in centre lathes.
Thread casting: Only a few threads over short length. This process provides less accurate and poor finished threads.
Example – cast iron pipes
Thread cutting by Single point and multipoint chasing: This process is slow but can provide high quality. Multipoint chasing gives more productivity but at the cost of quality to some extent.
28. Through holes of 10 mm diameter are to be drilled in a steel plate of 20 mm thickness. Drill spindle speed is 300 rpm, feed 0.2 mm/rev and drill point angle is 120^{0}. Assuming drill over travel of 2 mm, the time for producing a hole will be
(a) 4 seconds
(b) 25 seconds
(c) 100 seconds
(d) 110 seconds
(2 Mark, 2004)

Ans: b
29. The figure below shows a graph, which qualitatively relates cutting speed and cost per piece produced.
The three curves, 1, 2 and 3 respectively represent
(a) Machining cost, nonproductive cost, tool changing cost
(b) Nonproductive cost, machining cost, tool changing cost
(c) Tool changing cost, machining cost, nonproductive cost
(d) Tool changing cost, nonproductive cost, machining cost
(1 Mark, 2005)

Ans: a
machining cost (C) = machining time (t) \times machining cost per unit time (C_{u})
=> As the speed increases, (t) reduces but (C_{u}) remain constant. So, All total, (C) reduces.
=>Thus curve – 1 represents machining cost.As the speed increases, the wear and tear of the tool increases. So, tool changing cost increases.
=> Thus curve – 3 represents toolchanging cost.The non productive costs are independent of cutting speed.
=> Thus curve – 2 represents nonproductive costs.
30. A 600 mm × 300 mm flat surface of a plate is to be finish machined on a shaper. The plate has been fixed with the 600 mm side along the tool travel direction. If the tool overtravel at each end of the plate is 20 mm, average cutting speed is 8 m/min, feed rate is 0.3 mm/stroke and the ratio of return time to cutting time of the tool is 1:2, the time required for machining will be
(a) 8 minutes
(b) 12 minutes
(c) 16 minutes
(d) 20 minutes
(2 Mark, 2005)

Ans: b
Linked answer question 31 and 32.
A low carbon steel bar of 147 mm diameter with a length of 630 mm is being turned with uncoated carbide insert. The observed tool lives are 24 min. and 12 min. for cutting velocities of 90 m/min and 120 m/min. respectively. The feed and depth of cut are 0.2mm/rev and 2mm respectively. Use the unmachined diameter to calculate the cutting velocity.
31. When tool life is 20min, the cutting velocity in m/min is
(a) 87
(b) 97
(c) 107
(d) 114
(2 Mark, 2007)

Ans: b
32. Neglect overtravel or approach of the tool. When tool life is 20min., the machining time in min for a single pass is
(a) 5
(b) 10
(c) 15
(d) 20
(2 Mark, 2007)

Ans: c
33. The figure shows an incomplete schematic of a conventional lathe to be used for cutting threads with different pitches. The speed gear box U_{v} is shown and the feed gear box U_{s} is to be placed. P, Q, R and S denote locations and have no other significance. Changes in U_{v} should NOT affect the pitch of the thread being cut and changes in U_{s} should NOT affect the cutting speed.
The correct connections and the correct placement of U_{s} are given by
(a) Q and E are connected. U_{s} is placed between P and Q.
(b) S and E are connected. U_{s} is placed between R and S.
(c) Q and E are connected. U_{s} is placed between Q and E.
(d) S and E are connected. U_{s} is placed between S and E.
(2 Mark, 2008)

Ans: c
34. Internal gear cutting operation can be performed by
(a) Milling
(b) Shaping with rack cutter
(c) Shaping with pinion cutter
(d) Hobbing
(1 Mark, 2008)

Ans: c
Gear planning and gear hobbing are limited to external gears only whereas gear shaping with pinion cutter is for both external and internal gears.
35. In a single pass drilling operation, a through hole of 15 mm diameter is to be drilled in a steel plate of 50 mm thickness. Drill spindle speed is 500 rpm, feed is 0.2 mm/rev and drill point angle is 118°. Assuming 2 mm clearance at approach and exit, the total drill time (in seconds) is
(a) 35.1
(b) 32.4
(c) 31.2
(d) 30.1
(2 Mark, 2012)

Ans: a
36. A CNC vertical milling machine has to cut a straight slot of 10 mm width and 2 mm depth by a cutter of 10 mm diameter between points (0, 0) and (100, 100) on the XY plane (dimensions in mm). The feed rate used for milling is 50 mm/min. Milling time for the slot (in seconds) is
(a) 120
(b) 170
(c) 180
(d) 240
(1 Mark, 2012)

Ans: b
37. A steel bar 200 mm in diameter is turned at a feed of 0.25 mm/rev with a depth of cut of 4 mm. The rotational speed of the workpiece is 160 rpm. The material removal rate in mm^{3}/s is
(a) 160
(b) 167.6
(c) 1600
(d) 1675.5
(1 Mark, 2013)

Ans: d
38. Match the Machine Tools (Gr. A) with the probable Operations (Gr. B):
(a) P1, Q2, R4, S3
(b) P2, Q1, R4, S3
(c) P3, Q1, R4, S2
(d) P3, Q4, R2, S1
(1 Mark, 2014[2])

Ans: c
39. A hole of 20 mm diameter is to be drilled in a steel block of 40 mm thickness. The drilling is performed at rotational speed of 400 rpm and feed rate of 0.1 mm/rev. The required approach and over run of the drill together is equal to the radius of drill. The drilling time (in minute) is
(a) 1.00
(b) 1.25
(c) 1.50
(d) 1.75
(2 Mark, 2014[2])

Ans: b
40. Cutting tool is much harder than the workpiece. Yet the tool wears out during the toolwork interaction, because
(a) Extra hardness is imparted to the workpiece due to coolant used
(b) Oxide layers on the workpiece surface impart extra hardness to it
(c) Extra hardness is imparted to the workpiece due to severe rate of strain
(d) Vibration is induced in the machine tool
(1 Mark, 2014[3])

Ans: c
41. Two separate slab milling operations, 1 and 2, are performed with identical milling cutters. The depth of cut in operation 2 is twice that in operation 1. The other cutting parameters are identical. The ratio of maximum uncut chip thicknesses in operations 1 and 2 is _____.
(1 Mark, 2014[4])

Ans: 0.707
42. A cast iron block of 200 mm length is being shaped in a shaping machine with a depth of cut of4 mm, feed of 0.25 mm/stroke and the tool principal cutting edge angle of 30°. Number of cutting strokes per minute is 60. Using specific energy for cutting as 1.49 J/mm^{3}, the average power consumption (in watt) is _______.
(2 Mark, 2014[4])

Ans: 298
43. An orthogonal turning operation is carried out under the following conditions: rake angle = 5º, spindle rotational speed = 400 rpm; axial feed = 0.4 m/min and radial depth of cut = 5 mm. The chip thickness, t_{c}, is found to be 3 mm. The shear angle (in degrees) in this turning process is_____.
(2 Mark, 2015[1])

Ans: 18.88
Axial feed = 0.4 m/min = 400 mm/min
Feed in mm/rev = \frac { Axial.feed }{ N }
=> f = \frac { 400 }{ 400 } = 1 mm/rev
In orthogonal turning,
Uncut chip thickness (t) = feed = 1 mm
Chip thickness ratio (r)=\frac { t }{ { t }_{ c } } =\frac { 1 }{ 3 } = 0.33
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.33\times cos{ 5 }^{ 0 } }{ 10.33\times sin{ 5 }^{ 0 } }
=> \phi ={ 18.88 }^{ 0 }
44. In a machining operation, if the generatrix and directrix both are straight lines, the surface obtained is
(a) Cylindrical
(b) Helical
(c) Plane
(d) Surface of revolution
(1 Mark, 2015[3])
45. A shaft of length 90 mm has a tapered portion of length 55 mm. The diameter of the taper is 80 mm at one end and 65 mm at the other. If the taper is made by tail stock set over method, the taper angle and the set over respectively are
(a) 15°32′ and 12.16 mm
(b) 18°32′ and 15.66 mm
(c) 11°22′ and 10.26 mm
(d) 10°32′ and 14.46 mm
(2 Mark, 2015[3])

Ans: a
46. The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is 200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min) is ______.
(1 Mark, 2016[2])

Ans: 7.54
47. Internal gears are manufactured by
(a) Hobbing
(b) Shaping with pinion cutter
(c) Shaping with rack cutter
(d) Milling
(1 Mark, 2016[3])

Ans: b
Gear planning and gear hobbing are limited to external gears only whereas gear shaping with pinion cutter is for both external and internal gears.
48. A firm uses a turning center, a milling center and a grinding machine to produce two parts. The table below provides the machining time required for each part and the maximum machining time available on each machine. The profit per unit on parts I and II are Rs. 40 and Rs. 100, respectively. The maximum profit per week of the firm is Rs._____.
(2 Mark, 2016[3])

Ans: 40,000
49. A block of length 200 mm is machined by a slab milling cutter 34 mm in diameter. The depth of cut and table feed are set at 2 mm and 18 mm/minute, respectively. Considering the approach and the over travel of the cutter to be same, the minimum estimated machining time per pass is ______ minutes.
(2 Mark, 2017[1])

Ans: 12
50. Metric thread of 0.8 mm pitch is to be cut on a lathe. Pitch of the lead screw is 1.5 mm. If the spindle rotates at 1500 rpm, the speed of rotation of the lead screw (rpm) will be_____.
(1 Mark, 2017[1])

Ans: 800
We know,
\frac { { \left( Pitch \right) }_{ th } }{ { \left( Pitch \right) }_{ ls } } = \frac { { \left( Rotational\quad speed \right) }_{ ls } }{ { \left( Rotational\quad speed \right) }_{ th } }
=> \frac { 0.8 }{ 1.5 } =\frac { N }{ 1500 }
=> N = 800 rpm
51. A grinding ratio of 200 implies that the
(a) Grinding wheel wears 200 times the volume of the material removed
(b) Grinding wheel wears 0.005 times the volume of the material removed
(c) Aspect ratio of abrasive particles used in the grinding wheel is 200
(d) Ratio of volume of abrasive particle to that of grinding wheel is 200
(1 Mark, 2018[1])

Ans: b
Grinding ratio is the quick method of evaluating the grinding wheel performance.
Grinding ratio is defined as the ratio between the volume of the material removed from the workpiece to the wear of grinding wheel.
So, option (b) is correct.
52. Feed rate in slab milling operation is equal to
(a) Rotation per minute (rpm)
(b) Product of rpm and number of teeth in the cutter
(c) Product of rpm, feed per tooth and number of teeth in the cutter
(d) Product of rpm, feed per tooth and number of teeth in contact
(1 Mark, 2018[2])

Ans: c
Feed rate in slab milling = N\times Z\times f
Where, N = rpm
Z = number of teeth in cutter
f = feed per tooth
53. The preferred option for holding an oddshaped workpiece in a centre lathe is
(a) Live and dead centres
(b) Three jaw chuck
(c) Lathe dog
(d) Four jaw chuck
(1 Mark, 2018[2])

Ans: d
Four jaw chuck/ independent jaw chuck:
Here the jows can move independently, thus any irregular surface can be effectively cantered.
Limitation: More time is spent in fixturing a component compared to the three jaw chuck.
54. The length, width and thickness of a steel sample are 400 mm, 40 mm and 20 mm, respectively. Its thickness needs to be uniformly reduced by 2 mm in a single pass by using horizontal slab milling. The milling cutter (diameter: 100 mm, width: 50 mm) has 20 teeth and rotates at 1200 rpm. The feed per tooth is 0.05 mm. The feed direction is along the length of the sample. If the overtravel distance is the same as the approach distance, the approach distance and time taken to complete the required machining task are
(a) 14mm, 18.4 s
(b) 21 mm, 39.4 s
(c) 21 mm, 28.9s
(d) 14mm, 21.4 s
(1 Mark, 2019[1])

Ans: d
Depth of cut (d) = 2 mm
Diameter of milling cutter (D) = 100 mm
Approach (A) = \sqrt { d\left( Dd \right) }
= \sqrt { 2\left( 1002 \right) } = 14 mm
Total length (TL) = L + A + O = 400 + 14 + 14 = 428 mm
Table feed (f_{tab}) = Z * f * N
= 20 * 0.05 * 1200
= 1200 mm/min. = 20 mm/sec
Time required to complete the machining = \frac { TL }{ { f }_{ tab } } =\frac { 428 }{ 20 } = 21.4 s
55. A through hole is drilled in an aluminum alloy plate of 15 mm thickness with a drill bit of diameter 10 mm, at a feed of 0.25 mm/rev and a spindle speed of 1200 rpm. If the specific energy required for cutting this material is 0.7 Nm/mm^{3}, the power required for drilling is ___ W (round off to two decimal places).
(2 Mark, 2019[2])

Ans: 274.88
Power required for drilling
= MRR \times Specific energy consumption
= \left( \frac { \pi { d }^{ 2 } }{ 4 } \times f\times N \right) \times 0.7
= \left( \frac { \pi \times { 10 }^{ 2 } }{ 4 } \times 0.25\times \frac { 1200 }{ 60 } \right) \times 0.7
= 274.88 W