1. Cutting tools are provided with large positive rake angle mainly for
(a) Increasing the strength of the cutting edge
(b) Avoiding rubbing action with the finished surfaces
(c) Reducing the magnitude of the cutting force
(d) Better heat dissipation
(1 Mark, 1987)

Ans: c
Large positive rake angle offers sharp cutting edge and thus less chip deformation during machining action. So, the cutting force is reduced.
2. If in a turning operation both the feed rate and the nose radius are doubled the surface finish value will be
(a) Decreases by 50%
(b) Increases by 300%
(c) Increases by 100%
(d) Remain unaffected
(1 Mark, 1987)

Ans: c
Surface finish value (h) = \frac { { f }^{ 2 } }{ 8r }
When both feed rate and the nose radius doubled,
Surface finish value (h) = \frac { { \left( 2f \right) }^{ 2 } }{ 8\left( 2r \right) } = 2\times \frac { { f }^{ 2 } }{ 8r }
S0, the surface finish value doubles or it increases 100%.
3. Pure metal poses machinability problem in turning operations. The reason is the
(a) Increased length of contact due to the production of continuous chip
(b) Susceptibility to chemical reactions
(c) Tendency to from intense adhesion joint with the tool face
(d) Absence of inclusions which aids chip formation
(2 Mark, 1988)

Ans: a
Option ‘a’ is correct.
4. The size of BUE in metal cutting increases with
(a) Very high speed
(b) Large uncut chip thickness
(c) Use of cutting fluid
(d) Increase in positive rake angle
(1 Mark, 1989)

Ans: b
Large uncut chip thickness increases the size of BUE in metal cutting.
5. Crater wear always starts at some distance from the tool tip because at that point
(a) Cutting fluid does not penetrate
(b) Chip tool interface temperature is maximum
(c) Normal stress on rake face is maximum
(d) Tool strength is minimum
(1 Mark, 1989)

Ans: b
6. 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
7. Most of the metal cutting heat goes into the
(a) Moving chip
(b) Cutting tool
(c) Work material
(d) Machine tool
(1 Mark, 1990)
8. The effect of rake angle on the mean friction angle in machining can be explained by
(a) Sliding (coulomb) model of friction
(b) Sticking and then siding model of friction
(c) Sticking friction model
(d) Sliding and then sticking model of friction
(1 Mark, 1992)

Ans: a
Chips always slide over the tool.
9. Cutting power consumption in turning can be significantly reduced by
(a) Increasing rake angle of the tool
(b) Increasing the cutting angles of the tool
(c) Widening the nose radius of the tool
(d) Increasing the clearance angle
(1 Mark, 1995)

Ans: a
Large positive rake angle offers sharp cutting edge and thus less chip deformation during machining action. So, the cutting power is reduced.
10. Plain milling of mild steel plate produces
(a) Irregular shaped discontinuous chips
(b) Regular shaped discontinuous chips
(c) Continuous chips without built up edge
(d) Jointed chips
(1 Mark, 1995)

Ans: b

Ans: A – 3, B – 4, C – 2, D – 1
12. Machine tool structures are made ______ for high process capability. (tough/ strong/ rigid)
(2 Mark, 1995)

Ans: Rigid
13. The rake angle in a drill
(a) Increases from centre to periphery
(b) Decreases from centre to periphery
(c) Remains constant
(d) Is irrelevant to the drilling operation
(1 Mark, 1996)
14. In a typical metal cutting operation, using a cutting tool of positive rake α =10°, it was observed that the shear angle was 20°. The friction angle is
(a) 45°
(b) 30°
(c) 60°
(d) 40°
(2 Mark, 1997)

Ans: c
from merchant’s theory,
\phi = { 45 }^{ 0 }\left( \frac { \beta }{ 2 } \frac { \alpha }{ 2 } \right)
=> { 20 }^{ 0 }={ 45 }^{ 0 }\left( \frac { \beta }{ 2 } \frac { { 10 }^{ 0 } }{ 2 } \right)
=> \beta ={ 60 }^{ 0 }
15. A cutting tool has a nose radius of 1.8 mm. The feed rate for a theoretical surface roughness of R_{a} = 5μm is
(a) 0.36 mm/rev
(b) 0.187 mm/rev
(c) 0.036 mm/rev
(d) 0.0187 mm/rev
(2 Mark, 1997)

Ans: b
We know, { R }_{ a }=\frac { 8{ f }^{ 2 } }{ 18\sqrt { 3 } r }
=> 5\times { 10 }^{ 3 }=\frac { 8{ \times f }^{ 2 } }{ 18\sqrt { 3 } \times 1.8 }
=> f = 0.187 mm/rev
16. In an orthogonal machining operation, the chip thickness and the uncut thickness are equal to 0.45 mm. If the tool rake angle is 0°, the shear plane angle is
(a) 45°
(b) 30°
(c) 18°
(d) 60°
(2 Mark, 1998)

Ans: a
Chip thickness ratio (r)=\frac { t }{ { t }_{ c } } =\frac { 0.45 }{ 0.45 } = 1
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 1\times cos{ 0 }^{ 0 } }{ 11\times sin{ 0 }^{ 0 } }
=> \phi ={ 45 }^{ 0 }
17. What is approximate percentage change is the life, t, of a tool with zero rake angle used in orthogonal cutting when its clearance angle, a, is changed from 10° to 7°? (Hint: Flank wear rate is proportional to cotα)
(a) 30% increase
(b) 30% decrease
(c) 70% increase
(d) 70% decrease
(2 Mark, 1999)

Ans: b
Tool life \propto \frac { 1 }{ flank\quad wear }
=> T\propto \frac { 1 }{ cot\alpha }
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } = \frac { cot{ \alpha }_{ 1 } }{ cot{ \alpha }_{ 2 } }
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } = \frac { cot{ 10 }^{ 0 } }{ cot{ 7 }^{ 0 } }
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } = 0.7
=> { T }_{ 2 }=0.7{ T }_{ 1 }
=>Toll life decreased by 30%.
18. During orthogonal cutting of mild steel with a 10° rake angle tool the chip thickness ratio was obtained as 0.4. the shear angle (in degrees) evaluated from this data is
(a) 6.53
(b) 20.22
(c) 22.94
(d) 50.00
(2 Mark, 2001)

Ans: c
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.4\times cos{ 10 }^{ 0 } }{ 10.4\times sin{ 10 }^{ 0 } }
=> \phi ={ 22.94 }^{ 0 }
19. A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?
(a) 29
(b) 31
(c) 37
(d) 42
(2 Mark, 2003)

Ans: a
Given, 10 tools produce 500 components at 50 rpm.
So, 1 tool produce 50 components at 50 rpm.
Take, V_{1} = 50 rpm
and T_{1 }= 50 components
We know, VT^{n} = C
=> 50\times 50^{n }= C ……….. (1)
Again, 10 tools produce 122 components at 80 rpm.
So, 1 cutting tool produce 12.2 components at 80 rpm.
Take, V_{2} = 80 rpm
and T_{2 }= 12.2 components
80\times 12.2^{n }= C ……….. (2)
Solving equation (1) and (2),
n = 0.33, C = 181.81
The components can be produced with one cutting tool at 60 rpm,
60\times T^{0.33 }= 181.81
=> T \simeq 29 components
Data for Q.20 – 21 are given below.
A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at 200 rpm. The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4mm. the rake angle is 10°. In the analysis it is found that the shear angle is 27.75°.
20. The thickness of the produced chip is
(a) 0.511 mm
(b) 0.528 mm
(c) 0.818 mm
(d) 0.846 mm
(2 Mark, 2003)

Ans: a
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan{ 27.75 }^{ 0 }=\frac { rcos{ 10 }^{ 0 } }{ 1rsin{ 10 }^{ 0 } }
After solving, r = 0.488
Given axial feed = 0.25 mm/rev
So, Uncut chip thickness (t) = 0.25 mm/rev
Again, r = \frac { t }{ { t }_{ c } }
=> 0.488 = \frac { 0.25 }{ { t }_{ c } }
Chip thickness (t_{c}) = \frac { 0.25 }{ 0.488 } = 0.511 mm
21. In the above problem, the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is
(a) 0.18
(b) 0.36
(c) 0.71
(d) 0.98
( Mark, 2003)

Ans: d
According to Earnest and Merchant theory,
\phi = { 45 }^{ 0 }+\frac { 1 }{ 2 } \left( \alpha \beta \right)
=> { 27.75 }^{ 0 }={ 45 }^{ 0 }+\frac { 1 }{ 2 } \left( { 10 }^{ 0 }\beta \right)
=> { \beta =44.5 }^{ 0 }
Coefficient of chip = { tan\beta =tan44.5 }^{ 0 } = 0.98
22. In an orthogonal cutting test on mild steel, the following data were obtained
Cutting speed: 40 m/min,
Depth of cut: 0.3 mm,
Tool rake angle: + 5^{0},
Chip thickness: 1.5 mm,
Cutting force: 900 N,
Thrust force: 450 N
Using Merchant’s analysis, the Friction angle during the machining will be
(a) 26.6^{0}
(b) 31.5^{0}
(c) 45^{0}
(d) 63.4^{0}
(2 Mark, 2004)

Ans: b
We know, Friction force = F
Normal force = N
Cutting force = F_{c}
Thrust force = F_{t}Now, F={ F }_{ c }sin\alpha +{ F }_{ t }cos\alpha
=> F=900\times sin{ 5 }^{ 0 }+450\times cos{ 5 }^{ 0 }
=> F = 526.72 N…….(1)
Again, N = { F }_{ c }cos\alpha { F }_{ t }sin\alpha
=> N=900\times cos{ 5 }^{ 0 }450\times sin{ 5 }^{ 0 }
=> N = 857.355 N………(2)
Now, tan\beta =\frac { F }{ N }
=> tan\beta =\frac { 526.72 }{ 857.355 }
=> \beta ={ 31.5 }^{ 0 }
23. In a machining operation, doubling the cutting speed reduces the tool life to 1/8^{th} of the original value. The exponent n in Taylor’s tool life equation VT^{n} = C is
(a) 1/8
(b) 1/4
(c) 1/3
(d) 1/2
(2 Mark, 2004)

Ans: c
Given, V{ T }^{ n }=C
=> { V }_{ 1 }{ T }_{ 1 }^{ n }={ V }_{ 2 }{ T }_{ 2 }^{ n }
=> { V }_{ 1 }{ T }_{ 1 }^{ n }={ 2V }_{ 1 }\times \left( \frac { 1 }{ 8 } { T }_{ 1 } \right) ^{ n }
=> 1=2\times \left( \frac { 1 }{ 8 } \right) ^{ n }
=> { 8 }^{ n }=2
=> n = 1/3
24. Two tools P and Q have signatures 5° − 5° − 6° − 8° − 30° − 0 and 5° − 5° − 7° − 7° − 8° − 15° − 0 (both ASA) respectively. They are used to turn components under the same machining conditions. If h_{P} and h_{Q} denote the peak to valley heights of surfaces produced by the tools P and Q, the ratio h_{P}/h_{Q} will be
(a) \frac { { tan 8 }^{ 0 } + cot { 15 }^{ 0 } }{ { tan 8 }^{ 0 } + cot { 30 }^{ 0 } }
(b) \frac { { tan 15 }^{ 0 } + cot { 8 }^{ 0 } }{ { tan 30 }^{ 0 } + cot { 8 }^{ 0 } }
(c) \frac { { tan 15 }^{ 0 } + cot { 7 }^{ 0 } }{ { tan 30 }^{ 0 } + cot 7^{ 0 } }
(d) \frac { { tan 7 }^{ 0 } + cot { 15 }^{ 0 } }{ { tan 7 }^{ 0 } + cot 30^{ 0 } }
(2 Mark, 2005)

Ans: b
Tool signature in ASA system:
Bake rake angle – Side rake angle – End relief angle – Side relief angle – End cutting edge angle – Side cutting edge angle – Nose radius
Peak to valley surface roughness (h) = \frac { f }{ tanSCEA+cotECEA }
Now, { h }_{ P }=\frac { f }{ tan{ 30 }^{ 0 }+cot{ 30 }^{ 0 } }
Again, { h }_{ Q }=\frac { f }{ tan{ 15 }^{ 0 }+cot{ 8 }^{ 0 } }
So, \frac { { h }_{ P } }{ { h }_{ Q } } =\frac { tan{ 15 }^{ 0 }+cot{ 8 }^{ 0 } }{ tan{ 30 }^{ 0 }+cot{ 8 }^{ 0 } }
25. In orthogonal turning of a low carbon steel bar of diameter 150mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2mm. The chip thickness obtained is 0.48mm. If the orthogonal rake angle is zero and the principle cutting edge angle is 90^{o}, the shear angle in degree is
(a) 20.56
(b) 26.56
(c) 30.56
(d) 36.56
(1 Mark, 2007)

Ans: b
For turning,
Uncut chip thickness (t) = fsin\phi
=> t = 0.24\times sin{ 90 }^{ 0 } = 0.24
Chip thickness ratio (r)=\frac { t }{ { t }_{ c } }
=> r = \frac { 0.24 }{ 0.48 } = 0.5
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.5\times cos{ 0 }^{ 0 } }{ 10.5\times sin{ 0 }^{ 0 } }
=> \phi ={ 26.56 }^{ 0 }
Common data for questions 26, 27 and 28.
In an orthogonal machining operation:
Uncut thickness = 0.5 mm,
Cutting speed = 20 m/min
Width of cut = 5 mm,
Chip thickness = 0.7 mm,
Thrust force = 200 N,
Cutting force = 1200 N
Rake angle = 15°
Assume Merchant’s theory.
26. The values of shear angle and shear strain, respectively, are
(a) 30.3° and 1.98
(b) 30.3° and 4.23
(c) 40.2° and 2.97
(d) 40.2° and 1.65
(2 Mark, 2006)

Ans: d
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.5 }{ 0.7 } = 0.714
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.714\times cos{ 15 }^{ 0 } }{ 10.714\times sin{ 15 }^{ 0 } }
=> \phi ={ 40.2 }^{ 0 }
Shear strain \left( \varepsilon \right) =cot\phi +tan\left( \phi \alpha \right)
= cot{ 40.2 }^{ 0 }+tan\left( { 40.2 }^{ 0 }15^{ 0 } \right) = 1.65
27. The coefficient of friction at the toolchip interface is:
(a) 0.23
(b) 0.46
(c) 0.85
(d) 0.95
(2 Mark, 2006)

Ans: b
We know, Friction force = F
Normal force = N
Cutting force = F_{c}
Thrust force = F_{t}Now, F={ F }_{ c }sin\alpha +{ F }_{ t }cos\alpha
=> F=1200\times sin{ 15 }^{ 0 }+200\times cos{ 15 }^{ 0 }
=> F = 503.76 N…….(1)
Again, N = { F }_{ c }cos\alpha { F }_{ t }sin\alpha
=> N=1200\times cos{ 15 }^{ 0 }200\times sin{ 15 }^{ 0 }
=> N = 1107.34 N………(2)
Coefficient of friction at the tool chip:
tan\beta = \frac { F }{ N }
= \frac { 503.76 }{ 1107.34 } = 0.46
28. The percentage of total energy dissipated due to friction at the toolchip interface is:
(a) 30%
(b) 42%
(c) 58%
(d) 70%
(2 Mark, 2006)

Ans: a
Cutting force (F_{c}) = 1200 N
Cutting velocity (V_{c}) = 20 m/min
Let, Shear force = F_{s}
Shear velocity = V_{s}
Friction force = F
Chip velocity = VWe know, { F }_{ s }={ F }_{ c }cos\phi { F }_{ t }sin\phi
= 1200\times cos{ 40.2 }^{ 0 }200\times sin{ 40.2 }^{ 0 }
= 787.46 N
Again, \frac { { V }_{ c } }{ cos\left( \phi \alpha \right) } =\frac { { V }_{ s } }{ cos\alpha }
=> \frac { 20 }{ cos\left( 40.215 \right) } =\frac { { V }_{ s } }{ cos15 }
=> V_{s }= 21.35 m/min.
We know, F_{c}V_{c }= F_{s}V_{s }+ FV
Total energy dissipated due to friction:
FV = F_{c}V_{c }– F_{s}V_{s}
Percentage of total energy dissipated due to friction
= \frac { FV }{ { F }_{ c }{ V }_{ c } } =\frac { { F }_{ c }{ V }_{ c }{ F }_{ s }{ V }_{ s } }{ { F }_{ c }{ V }_{ c } }
= \frac { 1200\times 20787.46\times 21.35 }{ 1200\times 20 }
= 0.30 = 30%
29. In orthogonal turning of medium carbon steel, the specific machining energy is 2.0 J/mm^{3}. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev. and 2 mm respectively. The main cutting force in N is
(a) 40
(b) 80
(c) 400
(d) 800
(2 Mark, 2007)

Ans: d
Specific machining energy (e) = \frac { power }{ MRR }
=> e = \frac { { F }_{ C }\times V }{ V\times f\times d }
Where, F_{c }= Cutting force
V = Cutting velocity
f = feed, d = depth of cut=> e = \frac { { F }_{ C } }{ f\times d }
=> 2 = \frac { { F }_{ C } }{ 0.2\times 2 }
=> F_{c }= 0.8 J/mm = 800 J/m = 800 N
30. In orthogonal turning of low carbon steel pipe with principal cutting edge angle of 90^{0}, the main cutting force is 1000 N and the feed force is 800 N. The shear angle is 25^{0} and orthogonal rake angle is zero. Employing Merchant’s theory, the ratio of friction force to normal force acting on the cutting tool is
(a) 1.56
(b) 1.25
(c) 0.80
(d) 0.64
(2 Mark, 2007)

Ans: c
We know, Friction force = F
Normal force = N
Cutting force = F_{c}
Feed force = F_{t}Now, F={ F }_{ c }sin\alpha +{ F }_{ t }cos\alpha
=> F=1000\times sin{ 0 }^{ 0 }+800\times cos{ 0 }^{ 0 }
=> F = 800 N…….(1)
Again, N = { F }_{ c }cos\alpha { F }_{ t }sin\alpha
=> N=1000\times cos{ 0 }^{ 0 }800\times sin{ 0 }^{ 0 }
=> N = 1000 N………(2)
Ratio of friction force to normal force acting on the cutting tool
= \frac { F }{ N } = \frac { 800 }{ 1000 } = 0.8
Statement for Linked Answer Questions 31 and 32:
Orthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following conditions are used: cutting velocity is 180 m/min, feed is 0.20 mm/rev, depth of cut is 3 mm, chip thickness ratio = 0.5. The orthogonal rake angle is 7˚. Apply Merchant’s theory for analysis.
31. The shear plane angle (in degrees) and the shear force respectively are
(a) 52; 320N
(b) 52; 400N
(c) 28 ; 400 N
(d) 28 ; 320 N
(2 Mark, 2008)

Ans: d
Chip thickness ratio (r) = 0.5
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.5\times cos{ 7 }^{ 0 } }{ 10.5\times sin{ 7 }^{ 0 } }
=> \phi ={ 28 }^{ 0 }
Shear force (F_{s}) = { \tau }_{ s }{ A }_{ s }
= { \tau }_{ s }\times \frac { fd }{ sin\phi }
= 250\times \frac { 0.2\times 3 }{ sin{ 28 }^{ 0 } } \simeq 320 N
32. The cutting and frictional forces, respectively, are
(a) 568 N ; 387 N
(b) 565 N ; 447 N
(c) 440 N ; 342 N
(d) 480 N ; 356 N
(2 Mark, 2008)

Ans: b
From merchant’s theory:
2\phi +\beta \alpha = { 90 }^{ 0 }
=> 2\times 28+\beta 7={ 90 }^{ 0 }
=> \beta ={ 41 }^{ 0 }
From the merchant’s circle:
R = \frac { { F }_{ s } }{ cos\left( \phi +\beta \alpha \right) }
= \frac { 320 }{ cos\left( 28+417 \right) } = \frac { 320 }{ cos62 } = 681.61 N
Cutting force (F_{c}) = Rcos\left( \beta \alpha \right)
= 681.61\times cos\left( 417 \right) = 565 N
Frictional force (F) = Rsin\left( \beta \right)
= 681.61\times sin\left( 41 \right) = 447 N
33. In a single point turning tool, the side rake angle and orthogonal rake angle are equal. φ is the principal cutting edge angle and its range is 0° ≤ φ ≤ 90°. The chip flows in the orthogonal plane. The value of φ is closest to
(a) 0°
(b) 45°
(c) 60°
(d) 90°
(2 Mark, 2008)

Ans: d
In a single point turning tool, the side rake angle and orthogonal rake angle are equal if the principal cutting edge angle is 90^{0}.
34. Friction at the toolchip interface can be reduced by
(a) Decreasing the rake angle
(b) Increasing the depth of cut
(c) Decreasing the cutting speed
(d) Increasing the cutting speed
(1 Mark, 2009)

Ans: d
35. Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is
(a) 0.0
(b) 0.5
(c) 1.0
(d) 2.0
(2 Mark, 2009)

Ans: d
Shear strain \left( \epsilon \right) =cot\phi +tan\left( \phi \alpha \right)
If \alpha = 0 then cot\phi +tan\phi \ge 2
and when \phi ={ 45 }^{ 0 }, { \varepsilon }_{ min }=2
Statement for Linked Questions: 36 & 37.
In a machining experiment, tool life was found to vary with the cutting speed in the following manner:
36. The exponent (n) and constant (k) of the Taylor’s tool life equation are
(a) n = 0.5 and k = 540
(b) n = 1 and k = 4860
(c) n = −1 and k = 0.74
(d) n = −0.5 and k = 1.155
(2 Mark, 2009)

Ans: a
According to Taylor’s tool life equation:
V{ T }^{ n } = k
After using the table,
60\times { 81 }^{ n }=k….(1)
90\times { 36 }^{ n }=k….(2)
Solving equation (1) and (2),
\frac { 6 }{ 9 } \times { \left( \frac { 81 }{ 36 } \right) }^{ n } = 1
=> { \left( \frac { 81 }{ 36 } \right) }^{ n }=\frac { 9 }{ 6 }
=> n = 0.5
Putting the value n = 0.5 in equation (1),
k = 60\times { 81 }^{ 0.5 } = 540
37. What is the percentage increase in tool life when the cutting speed is halved?
(a) 50%
(b) 200%
(c) 300%
(d) 400%
(2 Mark, 2009)

Ans: c
The Taylor’s equation: V\times { T }^{ 0.5 }=540
=> \frac { { V }_{ 1 } }{ { V }_{ 2 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.5 }
=> \frac { { V }_{ 1 } }{ { 0.5V }_{ 1 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.5 }
=> { 2 }^{ 2 }=\frac { { T }_{ 2 } }{ { T }_{ 1 } }
=> { T }_{ 2 }=4{ T }_{ 1 }
The tool life becomes 400% or the tool life increases 300%.
38. For tool A, Taylor’s tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is
(a) 26.7
(b) 42.5
(c) 80.7
(d) 142.9
(2 Mark, 2010)

Ans: a
According to the question:
For tool A: { V }_{ A }{ T }_{ A }^{ 0.45 }=90………(1)
For tool B: { V }_{ B }{ T }_{ B }^{ 0.3 }=60……….(2)
The speed at which both the tools have same tool life means
So, { { T }_{ A }=T }_{ B }=T
and { { V }_{ A }=V }_{ B }=V
Now equation 1 an 2 becomes:
V{ T }^{ 0.45 }=90…….(3)
V{ T }^{ 0.3 }=60……..(4)
Solving equation 3 and 4 We get,
T = 14.926 min.
V = 26.7 m/min
39. A single–point cutting tool with 12^{0} rake angle is used to machine a steel work–piece. The depth of cut, i.e. uncut thickness is 0.81mm. The chip thickness under orthogonal machining condition is 1.8mm. The shear angle is approximately
(a) 22^{0}
(b) 26^{0}
(c) 56^{0}
(d) 76^{0}
(2 Mark, 2011)

Ans: b
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.81 }{ 1.8 } = 0.45
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.45\times cos{ 12 }^{ 0 } }{ 10.45\times sin{ 12 }^{ 0 } }
=> \phi ={ 26 }^{ 0 }
40. Details pertaining to an orthogonal metal cutting process are given below.
Chip thickness ratio = 0.4,
Undeformed thickness = 0.6 mm
Rake angle = +10°,
Cutting speed 2.5 m/s,
Mean thickness of primary shear zone = 25 microns
The shear strain rate in s^{–1} during the process is
(a) 0.1781×10^{5}
(b) 0.7754×10^{5}
(c) 1.0104×10^{5}
(d) 4.397×10^{5}
(2 Mark, 2012)

Ans: c
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.4\times cos10 }{ 10.4\times sin10 }
=> \phi ={ 22.94 }^{ 0 }
We know, \frac { V }{ Cos(\phi \alpha ) } =\frac { { V }_{ S } }{ Cos\alpha }
=> \frac { 2.5 }{ Cos(22.9410) } =\frac { { V }_{ S } }{ Cos10 }
=> Shear velocity = { V }_{ S } = 2.6 m/s
Shear strain rate = \frac { { V }_{ S } }{ { t }_{ S } }
Where, { t }_{ S } = Mean thickness of primary shear zone
=> Shear strain rate = \frac { 2.6 }{ 25\times { 10 }^{ 6 } }
\simeq 1.0104\times { 10 }^{ 5 }{ s }^{ 1 }
41. The main cutting force acting on a tool during the turning (orthogonal cutting) operation of a metal is 400 N. The turning was performed using 2 mm depth of cut and 0.1 mm/rev feed rate. The specific cutting pressure (in N/mm^{2}) is
(a) 1000
(b) 2000
(c) 3000
(d) 4000
(1 Mark, 2014[1])

Ans: b
The specific cutting pressure = \frac { F }{ d\times f }
= \frac { 400 }{ 2\times 0.1 }
= 2000 \frac { N }{ { mm }^{ 2 } }
42. During pure orthogonal turning operation of a hollow cylindrical pipe, it is found that the thickness of the chip produced is 0.5 mm. The feed given to the zero degree rake angle tool is 0.2 mm/rev. The shear strain produced during the operation is _______.
(2 Mark, 2014[1])

Ans: 2.9
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.2 }{ 0.5 } = 0.4
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.4\times cos{ 0 }^{ 0 } }{ 10.4\times sin{ 0 }^{ 0 } }
=> \phi ={ 21.8 }^{ 0 }
Shear strain \left( \varepsilon \right) =cot\phi +tan\left( \phi \alpha \right)
= cot{ 21.8 }^{ 0 }+tan\left( { 21.8 }^{ 0 }0^{ 0 } \right) = 2.9
43. If the Taylor’s tool life exponent n is 0.2, and the tool changing time is 1.5 min, then the tool life (in min) for maximum production rate is _______.
(2 Mark, 2014[1])

Ans: 6
Tool life for maximum production rate:
{ T }_{ 0 } = { T }_{ c }\left( \frac { 1n }{ n } \right)
=> { T }_{ 0 }=1.5\left( \frac { 10.2 }{ 0.2 } \right) = 8 min.
44. A straight turning operation is carried out using a single point cutting tool on an AISI 1020 steel rod. The feed is 0.2 mm/rev and the depth of cut is 0.5 mm. The tool has a side cutting edge angle of 60°. The uncut chip thickness (in mm) is _______.
(1 Mark, 2014[3])

Ans: 0.1
Side cutting edge angle = 60^{0}
principal cutting edge angle (\lambda)
= 90^{0 }– 60^{0} = 30^{0}Uncut chip thickness (t) = f.sin\lambda
=> t = 0.2sin30^{0 }= 0.1
45. Which pair of following statements is correct for orthogonal cutting using a singlepoint cutting tool?
P. Reduction in friction angle increases cutting force
Q. Reduction in friction angle decreases cutting force
R. Reduction in friction angle increases chip thickness
S. Reduction in friction angle decreases chip thickness
(a) P and R
(b) P and S
(c) Q and R
(d) Q and S
(2 Mark, 2014[3])

Ans: d
46. Better surface finish is obtained with a large rake angle because
(a) The area of shear plane decreases resulting in the decrease in shear force and cutting force
(b) The tool becomes thinner and the cutting force is reduced
(c) Less heat is accumulated in the cutting zone
(d) The friction between the chip and the tool is less
(1 Mark, 2014[4])

Ans: a
47. Under certain cutting conditions, doubling the cutting speed reduces the tool life to { \left( \frac { 1 }{ 16 } \right) }^{ th } of the original. Taylor’s tool life index (n) for this toolworkpiece combination will be ____.
(1 Mark, 2015[1])

Ans: 0.25
We know, V{ T }^{ n }=C
=> { V }_{ 1 }{ T }_{ 1 }^{ n }={ V }_{ 2 }{ T }_{ 2 }^{ n }
=> { V }_{ 1 }{ T }_{ 1 }^{ n }={ 2V }_{ 1 }\times \left( \frac { 1 }{ 16 } { T }_{ 1 } \right) ^{ n }
=> 1=2\times \left( \frac { 1 }{ 16 } \right) ^{ n }
=> { 16 }^{ n }=2
=> n = 0.25
48. A single point cutting tool with 0° rake angle is used in an orthogonal machining process. At a cutting speed of 180 m/min, the thrust force is 490 N. If the coefficient of friction between the tool and the chip is 0.7, then the power consumption (in kW) for the machining operation is ________.
(2 Mark, 2015[2])

Ans: 2.1
Given, cutting speed (V) = 180 m/min. = 3 m/s
If rake angle = 0^{o }then
Thrust force (F_{t}) = Friction force (F)
Cutting force (F_{c}) = Normal to frictional force (N)
Coefficient of friction = tan\beta = 0.7
=> \frac { F }{ N } = 0.7
=> \frac { 490 }{ N } = 0.7
=> N = 700 N = F_{c}
Power consumption (P) = F_{c }\times V
=> p = 700 \times 3 = 2100 W = 2.1 kW
49.Orthogonal turning of a mild steel tube with a tool of rake angle 10° is carried out at a feed of 0.14 mm/rev. If the thickness of the chip produced is 0.28 mm, the values of shear angle and shear strain will be respectively
(a) 28°20′ and 2.19
(b) 22°20′ and 3.53
(c) 24°30′ and 4.19
(d) 37°20′ and 5.19
(2 Mark, 2015[3])

Ans: a
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.14 }{ 0.28 } = 0.5
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.5\times cos{ 10 }^{ 0 } }{ 10.5\times sin{ 10 }^{ 0 } }
=> \phi ={ 28.33 }^{ 0 }
Shear strain \left( \varepsilon \right) =cot\phi +tan\left( \phi \alpha \right)
= cot{ 28.33 }^{ 0 }+tan\left( { 28.33 }^{ 0 }10^{ 0 } \right) = 2.186
50. The tool life equation for HSS tool is VT^{0.14}f^{0.7}d^{0.4 }= Constant. The tool life (T) of 30 min is obtained using the following cutting conditions: V = 45 m/min, f = 0.35 mm, d = 2.0 mm. If speed (V), feed (f) and depth of cut (d) are increased individually by 25%, the tool life (in min) is
(a) 0.15
(b) 1.06
(c) 22. 50
(d) 30.0
(2 Mark, 2016[1])

Ans: b
Given, VT^{0.14}f^{0.7}d^{0.4 }= Constant
=> { V }_{ 1 }{ T }_{ 1 }^{ 0.14 }{ f }_{ 1 }^{ 0.7 }{ d }_{ 1 }^{ 0.4 }={ V }_{ 2 }{ T }_{ 2 }^{ 0.14 }{ f }_{ 2 }^{ 0.7 }{ d }_{ 2 }^{ 0.4 }
=> { V }_{ 1 }{ T }_{ 1 }^{ 0.14 }{ f }_{ 1 }^{ 0.7 }{ d }_{ 1 }^{ 0.4 }=\left( 1.25{ V }_{ 1 } \right) { T }_{ 2 }^{ 0.14 }\left( 1.25{ T }_{ 1 } \right) ^{ 0.7 }\left( 1.25{ d }_{ 1 } \right) ^{ 0.4 }
=> { T }_{ 1 }^{ 0.14 }=\left( 1.25 \right) \left( 1.25 \right) ^{ 0.7 }\left( 1.25 \right) ^{ 0.4 }{ T }_{ 2 }^{ 0.14 }
=> \frac { 30^{ 0.14 } }{ \left( 1.25 \right) ^{ 2.1 } } ={ T }_{ 2 }^{ 0.14 }
=> { T }_{ 2 } = 1.06
51. In an orthogonal cutting process the tool used has rake angle of zero degree. The measured cutting force and thrust force are 500 N and 250 N, respectively. The coefficient of friction between the tool and the chip is _____.
(1 Mark, 2016[1])

Ans: 0.5
If rake angle = 0^{o }then
Thrust force (F_{t}) = Friction force (F)
Cutting force (F_{c}) = Normal to frictional force (N)
Coefficient of friction = tan\beta = \frac { F }{ N }
=> Coefficient of friction = \frac { 250 }{ 500 } = 0.5
52. For a certain job, the cost of metal cutting is Rs. 18C/V and the cost of tooling is Rs. 270C/(TV), Where C is a constant, V is the cutting speed in m/min. and and T is the tool life in minutes. The Taylor’s tool life equation is VT^{0.25 }= 150. The cutting speed (in m/min) for the minimum total cost is _______.
(2 Mark, 2016[2])

Ans: 57.9
Given, VT^{0.25 }= 150
=> T={ \left( \frac { 150 }{ V } \right) }^{ 4 }
Total cost (TC) = Metal cutting cost + Tooling cost
=> TC = 18C/V + 270C/(TV)
=> TC=18\frac { C }{ V } +270\frac { C }{ V{ \left( \frac { 150 }{ V } \right) }^{ 4 } }
=> TC=18\frac { C }{ V } +270\frac { C{ V }^{ 3 } }{ 150^{ 4 } }
Cutting speed for minimum total cost:
\frac { d(TC) }{ dV } = 0
=> 18\frac { C }{ { V }^{ 2 } } +270\times \frac { C }{ { 150 }^{ 4 } } \times 3{ V }^{ 2 }=0
=> V = 57.91 m/min
53. In a single point turning operation with cemented carbide tool and steel work piece, it is found that the Taylor’s exponent is 0.25. If the cutting speed is reduced by 50% then the tool life changes by ______ times.
(2 Mark, 2016[3])

Ans: 15
According to the Taylor’s equation:
=> \frac { { V }_{ 1 } }{ { V }_{ 2 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.25 }
=> \frac { { V }_{ 1 } }{ { 0.5V }_{ 1 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.25 }
=> { 2 }^{ 4 }=\frac { { T }_{ 2 } }{ { T }_{ 1 } }
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } =16
=> { T }_{ 2 }=16{ T }_{ 1 }
Changes in tool life = \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 } }
= \frac { { 16T }_{ 1 }{ T }_{ 1 } }{ { T }_{ 1 } } = 15
54. For an orthogonal cutting operation, tool material is HSS, rake angle is 22°, chip thickness is 0.8 mm, speed is 48 m/min and feed is 0.4 mm/rev. The shear plane angle (in degrees) is
(a) 19.24
(b) 29.70
(c) 56.00
(d) 68.75
(2 Mark, 2016[3])

Ans: b
Chip thickness ratio (r)=\frac { t }{ { t }_{ c } } =\frac { 0.4 }{ 0.8 } = 0.5
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.5\times cos{ 22 }^{ 0 } }{ 11\times sin{ 22 }^{ 0 } }
=> \phi ={ 29.7 }^{ 0 }
55. Two cutting tools with tool life equations given below are being compared
Tool 1: VT^{0.1} = 150
Tool 2: VT^{0.3} = 300
Where V is cutting speed in m/minute and T is tool life in minutes. The breakeven cutting speed beyond which Tool 2 have a higher tool life is ____ m/minute.
(2 Mark, 2017[1])

Ans: 106.07
According to the question:
For tool 1: { V }_{ 1 }{ T }_{ 1 }^{ 0.1 }=150………(1)
For tool 2: { V }_{ 2 }{ T }_{ 2 }^{ 0.3 }=300……….(2)
The speed at which both the tools have same tool life means
So, { { T }_{ 1 }=T }_{ 2 }=T
and { { V }_{ 1 }=V }_{ 2 }=V
Now equation 1 an 2 becomes:
V{ T }^{ 0.1 }=150…….(3)
V{ T }^{ 0.3 }=300……..(4)
Solving equation 3 and 4 We get,
T = 32 min.
V = 106.08 m/min
56. Assume that the surface roughness profile is triangular as shown schematically in the figure. If the peak to valley height is 20 μm, the central line average surface roughness R_{0} (in μm) is
(a) 5
(b) 6.67
(c) 10
(d) 20
(2 Mark, 2017[1])

Ans: a
Average surface roughness:
{ R }_{ a } = \frac { peak\quad to\quad vally\quad height }{ 4 } = \frac { 20 }{ 4 } = 5
57. In an orthogonal machining with a tool of 9^{0} orthogonal rake angle, the uncut chip thickness is 0.2 mm. The chip thickness fluctuates between 0.25 mm and 0.4 mm. The ratio of the maximum shear angle to the minimum shear angle during machining is_____.
(2 Mark, 2017[2])

Ans: 1.493
For chip thickness = 0.25 mm,
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } } = \frac { 0.2 }{ 0.25 } = 0.8
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.8\times cos{ 9 }^{ 0 } }{ 10.8\times sin{ 9 }^{ 0 } }
=> \phi ={ 42.077 }^{ 0 }
For chip thickness = 0.4 mm,
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } } = \frac { 0.2 }{ 0.4 } = 0.5
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.5\times cos{ 9 }^{ 0 } }{ 10.5\times sin{ 9 }^{ 0 } }
=> \phi ={ 28.177 }^{ 0 }
Ratio of maximum shear angle to minimum shear angle = \frac { 42.077 }{ 28.177 } = 1.493
58. During the turning of a 20 mm diameter steel bar at a spindle speed of 400 rpm, a tool life of 20 minute is obtained. When the same bar is turned at 200 rpm, the tool life becomes 60 minute. Assume that Taylor’s tool life equation is valid. When the bar is turned at 300 rpm, the tool life ( in minute) is approximately
(a) 25
(b) 32
(c) 40
(d) 50
(2 Mark, 2017[2])

Ans: b
According to Taylor’s tool life equation:
V{ T }^{ n } = k
So, 400\times { 20 }^{ n }=k….(1)
200\times { 60 }^{ n }=k….(2)
Solving equation (1) and (2),
\frac { 2 }{ 1 } \times { \left( \frac { 1 }{ 3 } \right) }^{ n } = 1
=> 3^{ n }=2
=> n = 0.63
Putting the value n = 0.63 in equation (1),
k = 400\times { 20 }^{ 0.63 } = 2640.638
Again, 300\times { T }^{ 0.63 }=2640.638
=> T \simeq 32
59. Using the Taylor’s tool life equation with exponent n = 0.5, if the cutting speed is reduced by 50%, the ratio of new tool life to original tool life is
(a) 4
(b) 2
(c) 1
(d) 0.5
(1 Mark, 2018[1])

Ans: a
According to the Taylor’s equation:
=> \frac { { V }_{ 1 } }{ { V }_{ 2 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.5 }
=> \frac { { V }_{ 1 } }{ { 0.5V }_{ 1 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 0.5 }
=> { 2 }^{ 2 }=\frac { { T }_{ 2 } }{ { T }_{ 1 } }
=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } =4
60. An orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15^{0} and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is ______ (correct to two decimal places).
(2 Mark, 2018[1])

Ans: 0.441
Cutting force (F_{c}) = 60 N
Thrust force (F_{t}) = 25 N
Cutting speed (V_{c}) = 130 m/min
Let, Friction force = F
Chip velocity = VWe know, { F }={ F }_{ c }sin\alpha { +F }_{ t }cos\alpha
= 60\times sin{ 15 }^{ 0 }+25\times cos{ 15 }^{ 0 }
= 39.677 N
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.010 }{ 0.015 } = 0.667
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.667\times cos{ 15 }^{ 0 } }{ 10.667\times sin{ 15 }^{ 0 } }
=> \phi ={ 37.9 }^{ 0 }
Again, \frac { { V }_{ c } }{ cos\left( \phi \alpha \right) } =\frac { { V } }{ sin\phi }
=> \frac { { V }_{ c } }{ cos\left( 37.915 \right) } =\frac { { V } }{ sin37.9 }
=> V_{ }= 86.7 m/min.
Ratio of friction energy to total energy”
= \frac { FV }{ { F }_{ c }{ V }_{ c } } =\frac { 39.677\times 86.7 }{ 60\times 130 }
= 0.441
61. Following data correspond to an orthogonal turning of a 100 mm diameter rod on a lathe. Rake angle: +15^{0}; Uncut chip thickness: 0.5 mm; nominal chip thickness after the cut: 1.25 mm. The shear angle (in degrees) for this process is _______ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 23.31
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.5 }{ 1.25 } = 0.4
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.4\times cos{ 15 }^{ 0 } }{ 10.4\times sin{ 15 }^{ 0 } }
=> \phi ={ 23.31 }^{ 0 }
62. Taylor’s tool life equation is used to estimate the life of a batch of identical HSS twist drills by drilling through holes at constant feed in 20 mm thick mild steel plates. In test 1, a drill lasted 300 holes at 150 rpm while in test 2, another drill lasted 200 holes at 300 rpm. The maximum number of holes that can be made by another drill from the above batch at 200 rpm is ______ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 253.54
According to Taylor’s tool life equation:
V{ T }^{ n } = k
So, 150\times { 300 }^{ n }=k….(1)
300\times { 200 }^{ n }=k….(2)
Solving equation (1) and (2),
\frac { 1 }{ 2 } \times { \left( \frac { 3 }{ 2 } \right) }^{ n } = 1
=> \frac { 1 }{ 2 } ={ \left( \frac { 2 }{ 3 } \right) }^{ n }
=> n = 1.71
Putting the value n = 1.71 in equation (1),
k = 150\times { 300 }^{ 1.71 } = 2582073.524
Again, 200\times { T }^{ 1.71 }=2582073.524
=> T = 253.54
63. In orthogonal turning of a cylindrical tube of wall thickness 5 mm, the axial and the tangential cutting forces were measured at 1259 N and 1601 N, respectively. The measured chip thickness after machining was found to be 0.3 mm. The rake angel was 10° and the axial feed was 100 mm/min. The rotational speed of the spindle was 1000 rpm. Assuming the material to be perfectly plastic and Merchant’s first solution, the shear strength of the martial is closest to
(a) 722 MPa
(b) 920 MPa
(c) 200 MPa
(d) 875 MPa
(2 Mark, 2019[1])

Ans:
Axial feed in mm/min. = 100
Rotational speed = 1000 rpm
Axial feed in mm/rev. (t) = 100/1000 = 0.1
Chip thickness (t_{c}) = 0.3 mm
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.1 }{ 0.3 } = 0.33
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.33\times cos{ 10 }^{ 0 } }{ 10.33\times sin{ 10 }^{ 0 } }
=> \phi ={ 19 }^{ 0 }
Given, Axial force / Thrust force (F_{t}) = 1259 N
Tangential cutting force (F_{c}) = 1601 N
Shear force (F_{s}) = { F }_{ c }cos\phi { F }_{ t }sin\phi
= 1601\times cos{ 19 }^{ 0 }1259\times sin{ 19 }^{ 0 }
= 1103.88 N
Shear strength of the material \left( { \tau }_{ s } \right) =\frac { { F }_{ s }\times sin\phi }{ bt }
= \frac { 1103.88\times sin{ 19 }^{ 0 } }{ 5\times 0.1 } \simeq 722 MPa
64. In ASA system, the side cutting and end cutting edge angles of a sharp turning tool are 45° and 10°, respectively. The feed during cylindrical turning is 0.1 mm/rev. The center line average surface roughness (in μm, round off to one decimal place) of the generated surface is _______.
(2 Mark, 2019[1])

Ans: 3.7
The centre line average surface roughness:
R_{cla }= \frac { 0.1 }{ 4\left( tanSCEA+cotECEA \right) }
= \frac { 0.1 }{ 4\left( tan45+cot10 \right) }
= 0.003747 mm
= 3.7 \mu m
65. Taylor’s tool life equation is given by V{ T }^{ n }=C, where V is in m/min and T is in min. In a turning operation, two tools X and Y are used. For tool X, n = 0.3 and C = 60 and for tool Y, n = 0.6 and C = 90. Both the tools will have the same tool life for the cutting speed (in m/min, round off to one decimal place) of _____.
(2 Mark, 2019[1])

Ans: 40
According to the question:
For tool x: { V }_{ x }{ T }_{ x }^{ 0.3 }=60………(1)
For tool y: { V }_{ y }{ T }_{ y }^{ 0.6 }=90……….(2)
The speed at which both the tools have same tool life means
So, { { T }_{ x }=T }_{ y }=T
and { { V }_{ x }=V }_{ y }=V
Now equation 1 an 2 becomes:
V{ T }^{ 0.3 }=60…….(3)
V{ T }^{ 0.6 }=90……..(4)
Solving equation 3 and 4 We get,
T = 3.863 min.
V = 40 m/min
66. In an orthogonal machining with a single point cutting tool of rake angle 10°, the uncut chip thickness and the chip thickness are 0.125 mm and 0.22 mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip tool interface is _____ (round off to two decimal places).
(2 Mark, 2019[2])

Ans: 0.73
Chip thickness ratio (r) = \frac { t }{ { t }_{ c } }
=> r = \frac { 0.125 }{ 0.22 } = 0.568
We know, tan\phi =\frac { rcos\alpha }{ 1rsin\alpha }
=> tan\phi =\frac { 0.568\times cos{ 10 }^{ 0 } }{ 10.568\times sin{ 10 }^{ 0 } }
=> \phi ={ 31.83 }^{ 0 }
According to Merchant’s theory:
\phi = \frac { \pi }{ 4 } +\frac { 1 }{ 2 } \left( \alpha \beta \right)
=> { 31.83 }^{ 0 }={ 45 }^{ 0 }+\frac { 1 }{ 2 } \left( 10\beta \right)
=> \beta ={ 36.34 }^{ 0 }
Coefficient of friction (\mu )= tan\beta
=> \mu ={ tan36.34 }^{ 0 }=0.73