1. For wire drawing operation, the work material should essentially be
(a) Ductile
(b) Tough
(c) Hard
(d) Malleable
(1 Mark, 1987)

Ans: a
2. In forging operation the sticking friction condition occurs near the …. (centre/ ends).
(1 Mark, 1987)

Ans: centre
4. The process of hot extrusion is used to produce
(a) Curtain rods made of aluminium
(b) Steel pipes for domestic water supply
(c) Stainless steel tubes used in furniture
(d) Large size pipes used in city water mains
(1 Mark, 1994)

Ans: c
5. A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield strength will
(a) Decrease
(b) Increase
(c) Remains same
(d) becomes equal to ultimate tensile strength
(1 Mark, 1995)

Ans: b
6. A wire of 0.1 mm diameter is drawn from a rod of 15 mm diameter. Dies giving reductions of 20%, 40% and 80% are available. For minimum error in the final size, the number of stages and reduction at each stage respectively would be
(a) 3 stages and 80% reduction for all three stages
(b) 4 stages and 80% reduction for first three stages followed by a finishing stage of 20% reduction
(c) 5 stages and reduction of 80%, 80%, 40%, 40%, 20% in a sequence
(d) None of the above
(2 Mark, 1996)

Ans: b
Explanation:
This question will be solved by checking all the options given in the question.
% reduction = \left( 1\frac { d }{ D } \right) \times 100%
If we check option (B) then,
Diameter of wire after first drawing = 0.2*15 = 3 mm
Diameter of wire after second drawing = 0.2*3 = 0.6 mm
Diameter of wire after third drawing = 0.2*0.6 = 0.12 mm
Diameter of wire after second drawing = 0.8*0.12 = 0.96 mm < 0.1 mm
So, option B is correct.
You can check the other options, but not in any option we get the final diameter equal or less then 0.1 mm.

Ans: A – 3, B – 4, C – 1, D – 2
8. A strip with a crosssection 150 mm × 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rolls. The angle subtended by the deformation zone at the roll centre is (in radian):
(a) 0.01
(b) 0.02
(c) 0.03
(d) 0.06
(2 Mark, 1998)

Ans: d
Explanation:
We know: \Delta h=D(1cos\alpha )
\Delta h = 4.5 – 0.8*4.5 = 0.9
Now, \alpha ={ cos }^{ 1 }\left( 1\frac { \Delta h }{ D } \right)
=> \alpha ={ cos }^{ 1 }\left( 1\frac { 0.9 }{ 450 } \right) = 0.0632

Ans: A – 3, B – 2, C – 1, D – 5
9. Hot rolling of mild steel is carried out
(a) At recrystallization temperature
(b) Between 100^{0} C to 150^{0} C
(c) Between recrystallization temperature
(d) Above recrystallization temperature
(1 Mark, 2002)

Ans: d
Explanation:
Hot rolling is always carried out above recrystallisation temperature.
10. The total area under the stressstrain curve of a mild steel specimen tested up to failure under tension is a measure of
(a) Ductility
(b) Ultimate strength
(c) Stiffness
(d) Toughness
(1 Mark, 2002)

Ans: d
11. A brass billet is to be extruded from its initial diameter of 100 mm to a final diameter of 50 mm. The working temperature of 700°C and the extrusion constant is 250 MPa. The force required for extrusion is
(a) 5.44 MN
(b) 2.72 MN
(c) 1.36 MN
(d) 0.36 MN
(2 Mark, 2003)

Ans: b
Explanation:
Extrusion force (F) = { { A }_{ o }\sigma }_{ o }ln\left( \frac { { A }_{ o } }{ { A }_{ f } } \right)
=> F = \frac { \pi }{ 4 } { \times 100 }^{ 2 }\times 250ln\left( \frac { \frac { \pi }{ 4 } { \times 100 }^{ 2 } }{ \frac { \pi }{ 4 } { \times 50 }^{ 2 } } \right)
=> F = 2.72 MN
12. Match the following.
(a) P1, Q4, R6, S3
(b) P4, Q5, R2, S3
(c) P1, Q5, R3, S2
(d) P5, Q1, R2, S2
(2 Mark, 2004)

Ans: b
13. In a rolling process, sheet of 25 mm thickness is rolled to 20 mm thickness. Roll is of diameter 600 mm and it rotates at 100 rpm. The roll strip contact length will be
(a) 5 mm
(b) 39 mm
(c) 78 mm
(d) 120 mm
(2 Mark, 2004)

Ans: b
Explanation:
Angle of bite = \alpha ={ cos }^{ 1 }\left( 1\frac { \Delta h }{ D } \right)
=> \alpha ={ cos }^{ 1 }\left( 1\frac { 5 }{ 600 } \right) = 0.129 rad
Roll strip contact length = \alpha R = 38.8 mm
14. A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any change in its width. The friction coefficient at the workroll interface is 0.1. The minimum possible thickness of the sheet that can be produced in a single pass is:
(a) 1.0 mm
(b) 1.5 mm
(c) 2.5 mm
(d) 3.7 mm
(2 Mark, 2006)

Ans: c
Explanation:
Maximum reduction possible = { \Delta h }_{ max }={ \mu }^{ 2 }R={ 0.1 }^{ 2 }\times 150=1.5mm
So, Minimum possible sheet thickness = 4 – 1.5 = 2.5 mm
15. In a wire drawing operation, diameter of a steel wire is reduced from 10 mm to 8 mm. the mean flow stress of the material is 400 MPa. The ideal force requied for drawing (ignoring friction and redundant work) is:
(a) 4.48 kN
(b) 8.97 kN
(c) 20.11 kN
(d) 31.41 kN
(2 Mark, 2006)

Ans: b
Explanation:
Ideal force required for drawing (F) = { A }_{ f }{ \sigma }_{ o }ln\frac { { { A }_{ o } } }{ { A }_{ f } }
=> F = \frac { \pi }{ 4 } \times { 8 }^{ 2 }{ \times }400\times ln\left( \frac { { \frac { \pi }{ 4 } \times { 10 }^{ 2 } } }{ \frac { \pi }{ 4 } \times { 8 }^{ 2 } } \right)
=> F = 8.97 kN
16. The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degree will be.
(a) 5.936
(b) 7.936
(c) 8.936
(d) 9.936
(2 Mark, 2007)

Ans: d
Explanation:
\Delta h=16 – 10 = 6 mm
We know,
\Delta h=D(1cos\alpha )=> 6 = 400(1cos\alpha)
=>\alpha = 9.936 degree
17. In a single pass rolling operation, a 20 mm thick plate with plate width of 100 mm, is reduced to 18 mm. The roller radius is 250 mm and rotational speed is 10 rpm. The average flow stress for the plate material is 300 MPa. The power required for the rolling operation in kW is closest to
(a) 15.2
(b) 18.2
(c) 30.4
(d) 45.6
(2 Mark, 2008)

Ans: a
Explanation:
Given, w = 100 mm = 0.1 m
\Delta h = 20 – 18 = 2 mm = 0.002 m
R = 250 mm = 0.25 m
n = 10 rpm
\sigma =300\quad MPaRoll strip contact length (L)=\sqrt { R\Delta h }
=> L = 0.02236 m
Rolling load (F) = \sigma Lw
=> F = 300\times { 10 }^{ 6 }\times 0.02236\times 0.1=670.8kN
Torque (T) = Fa (Where, a = 0.5L)
=> T = 670.8\times 0.01118 = 7.49 kN
Power (P) = 2T\omega =2\times T\times \frac { 2\pi n }{ 60 }
=> P = 2\times 7.49\times \frac { 2\times \pi \times 10 }{ 60 } =15.68kW
18. The maximum possible draft in cold rolling of sheet increases with the
(a) Increase in coefficient of friction
(b) Decrease in coefficient of friction
(c) Decrease in roll radius
(d) Increase in roll velocity
(1 Mark, 2011)

Ans: a
Explanation:
Maximum draft possible { (\Delta h) }_{ max }={ \mu }^{ 2 }R
=> { (\Delta h) }_{ max }\alpha \quad { \mu }^{ 2 }
19. Match the following metal forming processes with their associated stresses in the workpiece.
(b) 1S, 2P, 3R, 4Q
(c) 1P, 2Q, 3S, 4R
(d) 1P, 2R, 3Q, 4S
(1 Mark, 2011)

Ans: a
20. A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. The percentage change in diameter is
(a) 0
(b) 2.07
(c) 20.7
(d) 41.4
(1 Mark, 2012)

Ans: d
Explanation:
The volume remains same.
So, \frac { \pi }{ 4 } \times { D }_{ 1 }^{ 2 }\times { H }_{ 1 }=\frac { \pi }{ 4 } \times { D }_{ 2 }^{ 2 }\times { H }_{ 2 }
=> { 100 }^{ 2 }\times 50={ D }_{ 2 }^{ 2 }\times 25
=> { D }_{ 2 }=100\sqrt { 2 } =141.42mm
Percentage change in diameter = \frac { { D }_{ 2 }{ D }_{ 1 } }{ { D }_{ 1 } } \times 100 = \frac { 141.42100 }{ 100 } \times 100 = 41.4%
21. In a single pass rolling process using 410 mm diameter steel rollers, a strip of width 140 mm and thickness 8 mm undergoes 10% reduction of thickness. The angle of bite in radians is
(a) 0.006
(b) 0.031
(c) 0.062
(d) 0.600
(2 Mark, 2012)

Ans: c
Explanation:
Given, D = 410 mm
\Delta h=10%\times 8=0.8mmNow, \Delta h=D(1cos\alpha )
=> \Delta 0.8=410(1cos\alpha )
=> \alpha =0.062radian
22. In a rolling process, the state of stress of the material undergoing deformation is
(a) Pure compression
(b) Pure shear
(c) Compression and shear
(d) Tension and shear
(1 Mark, 2013)

Ans: c
23. With respect to metal working, match Group A with Group B:
(a) PII, QIII, RVI, SV
(b) PIII, QI, RVI, SV
(c) PIII, QI, RIV, SVI
(d) PI, QII, RV, SVI
(1 Mark, 2014[2])

Ans: b
24. A mild steel plate has to be rolled in one pass such that the final plate thickness is 2/3rd of the initial thickness, with the entrance speed of 10 m/min and roll diameter of 500 mm. If the plate widens by 2% during rolling, the exit velocity (in m/min) is _______.
(2 Mark, 2014[2])

Ans: 14.7
Explanation:
Given, { t }_{ 2 }=\frac { 2 }{ 3 } { t }_{ 1 }
{ w }_{ 2 }=1.02{ w }_{ 1 }and { v }_{ 1 }=10m/min
Since the volume rate remains same before and after rolling
{ v }_{ 1 }{ t }_{ 1 }{ w }_{ 1 }={ v }_{ 2 }{ t }_{ 2 }{ w }_{ 2 }=> { v }_{ 2 }=\frac { { v }_{ 1 }{ t }_{ 1 }{ w }_{ 1 } }{ { t }_{ 2 }{ w }_{ 2 } }
=> { v }_{ 2 }=\frac { 1 }{ 1.02 } \times \frac { 3 }{ 2 } \times 10=14.7m/min
25. In a rolling process, the maximum possible draft, defined as the difference between the initial and the final thickness of the metal sheet, mainly depends on which pair of the following parameters?
P: Strain
Q: Strength of the work material
R: Roll diameter
S: Roll velocity
T: Coefficient of friction between roll and work
(a) Q, S
(b) R, T
(c) S, T
(d)P, R
(1 Mark, 2014[4])

Ans: b
Explanation:
Maximum draft possible { (\Delta h) }_{ max }={ \mu }^{ 2 }R
So, the maximum possible draft depends upon Coefficient of friction and roll diameter.
26. Match the following products with preferred manufacturing processes:
(a) P4, Q3, R1, S2
(b) P4, Q3, R2, S1
(c) P2, Q4, R3, S1
(d) P3, Q4, R2, S1
(1 Mark, 2015[1])

Ans: b
27. In a slab rolling operation, the maximum thickness reduction (Δℎ_{max}) is given by Δℎ_{max}= { \mu }^{ 2 }R, where R is the radius of the roll and μ is the coefficient of friction between the roll and the sheet. If μ = 0.1, the maximum angle subtended by the deformation zone at the centre of the roll (bite angle in degrees) is ____.
(2 Mark, 2015[1])

Ans: 5.73
Explanation:
Given:
Maximum draft possible { (\Delta h) }_{ max }={ \mu }^{ 2 }R…….(1)
\Delta h=D(1cos\alpha )……(2)
From equation 1 and 2,
{ \mu }^{ 2 }R=D(1cos\alpha )Putting the values:
=> \alpha ={ 5.73 }^{ o }
28. In a twostage wire drawing operation, the fractional reduction (ratio of change in crosssectional area to initial crosssectional area) in the first stage is 0.4. The fractional reduction in the second stage is 0.3. The overall fractional reduction is
(a) 0.24
(b) 0.58
(c) 0.60
(d) 1.00
(2 Mark, 2015[2])

Ans: b
29. The strain hardening exponent n of stainless steel SS 304 with distinct yield and UTS values undergoing plastic deformation is
(a) n < 0
(b) n = 0
(c) 0 < n < 1
(d) n = 1
(1 Mark, 2015[3])

Ans: c
30. In a rolling operation using rolls of diameter 500 mm, if a 25 mm thick plate can’t be reduced to less than 20 mm in one pass, the coefficient of friction between the roll and the plate is _____.
(2 Mark, 2015[3])

Ans: 0.1414
31. A 300 mm thick slab is being cold rolled using roll of 600 mm diameter. If the coefficient of friction is 0.08, the maximum possible reduction (in mm) is ____.
(2 Mark, 2016[1])

Ans: 1.92
33. It is desired to make a product having Tshaped crosssection from a rectangular aluminium block. Which one of the following processes is expected to provide the highest strength of the product?
(a) Welding
(b) Casting
(c) Metal forming
(d) Machining
(1 Mark, 2017[2])

Ans: c
Explanation:
Highest strength is obtained in metal forming processes because work hardening occurs in metal forming.
34. A strip of 120mm width and 8mm thickness is rolled between two 300mm diameter rolls to get a strip of 120 mm width and 7.2 mm thickness. The speed of the strip at the exit is 30 m/min. There is no front or back tension. Assuming uniform roll pressure of 200 MPa in the roll bite and 100% mechanical efficiency, the minimum total power (in kW) required to drive the two rolls is _____.
(2 Mark, 2017[2])

Ans: 9.6
Explanation:
Given, w = 120 mm
\Delta h = 8 – 7.2 = 0.8 mm
Roll radius (R) = 300/2 = 150 mm
\sigma =200\quad MPaRoll strip contact length (L)=\sqrt { R\Delta h }
=> (L)=\sqrt { 150\times 0.8 } = 10.954 mm
Rolling load (F) = \sigma Lw
=> F = 200\times 10.954\times 120=262.89kN
Torque (T) = Fa (Where, a = 0.5L)
=> T = 262.89\times 0.005477 = 1.44 kNm
The speed of the strip at the exit = 30 m/min
For 100% mechanical efficiency:
R\omega =\frac { 30 }{ 60 } m/s
\omega =3.33rad/sPower (P) = 2T\omega
=> P = 2\times 1.44\times 3.33 =9.6kW
35. The maximum reduction in crosssectional area per pass (R) of a cold wire drawing process is, R = 1 e ^{(n+1)}
Where, n represents the strain hardening coefficient. For the case of a perfectly plastic material, R is
(a) 0.865
(b) 0.826
(c) 0.777
(d) 0.632
(2 Mark, 2018[1])

Ans: d
Explanation:
For the case of perfectly plastic material, strain hardening coefficient(n) is zero, hence we have
R=1{ e }^{ (n+1) }=1{ e }^{ 1 }=0.632
36. A steel wire is drawn from an initial diameter (d_{i}) of 10 mm to a final diameter (d_{f}) of 7.5 mm. The half cone angle (α) of the die is 5^{0} and the coefficient of friction (μ) between the die and the wire is 0.1. The average of the initial and final yield stress [(σ_{Y})_{avg}] is 350 MPa. The equation for drawing stress σ_{f}, (in MPa) is given as:
The drawing stress (in MPa) required to carry out this operation is _____ (correct to two decimal places).
(2 Mark, 2018[2])
37. Match the following products with the suitable manufacturing process.
(a) P4, Q3, R1, S2
(b) P2, Q1, R3, S4
(c) P4, Q1, R2, S3
(d) P1, Q3, R4, S2
(1 Mark, 2018[2])

Ans: c
38. The cold forming process in which a hardened tool is pressed against a workpiece (when there is relative motion between the tool and the workpiece) to produce a roughened surface with a regular pattern is
(a) Roll forming
(b) Strip rolling
(c) Knurling
(d) Chamfering
(1 Mark, 2019[2])

Ans: c
39. The thickness of a sheet is reduced by rolling (without any change in width) using 600 mm diameter rolls. Neglect elastic deflection of the rolls and assume that the coefficient of friction at the rollworkpiece interface is 0.05. The sheet enters the rotating rolls unaided. If the initial sheet thickness is 2 mm, the minimum possible final thickness that can be produced by this process in a single pass is _____ mm (round of to two decimal places).
(2 Mark, 2019[2])

Ans: 1.25
Explanation:
We know, { (\Delta h) }_{ max }={ \mu }^{ 2 }R
=> { h }_{ i }{ h }_{ f }={ \mu }^{ 2 }R
=>2{ h }_{ f }={ 0.05 }^{ 2 }\times 300
=> { h }_{ f }=1.25mm