1. Penetration is increased by
(a) Increasing welding current and welding speed
(b) Increasing welding current and decreasing welding speed
(c) Decreasing welding current and welding speed
(d) Decreasing welding current and increasing welding speed
(1 Mark, 1990)

Ans: b
2. At small variations of arc length at operating conditions, the manual metal arc welding transformer provides nearly
(a) Constant power
(b) Constant power factor
(c) Constant voltage
(d) Constant current
(1 Mark, 1990)

Ans: d (Reference: (Manual metal arc welding) NPTEL PDF)
3. For gas welding a particular job using a neutral oxyacetylene flame the acetylene consumption was 10 ltrs. The oxygen consumption from the cylinder in liters will be
(a) 5
(b) 10
(c) 15
(d) 20
(1 Mark, 1991)

Ans: b
Explanation:
In oxyacetylene welding, 1 unit acetylene requires 2.5 units of oxygen for complete burning. But entire oxygen is not provided from oxygen cylinder.
! unit of oxygen is provided from oxygen cylinder and other 1.5 units are compensated by atmosphere.
4. For resistance spot welding of 1.5 mm thick steel sheets, the current required is of the order
(a) 10 Amp
(b) 100 Amp
(c) 1000 Amp
(d) 10,000 Amp
(1 Mark, 1991)

Ans: d (Reference: P N Rao Book, Resistance spot welding)

Ans: A – 2, B – 4, C – 6, D – 5
6. In DC welding, the straight polarity (electrode negative) results in
(a) Lower penetration
(b) Lower deposition rate
(c) Less heating of work piece
(d) Smaller weld pool
(1 Mark, 1992)

Ans: b
7. In an explosive welding process, the ___ (maximum/minimum) velocity of impact is fixed by the velocity of sound in the ____ (flyer/target) plate material.
(1 Mark, 1992)

Ans: maximum, flyer
8. The electrodes used in arc welding are coated. This coating is not expected to
(a) Provide protective atmosphere to weld
(b) Stabilize the arc
(c) Add alloying elements
(d) Prevents electrode from contamination
(1 Mark, 1993)

Ans: d
9. The ratio of acetylene to oxygen is approximately ______ a neutral flames used in gas welding.
(1 Mark, 1994)

Ans: 1
10. Preheating before welding is done to
(a) Make the steel softer
(b) Burn away oil, grease, etc., from the plate surface
(c) Prevent cold cracks
(d) Prevent plate distortion
(1 Mark, 1996)

Ans: c (Reference: P N Rao Book, Defects in welding)
12. For butt – welding 40 mm thick steel plates, when the expected quantity of such jobs is 5000 per month over a period of 10 years, choose the best suitable welding process out of the following available alternatives.
(a) Submerged arc welding
(b) Oxy – acetylene gas welding
(c) Electron beam welding
(d) MIG welding
(1 Mark, 1999)

Ans: a
13. Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001Ω, the heat generated to form the weld is
(a) 5625 Wsec
(b) 8437 Wsec
(c) 22500 Wsec
(d) 33750 Wsec
(2 Mark, 2001)

Ans: a
Explanation:
Heat generated = { I }^{ 2 }Rt={ 15000 }^{ 2 }\times 0.0001\times 0.25=5625\quad Wsec
14. Two plates of the same metal having equal thickness are to be butt welded with electric arc. When the plate thickness changes, welding is achieved by
(a) Adjusting the current
(b) Adjusting the duration of current
(c) Changing the electrode size
(d) Changing the electrode coating
(1 Mark, 2002)

Ans: a
15. The temperature of a carburising flame in gas welding is _______ that of a neutral or an oxidizing flame.
(a) Lower than
(b) Higher than
(c) Equal to
(d) Unrelated to
(1 Mark, 2002)

Ans: a
16. Which of the following arc welding processes does not use consumable electrodes
(a) GMAW
(b) GTAW
(c) Submerged Arc Welding
(d) None of these
(1 Mark, 2002)

Ans: b
17. Match the following.
(a) P – 2, Q – 5, R – 1, S – 3
(b) P – 6, Q – 3, R – 4, S – 4
(c) P – 4, Q – 1, R – 6, S – 2
(d) P – 5, Q – 4, R – 2, S – 6
(2 Mark, 2003)

Ans: d
18. In Oxyacetylene gas welding, temperature at the inner cone of the flame is around
(a) 3500°C
(b) 3200°C
(c) 2900°C
(d) 2550°C
(1 Mark, 2003)

Ans: b
19. Two 1 mm thick steel sheets are to be spot welded at a current of 5000 A. Assuming effective resistance to be 200 microohms and current flow time of 0.2 second, heat generated during the process will be
(a) 0.2 Joule
(b) 1 Joule
(c) 5 Joule
(d) 1000 Joules
(1 Mark, 2004)

Ans: d
Explanation:
Heat generated = { I }^{ 2 }Rt={ 5000 }^{ 2 }\times 200\times { 10 }^{ 6 }\times 0.2=1000J
20. The strength of a brazed joint
(a) Decreases with increase in gap between the two joining surfaces
(b) Increases with increase in gap between the two joining surfaces
(c) Decreases up to certain gap between the two joining surfaces beyond which it increases
(d) Increases up to certain gap between the two joining surfaces beyond which it decreases
(1 Mark, 2005)

Ans: d
21. Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m^{3}) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μΩ, the current passing through the electrodes is approximately
(a) 1480 A
(b) 3300 A
(c) 4060 A
(d) 9400 A
(2 Mark, 2005)

Ans: c
Explanation:
Heat require to melt the nugget = \frac { \pi }{ 4 } \times { d }^{ 2 }\times t\times \rho \times Latent\quad Heat
=\frac { \pi }{ 4 } \times { \left( 5\times { 10 }^{ 3 } \right) }^{ 2 }\times 1.5\ { 10 }^{ 3 }\times 8000\times 1400\times { 10 }^{ 3 } = 329.8 J
Now,{ I }^{ 2 }Rt = 329.8 J
=> { I }^{ 2 }\times 200\times { 10 }^{ 6 }\times 0.1=329.8
=> I = 4060 A
22. In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/sec. the net heat input (in J/mm) is:
(a) 64
(b) 797
(c) 1103
(d) 79700
(2 Mark, 2006)

Ans: b
Explanation:
Heat input = \frac { V\times I }{ v } \times \eta =\frac { 25\times 300 }{ 8 } \times 0.85=796.87\quad J/mm
23. A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm. The linear voltage (E) arc length (L) characteristic of the welding arc can be given as (where E is in volt and L in in mm)
(a) E = 20 + 2L
(b) E = 20 + 8L
(c) E = 80 + 2L
(d) E = 80 + 8L
(2 Mark, 2007)

Ans: a
Explanation:
Given: V_{o }= 80 V, I_{o }= 800 A
We know:
E={ V }_{ o }\left( 1\frac { I }{ { I }_{ o } } \right) …..(1)
E=a+bL….(2)
Given when L= 5 mm, I= 500 A
So, From equation (1), E = 30
Equation (2) becomes 30 = a+5b……(3)
Agian, When L = 7 mm, I = 460 A
From equation (1), E = 34
Equation (2) becomes 34 = a+7b……(4)
From equation (3) and (4), a = 20 and b= 2
So, E = 20+2L
24. Which one of the following is a solid state joining process?
(a) Gas tungsten arc welding
(b) Resistance spot welding
(c) Friction welding
(d) Submerged arc welding
(1 Mark, 2007)

Ans: c
25. Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 millisecond. A spherical fusion zone extending up to full thickness of each sheet is formed. The properties of the metallic sheets are given as : Ambient temperature = 293 K, Melting temperature = 1793 K, Density = 7000 kg/m^{3}, Latent heat of fusion = 300 kJ/kg, Specific heat = 800 J/kgK
Assume:
(i) Contact resistance along sheet interface is 500 microohm and along electrodesheet interface is zero;
(ii) No conductive heat loss through the bulk sheet materials; and
(iii) The complete weld fusion zone is at the melting temperature.
The melting efficiency (in %) of the process is
(a) 50.37
(b) 60.37
(c) 70.37
(d) 80.37
(2 Mark, 2007)

Ans: c
Explanation:
\eta =\frac { Heat\quad utilised\quad to\quad melt }{ Heat\quad provided }=> \eta =\frac { m(CdT+Latent\quad Heat) }{ { I }^{ 2 }Rt } =\frac { 7000\times \frac { 4 }{ 3 } \pi \times { .002 }^{ 3 }\times \left( 800\times (1793293)+(300\times { 10 }^{ 3 }) \right) }{ { 10000 }^{ 2 }\times (500\times { 10 }^{ 6 })\times (10\times { 10 }^{ 3 }) } =70.37%
26. In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm^{2} and the unit energy required to melt the metal is 10 J/mm^{3}. If the welding power is 2 kW, the welding speed in mm/s is closest to
(a) 4
(b) 14
(c) 24
(d) 34
(2 Mark, 2008)

Ans: b
Explanation:
From energy balance:
10J/{ mm }^{ 3 }\times 5{ mm }^{ 2 }\times v=2000W\times 0.5\times 0.7=> V = 14 mm/s
27. Two pipes of inner diameter 100mm and outer diameter 110mm each joined by flash butt welding using 30V power supply. At the interface, 1mm of material melts from each pipe which has a resistance of 42.4Ω. If the unit melt energy is 64.4 MJm^{3}, then time required for welding in seconds is
(a) 1
(b) 5
(c) 10
(d) 20
(2 Mark, 2010)

Ans: c
Explanation:
Energy required for melting = \frac { \pi }{ 4 } \left( { d }_{ o }^{ 2 }{ d }_{ i }^{ 2 } \right) t\times 64.4\times { 10 }^{ 6 }=\frac { \pi }{ 4 } \left( 0.11^{ 2 }{ 0.1 }^{ 2 } \right) 0.002\times 64.4\times { 10 }^{ 6 }=212.32J
Now, \frac { { V }^{ 2 } }{ R } t=212.32J
=> \frac { { 30 }^{ 2 } }{ 42.4 } t=212.32
=> t = 10 s
28. Which one among the following welding processes uses nonconsumable electrode?
(a) Gas metal arc welding
(b) Submerged arc welding
(c) Gas tungsten arc welding
(d) Flux coated arc welding
(1 Mark, 2011)

Ans: c
29. In a DC arc welding operation, the voltagearc length characteristic was obtained as V_{arc} = 20 + 5l where the arc length l was varied between 5 mm and 7 mm. Here V_{arc} denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are
(a) 45 V, 450 A
(b) 75 V, 750 A
(c) 95 V, 950 A
(d) 150 V, 1500 A
(2 Mark, 2012)

Ans: c
Explanation:
Given: V_{arc} = (20 + 5l)……..(1)
We know: { V }={ V }_{ o }\left( 1\frac { I }{ { I }_{ o } } \right) ……(2)
Where, V_{o} = Open circuit voltage
I_{o }= Short circuit current
According to the question:
When the arc length is 5 mm then the welding current = 500 A
and V = 20 + 5*5 = 45
Equation (2) becomes
45={ V }_{ o }\left( 1\frac { 500 }{ { I }_{ o } } \right) ……….(3)
When the arc length is 7 mm then the welding current = 400 A
V = 20 + 5*7 = 55
Equation (2) becomes
55={ V }_{ o }\left( 1\frac { 400 }{ { I }_{ o } } \right) ………..(4)
Now solving Equation (3) and (4)
we get I_{o }= 950 A and V_{o }= 95 V
30. Match the CORRECT pairs
(b) P4, Q2, R3, S1
(c) P2, Q3, R4, S1
(d) P2, Q4, R1, S3
(1 Mark, 2013)

Ans: a
31.The major difficulty during welding of aluminium is due to its
(a) High tendency of oxidation
(b) High thermal conductivity
(c) Low melting point
(d) Low density
(1 Mark, 2014[1])

Ans: a
32.In solidstate welding, the contamination layers between the surfaces to be welded are removed by
(a) Alcohol
(b) Plastic deformation
(c) Water jet
(d) Sand blasting
(1 Mark, 2014[1])

Ans: b
33. For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of 10000 A is applied for 0.2 s. The heat dissipated to the base metal is 1000 J. Assuming that the heat required for melting 1 mm^{3} volume of steel is 20 J and interfacial contact resistance between sheets is 0.0002 Ω, the volume (in mm^{3}) of weld nugget is ___.
(2 Mark, 2014[3])

Ans: 150
Explanation:
Total heat generated = { I }^{ 2 }Rt={ 10000 }^{ 2 }\times 0.0002\times 0.2=4000J
Heat used to melt the nugget = 4000 – 1000 = 3000 J
Given, Heat required for melting = 20 J/mm^{3}
The volume of weld nugget is = \frac { 3000 }{ 20 } =150{ mm }^{ 3 }
34. Within the Heat Affected Zone (HAZ) in a fusion welding process, the work material undergoes
(a) Microstructural changes but does not melt
(b) Neither melting nor microstructural changes
(c) Both melting and microstructural changes after solidification
(d) Melting and retains the original microstructure after solidification
(1 Mark, 2014[4])

Ans: a
35. A DC welding power source has a linear voltagecurrent (VI) characteristic with open circuit voltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (in Amperes) should be set as_________.
(2 Mark, 2015[1])

Ans: 150
Explanation:
We know: { V }={ V }_{ o }\left( 1\frac { I }{ { I }_{ o } } \right)
Again Arc Power (P) = VI = { V }_{ o }\left( I\frac { { I }^{ 2 } }{ { I }_{ o } } \right)
For maximum arc power \frac { dP }{ dI } =0
=> \frac { d{ V }_{ o }\left( I\frac { { I }^{ 2 } }{ { I }_{ o } } \right) }{ dI } =0
By solving: I=\frac { { I }_{ o } }{ 2 } =\frac { 300 }{ 2 } =150A
36. In a linear arc welding process, the heat input per unit length is inversely proportional to
(a) Welding current
(b) Welding voltage
(c) Welding speed
(d) Duty cycle of the power source
(1 Mark, 2015[1])

Ans: c
37. During a TIG welding process, the arc current and arc voltage were 50 A and 60 V, respectively, when the welding speed was 150 mm/min. In another process, the TIG welding is carried out at a welding speed of 120 mm/min at the same arc voltage and heat input to the material so that weld quality remains the same. The welding current (in A) for this process is
(a) 40.00
(b) 44.72
(c) 55.90
(d) 62.25
(2 Mark, 2015[2])

Ans: a
Explanation:
heat input is given by Q = VIt
We know: Time (t) is inversely proportional to welding speed (v).
So, \frac { { V }_{ 1 }{ I }_{ 1 } }{ { v }_{ 1 } } =\frac { { V }_{ 2 }{ I }_{ 2 } }{ { v }_{ 2 } }
=> \frac { 50\times 60 }{ 150 } =\frac { 60\times { I }_{ 2 } }{ 120 }
=> I = 40A
38. Which two of the following joining processes are autogeneous?
(i) Diffusion welding
(ii) Electroslag welding
(iii) Tungsten inert gas welding
(iv) Friction welding
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iii)
(1 Mark, 2015[3])

Ans: a
Explanation:
Autogenous Welding is a process that coalesces two or more metals without the addition of filler metal.
40.Under optimal conditions of the process the temperatures experienced by a copper work piece in fusion welding, brazing and soldering are such that
(a) T_{welding} > T_{soldering }> T_{brazing}
(b) T_{soldering} > T_{welding} > T_{brazing}
(c) T_{brazing} >T_{welding} > T_{soldering}
(d) T_{welding} > T_{brazing} > T_{soldering}
(1 Mark, 2016[1])

Ans: d
41. The welding process which uses a blanket of fusible granular flux is
(a) Tungsten inert gas welding
(b) Submerged arc welding
(c) Electroslag welding
(d) Thermit welding
(1 Mark, 2016[2])

Ans: b
42. The voltage – length characteristic of a direct current arc in an arc welding process is V = (100 + 40l), where l is the length of the arc in mm and V is arc voltage in volts. During a welding operation, the arc length varies between 1 and 2 mm and the welding current is in the range 200 – 250 A. Assuming a linear power source, the short circuit current is _________ A.
(2 Mark, 2016[2])

Ans: 425
Explanation:
Given: V = (100 + 40l)……..(1)
We know: { V }={ V }_{ o }\left( 1\frac { I }{ { I }_{ o } } \right) ……(2)
Where, V_{o} = Open circuit voltage
I_{o }= Short circuit current
According to the question:
When the arc length is 1 mm then the welding current = 250 A
and V = 100 + 40*1 = 140
Equation (2) becomes
140={ V }_{ o }\left( 1\frac { 250 }{ { I }_{ o } } \right) ……….(3)
When the arc length is 2 mm then the welding current = 200 A
V = 100 + 40*2 = 180
Equation (2) becomes
180={ V }_{ o }\left( 1\frac { 200 }{ { I }_{ o } } \right) ………..(4)
Now solving Equation (3) and (4) we get I_{o }= 425 A
43. Spot welding of two steel sheets each 2 mm thick is carried out successfully by passing 4 kA of current for 0.2 seconds through the electrodes. The resulting weld nugget formed between the sheets is 5 mm in diameter. Assuming cylindrical shape for the nugget, the thickness of the nugget is ________ mm.
(2 Mark, 2016[3])

Ans: 2.91
Explanation:
Energy supplied = { I }^{ 2 }Rt={ 4000 }^{ 2 }\times 200\times { 10 }^{ 6 }\times 0.2=640J
Energy required for melting = Density*Volume*Latent heat of fusion = (8000\times \frac { \pi }{ 4 } \times { 0.005 }^{ 2 }\times Thickness\times 1400\times { 10 }^{ 3 }) J
Equating both energies,
Thickness = 2.91 mm
44. In an arc welding process, welding speed is doubled. Assuming all other process parameters to be constant, the cross sectional area of the weld bead will
(a) Increase by 25%
(b) Increase by 50%
(c) Reduce by 25%
(d) Reduce by 50%
(1 Mark, 2017[1])

Ans: d
Explanation:
Since all the process parameters are constant.
Material deposition rate (MDR) = constant
=> Av = constant
=> A\alpha \frac { 1 }{ v }
=> \frac { { A }_{ 2 } }{ { A }_{ 1 } } =\frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { v }{ 2v } =\frac { 1 }{ 2 }
=> { A }_{ 2 }=\frac { { A }_{ 1 } }{ 2 }
That means the cross sectional area of the weld bea will reduced by 50%.
45. The type of weld represented by the shaded region in the figure is
(a) Groove
(b) Spot
(c) Fillet
(d) Plug
(1 Mark, 2018[1])

Ans: c
46. A welding operation is being performed with voltage = 30 V and current = 100 A. The crosssectional area of the weld bead is 20 mm^{2}. The workpiece and filler are of titanium for which the specific energy of melting is 14 J/mm^{3}. Assuming a thermal efficiency of the welding process 70%, the welding speed (in mm/s) is __________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 7.5
Explanation:
Voltage(V) = 30 V
Current(I) = 100 A
Crosssectional area of the weld bead(A) = 20 mm^{2}
Specific energy of melting (E) = 14 J/mm^{3}
Thermal efficiency of welding process = \frac { EAv }{ VI }
=> 0.7 = \frac { 14\times 20\times v }{ 30\times 100 }
=> v = 7.5 mm/s
47. Which one of the following welding methods provides the highest heat flux (W/mm^{2})?
(a) Plasma are welding
(b) Tungsten inert gas welding
(c) Oxyacetylene gas welding
(d) Laser beam welding
(1 Mark, 2019[1])

Ans: d
Explanation:
Heat flux in LBW > PAW > TIGW > Oxyacetylene gas welding
48. A gas tungsten arc welding operation is performed using a current of 250 A and an arc voltage of 20 V at a welding speed of 5 mm/s. Assuming that the arc efficiency is 70% the net heat input per unit length of the weld will be______ kJ/mm (round off to one decimal place).
(2 Mark, 2019[2])

Ans: 0.7
Explanation:
Heat input per unit length of the weld = { H }_{ t }=\frac { VI }{ v } \times \eta =\frac { 20\times 250 }{ 5 } \times 0.70=0.7kJ/mm