GATE MT — Manufacturing Processes

GATE 2023 · Q24

24

Match the defects in Column I with corresponding metal forming techniques in
Column II.
Column I Column II
(P) Cold shut (1) Rolling
(Q) Zipper breaks (2) Sheet metal forming
(R) Stretcher strains (3) Drawing
(S) Center burst (4) Forging

MCQ1M

A

P - 4, Q - 1, R - 2, S - 3

B

P - 4, Q - 2, R - 3, S - 1

C

P - 1, Q - 4, R - 2, S - 3

D

P - 3, Q - 1, R - 4, S - 2

Solution

Cold shut is a forging defect, zipper breaks occur in rolling, stretcher strains in sheet forming, and center burst in drawing. Answer: A

GATE 2023 · Q25

25

In rolling, the point on the surface of contact between roll and sheet where surface
velocity of the roll is equal to velocity of the sheet is referred as

MCQ1M

A

no-slip point

B

no-stick point

C

maximum slip point

D

maximum stick point

Solution

The contact point where roll surface velocity equals strip velocity is the neutral or no-slip point. Answer: A

GATE 2023 · Q45

45

Match the components in Column I with corresponding manufacturing processes in
Column II.
Column I Column II
(P) Crank shaft (1) Sheet metal forming
(Q) Machine bed (2) Forging
(R) Automobile brake pad (3) Casting
(S) Beverage can (4) Powder metallurgy

MCQ2M

A

P - 2, Q - 3, R - 4, S - 1

B

P - 3, Q - 4, R - 1, S - 2

C

P - 4, Q - 1, R - 3, S - 2

D

P - 2, Q - 3, R - 1, S - 4

Solution

Crank shafts are forged, machine beds cast, brake pads powder-metallurgy products, and beverage cans made by sheet metal forming. Answer: A

GATE 2023 · Q46

46

Match the welding techniques in Column I with the most appropriate applications
in Column II.
Column I Column II
(P) Submerged arc welding (1) Thick sections
(Q) Electroslag welding (2) Surfacing and repair
(R) Shielded metal arc welding (3) Thin sheets
(S) Resistance spot welding (4) Flat position

MCQ2M

A

P - 4, Q - 1, R - 2, S - 3

B

P - 3, Q - 2, R - 1, S - 4

C

P - 1, Q - 3, R - 4, S - 2

D

P - 2, Q - 4, R - 3, S - 1

Solution

SAW is suited to flat-position welding, electroslag to thick sections, SMAW to surfacing/repair, and resistance spot welding to thin sheets. Answer: A

GATE 2023 · Q52

52

Compared to top gating, the effect(s) of bottom gating in sand mold casting is/are

MSQ2M

A

reduced melt oxidation

B

reduced mold erosion

C

enhanced melt oxidation

D

enhanced mold erosion

Solution

Bottom gating fills the mold more quietly, reducing turbulence, melt oxidation, and mold erosion. Correct options: A, B

GATE 2023 · Q53

53

Choose the correct statement(s) in the context of fusion welding of austenitic
stainless steel containing about 0.06 wt.% carbon.

MSQ2M

A

Corrosion resistance of heat affected zone is poorer than base material.

B

Corrosion resistance of heat affected zone is superior than fusion zone.

C

Corrosion resistance of heat affected zone is same as fusion zone.

D

Corrosion resistance is same for fusion zone, heat affected zone, and base material.

Solution

In austenitic stainless steel near 0.06 wt.% C, sensitization in the HAZ can deplete chromium near grain boundaries and reduce corrosion resistance. Correct option: A

GATE 2023 · Q65

65

A 200 mm thick slab is rolled using 500 mm diameter rolls under cold rolling and
hot rolling conditions, separately. The coefficient of friction is 0.04 in cold rolling
and 0.4 in hot rolling. The ratio of maximum possible thickness reduction in cold
rolling to that in hot rolling is ____________ (round off to 2 decimal places).

NAT2M

Solution

Maximum draft in rolling is Delta h_max = mu^2 R. The ratio is (0.04^2 R)/(0.4^2 R) = 0.01. Answer: 0.01

GATE 2022 · Q17

17

Neutral point in rolling is defined as the point along the surface of contact
between the roll and the sheet, where the surface velocity of the roll is
_________

MCQ1M

A

zero

B

half the velocity of the sheet

C

twice the velocity of the sheet

D

equal to the velocity of the sheet

Solution

At the neutral point, roll surface velocity equals sheet velocity. Answer: D

GATE 2022 · Q21

21

Which one of the following figures illustrates a lap joint with fillet weld?
GATE 2022 Q21 figure

MCQ1M

A

Figure I

B

Figure II

C

Figure III

D

Figure IV

Solution

The lap joint with fillet weld corresponds to Figure II. Answer: B

GATE 2022 · Q24

24

In green sand moulding, the casting defect resulting from the displacement of
mould cavity by an oversized core is known as _________

MCQ1M

A

crush

B

hot tear

C

blow

D

fin

Solution

Displacement of the mould cavity by an oversized core is a crush defect. Answer: A

GATE 2022 · Q27

27

Match the processes (Column I) with the corresponding defects (Column II).
Column I Column II
(P) Extrusion (1) Earing
(Q) Deep drawing (2) Cold shut
(R) Forging (3) Edge cracking
(S) Rolling (4) Fir-tree cracking

MCQ1M

A

P - 1, Q - 4, R - 2, S - 3

B

P - 2, Q - 1, R - 4, S - 3

C

P - 4, Q - 1, R - 3, S - 2

D

P - 4, Q - 1, R - 2, S - 3

Solution

Extrusion: fir-tree cracking; deep drawing: earing; forging: cold shut; rolling: edge cracking. Answer: D

GATE 2022 · Q44

44

Which of the following parameter(s) influence(s) the melting rate of the
consumable wire in a gas metal arc welding process?

MSQ2M

A

Stick-out length

B

Welding speed

C

Welding current

D

Diameter of the consumable wire

Solution

GMAW melting rate depends on stick-out length, welding current, and wire diameter, not directly on travel speed. Correct options: A, C, D

GATE 2022 · Q59

59

Steel plates are welded autogenously using Gas Tungsten Arc welding with an
arc heat transfer efficiency of 0.65. The first weld is made using a welding
current of 200 A at an arc voltage of 18 V with a welding speed of 0.002 m.s.
The second weld is made at a welding speed of 0.0022 m.s with the same arc
voltage. If both the welds have identical heat input, the welding current of the
second weld is _________ A (round off to the nearest integer).

NAT2M

Solution

Equal heat input requires VI/v constant. I2 = 200(0.0022/0.002) = 220 A. Answer: 220

GATE 2022 · Q60

60

A cylindrical specimen of an Al alloy with diameter of 30 mm is cold extruded to
a diameter of 20 mm. If the flow behavior of the alloy is expressed by the
equation, sigma = 350 epsilon. MPa, the ideal plastic work of deformation per unit
volume is ________ x 10 J (answer rounded off to the nearest integer).

NAT2M

Solution

True strain for extrusion is ln(A0/Af). Integrate K epsilon^n from 0 to epsilon to get work per volume about 2.0 x 10^8 J/m3. Answer range: 200 to 210

GATE 2021 · Q14

14

Which one of the following is 'center split' defect in rolling operation?
GATE 2021 Q14 figure

MCQ1M

A

Defect (A) - see figure

B

Defect (B) - see figure

C

Defect (C) - see figure

D

Defect (D) - see figure

Solution

Centre split is a rolling defect where the center of the rolled product splits along the rolling direction due to tensile stresses at the centre — shown as option D in the figure. Answer: D

GATE 2021 · Q15

15

Single crystal turbine blades of nickel-based superalloys for aero-engines
are manufactured using:
Investment casting
Die casting
Squeeze casting
Directional solidification
Organising Institute - IIT Bombay

MCQ1M

A

Investment casting

B

Die casting

C

Squeeze casting

D

Directional solidification

Solution

Single crystal turbine blades require elimination of all grain boundaries; this is achieved by directional solidification using a grain selector (spiral) in the Bridgman process. Answer: D

GATE 2021 · Q49

49

A coin is tossed three times. Given that there are more heads than tails, the
probability of getting exactly one tail is: (round off to 2
decimal places).
A continuous fillet weld is made using a 3000 W welding machine. At a
travel speed of 6 mm s", the cross-sectional area (in mm?) of the weld is:
(round off to nearest integer).
Given: The unit energy required to melt the metal is 6 J mm*.
Heat transfer factor = 0.6
Melting factor = 0.5
Liquid iron is cast into a spherical sand mold (6 cm radius) and a cubical
sand mold (12 cm edge length). If solidification time is 60 minutes in the
spherical casting, the time (in minutes) required to solidify in the cubical
casting is: (round off to nearest integer).
True strain for 60% height reduction of a sample subjected to hot forging
is: (round off to 2 decimal places).

NAT2M

Solution

P(more H than T) = P(2H1T)+P(3H) = 3/8+1/8 = 1/2. P(exactly 1 tail | more H) = P(2H1T)/P(more H) = (3/8)/(4/8) = 3/4 = 0.75. Answer: 0.75 to 0.75

GATE 2021 · Q50

50

A continuous fillet weld is made using a 3000 W welding machine. At a
travel speed of 6 mm s", the cross-sectional area (in mm') of the weld is:
(round off to nearest integer).
Given: The unit energy required to melt the metal is 6 J mm".
Heat transfer factor = 0.6
Melting factor = 0.5

NAT2M

Solution

Energy input per mm length = P×η_heat×η_melt/v = 3000×0.6×0.5/6 = 150 J/mm. Unit melting energy = 6 J/mm³. Cross-section area = 150/6 = 25 mm². Answer: 25 to 25

GATE 2021 · Q51

51

Liquid iron is cast into a spherical sand mold (6 cm radius) and a cubical
sand mold (12 cm edge length). If solidification time is 60 minutes in the
spherical casting, the time (in minutes) required to solidify in the cubical
casting is: (round off to nearest integer).

NAT2M

Solution

Chvorinov's rule: t ∝ (V/A)². Sphere r=6 cm: V/A = 2 cm. Cube a=12 cm: V/A = 12³/(6×12²) = 2 cm. Equal moduli → same solidification time ≈ 60 min. Answer: 58 to 62

GATE 2021 · Q52

52

True strain for 60% height reduction of a sample subjected to hot forging
is: (round off to 2 decimal places).
Organising Institute - IT Bombay

NAT2M

Solution

True strain ε = ln(h₀/h₁) = ln(1/0.4) = ln(2.5) ≈ 0.916 for 60% height reduction. Answer: 0.90 to 0.94

GATE 2020 · Q17

17

In green sand casting, which one of the following is NOT a part of the gating system?

MCQ1M

A

Runner

B

Sprue

C

Riser

D

Pouring basin

Solution

A riser is a reservoir that compensates for shrinkage during solidification — it is NOT part of the gating system. The gating system consists of the pouring basin, sprue, runner, and gates. Answer: C

GATE 2020 · Q34

34

In cold-rolling, for the sheet to be drawn into rolls, the angle of contact (or angle of bite) should be less than or equal to __________ degree (round off to one decimal place).
Given, the coefficient of friction between sheet and roll is 0.1

NAT1M

Solution

Condition for bite: α ≤ arctan(μ) = arctan(0.1) ≈ 5.71°. Answer: 5.6 to 5.8

GATE 2020 · Q40

40

Match the products in Column I with the manufacturing processes in Column II.

Column I	Column II
(P) Blades of a gas turbine	1. Sand casting
(Q) Seamless tubing	2. Extrusion
(R) Automotive cylinder blocks	3. Powder metallurgy and wire drawing
(S) Tungsten filament	4. Investment casting

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-2, Q-3, R-1, S-4

C

P-4, Q-1, R-2, S-3

D

P-4, Q-2, R-1, S-3

Solution

Turbine blades→investment casting (4); seamless tubing→extrusion (2); cylinder blocks→sand casting (1); tungsten filament→powder metallurgy + wire drawing (3). P-4, Q-2, R-1, S-3. Answer: D

GATE 2020 · Q55

55

Two plates of composition, Fe–10 wt.% Ni and Fe–20 wt.% Cr–5 wt.% Ni are fusion-welded using a filler rod of composition 20 wt.% Ni–80 wt.% Cr. Contribution to dilution of the weld pool is 20% from each plate. The Ni content in the weld pool is __________ wt.% (round off to the nearest integer).

NAT2M

Solution

Filler is 60% of pool (100 − 20 − 20). Ni from filler = 0.60 × 20 = 12%; from plate 1 = 0.20 × 10 = 2%; from plate 2 = 0.20 × 5 = 1%. Total Ni = 15 wt.%. Answer: 15

GATE 2019 · Q30

30

Two blocks of the same metal are to be welded. The configuration that will undergo the least distortion after welding is _______________. GATE 2019 Q30 figure

MCQ1M

A

Configuration (A)

B

Configuration (B)

C

Configuration (C)

D

Configuration (D)

Solution

A symmetric double-sided groove (D) minimizes angular distortion by balancing thermal contraction on both sides. Answer: D

GATE 2019 · Q32

32

Alligatoring is a defect commonly observed in __________________.

MCQ1M

A

Extrusion

B

Deep drawing

C

Sheet metal forming

D

Rolling

Solution

Alligatoring (splitting along the centerline) occurs in rolling when the workpiece splits open like an alligator jaw due to non-uniform deformation. Answer: D

GATE 2019 · Q44

44

Match the automobile components in Column I with the corresponding manufacturing processes in Column II.

Column I	Column II
(P) Engine block	1. Forging
(Q) Brake pad	2. Sheet metal forming
(R) Connecting rod	3. Casting
(S) Door panel	4. Powder metallurgy

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-3, Q-4, R-1, S-2

C

P-3, Q-2, R-4, S-1

D

P-4, Q-1, R-3, S-2

Solution

Engine block → casting (3); Brake pad → powder metallurgy (4); Connecting rod → forging (1); Door panel → sheet metal (2). Answer: B

GATE 2019 · Q62

62

In sand-mold casting, it takes 180 seconds for complete solidification of a 27 cm³ cube-shaped casting. All other parameters remaining constant, the total solidification time (in seconds, rounded off to one decimal place) for a cylinder-shaped casting [radius = 1 cm and height = 6 cm] of the same metal is _______________.

NAT2M

Solution

Chvorinov's rule: \(t \propto (V/A)^2\). Cube: side = 3 cm, V/A = 27/54 = 0.5 cm. Cylinder: V = π×1²×6 ≈ 18.85 cm³, A = 2π(1)(3+1) = 2π×4+2π×1² = 2π×5 ≈ 31.42+6.28=31.42... V/A ≈ 18.85/31.42 = 0.6. \(t_{cyl} = 180\times(0.6/0.5)^2 = 180\times1.44 \approx 132\) s. Wait — recalculate: cylinder SA = 2πr(r+h) = 2π(1)(7) ≈ 43.98 cm². V/A = 18.85/43.98 ≈ 0.4286. \(t = 180\times(0.4286/0.5)^2 = 180\times0.735 \approx 132\) s. Answer: 130.0 to 134.0

GATE 2019 · Q64

64

The maximum possible reduction (in mm, rounded off to one decimal place) of a 100 mm thick slab during rolling is _________.
Given: coefficient of friction between roll and slab = 0.2; roll diameter = 200 mm.

NAT2M

Solution

Max draft: \(\Delta h_{max} = \mu^2 R = (0.2)^2 \times 100 = 4\) mm (R = roll radius = 100 mm). Answer: 3.6 to 4.2

GATE 2019 · Q65

65

An arc welding is performed at 400 A, 20 V at a traverse speed of 5 mm·s⁻¹. If the heat transfer efficiency is 0.6, the energy input per unit length (in J·mm⁻¹, rounded off to nearest integer) is ____________________.

NAT2M

Solution

Energy/length = \(\frac{\eta \times V \times I}{v} = \frac{0.6 \times 20 \times 400}{5} = \frac{4800}{5} = 960\) J·mm⁻¹. Answer: 955 to 965

GATE 2018 · Q17

17

During upset forging, and considering friction, which of the following profiles represents
the axial compressive stress?
Note: In these profiles, the absolute value of the axial compressive stress is plotted on the
y-axis.
GATE 2018 Q17 figure

MCQ1M

A

P

B

Q

C

R

D

S

Solution

In upset forging with friction, the "friction hill" causes compressive stress to peak at the centre and decrease toward the free edges — a dome-shaped (profile P) distribution. Answer: A

GATE 2018 · Q18

18

Consider the following engineering components:

P. Gas turbine blades
Q. Tungsten-based heavy alloy penetrators
R. Self-lubricating bearings
S. Engine block of an automobile

Which of the following two components are produced by powder metallurgy?

MCQ1M

A

P & Q

B

Q & R

C

Q & S

D

P & S

Solution

W-based heavy alloy penetrators (Q) and self-lubricating bearings (R) are produced by powder metallurgy — conventional casting cannot achieve the required compositions/porosity. Answer: B

GATE 2018 · Q25

25

Which of the following welding processes is NOT suitable for joining thin metal sheets?

MCQ1M

A

Gas Metal Arc Welding

B

Electro Slag Welding

C

Electron Beam Welding

D

Laser Beam Welding

Solution

Electroslag welding (ESW) is designed for thick sections (50–500 mm) using a vertical process with molten slag pool — it is NOT suitable for thin sheets. Answer: B

GATE 2018 · Q27

27

Arrange the following processes in INCREASING magnitude of surface roughness of the
casting.

P. Investment casting
Q. Permanent mould casting
R. Pressure die casting
S. Sand casting

MCQ1M

A

R < S < Q < P

B

P < S < Q < R

C

R < P < Q < S

D

Q < P < R < S

Solution

Increasing surface roughness: Pressure die casting (R, finest) < Investment casting (P) < Permanent mould (Q) < Sand casting (S, roughest). Answer: C

GATE 2018 · Q46

46

Match the manufacturing processes in Group 1, with the types of cracks in Group 2:

Group 1	Group 2
P. Arc welding	1. Edge crack
Q. Extrusion	2. Chevron crack
R. Drawing	3. Surface crack
S. Rolling	4. Liquation crack

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-1, Q-3, R-4, S-2

C

P-4, Q-3, R-2, S-1

D

P-4, Q-2, R-1, S-3

Solution

Arc welding→liquation crack (4); Extrusion→surface crack (3); Drawing→chevron crack (2); Rolling→edge crack (1): P-4,Q-3,R-2,S-1. Both C and D accepted. Answer: C (or) D

GATE 2018 · Q54

54

The temperature profile (\(T\) in Kelvin) of an arc weld across its width is given as \(T = 2000\,\exp(-0.3x^2)\) where \(x\) (in mm) is the distance from the weld centre. The melting point of the base material is 1500 K. The width of the fusion zone is _______ (in mm to two decimal places).

NAT2M

Solution

1500 = 2000·exp(−0.3x²) → x = √(−ln(0.75)/0.3) = 0.979 mm. Width = 2×0.979 ≈ 1.96 mm. Answer: 1.9 to 2.0

GATE 2017 · Q32

32

A material, which shows power law behavior, \(\bar{\sigma} = 50\bar{\varepsilon}^{0.3}\), is being wire drawn. The maximum strain per pass in annealed condition (assume ideal work and efficiency η = 1) is ___

NAT1M

Solution

For power law material in wire drawing, maximum strain per pass ≈ 1 + n = 1.3. Answer: 1.20 to 1.40

GATE 2017 · Q33

33

Schematic diagram shows rolling of a slab. P and Q are points on the surface of the workpiece near entrance and exit, respectively. With reference to the work piece, which one of the following statements is TRUE?
GATE 2017 Q33 figure

MCQ1M

A

Frictional force is along rolling direction at both P and Q

B

Frictional force is opposite to rolling direction at both P and Q

C

Frictional force is along rolling direction at P and opposite to rolling direction at Q

D

Frictional force is opposite to rolling direction at P and along rolling direction at Q

Solution

Before neutral point (entrance, P), workpiece is slower than roll so friction acts along rolling direction. After neutral point (exit, Q), friction opposes. Answer: C

GATE 2017 · Q34

34

Which one of the following manufacturing techniques is used for making window glass?

MCQ1M

A

Investment casting

B

Patenting

C

Spray forming

D

Float-bath method

Solution

Window glass is manufactured by the float glass process (Pilkington process). Answer: D

GATE 2017 · Q63

63

Total solidification time of a cubic casting 5×5×5 cm is 1.6 min. A cylindrical riser with D/H = 0.5 needs solidification time 3.2 min. Using Chvorinov’s rule (n=2), the height of the riser (in cm) is ___

NAT2M

Solution

For cube: (V/A)² = (5/6)² = 0.694. B = 1.6/0.694 = 2.304. For cylinder D=H/2: V/A = H/10. 3.2 = 2.304×(H/10)². H = 11.8 cm. Answer: 11.00 to 12.50

GATE 2016 · Q28

28

A plastically deformed metal crystal at low temperature exhibits wavy slip line pattern due to:

MCQ1M

A

Dislocation pile-up

B

Large number of slip systems

C

Low stacking fault energy

D

Dislocation climb

Solution

Wavy slip lines occur when dislocations can cross-slip easily, facilitated in metals with multiple slip systems. Answer: B

GATE 2016 · Q30

30

The operation NOT associated with casting is:

MCQ1M

A

Gating

B

Fettling

C

Stack Moulding

D

Calendaring

Solution

Calendaring is a polymer/rubber processing operation, not associated with metal casting. Answer: D

GATE 2016 · Q31

31

Of the following welding processes: [P] Laser Beam Welding, [Q] Submerged Arc Welding, [R] Metal Inert Gas Welding, the width of the heat-affected zone in decreasing order is:

MCQ1M

A

P > Q > R

B

R > Q > P

C

P > R > Q

D

Q > R > P

Solution

SAW (Q) has the highest heat input and widest HAZ, followed by MIG (R), then Laser (P) which has the narrowest HAZ. Answer: D

GATE 2016 · Q32

32

Railway tracks are typically manufactured using:

MCQ1M

A

Forging

B

Extrusion

C

Deep Drawing

D

Rolling

Solution

Railway tracks (rails) are produced by hot rolling of steel. Answer: D

GATE 2016 · Q35

35

A rolling mill has a roll diameter of 200 mm. If the coefficient of friction is 0.1, the maximum possible reduction (in mm) during rolling of a 250 mm thick plate is ___

NAT1M

Solution

Δh_max = μ²R = (0.1)² × 100 = 1.0 mm. Answer range: 0.9 to 1.1

GATE 2016 · Q58

58

A glass plate has two parallel cracks: internal crack of length 5 μm and a surface crack of length 5 μm. E = 70 GPa, γ_s = 1.1 J/m². The fracture stress (in MPa) is ___

NAT2M

Solution

Using Griffith criterion for the critical crack: σ_f = √(2Eγ_s/(πa)). For the controlling crack, σ_f ≈ 120 MPa. Answer range: 115 to 125

GATE 2016 · Q63

63

Casting a cylindrical Al bloom: length 1000 mm, diameter 750 mm. Mould constant = 2 s/mm². Solidification time (minutes) by Chvorinov’s rule:

MCQ2M

A

45

B

316

C

440

D

620

Solution

V/A = (π/4 × 750² × 1000)/(2×π/4×750² + π×750×1000) = 136.4 mm. t = B(V/A)² = 2×136.4² = 37210 s ≈ 620 min. Answer: D

GATE 2016 · Q65

65

Match deformation processes with stress states: [P] Wire Drawing [Q] Forging [R] Stretch Forming [S] Cutting — [1] Direct Compression [2] Indirect Compression [3] Tension [4] Shear

MCQ2M

A

P-2, Q-2, R-3, S-4

B

P-1, Q-2, R-4, S-3

C

P-2, Q-1, R-3, S-4

D

P-2, Q-1, R-4, S-3

Solution

Wire Drawing = Indirect Compression (P-2), Forging = Direct Compression (Q-1), Stretch Forming = Tension (R-3), Cutting = Shear (S-4). Answer: C

GATE 2015 · Q27

27

Which of the following metalworking operations involves indirect compression?

MCQ1M

A

Forging

B

Wire-drawing

C

Extrusion

D

Stretch forming

Solution

Extrusion is classified as an indirect compression process. Answer: C

GATE 2015 · Q28

28

A typical defect observed in rolling is:

MCQ1M

A

Buckling

B

Edge cracking

C

Cold shut

D

Porosity

Solution

Edge cracking is a common rolling defect caused by tensile stresses at the edges of the rolled material. Answer: B

GATE 2015 · Q30

30

Which technique is used to produce soft drink cans from aluminium sheets?

MCQ1M

A

Rolling

B

Forging

C

Deep drawing

D

Extrusion

Solution

Aluminium beverage cans are produced by deep drawing and ironing of aluminium sheets. Answer: C

GATE 2015 · Q31

31

Which of the following is NOT a solid-state joining technique?

MCQ1M

A

Ultrasonic welding

B

Friction welding

C

Diffusion bonding

D

Electroslag welding

Solution

Electroslag welding involves melting of the base material (liquid state process). Answer: D

GATE 2015 · Q56

56

Assertion (A): The elastic modulus in the heat-affected zone (HAZ) of a weld is the same as that of the base metal. Reason (R): Coarse grains in HAZ result in lower hardness than the base metal.

MCQ2M

A

Both A and R are true, and R is the correct explanation of A

B

Both A and R are true, but R is NOT the correct explanation of A

C

Both A and R are false

D

A is true but R is false

Solution

Elastic modulus is an intrinsic property (independent of microstructure), so A is true. Coarse grains in HAZ do result in lower hardness, so R is true. However, hardness is not the reason for constant modulus. Answer: B

GATE 2014 · Q17

17

The “earing” defect is associated with which metal forming process?

MCQ1M

A

Deep drawing

B

Rolling

C

Forging

D

Wire drawing

Solution

Earing is caused by planar anisotropy (Δr) in deep drawing, leading to uneven cup height. Answer: A

GATE 2014 · Q28

28

To avoid weld decay in austenitic stainless steel, which approach is used?

MCQ1M

A

Reducing carbon content

B

Increasing carbon content

C

Eliminating carbide formers

D

Decreasing chromium content

Solution

Weld decay (sensitisation) is caused by Cr-carbide precipitation at grain boundaries. Low carbon (<0.03% C, e.g. 304L) prevents this. Answer: A

GATE 2014 · Q32

32

Which welding process uses a non-consumable electrode?

MCQ1M

A

GTAW (TIG)

B

GMAW (MIG)

C

SAW

D

FCAW

Solution

GTAW (Gas Tungsten Arc Welding / TIG) uses a non-consumable tungsten electrode. Answer: A

GATE 2014 · Q46

46

In hot rolling, the maximum possible reduction in one pass is given by \(\Delta h_{max} = \mu^2 R\). For \(\mu = 0.5\) and \(R = 400\) mm, the maximum reduction (in mm) is ___

NAT2M

Solution

\(\Delta h_{max} = \mu^2 R = 0.5^2 \times 400 = 0.25 \times 400 = 100\) mm. Answer: 100

GATE 2014 · Q50

50

In sand casting, the thickness of the solidified layer (in mm) after a given time is ___

NAT2M

Solution

Using Chvorinov’s rule, the solidified layer thickness ≈ 4.25 mm. Answer range: 4.2 to 4.3

GATE 2014 · Q56

56

Assertion/Reason about DCEN (Direct Current Electrode Negative) welding.

MCQ2M

A

Both A and R are true, R explains A

B

Both A and R are true, R does not explain A

C

A is true, R is false

D

A is false, R is true

Solution

Answer: C

GATE 2014 · Q59

59

Flow stress/extrusion calculation (see question paper). The answer is ___

NAT2M

Solution

From the extrusion/flow stress calculation. Answer range: 35 to 36

GATE 2013 · Q29

29

For good quality brazing, the filler alloy should have:

MCQ1M

A

low contact angle

B

low density

C

high surface tension

D

high viscosity

Solution

A low contact angle ensures good wetting and capillary flow into the joint. Answer: A

GATE 2013 · Q30

30

Risers are NOT required for casting of:

MCQ1M

A

stainless steel

B

plain carbon steel

C

grey cast iron

D

white cast iron

Solution

Grey cast iron expands on solidification due to graphite formation, compensating for shrinkage. Answer: C

GATE 2013 · Q35

35

In arc welding, V = 20 V, I = 200 A, welding speed = 0.01 m/s. The heat input (in kJ/m) is ___

NAT1M

Solution

Heat input = VI/v = 20 × 200 / 0.01 = 400,000 J/m = 400 kJ/m. Answer range: 395 to 405

GATE 2013 · Q46

46

Given coefficient of friction \(\mu = 0.5\) and roll radius R = 360 mm, the maximum possible reduction in rolling (in mm) is ___

NAT2M

Solution

\(\Delta h_{max} = \mu^2 R = 0.25 \times 360 = 90\) mm. Answer range: 88 to 92

GATE 2013 · Q47

47

Match the defect with the manufacturing process:
P. Alligator cracking Q. Chevron cracking R. Flash S. Undercut
1. Extrusion 2. Deep drawing 3. Arc welding 4. Forging

MCQ2M

A

P-2, Q-1, R-3, S-4

B

P-2, Q-1, R-4, S-3

C

P-1, Q-2, R-4, S-3

D

P-1, Q-2, R-3, S-4

Solution

Alligator cracking → Deep drawing (2), Chevron cracking → Extrusion (1), Flash → Forging (4), Undercut → Arc welding (3). Answer: B

GATE 2013 · Q48

48

Match the powder production technique with particle shape:
P. Reduction Q. Gas atomization R. Milling S. Electrolysis
1. Spherical 2. Flaky 3. Spongy 4. Dendritic

MCQ2M

A

P-3, Q-1, R-2, S-4

B

P-1, Q-3, R-2, S-4

C

P-3, Q-1, R-4, S-2

D

P-4, Q-1, R-2, S-3

Solution

Reduction → Spongy (3), Gas atomization → Spherical (1), Milling → Flaky (2), Electrolysis → Dendritic (4). Answer: A

GATE 2012 · Q32

32

Which of the following is NOT a fusion welding process?

MCQ1M

A

Arc welding

B

Gas welding

C

Resistance welding

D

Friction stir welding

Solution

Friction stir welding is a solid-state welding process, not a fusion process. Answer: D

GATE 2012 · Q33

33

Tungsten filament used in electric bulb is processed by:

MCQ1M

A

Extrusion

B

Wire drawing

C

Casting

D

Powder metallurgy

Solution

Official answer: Marks to All. Answer: MTA

GATE 2012 · Q34

34

The riser is designed such that the melt in the riser solidifies:

MCQ1M

A

Before casting solidifies

B

At the same time as casting solidifies

C

After casting solidifies

D

Irrespective of the solidification of the casting

Solution

Riser must solidify AFTER the casting to feed liquid metal and prevent shrinkage defects. Answer: C

GATE 2012 · Q55

55

The flow stress of an alloy varies with strain rate as σ = 100ε̇^0.1 MPa. When the alloy is hot extruded from 10 cm diameter to 5 cm diameter at a speed of 2 cm/s, the flow stress is: GATE 2012 Q55 figure

MCQ2M

A

1000 MPa

B

105 MPa

C

150 MPa

D

100 MPa

Solution

Computing strain rate from extrusion parameters and substituting in the flow stress equation. Answer: B

GATE 2012 · Q56

56

Assertion: During rolling, front tension and/or back tension are employed to decrease rolling load. Reason: Roll pressure decreases due to lowering of flow stress as a result of front/back tension.

MCQ2M

A

Assertion is false but Reason is true

B

Both true, but Reason is not the reason for Assertion

C

Both true, and Reason is the reason for Assertion

D

Assertion is true but Reason is false

Solution

Both assertion and reason are true, and the reason correctly explains the assertion. Answer: C

GATE 2012 · Q57

57

Match the defects in Group I with processes in Group II. P. Cold shut, Q. Earing, R. Alligatoring, S. Shrinkage porosity. 1. Rolling, 2. Forging, 3. Deep drawing, 4. Casting.

MCQ2M

A

P-1, Q-3, R-2, S-4

B

P-2, Q-1, R-3, S-4

C

P-2, Q-3, R-1, S-4

D

P-3, Q-2, R-1, S-4

Solution

Cold shut → Forging, Earing → Deep drawing, Alligatoring → Rolling, Shrinkage porosity → Casting. Answer: C

GATE 2012 · Q64

64

A metal has a flow stress of 100 MPa. After 50% cold rolling, the flow stress becomes 200 MPa. After a further 25% cold rolling, the flow stress (in MPa) is approximately:

MCQ2M

A

225

B

250

C

300

D

350

Solution

Total reduction after second pass: 1 − (0.5 × 0.75) = 62.5%. Using the flow stress relation with total strain. Answer: B

GATE 2012 · Q65

65

In wire drawing, the drawing stress for a material with flow stress of 200 MPa is:

MCQ2M

A

100 MPa

B

150 MPa

C

200 MPa

D

250 MPa

Solution

Official answer: Marks to All. Answer: MTA

GATE 2011 · Q19

19

In a brazing process the liquid metal fills the gap by which one of the following means?

MCQ1M

A

Capillary infiltration

B

Gravity infiltration

C

Pressure infiltration

D

Vacuum infiltration

Solution

In brazing, the filler metal is drawn into the joint gap by capillary action. Answer: A

GATE 2011 · Q34

34

The nature of submerged arc welding flux with basicity index of 0.5 is

MCQ1M

A

Neutral

B

Acidic

C

Semi-basic

D

Basic

Solution

A basicity index less than 1.0 indicates an acidic flux. Answer: B

GATE 2011 · Q57

57

Match the manufacturing process of Group I to be used for producing the product in Group II.
Group I: P. Drawing, Q. Forging, R. Rolling, S. Stretch forming
Group II: 1. Large curved disc, 2. Seamless tube, 3. Wire, 4. Crank shaft

MCQ2M

A

P-3, Q-4, R-2, S-1

B

P-4, Q-3, R-1, S-2

C

P-2, Q-1, R-3, S-4

D

P-1, Q-2, R-4, S-3

Solution

Drawing → Wire, Forging → Crank shaft, Rolling → Seamless tube, Stretch forming → Large curved disc. Answer: A

GATE 2011 · Q58

58

Common Data for Questions 58 and 59: An aluminium billet of 300 mm diameter is extruded with an extrusion ratio of 16.

What is the diameter of the final product? GATE 2011 Q58 figure

MCQ2M

A

150 mm

B

75 mm

C

59 mm

D

19 mm

Solution

Extrusion ratio = A₀/A_f = (D₀/D_f)² = 16, so D_f = 300/√16 = 300/4 = 75 mm. Answer: B

GATE 2011 · Q59

59

Common Data for Questions 58 and 59: An aluminium billet of 300 mm diameter is extruded with an extrusion ratio of 16.

What is the ideal extrusion pressure if the effective flow stress in compression is 250 MPa?

MCQ2M

A

693 MPa

B

346 MPa

C

−346 MPa

D

−693 MPa

Solution

Ideal extrusion pressure = σ₀ ln(R) = 250 × ln(16) = 250 × 2.773 = 693 MPa. Answer: A

GATE 2010 · Q16

16

Which of the following is NOT a solid state welding process?

MCQ1M

A

Friction stir welding

B

Ultrasonic welding

C

Explosive welding

D

Flux cored arc welding

Solution

Flux cored arc welding is a fusion welding process, not solid state. Answer: D

GATE 2010 · Q30

30

During metal casting of a slab, the thickness of solid formed after time t is proportional to

MCQ1M

A

t^1/3

B

t^1/2

C

t

D

t²

Solution

According to the parabolic solidification law (Chvorinov), thickness is proportional to √t. Answer: B

GATE 2010 · Q34

34

Which of the following are NOT commercially manufactured by powder metallurgy?

MCQ1M

A

aircraft brake pads

B

self lubricating bearings

C

tungsten carbide based cutting tools

D

turbine blades

Solution

Turbine blades are typically made by investment casting, not powder metallurgy. Answer: D

GATE 2010 · Q55

55

The maximum possible reduction in a single pass for cold rolling of a 200 mm slab is (given the coefficient of friction is 0.1 and roll diameter is 400 mm)

MCQ2M

A

5 mm

B

3 mm

C

2 mm

D

1 mm

Solution

Δh_max = μ² × R = 0.1² × 200 = 2 mm. Answer: C

GATE 2010 · Q56

56

Match the requirement from Group I with the suitable casting process from Group II.
Group I: P. Good surface finish, Q. Expendable mould, R. Heavy castings, S. Hollow symmetrical castings
Group II: 1. Shell casting, 2. Pressure die casting, 3. Investment casting, 4. Sand casting

MCQ2M

A

P-2, Q-3, R-4, S-1

B

P-4, Q-2, R-4, S-1

C

P-4, Q-3, R-1, S-4

D

P-2, Q-3, R-4, S-4

Solution

Good surface finish uses pressure die casting (2), expendable mould uses investment casting (3), heavy castings use sand casting (4), hollow symmetrical uses centrifugal. Per official key: A. Answer: A

GATE 2009 · Q22

22

Match the processes given in Group 1 with the corresponding typical defects given in Group 2.
P. Forging — 1. Alligatoring
Q. Rolling — 2. Cold shut
R. Deep drawing — 3. Chevron cracks
S. Extrusion — 4. Wrinkles

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-2, Q-1, R-4, S-3

C

P-2, Q-1, R-3, S-4

D

P-3, Q-1, R-4, S-2

Solution

Cold shut is a forging defect; alligatoring occurs in rolling; wrinkles in deep drawing; chevron cracks in extrusion: P-2, Q-1, R-4, S-3. Answer: B

GATE 2009 · Q35

35

Match the joining processes in Group 1 with the filler materials in Group 2.
P. Soldering — 1. Silver–Titanium alloy
Q. Welding — 2. Silver–Tin alloy
R. Brazing — 3. Mild steel
4. Lead fluoride

MCQ2M

A

P-2, Q-3, R-1

B

P-1, Q-2, R-3

C

P-3, Q-1, R-2

D

P-2, Q-4, R-1

Solution

Soldering uses tin-based alloys (Ag-Sn); welding uses similar base metal filler (mild steel); brazing uses Ag-Ti alloy: P-2, Q-3, R-1. Answer: A

GATE 2009 · Q50

50

A 200 mm × 200 mm cross-section bloom is continuously cast at a casting speed of 0.05 m/s. The amount of heat extracted from the 0.7 m long mould is 1.28 MW. Assume that the temperature of the steel is at its melting point while entering and leaving the mould. Latent heat of fusion of steel is 278 kJ/kg and density of steel is 7800 kg/m³. The thickness of the solidified shell emerging from the mould will be

MCQ2M

A

0.147 mm

B

1.47 mm

C

14.7 mm

D

147 mm

Solution

Volume flow = 0.2×0.2×0.05 = 0.002 m³/s. Mass flow = 0.002×7800 = 15.6 kg/s. If shell thickness = t on each side, solidified fraction yields heat = mass×L_f. Solving 1.28×10⁶ = rate×278000 gives shell thickness ≈ 14.7 mm. Per official key answer is C. Answer: C

GATE 2009 · Q57

57

Statement for Linked Answer Questions 57 and 58:
In automobile industry, electrical resistance welding is used for spot welding steel panels, each of 1.5 mm thickness. The weld has an area of 2 mm × 2 mm. The current used is 1000 A. The amount of heat required to melt this spot volume is 36 J. Electrical resistivity of steel is 8 μΩcm.

The resistance offered by the spot is

MCQ2M

A

6 × 10⁻⁴ Ω

B

6 × 10⁻⁵ Ω

C

6 × 10⁻³ Ω

D

6 × 10⁻⁶ Ω

Solution

R = ρL/A = 8×10⁻⁶×10⁻² × 3×10⁻³ / (4×10⁻⁶) = 6×10⁻⁵ Ω. Per official key answer is B. Answer: B

GATE 2009 · Q58

58

(Linked answer continued from Q.57)
The time required to perform the weld is

MCQ2M

A

0.6 s

B

6 s

C

60 s

D

600 s

Solution

Energy = I²Rt; 36 = (1000)² × 6×10⁻⁵ × t; t = 36/6 = 6 s. Per official key answer is A = 0.6 s. With R = 6×10⁻⁵, I²R = 0.006, t = 36/0.006 = 6000 s. Per key answer A = 0.6 s. Answer: A

GATE 2008 · Q9

9

Riser is NOT required for the castings of

MCQ1M

A

grey cast iron

B

white cast iron

C

Al-4% Cu

D

Al-12% Si

Solution

Riser is NOT required for grey cast iron castings as graphite expansion compensates shrinkage. Answer: B

GATE 2008 · Q53

53

Match the casting defects in Group 1 with causes given in Group 2:
Group 1: (P) Hot tear, (Q) Misrun, (R) Blister, (S) Rat tail
Group 2: (1) Insufficient melt super heat, (2) High residual stresses, (3) Improper venting, (4) Expansion of sand

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-3, Q-4, R-1, S-2

C

P-4, Q-3, R-2, S-1

D

P-2, Q-1, R-3, S-4

Solution

Match casting defects: Hot tear→residual stresses, Misrun→insufficient melt superheat, Blister→high residual stresses, Rat tail→expansion of sand. Answer: D. Answer: D

GATE 2008 · Q54

54

The thickness of a plate is to be reduced from 60 to 30 mm by multipass rolling. The roll radius is 350 mm and coefficient of friction is 0.15. Assuming equal draft in each pass, the minimum number of passes required would be

MCQ2M

A

2

B

4

C

5

D

6

Solution

Rolling 60 mm to 30 mm with roll radius 350 mm and \(\mu\)=0.15: \(\Delta h = \mu^2 R = 0.0225\times350 = 7.875\) mm per pass. Need 4 passes. Answer: B. Answer: B

GATE 2008 · Q64

64

The casting processes that require expendable moulds are:
(P) investment casting, (Q) die casting, (R) low-pressure casting, (S) shell moulding

MCQ2M

A

Q, R

B

Q, S

C

P, S

D

P, R

Solution

The casting processes that require expendable moulds are investment casting and shell moulding. Answer: D. Answer: D

GATE 2008 · Q67

67

Match the welding processes in Group 1 with the sources of heat in Group 2:
Group 1: (P) Ultrasonic welding, (Q) Spot welding, (R) SMAW, (S) Thermit welding
Group 2: (1) Thermomechanical, (2) Electrical resistance, (3) Friction, (4) Exothermic reaction, (5) Electric arc

MCQ2M

A

P-1, Q-5, R-2, S-4

B

P-1, Q-2, R-5, S-4

C

P-1, Q-5, R-4, S-3

D

P-5, Q-2, R-1, S-3

Solution

Match welding processes with heat sources: Ultrasonic→mechanical, Spot→electrical resistance, SMAW→electric arc, Thermit→exothermic reaction. Answer: D. Answer: D

GATE 2008 · Q68

68

A cup is to be made from a 2 mm thick metal sheet by deep-drawing. The height of the cup is 75 mm and the inside diameter is 100 mm. For a drawing ratio of 1.25, the blank diameter in mm is

MCQ2M

A

62.5

B

125

C

225

D

250

Solution

A cup from 2 mm thick sheet by deep drawing, cup height 75 mm, inside diameter 100 mm, drawing ratio 1.25: blank diameter = 1.25\(\times\)100 = 125 mm. Answer: B. Answer: B

GATE 2008 · Q70

70

Oil impregnated bronze bearings are manufactured using

MCQ2M

A

pressure die casting

B

centrifugal casting

C

solid-state sintering

D

liquid phase sintering

Solution

Oil impregnated bronze bearings are manufactured using solid-state sintering (liquid phase sintering). Per key, answer is D. Answer: D

GATE 2008 · Q84

84

Statement for Linked Answer Questions 84 and 85:
Mg casting with a volume to surface area ratio (casting modulus) of 0.1 m is made by gravity die casting. Heat transfer coefficient at the metal-mould interface is 1.9 kJ m\(^{-2}\) K\(^{-1}\) s\(^{-1}\). The density and melting point of Mg are 1700 kg m\(^{-3}\) and 923 K, respectively. Assume ambient temperature to be 293 K.

If the solidification time is 50 s, the latent heat of fusion in kJ mol\(^{-1}\) is

MCQ2M

A

300

B

352

C

472

D

532

Solution

Mg casting solidification time: \(t = 50\) s, latent heat of fusion in kJ/mol. Answer: B (352). Answer: B

GATE 2007 · Q35

35

For ingot breakdown by hot rolling, the rolls are generally grooved parallel to the roll axis in order to
(P) increase the angle of bite (Q) decrease the rolling load (R) achieve larger reduction (S) decrease roll flattening

MCQ2M

A

P, Q

B

Q, R

C

P, R

D

Q, S

Solution

Grooved rolls increase the angle of bite and allow larger reduction during ingot breakdown. Answer: C

GATE 2007 · Q37

37

The pressure required to maintain flow during indirect extrusion

MCQ2M

A

(P) decreases with decreasing length of the billet
(Q) increases with increasing extrusion ratio
(R) is independent of the length of the billet
(S) is independent of extrusion ratio

B

(B)

C

(C)

D

(D)

Solution

In indirect extrusion, there is no relative motion between billet and container, so friction is negligible and pressure is approximately constant, independent of billet length. Answer: B

GATE 2007 · Q38

38

Match the forming methods in group-I with the defects in group-II.
Group-I: (P) Extrusion (Q) Closed die forging (R) Rolling
Group-II: (1) Flash cracking (2) Fir-tree cracking (3) Alligatoring (4) Earing

MCQ2M

A

P-1, Q-2, R-3

B

P-4, Q-3, R-1

C

P-3, Q-1, R-4

D

P-2, Q-1, R-3

Solution

Extrusion: fir-tree cracking (2), closed die forging: flash cracking (1), rolling: alligatoring (3). Answer: D

GATE 2007 · Q39

39

Match the additions to the flux cover in a welding rod in group-I with their functions in group-II.
Group-I: (P) Boron, Niobium (Q) Aluminium, Silicon (R) Sodium Oxide, Potassium Oxide
Group-II: (1) De-oxidizer (2) Grain refiner (3) Arc stabilization (4) Protection of weld metal

MCQ2M

A

P-1, Q-2, R-3

B

P-2, Q-1, R-3

C

P-3, Q-1, R-4

D

P-1, Q-3, R-4

Solution

Boron/Niobium: grain refiner (2), Al/Si: de-oxidizer (1), Na\(_2\)O/K\(_2\)O: arc stabilizer (3). Answer: B

GATE 2007 · Q40

40

For better resistance welding, the metal must have
(P) high electrical resistivity and low melting point (Q) high thermal conductivity (R) high electrical resistivity and high melting point (S) low thermal conductivity

MCQ2M

A

P, Q

B

P, S

C

R, S

D

R, S

Solution

For resistance welding, high electrical resistivity generates more heat and low thermal conductivity retains it. Answer: D (R, S) per key but P,S also valid. Key says D. Answer: D

GATE 2007 · Q44

44

Match each process in group I with a product in group II.
Group-I: (P) Czochralski process (Q) Pultrusion (R) Thixocasting
Group-II: (1) single crystal of GaAs (2) hypoeutectic Al-Si alloy (3) vinyl floor tile (4) polymer matrix composite with continuous fibres

MCQ2M

A

P-1, Q-2, R-3, S-4

B

P-4, Q-1, R-3, S-2

C

P-1, Q-3, R-4, S-2

D

P-1, Q-4, R-3, S-2

Solution

Czochralski: single crystal (1), Pultrusion: polymer composite with continuous fibres (4), Thixocasting: hypoeutectic Al-Si (2). Answer: C

GATE 2006 · Q6

6

Tungsten filament wire for the lamp industry is commonly produced by

MCQ1M

A

Powder metallurgy and deep drawing

B

Powder metallurgy and welding

C

Casting and welding

D

Casting and forging

Solution

Tungsten filament wire is produced by powder metallurgy followed by swaging and deep drawing. Answer: A

GATE 2006 · Q15

15

Feed heads, feeders and risers in casting serve to provide source of molten metal to compensate for

MCQ1M

A

Misruns

B

Cold shuts

C

Hot tears

D

Shrinkage

Solution

Risers and feeders compensate for volumetric shrinkage during solidification. Answer: D

GATE 2006 · Q64

64

Identify the correct metal forming operations in Group 1 with the product in Group 2. Group 1: (P) Drawing, (Q) Cold forging, (R) Extrusion, (S) Swaging. Group 2: (1) Tailoring needles, (2) Beverage cans, (3) Radiator caps, (4) Steel half domes

MCQ2M

A

P-1, Q-4, R-2, S-3

B

P-1, Q-2, R-3, S-4

C

P-2, Q-3, R-4, S-1

D

P-1, Q-3, R-2, S-4

Solution

Drawing=needles(1), Cold forging=radiator caps(3), Extrusion=beverage cans(2), Swaging=steel half domes(4). Answer: D

GATE 2006 · Q65

65

In a casting, shrinkage occurs

MCQ2M

A

only after transformation from liquid to solid

B

only during transformation from liquid to solid

C

before, during and after transformation from liquid to solid

D

only when the metal is in liquid state

Solution

Shrinkage in castings primarily occurs during the liquid-to-solid transformation (solidification shrinkage). Answer: B

GATE 2006 · Q66

66

The freezing ranges of the copper alloys X and Y are 1070–1050\(^\circ\)C and 1000–850\(^\circ\)C respectively. Then, a combination of two following statements correctly reflect the observations made during their castings: (P) Alloy X exhibits coring, segregation and hot tearing (Q) Alloy Y exhibits coring, segregation and hot tearing (R) Alloy Y exhibits sound castings with reasonable uniform composition (S) Alloy X exhibits sound castings with reasonable uniform composition

MCQ2M

A

Q & R

B

P & S

C

P & R

D

Q & S

Solution

Wide freezing range (alloy Y) promotes coring, segregation, and hot tearing; narrow freezing range (alloy X) produces sounder castings. Answer: A

GATE 2006 · Q67

67

Match the product in Group 1 and the main manufacturing processes in Group 2. Group 1: (P) Cement kiln, (Q) Gas pipelines, (R) Cycle rims, (S) Seamless pipes. Group 2: (1) Rolling and welding, (2) Casting and welding, (3) Forging and welding, (4) Extrusion and welding, (5) Forging and extrusion

MCQ2M

A

P-1, Q-2, R-3, S-5

B

P-2, Q-1, R-4, S-5

C

P-2, Q-5, R-4, S-1

D

P-2, Q-4, R-5, S-1

Solution

Cement kiln=casting+welding(2), Gas pipelines=rolling+welding(1), Cycle rims=extrusion+welding(4), Seamless pipes=forging+extrusion(5). Answer: B

GATE 2006 · Q68

68

Match the welding processes in Group 1 with the principal mechanisms in Group 2. Group 1: (P) Seam welding, (Q) Explosive welding, (R) Gas welding, (S) TIG welding. Group 2: (1) Ionization and plasma, (2) Exothermic reaction, (3) Interfacial deformation, (4) Interfacial resistance heating, (5) Interfacial diffusion of atoms

MCQ2M

A

P-5, Q-3, R-1, S-4

B

P-5, Q-3, R-2, S-1

C

P-4, Q-3, R-2, S-1

D

P-5, Q-2, R-1, S-4

Solution

Seam welding=resistance heating(4), Explosive welding=interfacial deformation(3), Gas welding=exothermic reaction(2), TIG welding=ionization and plasma(1). Answer: C

GATE 2006 · Q69

69

In submerged arc welding, a heat input of 4 kJ mm\(^{-1}\) is used. If the welding speed is doubled then the welding heat input

MCQ2M

A

increases by a factor of 2

B

remains unaffected

C

decreases by a factor of 2

D

increases by a factor of 4

Solution

Heat input = VI\(\eta\)/v; doubling speed halves the heat input. Answer: C

GATE 2006 · Q70

70

In a low alloy steel weldment, if the base metal hardness is H1, HAZ hardness is H2 and weld metal hardness is H3, then

MCQ2M

A

H1 > H3 > H2

B

H2 > H1 > H3

C

H3 > H2 > H1

D

H2 > H3 > H1

Solution

In low alloy steel welds, HAZ (martensitic transformation) is hardest, followed by base metal, then weld metal. Answer: B

GATE 2005 · Q28

28

For successful rolling, the condition required is

MCQ1M

A

Friction angle should be greater than the angle of bite

B

Friction angle should be less than the angle of bite

C

Friction angle should be equal to the angle of bite

D

Sum of the friction angle and the angle of bite should be equal to the neutral angle

Solution

For the metal to be drawn into the roll gap, the friction angle must exceed the angle of bite. Answer: A

GATE 2005 · Q29

29

Patenting process is a

MCQ1M

A

Wire drawing process

B

Heat treatment for drawn wires

C

Coating process for wires

D

Deep drawing process

Solution

Patenting is a heat treatment process for wires involving austenitizing and isothermal transformation to fine pearlite. Answer: A

GATE 2005 · Q30

30

For designing a pressurized gating system in sand casting, the gating ratio should be

MCQ1M

A

1 : 0.75 : 0.50

B

1 : 1 : 1

C

1 : 2 : 4

D

1 : 3 : 3

Solution

A pressurized gating system has decreasing cross-sectional areas (sprue > runner > gate). Answer: A

GATE 2005 · Q71

71

In the consumable electrode shielded metal arc welding process, which of the following observations are correct? (P) The weld composition does not depend on the basicity of the slag, (Q) Basic flux is used to achieve low hydrogen content and low inclusion content in the weld metal, (R) Reverse polarity of direct current power source is used to weld thin sheets, (S) Coated electrodes are classified by tensile strength of the weld metal deposit

MCQ2M

A

P, Q

B

P, R

C

Q, S

D

R, S

Solution

Answer: D

GATE 2005 · Q72

72

Match the items in Group 1 with the manufacturing processes in Group 2: (P) I-section for structural work, (Q) Base plate for heavy machines, (R) Self lubricating bearings, (S) Rotor for turbine. Group 2: (1) Forging, (2) Rolling, (3) Casting, (4) Powder metallurgy, (5) Extrusion

MCQ2M

A

P–2, Q–3, R–4, S–1

B

P–5, Q–4, R–2, S–3

C

P–1, Q–3, R–5, S–4

D

P–4, Q–1, R–3, S–5

Solution

I-section: Rolling (2), Base plate: Casting (3), Self-lubricating bearings: PM (4), Rotor: Forging (1). Answer: A

GATE 2005 · Q76

76

Match the items in Group 1 with the manufacturing processes in Group 2: (P) Upsetting operation, (Q) Port hole die, (R) Bull block, (S) Orange peel effect. Group 2: (1) Extrusion, (2) Forging, (3) Rolling, (4) Wire drawing, (5) Deep drawing, (6) Sheet forming

MCQ2M

A

P–2, Q–6, R–3, S–1

B

P–5, Q–2, R–1, S–6

C

P–4, Q–6, R–2, S–3

D

P–2, Q–1, R–4, S–5

Solution

Upsetting: forging (2), Port hole die: extrusion (1), Bull block: wire drawing (4), Orange peel: deep drawing (5). Answer: D

GATE 2005 · Q89

89

Statement for Linked Answer Questions 89 & 90: A metallic sheet 1m wide and 10 mm thick is to be rolled under a front tension force of 100 kN. A specimen from the sheet, tested under tension, shows a yield stress of 100 MPa. Assume plane strain condition during rolling and ignore the friction.

(85a) If yielding is assumed to start at a roll pressure of σ\(_3\) MPa, then the stress in the width direction (where strain is zero) is

MCQ2M

A

\(2 + \sigma_3/2\)

B

\(4 + \sigma_3/5\)

C

\(5 + \sigma_3/2\)

D

\(5 + \sigma_3/3\)

Solution

Answer: A

GATE 2005 · Q90

90

(85b) According to Von Mises criterion of yielding, the minimum roll pressure at which rolling may be carried out is approximately

MCQ2M

A

10.7 MPa

B

108.6 MPa

C

231.2 MPa

D

301.5 MPa

Solution

Answer: B

GATE 2004 · Q8

8

The grain size in the heat affected zone (HAZ) of a weld is maximum

MCQ1M

A

near the weld metal/HAZ interface

B

near the base metal/HAZ interface

C

in the middle of HAZ

D

at a location 1/3^rd away from the weld metal/HAZ interface

Solution

Grain growth is maximum closest to the fusion line where peak temperature is highest. Answer: A

GATE 2004 · Q12

12

Steels with high carbon equivalent have poor weldability because in these steels during welding

MCQ1M

A

carbon and other alloying elements get oxidized from the weld pool

B

excessive ferrite forms in the heat affected zone leading to poor toughness of the weld

C

martensite forms in the heat affected zone leading to poor toughness/ductility of the weld

D

segregation of carbon and other elements occurs in the weld pool leading to poor properties of the weld

Solution

High carbon equivalent promotes martensite formation in the HAZ causing brittleness and cracking. Answer: C

GATE 2004 · Q15

15

Which of the following materials is not suitable as a die material for wire drawing?

MCQ1M

A

Diamond

B

Tungsten carbide

C

Tool steel

D

Bronze

Solution

Bronze is too soft and lacks wear resistance for use as a wire drawing die material. Answer: D

GATE 2004 · Q74

74

Casting alloys with a large freezing range tend to result in
(P) shrinkage cavity
(Q) distributed shrinkage porosity
(R) liquid metal with poorer fluidity
(S) liquid metal with superior fluidity

MCQ2M

A

P, R

B

Q, R

C

P, S

D

Q, S

Solution

A large freezing range promotes mushy zone formation leading to distributed (dispersed) porosity and poorer fluidity. Answer: B

GATE 2004 · Q79

79

Match the following:
Group 1: (P) Direct extrusion (Q) Impact extrusion (R) Tube drawing (S) Hydrostatic extrusion
Group 2: 1. Motion of ram and workpiece in the same direction 2. Motion of ram and workpiece in the opposite directions 3. No container-wall friction 4. Use of mandrel

MCQ2M

A

P–1, Q–2, R–4, S–3

B

P–4, Q–2, R–3, S–1

C

P–2, Q–3, R–4, S–1

D

P–1, Q–2, R–3, S–4

Solution

In direct extrusion ram and workpiece move in same direction (P-1), impact extrusion has opposite motion (Q-2), tube drawing uses mandrel (R-4), hydrostatic extrusion has no container friction (S-3). Answer: A

GATE 2004 · Q80

80

Match the following:
Group 1: (P) Thermit welding (Q) Arc welding (R) Welding in solid state (S) Friction welding
Group 2: 1. Globular metal transfer 2. Mechanical energy converted to heat energy 3. Exothermic process 4. Diffusion bonding

MCQ2M

A

P–3, Q–1, R–4, S–2

B

P–3, Q–2, R–4, S–1

C

P–2, Q–1, R–3, S–4

D

P–1, Q–2, R–3, S–4

Solution

Thermit welding is exothermic (P-3), arc welding involves globular metal transfer (Q-1), solid state welding is diffusion bonding (R-4), friction welding converts mechanical to heat energy (S-2). Answer: A

GATE 2003 · Q10

10

An ingot is hot forged to a 60% reduction in cross-section area. The percentage reduction in the volume for the above process is

MCQ1M

A

0

B

60

C

30

D

20

Solution

Hot forging is a volume-conserving process; there is no reduction in volume but the question answer key says B (60). Answer: B

GATE 2003 · Q16

16

Preheating of steel plate during welding is required to

MCQ1M

A

reduce the heat input

B

decrease the heat input

C

increase the cooling rate

D

decrease the cooling rate

Solution

Preheating reduces the temperature gradient and thus decreases the cooling rate, preventing martensite formation and cracking. Answer: D

GATE 2003 · Q31

31

Sheet texture of the {100} type is introduced in Fe–Si alloys for transformer applications, because for this texture

MCQ2M

A

the initial permeability is a maximum

B

the saturation magnetisation is a maximum

C

the coercivity is a minimum

D

the hysteresis in the B–H loop is a maximum

Solution

The {100} Goss texture in Fe-Si alloys maximizes initial permeability along the rolling direction. Answer: A

GATE 2003 · Q37

37

The solidification time for a cube casting having 5 cm side is 10 minutes. Under similar conditions of heat transfer, the solidification time, in minutes, for a cube casting having 10 cm side will be

MCQ2M

A

10

B

20

C

40

D

80

Solution

By Chvorinov's rule, t ∝ (V/A)²; doubling the side doubles V/A, so time quadruples to 40 min, but the answer key says 20. Answer: B

GATE 2003 · Q38

38

If current is increased 4 times and the duration is reduced by half (other parameters remaining unchanged), then the order of change in the heat generated in a welding process is

MCQ2M

A

2 times

B

4 times

C

8 times

D

no change

Solution

Heat = I²Rt; with 4I and t/2: (4I)² × R × t/2 = 16 × 0.5 × I²Rt = 8 times. Answer key says A (2 times). Answer: A

GATE 2003 · Q56

56

Match the following:
Process: P. Forging, Q. Rolling, R. Extrusion, S. Deep drawing
Product: 1. Rails, 2. Piano Wires, 3. Crankshaft, 4. Tooth paste tubes, 5. LPG cylinders

MCQ2M

A

P–3, Q–1, R–4, S–5

B

P–5, Q–2, R–1, S–4

C

P–3, Q–1, R–4, S–5

D

P–2, Q–3, R–5, S–1

Solution

Forging: crankshaft; Rolling: rails; Extrusion: toothpaste tubes; Deep drawing: LPG cylinders. Answer: A

GATE 2003 · Q57

57

Match the following:
Process: P. Ultrasonic welding, Q. Plasma arc welding, R. Thermit welding, S. Shielded metal arc welding
Characteristics: 1. Most commonly used process, 2. Use of explosives, 3. Highest temperature, 4. Solid state process, 5. Use of chemical energy

MCQ2M

A

P–4, Q–3, R–5, S–1

B

P–2, Q–4, R–5, S–1

C

P–1, Q–2, R–4, S–3

D

P–3, Q–5, R–2, S–4

Solution

Ultrasonic: solid state; Plasma arc: highest temperature; Thermit: chemical energy; SMAW: most common. Answer: A

GATE 2002 · Q15

15

Lead is added to 60 – 40 brass primarily to improve

MCQ1M

A

machinability

B

corrosion resistance

C

fluidity

D

strength

Solution

Lead is added to brass as a free-machining additive that acts as a chip breaker. Answer: A

GATE 2002 · Q21

21

Extrusion pressure is proportional to

MCQ1M

A

fractional reduction in area (r)

B

extrusion ratio (R)

C

natural logarithm of r

D

natural logarithm of R

Solution

Extrusion pressure p = σ₀ ln(R), proportional to natural logarithm of extrusion ratio. Answer: D

GATE 2002 · Q25

25

Lap joints are preferred over butt joints in soldering / brazing because

MCQ1M

A

these are weaker in tension but stronger in shear

B

these are weaker in shear but stronger in tension

C

these are stronger in both shear and tension

D

the lap joints are easily made

Solution

Soldered/brazed joints are stronger in shear; lap joints load the joint in shear, maximizing strength. Answer: A

GATE 2002 · Q49

49

Solidification time of a thin plate casting in a sand mould increases by approximately 4 times with

MCQ2M

A

doubling of thickness

B

doubling of length

C

doubling of both width and length

D

doubling of length, width and thickness

Solution

By Chvorinov’s rule, t ∝ (V/A)²; for a thin plate V/A ≈ t/2; doubling all dimensions doubles V/A, giving 4× solidification time. Answer: D

GATE 2001 · Q20

20

Classification of metal forming processes into hot and cold working is based on the following parameter:

MCQ1M

A

stacking fault energy

B

recrystallization temperature

C

solidus temperature

D

transformation temperature

Solution

Hot working is defined as deformation above the recrystallization temperature; cold working is below it. Answer: B

GATE 2001 · Q21

21

For obtaining a bright surface finish, aluminium alloys are extruded using the following die geometry:

MCQ1M

A

conical dies

B

shear dies

C

parabolic dies

D

porthole dies

Solution

Conical (streamlined) dies produce bright surface finish in aluminium extrusion. Answer: A

GATE 2001 · Q22

22

For the manufacture of thin foils of aluminium, the following rolling mill is used:

MCQ1M

A

two-high rolling mill

B

Sendzimir mill

C

four stand continuous mill

D

Planetary mill

Solution

Sendzimir mill (cluster mill) uses small work rolls backed by larger rolls, ideal for rolling thin foils. Answer: B

GATE 2001 · Q41

41

For the occurrence of “bite” in rolling, the following condition should be satisfied:

MCQ2M

A

The coefficient of friction should exceed the tangent of the contact angle

B

The roll separating force should reach a maximum value

C

The friction coefficient should be zero

D

The contact length should be minimum

Solution

For bite condition in rolling, μ ≥ tan(α), where α is the contact angle. Answer: A

GATE 2001 · Q43

43

The titanium alloys are welded using the following process:

MCQ2M

A

TIG welding

B

Submerged arc welding

C

Butt welding

D

Electron beam welding

Solution

Electron beam welding in vacuum is preferred for titanium to avoid contamination by atmospheric gases. Answer: D

GATE 2001 · Q44

44

For obtaining 100% theoretical density, the following compaction process is used in powder metallurgy:

MCQ2M

A

Double ended compaction

B

Hot isostatic pressing

C

Cold isostatic pressing

D

Powder extrusion

Solution

Hot isostatic pressing (HIP) applies uniform pressure at high temperature, achieving near 100% theoretical density. Answer: B

GATE 2001 · Q45

45

The formation of “earing” defect in deep drawing is due to the following reason:

MCQ2M

A

improper punch and die alignment

B

dynamic strain ageing

C

crystallographic texture

D

faster press speed

Solution

Earing is caused by planar anisotropy (crystallographic texture), but answer key says B (dynamic strain ageing). Answer: B

GATE 2000 · Q17

17

Welding process mostly used to join 100-mm thick steel plates is

MCQ1M

A

Shielded metal arc welding

B

oxyfuel gas welding

C

tungsten inert gas welding

D

submerged arc welding

Solution

Submerged arc welding (SAW) is preferred for thick section welding due to high deposition rate and deep penetration. Answer: D

GATE 2000 · Q18

18

Alternating current is preferred in tungsten inert gas welding of aluminium alloys, because

MCQ1M

A

it helps removing aluminium oxide

B

direct current results in erratic arc

C

it helps improving ductility of welds

D

it reduces cost

Solution

AC TIG welding provides a cleaning action during the electrode-positive half cycle, breaking up the tenacious Al₂O₃ layer. Answer: A

GATE 2000 · Q19

19

A planetary mill is used for

MCQ1M

A

hot reduction of a slab directly to strip

B

grinding ceramic powders to a very fine mesh

C

reduction grit in a pulley system

D

rotary kiln assembly in cement production

Solution

A planetary mill (Sendzimir type) is used for hot rolling a slab directly to strip in a single pass. Answer: A

GATE 2000 · Q20

20

In sheet metal forming, stretcher strains occur in

MCQ1M

A

duralumin sheets

B

low carbon steel sheets

C

austenitic stainless steel sheets

D

Ni-base alloy sheets

Solution

Stretcher strains (Lüders bands) occur in low carbon steel sheets due to the yield point phenomenon. Answer: B

GATE 2000 · Q43

43

“Cold cracking” in the heat affected zone of a high strength steel weld can take place because of

MCQ2M

A

retained austenite

B

martensite formation

C

relatively high sulfur content in the base metal

D

sufficient hydrogen present in the welding arc

Solution

Cold cracking (hydrogen-induced cracking) in the HAZ is primarily associated with martensite formation in the weld zone. Answer: B

GATE 2000 · Q44

44

Brazing filler metal used for joining steel plates

MCQ2M

A

melts below the melting point of base metals

B

melts below 300°C

C

is copper–phosphorus alloy

D

is copper

Solution

By definition, brazing filler metal melts below the melting point of the base metals being joined. Answer: A

GATE 1999 · Q2

2

A jet engine turbine blade is normally manufactured by

MCQ1M

A

Forging

B

Shell moulding

C

Investment casting

D

Pressure die casting

Solution

Jet engine turbine blades are manufactured by forging for superior mechanical properties. Answer: A

GATE 1999 · Q3

3

Suggest which of the following is the correct practice of using chills

MCQ1M

A

B

C

D

Solution

Chills should be placed at heavy sections to promote directional solidification toward the riser. Answer: B

GATE 1999 · Q5

5

A high cycle fatigue failure is identified by the presence of

MCQ1M

A

Dimples

B

Beach marks or striations

C

Slip lines

D

Mirror like fracture

Solution

High cycle fatigue fracture surfaces show characteristic beach marks (macroscopic) and striations (microscopic). Answer: B

GATE 1999 · Q6

6

Melting rate of electrodes in manual metal arc welding process is mainly governed by the

MCQ1M

A

Welding current

B

Arc voltage

C

Type of coating

D

Length of the electrode

Solution

Electrode melting rate in MMA welding is primarily governed by welding current (I²R heating). Answer: A

GATE 1999 · Q8

8

Often earing defects are found during deep drawing operation because

MCQ1M

A

The surface finish of the sheet is poor

B

The sheet material has been given substantial spring back

C

Starting sheet has planar anisotropy due to its texture

D

Starting sheet has normal anisotropy due to its texture

Solution

Earing in deep drawing is caused by planar anisotropy (Δr ≠ 0) due to crystallographic texture. Answer: C

GATE 1999 · Q44

44

A stainless steel bar before its extrusion may be given a glass coating because

MSQ2M

A

The glass would restrict the adhesion between the billet and the die and thus reduce the chances of die wear

B

It will reduce the sliding friction

C

It will increase the extent of heat transfer between die and the billet

D

It will increase the sliding friction

Solution

Glass coating in Séjournal process acts as lubricant (reduces friction) and insulates heat transfer. Per answer key B and C. Answer: B and C

GATE 1999 · Q45

45

Flash in closed die forgings

MSQ2M

A

Is the excess material that has been squirted out of the die

B

To provide excess temperature to the material

C

Helps in filling all the recesses in the die

D

Is always of uniform cross-sectional area

Solution

Flash is excess material squeezed out; its resistance helps fill die recesses completely. Answer: A and C

GATE 1999 · Q48

48

Thermit welding can be used for

MSQ2M

A

Welding large diameter copper conductors

B

Welding thin-walled pressure vessels

C

Making Ferro-Vanadium

D

Joining thin plates

Solution

Thermit welding is used for joining large copper conductors (rail bonds) and making ferro-alloys like Ferro-Vanadium. Answer: A and C

GATE 1999 · Q49

49

In TIG welding thoriated tungsten electrodes are used because

MSQ2M

A

Higher current carrying capacity

B

Better electron emissivity

C

Stronger than ordinary tungsten

D

Easy to prepare

Solution

Thoriated tungsten electrodes have higher current capacity and better electron emissivity (lower work function). Answer: A and B

GATE 1999 · Q50

50

In a casting cold shut defect is caused by

MSQ2M

A

Low pouring temperature

B

Melt-mould reactions

C

Faulty gating system

D

Very low mould permeability

Solution

Cold shuts occur when two metal streams meet but don’t fuse, caused by low pouring temperature and faulty gating. Answer: A and C

GATE 1998 · Q4

4

The rolling process cannot be used to produce

MCQ1M

A

plates

B

rods

C

tubes

D

wires

Solution

Tubes are typically produced by extrusion or seamless tube-making processes, not by conventional rolling. Answer: C

GATE 1998 · Q10

10

Powder metallurgy is used to produce

MCQ1M

A

high precision components with complex cavities and sharp features

B

components of large size

C

porosity free components

D

components of such alloys whose constituents do not form alloys readily

Solution

Powder metallurgy is ideal for producing components from alloys whose constituents are immiscible or don’t alloy readily (e.g., W-Cu, Cu-graphite). Answer: D

GATE 1998 · Q22

22

The swift cup test evaluates the following property of a sheet metal

MCQ1M

A

stretchability

B

drawability

C

bendability

D

none of these

Solution

The Swift cup test measures drawability of sheet metal by determining the limiting drawing ratio (LDR). Answer: B

GATE 1998 · Q29

29

If two parts of the mould are not joined properly, the casting may have the following defect

MCQ1M

A

scab

B

blowhole

C

gas porosity

D

flash

Solution

When mould halves are not joined properly, molten metal seeps through the parting line gap producing flash. Answer: D

GATE 1998 · Q36

36

Thermit welding uses the following energy source:

MCQ1M

A

electrical energy

B

chemical energy

C

energy of high velocity electrons

D

heat generated by friction

Solution

Thermit welding uses the exothermic chemical reaction between aluminium powder and iron oxide to generate heat. Answer: B

GATE 1998 · Q37

37

Most important property of steels for use in automobile bodies is

MCQ1M

A

formability

B

yield strength

C

toughness

D

resilience

Solution

Automobile body panels require extensive sheet metal forming, so formability is the most important property. Answer: A

GATE 1997 · Q39

39

An aluminium rod is joined to a mild steel rod of similar diameter by

MCQ1M

A

explosive welding

B

electric resistance welding

C

friction welding

D

flash butt welding

Solution

Flash butt welding is suitable for joining dissimilar metals like aluminium and mild steel rods of similar diameter end-to-end. Answer: D

GATE 1997 · Q45

45

Maximum possible reduction per pass during wire drawing is

MCQ1M

A

33%

B

63%

C

82%

D

unlimited

Solution

The maximum theoretical reduction per pass in wire drawing (for an ideal material with no friction) is 63% (1 − 1/e ≈ 0.632). Answer: B

GATE 1996 · Q13

13

Preheating prior to welding is done for the purpose of

MCQ1M

A

decreasing cooling rate

B

facilitating fusion of high melting metals

C

preventing hot cracking

D

ensuring full penetration

Solution

Preheating reduces the temperature gradient and cooling rate, minimizing residual stress and the risk of hydrogen-induced cracking in the HAZ. Answer: A

GATE 1996 · Q22

22

In an arc welding experiment the current, voltage and electrode travel speed were 140 A, 22 V and 15 cm/min respectively. The heat input per unit length of the weld is approximately

MCQ2M

A

2.464 kJ/cm

B

4.62 kJ/cm

C

9.24 kJ/cm

D

12.32 kJ/cm

Solution

Heat input = V × I / speed = 22 × 140 / (15/60) = 3080 / 0.25 = 12320 J/cm ≈ 12.32 kJ/cm. Answer: D

GATE 1996 · Q23

23

The coefficient of friction increases by 5 times when a slab is hot rolled instead of cold rolling in a specific rolling mill. The maximum reduction in the thickness of the slab will increase by a factor of

MCQ2M

A

25

B

5

C

0.04

D

0.2

Solution

Maximum draft Δh_max = μ²R; if μ increases 5×, Δh_max increases 25×. However the answer key gives B (5), suggesting direct proportionality is assumed. Answer: B

GATE 1995 · Q10

10

Riser in a casting compensates for

MCQ1M

A

liquid state shrinkage

B

solidification shrinkage

C

liquid state shrinkage, solidification shrinkage and solid state shrinkage

D

liquid state shrinkage and solidification shrinkage

Solution

A riser feeds liquid metal to compensate for liquid-state and solidification shrinkage; solid-state shrinkage occurs after the metal has solidified and cannot be fed. Answer: D

GATE 1995 · Q15

15

Stainless steel is welded using

MCQ1M

A

oxy-acetylene flame

B

oxy-hydrogen flame

C

arc welding

D

inert gas arc welding

Solution

Stainless steel requires inert gas shielding (TIG/MIG) during welding to prevent oxidation and preserve corrosion resistance. Answer: D

GATE 1995 · Q32

32

True or False: Friction is essential to rolling of metals.

MCQ1M

A

True

B

False

Solution

Friction between the rolls and the workpiece is necessary to draw the metal into the roll gap; without it rolling cannot proceed. Answer: True

GATE 1994 · Q10

10

A weldment consists of

MCQ1M

A

Fused part of the weld

B

Fused part and the heat affected zone (HAZ)

C

Fused part + HAZ + base metal

D

HAZ

Solution

As per the answer key, the answer is D (HAZ); however, a weldment typically includes the fused zone, HAZ, and base metal. Answer: D

GATE 1994 · Q12

12

The typical dislocation density (lines/cm²) of a hot rolled material is

MCQ1M

A

10²

B

10¹²

C

10⁶

D

10⁸

Solution

Hot rolled metals typically have a dislocation density of ~10¹⁰–10¹² lines/cm² due to work hardening during rolling. Answer: B

GATE 1994 · Q14

14

As the % reduction increases, the flow stress during hot isothermal forging of a metal

MCQ1M

A

increases linearly

B

decreases exponentially

C

decreases linearly

D

remains almost constant

Solution

During hot isothermal forging, increasing reduction raises strain, and flow stress increases approximately linearly with strain. Answer: A

GATE 1993 · Q38

38

Lamellar tearing in weldments occurs

MCQ2M

A

parallel to the cementite lamellae

B

parallel to the plane of non-metallic inclusions

C

parallel to the heat affected zone

D

across the surface of cross fillet welds

Solution

Lamellar tearing occurs parallel to the rolling plane along elongated non-metallic inclusions (MnS) under through-thickness stress. Answer: B

GATE 1993 · Q49

49

True or False: Headstock in a lathe is used for controlling the roll separating force.

MCQ2M

A

True

B

False

Solution

A headstock is a lathe component housing the spindle and drive mechanism; roll separating force is a rolling mill concept, not related to lathes. Answer: False

GATE 1993 · Q50

50

True or False: Weld decay in stainless steel occurs due to chromium carbide precipitation at grain boundaries.

MCQ2M

A

True

B

False

Solution

Weld decay (sensitisation) in austenitic stainless steel results from Cr₂₃C₆ precipitation at grain boundaries, depleting adjacent zones of Cr below the passivation threshold. Answer: True

GATE 1993 · Q75

75

Zn is removed from molten lead by allowing droplets of impure lead to pass through a molten salt mixture containing 10³ g moles/m³ of PbCl₂ according to the reaction:
{Zn} + (PbCl₂) = [Pb] + (ZnCl₂)
Calculate the rate of removal of Zn from molten lead droplets assuming that the rate controlling step is mass transfer of PbCl₂ from the bulk of the salt mixture to the molten lead/salt interface.
Data: Average diameter of lead droplet = 2 × 10⁻³ m. Mass transfer coefficient of PbCl₂ in salt mixture = 2.5 × 10⁻⁴ m/sec.

SUB5M

Solution

Rate of Zn removal = mass transfer coefficient × surface area × bulk concentration of PbCl₂. Surface area of droplet = πd² = π(2 × 10⁻³)². Rate = k_m × C_bulk × A = 2.5 × 10⁻⁴ × 10³ × π(2 × 10⁻³)² g-moles/sec per droplet.

GATE 1992 · Q10

10

True or False: During hot rolling dislocation density increases.

T/F2M

A

True

B

False

Solution

During hot rolling, dynamic recovery and recrystallisation occur simultaneously with deformation, so the dislocation density does not accumulate — it remains relatively low compared to cold rolling. Answer: False

GATE 1992 · Q25

25

Eutectic Al–Si castings are modified by

MSQ2M

A

Na

B

P

C

S

D

None

Solution

Sodium (Na) is used for modification of eutectic Al–Si alloys (also Sr), converting coarse plate-like Si to a fine fibrous morphology. Phosphorus is used for refinement of primary Si in hypereutectic alloys. Answer: A

GATE 1992 · Q27

27

For preparation of porous bearings by powder metallurgy preferred particle shape is

MSQ2M

A

Spherical

B

Nodular

C

Irregular

D

No preferred shape

Solution

Spherical and nodular particles pack with controlled, interconnected porosity ideal for self-lubricating porous bearings. Irregular particles interlock too tightly, reducing porosity. Answer: A, B

GATE 1992 · Q43

43

Modulus of feeder is ______ than the modulus of the casting.

FIB1M

Solution

Answer: higher
The feeder (riser) must solidify after the casting to supply liquid metal. A higher modulus (V/SA ratio) ensures slower solidification (Chvorinov’s rule).

GATE 1992 · Q44

44

Heat diffusivity of the mould influences ______ of the mould.

FIB1M

Solution

Answer: cooling rate (chilling power)
Heat diffusivity (α = k/ρc_p) determines how quickly the mould absorbs and conducts heat away, controlling the solidification/cooling rate.

GATE 1992 · Q56

56

What is autogenous welding?

SUB3M

Solution

Autogenous welding is fusion welding without the addition of filler metal. The weld joint is formed by melting and fusing the base metal of the two pieces being joined, without any external filler material. Examples include certain TIG welding operations on thin sheets.

GATE 1992 · Q57

57

Why do gray iron castings need very little feeding?

SUB3M

Solution

During solidification of gray iron, carbon precipitates out in the form of graphite which has lower density than iron and therefore higher specific volume. This partly compensates for the volumetric shrinkage of iron that occurs during solidification, resulting in very little net shrinkage and hence minimal feeding requirement.

GATE 1992 · Q60

60

Why are aluminium castings not prone to reaction unsoundness (due to reaction between two solvents)?

SUB3M

Solution

Reaction unsoundness (gas porosity due to reaction between dissolved gases) requires the reaction product to be a gas that is insoluble in the metal. In copper castings, dissolved oxygen and hydrogen can react to form steam (H₂O) causing porosity. In aluminium, the strong affinity of Al for oxygen means dissolved oxygen forms Al₂O₃ (a solid), not a gaseous product. Hence, there is no reaction between dissolved gases to form gas bubbles, and aluminium castings are not prone to reaction unsoundness.

GATE 1992 · Q64

64

What are the basic units of a continuous casting facility?

SUB4M

Solution

The basic units of a continuous casting facility are:
• Tundish — for guiding and controlling the metal flow
• Water-cooled mould — primary cooling to form a solidified shell
• Water spraying arrangement on moulds — secondary cooling
• Withdrawal rollers — to pull the solidifying strand
• Bending roller — for changing the direction (vertical to horizontal)
• Shearing arrangement — for cutting billets/slabs to length

GATE 1992 · Q76

76

(a) For 800 mm diameter roll, initial height of 200 mm, coefficient of friction is 0.4; determine the angle of bite.

(b) Given mean flow stress of 20 MPa, diameter of extrusion chamber of 150 mm and diameter of extruded rod of 15 mm; determine the thrust required for extrusion. Neglect redundant work and coefficient of friction.

SUB6M

Solution

(a) D = 800 mm ⇒ R = 400 mm, h₀ = 200 mm, f = 0.4.

For rolling, the condition for bite is: f ≥ √(Δh/R), i.e. Δh ≤ f² × R.
Maximum Δh = (0.4)² × 400 = 0.16 × 400 = 64 mm.

Angle of bite α = √(Δh/R) = √(64/400) = √0.16 = 0.4 rad ≈ 22.9°
Also, tan α ≈ α = f = 0.4 rad for the maximum bite condition.

(b) Extrusion ratio R_e = (D₀/D_f)² = (150/15)² = 100
Extrusion pressure = σ₀ ln(R_e) = 20 × ln(100) = 20 × 4.605 = 92.1 MPa
Thrust = Pressure × Area = 92.1 × (π/4)(150)² = 92.1 × 17671 ≈ 1.628 MN

GATE 1991 · Q8

8

Earing is a defect found in steels after the following metal working operation(s):
(A) Deep drawing (B) Rolling
(C) Extrusion (D) Wire drawing

MSQ2M

A

B

C

D

Solution

Earing is a defect caused by planar anisotropy (crystallographic texture) in sheet metal, observed specifically during deep drawing where uneven cup heights form. Answer: A

GATE 1991 · Q13

13

In extrusion of metals, which of the following statement(s) is [are] true:
(A) speed of the extruded material is same as that of ram speed
(B) redundant work is a function of die angle
(C) relative motion between the billet and the container wall is always present
(D) hollow ram is used for indirect extrusion.

MSQ2M

A

B

C

D

Solution

Redundant work depends on die angle (larger angles increase redundant deformation). In direct extrusion, there is relative motion between billet and container wall. The extruded material speed differs from ram speed due to area reduction. In indirect extrusion, a hollow ram carries the die. Answer: B, C

GATE 1991 · Q14

14

Powder metallurgy processing can be used for making:
(A) Dispersion strengthened copper rod (B) Self lubricating bearing
(C) Connecting rod (D) Cemented carbide.

MSQ2M

A

B

C

D

Solution

Self-lubricating (porous) bearings and cemented carbides (WC-Co) are classic powder metallurgy products. These cannot be made economically by conventional casting/forming routes. Answer: B, D

GATE 1991 · Q25

25

True or False: The serious effect of fibering upon mechanical properties of rolled products is its tendency to produce poor ductility transverse to the direction of fibers.

T/F2M

A

True

B

False

Solution

The fibers give anisotropy of properties. Fibering in rolled products leads to directionality — ductility and toughness are significantly lower in the transverse (short transverse) direction compared to the rolling direction. Answer: True

GATE 1991 · Q27

27

True or False: It is easy to cast an alloy having long freezing range.

T/F2M

A

True

B

False

Solution

Alloys with long freezing range possess poor fluidity, and they are more difficult to feed. Such alloys form a mushy zone during solidification, leading to dispersed porosity and hot tearing. Answer: False

GATE 1991 · Q29

29

True or False: In wire drawing the applied stress on the wire is below its flow stress.

T/F2M

A

True

B

False

Solution

In wire drawing, the drawing stress (pull stress) must be below the yield/flow stress of the drawn wire, otherwise the wire would deform plastically outside the die and neck or break. The actual deformation occurs inside the die where the combined stress state (pull + die pressure) causes plastic flow. Answer: True

GATE 1991 · Q40

40

The application of ______ tension lowers the load during cold rolling.

FIB1M

Solution

Answer: Front & back (applying front and/or back tension to the strip reduces the rolling load by superimposing a tensile stress component)

GATE 1991 · Q43

43

Define: MII modulus

SUB2M

Solution

MII modulus (also called the modulus of resilience or Mill modulus) is the ratio of volume to surface area of a casting, used in Chvorinov’s rule to estimate solidification time: t = B(V/A)², where B is a mould constant. A higher modulus indicates slower solidification.

GATE 1991 · Q50

50

Distinguish between: Conventional forging and powder forging.

SUB4M

Solution

Conventional forging: Wrought stock (cast ingot or billet) is heated and plastically deformed between dies. Material has continuous grain flow. Can produce parts of virtually any size. High material waste (flash). Full density starting material.
Powder forging: A sintered powder metallurgy preform is heated and forged in a single blow in a closed die. Near-net shape with minimal flash and material waste. Achieves full density from porous preform. Lower energy consumption per part. Limited to smaller parts. Eliminates machining in many cases.

GATE 1991 · Q51

51

Explain Matano interface with the help of a diagram. Also state its utility.

SUB4M

Solution

The Matano interface is a reference plane in a diffusion couple, defined such that equal amounts of material have crossed it from both sides (area balance). It is found by drawing the concentration–distance profile after diffusion and locating the plane where the areas on either side of the curve are equal. GATE 1991 Q51 figure

Utility: The Matano–Boltzmann analysis uses this interface as the origin (x = 0) to calculate the interdiffusion coefficient as a function of concentration from a single diffusion couple experiment.

GATE 1991 · Q57

57

Describe the salient features of conventional methods used for welding steel rails.

SUB4M

Solution

Conventional methods for welding steel rails include: (1) Thermit welding: Most common field method. An exothermic reaction between aluminium powder and iron oxide produces molten steel that fills the gap between rail ends. Self-contained, no external power needed. (2) Flash butt welding: Factory method using resistance heating. Rail ends are brought together under pressure while electric current causes flashing and forging. Produces consistent high-quality joints. (3) Gas pressure welding: Rail ends heated by oxy-acetylene flames and forged together under axial pressure. Less common but used in some railways.

GATE 1991 · Q65

65

Explain the mechanism of crack initiation and growth when a metal is subjected to cyclic stress.

SUB4M

Solution

Crack initiation: Under cyclic loading, persistent slip bands (PSBs) form on the surface due to irreversible to-and-fro slip. These create intrusions and extrusions (surface roughening) that act as stress concentrators. Microcracks nucleate at these surface irregularities, typically at ~45° to the tensile axis (Stage I, shear mode). Crack growth: Once the microcrack reaches a critical size, it changes direction to propagate perpendicular to the maximum tensile stress (Stage II, tensile mode). Growth occurs by repeated blunting and resharpening of the crack tip during each loading cycle, producing characteristic fatigue striations on the fracture surface. The crack grows incrementally each cycle until the remaining cross-section can no longer support the load, leading to final fast fracture.

GATE 1991 · Q69

69

Calculate the approximate load necessary to cold roll an annealed aluminium sheet from 3 mm to 2.5 mm by cold rolling with no spread. The material follows the following uniaxial flow curve relationship: σ = 160 ε^0.25 N/mm². The radius of the roll is 200 mm. Assume 20% margin for friction.

SUB8M

Solution

Reduction: Δh = 3 − 2.5 = 0.5 mm. True strain: ε = ln(3/2.5) = ln(1.2) = 0.1823.
Mean flow stress: σ̄ = Kεⁿ/(n+1) × (n+1) → Using average: σ̄ = 160 × (0.1823)^0.25 = 160 × 0.6534 = 104.5 N/mm². For plane strain: σ̄′ = σ̄ × 2/√3 = 104.5 × 1.155 = 120.7 N/mm².
Contact length: L = √(R × Δh) = √(200 × 0.5) = √100 = 10 mm.
Rolling load per unit width: P = σ̄′ × L = 120.7 × 10 = 1207 N/mm.
With 20% friction margin: P = 1207 × 1.2 = 1448 N/mm width.
For a 1000 mm wide sheet: total load ≈ 1448 kN ≈ 1.45 MN.

GATE 1990 · Q12

12

True or False: Phosphor-copper deoxidizer can be used for complete deoxidation of copper melt for high conductivity copper castings.

T/F2M

A

True

B

False

Solution

While phosphor-copper is an effective deoxidizer, residual phosphorus severely reduces the electrical conductivity of copper. For high-conductivity copper castings, deoxidation must be done with agents that do not leave harmful residuals (e.g. lithium or boron). Answer: False

GATE 1990 · Q15

15

True or False: Work input in a ball mill increases linearly with speed.

T/F2M

A

True

B

False

Solution

Work input in a ball mill does not increase linearly with speed. It increases up to a certain speed, but beyond the critical speed the balls centrifuge against the shell and grinding ceases. The relationship is non-linear. Answer: False

GATE 1990 · Q26

26

Alligatoring occurs in ______.

FIB1M

Solution

Answer: rolling (Alligatoring is a defect in rolling where the slab splits along a horizontal plane)

GATE 1990 · Q48

48

Self-lubricating bearings are produced by ______ technique.

FIB1M

Solution

Answer: powder metallurgy

GATE 1990 · Q60

60

What are the factors responsible for the occurrence of “hot tear” defects in castings?

SUB4M

Solution

Hot tears (hot cracks) in castings are caused by: (1) Thermal contraction stresses — the casting contracts during solidification but is restrained by the mould, generating tensile stresses; (2) Wide freezing range of the alloy, producing a mushy zone with poor feeding; (3) Poor mould design — sharp corners, abrupt section changes, and inadequate fillets concentrate stress; (4) Low hot strength of the alloy near the solidus temperature where liquid films at grain boundaries cannot resist tensile stress; (5) Improper gating and risering leading to inadequate liquid metal feeding during the last stages of solidification.

GATE 1990 · Q61

61

Why should flaskless forgings be preferred to castings in case of production of small near-net-shape components?

SUB4M

Solution

Flaskless forgings are preferred over castings for small near-net-shape components because: (1) Forgings have superior mechanical properties — the wrought grain flow follows the component contour, giving higher strength, ductility, and fatigue resistance; (2) No casting defects such as porosity, shrinkage cavities, or inclusions; (3) Finer grain structure from dynamic recrystallization during forging; (4) Better dimensional accuracy and surface finish, reducing machining allowance; (5) Higher production rates for small components with less material waste.

GATE 1990 · Q73

73

Give the flow sheet of a process used for the production of iron powder based on hydrogen reduction of mill scale.

SUB8M

Solution

The Höganäs-type process for iron powder production from mill scale:
1. Collection of mill scale: Iron oxide scale (Fe₃O₄/Fe₂O₃) from hot rolling mills is collected.
2. Crushing and grinding: Mill scale is crushed and ground to desired particle size.
3. Magnetic separation: Removal of non-magnetic impurities.
4. Reduction: Ground mill scale is reduced in a continuous belt furnace at 900–1100°C in a hydrogen atmosphere: Fe₃O₄ + 4H₂ → 3Fe + 4H₂O. The sponge iron cake is formed.
5. Crushing of sponge cake: The reduced cake is crushed and ground.
6. Annealing: Final anneal in hydrogen at ~800°C to remove residual carbon and oxygen, and to soften the powder.
7. Screening and classification: Powder is screened to desired size fractions.
8. Blending: Different batches are blended for uniformity.