UNIT 1.

1. Lead is a metallic crystal having a _______ structure.

a) FCC
b) BCC
c) HCP
d) TCP

Answer: a
Explanation: Crystalline solids are classified as either metallic or non-metallic. Pb, along with
Cu, Ag, Al, and Ni, has a face-centered cubic structure.
2. Which of the following has a HCP crystal structure?
a) W
b) Mo
c) Cr
d) Zr

Answer: d
Explanation: Crystalline solids are classified as either metallic or non-metallic. W, Mo, and Cr
are examples of the body-centered cubic structure of crystals. The HCP structure is found in Mg,
Zn, Ti, Cd, Zr, and others.
3. Amorphous solids have _______ structure.
a) Regular
b) Linear
c) Irregular
d) Dendritic
.
Answer: c
Explanation: Materials in which the molecule is the basic structural solid and has an irregular
structure is known as amorphous solid. Crystalline solids, on the other hand, usually are arranged
in a regular manner.
4. At ________ iron changes its BCC structure to FCC.
a) 308oC
b) 568oC
c) 771oC
d) 906oC
.
Answer: d
Explanation: Similar to metallic crystals, a few non-metallic crystals also change form due to
temperature and pressure differences. This process is termed as polymorphism. Iron changes
from BCC at room temperature to FCC form at 906oC.
5. At room temperature, tin is formed into _________
a) Gray tin
b) White tin
c) Red tin
d) Yellow tin
.
Answer: b
Explanation: Similar to metallic crystals, a few non-metallic crystals also change form due to
temperature and pressure differences. Tin crystallizes in a non-metallic diamond structure (gray
tin) at low temperatures. At room temperature, it forms a metallic structure (white tin).
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6. Which of the following is a property of non-metallic crystals?

a) Highly ductile
b) Less brittle
c) Low electrical conductivity
d) FCC structure
.
Answer: c
Explanation: Non-metallic crystals are less ductile and have low electrical conductivity. On the
other hand, metallic crystals are differing since they are more ductile and have high electrical
conductivity.
7. Which of the following is not an amorphous material?
a) Glass
b) Plastics
c) Lead
d) Rubbers
.
Answer: c
Explanation: Materials in which the molecule is the basic structural solid and has an irregular
structure are known as amorphous solid. Most amorphous materials are polymers such as plastics
and rubbers. The most common amorphous material is glass.
8. The crystal lattice has a _________ arrangement.
a) One-dimensional
b) Two-dimensional
c) Three-dimensional
d) Four-dimensional
.
Answer: c
Explanation: Lattice is defined as the regular geometrical arrangement of points in crystal space.
Space or crystal lattice is a three-dimensional network of imaginary lines connecting the atoms.
9. The smallest portion of the lattice is known as __________
a) Lattice structure
b) Lattice point
c) Bravais crystal
d) Unit cell
.
Answer: d
Explanation: Lattice is defined as the regular geometrical arrangement of points in crystal space.
The unit cell is the smallest portion of the lattice, which when repeated in all directions gives rise
to a lattice structure.
10. Bravais lattice consists of __________ space lattices.
a) Eleven
b) Twelve
c) Thirteen
d) Fourteen
.
Answer: d
Explanation: There are fourteen ways in which points can be arranged in a space so that each has
identical surroundings. These fourteen space lattices constitute the Bravais space lattices.
11. A unit cell that contains lattice points only at the corners is known as _________
a) Primitive unit cell
b) Secondary unit cell
c) Layered unit cell
d) Derived unit cell
.
Answer: a
Explanation: If a unit cell chosen contains lattice points only at its corners, it is called a primitive
or simple unit cell. It contains only one lattice point since each point at the eight corners is
shared equally with adjacent unit cells.
12. The axial relationship of a monoclinic crystal system is given as ___________
a) a = b = c
b) a = b ≠ c
c) a ≠ b = c
d) a ≠ b ≠ c
.
Answer: d
Explanation: The crystal system is a format by which crystal structures are classified. Each
crystal system is defined by the relationship between edge lengths a, b, and c. For monoclinic,
orthorhombic, and triclinic systems, the axial relationship is given by a ≠ b ≠ c.
13. The axial relationship of a rhombohedral crystal system is given as ___________
a) a = b = c
b) a = b ≠ c
c) a ≠ b = c
d) a ≠ b ≠ c
.
Answer: a
Explanation: The crystal system is a format by which crystal structures are classified. Each
crystal system is defined by the relationship between edge lengths a, b, and c. For cubic and
rhombohedral systems, the axial relationship is given by a = b = c.
14. The interracial angles of a hexagonal crystal system are given by __________
a) α = β = ϒ = 90o
b) α = β = 90o ϒ = 120o
c) α = β = ϒ ≠ 90o
d) α ≠ β ≠ ϒ ≠ 90o
.
Answer: b
Explanation: The crystal system is a system by which crystal structures are classified. Each
crystal system is defined by the relationship between edge lengths a, b, and c and interaxial
angles α, β, and ϒ. For hexagonal system, the interaxial angles are given by α = β = 90o ϒ =
120o.
15. The interracial angles of a triclinic crystal system are given by __________
a) α = β = ϒ = 90o
b) α = β = 90o ϒ = 120o
c) α = β = ϒ ≠ 90o
d) α ≠ β ≠ ϒ ≠ 90o
.
Answer: d
Explanation: The crystal system is a system by which crystal structures are classified. Each
crystal system is defined by the relationship between edge lengths a, b, and c and interaxial
angles α, β, and ϒ. For the triclinic system, the interaxial angles are given by α ≠ β ≠ ϒ ≠ 90o.
16. What is the atomic radius of a BCC crystal structure?
a) a/2
b) a/4
c) a√2/4
d) a√3/4
.
Answer: d
Explanation: Atomic radius is defined as half the distance between the centers of two
neighboring atoms. The atomic radius of a simple cube and HCP is a/2 respectively, whereas it is
a√2/4 and a√3/4 for FCC and BCC respectively.
17. What is the coordination number of a simple cubic structure?
a) 6
b) 8
c) 10
d) 12
.
Answer: a
Explanation: Coordination number is defined as the number of nearest neighboring atoms in
crystals. The coordination number for the simple cubic structure is 6, whereas it is 8 and 12 for
BCC and FCC respectively.
18. What is the atomic packing factor of BCC structure?
a) 0.54
b) 0.68
c) 0.74
d) 0.96
.
Answer: b
Explanation: The density of packing in a crystal is determined using the atomic packing factor
(APF). The APF of FCC and HCP structures is 0.74, and 0.54 for simple cubic structure,
whereas it is 0.68 for BCC structure.
1. The tendency of a deformed solid to regain its actual proportions instantly upon unloading
known as ______________
a) Perfectly elastic
b) Delayed elasticity
c) Inelastic effect
d) Plasticity
.
Answer: a
Explanation: If the recovery of the solid is instantaneous and complete, it is known as perfectly
elastic. Delayed elasticity is defined as the steady mode of solid recuperation, while the inelastic
effect is when the recuperation of deformed solid is partial. Plasticity is the contradictory
occurrence of elasticity.
2. How is Young’s modulus of elasticity defined?

a) 

b) 

c)

d) 
.
Answer: c

Explanation: The ratio of load to the area   is the description of stress on the solid. The ratio

of alteration in length to its original length   is the description of strain on the solid.
Young’s modulus of elasticity is defined as the ratio of stress on the solid to the strain of the

solid   The ratio of the mass of the solid to its volume  is defined as
density.
3. The permanent mode of deformation of a material known as _____________
a) Elasticity
b) Plasticity
c) Slip deformation
d) Twinning deformation
.
Answer: b
Explanation: Plasticity is defined as the property of a material due to which it is permanently
deformed due to loading. Elasticity is the temporary form of deformation. Twinning and Slip are
mechanisms of Plastic deformation.
4. The ability of materials to develop a characteristic behavior under repeated loading known as
___________
a) Toughness
b) Resilience
c) Hardness
d) Fatigue
.
Answer: d
Explanation: Toughness is the ability of a material to absorb energy during deformation, while
resilience is its capacity to absorb the energy. Hardness is the knack of material to defy
indentation. The ability of a material to develop a characteristic behavior under repeated loading
is known as fatigue.
5. What is the unit of tensile strength of a material?

a) 
b) kg/cm2

c) 
d) cm2/kg
.
Answer: b
Explanation: Tensile strength is defined as the ratio of maximum load (kg) applied to its cross-
sectional area (cm2). Young’s modulus of elasticity is defined as the ratio of stress on the solid to
the strain of the solid. The remaining choices given are reciprocals of same.
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6. Which of the following factors affect the mechanical properties of a material under applied
loads?
a) Content of alloys
b) Grain size
c) Imperfection and defects
d) Shape of material
.
Answer: d
Explanation: Contents of alloys improve or decrease the hardness and strength of materials. Finer
grain sizes improve the strength of the material. Imperfection and defects reduce the strength of
the material. Shape, however, has little or no effect on the material.
7. The ability of a material to resist plastic deformation known as _____________
a) Tensile strength
b) Yield strength
c) Modulus of elasticity
d) Impact strength
.
Answer: b
Explanation: The point of stretching where it increases suddenly is known as yield strength, i.e.
the region where the stretch is elastic. Tensile strength is the force needed to fracture the
material. Impact strength is the capacity of a material to resist shock energy before a fracture.
8. The ability of a material to be formed by hammering or rolling is known as _________
a) Malleability
b) Ductility
c) Harness
d) Brittleness
.
Answer: a
Explanation: The capacity of a material to withstand deformation under compression, i.e.
hammering is known as malleability. The capacity of a material to withstand deformation under
tension, i.e. wire drawing is known as ductility. Hardness is the knack of a material to oppose
indentation. Brittleness is the tendency to fracture without deformation of the material.
9. What type of wear occurs due to an interaction of surfaces due to adhesion of the metals?
a) Adhesive wear
b) Abrasive wear
c) Fretting wear
d) Erosive wear
.
Answer: a
Explanation: Adhesive wear occurs due to frictional contact between metals. Abrasive wear
occurs due to the descending of the surface of the material over a different harder surface.
Fretting wear is the wear which causes the deduction of material from both the surfaces in
contact over an extended period of time. Erosive wear causes wear of the material due to the
effect of solid or liquid particles over a short period of time.
10. Deformation that occurs due to stress over a period of time is known as ____________
a) Wear resistance
b) Fatigue
c) Creep
d) Fracture
.
Answer: c
Explanation: Creep is the time-dependent deformation of the material under stress. Wear
resistance, fatigue, and fracture deal with deformation under stress without a time factor, i.e. they
are time-independent.

1. What is the attribute of a material which resists the flow of electricity known?
a) Conductivity
b) Thermoelectricity
c) Dielectric strength
d) Resistivity
.
Answer: d
Explanation: Resistivity is that attribute of a material which resists the flow of electricity, while
conductivity is it’s reciprocal (or) opposite. The process by which two dissimilar metals join and
produce a voltage is known as thermoelectricity. The dielectric strength of a material is its
insulating capacity against voltage.
2. How is conductivity of a material defined?

a) 

b) 

c) 

d) 
.
Answer: c

Explanation: Electrical conductivity is defined by   whereas its reciprocal defines the

electrical resistivity given by   or the ratio of magnitudes of an electrical field to the

electrical density  . Permeability is a magnetic property which is defined by 

3. What is the electrical conductivity of Aluminum?
a) 6.3 * 107
b) 5.9 * 107
c) 3.5 * 107
d) 1 * 107
.
Answer: c
Explanation: The electrical conductivity of Aluminum is 3.5 * 107. The electrical conductivity of
Silver, Copper, and Iron is 6.3 * 107, 5.9 * 107, and 1 * 107 in that order. The electrical
conductivity is generally measured at 20oC.
4. What is the electrical resistivity of Copper?
a) 1.59 * 10-8
b) 1.68 * 10-8
c) 2.65 * 10-8
d) 5.9 * 10-8
.
Answer: b
Explanation: The electrical resistivity of Copper is 1.68 * 10-8. The electrical resistivity of Silver,
Aluminum, and Zinc is 1.59 * 10-8, 2.65 * 10-8, and 5.9 * 10-8 in that order. The electrical
resistivity is generally measured at 20oC.
5. Measured using an Electrical Conductivity meter, what is the order of resistivity of
superconducting materials?
a) 10-8
b) 1016
c) ∞
d) 0
.
Answer: d
Explanation: The resistivity of superconducting materials is 0 since they conduct all electricity
and resist none of it. Metals have the resistivity of the order of 10-8 while that of insulators is of
the order of 1016. Since maximum electricity is refused by super-insulators, their resistivity is ∞.
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6. Which of the following represents the energy band diagram of a semiconducting material?

a) 

b) 

c) 

d) 
.
Answer: b
Explanation: The diagram with the short energy gap represents the energy band diagram of
semiconductor, which in this case is silicon. The diagram with multiple valence bands denotes
that of a conductor, which in this case is sodium. The diagram with a wide energy gap represents
an insulator like a diamond. The diagram having no energy gap is not an energy band diagram.
7. Which of the following processes is not an application of thermoelectric effect?
a) Seebeck effect
b) Peltier effect
c) Thomson effect
d) Ettingshausen effect
.
Answer: d
Explanation: Seebeck, Peltier, and Thomson are three distinct effects which make up the
thermoelectric effect. Ettingshausen effect, however, is a separate thermoelectric phenomenon
like the Nernst effect.
8. The insulating capacity of material against high voltages is known as _______
a) Dielectric strength
b) Thermoelectricity
c) Electromechanical effect
d) Electrochemical effect
.
Answer: a
Explanation: The dielectric strength of a material is its insulating capacity against a high voltage.
The process by which two dissimilar metals join and produce a voltage is known as
thermoelectricity. The electrical effect in a material is the relation between electrical energy and
chemical change (as in batteries) and that of electromechanical effect is in an electrically
operated mechanical device (as radar).
9. What is the nature of the coefficient of resistance of an insulator?
a) Positive
b) Negative
c) Zero
d) Infinite
.
Answer: d
Explanation: The coefficient of resistance of an insulator and a semiconductor material is
generally negative. It is usually positive for pure metals. Zero coefficient values can be obtained
by alloying with a specific type of metals.
10. What is the dielectric strength of mica?
a) 118 MV/m
b) 2000 MV/m
c) 3 MV/m
d) 1012 MV/m
.
Answer: a
Explanation: The dielectric strength of mica is 118 MV/m. It is generally measured at 20oC,
where 1 MV/m equals 10 V/m. The dielectric strengths of diamond, air, and vacuum are 2000,
and 10 in MV/m in that order.

1. Which of the following represents the slip mode of plastic deformation?

a) 
b) 

c) 

d) 
.
Answer: b
Explanation: The first image shows an ordinary crystal lattice. The second image shows the
mechanism of slip where a part of the crystal slides over another. The third image shows the
twinning mode of plastic deformation in which the atoms rearrange themselves to form a mirror
image. The last image shows a defect.
2. What is the SI unit for the stress of a material?
a) N
b) N m
c) N m-1
d) N m-2
.
Answer: d
Explanation: Stress is defined as the internal forces acting in a material. The various types of
stresses are shear, tensile, yield, and ultimate tensile strength. All these stresses are defined as
Newton/meganewton per square meter.
3. The process of increasing strength of a material by changing grain size known as __________
a) Grain boundary strengthening
b) Work hardening
c) Solid solution hardening
d) Precipitation hardening
.
Answer: a
Explanation: Grain-boundary strengthening is the technique of increasing the strength of
materials by changing their average crystallite size. This is otherwise known as Hall-Petch
method. The Hall-Petch constant is written in the form of MN m-3/2.
4. What is the value of Boltzmann’s constant?
a) 8.314
b) 1.38 * 10-23
c) 1.38
d) 8.314 * 10-23
.
Answer: b
Explanation: Boltzmann’s constant is associated with the kinetic energy of gas particles. It is
denoted as gas constant divided by Avogadro’s number. The value of Boltzmann’s constant is
1.38 * 10-23 J K-1 and that of the gas constant is 8.314 J mol-1 K-1.
5. At what value of the index of strain rate sensitivity does the material behave superplastically?
a) 0
b) 0.2
c) 0.4 – 0.9
d) 1.0 – 1.2
.
Answer: c
Explanation: The relationship between stress and strain rate can be expressed as σ = A ϵm. Here,
m represents the index of strain rate sensitivity. For m = 0.4 – 0.9, the material exhibits
superplastic behavior.
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6. What is the slip plane for an FCC crystal?
a) {110}
b) {111}
c) {112}
d) {123}
.
Answer: b
Explanation: A slip plane and a slip direction constitute a slip system of the material. FCC
crystal exhibits slip plane {111} in the slip direction <110>.
7. How many slip systems are present in a NaCl crystal?
a) 3
b) 6
c) 12
d) 18
.
Answer: b
Explanation: A slip plane and a slip direction constitute a slip system of the material. Both FCC
and BCC crystals have 12 slip systems, while HCP crystals have only 3 slip systems. A NaCl
crystal has 6 slip systems.
8. According to ________ five independent systems are required to maintain the reliability of
grain boundaries.
a) Hal-Petch principle
b) Mohr-Coulomb theory
c) Tresca criterion
d) Von Mises criterion
.
Answer: d
Explanation: In polycrystalline materials, a slip must be accommodated by a slip in adjacent
crystals so that the grain boundaries remain continuous. According to the Von Mises criterion, at
least five independent slip systems are required to maintain the integrity of grain boundaries
during plastic deformation.
9. What is the CRSS value of a copper crystal?
a) 0.5
b) 0.75
c) 5
d) 15
.
Answer: a
Explanation: CRSS is defined as the critical resolved shear stress of a material. For a copper
crystal, it has an FCC structure and a CRSS value of 0.5. The CRSS values for aluminum, nickel,
and iron and 0.75, 5, and 15 correspondingly.
10. What is the Hall-Petch constant for Iron?
a) 0.068
b) 0.11
c) 0.71
d) 1.77
.
Answer: c
Explanation: The grain-boundary strengthening method is otherwise known as Hall-Petch
method. The Hall-Petch constant is denoted as k and has a constant value of 0.71 for Iron. This
constant for aluminum, copper, and molybdenum is 0.068, 0.11, and 1.77 in that order.

. Creep occurs at a temperature above ________

a) 0.16 Tm
b) 0.22 Tm
c) 0.4 Tm
d) 0.91 Tm
.
Answer: c
Explanation: Creep is defined as the permanent deformation of a material due to the application
of steady load. It is appreciable only at temperatures above 0.4 Tm.
2. __________ experiences creep at room temperature.
a) Iron
b) Copper
c) Nickel
d) Lead
.
Answer: d
Explanation: Room temperature is about the same as that of the melting point of lead. Therefore,
nickel experiences creep at that temperature under its own load. The room temperature of iron
and copper is 0.16 Tm and 0.22 Tm respectively, at which creep does not occur.
3. _______ is known as steady-state creep.
a) Primary creep
b) Secondary creep
c) Tertiary creep
d) Quaternary creep
.
Answer: b
Explanation: Creep occurs in three stages known as primary, secondary, and tertiary creep.
Secondary creep is also known as steady-state creep since the rate of work and recoveries are
equal.
4. Which of the following does not affect creep?
a) Grain size
b) Thermal stability
c) Chemical reactions
d) Crystal structure
.
Answer: d
Explanation: Coarse-grained materials and those having higher thermal stability generally affect
creep resistance of a material. The chemical reaction involved and work hardening also influence
the creep resistance.
5. The time required for the occurrence of creep is known as _________
a) Creep resistance
b) Creep life
c) Creep limit
d) Creep recurrence
.
Answer: b
Explanation: Creep life is defined as the time required for the occurrence of creep fracture in a
material under static load. Creep limit is defined as the maximum static stress that will result in
creep. Creep resistance is known as the resistance offered by the material over creep.
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6. Which of the following will be rendered useless for prevention of creep?

a) Coarse-grained material
b) Strain hardening
c) Precipitation hardening
d) Quenching
.
Answer: d
Explanation: Creep fracture can be prevented by using coarse-grained materials and treatment by
strain hardening a precipitation hardening. The material must be free from residual stresses and
can also be heat treated.
7. MgO and Al2O3 can be used at high temperature creep resistance due to _________
a) High melting point
b) High oxygen concentration
c) Low density
d) Chemical stability
.
Answer: a
Explanation: Creep starts to occur at a temperature higher than 0.4 Tm. Therefore, high melting
point materials like MgO and Al2O3 are used for high-temperature operations.
8. Iron base alloys have melting point around __________
a) 900oC
b) 1500oC
c) 1900oC
d) 2400oC
.
Answer: b
Explanation: The most commonly used high-temperature alloys are iron base, nickel base, and
cobalt base alloys. Each of these alloys possesses a high melting point temperature of around
1500oC.
9. Strengthening of an iron base, nickel bade, and cobalt base alloys is done by __________
a) Precipitation hardening
b) Grain boundary hardening
c) Dispersion hardening
d) Transformation hardening
.
Answer: c
Explanation: Iron base, nickel base, and cobalt base alloys have creep resistance around a
temperature of 0.5 Tm. Their creep resistance can be improved by a heat treatment process
known as dispersion hardening.
10. Why can’t cold working be used for strengthening in case of plastic deformation?
a) Recrystallization
b) Thermal instability
c) Lowers tensile strength
d) Introduces coarsening
.
Answer: a
Explanation: At temperatures above 0.4 Tm, recrystallization occurs readily. This results in the
loss of strength of the cold-worked material. Therefore, for plastic deformation, cold working is
generally avoided for creep resistance.

1. Which of the following is a point defect in crystals?

a) Edge dislocation
b) Interstitialcies
c) Grain boundaries
d) Cracks
.
Answer: b
Explanation: Crystal defects are classified as point defects, line defects, and boundary defects.
Point defects include vacancies, impurities, interstitialcies, and electronic defects.
2. How can the number of defects be determined?

a) 

b) 

c) 

d) 
.
Answer: a
Explanation: The number of defects at equilibrium at a definite temperature can be determined
using the equation Ne(-Ed⁄kT). It is denoted by nd. Here, N stands for the total number of atomic
spots and Ed is the activation energy.
3. The defect that occurs due to a displacement of an ion is known as __________
a) Vacancy defect
b) Schottky defect
c) Frankel defect
d) Interstitial defect
.
Answer: c
Explanation: Frankel defect occurs due to a displacement of an ion from the crystal lattice. It is
related to the interstitial defect, where an ion simply occupies a position between regular atoms.
4. Which defect does the following figure depict?

a) Vacancy defect
b) Schottky defect
c) Frankel defect
d) Interstitial defect
.
Answer: b
Explanation: When a pair of positive and negative ions both disappear from a crystal lattice, the
effect is called a Schottky defect. It is closely related to vacancy defects where simply an ion is
missing.
5. Which of these is a Frankel defect?

a) 

b) 

c) 
d) 
.
Answer: a
Explanation: Frankel defect occurs due to a displacement of an ion from the crystal lattice while
retaining all of the ions, unlike other defects. It is related to the interstitial defect, where an ion
simply occupies a position between regular atoms.
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6. Which defect does the following diagram represent?

a) Vacancy defect
b) Schottky defect
c) Frankel defect
d) Interstitial defect
.
Answer: d
Explanation: Interstitial defects occur when an atom occupies an empty position in a crystal
lattice. Self-interstitial effects occur due to their own atoms, while others occur due to a foreign
substance.
7. _______ occurs when a foreign substance replaces an atom in a crystal.
a) Vacancy defect
b) Substitutional impurity
c) Frankel defect
d) Interstitial impurity
.
Answer: b
Explanation: A substitutional impurity occurs due to the occupation of a foreign atom in place of
an atom in a crystal. On the other hand, interstitial impurities occur when a regular atom
occupies a random space in the crystal lattice.
8. A disturbance in a region between two ideal parts of a crystal is known as ________
a) Boundary defect
b) Point defect
c) Line defect
d) Volume defect
.
Answer: c
Explanation: Line defect is regarded as a disturbed region between two perfect parts of a crystal.
They may be of either edge dislocation type or screw dislocation.
9. In screw dislocation, the Burger’s vector lies _________ to the dislocation line.
a) Perpendicular
b) Parallel
c) At an angle
d) Sideways
.
Answer: b
Explanation: The Burger’s vector in screw dislocation lies parallel to the dislocation line along
the axis of a line of atoms in the same plane. On the other hand, it lies at an angle for edge
dislocation.
10. Generation of dislocations can be identified using _______
a) Schottky mechanism
b) Burger’s vector
c) Twist
d) Frank-Read mechanism
.
Answer: d
Explanation: The Frank-Read mechanism uses the Frank-Read source and its operation as a
dislocation multiplier. The Frank-Read source contains a fixed lined at the X and Y nodes. Under
the application of stress, the dislocation line expands and is further operated until it becomes
readable.
11. What are one-dimensional defects?
a) Boundary defect
b) Point defect
c) Line defect
d) Volume defect
.
Answer: c
Explanation: When compared geometrically, line defects are seen as one-dimensional defects.
Line defects are also known as dislocations, with common types as edge and screw dislocations.
12. What are two-dimensional defects?
a) Boundary defect
b) Point defect
c) Line defect
d) Volume defect
.
Answer: a
Explanation: The defects that occur on the surface of a material are known as surface or
boundary defects. Geometrically, they are regarded as two-dimensional defects.
13. How is the dislocation energy defined?
a) J m-1
b) J m-2
c) m-2
d) N m-1
.
Answer: a
Explanation: Dislocation energy is defined ad joule per meter and is denoted by E. Dislocation
density is defined as meter per cubic meter or simply as per meter square.

1. How is fracture stress defined?

a) MN m-2
b) J m-2
c) mm
d) GN m-1
.
Answer: a
Explanation: Fracture is defined as the breaking of a material into multiple parts. The fracture
stress is defined as meganewton per square meter or MN m-2 and is denoted by σf.
2. What is the SI unit of fracture toughness?
a) N m-2
b) J m-2
c) J m-1
d) N m-1
.
Answer: b
Explanation: Fracture toughness is regarded as the rate of strain energy released. Its SI unit is
joule per meter square (J m-2) and is denoted as Gc.
3. The tensile strength required to break interatomic bonds across two adjacent planes is
________
a) Y
b) Y/2
c) Y/4
d) Y/6
.
Answer: d
Explanation: Calculations for shear strength of perfect crystal has been found to be similar to that
of two adjacent planes. It was determined that the tensile strength necessary to rupture
interatomic bonds of two adjacent atomic planes is of the order of Y/6.
4. What is the breaking stress of brittle materials?
a) Y/6
b) Y/10
c) Y/100
d) Y/1000
.
Answer: d
Explanation: It was found that the tensile strength of two adjacent atomic planes was Y/6.
However, brittle materials may break at a stress of the order of Y/1000. Here, Y is Young’s
modulus.
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5. In crystalline materials, fracture occurs normal to crystallographic planes called ___________
a) Fracture point
b) Cleavage plane
c) Brittle plane
d) Crystallographic point
.
Answer: b
Explanation: In crystalline materials, fracture usually occurs after some deformation. In these
materials, the fracture occurs normal to crystallographic planes called cleavage planes.
6. What is the surface energy of a crack?
a) ϒ
b) 2 ϒ c
c) 4 ϒ c
d) 12 ϒ c
.
Answer: c
Explanation: Griffith suggested the norm for the formation and spreading of cracks in brittle
materials. He found that the surface energy of the crack is 4 ϒ c. Here, ϒ is the surface energy
per unit area of the material.
7. The maximum stress at the tip of the crack is defined by _________

a) 

b) 

c) 

d) 
.
Answer: c
Explanation: When tensile stress is applied to a material, the maximum stress occurs at the tips.

This maximum stress is denoted as σmax and is given by   .

8. Griffith theory is applicable for ______ materials.
a) All brittle
b) Perfect brittle
c) All ductile
d) Perfect ductile
.
Answer: b
Explanation: The stress required to cause brittle fracture varies inversely as the square root of
crack length. From observation, we understand that Griffith theory is applicable only to perfect
brittle materials like glass.

UNIT 2

1. Which among the following is not a type of Non-destructive testing?

a) Compression test
b) Visual testing
c) Ultrasonic testing
d) Eddy current testing
.
Answer: a
Explanation: Compression test is a type of destructive testing. This test is used to determine
behavior of metals under compressive load. Visual testing, ultrasonic testing, eddy current
testing are types of non-destructive testing.
2. Identify the type of destructive testing ______________
a) Radiographic test
b) Dye penetrant test
c) Creep test
d) Visual testing
.
Answer: c
Explanation: Creep test is a type of destructive test. It is defined as slow plastic deformation at
high temperatures for a longer time under constant stresses. Creep occurs at room temperature
and at high temperatures.
3. Which among the following is the last step in magnetic particle test method?
a) Observation and inspection
b) Circular magnetization
c) Demagnetization
d) Magnetization
.
Answer: c
Explanation: Different steps involved in magnetic particle test are cleaning the surface,
magnetizing the metallic component, application of ferromagnetic powder, observation and
inspection and demagnetization.
4. Which of the following statements is/are true for the ultrasonic test?
a) Equipment used for ultrasonic testing is portable
b) Complicated shapes can be easily scanned
c) Waves generated are health hazardous
d) Waves generated are health hazardous and complicated shapes can be easily scanned
.
Answer: a
Explanation: Ultrasonic test uses sound waves of high frequency to detect discontinuities. This
method is used to detect flaws on the surface and also deep inside the component. The waves
travel in straight line and are reflected from metal gas interface or discontinuities in their patch.
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5. Which test can be performed without skilled labour?

a) Probe test
b) Bend liquid test
c) Dye penetrant test
d) Torsion test
.
Answer: c
Explanation: Dye penetrant test does not require any skilled labour. This method only detects
surface discontinuities and this test needs to be observed with naked eyes or with a low
magnifying glass.
6. What is nondestructive test?
a) Nondestructive tests are applications for detecting flaws in materials without impairing their
usefulness
b) Nondestructive tests are applications for detecting flaws that impair the use of the materials
such as pressure testing
c) Nondestructive tests are applications for detecting flaws in materials with impairing their
usefulness
d) Nondestructive tests are applications for detecting flaws that do not impair the use of the
materials such as pressure testing
.
Answer: a
Explanation: Nondestructive tests are applications for detecting flaws in materials without
impairing their usefulness.
7. What is a destructive test?
a) Destructive tests are applications for detecting flaws in materials without impairing their
usefulness
b) Destructive tests are applications for detecting flaws that impair the use of the materials such
as pressure testing
c) Destructive tests are applications for detecting flaws in materials with impairing their
usefulness
d) Destructive tests are applications for detecting flaws that do not impair the use of the materials
such as pressure testing
.
Answer: b
Explanation: Destructive tests are applications for detecting flaws that impair the use of materials
such as pressure testing.
1. Destructive tests are generally much easier to interpret than the non-destructive tests.
a) True
b) False
.
Answer: a
Explanation: Destructive tests are mainly carried out to the specimen’s failure, in order to
evaluate a specimen’s performance or material’s behavior under different loads. These tests are
much easier to carry out, give more information, and are easier to interpret than non-destructive
tests.
2. Destructive testing is not economical for mass production as this method destroys material for
the inspection.
a) True
b) False
.
Answer: b
Explanation: Destructive testing is the most suitable and economic method for objects which will
be mass-produced, it only involves the cost of destroying a small number of specimens which is
quite negligible. Destructive test is not economical where only one or two things are to be
produced.
3. In destructive testing, all the operations are performed manually, thus it does not require any
technologies or electronic devices.
a) True
b) False
.
Answer: b
Explanation: In destructive testing, analyzing of destructive failure mode is often completed
using a high-speed camera, which records continuously until the failure is detected. Detection of
the failure can be accomplished by using a sound detector which produces a signal to start the
high-speed camera.
4. High speed camera used in destructive testing can give very precise information about the
failure of casting or material.
a) True
b) False
.
Answer: a
Explanation: The high-speed cameras used in destructive testing have advance recording modes
to capture almost any kind of destructive failure. After the failure of casting, camera stops
recording and the captured images can be played back in slow motion which shows precise
information, image by image during and after the destructive test.
5. Destructive testing method can also be economical for large casting or structure.
a) True
b) False
.
Answer: b
Explanation: Destructive testing method can be applicable to the large structures but it is not
economical because destruction of large casting or structure will cause steep downing of
productivity. So for large structures or castings, other methods are used for the inspection like
modeling by finite element software.
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6. Which of the following tests is not the type of destructive testing of materials?
a) Stress test
b) Crash test
c) Hardness test
d) Pressure test
.
Answer: d
Explanation: Pressure test is basically a type of non-destructive test in which a gas is injected in
a closed casting, and then the casting is dipped into water and if it produces any leak of gas in the
water, that basically shows presence of defects in the casting that can be simply observed by
visual inspection. And the test like a stress test, hardness test and crash test, these are related to
the destructive testing of materials.
7. Which of the following tests uses a heavy pendulum for the breaking of a specimen while
inspection?
a) Pressure test
b) Static tear test
c) Stress test
d) Charpy v-notch test
.
Answer: d
Explanation: In the Charpy v-notch test, the specimen is first heated to a specified temperature.
Then it is placed in a machine and broken within five seconds. This testing machine uses a heavy
pendulum to break the specimen. This test indirectly provides fracture toughness of the material
or casting.
8. Charpy v-notch test can only estimate the energy absorbed by the fracturing of material.
a) True
b) False
.
Answer: b
Explanation: In charpy test, not only the absorbed energy, but the percentage of shear fracture
area can also be measured. This is generally evaluated by observing the fracture surface of the
specimen or work-piece and this estimates the relative amount of shear fracture, which looks
rough or torn.
9. The dynamic tear test is mainly used for evaluating fracture properties of ultrahigh strength
steel castings.
a) True
b) False
.
Answer: a
Explanation: The dynamic tear test is used to characterize the fracture properties of ultrahigh-
strength steels and also for the aluminum and titanium alloys which generally do not exhibit any
sharp transition temperature behavior. This test is performed by impact loading and results into
absorption of energy at various testing temperatures.
10. In impact testing of materials, the notches are made to reduce the stress concentration.
a) True
b) False
.
Answer: b
Explanation: In impact testing of materials, the notches in specimen are made intentionally to
increase the stress concentration so as to increase the tendency to fracture as most of the
mechanical components have stress raisers. The notch material should be tough enough to
withstand the impact force.
11. The fatigue performance of metallic casting is generally determined by endurance limit of the
material.
a) True
b) False
.
Answer: a
Explanation: The fatigue performance of the material is generally evaluated by endurance limit
of the metal which indicates maximum stress, stress range or stress amplitude. It is also
determined by a number of load cycles which are applied to the material or casting for the
evaluation of properties.
12. Which of the following parameters is also known as stress intensity factor (K)?
a) Creep
b) Proof resilience
c) Fracture toughness
d) Endurance
.
Answer: c
Explanation: The fracture toughness parameter is also called as stress intensity factor (K) which
is used for heavy sections of high strength and low ductility material of casting developing plain
strain conditions, and other energy based methods are generally used for comparatively thinner
sections made of low strength and high ductility material

1. In non-destructive testing, sound test used is a very fine and accurate method of detecting
flaws in the castings.
a) True
b) False
.
Answer: b
Explanation: Sound test is a very rough test to indicate any flaws or discontinuities in the casting.
The casting is suspended by suitable support and tapping is done at the surface of the casting
with a hammer that makes a variation in the tone which indicates the existence of flaws. This
method does not indicate the exact location and extent of the discontinuity in the casting.
2. Impact test for detection of defects in the casting is the most crude and unreliable method of
non-destructive testing.
a) True
b) False
.
Answer: a
Explanation: In the impact test, the casting is subjected to a blow by the help of a hammer of
known weight, which basically strikes or falls on the surface of the casting. Defective castings
fail due to the impact of the blow, but this method is very crude and unreliable to the material or
casting which is inspected.
3. Which of the following methods of NDT requires leak proofing of casting before inspection?
a) Impact test
b) Visual inspection
c) Sound test
d) Pressure test
.
Answer: d
Explanation: Pressure test used on castings required to be leak proof. In this method, all the
openings of the casting are closed and then gas with high pressure is introduced in it. If the
casting is having a porosity or another defect, then it can be detected by leaking of gas in the
water when the casting is submerged into the water.
4. Which of the following types of rays is used in radiography for the inspection of castings?
a) X- rays
b) Infrared rays
c) Ultraviolet rays
d) Visible rays
.
Answer: a
Explanation: Radiography uses X-rays, these rays penetrate through the castings and makes a
shadow picture on a film which is placed behind the material. These rays have a very short wave
length of the order of 0.001 Angstrom. And sometimes gamma rays are also used for the
inspection of castings.
5. In radiography, the penetration of rays is much easier with the less density of metal or casting.
a) True
b) False
.
Answer: a
Explanation: The ability of rays to penetrate through the metal mainly depends on the density of
metal and they can penetrate more easily where less density of metal is present and it leads to the
formation of shadow picture on the film. And any defects in the casting can easily be identified
from the shadow picture.
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6. In penetrant testing of NDT, a liquid is penetrated into the cracks of metal by the application
of pressure.
a) True
b) False
.
Answer: b
Explanation: Penetrant testing method is generally used for detecting very small surface cracks
and it has an advantage over the magnetic particle method that it can be used for any material. A
penetrant liquid is used which is drawn into the cracks or voids by means of capillary action. In
this method, there is no requirement of pressure.
7. Which of the following methods of inspection uses high frequency of sound waves for the
detection of flaws in the castings?
a) Penetrant test
b) Radiography
c) Pressure test
d) Ultrasonic inspection
.
Answer: d
Explanation: Ultrasonic inspection is used to detect defects like cracks and porosity within the
interior of the casting or material. This method uses reflection and transmission of high
frequency sound waves, which are much higher than the audible range and then these waves are
made pass through the casting for inspection.
8. In ultrasonic inspection, a signal processing technique is used for the accurate indication of
porosity in the castings.
a) True
b) False
.
Answer: a
Explanation: A signal processing technique is used for accurately verifying the presence of
porosity and also to determine the size and volume fraction. This is mainly done by measuring
the changes in frequency by using a broad band transducer, in which the front and the back
surface signals of the casting are acquired and digitized.
9. Which of the following terms changes in the eddy current testing method for the detection of
defects in the castings?
a) Resistance
b) Impedance
c) Conductivity
d) Capacitance
.
Answer: b
Explanation: In the method of eddy current testing, the impedance of a coil is changes and the
coil is brought close to a conductive material which indicates the eddy current induced by the
coil and thereby indicates various properties and also the defects of the casting. This method can
be used effectively with both ferromagnetic and non-ferromagnetic materials.
10. There are no restrictions in the eddy current testing method; it can detect defects up to high
depth in the castings.
a) True
b) False
.
Answer: b
Explanation: Eddy current inspection is generally restricted to the depth less than 6mm, that’s
why it is not as sensitive to small open defects of high depth as liquid penetrant testing (LPT) or
magnetic particle inspection (MPI). But it can replace LPT method for detection of surface
connected discontinuities.
11. Eddy current testing method can also be used for the evaluation of heat damage to the metal
alloys.
a) True
b) False
.
Answer: a
Explanation: Eddy current conductivity testing method is commonly used to assess heat damage
of various heat treatable alloys. In order to have a quantitative assessment of heat damaged
material, the establishment of conductivity, hardness and strength, a (CHS) relationship is
required for each alloy.
12. Acoustic emission testing method is basically employed for the detection of surface
discontinuities on the castings.
a) True
b) False
.
Answer: b
Explanation: Acoustic emission testing method, in which solid materials emit sound or acoustic
emission when they are stressed mechanically or thermally to the point where deformation or
fracturing occurs. This creates elastic waves which can be analyzed by an acoustic emission test
system to monitor the condition of the material or casting under stress

16. Macroscopy is used to obtain -------------- of a specimen

A)  presence of defects       B) examination of fractures        
C) flowlines in extruded, forged and drawn parts D) all the above
17. -------------- indicates the direction in which the steel was mechanically work.
A)  sulphar printing           B)  flow lines       C) etching D) none of the above
18. in a macroscopic examination of components ---------------------component do not show flow
lines
A)  cast           B)  rolled     C) machined D) forged

18. In a macroscopic examination, --------------- components show broken and discontinues flow
lines
A)  cast           B)  rolled     C) machined D) forged
19. Air cooling is ----------------
A)  Equilibrium cooling           B) non equilibrium cooling               C) both D) none
20. Optical microscope works on the principle of reflection of
A)  light rays           B)  X-rays          C) gamma rays  D) all

1. Necking occurs in which of the following fractures?

a) Ductile fractures
b) Brittle fracture
c) Fatigue
d) It doesn’t occur during fracture
.
Answer: a
Explanation: Necking occurs when tensile force is applied on metals and it occurs when ductile
failure is occurring. In brittle fracture the material just breaks before breaking.
2. In ductile fracture, at what angle to the tensile axis does the crack propagate outwardly to the
surface of the material?
a) 90°
b) 30°
c) 45°
d) 15°
.
Answer: c
Explanation: The crack propagates outwardly to the surface of the material at an angle of 45° to
the tensile axis and this type of fracture is called “cup and cone fracture”.
3. At the point of necking which of the following is false?
a) Strain hardening occurs at this point
b) Cross sectional area increases
c) Cross sectional area decreases
d) Strength increases
.
Answer: b
Explanation: At the point of the necking the strength increases due to strain hardening and the
cross-sectional areas decreases at the same time. At the time where cross sectional area decreases
faster than necking starts.
4. Ductile fracture happens after extensive plastic deformations.
a) True
b) False
.
Answer: a
Explanation: Unlike brittle fracture, ductile fracture happens after extensive plastic deformation
and goes through multiple visible stages before the final fracture is observed.
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5. Which of the following cannot be a reason for ductile failure?

a) Error in design
b) Improper material
c) Improper manufacturing techniques
d) Unforeseen operating condition
.
Answer: d
Explanation: Ductile fracture is a slow process with many steps and under unforeseen operating
conditions if the fracture is going to occur it will be noticeable and can be avoidable.
6. After which point is necking observed?
a) Ultimate strength
b) Yield strength
c) Elastic point
d) Fracture point
.
Answer: a
Explanation: Necking in a ductile material is observed after the body crosses the point of
ultimate strength on the stress-strain curve.
7. What is the amount of strain in a stable neck is called?
a) Natural draw ratio
b) Forced draw ratio
c) Poisson’s ratio
d) Physical activity ratio
.
Answer: a
Explanation: Natural draw ratio is the name given to the amount of strain in a stable neck during
necking of a material under load.
8. Material subjected to only tensile force even during necking.
a) True
b) False
.
Answer: b
Explanation: Material during necking due to high dislocation density is subjected to a complex
stress. At this point, the material isn’t subjected only a single tensile force.

UNIT 3
1. Why are Hume Rothery’s rules followed?
a) Extensive solid solution
b) Liquid solution
c) Weak solid solution
d) Extensive liquid solution
.
Answer: a
Explanation: To form an extensive solid solution, Hume Rothery’s rules are obeyed. An
extensive solid solution is generally considered as one that is greater than 10 atomic percent
soluble.
2. According to Hume Rothery’s rules, size of atoms must not differ by more than ________
a) 5%
b) 15%
c) 35%
d) 55%
.
Answer: b
Explanation: Hume Rothery’s rules state that the atomic radius or size of solute and solvent must
not differ by more than 15%. This must be in order to minimize the lattice strain.
3. What is formed when the electronegativities of atoms differ?
a) Solid solutions
b) Liquid solution
c) Intermetallic compound
d) Maraging steel
.
Answer: c
Explanation: According to Hume Rothery’s rules, the solute and solvent mist have similar
electronegativities. If this difference is too large, the metals are inclined to form intermetallic
compounds in place of solid solutions.
4. For interstitial solid solutions, atomic radii difference must not differ by more than ________
a) 25%
b) 37%
c) 59%
d) 73%
.
Answer: c
Explanation: For interstitial solid solutions, Hume Rothery’s rules state that the radius of solute
atoms must not be larger than solvent atoms by more than 59%. Moreover, they should have
similar electronegativities and valencies.
5. Dissolution of limited amount of solute in solvent, the solution is a __________
a) Saturated solution
b) Unsaturated solution
c) Supersaturated solution
d) Oversaturated solution
.
Answer: a
Explanation: If the solvent is dissolving a limited quantity of solute, it is known as a saturated
solution. An unsaturated solution is considered one where a small quantity of solute is dissolved
in a solvent, whereas for a supersaturated solution, the amount of solute in a solvent is more.
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6. A solution of exchange of impurities for solvent atoms is called a _________
a) Interstitial solid solution
b) Substitutional solid solution
c) Saturated solution
d) Unsaturated solution
.
Answer: b
Explanation: When the solute atoms (impurities) are substituted for parent solvent atoms in a
crystal lattice, the atoms are called substitutional atoms. Such a mixture of two elements is called
a substitutional solid solution.
7. What kind of solid solution is found in a Cu-Ni crystal?
a) Interstitial solid solution
b) Substitutional solid solution
c) Supersaturated solution
d) Unsaturated solution
.
Answer: b
Explanation: The Cu-Ni system obeys Hume Rothery’s laws of similar atomic radii (1.28 and
1.25), same FCC crystal structure, similar valencies (+1 and +2), and similar electronegativities
(1.9 and 1.8). These elements are completely soluble in one another and form a substitutional
solid solution.
8. Which of the following is a random substitutional solid solution?
a) Cu-Zn
b) Au-Cu
c) Cu2MnAl
d) Carbon in ϒ iron
.
Answer: a
Explanation: A random substitutional solid solution is one in which the solute and solvent atoms
occupy random positions in the crystal lattice. Cu-Zn, or brass, is such a random substitutional
solid solution.
9. Which type of solid solution does this figure illustrate?

a) Interstitial solid solution

b) Substitutional solid solution
c) Supersaturated solution
d) Unsaturated solution
.
Answer: a
Explanation: An interstitial solid solution is one which solute atoms fit into spaces between
solvent atoms. These spaces are known as interstices. In the figure, a system of carbon in FCC ϒ
iron is shown.
10. Au-Cn is an example of _________ solid solution.
a) Interstitial solid solution
b) Random substitutional solid solution
c) Supersaturated solution
d) Ordered substitutional solid solution
.
Answer: d
Explanation: An ordered substitutional solid solution is one in which the solute and solvent
atoms occupy specific or preferred positions in the crystal lattice. Au-Cu and Cu2MnAl are
examples of such ordered substitutional solid solution.
11. Interstitial solutions have a _________ distribution.
a) Random
b) Linear
c) Alternating
d) Dendritic
.
Answer: a
Explanation: In interstitial solutions, the solute atoms are randomly distributed throughout the
solvent. In interstitial compounds, however, the pattern is regular in the specific compound.
12. Cu3Al and NiAl are examples of __________
a) Interstitial solutions
b) Interstitial compounds
c) Electron compounds
d) Valency compounds
.
Answer: c
Explanation: If two metals consist of atoms of similar valencies, then the compounds formed are
known as electron compounds. Cu3Al, CuZn, Cu3Sn and NiAl are examples of such electron
compounds.

1. What is/are the phenomenon involved in phase transformation?

a) Nucleation
b) Growth
c) Fission
d) Nucleation and growth
.
Answer: d
Explanation: Formation of a nucleus or tiny particles of the
new phase is nucleation and increase in size of the nucleus at the expense of the parent phase is
growth.
2. How many types of nucleation process are there and what are they?
a) 2 and (fusion and fission)
b) 2 and (Heterogeneous and Homogeneous)
c) 2 and (Heterogeneous and fusion)
d) 4 and (fusion, fission, Heterogeneous and Homogeneous)
.
Answer: b
Explanation: Two types of nucleation – Heterogeneous and Homogeneous whereas fusion and
fission are nuclear reactions.
3. What reactions come under supercooling?
a) Peritectic
b) Eutectic and Peritectic
c) Eutectic and Eutectoid
d) Peritectic and Eutectoid
.
Answer: c
Explanation: Driving force to nucleate increases as we increase T.
4. What are the characteristics of large supercooling in nucleation?
a) Few nuclei, large crystals
b) Rapid nucleation, many nuclei, small crystals
c) Rapid nucleation, many nuclei, large crystals
d) Few nuclei, small crystals
.
Answer: b
Explanation: Nuclei (seeds) act as a template to grow crystals and for nucleus will help in a rate
of the addition of atoms to the nucleus must be faster than the rate of loss.
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5. At what temperature does supercooling is needed in homogeneous nucleation?
a) 0.1-10°C
b) 100- 250°C
c) 80-300°C
d) 10-15°C
.
Answer: c
Explanation: Nuclei form in the bulk of liquid metal so requires supercooling (typically 80-
300°C max).
6. From the following figure if blue curve represent growth rate what does red and green curve
represent respectively?

a) Nucleation rate, over all transformation rate

b) Over all transformation rate, Nucleation rate
c) Nucleation rate, finer grain size
d) Over all transformation rate, finer grain size
.
Answer: a
Explanation: The overall transformation rate is the product of nucleation and growth rates.
7. Which is the most important thermodynamic parameter in Homogenous nucleation?
a) Free energy G
b) Enthalpy H
c) Entropy S
d) Free energy G, Enthalpy H, Entropy S
.
Answer: a
Explanation: G is important as a phase transformation will occur immediately only when G has a
negative value.
8. What does phase transformation involve?
a) Transformation rates kinetics
b) Movement/rearrangement of atoms
c) Changes in microstructure
d) Transformation rate kinetics, rearrangement of Atoms, Changes in microstructure
.
Answer: d
Explanation: Formation of a new phase having a
distinct physical/chemical character and a different structure than that of the parent phase.
9. Below the critical radius the tiny particles are _______ and are called _______
a) unstable, grains
b) stable, grains
c) unstable, embryo
d) stable, embryo
.
Answer: c
Explanation: The tiny particle of the solid that forms first will get stabilized only when it
achieves a critical radius (r*). Below the critical radius, it is considered unstable and is said to be
an embryo

1. Which of the following is/are the components of a brass alloy?

a) Brass, Copper, Zinc
b) Copper only
c) Zinc only
d) Both Copper and Zinc
.
Answer: d
Explanation: Components are pure metals and/or compounds of which an alloy is composed.
Brass is an alloy and Copper and zinc are components.
2. At certain temperature the maximum concentration of solute atoms that dissolve in solvent to
form solid solution, this condition is called __________
a) Solubility
b) Formation of phase
c) Solubility limit
d) Formation of super saturated solution
.
Answer: c
Explanation: At this limit there will maximum solubility o solution and solution formed is
saturated above this there will be formation of super saturated solution.
3. Analyze the following graph and the following question regarding solubility of sugar in water.

a) Decrease with increase in temperature

b) Increase with increase in temperature
c) Increase with decrease in temperature
d) Remains constant with increase or decrease in temperature
.
Answer: b
Explanation: The solubility of sugar increases when the solution is heated.
4. What is a phase?
a) The substance which is physically distinct
b) The substance which is homogenous chemically
c) The substance which is both physically distinct and chemically homogenous
d) The substance which is both physically distinct and chemically heterogeneous
.
Answer: c
Explanation: A phase can be defined as a physically distinct and homogeneous chemically
portion of a system that has a particular chemical composition and structure.
5. Which of the following portion(s) of the given diagram contains only gas?

a) 1
b) 4
c) 5
d) 4, 5
.
Answer: c
Explanation: In the figure pure substances exists at 1, 3, 5 out of which 1 is given solid so 5 is
pure gas.
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6. Which of the following portion(s) of the given diagram contains liquid in it?

a) 2, 3
b) 2, 4
c) 3, 4
d) 2, 3, 4
.
Answer: d
Explanation: From figure in process 2 the substance solid is being converted to liquid. In 3 only
pure liquid exists. In 4 liquid is converting into solid. These all stages contain liquid in it.
7. Which of the following portion(s) of given diagrams contains only pure substances?

a) 1, 3, 5
b) 1, 5
c) only 1
d) 3, 5
.
Answer: a
Explanation: Pure substances are substances that contain only a single phase such as solid or
liquid or gas. These phases exist in 1, 3 and 5 portions.
.
8. Which of the following portion(s) of the given diagram has a freezing section?

a) 1
b) 2
c) 3
d) 4
.
Answer: b
Explanation: Since this is a warming curve, this section indicates the melting temperature, but
freezing point is identical to melting temperature. It can be noted as freezing range in a cooling
curve.
9. Which of the following portion(s) of the given graph has a boiling point?

a) 2
b) 3
c) 4
d) 5
.
Answer: c
Explanation: At this point the substance gets converted from liquid to gas.
10. Which of the following section(s) given graph in which we use (Specific Heat) * (Mass) *
(Change in temperature) to calculate the energy change?

a) 2
b) 1, 3
c) 1, 3, 5
d) 2, 3, 5
.
Answer: c
Explanation: Since the phase of substance remains constant and there is change in temperature of
substance
1. What is Gibbs phase rule for general system?
a) P = C – 1 – F
b) P = C + 1 – F
c) P + F = C – 2
d) P + F = C + 2
.
Answer: d
Explanation: The number of degrees of freedom, F (no. of independently variable factors),
number of components, C, and number of phases in equilibrium, P.
2. What is Gibbs phase rule for metallurgical system?
a) F = C – 1 – P
b) F = C + 1 – P
c) P + F = C – 2
d) P + F = C + 2
.
Answer: b
Explanation: For metallurgical system pressure has no appreciable effect on phase equilibrium
and hence, F = C – P + 1.
3. In a single – component condensed system, if degree of freedom is zero, maximum number of
phases that can co – exist _________
a) 2
b) 3
c) 0
d) 1
.
Answer: a
Explanation: Given F = 0
Then p = c + 1, c = 1
.: P = 2.
4. The degree of freedom at a triple point in the unary diagram for water is ________
a) 2
b) 3
c) 0
d) 1
.
Answer: c
Explanation: For three phase system degree of freedom is 0.
5. What is degree of freedom for single – phase fields on the phase diagram?
a) 2
b) 3
c) 0
d) 1
.
Answer: a
Explanation: F = C + 1 – P
F = 3 – P (C = 2)
.:F = 2.
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6. What is degree of freedom when two phases co – exist?

a) 2
b) 3
c) 0
d) 1
.
Answer: d
Explanation: F = C + 1 – P
F = 3 – P (C = 2)
F = 3 – 2 = 1.
7. What is degree of freedom when three phases co – exist?
a) 2
b) 3
c) 0
d) 1
.
Answer: c
Explanation: F = C + 1 – P
F = 3 – P (C = 2)
F = 3 – 3 = 0.
8. For single component system when degree of freedom is 1(one) then number of phases
_______
a) 2
b) 3
c) 0
d) 1
.
Answer: d
Explanation: F = C + 1 – P
F = 2 – P (C = 1)
→ p = 2 – F = 2 – 1 = 1.
9. When α, L and β phase fields touch the isotherm line what are the respective phase
compositions?
a) 8.0 wt%, 71.9 wt%, 91.2 wt% of Ag
b) 8.0 wt%, 91.2 wt%, 71.9 wt% of Ag
c) 71.9 wt%, 91.2 wt%, 8.0 wt% of Ag
d) 91.2 wt%, 8.0 wt%, 71.9 wt% of Ag
.
Answer: a
Explanation: For binary systems, when three phases are present, there will be F = 0, so
composition is fixed.
10. For binary alloy consisting of three phases of non – equilibrium one, What will be the
temperature of these phases?
a) Different
b) Constant
c) Same
d) Two of them will be with one temperature
.
Answer: c
Explanation: One use of the Gibbs phase rule is in analyzing for non – equilibrium conditions by
analyzing with above method we come to know (under these Circumstances, three phases will
exist only at a single temperature).

1. Which of the following is associated with minimum plastic deformation?

a) Ductile fracture
b) Brittle fracture
c) Fatigue
d) It doesn’t occur during fracture
.
Answer: b
Explanation: In brittle fracture, the plastic deformation occurring in the body is minimum and
there is a rapid movement of crack with a very little amount of plastic deformation.
2. Which of the following fracture mechanisms in which the crack propagates more rapidly?
a) Ductile fracture
b) Brittle fracture
c) Fatigue
d) It doesn’t occur during fracture
.
Answer: b
Explanation: In brittle fracture, after a little plastic deformation the material fractures and then
the crack propagates rapidly, making the material break.
3. Below which point does brittle fracture occur?
a) Ultimate tensile strength
b) Fracture point
c) Elastic limit
d) Yield point
.
Answer: d
Explanation: Fatigue is one of the mechanical failures of the metals that occur due to cyclic
loading. It occurs when the stress applied is less than the yield stress or yield point.
4. Cleavage planes are a term associated with ductile fracture.
a) True
b) False
.
Answer: b
Explanation: Brittle fracture occurs generally along crystal planes called cleavage planes where
the number of atomic bonds are relatively low.
5. The tendency of brittle fracture increases with:
a) Decreasing temperature
b) Increasing temperature
c) Decrease in strain rate
d) It doesn’t depend on temperature or strain rate
.
Answer: a
Explanation: The tendency of brittle fracture to be observed increases as the temperature
decreases.
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6. The tendency of brittle fracture increases with:
a) Increasing temperature
b) Increase in strain rate
c) Decrease in strain rate
d) It doesn’t depend on temperature or strain rate
.
Answer: b
Explanation: The tendency of brittle fracture to be observed increases as the strain rate of the
body increases. As the strain rate increases the body dislocations start piling up and ultimately
leading to failure.
7. Which of the following statement is true regarding the toughness of a metal?
a) Energy consumed is less in ductile fracture than brittle fracture
b) Energy consumed is more in ductile fracture than brittle fracture
c) Energy consumed in the brittle fracture is same as the ductile fracture
d) Energy consumed is more in brittle fracture than ductile fracture
.
Answer: b
Explanation: As brittle fracture less plastic deformation and absorbs less amount of energy
before breaking, the energy consumed by brittle fracture is less than the ductile fracture.
8. Which of the theory is related to brittle fracture?
a) Laundau theory
b) Dirac hole theory
c) Valence bond theory
d) Griffith’s theory
.
Answer: d
Explanation: Griffith’s theory is a theory which was postulated by Griffith and is related to
brittle fracture. It is known as Griffith’s theory of brittle fracture.
9. The graph for Griffith’s crack is ____________
a) An ellipse
b) A circle
c) A straight line
d) A hyperbola
.
Answer: a
Explanation: The graph for Griffith’s crack is an ellipse. The graph tensile stress applied
perpendicular to the length of crack and the length of the crack.
10. Surface energy of the specimen increases when the crack lengthens.
a) True
b) False
.
Answer: a
Explanation: When the crack lengthens the surface area increases and hence the surface energy
of the specimen increases

1. Which of the following is associated with minimum plastic deformation?

a) Ductile fracture
b) Brittle fracture
c) Fatigue
d) It doesn’t occur during fracture
.
Answer: b
Explanation: In brittle fracture, the plastic deformation occurring in the body is minimum and
there is a rapid movement of crack with a very little amount of plastic deformation.
2. Which of the following fracture mechanisms in which the crack propagates more rapidly?
a) Ductile fracture
b) Brittle fracture
c) Fatigue
d) It doesn’t occur during fracture
.
Answer: b
Explanation: In brittle fracture, after a little plastic deformation the material fractures and then
the crack propagates rapidly, making the material break.
3. Below which point does brittle fracture occur?
a) Ultimate tensile strength
b) Fracture point
c) Elastic limit
d) Yield point
.
Answer: d
Explanation: Fatigue is one of the mechanical failures of the metals that occur due to cyclic
loading. It occurs when the stress applied is less than the yield stress or yield point.
4. Cleavage planes are a term associated with ductile fracture.
a) True
b) False
.
Answer: b
Explanation: Brittle fracture occurs generally along crystal planes called cleavage planes where
the number of atomic bonds are relatively low.
5. The tendency of brittle fracture increases with:
a) Decreasing temperature
b) Increasing temperature
c) Decrease in strain rate
d) It doesn’t depend on temperature or strain rate
.
Answer: a
Explanation: The tendency of brittle fracture to be observed increases as the temperature
decreases.
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6. The tendency of brittle fracture increases with:

a) Increasing temperature
b) Increase in strain rate
c) Decrease in strain rate
d) It doesn’t depend on temperature or strain rate
.
Answer: b
Explanation: The tendency of brittle fracture to be observed increases as the strain rate of the
body increases. As the strain rate increases the body dislocations start piling up and ultimately
leading to failure.
7. Which of the following statement is true regarding the toughness of a metal?
a) Energy consumed is less in ductile fracture than brittle fracture
b) Energy consumed is more in ductile fracture than brittle fracture
c) Energy consumed in the brittle fracture is same as the ductile fracture
d) Energy consumed is more in brittle fracture than ductile fracture
.
Answer: b
Explanation: As brittle fracture less plastic deformation and absorbs less amount of energy
before breaking, the energy consumed by brittle fracture is less than the ductile fracture.
8. Which of the theory is related to brittle fracture?
a) Laundau theory
b) Dirac hole theory
c) Valence bond theory
d) Griffith’s theory
.
Answer: d
Explanation: Griffith’s theory is a theory which was postulated by Griffith and is related to
brittle fracture. It is known as Griffith’s theory of brittle fracture.
9. The graph for Griffith’s crack is ____________
a) An ellipse
b) A circle
c) A straight line
d) A hyperbola
.
Answer: a
Explanation: The graph for Griffith’s crack is an ellipse. The graph tensile stress applied
perpendicular to the length of crack and the length of the crack.
10. Surface energy of the specimen increases when the crack lengthens.
a) True
b) False
.
Answer: a
Explanation: When the crack lengthens the surface area increases and hence the surface energy
of the specimen increases.
1. What is maximum solubility % of carbon and temperature range that α-ferrite exist?

a) 0.05% and 273°c to 910°c

b) 0.025% and 273°c to 910°c
c) 2.1% and 910°c to 1394°c
d) 0.05% and 910°c to 1124°c
.
Answer: b
Explanation: Interstitial solid solution of C in BCC iron. Max solubility of C is 0.025% and it
exists from 273°C to 910°C.
2. What is maximum solubility % of carbon and temperature range that Austenite (Ɣ) exist?

a) 0.05% and 273°c to 910°c

b) 0.025% and 273°c to 910°c
c) 2.1% and 910°c to 1394°c
d) 0.09% and 1394°c to 1539°c
.
Answer: c
Explanation: Interstitial solid solution of C in FCC iron has a 2.1% max solubility of C and exists
from 910°C – 1394°C.
3. What is maximum solubility % of carbon and temperature range that δ-ferrite exist?

a) 0.05% and 273°c to 910°c

b) 0.025% and 273°c to 910°c
c) 2.1% and 910°c to 1394°c
d) 0.09% and 1394°c to 1539°c
.
Answer: d
Explanation: δ-ferrite (BCC) exists over the temperature range of 1394°C to 1539°C and the
maximum solubility of C is 0.09%.
4. What is the % C content in Cementite (Fe3C)?

a) 6.67%
b) 0.025%
c) 2.1%
d) 0.09%
.
Answer: a
Explanation: It is an intermetallic compound. C content in Fe3C is 6.67%.
5. What is the hardest phase of Fe-C system?

a) Graphite
b) Bainite
c) Martensite
d) Cementite
.
Answer: d
Explanation: Cementite is the hardest phase in the Iron-Carbon system whereas martensite is not
a phase because it is metastable in nature.
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6. What is eutectoid temperature?

a) 727°c
b) 768°c
c) 1146°c
d) 1495°c
.
Answer: a
Explanation: The eutectoid temperature (727°C) during heating and cooling is Ac1 and Ar1
respectively. Eutectoid reaction is a reaction that transforms one solid into two different solids on
cooling.
7. At what temperature Fe turns paramagnetic while heating?

a) 727°c
b) 768°c
c) 1146°c
d) 1495°c
.
Answer: b
Explanation: At 768°C. Iron converts from ferromagnetic to paramagnetic and this temperature
is known as curie temperature. It is also called as A1 isotherm in iron carbon phase diagram.
8. What is the value of eutectic temperature?

a) 727°c
b) 768°c
c) 1146°c
d) 1495°c
.
Answer: c
Explanation: The eutectic temperature for Fe-C system is 1146°C. The eutectic reaction is a
reaction that transforms one liquid phase into two solid phases on cooling.
9. What is the value of peritectic temperature?

a) 727°c
b) 768°c
c) 1146°c
d) 1495°c
.
Answer: d
Explanation: The peritectic temperature is at 1495°C for the Fe-C system. The peritectic reaction
is a reaction that transforms one liquid and solid to another solid on cooling.
10. Carbon steel containing 9,1% ferrite is cooled from the region of austenitic (727 °c). What is
the percentage C content in the steel?
a) 0.05% C
b) 0.1% C
c) 0.2% C
d) 0.3% C
.
Answer: b
Explanation: Let c be Carbon content. Applying the lever rule
0.091 = (6.67 – c)/(6.67 – 0.025)
c = 0.1% C.

1. What is the Peritectic reaction at 1495°C?

a) L (0.53% C) + δ(0.09% C) → γ(0.17% C)
b) L (4.3% C) → γ(2.1 % C) + Fe3C (6.67% C)
c) γ (0.8 % C) → α (0.025% C) + Fe3C (6.67% C)
d) L (0.53% C) + δ(0.09% C) → γ (0.8 % C)
.
Answer: a
Explanation: Peritectic reaction at 1495°C is
L (0.53% C) + δ(0.09% C) → γ(0.17% C).
2. What is the Eutectic reaction at 1146°C?
a) L (0.53% C) + δ(0.09% C) → γ(0.17% C)
b) L (4.3% C) → γ(2.1 % C) + Fe3C (6.67% C)
c) γ (0.8 % C) → α (0.025% C) + Fe3C (6.67% C)
d) L (0.53% C) + δ(0.09% C) → γ (0.8 % C)
.
Answer: b
Explanation: Eutectic reaction at 1146°C is
L (4.3% C) → γ(2.1 % C) + Fe3C (6.67% C).
3. What is Eutectoid reaction at 727°C?
a) L (0.53% C) + δ(0.09% C) → γ(0.17% C)
b) L (4.3% C) → γ(2.1 % C) + Fe3C (6.67% C)
c) γ (0.8 % C) → α (0.025% C) + Fe3C (6.67% C)
d) L (0.53% C) + δ(0.09% C) → γ (0.8 % C)
.
Answer: c
Explanation: Eutectoid reaction at 727°C is
γ (0.8 % C) → α (0.025% C) + Fe3C (6.67% C).
4. The eutectic mixture of austenite (γ) and cementite (Fe3C) is called __________
a) Ledeburite
b) Pearlite
c) Hyper and hypo eutectoid steel
d) Cast iron
.
Answer: a
Explanation: Eutectic reaction at 1146°C is
L (4.3% C) → γ(2.1 % C) + Fe3C (6.67% C).
The eutectic mixture of austenite (γ) and cementite (Fe3C) is called Ledeburite.
5. The eutectoid mixture of ferrite (α) and cementite (Fe3C) is called __________
a) Ledeburite
b) Pearlite
c) Hyper and hypo eutectoid steel
d) Cast iron
.
Answer: b
Explanation: The pearlite is formed under equilibrium
Conditions which consists of alternate lamellas or layers of ferrite and Fe3C.
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6. Compositions right and left of 0.8% C of Pearlite are called __________
a) Ledeburite
b) Ferrite
c) Hyper and Hypoeutectoid steel
d) Cast iron
.
Answer: c
Explanation: Hypoeutectoid steels – α+ p; hypereutectoid – Fe3C + p.
7.Compositions above 2.1% C is called as _______
a) Ledeburite
b) Ferrite
c) Hyper and Hypoeutectoid steel
d) Cast iron
.
Answer: d
Explanation: Compositions up to 2.1% of Carbon are considered steels and beyond this it is
considered as cast iron.
8. From the following figure if Diffusion rates below Ms is so low that γ → M transformation is
a diffusionless process then what is change in crystal structure

a) FCC → BCT
b) FCC → BCC
c) BCC → BCT
d) BCT → BCC
.
Answer: a
Explanation: Diffusion rates below Ms is so low that γ → M transformationis a diffusion less
process (the C content remains same). However, the crystal structure changes from FCC (γ) to
body centered tetragonal (BCT).
9.Finer size pearlite is called _______
a) Troostite
b) Ledeburite
c) Ferrite
d) Sorbite
.
Answer: d
Explanation: Finer size pearlite is called sorbite and very fine size pearlite is called troostite.
10. A eutectoid steel is slowly cooled from a temperature of 750°C to a temperature just below
727°C. Calculate the percentage of ferrite and cementite.
a) 88.3% and 11.7%
b) 70% and 30 %
c) 85.4% and 14.6%
d) 20% and 20%
.
Answer: a
Explanation: Eutectoid composition – 0.8% of Carbon,
Ferrite composition- 0.025% of Carbon and
cementite – 6.67% C.
Apply the lever rule to get the percentages as
% of Ferrite = 100(6.67 – 0.80)/(6.67 – 0.025)=88.3% ferrite
%of Cementite = 100(0.80 – 0.025)/(6.67 – 0.025) =11.7%.

1. In an iron carbon alloy, the mechanical properties are dependent on the micro structure.
a) True
b) False
.
Answer: a
Explanation: The above statement is true. In an iron carbon alloy, the mechanical properties are
dependent on the micro structure.
2. Which of the following is present in pearlite?
a) Spherodite
b) Bainite
c) Ledeburite
d) Ferrite
.
Answer: d
Explanation: Pearlite consists of alternate layer of ferrite and cementite in which cementite is
harder and more brittle as compared to ferrite.
3. Which of the following is true?
a) Fine pearlite is harder than coarse pearlite
b) Fine pearlite is softer than coarse pearlite
c) Fine pearlite is more ductile than coarse pearlite
d) Toughness of steel decreases with increase in carbon percentage
.
Answer: a
Explanation: Fine pearlite is harder than coarse pearlite, but coarse pearlite is more ductile than
fine pearlite.
4. Which is the most tough among the steels given their carbon composition?
a) 0.1%
b) 0.2%
c) 1.5%
d) 2.5%
.
Answer: a
Explanation: The toughness of the steel decreases as we increase the carbon content and
therefore 0.1% carbon which is lower than 1.5%, 0.2% and 2.5% is most tough among the
options.
5. Which of the following is present in spherodite?
a) Pearlite
b) Bainite
c) Ledeburite
d) Ferrite
.
Answer: d
Explanation: Spherodite also consists of ferrite and cementite. In spherodite it cementite is
present in a sphere structure in a matrix of ferrite.
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6. Which of the following is present in bainite?

a) Ferrite
b) Bainite
c) Ledeburite
d) α-Ferrite
.
Answer: d
Explanation: Bainite steels consists of α-Ferrite and cementite and have a finer structure which
makes then stronger than pearlite steel.
7. Which among the following is the strongest?
a) Austenite
b) Pearlite
c) Bainite
d) Martensite
.
Answer: d
Explanation: Among all the structure martensite has the strongest structure. It is the hardest and
most brittle among the microstructure.
8. In which of the following steels there is a chance of quenching cracks being formed?
a) 0.2% carbon
b) 0.4% carbon
c) 0.5% carbon
d) 0.6% carbon
.
Answer: d
Explanation: If the carbon content in a steel is higher than 0.5 %, during quenching there can be
a formation of quenching cracks.
9. At what temperature is martensite heated in tempering? (in degree Celsius)
a) 727
b) 627
c) 327
d) 927
.
Answer: a
Explanation: The martensitic steel is tempered in the temperature of 650–800°C to get optimum
mechanical properties that includes ductility and toughness.
10. Tempered martensite has better ductility than martensite.
a) True
b) False
.
Answer: a
Explanation: Tempered martensite has better ductility than martensite due to tempering. The
hardness of the two is nearly the same

UNIT 4

1. Which of the following is the hardest constituent of steel?

a) Ledeburite
b) Austenite
c) Bainite
d) Martensite
.
Answer: d
Explanation: Martensite is the hardest constituent of steel. The primary reasons accounting for
this could be, the internal strains within BCC iron due to the excess carbon presence and due to
the plastic deformation of parent FCC iron (austenite) surrounding the martensitic plate. Rate of
cooling and the amount of carbon percentage in steel are directly proportional to the amount of
hardness achieved in martensitic transformation.
2. Iron possesses BCC crystal structure up to (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
.
Answer: b
Explanation: Pure iron possess either BCC or FCC crystal structure as its temperature is
increased from room temperature to its melting point. At room temperature to 910oC, it is having
BCC, between 910oC and 1410oC it is having face centered cubic, and from 1410oC to its melting
point (1539oC) it returns to its BCC crystal structure.
3. Iron possesses BCC crystal structure above (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
.
Answer: d
Explanation: From 1410oC to its melting point (1539oC) iron is having BCC crystal structure.
4. Iron possesses FCC crystal structure above (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
.
Answer: c
Explanation: Between 910oC and 1410oC iron is having face centered cubic crystal structure.
5. Which of the following form of iron is magnetic in nature?
a) α
b) δ
c) γ
d) λ
.
Answer: a
Explanation: The alpha form of iron is magnetic and stable at all temperatures below 910oC.
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6. For steel, which one of the following properties can be enhanced upon annealing?
a) Hardness
b) Toughness
c) Ductility
d) Resilience
.
Answer: c
Explanation: A furnace cooling technique, annealing will enhance the ductility of steel, due to
the formation of coarse pearlite.
7. In Annealing, cooling is done in which of the following medium?
a) Air
b) Water
c) Oil
d) Furnace
.
Answer: d
Explanation: In annealing, after solutionising, material is used to furnace cool, means furnace is
switched off and the steel sample inside is let cool down.
8. In normalizing, cooling is done in which of the following medium?
a) Air
b) Water
c) Oil
d) Furnace
.
Answer: a
Explanation: In normalizing, steel is heat treated above its critical temperature, solutionised, and
then allowed to cool for a long time by keeping it in air. In steel, it forms fine pearlite, which
imparts strength to steel.
9. Mild steel can be converted into high carbons steel by which of the following heat treatment
process?
a) Annealing
b) Normalizing
c) Case hardening
d) Nitriding
.
Answer: c
Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus,
imparting hardness to steel.
10. Upon annealing, eutectoid steel converts to which of the following?
a) Perlite
b) Cementite
c) Austenite
d) Martensite
.
Answer: a
Explanation: Eutectoid steels upon annealing produces pearlite (coarse pearlite). Pearlite is an
alternate lamellae of ferrite and cementite

1. Composition up to 0.008% carbon is considered as _________

a) Pure iron
b) Steel
c) Cast iron
d) Annealed steel
.
Answer: a
Explanation: In general, the different types of ferrous alloys are steel, iron, and cast iron. Pure
iron is composed of 0.008% carbon, whereas steel and cast iron contain 0.008-2.0% and over 2%
respectively.
2. Steels containing 0.8% C are called ________
a) Eutectoid
b) Hypoeutectoid
c) Hypereutectoid
d) Mild eutectoid
.
Answer: a
Explanation: In general, the different types of ferrous alloys are steel, iron, and cast iron. Steels
are further subdivided as eutectoid, hypoeutectoid, and hypereutectoid. Eutectoid steels contain
0.8% of carbon content.
3. How much carbon does hypoeutectoid steel contain?
a) <0.8%
b) 0.8%
c) >0.8%
d) 8%
.
Answer: a
Explanation: In general, the different types of ferrous alloys are steel, iron, and cast iron. Steels
are further subdivided as eutectoid, hypoeutectoid, and hypereutectoid. Hypoeutectoid steels
contain less than 0.8% of carbon content, whereas hypereutectoid steels contain over 0.8%.
4. Eutectoid steel is heated at __________
a) 150oC
b) 450oC
c) 750oC
d) 950oC
.
Answer: c
Explanation: For the transformation of eutectoid steel, a sample of 0.8% steel is heated at 750oC.
It is held at this temperature until a transformation begins.
5. Transformation in eutectoid steels is named as _________
a) Austenitizing
b) Carburizing
c) Decarburizing
d) Nitriding
.
Answer: a
Explanation: For the transformation of eutectoid steel, a sample of 0.8% steel is heated at 750oC
and held at that temperature. Due to this, the structure becomes homogeneous austenite. This
process is known as austenitizing.
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6. Cooling of hypoeutectoid steel post heating forms __________ ferrite.
a) Proeutectoid
b) Posteutectoid
c) Hypereutectoid 2.08%
d) Mild eutectoid
.
Answer: a
Explanation: 0.022-0.8% carbon steel is heated at around 875oC and held at this temperature.
This results in the formation of homogeneous austenite. Then the steel is cooled to about 775oC,
due to which pro-eutectoid ferrite will nucleate and grow at austenite grain boundaries.
7. Slow cooling of hypereutectoid steel results in formation of __________
a) Proeutectoid cementite
b) Posteutectoid cementite
c) Proeutectoid ferrite
d) Proeutectoid austenite
.
Answer: a
Explanation: Over 0.8% carbon steel is heated at around 925oC and held at this temperature. This
results in the formation of austenite. Then the steel is slowly cooled to form proeutectoid
cementite which will nucleate and grow at austenite grain boundaries.
8. A mixture of austenite and cementite is called ___________
a) Ferrite
b) Ledeburite
c) Pearlite
d) Bainite
.
Answer: b
Explanation: Slow cooling of eutectic cast iron leads to precipitation of austenite of cementite.
This combination of austenite and cementite is called ledeburite. This reaction occurs at 1148oC.
9. Cooling of austenite of eutectoid composition at 723oC results in formation of ________
a) Ledeburite
b) Sphereoidite
c) Pearlite
d) Bainite
.
Answer: c
Explanation: The eutectic reaction of cast iron occurs at 1148oC. Below the eutectic temperature,
the cast iron consists of austenite crystals and ledeburite. Cooling of this alloy below 723oC
transforms austenite into pearlite.
10. Heating of 4.8% cast iron at _______ results in the formation of liquid metal.
a) 450oC
b) 650oC
c) 950oC
d) 1450oC
.
Answer: d
Explanation: If a sample of 4.8% cast iron is heated to about 1450oC for a sufficient time, its
structure becomes a homogeneous liquid metal. If this cast iron is slowly cooled, cementite
crystals nucleate and grow in the liquid phase.
11. Transformation of hypoeutectic cast iron is applicable for the composition of cast iron of
________ carbon.
a) 2.0-4.3%
b) 4.3%
c) 4.3-5.0%
d) Over 5%
.
Answer: a
Explanation: Transformation of hypoeutectic cast iron can be seen for the composition of cast
iron between 2.0-4.3%. For eutectic cast iron, this composition is 4.3% carbon, whereas, for
hypereutectic cast irons, it is 4.3-5.0% carbon.
1. Non-equilibrium phases are shown for their time and transformation using _________
a) Fe-Fe3C diagram
b) TTT diagram
c) CCT diagram
d) TTT and CCT diagram
.
Answer: d
Explanation: Ferrite, cementite, pearlite, and austenite are equilibrium phases which are based on
the iron-iron carbide equilibrium diagram. For other non-equilibrium phases like martensite and
bainite, the Fe-Fe3C diagram cannot be used. In such cases, we use the TTT and CCT diagrams.
2. The CCT or the TTT diagrams are used for _________
a) One steel of specific composition
b) A family of various steels
c) Alloy system of various compositions
d) Combination of all alloys and steels with various compositions
.
Answer: a
Explanation: An individual equilibrium diagram like the Fe-Fe3C can be used to depict an entire
alloy system of various compositions. However, a TTT or a CCT diagram can only be used for
one steel of a specified composition.
3. What does CCT diagram stand for?
a) Constant-critical-temperature
b) Constant-cooling-temperature
c) Continuous-cooling-transformation
d) Continuous-creep-transformation
.
Answer: c
Explanation: Martensite and bainite are non-equilibrium phases which cannot be depicted on a
Fe-Fe3C equilibrium diagram. In such cases, either the time-temperature-transformation or the
continuous-cooling-transformation diagrams can be used.
4. __________ is used to predict quenching reactions in steels.
a) Isothermal transformation diagram
b) Iron-iron carbide equilibrium diagram
c) Continuous cooling transformation diagram
d) Logarithm scale
.
Answer: a
Explanation: An isothermal transformation (IT) is a tool used by heat treaters to calculate
quenching reactions in steels. The IT diagram can also be called as a time-temperature-
transformation diagram.
5. Which of the following is not an alternative name for the TTT diagram?
a) S curve
b) C curve
c) Adiabatic curve
d) Bain’s curve
.
Answer: c
Explanation: Owing to the shape of the diagram, TTT diagrams may be called S curves or C
curves. Due to their nature and type of work, they may also be known as isothermal
transformation diagrams or Bain’s curves.
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6. The first step in constructing a TTT diagram involves _________ the sample.
a) Annealing
b) Normalising
c) Quenching
d) Austenising
.
Answer: d
Explanation: To construct a TTT diagram, a large number of the small specimen are collected
and austenised in a furnace. Next, heat treatment and quenching stages are carried out. After each
stage, the temperature and time are plotted as curves.
7. Austenising of samples for TTT diagram is done __________ temperature.
a) At room temperature
b) Below melting point
c) Above eutectoid temperature
d) Above boiling temperature
.
Answer: c
Explanation: A large number of small samples of the same material are collected and austenized
in a furnace above the eutectoid temperature. Then it is rapidly cooled at a desired temperature
below the eutectoid temperature.
8. Examination of transformation time after quenching is done ___________
a) At room temperature
b) Below melting point
c) Above eutectoid temperature
d) Above boiling temperature
.
Answer: a
Explanation: Austenising of the sample is done above eutectoid temperature, whereas quenching
is done below the eutectoid temperature. After each transformation time, the microstructure is
examined at room temperature and plotted in the form of curves.
9. Isothermal transformations of eutectoid steel between 723oC and 550oC produces
__________ microstructure.
a) Pearlite
b) Bainite
c) Ferrite
d) Cementite
.
Answer: a
Explanation: Isothermal transformations of eutectoid steel between 723oC and 550oC bring into
being a pearlitic microstructure. The pearlite changes from coarse structure to fine structure as
the transformation is decreased in this range.
10. Rapid quenching of eutectoid steel ___________ transforms the austenite into martensite.
a) At room temperature
b) Below 320oC
c) At 550oC
d) Above 723oC
.
Answer: d
Explanation: An isothermal transformation of eutectoid steel between 723oC and 550oC gives a
pearlitic microstructure. Rapid quenching above 723oC transforms the austenitic condition into
martensitic condition.
11. Hot-quenching of eutectoid steels in austenitic condition results in formation of
___________
a) Pearlite
b) Bainite
c) Ferrite
d) Cementite
.
Answer: b
Explanation: If eutectoid steels in austenitic condition are hot-quenched in a 550oC to 250oC
range, they are isothermally transformed into bainite. This bainite formed is an intermediate
structure between pearlite and martensite.
12. Bainite in iron-carbon alloys has a ___________ structure.
a) Dendritic
b) Non-lamellar
c) Linear
d) Hexahedral
.
Answer: b
Explanation: Bainite in iron-carbon alloys can be defined as an austenitic decomposition product.
It has a non-lamellar eutectoid structure containing α ferrite and cementite (Fe3C).
13. Lower bainite is formed at ___________ temperature range.
a) 750-550oC
b) 550-350oC
c) 350-250oC
d) 250-150oC
.
Answer: c
Explanation: For eutectoid steels, bainite exists in two forms known as upper and lower bainite.
Upper bainite is formed by the isothermal transformation between 550-350oC. Lower bainite is
formed between 350-250oC which has much finer cementite particles.
14. Which of the following factors do not affect the critical cooling rate?
a) Chemical composition
b) Hardening temperature
c) Number or nature of grains
d) Purity of steel
.
Answer: c
Explanation: The slowest rate of cooling of austenite that results in 100% martensite
transformation is called a critical cooling rate. This depends on the chemical composition,
hardening temperature, and metallurgical nature (purity) of steel

1. Which of the following factors does not influence the variety and quality of metal?
a) Rate of heating and cooling
b) Quenching medium
c) Furnace
d) Grain size
.
Answer: d
Explanation: The variety of metal and various metallurgical processes depend upon the method
and rate of heating and cooling, furnaces used, and quenching medium. Grain size is one of the
effects of heat treatment, not a cause for any change.
2. How does the rate of cooling affect the hardness of the metal?
a) Slow cooling, hard material
b) Slow cooking, soft material
c) Rapid cooling, soft material
d) No effective change
.
Answer: b
Explanation: The rate of cooling is the controlling factor in developing either a hard or soft
structure. Rapid cooling from critical range results in a hard structure, whereas very slow cooling
gives a soft structure.
3. Which of the following is not a stage of annealing?
a) Heating
b) Soaking
c) Tempering
d) Quenching
.
Answer: c
Explanation: Annealing is defined as the softening process which involves heating the material at
an elevated temperature and then slowly cooling it. The annealing cycle consists of three stages
of heating to the desired temperature, holding at that temperature (soaking), and then slowly
cooling (quenching).
4. Full annealing is applied to which kind of materials?
a) Steel castings
b) Steel wires
c) High carbon steels
d) Sheet products
.
Answer: a
Explanation: When we refer to annealing, usually we only talk about full annealing. This method
is used to soften the material, to refine crystalline structure, and relieve stresses. This method is
applied to steel castings and steel ingots.
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5. For full annealing of hypoeutectoid steels, they are heated in a range above __________
a) 153-250oC
b) 273-350oC
c) 551-770oC
d) 723-910oC
.
Answer: d
Explanation: Hypoeutectoid steels contain carbon content which is less than 0.77%. For full
annealing of this steel, it is heated 30-60oC above the A3 line. It is held at this temperature for a
period of time, and then slowly cooled to room temperature.
6. Cooling of hypoeutectoid steels in done is furnace by decreasing the temperature to at least
________ below the A1 line.
a) 4oC
b) 10oC
c) 16oC
d) 30oC
.
Answer: d
Explanation: For full annealing of this steel, it is heated 30-60oC above the A3 line. Then it is
cooled within the furnace by decreasing the temperature 10-30oC per hour, to at least 30oC below
A1 line.
7. What is the result of full annealing of hypoeutectoid steels?
a) Coarse pearlite
b) Bainite
c) Cementite
d) Silicon
.
Answer: a
Explanation: Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace.
Then it removed from the furnace and then cooled at room temperature. This results in coarse
pearlite with excess ferrite.
8. How does heating for hypereutectoid steels differ from hypoeutectoid steels?
a) Heated 30-60oC above A1 line
b) Heated 30-60oC below A1 line
c) Heated 30-60oC below A2 line
d) Heated 30-60oC below A3 line
.
Answer: a
Explanation: Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace.
On the other hand, hypereutectoid steels are heated 30-60oC above the A1 line and then cooled
similar to hypoeutectoid steels.
9. Heating of hypereutectoid steels results in formation of coarse pearlite with excess
__________
a) Ferrite
b) Bainite
c) Cementite
d) Austenite
.
Answer: c
Explanation: Similar to hypoeutectoid steels, hypereutectoid steels are heated 30-60oC above the
A1 line. This results in coarse pearlite with excess cementite in a dispersed spheroidal form. This
improves mechanical properties, ductility, and toughness.

1. Process annealing is applied to which kind of materials?

a) Steel castings
b) Low carbon steels
c) High carbon steels
d) Medium carbon steels
.
Answer: b
Explanation: Process annealing is a method used to soften and increase the ductility of a
previously strain-hardened metal. It is mainly used in case of steel wires and sheet products,
especially low carbon steels.
2. For process annealing, the steel is heated at a temperature range of _________
a) 550-650oC
b) 723-910oC
c) 723-1138oC
d) 750-800oC
.
Answer: a
Explanation: In process annealing, the low carbon steels are heated to a temperature slightly
below the A1 line, which ranges between 550oC and 650oC. Full annealing is carried out at 723-
910oC (hypoeutectoid) or 723-1138oC (hypereutectoid).
3. Which among the following is not applicable to process annealing of steels?
a) Low scaling
b) Rapid
c) Expensive
d) Single phase morphology
.
Answer: c
Explanation: Process annealed steels are heated around 550-650oC, held for some time for
softening, then cooled at any desired rate. This process induces a single phase morphology. This
process is cheaper, more rapid, and produces less scaling as the material is not heated at a high
temperature like for full annealing.
4. What happens when internal residual stresses are not removed?
a) Coarse structure
b) Reduction of grain size
c) Distortion
d) Recrystallization
.
Answer: c
Explanation: When internal residual stresses are not removed, distortion or warpage of the
material may occur. In order to remove these stresses, stress relief annealing is carried out.
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5. Subcritical annealing is another name for ____________

a) Process annealing
b) Stress relief annealing
c) Recrystallization annealing
d) Spheroidise annealing
.
Answer: a
Explanation: Process annealed steels are heated around 550-650oC, held for some time for
softening, then cooled at any desired rate. This process induces a single phase morphology. This
process is otherwise known as subcritical annealing.
6. Commercial annealing is another name for ____________
a) Process annealing
b) Stress relief annealing
c) Recrystallization annealing
d) Spheroidise annealing
.
Answer: b
Explanation: When internal residual stresses are not removed, distortion or warpage of the
material may occur. Stress relief annealing is carried out to remove these stresses caused by
castings, quenching, machining, welding etc. Stress relief annealing is otherwise also known as
commercial annealing.
7. Removal of internal residual stresses at low temperatures is known as ________
a) Recrystallization
b) Recovery
c) Morphology
d) Phase transformation
.
Answer: b
Explanation: Recovery is a low-temperature phenomenon used to remove internal residual
stresses. This is done with little change in mechanical properties and no major changes in
microstructure. Stress relief is also known as recovery.
8. What kind of annealing does this graph illustrate?

a) Process annealing
b) Stress relief annealing
c) Recrystallization annealing
d) Spheroidise annealing
.
Answer: d
Explanation: Stress relief annealing is carried out to remove the stresses caused by castings,
quenching, machining, welding etc. This can be done by prolonged heating of the material
slightly above and below the A1 line. Full annealing occurs above the A1 and A3 lines, whereas
process annealing occurs slightly below the A1 line.

1. Which of the following factors does not influence the variety and quality of metal?
a) Rate of heating and cooling
b) Quenching medium
c) Furnace
d) Grain size
.
Answer: d
Explanation: The variety of metal and various metallurgical processes depend upon the method
and rate of heating and cooling, furnaces used, and quenching medium. Grain size is one of the
effects of heat treatment, not a cause for any change.
2. How does the rate of cooling affect the hardness of the metal?
a) Slow cooling, hard material
b) Slow cooking, soft material
c) Rapid cooling, soft material
d) No effective change
.
Answer: b
Explanation: The rate of cooling is the controlling factor in developing either a hard or soft
structure. Rapid cooling from critical range results in a hard structure, whereas very slow cooling
gives a soft structure.
3. Which of the following is not a stage of annealing?
a) Heating
b) Soaking
c) Tempering
d) Quenching
.
Answer: c
Explanation: Annealing is defined as the softening process which involves heating the material at
an elevated temperature and then slowly cooling it. The annealing cycle consists of three stages
of heating to the desired temperature, holding at that temperature (soaking), and then slowly
cooling (quenching).
4. Full annealing is applied to which kind of materials?
a) Steel castings
b) Steel wires
c) High carbon steels
d) Sheet products
.
Answer: a
Explanation: When we refer to annealing, usually we only talk about full annealing. This method
is used to soften the material, to refine crystalline structure, and relieve stresses. This method is
applied to steel castings and steel ingots.
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5. For full annealing of hypoeutectoid steels, they are heated in a range above __________
a) 153-250oC
b) 273-350oC
c) 551-770oC
d) 723-910oC
.
Answer: d
Explanation: Hypoeutectoid steels contain carbon content which is less than 0.77%. For full
annealing of this steel, it is heated 30-60oC above the A3 line. It is held at this temperature for a
period of time, and then slowly cooled to room temperature.
6. Cooling of hypoeutectoid steels in done is furnace by decreasing the temperature to at least
________ below the A1 line.
a) 4oC
b) 10oC
c) 16oC
d) 30oC
.
Answer: d
Explanation: For full annealing of this steel, it is heated 30-60oC above the A3 line. Then it is
cooled within the furnace by decreasing the temperature 10-30oC per hour, to at least 30oC below
A1 line.
7. What is the result of full annealing of hypoeutectoid steels?
a) Coarse pearlite
b) Bainite
c) Cementite
d) Silicon
.
Answer: a
Explanation: Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace.
Then it removed from the furnace and then cooled at room temperature. This results in coarse
pearlite with excess ferrite.
8. How does heating for hypereutectoid steels differ from hypoeutectoid steels?
a) Heated 30-60oC above A1 line
b) Heated 30-60oC below A1 line
c) Heated 30-60oC below A2 line
d) Heated 30-60oC below A3 line
.
Answer: a
Explanation: Hypoeutectoid steels are heated above the A3 line and then cooled in the furnace.
On the other hand, hypereutectoid steels are heated 30-60oC above the A1 line and then cooled
similar to hypoeutectoid steels.
9. Heating of hypereutectoid steels results in formation of coarse pearlite with excess
__________
a) Ferrite
b) Bainite
c) Cementite
d) Austenite
.
Answer: c
Explanation: Similar to hypoeutectoid steels, hypereutectoid steels are heated 30-60oC above the
A1 line. This results in coarse pearlite with excess cementite in a dispersed spheroidal form. This
improves mechanical properties, ductility, and toughness.

1. How is cooling of the material done is normalising process?

a) Furnace
b) Cooling tower
c) Still air
d) Liquid chamber
.
Answer: c
Explanation: Normalising is a process similar to annealing, which helps to refine grain structure,
increase strength, and relieve stresses and other properties. Cooling for annealing is done in a
furnace, whereas it is done in still air for normalising.
2. Normalising is best used for is what kind of materials?
a) Steel castings
b) Steel wires
c) High carbon steels
d) Low and medium carbon steels
.
Answer: d
Explanation: Normalising is a process similar to annealing, where the material is subjected to
temperature changes to alter their properties. This process is particularly employed for low and
medium carbon steels, as well as alloy steels.
3. For normalising, the steel is heated ___________ its upper critical temperature.
a) 30-40oC above
b) 30-40oC C below
c) 50-60oC C above
d) 50-60oC C below

Answer: c
Explanation: For hypoeutectoid steels, normalising is done by heating the steel 50-60oC C above
its upper critical temperature (A3 line). For hypereutectoid steels, it is heated above the Acm
line. The steel is held at that temperature for some time and then allowed to cool in still air.
4. Which of the following is not applicable for normalising process?
a) Economical
b) Time-consuming
c) Fine grain structure
d) Variable properties

Answer: b
Explanation: Normalising is carried out by heating of the steel at higher temperatures (compared
to annealing) and then cooling it in still air. Since cooling is different at different locations, their
properties will also vary. It also provides a fine grain structure and is less time-consuming.
5. The structure obtained by normalising depends on __________
a) Stresses induced
b) Toughness of material
c) Thickness of cross section
d) Strength of weld

Answer: c
Explanation: The type of structure obtained by normalising largely depends on the thickness of
the cross-section of the material, as this also affects the rate of cooling. Thin sections tend to give
a finer grain structure than thick sections.
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6. Microstructure containing ferrite is termed as _________

a) Dark area
b) White network
c) Red area
d) Blue zone
.
Answer: b
Explanation: Normalising produces microstructure containing ferrite (white network) and
pearlite (dark area) for hypoeutectoid steels. For eutectoid steels, the microstructure is only
pearlite. Meanwhile, for hypereutectoid steels, it is a pearlite and cementite microstructure.
7. Which among the following media of quenching the slowest?
a) Caustic soda
b) Sodium chloride
c) Mineral oil
d) Air
.
Answer: d
Explanation: The rate of cooling heavily depends on the quenching medium used. 5-10% caustic
soda and 5-20% NaCl are the two most popular media in terms of severity, followed by cold and
warm water. This is followed by mineral, animal, and vegetable oils. Finally, air has the lowest
severity of the preferred media.
8. Which of the following is not a preferred vegetable oil for quenching medium?
a) Linseed
b) Brine
c) Rapeseed
d) Cottonseed
.
Answer: b
Explanation: Oils are generally more effective as quenching media as compared to air. In
decreasing order, mineral oil, animal oil, and vegetable oil are widely used. The commonly used
vegetable oils are linseed, cottonseed, and rapeseed.
9. Mineral oils are used in the hardening process of ____________
a) Pipes
b) Carbon
c) Alloy steel
d) Heavy forging
.
Answer: c
Explanation: Mineral oils are obtained during the refining of crude petroleum. These are widely
used as quenching media due to their severity of quench. Mineral oils are used in hardening alloy
steels.
10. Which quenching medium is used for quenching of carbon and low alloy steels?
a) Vegetable oil
b) Water
c) Air
d) Animal oil
.
Answer: b
Explanation: Water produces the most severe quench followed by oils and air. For the quenching
of carbon and low alloy steels, water or an aqueous solution of NaOH or NaCl is used.
11. Which stage of quenching is the slowest?
a) Vapour-Jacket
b) Vapour-Transport cooling
c) Liquid-Cooling
d) They are all equally slow
.
Answer: c
Explanation: In general, both vapor-jacket and liquid-cooling are slow cooking stages. However,
liquid-cooling is the slowest as all the heat transfer occurs through conduction across the solid-
liquid interface.
12. Vapour-jacket cooling stage of quenching process occurs _________
a) Below boiling point
b) Above boiling point
c) Below melting point
d) At recrystallization temperature
.
Answer: b
Explanation: The vapor-jacket stage of quenching is a slow cooling stage since all the heat is
transported through a gas by conduction and radiation. This stage occurs when the metal is above
the boiling point of the quenching

1. Non-equilibrium phases are shown for their time and transformation using _________
a) Fe-Fe3C diagram
b) TTT diagram
c) CCT diagram
d) TTT and CCT diagram
.
Answer: d
Explanation: Ferrite, cementite, pearlite, and austenite are equilibrium phases which are based on
the iron-iron carbide equilibrium diagram. For other non-equilibrium phases like martensite and
bainite, the Fe-Fe3C diagram cannot be used. In such cases, we use the TTT and CCT diagrams.
2. The CCT or the TTT diagrams are used for _________
a) One steel of specific composition
b) A family of various steels
c) Alloy system of various compositions
d) Combination of all alloys and steels with various compositions
.
Answer: a
Explanation: An individual equilibrium diagram like the Fe-Fe3C can be used to depict an entire
alloy system of various compositions. However, a TTT or a CCT diagram can only be used for
one steel of a specified composition.
3. What does CCT diagram stand for?
a) Constant-critical-temperature
b) Constant-cooling-temperature
c) Continuous-cooling-transformation
d) Continuous-creep-transformation
.
Answer: c
Explanation: Martensite and bainite are non-equilibrium phases which cannot be depicted on a
Fe-Fe3C equilibrium diagram. In such cases, either the time-temperature-transformation or the
continuous-cooling-transformation diagrams can be used.
4. __________ is used to predict quenching reactions in steels.
a) Isothermal transformation diagram
b) Iron-iron carbide equilibrium diagram
c) Continuous cooling transformation diagram
d) Logarithm scale
.
Answer: a
Explanation: An isothermal transformation (IT) is a tool used by heat treaters to calculate
quenching reactions in steels. The IT diagram can also be called as a time-temperature-
transformation diagram.
5. Which of the following is not an alternative name for the TTT diagram?
a) S curve
b) C curve
c) Adiabatic curve
d) Bain’s curve
.
Answer: c
Explanation: Owing to the shape of the diagram, TTT diagrams may be called S curves or C
curves. Due to their nature and type of work, they may also be known as isothermal
transformation diagrams or Bain’s curves.
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6. The first step in constructing a TTT diagram involves _________ the sample.
a) Annealing
b) Normalising
c) Quenching
d) Austenising
.
Answer: d
Explanation: To construct a TTT diagram, a large number of the small specimen are collected
and austenised in a furnace. Next, heat treatment and quenching stages are carried out. After each
stage, the temperature and time are plotted as curves.
7. Austenising of samples for TTT diagram is done __________ temperature.
a) At room temperature
b) Below melting point
c) Above eutectoid temperature
d) Above boiling temperature
.
Answer: c
Explanation: A large number of small samples of the same material are collected and austenized
in a furnace above the eutectoid temperature. Then it is rapidly cooled at a desired temperature
below the eutectoid temperature.
8. Examination of transformation time after quenching is done ___________
a) At room temperature
b) Below melting point
c) Above eutectoid temperature
d) Above boiling temperature
.
Answer: a
Explanation: Austenising of the sample is done above eutectoid temperature, whereas quenching
is done below the eutectoid temperature. After each transformation time, the microstructure is
examined at room temperature and plotted in the form of curves.
9. Isothermal transformations of eutectoid steel between 723oC and 550oC produces
__________ microstructure.
a) Pearlite
b) Bainite
c) Ferrite
d) Cementite
.
Answer: a
Explanation: Isothermal transformations of eutectoid steel between 723oC and 550oC bring into
being a pearlitic microstructure. The pearlite changes from coarse structure to fine structure as
the transformation is decreased in this range.
10. Rapid quenching of eutectoid steel ___________ transforms the austenite into martensite.
a) At room temperature
b) Below 320oC
c) At 550oC
d) Above 723oC
.
Answer: d
Explanation: An isothermal transformation of eutectoid steel between 723oC and 550oC gives a
pearlitic microstructure. Rapid quenching above 723oC transforms the austenitic condition into
martensitic condition.
11. Hot-quenching of eutectoid steels in austenitic condition results in formation of
___________
a) Pearlite
b) Bainite
c) Ferrite
d) Cementite
.
Answer: b
Explanation: If eutectoid steels in austenitic condition are hot-quenched in a 550oC to 250oC
range, they are isothermally transformed into bainite. This bainite formed is an intermediate
structure between pearlite and martensite.
12. Bainite in iron-carbon alloys has a ___________ structure.
a) Dendritic
b) Non-lamellar
c) Linear
d) Hexahedral
.
Answer: b
Explanation: Bainite in iron-carbon alloys can be defined as an austenitic decomposition product.
It has a non-lamellar eutectoid structure containing α ferrite and cementite (Fe3C).
13. Lower bainite is formed at ___________ temperature range.
a) 750-550oC
b) 550-350oC
c) 350-250oC
d) 250-150oC
.
Answer: c
Explanation: For eutectoid steels, bainite exists in two forms known as upper and lower bainite.
Upper bainite is formed by the isothermal transformation between 550-350oC. Lower bainite is
formed between 350-250oC which has much finer cementite particles.
14. Which of the following factors do not affect the critical cooling rate?
a) Chemical composition
b) Hardening temperature
c) Number or nature of grains
d) Purity of steel
.
Answer: c
Explanation: The slowest rate of cooling of austenite that results in 100% martensite
transformation is called a critical cooling rate. This depends on the chemical composition,
hardening temperature, and metallurgical nature (purity) of steel.

1. For hardening of steel by quenching, the steel is cooled in __________

a) Furnace
b) Still air
c) Oil bath
d) Cooling tower
.
Answer: c
Explanation: After heating and soaking of the steel, it must be properly cooled. The steel is
quenched to room temperature in a water or oil bath.
2. The cooling rate must be _________ the critical cooling rate for hardening of steel by
quenching.
a) Higher than
b) Lower than
c) Equal to
d) Half of
.
Answer: a
Explanation: Post heating and soaking, the steels must be cooled. The cooling rate must be
higher than the critical cooling rate to get the completely martensitic structure. The steel is
quenched to room temperature with the help of a water bath or oil bath.
3. Phase transformation during hardening transforms _________
a) BCC to FCC
b) FCC to BCT
c) BCT to HCP
d) FCC to HCP
.
Answer: b
Explanation: Due to rapid cooking, austenite is supercooled by nearly 500oC. The large force
helps convert the FCC into BCT structure. The resulting structure is called martensite.
4. The slip does not occur in martensite due to the presence of _______ in the lattice.
a) Silicon
b) Germanium
c) Carbon
d) Tin
.
Answer: c
Explanation: Martensite is a super-saturated solution of carbon in α-iron. Due to the presence of
carbon in the lattice, slip does not occur. As a result, martensite is strong, hard, and brittle.
5. The hardening process is carried out on ________ steel.
a) No carbon
b) Low carbon
c) Medium carbon
d) High carbon
.
Answer: d
Explanation: As the carbon content increases, the hardness also increases. Due to this, the
hardening process is carried out on high carbon steels containing 0.35-0.50% C.
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6. How does the rate of cooling affect the hardness of steel?
a) Faster cooling results in low hardness
b) Slow cooling results in high hardness
c) Fast cooling results in high hardness
d) No change is found
.
Answer: c
Explanation: We know that hardness depends on the nature and properties of the quenching
medium. It was found that faster cooling resulted in greater hardness of the steel, and slow
cooling lowers the hardness.
7. How does the size of the specimen affect the hardness of steel?
a) Smaller size results are high hardness
b) Smaller size results in low hardness
c) Larger size results in high hardness
d) No change is found
.
Answer: a
Explanation: The size of the specimen also greatly affects the hardness of the steel during the
hardening process. As the size of the specimen increases, its hardness decreases. For example,
hardness with a 50 mm diameter steel bar will be higher than a similar one of 100 mm diameter.
8. ___________ is defined as the ease of forming martensite.
a) Hardness
b) Hardenability
c) Toughness
d) Strength
.
Answer: b
Explanation: Hardenability is defined as a measure or ease with which hardness is achieved. It
can also be said that hardenability is the ease with which martensite is formed. Hardness, on the
other hand, is the ability of a metal to resist abrasion, indentation, and scratching.
9. Hardenability of a material can be measured using __________ test.
a) Jominy end-quench
b) Charpy
c) Rockwell
d) Izod
.
Answer: a
Explanation: Hardenability is the measure of inclination of a material to achieve hardness. It is
affected by the alloying elements in the material and grain sizes. The hardenability of the
material can be measured using the Jominy end-quench test method. Charpy and Izod impact
testing methods, whereas Rockwell is a hardness testing scale.
10. Which of the following factors affect the hardenability of a material?
a) Composition of steel
b) Grain size
c) Temperature of specimen
d) Quenching medium
.
Answer: c
Explanation: The hardenability of the material can be measured using the Jominy end-quench
test method. The composition of steel, austenitic grain size, structure before quenching, and
quenching medium and method affect the hardenability of the steel.
11. Quenching of the sample in Jominy end-quench method is done at _______
a) 0oC
b) 25oC
c) 125oC
d) 200oC
.
Answer: b
Explanation: The given sample of material is heated above the upper critical temperature and
dropped into position in the frame of the apparatus. It is then quenched on one end at 25oC to
achieve different rates of cooling along the length of the sample.
12. Hardness readings are taken every ________ after quenching in Jominy end-quench test.
a) 0.01 mm
b) 0.1 mm
c) 1.6 mm
d) 2.5 mm
.
Answer: c
Explanation: The given sample of material is heated above the upper critical temperature. It is
then quenched at one end at 25oC. After cooling, a flat is ground along the length of the bar.
Rockwell C hardness readings are then taken for every 1.6 mm along the length from the
quenched end, which are then plotted in the form of a graph.

1. Which of the following is not a result of tempering?

a) Increased ductility
b) Improved toughness
c) Increased electrical conductivity
d) Internal stresses are relieved
.
Answer: c
Explanation: Martensite formed during the hardening process is too brittle and lacks toughness
and ductility. This makes it unusable for many applications. Therefore, a heat treatment process
called tempering is done to bring about these changes.
2. Tempering of martensite steel is done _________
a) Below eutectic temperature
b) Below eutectoid transformation temperature
c) At room temperature
d) At 1000oC
.
Answer: b
Explanation: Tempering is a process of heating martensite steel below the eutectoid
transformation temperature (250-650oC). It is held there for some time and slowly cooled to
room temperature.
3. Internal stresses in martensite steel are relieved by heating at __________
a) 200oC
b) 500oC
c) 800oC
d) 1000oC
.
Answer: a
Explanation: Tempering usually involves heating the martensite steel below the eutectoid
transformation temperature (250-650oC). When this martensite steel is heated at temperatures as
low as 200oC, internal stresses are relieved.
4. What kind of transformation occurs during tempering of martensite steel?
a) BCC to FCC
b) BCT to HCP
c) FCC to BCT
d) BCT to α+Fe3C
.
Answer: d
Explanation: Tempering is a process of heating martensite steel below the eutectoid
transformation temperature. This heat treatment process transforms martensite (BCT, single
phase) into tempered martensite (α+Fe3C) by the diffusional process.
5. How is the tensile strength of the material affected due to tempering?
a) Increases with an increase in temperature
b) Decreases with an increase in temperature
c) Increases constantly
d) Does not change
.
Answer: b
Explanation: We know that hardness of the material decreases with an increase in tempering
time. Similarly, the tensile and yield strengths of the material also decrease with increase in
tempering temperature.
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Java Tricky Program 23 - Main method signature

6. How is the ductility of the material affected due to tempering?

a) Increases with an increase in temperature
b) Decreases with an increase in temperature
c) Increases constantly
d) Does not change
.
Answer: a
Explanation: The hardness and tensile and yield strengths of the material decrease with
increasing tempering temperature. On the other hand, however, it is noted that the ductility of the
material increases with an increase in tempering temperature.
7. Low temperature tempering occurs at __________
a) 50-150oC
b) 150-250oC
c) 350-450oC
d) 500-650oC
.
Answer: b
Explanation: Tempering temperature is classified into three classes. Low temperature tempering
occurs at 150-250oC. Medium and high temperature tempering occur at 350-450oC and 500-
650oC correspondingly.
8. Medium temperature tempering develops a ________ structure.
a) Non-lamellar
b) Sorbite
c) Tempered-troostite
d) Bainite
.
Answer: c
Explanation: Medium temperature tempering is usually carried out in a temperature range
between 350oC and 450oC. This process develops a tempered-troostite structure.
9. Tempering that results in a reduction of toughness is known as __________
a) Caustic embrittlement
b) Temper embrittlement
c) End-temper
d) Quench-temper embrittlement
.
Answer: b
Explanation: Tempering of some steels and alloys may result in the reduction of toughness. This
phenomenon is known as temper embrittlement.
10. High temperature tempering develops a ________ structure.
a) Non-lamellar
b) Sorbite
c) Tempered-troostite
d) Pearlite
.
Answer: b
Explanation: High-temperature tempering is usually carried out in a temperature range between
500oC and 650oC. This process develops a sorbite structure. This is used to completely eliminate
internal stresses in the material.
11. Temper embrittlement occurs at a temperature __________
a) Below 0oC
b) Above 250oC
c) Above 575oC
d) Above 1000oC
.
Answer: c
Explanation: Tempering of some steels and alloys may result in the reduction of toughness. This
phenomenon is known as temper embrittlement. Temper embrittlement occurs when the steel is
tempered above 575oC which is followed by slow cooling to room temperature.
12. Which tempering process is used to increase the endurance and elastic limit of the material?
a) Low-temperature tempering
b) Medium temperature tempering
c) High-temperature tempering
d) Endurance limit cannot be increased
.
Answer: b
Explanation: Medium temperature tempering is performed between 350oC and 450oC. This
process develops a tempered-troostite structure and increases the endurance and elastic limits of
the material.
13. What are the applications of medium temperature tempering?
a) Spring and die steels
b) Structural steels
c) Low alloy steels
d) Surface hardened regions
.
Answer: a
Explanation: Medium temperature tempering is applied on materials like spring steels and die
steels. Likewise, high-temperature tempering is used on structural steels. Low-temperature
tempering is applied to cutting and measuring tools of carbon and low alloy steels, as well as for
surface-hardened and case-carburized parts.

1. Quenching of austenite steel for martempering is done using __________

a) Furnace
b) Still air
c) Cooling tower
d) Hot oil
.
Answer: d
Explanation: For martempering, the steel is first heated above its critical range to make it
austenitic. Then it is quenched in hot oil or molten salt above martensite start temperature.
2. How can temperature differences between center and surface be prevented?
a) Refined grain structure
b) Cooling to room temperature
c) Hot working
d) Addition of silicon
.
Answer: b
Explanation: Cooling the steel at moderate to room temperature prevents temperature differences
between center and surface. This is usually done in air. The resulting microstructure of
martempered steel is untempered martensite.
3. Martempering process is usually employed in _________
a) Alloy steels
b) Cast irons
c) Composites
d) Ceramics
.
Answer: a
Explanation: Martempering is an interrupted quenching process which eliminates some
disadvantages of rapid cooling. It is used to minimize stresses, distortion, and cracking of steels.
This process is usually used in alloy steels.
4. Martempering is otherwise known as __________
a) Interrupted quenching
b) Marquenching
c) Austempering
d) Isothermal quenching
.
Answer: b
Explanation: Interrupted quenching is a technique of eliminating the disadvantages caused by
rapid cooling. This consists of two methods, martempering and austempering. Martempering is
also known as marquenching, and austempering is also known as isothermal quenching.
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5. Austempering forms a _________ structure.
a) Pearlite
b) Bainite
c) Spherodite
d) Cementite
.
Answer: b
Explanation: Austempering is an isothermal heat treatment process used to reduce distortion
caused by quenching and to make tough and strong steels. It forms a bainite structure and is
otherwise known as isothermal quenching.
6. It is necessary to carry out _________ after martempering.
a) Refining
b) Tempering
c) Surface hardening
d) Cyaniding
.
Answer: b
Explanation: Untempered martensite structure is transformed into tempered martensite structure
by conventional tempering heat treatment processing rapid quenching. This is rarely needed in
austempering.
7. Which of the following is not a disadvantage of austempering?
a) Needs to be cold worked
b) Needs special bath
c) Can be used for limited steels
d) Can only be used for small sections
.
Answer: a
Explanation: Austempering experiences a number of disadvantages compared to quenching and
tempering. It needs a special molten salt bath and can be used only for a limited number of steels.
Furthermore, only small sections up to 9 mm thickness can be used.
8. Why are bigger sections not used in austempering?
a) Unable to cool
b) Does not fit in the working apparatus
c) Decreased strength
d) Cannot be cut
.
Answer: a
Explanation: Big sections cannot be used for austempering as they cannot be cooled rapidly to
avoid the formation of pearlite. As a result, only small sections up to 9 mm thickness are suitable
for this operation.

1. The treatment of steel to get a stronger casing while maintaining a soft core is called
_________
a) Surface hardening
b) Tempering
c) Sintering
d) Surface lining
.
Answer: a
Explanation: In many applications, it is required for the surface to be harder while the core of the
material remains soft. The treatment of steels to achieve this is known as surface heat treatment
or surface hardening.
2. Which of the following is not a diffusion method of surface heat treatment?
a) Carburizing
b) Cyaniding
c) Induction hardening
d) Carbonitriding
.
Answer: c
Explanation: The treatment of steels to achieve a hard casing and softer core is known as surface
heat treatment or surface hardening. The diffusion methods of surface heat treatment are
classified as carburizing, cyaniding, nitriding, and carbonitriding. Thermal methods include
flame hardening and induction hardening.
3. Steels containing ________ carbon are used for carburizing.
a) 0.1-0.2%
b) 0.2-0.35%
c) 0.4-0.6%
d) 0.7-0.9%
.
Answer: a
Explanation: Carburizing the process by which carbon atoms are introduced onto the surface of
low-carbon steels to produce a hard casing while the core remains soft. Steels used in carburizing
contain carbon content of 0.10-0.20%.
4. Which of the following is not a method of carburizing?
a) Pack carburizing
b) Gas carburizing
c) Cyaniding
d) Nitriding
.
Answer: d
Explanation: Carburizing the process by which carbon atoms are introduced onto the surface of
steels to produce a hard surface while the interior remains soft. When a piece of low-carbon steel
is placed in a carbon saturated temperature, the carbon diffuses into the steel and causes
carburizing. Pack carburizing, gas carburizing, and liquid carburizing (cyaniding) are three
methods of carburizing.
5. The energizer in pack carburizing makes up ________ of the total composition.
a) 10%
b) 25%
c) 40%
d) 80%
.
Answer: c
Explanation: The carburizing mixture contains a carbon-rich material along with an energizer
making up 40% of the total composition. The energizer is composed of soda ash and barium
carbonate. It is used to accelerate the carburizing process.
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6. What kind of carbon-rich material is used in the carburizing mixture?
a) Charred leather
b) Sodium carbonate
c) Barium carbonate
d) Carbon black
.
Answer: a
Explanation: The carburizing mixture contains a carbon-rich material like charcoal or charred
leather, along with an energizer. The energizer is composed of sodium carbonate (soda ash) and
barium carbonate.
7. What is the carburizing temperature?
a) 150-250oC
b) 400-450oC
c) 650-700oC
d) 900-950oC
.
Answer: d
Explanation: The components to be heat treated and the carburizing mixture are packed in steel
boxes and their lids are fixed on the boxes. This is then heat treated at 900-950oC and maintained
at this temperature for up to six hours. After carburizing, the components are quenched slowly in
the box.
8. Which of the following holds true for pack carburizing?
a) Efficient heating
b) Uniform temperature
c) Easy to handle
d) Not readily adoptable
.
Answer: d
Explanation: Pack carburizing is a process of carburizing which has several drawbacks. It has
inefficient heating, the temperature is not uniform, and is difficult to handle. Furthermore, pack
carburizing is not readily available to continuous operation.
9. Prevention of carburizing of components can be done by ___________
a) Adding filler
b) Electroplating
c) Cold working
d) Adding water
.
Answer: b
Explanation: Sometimes it is desired that a few parts of the surface of the component are
prevented from carburizing. In such cases, electroplating the area with copper achieves the
desired result. The thickness of electroplating with copper is usually between 0.07 mm to 0.1
mm.
10. Gas carburizing differs from pack carburizing due to the use of ________
a) Silicon-providing gas
b) Mercury-providing gas
c) Carbon-providing gas
d) Argon
.
Answer: c
Explanation: Gas carburizing overcomes the drawbacks of pack carburizing by replacing the
solid carburizing mixture with a carbon-providing gas. In general, natural gas, propane, or
generated gas atmospheres are used in these cases.
11. The molten baths used in liquid carburizing are composed of _________
a) 20-50% sodium cyanide, 40% sodium carbonate
b) 40% sodium cyanide, 20-50% sodium carbonate
c) 10% sodium cyanide, 20% sodium carbonate
d) 20% sodium cyanide, 10% sodium carbonate
.
Answer: a
Explanation: Liquid carburizing is carried out in molten salt baths. These baths contain 20-50%
sodium cyanide, 40% sodium carbonate, along with varying quantities of sodium or barium
chloride. It is also known as salt carburizing.
12. Liquid carburizing temperature range lies between ________
a) 120-250oC
b) 300-375oC
c) 425-660oC
d) 870-950oC
.
Answer: d
Explanation: Liquid carburizing is carried out in baths of molten salt containing 20-50% sodium
cyanide, 40% sodium carbonate, and with varying quantities of sodium or barium chloride. The
cyanide-rich mixture is heated in iron pots to a temperature of 870-950oC. The workpiece is
immersed for a little longer than 5-minute periods according to the depth of the case required.
13. Liquid carburizing can be used for producing ________ cases.
a) 0.05-0.1 mm
b) 0.1-0.25 mm
c) 0.3-0.6 mm
d) 0.7-1.0 mm
.
Answer: b
Explanation: Liquid carburizing is carried out in baths of molten salt baths containing 20-50%
sodium cyanide, 40% sodium carbonate, and with varying quantities of sodium or barium
chloride. The cyanide-rich mixture is heated in iron pots to a temperature of 870-950oC. This
process is suitable for producing shallow cases of 0.1-0.25 mm.
14. Which of the following is not applicable for liquid carburizing?
a) Uniform temperature
b) Dirty work surface
c) Temperature controlled by pyrometers
d) High cost of carburizing
.
Answer: b
Explanation: Liquid carburizing takes place due to the decomposition of sodium cyanide at the
surface of the steel. It has major advantages that the temperature of the salt bath is uniform, the
surface of work remains cleans, and that the temperature can be accurately controlled by
pyrometers. It also has the disadvantage that the salt pots require batch processing and the cost of
carburizing salt is high

1. Nitriding involves the addition of ___________ for the hardening of surface.

a) Nitrogen
b) Nichrome
c) Neon
d) Niobium
.
Answer: a
Explanation: Nitriding is a diffusion process for surface hardening of steels. It involves the
addition of nitrogen atoms to obtain the desired hardness of steel components.
2. Which of the following is required for the nitriding process?
a) Carbon
b) Nitralloy
c) Niobium
d) Zirconium
.
Answer: b
Explanation: Nitriding differs from case-hardening as they contain nitrogen instead of carbon.
Nitriding requires special steels called ‘Nitralloy’. This is because hardness depends on the
formation of very hard compounds of nitrogen and metal such as aluminum, chromium, and
vanadium present in the steel.
3. Why is nitriding not used for plain carbon steels?
a) Formation of oxides
b) Formation of cracks
c) Formation of iron nitrides
d) Formation of spots
.
Answer: c
Explanation: Nitriding is generally not preferred for plain carbon steels. This is due to the
formation of iron nitrides formed to a considerable depth below the surface of the steel. This
causes embrittlement of the material.
4. What is necessary to be done before the nitriding process?
a) Heat Treatment
b) Washing the material
c) Chamfering of sample
d) Shaping of material
.
Answer: a
Explanation: Before nitriding is carried out, it is required that the workpieces are heat treated.
This is done in order to produce the required properties in the core. The parts are heat treated at
just about 500oC for several hours.
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5. How long should a steel component be heat treated before nitriding?
a) 5-20 hours
b) 40-100 hours
c) 100-200 hours
d) 300-500 hours
.
Answer: b
Explanation: Before nitriding is carried out, it is required that the workpieces are heat treated to
produce the required properties in the core. The parts are heat treated at around 500oC for a
period of 40 to 100 hours. This process takes place in a gas-tight chamber through which
ammonia gas is allowed to circulate.
6. Why is atmospheric nitrogen inappropriate for use in the nitriding process?
a) Cannot be heated
b) Insufficient density
c) Inability to be absorbed
d) Expensive process
.
Answer: c
Explanation: Ordinary atmospheric nitrogen exists in the form of molecules (N2). As a result,
these molecules cannot be absorbed by the steel, which is necessary for the nitriding process.
Therefore, atmospheric nitrogen is not suitable for nitriding process.
7. Which of the following is not applicable for nitriding process?
a) Surface formed is extremely hard
b) Minimized cracking and distortion
c) Expensive
d) High-temperature treatment
.
Answer: d
Explanation: Nitriding has an edge over case-hardening as the surface formed is extremely hard.
The treatment is conducted at comparatively low temperatures due to which the cracking and
distortion effects are minimized. Nitriding is generally an expensive process. However, it may be
cheaper when a large number of components are to be treated.
8. Cyaniding is usually done for _________
a) Plain carbon steels
b) Cast irons
c) Stainless steels
d) Pig iron
.
Answer: a
Explanation: Cyaniding is a process of hardening the surface of steel components through the
addition of nitrogen and carbon. It is done by immersing the workpiece is a cyanide bath.
Usually, plain carbon steels and alloying steels containing 0.2% carbon are hardened using this
process.
9. Cyaniding is carried out at a temperature of ___________
a) 175oC
b) 500oC
c) 950oC
d) 1300oC
.
Answer: c
Explanation: Cyaniding is a process of hardening the surface of steel components through the
addition of nitrogen and carbon. It is done by immersing the workpiece is a bath of molten
sodium cyanide and sodium carbonate. This process is conducted at a temperature of 950oC. This
causes the formation of hard iron nitrides which leads to surface hardening of the material.

1. Which of the following does not apply to cyaniding?

a) Requires less time
b) Operated at lower temperatures
c) Low corrosion resistance
d) Costly process
.
Answer: c
Explanation: The cyaniding process is an expensive process of surface hardening of steels which
is applied to produce automobile parts. It requires lesser time and lower temperatures than
carburizing. It also lowers warping and distortion while increasing the corrosion and wear
resistance.
2. Gas cyaniding is otherwise known as __________
a) Nitriding
b) Carbonitriding
c) Induction Hardening
d) Flame Hardening
.
Answer: b
Explanation: Carbonitriding is a surface hardening process that involves the diffusion of nitrogen
and carbon into the steel. Carbonitriding may also be called as gas-cyaniding or dry-cyaniding.
3. How long is the heating process for carbonitriding done for?
a) 1-2 hours
b) 2-10 hours
c) 12-20 hours
d) 24-30 hours
.
Answer: b
Explanation: The carbonitriding process is carried out in a gas atmosphere furnace using a
carburizing gas such as propane or methane mixed with ammonia. The workpiece is heated to
850oC in the mixture of gases for 2-10 hours followed by quenching and tempering.
4. At what temperature is the workpiece tempered in carbonitriding process?
a) 180oC
b) 360oC
c) 540oC
d) 720oC
.
Answer: a
Explanation: The carbonitriding process is carried out in a gas atmosphere furnace using a
carburizing gas.The workpiece is heated to 850oC in the mixture of gases for 2-10 hours. This is
followed by quenching to increase hardness, then by tempering at 180oC for reducing the
brittleness.
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5. What is the case depth obtained in the cyaniding process?
a) 0.38 mm
b) 0.25 mm
c) 0.5 mm
d) 1.25 mm
.
Answer: b
Explanation: Of all the diffusion surface hardening methods, carburizing provides the highest
cast depth of 1.38 mm. This is followed by carbonitriding, nitriding, and cyaniding at 0.5 mm,
0.38 mm, and 0.25 mm correspondingly.
6. What is the hardness achieved in the carbonitriding process?
a) HRC 35
b) HRC 65
c) HRC 105
d) HRC 140
.
Answer: b
Explanation: Carbonitriding process of surface hardening results in a case depth of 0.5 mm and a
hardness after heat treatment of HRC 65. Interestingly, carburizing and cyaniding also result in
equal hardness after heat treatment.
7. What are the applications of nitriding?
a) Gears, camshafts
b) Valve guides and seatings
c) Chain links, nuts, bolts, and screws
d) Gears, nuts, bolts
.
Answer: b
Explanation: Nitriding process provides an extreme hardness of the surface and a case depth of
0.38 mm. It is typically used for valve guides and seatings, and for gears.
8. Nitriding steels are tempered at about ________
a) 100-200oC
b) 300-450oC
c) 600-700oC
d) 800-950oC
.
Answer: c
Explanation: Nitralloy steels are commonly used in nitriding process. These steels are hardened
by oil quenching from 900oC and tempered at 600-700oCbefore nitriding

1. Flame hardening can only be performed on steels with a minimum of ______ carbon.
a) 0.4%
b) 0.8%
c) 1.2%
d) 1.8%
.
Answer: a
Explanation: Flame hardening is a selective hardening process with a combustible gas flame as
the source of heat for austenizing. This process can only be performed with steels with a high
carbon content of at least 0.4%. This is in order to allow hardening of the surface.
2. How is the heating of surface done in flame hardening technique?
a) Oxy-acetylene torch
b) Carburizing flame
c) Electrode
d) Direct sunlight
.
Answer: a
Explanation: Flame hardening is a selective hardening process with a combustible gas flame as
the source of heat for austenizing. The surface to be hardened is heated to a temperature above its
upper critical temperature. This is done using an oxy-acetylene torch.
3. Flame hardening technique is suitable for ____________
a) Plain carbon steels
b) Alloy steels
c) Cast irons
d) Pig iron
.
Answer: a
Explanation: Flame hardening is a selective hardening process with a combustible gas flame as
the source of heat for austenizing. It is more suitable for plain carbon steels having a carbon
composition of 0.4-0.95%. It is used to improve wear resistance and surface hardness of gears,
wheels, sheaves, shafts, mills, and other components.
4. Which of the following is an application for flame hardening?
a) Distortion occurs
b) Expensive equipment
c) Suitable for larger quantities
d) Overheating causes cracks
.
Answer: d
Explanation: The advantage of flame hardening lies in the fact that there is no distortion of
workpiece since only a small section of the steel is heated. This process is efficient and
economical as the cost of equipment is low, thereby making it more suitable for small quantities.
It has a disadvantage that overheating of the steel may cause cracks.
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5. How does induction hardening differ from flame hardening?
a) Electric current is used
b) Higher quenching time
c) Low-temperature operation
d) Results In fine grain structure
.
Answer: a
Explanation: The basic mechanism of both induction and flame hardening techniques remains
the same. In induction hardening, however, the heat source is derived from the resistance to
induced eddy currents.
6. What is the frequency used for depths of 2 to 3 mm in induction hardening?
a) 1200 Hz
b) 3000 Hz
c) 5400 Hz
d) 9600 Hz
.
Answer: d
Explanation: The depth to which the heating occurs varies inversely as the square root of the
frequency in case of induction hardening. This means that as higher frequencies are used, the
depths achieved are shallower. The typical frequencies used are 3000 Hz for 3-6 mm depth and
9600 Hz for 2-3 mm depth.
7. Which of the following is true for induction hardening?
a) Requires less time
b) Hardness and depth is difficult to control
c) Only external surfaces can be hardened
d) It gives a scaling effect
.
Answer: a
Explanation: Induction hardening has a major advantage that it requires very less time, typically
of the order of 10 seconds. It is easy to control the surface hardness and depth and can be
automated. Here, both external and internal surfaces can be hardened. Since it does not give any
scaling effect, the machining is reduced.
8. What is the hardness achieved by flame hardening?
a) HRC 30-40
b) HRC 50-60
c) HRC 80-100
d) HRC 120-140
.
Answer: b
Explanation: Both flame hardening and induction hardening experience similar results. The
hardened layer is about 3 mm thick and has a hardness of about HRC 50-60. Distortion can be
often minimized in both processes. Additionally, the surface remains clean in induction
hardening.
9. ___________ is a method of obtaining diverse properties by varying thermal histories of
various regions.
a) Selective Hardening
b) Carbonitriding
c) Annealing
d) Austenizing
.
Answer: a
Explanation: Selective hardening is a technique by which different properties are obtained by
varying thermal histories of various regions. This is otherwise considered as the thermal method
of surface hardening. Flame hardening and induction hardening are the two methods of selective
hardening.

UNIT 5

1. Pig iron is a product of ____________

a) Cupola
b) Bessemer converter
c) Open hearth furnace
d) Blast furnace
.
Answer: d
Explanation: Blast furnace when smelted iron ore in it, produces pig iron. Pig iron has a carbon
content, typically 3.5-4.5%, which makes it brittle, thus of less industrial use.
2. Cast iron is a product of ___________
a) Cupola
b) Bessemer converter
c) Open hearth furnace
d) Blast furnace
.
Answer: a
Explanation: The cupola furnace (a modified blast furnace) can be used to melt cast irons,
bronzes, etc. Cast irons have a carbon content of 2-4%, and have low melting temperatures,
which makes them easily castable.
3. Wrought iron is a product of ___________
a) Cupola
b) Bessemer converter
c) Puddling furnace
d) Blast furnace
.
Answer: c
Explanation: The puddling furnace creates wrought iron (nearly pure iron) from the pig iron. The
wrought iron is tougher and malleable.
4. Steel is a product of ___________
a) Cupola
b) Blast furnace
c) Puddling furnace
d) Bessemer converter
.
Answer: d
Explanation: The Bessemer converter was the first inexpensive furnace to make steel in olden
days. The fundamental idea is, removal of impurities from iron and making steel by oxidizing.
Open hearth furnace has taken over Bessemer process due to its nitrogen control in steel.
5. Red hardness of an alloy steel can be improved by adding ____________
a) Tungsten
b) Vanadium
c) Manganese
d) Titanium
.
Answer: a
Explanation: Red hardness means capability of material to retain hardness at high temperature. It
can be achieved for steel when alloyed high high melting metals like, tungsten, molybdenum,
vanadium and chromium, etc.
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6. Abrasion resistance of an alloy steel can be improved by adding ___________
a) Tungsten
b) Vanadium
c) Manganese
d) Chromium
.
Answer: d
Explanation: Chromium is generally added to steel to increase corrosion resistance and
oxidation, to increase hardenability, to improve high-temperature strength, and to improve
abrasion resistance in high-carbon compositions. The formation of hexagonal Cr7C3 is
responsible for this abrasion resistance to steel.
7. Wear resistance of an alloy steel can be improved by adding ___________
a) Tungsten
b) Vanadium
c) Manganese
d) Titanium
.
Answer: c
Explanation: Manganese is normally present in all commercial steels. High levels of manganese
presence produces an austenitic steel with improved wear and abrasion resistance.
8. Corrosion resistance of an alloy steel can be improved by adding ___________
a) Tungsten
b) Vanadium
c) Chromium
d) Titanium
.
Answer: c
Explanation: Chromium when added in the range 10.5%-18% in weight to steel, forms a passive
oxide layer (Cr2O3), thus transforming steel to a corrosion resistant steel (stainless steel).
9. Tensile strength of an alloy steel can be improved by adding ____________
a) Nickel
b) Vanadium
c) Manganese
d) Titanium
.
Answer: a
Explanation: Nickel can improve tensile strength as well as toughness of alloy steel. Small
additions of niobium (Nb) also increases the tensile strength of carbon steel.
10. Which of the following induces fine grain distribution in alloy steel?
a) Nickel
b) Vanadium
c) Manganese
d) Titanium
.
Answer: b
Explanation: After Al, vanadium if by far the mostly used grain refiner in steel. It forms a
microscopic precipitate particle in steel, which acts as pinning agents, thus obstruct the grain
growth at higher temperatures, encouraging new grains to nucleate.

1. Which of the following is the property of high carbon steel?

a) high toughness
b) reduced ductility
c) high strength
d) reduced strength
.
Answer: b
Explanation: High carbon steel contains high carbon content. Hence it has reduced ductility,
toughness and weldability.
2. High carbon steel is used in ______________
a) transmission lines and microwave towers
b) structural buildings
c) fire resistant buildings
d) for waterproofing
.
Answer: a
Explanation: High carbon steel is used in transmission lines and microwave towers where
relatively light members are joint by bolting.
3. What is the permissible percentage of micro-alloys in medium and high strength micro-alloyed
steel?
a) 0.1%
b) 0.5%
c) 0.25%
d) 1.0%
.
Answer: c
Explanation: Medium and High strength micro-alloyed steel have low carbon content, but alloys
such as niobium, vanadium, titanium or boron are added to achieve high strength.
4. Fire resistant steels are also called as ____________
a) Stainless steel
b) Weathering steel
c) High strength steel
d) Thermomechanically treated steel
.
Answer: d
Explanation: Fire resistant steels are also called as thermomechanically treated steel. They
perform better than ordinary steel under fire.
5. What is the minimum percentage of chromium and nickel added to stainless steel?
a) 0.5%, 10.5%
b) 2%, 20%
c) 10.5%, 0.5%
d) 30%, 50%
.
Answer: c
Explanation: Stainless steel are low carbon steels to which a minimum of 10.5% chromium
(maximum 20%) and 0.5% nickel is added.
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Java Tricky Program 23 - Main method signature

6. Match the pair of Type of steel with its ultimate tensile strength :
TYPE OF STEEL ULTIMATE TENSILE CAPACITY
(A) Carbon Steel (i) 700-950 MPa
(B) High Strength Carbon Steel (ii) 440-590 MPa
(C) Weathering Steel (iii) 410-440 MPa
(D) High Strength quenched & tempered steel (iv) 480 MPa
(E) Medium and High strength microalloyed steel (v) 480-550 MPa
a) A-i, B-ii, C-iii, D-iv, E-v
b) A-v, B-iv, C-iii, D-ii, E-i
c) A-iii, B-v, C-iv, D-i, E-ii
d) A-ii, B-iii, C-v, D-i, E-iv
.
Answer: c
Explanation: Ultimate tensile strength is the capacity of material to withstand loads tending to
elongate. It is the maximum stress that a material can withstand while being stretched or pulled.
The ultimate tensile strength for Carbon Steel is 410-440 MPa, 480-550 MPa for High Strength
Carbon Steel, 480 MPa for Weathering Steel, 700-950 MPa for High Strength quenched &
tempered steel, 440-590 MPa for Medium and High strength microalloyed steel.
7. What is weathering steel?
a) low-alloy atmospheric corrosion-resistant steel
b) low-carbon steel
c) high strength quenched and tempered steel
d) fire resistant steel
.
Answer: a
Explanation: Weathering steel are low-alloy atmospheric corrosion-resistant steel. They are often
left unpainted. They have an ultimate tensile strength of 480 MPa.
8. Match the pair of Type of steel with its yield strength:
TYPE OF STEEL YIELD STRENGTH
(A) Carbon Steel (i) 300-450 MPa
(B) High Strength Carbon Steel (ii) 350 MPa
(C) Weathering Steel (iii) 350-400 MPa
(D) High Strength quenched & tempered steel (iv) 230-300MPa
(E) Medium and High strength microalloyed steel (v) 550-700 MPa
a) A-i, B-ii, C-iii, D-iv, E-v
b) A-i, B-iii, C-v, D-iv, E-ii
c) A-v, B-iv, C-iii, D-ii, E-i
d) A-iv, B-iii, C-ii, D-v, E-i
.
Answer: d
Explanation: Yield Strength is the stress that a material can withstand without any permanent
deformation i.e. the point of stress at which any material starts to deform plastically. The yield
strength of carbon steel is 230-300MPa, 350-400 MPa for High Strength Carbon Steel, 350 MPa
for Weathering Steel, 550-700 MPa for High Strength quenched & tempered steel, 300-450 MPa
for Medium and High strength microalloyed steel

1. What kind of steel requires definite amounts of other alloying elements?

a) Carbon steel
b) Alloying steel
c) Stainless steel
d) Tool steel
.
Answer: b
Explanation: Alloy steels are those steels which require a specific amount of the elements
making up its composition. Alloy steels consist of manganese, silicon, and copper as primary
elements whose quantities are equal to or more than 1.65%, 0.60%, and 0.60% respectively.
2. Which of these is not a function of alloy steels?
a) Increases strength
b) Improves ductility
c) Reduces cost
d) Improves machinability
.
Answer: c
Explanation: Alloy steels are used to improve properties such as strength, hardness, ductility,
grain structure, and machinability, among others. This, however, results in increased costs due to
multiple elements involved in the process.
3. Steels containing up to 3% to 4% of one or more alloying elements are known as ________
a) Low alloy steels
b) HSLA steels
c) High alloy steels
d) Stainless steels
.
Answer: a
Explanation: Low alloy steels consist of 3% to 4% of alloying elements making up its
composition. They have similar microstructure and heat treatments as plain carbon steels. HSLA
and AISI steels are the types of low alloy steels.
4. What does AISI steel stand for?
a) American-Indian Steel Institute
b) American-Indian Society of Iron
c) American Iron and Steel Institute
d) Alloys, Iron and Steel Institute
.
Answer: c
Explanation: American Iron and Steel Institute (AISI) is an association, established in America,
which produces steel. AISI steels are used in machine construction. They are otherwise known as
construction steels or structural steels.
5. Which of these is not an application of HSLA steels?
a) Bridges
b) Automobiles and trains
c) Building columns
d) Leaf and coil springs
.
Answer: d
Explanation: HSLA steels are known as high-strength-low-alloy steels. These high strength
steels are primarily used as structural materials or construction alloys. They are used to reduce
weight on bridges, automobiles, pressure vessels, building columns, etc.
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6. Steels containing more than 5% of one or more alloying elements are known as ________
a) HSLA steels
b) High alloy steels
c) Tool and die steels
d) Stainless steels
.
Answer: b
Explanation: High alloy steels are composed of more than 5% of alloying elements. They have
different microstructure and heat treatments than those of plain carbon steels. Tool and die, and
stainless steels are the types of high alloy steels.
7. Which of the following groups of alloying elements stabilize austenite?
a) Ni, Mn, Cu, and Co
b) Cr, W, Mo, V, and Si
c) Cr, W, Ti, Mo, Nb, V, and Mn
d) Co, Al, and Ni
.
Answer: a
Explanation: The alloying elements such as Ni, Mn, Cu, and Co have a tendency to alleviate
austenite, whereas Cr, W, Mo, V, and Si tend to stabilize ferrites. Alloying elements such as Cr,
W, Ti, Mo, Nb, V, and Mn tend to form carbides. Other elements like Co, Al, and Ni help to
weaken carbides and thereby form graphite.
8. Which family of steels are referred to as chromoly?
a) 40xx
b) 41xx
c) 43xx
d) 44xx
.
Answer: b
Explanation: The family of 41xx steel is usually called as chromoly or chrome-moly due to its
primary alloying elements, chromium and molybdenum. Steels such as 4130 and 4140 are
generally used for bicycle frames, and as parts of firearms, flywheels etc.
9. What is the common name of COR-TEN steel?
a) Weathering steel
b) Control-rolled steel
c) Pearlite-reduced steel
d) Microalloyed steel
.
Answer: a
Explanation: Weathering steels are otherwise also known as COR-TEN or corten steels. These
steels produce a corrosion resistance, which makes them ideal for eliminating the need to paint.
All choices¸ including corten steels are classifications of HSLA steels.
10. Alloy steels containing 0.05% to 0.15% of alloying elements are called _______
a) Weathering steel
b) Stainless steel
c) Tool and die steel
d) Microalloyed steel
.
Answer: d
Explanation: Microalloyed steels contain alloying elements in small quantities (0.05% to 0.15%).
These elements include niobium, vanadium, titanium, molybdenum, rare earth metals, among
others. They are used to refine the microstructure of the grain or for precipitate hardening
process

1. How much carbon is present in cast irons?

a) Less than 0.05%
b) Up to 1.5%
c) 1.5% to 2%
d) More than 2%
.
Answer: d
Explanation: Cast Irons are ferrous alloys containing more than 2% of carbon, along with silicon,
sulphur, manganese, and phosphorus. Maraging steels contain less than 0.05% carbon, while
plain carbon steels contain up to 1.5% carbon.
2. Cast iron is a _____ alloy.
a) Eutectic
b) Eutectoid
c) Peritectic
d) Peritectoid
.
Answer: a
Explanation: Cast irons are eutectic alloys made of iron and carbon. Since iron is available in
abundance, they are easy to find and are less expensive. Therefore cast irons are the cheapest of
all metals.
3. Iron obtained from broken ______ is known as white iron.
a) Cementite
b) Graphite
c) Pearlite
d) Bainite
.
Answer: a
Explanation: Upon, cementite becomes a silvery-white compound. This is due to the brittle
nature of cementite. This fracture results in white cast iron.
4. If the iron surface contains graphite, it is known as ________
a) Alloy cast iron
b) White iron
c) Grey iron
d) Spheroidal graphite
.
Answer: c
Explanation: When an iron containing graphite is fractured, the surface becomes grey. This is
due to the presence of graphite flakes in the iron structure. Hence, it is known as grey iron.
5. Which element causes cementite to behave in a stable manner?
a) Silicon
b) Sulphur
c) Manganese
d) Carbon
.
Answer: b
Explanation: Sulphur has a stabilizing effect in cementite. This helps it easily break the
cementite and produce white iron.
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6. An iron with high-silicon content is a _______
a) White iron
b) Grey iron
c) Malleable iron
d) Pig iron
.
Answer: b
Explanation: Silicon causes cementite to behave in an unstable manner, which decomposes and
releases graphite. If there is a high-silicon presence, more graphite is released and makes it a
grey iron. If low-silicon is available, then it becomes a white iron.
7. What is the effect of phosphorus and sulphur in cast irons?
a) Induces brittleness
b) Increases strength
c) Destabilizes cementite
d) No effect
.
Answer: a
Explanation: Although sulphur stabilizes cementite, it causes brittleness. The same is true for
phosphorus, due to which both elements are limited to a minimum quantity. Manganese is used
to increase the strength of an iron, which also controls the harmful effects of sulphur.
8. Decomposition of cementite to form ferrite and graphite is known as _______
a) Decomposition of cast irons
b) Production of cast irons
c) Growth of cast irons
d) Prevention of growth of cast irons
.
Answer: c
Explanation: When cementite is heated at 700C it decomposes into ferrite and graphite. This
causes it to occupy more space than the original structure. This phenomenon is known as the
growth of cast irons.
9. Which of these are applications of grey cast iron?
a) Camshafts, engine blocks
b) Wear plates, pump linings
c) Brake shoes, pedals
d) Gears, rocker arms
.
Answer: a
Explanation: Grey cast irons have good strength and corrosion resistance. Therefore, they are
used in camshafts and engine blocks. White cast irons are used for pump linings, whereas
malleable cast iron is used for pedals and levers. Spheroidal graphite is used in gears and rocker
arms.
10. Which of the following cast irons cannot be machined?
a) White cast iron
b) Grey cast iron
c) Malleable cast iron
d) Spheroidal graphite cast iron
.
Answer: a
Explanation: White cast irons are used where hardness and wear resistance matter, like grinding
and crushing. Since it is hard, it cannot be machined. Grey and malleable cast irons, on the other
hand, have excellent machinability. Spheroidal graphite cast irons also possess good
machinability.
11. How are malleable cast irons designated for different grades?
a) By tensile strength
b) By six or seven-digit numbers
c) By five-digit numbers
d) By alphabets
.
Answer: c
Explanation: ASTM has designated the grading system for malleable cast irons in five-digit
numbers. A grade 32510 has a yield strength of 32500 psi. Six or seven-digit numbers are used
to designate spheroidal graphite cast irons (60-40-18 with ferritic structure). Grey cast irons are
designated by their tensile strength (grade 20 denotes a tensile strength of 20000 psi).
12. What is the effect of Nickel on cast irons?
a) Stabilizes carbides
b) Increases hardness
c) Refines grain structure
d) Improves corrosion resistance
.
Answer: c
Explanation: Addition of nickel in cast irons refines the grain structure, and has a graphitizing
effect on cementite. Molybdenum and Vanadium are used to increase the hardness of cast irons,
while Chromium stabilizes carbides. Copper improves the resistance to corrosion.
13. What is the defining property of Wrought Irons?
a) High carbon
b) Low carbon
c) No carbon
d) Completely carbon-filled
.
Answer: b
Explanation: Wrought irons are iron alloys contain a very low amount of carbon (less than
0.08%). They were used for producing nails, chains, and bolts initially, but aren’t used these
days.

UNIT 6

1. Alloy of copper and zinc is known as __________

a) Brass
b) Bronze
c) Duralumin
d) Nichrome
.
Answer: a
Explanation: Brass is a substitutional alloy of copper and zinc. It has many useful properties,
such as low melting point, workability, electrical conductivity and thermal conductivity,
corrosion resistance, etc.
2. Alloy of Ni and Fe is termed as ___________
a) Brass
b) Bronze
c) Duralumin
d) Invar
.
Answer: d
Explanation: Invar is an alloy of nickel and iron generically known as FeNi36. It has low thermal
expansion coefficient.
3. Major constituent of the gun metal is ____________
a) Copper
b) Nickel
c) Iron
d) Zinc
.
Answer: a
Explanation: Gun metal (also known as red brass) contains 83% Cu and 2% Zn and 10% Sn.
4. Major constituent of Muntz metal is _____________
a) Copper
b) Nickel
c) Iron
d) Zinc
.
Answer: a
Explanation: Mutz metal contains 60% Cu and 40% Zn. It is commonly used in architectural
applications.
5. Major constituent of the Nichrome is _____________
a) Copper
b) Nickel
c) Iron
d) Zinc
.
Answer: b
Explanation: Nichrome contains 80% Ni and 20% Cr. Nichrome mostly used as a resistant wire
and in dental fillings.
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6. Major constituent of Constantan alloy is ____________

a) Copper
b) Nickel
c) Iron
d) Zinc
.
Answer: a
Explanation: Constantan (also known as Eureka/Ferry) contains 60% Cu and 40% Ni. It finds its
mainly in thermocouples and has strong negative seebeck coefficient above 0oC.
7. Major constituent of Elektron alloy is ____________
a) Copper
b) Nickel
c) Magnesium
d) Zinc
.
Answer: c
Explanation: Elektron contains 3-12% aluminum and 2% zinc, 0.03% manganese and rest is
magnesium. It is very light alloy and used where weight is the major consideration in design.
8. Which of the following alloy is widely used in thermo couples?
a) Brass
b) Bronze
c) Duralumin
d) Nichrome
.
Answer: d
Explanation: Nichrome contains 80% Ni and 20% Cr. It is used mostly in thermocouples and in
strain gauges.
9. Major constituent of Duralumin alloy is ____________
a) Copper
b) Nickel
c) Iron
d) Aluminum
.
Answer: d
Explanation: Duralumin is the age hardenable alloy of aluminium alloyed mainly with copper
and with manganese and magnesium. Being a lightweight alloy, duralumin finds its use widely in
aerospace industry.
10. What is the approximate percentage of Lead in soft solder?
a) 60
b) 50
c) 90
d) 99.02
.
Answer: b
Explanation: Soft solder is the eutectic alloy of 50% Pb and 50% Sn, having a melting point of
450oC.

1. How much copper is present in deoxidized copper?

a) > 99.9%
b) > 99.85%
c) > 99.5%
d) > 99.35%
.
Answer: b
Explanation: Deoxidized copper contains over 99.85% Cu, along with As, Fe, and Bi. It is
mainly used to remove cuprous oxides and reduce porosity. Arsenic deoxidized copper and
arsenic touch pitch copper contain over 99.5% and 99.35% copper respectively.
2. High conductivity copper is used ______
a) In electrical engineering
b) To reduce porosity
c) To raise softening temperature
d) To manufacture semiconductor elements
.
Answer: a
Explanation: High conductivity copper contains over 99.9% copper, less than 0.005% of both
lead and iron, along with low oxygen content. This copper finds its applications in the field of
electrical engineering as it is a great conductor of electricity.
3. Which copper grade is used to manufacture semiconductors and particle accelerator
components?
a) High conductivity copper
b) Deoxidised copper
c) Oxygen-free copper
d) Arsenic deoxidized copper
.
Answer: c
Explanation: Semiconductor and superconductor components require high chemical purity.
Oxygen-free copper serves this purpose, as well as for high vacuum devices like particle
accelerators. This copper is valued for its purity, rather than its electrical conductivity.
4. What is the melting point of Copper?
a) 419
b) 600
c) 1084
d) 2562
.
Answer: c
Explanation: Copper is a soft and ductile material with good electrical and thermal
conductivities. It’s melting point is 1084 and the boiling point is 2562.
5. Brass is an alloy of copper and ______
a) Zinc
b) Tin
c) Tin and zinc
d) Nickel
.
Answer: a
Explanation: Brasses are alloys which are stronger than plain copper. They are primarily
composed of copper and zinc, and may additionally contain tin, lead, aluminum etc. They can be
used to cast molds, draw wires, roll sheets, and turned into tubes.
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6. α brasses contain ______ of zinc.
a) 0%
b) <=36%
c) >36%
d) 100%
.
Answer: b
Explanation: Brasses are classified into two types depending on the amount of zinc contained in
them. Brasses up to 36% zinc are known as brasses. They are relatively soft, ductile, and are
easily cold worked. On the other hand, brasses with over 36% zinc are available in two different
phases; they are stronger than brasses.
7. What is the appearance of copper?
a) Gold
b) Blue
c) Yellow-green
d) Red-orange
.
Answer: d
Explanation: Pure copper generally appears in a reddish-orange color. It is a metal with a
lustrous property. Brass, an alloy of copper, may appear red or white, depending on the amount
of zinc.
8. Addition of tellurium to copper results in ________
a) High strength
b) Decreased electrical conductivity
c) Increased machinability
d) No effect
.
Answer: c
Explanation: When copper is added with tellurium, it gives good machinability. Beryllium,
chromium, or 1-3% lead can be added to increase strength. Cadmium also increases strength but
reduces electrical conductivity.
9. _______ is an alloy of copper and tin.
a) Brass
b) Bronze
c) Gunmetal
d) Cupro-nickel
.
Answer: b
Explanation: Bronze is a high strength an alloy made of copper and tin. It has a better corrosion
resistance compared to brasses.
10. Yellow metal is more commonly known as ______
a) Cartridge brass
b) Naval brass
c) Admiralty brass
d) Muntz metal
.
Answer: d
Explanation: Muntz metal is a brass alloy, which is otherwise also known as yellow metal. It
consists of 60% copper and 40% zinc. They are commonly used for extruding rods and tubes,
and for making condenser or heat exchanger plates.
11. Which brass alloy has high tensile strength and can be used for cast molding?
a) Manganese brass
b) Free cutting brass
c) Standard brass
d) Gilding metal
.
Answer: a
Explanation: Manganese brass has a good tensile strength compared to other brass alloys. It is
also known as high tensile brass. They can be hot worked, and are commonly used for making
pump-rods, and marine castings.
12. Which brass alloy is used to make imitation jewelry and decorative work?
a) Standard brass
b) Admiralty brass
c) Free cutting brass
d) Gilding metal
.
Answer: d
Explanation: Gilding metal, which is also known as commercial metal, is used for making coins,
medals, artificial jewelry, and decorative work. It consists of 90% copper and 10% zinc. It is
gold in appearance and had a good ductility.
13. Which brass alloy is suitable for high-speed machining?
a) Gilding metal
b) High tensile brass
c) Leaded brass
d) Muntz metal
.
Answer: c
Explanation: Leaded brass is used for high-speed machining due to its excellent strength and heat
resistance. It contains 59% Cu, 39% Zn, and 2% Pb. It is also known as free cutting brass.

1. What is the melting point of pure aluminum?

a) 520-600oC
b) 600oC
c) 800oC
d) 950oC
.
Answer: b
Explanation: Aluminum is an element which is available in abundance on earth. Its melting point
is 600oC. Pure aluminum has low strength, due to which it is added with alloying elements.
2. What is the tensile strength of aluminum?
a) 122.5 GPa
b) 220 MPa
c) 70.5 GPa
d) 45 MPa
.
Answer: d
Explanation: Pure aluminum has a low tensile strength of 45 MPa and Young’s modulus of 70.5
GPa. Copper has a tensile strength of 220 MPa and Young’s modulus of 122.5 GPa.
3. Compared to copper, how is the electrical conductivity of aluminum?
a) Higher
b) Lower
c) Equal
d) Zero
.
Answer: a
Explanation: When equal weights are compared, aluminum is a better conductor than copper. It
was found that aluminum conducts 201% more current than copper.
4. _______ is coated onto aluminum to improve its soldering ability.
a) Magnesium
b) Lithium
c) Tin
d) Copper
.
Answer: c
Explanation: To improve soldering or joining ability of aluminum, it is coated with tin and then
plated with other metals. Magnesium, lithium, and copper are better suited to form alloys to
serve various purposes.
5. Which aluminum alloy is known as aircraft aluminum?
a) 6061
b) 6063
c) 7068
d) 7075
.
Answer: d
Explanation: Aluminum alloy 7075 consists of zinc as its primary element, along with a bit of
magnesium and copper. Due to its high strength and good corrosion resistance, it is often used in
aircraft, and also known as aerospace aluminum. However, the other alloys are also employed in
a few aerospace functions and structures.
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6. Which of the following is not a classification of aluminum alloys?
a) Crucible alloys
b) Wrought alloys
c) Cast alloys
d) Heat-treatable alloys
.
Answer: a
Explanation: Aluminum alloys are categorized into wrought alloys, cast alloys, heat-treatable,
and non heat-treatable alloys. An example of each is 2.5% Mg, 0.25% Cr, and rest Al (wrought),
12% Si and rest Al (cast), 0.4-0.9% Mg, 0.3-0.7% Si, and rest Al (heat-treatable), and (0.8-1.5%
Mn and rest Al (non heat-treatable).
7. Which of these is not a property of duralumin?
a) High strength
b) 1/3 the weight of steel
c) Excellent casting and forging abilities
d) Poor machinability
.
Answer: d
Explanation: Duralumin contains 4% Cu along with 0.4-0.7% of both Mn and Mg. This provides
it excellent casting and forging abilities and high strength while maintaining only 1/3 of the
weight. It also possesses high machinability.
8. Which among the following is an example of a non heat-treatable alloy?
a) Al-Cu
b) Al-Li
c) Al-Mg-Si
d) Al-Mn
.
Answer: d
Explanation: Non heat-treatable alloys can be strengthened by cold-working operations. Al-Mn
alloy is an example of a non heat-treatable alloy. The leftover choices are examples of heat-
treatable alloys. Duralumin and Y-alloy are the important Al-Cu alloys.
9. Artificial aging process takes place at a temperature range of ________
a) 190-260oC
b) 260-300oC
c) 300-350oC
d) Room temperature
.
Answer: a
Explanation: Ageing of alloys at 190-260oC accelerates the aging process and reduces the overall
time. This phenomenon is known as artificial aging. Natural aging occurs at room temperature.
10. What happens when the maximum strength is achieved by the aging process?
a) Precipitate hardening
b) Age hardening
c) Over-aging
d) Natural ageing
.
Answer: c
Explanation: As the aging temperature decreases, the maximum strength increases and reaches
its peak. At the peak strength, the strength starts to decrease. This is referred to as over-aging.
11. Which of these is not a stage in precipitation hardening treatment?
a) Solution treatment
b) Tempering
c) Quenching
d) Ageing
.
Answer: b
Explanation: The alloy is heated to a certain temperature and treated at that temperature. Then it
is quenched (rapid cooling) and finally heated below solvus temperature; this is known as aging.
12. How much copper is present in Y-alloys?
a) 4%
b) 2%
c) 1.5%
d) 1%
.
Answer: a
Explanation: Y-alloy is an important type of Al-Cu alloy. Similar to duralumin, it contains 4%
copper. Addition of 2% nickel and 1.5% magnesium is done to make Y-alloys.
13. What is the melting point of zinc?
a) 419oC
b) 600oC
c) 907oC
d) 950oC
.
Answer: a
Explanation: Zinc has a relatively low melting point of 419. 5oC. Its boiling point, on the other
hand, is at 907oC. The melting point of aluminum is 600oC.
14. With the addition of which element, does zinc create resistance to creep?
a) Pb
b) Cd
c) Mg
d) Cu
.
Answer: c
Explanation: When magnesium is added to copper, it increases its resistance to creep. Lead
affects intercrystalline corrosion, while cadmium improves hardening effect. Copper makes the
Cu-Zu alloys more ductile.
15. The most common casting process for zinc alloys is _______
a) Sand casting
b) Die casting
c) Investment casting
d) Centrifugal casting
.
Answer: b
Explanation: Die casting is generally employed for zinc alloys. These die-castings range from a
few grams to 20 kilograms. Zinc base dies casting alloys to find applications in hardware items
like car body, grills etc.
16. What is the appearance of zinc?
a) Bluish-grey
b) Red
c) Yellow
d) Green
.
Answer: a
Explanation: Zinc is a metallic element whose color looks blue and grey. It has a lustrous
property and is diamagnetic in nature.
17. What is the temperature at which zinc become malleable
a) <100 oC
b) 100-150 oC
c) >210 oC
d) 419 oC
.
Answer: b
Explanation: Zinc is brittle at most temperatures. However, it becomes malleable between 100
and 150 oC. Above 210 oC, it becomes brittle again, reaching its melting point at 41 oC.

1. How much copper is present in deoxidized copper?

a) > 99.9%
b) > 99.85%
c) > 99.5%
d) > 99.35%
.
Answer: b
Explanation: Deoxidized copper contains over 99.85% Cu, along with As, Fe, and Bi. It is
mainly used to remove cuprous oxides and reduce porosity. Arsenic deoxidized copper and
arsenic touch pitch copper contain over 99.5% and 99.35% copper respectively.
2. High conductivity copper is used ______
a) In electrical engineering
b) To reduce porosity
c) To raise softening temperature
d) To manufacture semiconductor elements
.
Answer: a
Explanation: High conductivity copper contains over 99.9% copper, less than 0.005% of both
lead and iron, along with low oxygen content. This copper finds its applications in the field of
electrical engineering as it is a great conductor of electricity.
3. Which copper grade is used to manufacture semiconductors and particle accelerator
components?
a) High conductivity copper
b) Deoxidised copper
c) Oxygen-free copper
d) Arsenic deoxidized copper
.
Answer: c
Explanation: Semiconductor and superconductor components require high chemical purity.
Oxygen-free copper serves this purpose, as well as for high vacuum devices like particle
accelerators. This copper is valued for its purity, rather than its electrical conductivity.
4. What is the melting point of Copper?
a) 419
b) 600
c) 1084
d) 2562
.
Answer: c
Explanation: Copper is a soft and ductile material with good electrical and thermal
conductivities. It’s melting point is 1084 and the boiling point is 2562.
5. Brass is an alloy of copper and ______
a) Zinc
b) Tin
c) Tin and zinc
d) Nickel
.
Answer: a
Explanation: Brasses are alloys which are stronger than plain copper. They are primarily
composed of copper and zinc, and may additionally contain tin, lead, aluminum etc. They can be
used to cast molds, draw wires, roll sheets, and turned into tubes.
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6. α brasses contain ______ of zinc.

a) 0%
b) <=36%
c) >36%
d) 100%
.
Answer: b
Explanation: Brasses are classified into two types depending on the amount of zinc contained in
them. Brasses up to 36% zinc are known as brasses. They are relatively soft, ductile, and are
easily cold worked. On the other hand, brasses with over 36% zinc are available in two different
phases; they are stronger than brasses.
7. What is the appearance of copper?
a) Gold
b) Blue
c) Yellow-green
d) Red-orange
.
Answer: d
Explanation: Pure copper generally appears in a reddish-orange color. It is a metal with a
lustrous property. Brass, an alloy of copper, may appear red or white, depending on the amount
of zinc.
8. Addition of tellurium to copper results in ________
a) High strength
b) Decreased electrical conductivity
c) Increased machinability
d) No effect
.
Answer: c
Explanation: When copper is added with tellurium, it gives good machinability. Beryllium,
chromium, or 1-3% lead can be added to increase strength. Cadmium also increases strength but
reduces electrical conductivity.
9. _______ is an alloy of copper and tin.
a) Brass
b) Bronze
c) Gunmetal
d) Cupro-nickel
.
Answer: b
Explanation: Bronze is a high strength an alloy made of copper and tin. It has a better corrosion
resistance compared to brasses.
10. Yellow metal is more commonly known as ______
a) Cartridge brass
b) Naval brass
c) Admiralty brass
d) Muntz metal
.
Answer: d
Explanation: Muntz metal is a brass alloy, which is otherwise also known as yellow metal. It
consists of 60% copper and 40% zinc. They are commonly used for extruding rods and tubes,
and for making condenser or heat exchanger plates.
11. Which brass alloy has high tensile strength and can be used for cast molding?
a) Manganese brass
b) Free cutting brass
c) Standard brass
d) Gilding metal
.
Answer: a
Explanation: Manganese brass has a good tensile strength compared to other brass alloys. It is
also known as high tensile brass. They can be hot worked, and are commonly used for making
pump-rods, and marine castings.
12. Which brass alloy is used to make imitation jewelry and decorative work?
a) Standard brass
b) Admiralty brass
c) Free cutting brass
d) Gilding metal

Answer: d
Explanation: Gilding metal, which is also known as commercial metal, is used for making coins,
medals, artificial jewelry, and decorative work. It consists of 90% copper and 10% zinc. It is
gold in appearance and had a good ductility.
13. Which brass alloy is suitable for high-speed machining?
a) Gilding metal
b) High tensile brass
c) Leaded brass
d) Muntz metal

Answer: c
Explanation: Leaded brass is used for high-speed machining due to its excellent strength and heat
resistance. It contains 59% Cu, 39% Zn, and 2% Pb. It is also known as free cutting brass