Question 1

Effect of carbon state in iron: When carbon is present as combined carbon (primarily as iron carbide, cementite), how does it influence the hardness and grain character of iron?

Accepted Answer

Introduction / Context: Carbon in iron can exist in free (graphitic) or combined form (cementite, Fe3C). The state of carbon strongly affects mechanical properties and microstructure. Understanding this distinction is essential for cast irons and steels alike. Given Data / Assumptions: Combined carbon refers mainly to iron carbide formation not free graphite. Microstructure may include pearlite (lamellae of ferrite and cementite) and cementite networks. Cooling rate and composition can further refine the grain size. Concept / Approach: Cementite is a hard, brittle phase. Its presence increases hardness and strength. In steels, finer pearlite spacing and refined grains enhance strength and hardness further. Combined carbon tends to promote a harder, finer microstructure compared with graphitic irons that are softer and machinable. Step-by-Step Solution: Recognize that cementite hardens the matrix.Finer distribution (e.g., fine pearlite) leads to fine-grained appearance and increased hardness.Therefore, combined carbon generally makes iron harder and the structure finer.Select the option “hard and gives a fine-grained crystalline structure”. Verification / Alternative check: Hardness tests show higher values with increased combined carbon; microscopy reveals refined lamellar spacing with faster cooling. Why Other Options Are Wrong: Soft variants align with graphitic or annealed structures, not cementite-rich ones. “Hard and coarse-grained” ignores typical refinement associated with increased pearlite and controlled cooling. Common Pitfalls: Assuming carbon always refines grains regardless of cooling history; overlooking that graphite formation softens the material. Final Answer: hard and gives a fine-grained crystalline structure

Question 2

Improving machinability of steels: The machinability (ease of cutting, chip breaking, and surface finish at given tool life) of plain carbon steels is most effectively improved by adding which elements?

Accepted Answer

Introduction / Context: Free-machining steels are engineered to break chips easily, reduce cutting forces, and improve surface finish. This is achieved by adding specific elements that form inclusions or low-melting constituents at the tool–chip interface. Given Data / Assumptions: Base material: plain carbon steel for general machining. Objective: improved chip control and lower tool wear. Additions must not primarily target high-temperature strength or corrosion resistance. Concept / Approach: Sulphur forms manganese sulphide (MnS) inclusions which act as chip breakers and lubricants. Lead in small amounts smears at the cutting interface, improving lubricity and finish. Phosphorus can increase strength and machinability at the cost of toughness. Together, these form the basis of many free-cutting grades (e.g., 12L14). Nickel and chromium, while beneficial for strength and corrosion resistance, generally do not improve machinability and can make cutting more difficult. Step-by-Step Solution: Relate machinability to inclusion engineering: MnS and Pb improve chip breakage and lubrication.Assess candidate elements: Ni and Cr raise strength/hardness without enhancing chip control.Select the trio most associated with free-machining behavior: S, Pb, P.Confirm that these additions are standard in free-cutting steels. Verification / Alternative check: Machining handbooks list leaded and resulfurized steels with substantially higher machinability ratings than Ni/Cr-alloyed counterparts. Why Other Options Are Wrong: Nickel and chromium (alone or together) are not the primary additions for machinability; they target mechanical properties and corrosion resistance. Boron/niobium are microalloying elements for hardenability and grain control, not machinability. Common Pitfalls: Assuming that stronger alloys cut better; in reality, controlled inclusions and lubricious phases drive machinability. Final Answer: sulphur, lead and phosphorus

Question 3

Commercial aluminium alloys for utensils: Which aluminium alloy is most widely associated with anodized utensil manufacture due to its forming properties and anodizing response?

Accepted Answer

Introduction / Context: Household utensils and cookware require alloys that combine good formability, moderate strength, corrosion resistance, and an excellent anodizing finish. In the Indian market, “Hindalium” is a familiar trade name for such aluminium alloys used in utensils. Given Data / Assumptions: Target application: anodized utensils and cookware. Requirements: good surface finish after anodizing, reasonable strength, and deep drawability. Multiple candidate alloys are listed, including structural and high-temperature grades. Concept / Approach: Duralumin (Al–Cu–Mg) and Y-alloy (Al–Cu–Ni) are aerospace/high-temperature oriented and less corrosion-resistant without cladding. Magnalium denotes Al–Mg alloys broadly, used structurally as well, but “Hindalium” specifically denotes utensil-grade Al–Mg–Mn family alloys tailored for anodizing and forming. Hence, the trade name most strongly linked to anodized utensils is Hindalium. Step-by-Step Solution: Identify the trade name used for utensil-specific aluminium alloys: Hindalium.Eliminate high-strength structural alloys (duralumin, Y-alloy) better suited to aerospace/engine parts.Magnalium is generic and not specifically tied to utensil anodizing in common usage.Select Hindalium as the correct answer. Verification / Alternative check: Market references and consumer products label anodized cookware as Hindalium, emphasizing finish quality and durability. Why Other Options Are Wrong: Duralumin and Y-alloy are copper-rich systems prone to corrosion without proper protection; Al–Cu–Mg 2xxx alloys are not typical for utensils. Magnalium is a generic term and not the specific trade association sought here. Common Pitfalls: Assuming high-strength aerospace alloys are suitable for food utensils; prioritizing anodizing quality and corrosion resistance is key. Final Answer: hindalium

Question 4

Cutting tools — saw blade material choice: The blade of a power saw (for metal cutting) is commonly manufactured from which material to retain hardness at elevated cutting temperatures?

Accepted Answer

Introduction / Context: Power saw blades for metal cutting operate at high tip speeds and experience frictional heating. The tool material must maintain hardness (red hardness) and resist wear at elevated temperatures. High-speed steels are the traditional choice for such blades and many other cutting tools. Given Data / Assumptions: Application: metal-cutting power saw blades (e.g., hacksaw, bandsaw). Service: intermittent or continuous cutting with localized heating. Requirement: hot hardness and toughness to withstand intermittent loads. Concept / Approach: High-speed steels (e.g., M and T series) retain significant hardness above 500°C due to strong carbide formers like W, Mo, V, and Cr. Stainless steels prioritize corrosion resistance, not hot hardness. Malleable cast iron and spring steels lack the necessary hot hardness and wear resistance for sustained cutting edges, while boron steels are typically used for wear-resistant structural parts (e.g., agricultural blades) rather than fine cutting edges at high temperatures. Step-by-Step Solution: Identify performance need: hot hardness and wear resistance under cutting heat.Match material class: HSS designed for cutting at high speed.Eliminate alternatives lacking red hardness or edge stability.Select “high speed steel”. Verification / Alternative check: Industrial catalogs list HSS or bimetal blades with HSS tooth edges for metal cutting, confirming widespread use. Why Other Options Are Wrong: Stainless focuses on corrosion; spring steels emphasize elasticity; cast iron is too brittle; boron steel applications differ from precision cutting edges. Common Pitfalls: Assuming corrosion-resistant or springy materials make good cutting edges without considering high-temperature hardness. Final Answer: high speed steel

Question 5

Substitution in high-speed steels: Which alloying element can effectively replace tungsten in certain tungsten-based high-speed steel grades while retaining comparable hot-hardness performance?

Accepted Answer

Introduction / Context: Tungsten-based high-speed steels (T-series) are renowned for their red hardness but are heavy and costly due to the high tungsten content. Molybdenum (M-series) grades were developed to achieve similar performance with reduced tungsten dependency. Given Data / Assumptions: Objective: identify a practical, widely used substitute for tungsten in HSS. Performance target: maintain hot hardness and cutting capability. Alloying context: presence of carbide formers and solid-solution strengtheners. Concept / Approach: Molybdenum provides comparable strengthening and carbide formation effects in HSS, enabling the M-series (e.g., M2, M42) to deliver red hardness similar to tungsten-rich T-series (e.g., T1), often with added vanadium and cobalt to fine-tune wear resistance and hot hardness. Nickel and silicon are not direct substitutes for tungsten in HSS; cobalt boosts hot hardness but does not replace tungsten's role in the matrix and carbide balance by itself. Vanadium primarily refines carbides and improves wear, not the bulk replacement of tungsten. Step-by-Step Solution: Identify key property to preserve: red hardness at cutting temperatures.Recognize established M-series steels as Mo-based replacements.Eliminate elements that do not replicate tungsten's function on their own (Ni, Si, V, Co alone).Select molybdenum as the correct replacement element. Verification / Alternative check: Materials handbooks list M-series HSS as molybdenum-bearing analogs to tungsten-based T-series, with similar application ranges. Why Other Options Are Wrong: Nickel improves toughness; silicon aids deoxidation; vanadium and cobalt are important additives but not primary replacements for tungsten. Common Pitfalls: Assuming cobalt alone replaces tungsten; in practice, cobalt augments hot hardness but Mo substitutes for W in base composition. Final Answer: molybdenum

Question 6

Al–Si piston alloys and casting method: Assess the statement: “An aluminium alloy containing approximately 11% silicon is used for manufacturing engine pistons by the die-casting process.”

Accepted Answer

Introduction / Context: Aluminium–silicon (Al–Si) alloys are staple materials for automotive pistons due to their castability, low density, reasonable strength, and good wear resistance. Silicon content around the eutectic range confers fluidity and low shrinkage, enabling complex shapes in casting. Given Data / Assumptions: Silicon content ~11%, which is near the Al–Si eutectic range (~11–13%). Manufacturing route: die casting (pressure die casting or gravity die casting depending on design). Piston application demands dimensional stability, wear resistance, and good thermal conductivity. Concept / Approach: Near-eutectic Al–Si alloys exhibit excellent fluidity and feed thin sections well, which suits die-cast piston crowns and skirts. Silicon particles improve scuff resistance and reduce thermal expansion. Many piston alloys also incorporate small amounts of Cu, Ni, or Mg for strength at temperature, but the base premise of using ~11% Si is correct. Step-by-Step Solution: Relate required properties (wear, castability) to the Al–Si system.Note that ~11% Si aligns with eutectic composition, promoting sound die castings.Acknowledge optional alloying (Cu, Ni, Mg) without negating the core statement.Conclude: the statement is correct. Verification / Alternative check: Automotive materials data commonly list eutectic or hypereutectic Al–Si alloys (about 11–18% Si) for pistons, via gravity or pressure die casting. Why Other Options Are Wrong: “No” and variants contradict well-established practice; saying pistons avoid silicon is incorrect; limiting to sand casting ignores industry die-casting routes for certain piston designs. Common Pitfalls: Confusing die casting with forging routes used for high-performance pistons; both exist, but Al–Si die-cast pistons are widespread in mass production. Final Answer: Yes

Question 7

Materials science — definition check In mechanical metallurgy, the ability of a material to undergo large permanent (plastic) deformation under the application of a tensile force without fracturing is called ductility. State whether this statement i…

Accepted Answer

Introduction / Context: Ductility is a core mechanical property used to characterize metals and engineering materials. It indicates how much plastic deformation a material can sustain in tension before fracture. This property is critical for processes such as wire drawing, sheet metal forming, and for assessing failure modes in design and fracture prevention. Given Data / Assumptions: Tensile loading is the primary mode considered when defining ductility. Permanent deformation means plastic deformation, not elastic. Standard measures include percentage elongation and percentage reduction in area on a tensile specimen. Concept / Approach: Ductility relates to the ability to accommodate plastic strain under tensile stress. In a stress–strain test, a ductile material shows a long plastic region after yield, with significant elongation and necking prior to fracture. High ductility often correlates with good formability and the ability to redistribute stresses around flaws, delaying catastrophic fracture. Step-by-Step Solution: Recognize the definition: “large permanent deformation under tensile force” matches the standard meaning of ductility.Differentiate from toughness: toughness concerns energy absorption to fracture (area under the stress–strain curve), not simply large plastic strain in tension.Differentiate from malleability: malleability is the ability to undergo plastic deformation under compressive loading (e.g., rolling into sheet), while ductility emphasizes tensile loading.Therefore the statement provided is correct. Verification / Alternative check: Handbook definitions express ductility through % elongation and % reduction in area after a standardized tensile test (e.g., gauge length elongation and necked diameter). These quantify the “large permanent deformation” described. Why Other Options Are Wrong: “Incorrect; that is toughness” confuses two different properties. “Incorrect; that is malleability only” ignores the compression versus tension distinction. “Correct only for non-ferrous metals” is false; steels can be highly ductile depending on composition and heat treatment. Common Pitfalls: Equating high strength with low ductility automatically—some alloys achieve useful combinations. Also, using elastic elongation as evidence of ductility is incorrect; ductility is about plastic strain. Final Answer: Correct

Question 8

Steelmaking — Bessemer converter operation During a Bessemer converter blow, the appearance of brown smoke (brown fumes) indicates which stage of oxidation inside the converter?

Accepted Answer

Introduction / Context: The Bessemer converter refines pig iron into steel by blowing air through the molten bath. The sequence of oxidation reactions can be judged, historically, by flame and fume color. Recognizing these visual cues is part of traditional process control and remains a useful learning tool for steelmaking history and fundamentals. Given Data / Assumptions: Initial bath composition typical of pig iron: high C, and notable Si and Mn. Air is blown through tuyeres at the converter bottom. Visual observation of fumes and flame color relates to which elements are oxidizing. Concept / Approach: Oxidation proceeds preferentially for elements with higher oxygen affinity at the prevailing temperature. Silicon and manganese oxidize early, producing oxides and characteristic brownish smoke in the early stage. Carbon oxidation becomes dominant later, seen as a bright, voluminous flame due to carbon monoxide burning at the mouth. Understanding this order helps interpret the converter’s progress without modern instrumentation. Step-by-Step Solution: At blow start: strong oxidation of Si and Mn occurs first.Their oxides and accompanying particulates create a brown smoke/fume—visual indicator of early refining.As Si and Mn diminish, carbon oxidation (to CO/CO2) becomes dominant, altering flame character and intensity.Therefore brown smoke indicates the stage where air is burning out silicon and manganese. Verification / Alternative check: Classical metallurgical texts describe the “brown fumes” at the start of the blow and a later “carbon period” marked by a roaring flame. This aligns with the selected option. Why Other Options Are Wrong: Option (b) shifts to the carbon period, which is later. (c) is an operational instruction not tied to the fume color; tapping occurs at the proper endpoint. (d) contradicts historical practice. (e) deoxidation is a finishing step using additions, not an air-blow stage. Common Pitfalls: Assuming carbon oxidizes first because it is abundant; oxygen affinity and kinetics dictate the real order under converter conditions. Final Answer: Air is burning out silicon and manganese

Question 9

Fe–C microstructure — composition of pearlite Pearlite (the eutectoid mixture in plain-carbon steel near 0.8% C) consists approximately of which constituents by weight?

Accepted Answer

Introduction / Context: Pearlite is the lamellar mixture formed when austenite of eutectoid composition transforms at about 727°C in plain-carbon steel. Understanding its phase fractions is essential for interpreting microstructures, mechanical properties, and heat-treatment responses in steels. Given Data / Assumptions: Pearlite is made of alternating layers of ferrite (α-iron) and cementite (Fe3C). Cementite contains 6.67% carbon by weight; ferrite contains very little carbon at room temperature. Weight fractions derive from the lever rule at the eutectoid composition. Concept / Approach: At the eutectoid composition (~0.76–0.8% C), the austenite transforms to pearlite. Applying the lever rule between ferrite (≈0.02% C at room temperature) and cementite (6.67% C) yields approximately 12–13% cementite and 87–88% ferrite by weight in pearlite. These fractions govern hardness and strength; more cementite increases hardness and brittleness, while ferrite provides ductility. Step-by-Step Solution: Identify phases: ferrite + cementite in lamellae.Use eutectoid composition and end-member carbon contents to estimate fractions.Lever rule leads to about 12.5–13% cementite and 87–87.5% ferrite by weight.Therefore, the best answer is 13% cementite and 87% ferrite. Verification / Alternative check: Metallography references often quote pearlite as roughly 1/8 cementite and 7/8 ferrite by weight—consistent with option (b). Why Other Options Are Wrong: (a) confuses carbon content with cementite fraction. (c) inverts the correct proportions. (d) states cementite’s stoichiometric carbon, not pearlite composition. (e) 50/50 by weight is not supported by the lever rule. Common Pitfalls: Mixing up weight and volume fractions; the lamellar spacing affects hardness but not the overall phase fractions at a given composition. Final Answer: 13% cementite and 87% ferrite

Question 10

Ironmaking — typical carbon range in pig iron Select the most appropriate range for total carbon content in pig iron produced from a blast furnace (by weight %).

Accepted Answer

Introduction / Context: Pig iron is the immediate product of the blast furnace, later refined into steel or cast iron. Knowing its typical carbon range is foundational for distinguishing pig iron from steels and cast irons and for understanding downstream refining requirements (e.g., basic oxygen or electric arc processes). Given Data / Assumptions: Pig iron chemistry varies with burden, fluxes, and furnace practice. Carbon is present both dissolved and as carbides; silicon, manganese, phosphorus, and sulfur are also present to varying levels. We seek the commonly cited carbon range for standard blast-furnace pig iron. Concept / Approach: Pig iron typically contains high carbon, substantially more than steels. Most references place pig iron carbon between about 3.0% and 4.5% by weight, with special-purpose irons sometimes slightly outside this range. This high carbon content must be reduced during steelmaking to reach levels appropriate for structural or tool steels. Step-by-Step Solution: Identify the characteristic: pig iron >> steel in carbon content.Compare the options: 3.0–4.5% matches the standard textbook range.Other ranges are either too low (typical of steels) or unrealistically high for normal pig iron practice.Therefore choose 3.0 to 4.5%. Verification / Alternative check: Materials handbooks list typical pig iron C ≈ 3–4.5%, Si ≈ 1–3%, Mn ≈ 0.5–1.5%, with P and S controlled depending on grade. Why Other Options Are Wrong: 0.1–1.0% are steel-like carbon levels. 5–10% is far above practical solubility and process norms. 1–5% is too broad and includes non-representative low carbon values. Common Pitfalls: Confusing pig iron with cast iron; while both are high in carbon, pig iron refers to the crude blast-furnace product prior to refining, not the final casting alloy. Final Answer: 3.0 to 4.5%

Question 11

Heat treatment — classification of induction hardening Induction hardening of steels is basically which type of heat-treatment process?

Accepted Answer

Introduction / Context: Induction hardening uses alternating electromagnetic fields to rapidly heat the surface layer of a steel component above the austenitising temperature, followed by immediate quenching to form martensite at the surface. The core remains relatively unaffected, preserving toughness. This places induction hardening squarely within the family of surface hardening processes. Given Data / Assumptions: Steel composition contains sufficient carbon (typically >= 0.35% C) to form martensite upon quenching. Induction frequency and power are set to target a specific case depth. Rapid heating and quenching occur locally at the surface. Concept / Approach: Unlike carburising (which adds carbon to the surface at high temperature over hours), induction hardening does not change composition. It only transforms the existing surface microstructure by localized austenitising followed by quench. The result is a hard, wear-resistant martensitic case with a tough, ductile core—ideal for shafts, gears, and cams. Step-by-Step Solution: Apply alternating magnetic field → eddy currents heat the surface layer.Surface reaches austenitising temperature rapidly while the core stays cooler.Immediate quench converts surface austenite to martensite (hard case).Core retains original microstructure (toughness), confirming surface hardening classification. Verification / Alternative check: Case depth profiles and hardness gradients measured after induction confirm the hardened surface and softer core. No carbon enrichment is detected, differentiating from carburised cases. Why Other Options Are Wrong: Carburising adds carbon; induction does not. Core hardening implies through-hardening, which is not the objective here. Tempering is a secondary process used after quench to adjust hardness—not the primary mechanism of induction hardening. Common Pitfalls: Attempting to induction-harden very low carbon steel (insufficient martensite potential); mis-selecting frequency and power leading to too shallow or too deep a case. Final Answer: Surface hardening process (through rapid surface austenitising and quench)

Question 12

Basic properties — definition of strength Which statement correctly defines the mechanical property “strength” for engineering materials?

Accepted Answer

Introduction / Context: Strength is one of the most frequently cited mechanical properties in design. It determines the maximum stress a material can withstand before yielding or fracturing, guiding the sizing of components and choice of safety factors in mechanical, civil, and aerospace engineering. Given Data / Assumptions: Strength is evaluated relative to specific modes: tensile, compressive, shear, bearing, etc. Yield strength and ultimate strength are the most common metrics. Loading is within the intended service environment and rate. Concept / Approach: Strength relates to resistance to failure under load. When stress exceeds yield strength, permanent deformation occurs; when it exceeds ultimate strength, fracture can occur. The correct definition emphasizes withstanding external forces without breakdown (fracture) or yielding (permanent deformation). This differs from stiffness (resistance to elastic deformation) and toughness (energy absorption to fracture). Step-by-Step Solution: Identify terminology: “breakdown or yielding” covers both fracture and plastic flow.Map to standard measures: yield strength (e.g., 0.2% offset) and ultimate tensile strength.Exclude other properties: stiffness belongs to elasticity modulus; toughness to impact energy; elasticity to shape recovery.Thus, choose the statement that best captures failure resistance under force. Verification / Alternative check: Design codes specify allowable stresses using fractions of yield or ultimate strengths, confirming that “strength” is fundamentally about limits before yielding or breaking. Why Other Options Are Wrong: (a) defines stiffness; (c) defines toughness; (e) defines elasticity; (d) is invalid since a correct definition is provided. Common Pitfalls: Using high strength materials where stiffness controls deflection; strength does not guarantee low deflection. Also, conflating hardness with strength—related but distinct. Final Answer: Ability to resist externally applied forces without breakdown or yielding

Question 13

Cast irons — effect of alloying additions When elements such as nickel, chromium, copper, and molybdenum are added intentionally to molten cast iron, the product is termed:

Accepted Answer

Introduction / Context: Cast irons are iron–carbon–silicon alloys with varied microstructures (grey, white, malleable, nodular). Alloying additions are used to improve specific properties such as high-temperature strength, wear resistance, and corrosion resistance. Correct terminology helps link composition to expected performance and applications. Given Data / Assumptions: Alloying elements include Ni, Cr, Cu, Mo in purposeful amounts. Base matrix could be pearlitic, ferritic, or martensitic depending on heat treatment and composition. Graphite morphology (flakes, nodules) depends on inoculation and treatment (e.g., Mg for nodular iron). Concept / Approach: When cast iron contains deliberate additions of alloying elements to achieve enhanced properties beyond plain grey or white iron, the material is termed “alloy cast iron.” These elements can stabilize carbides (Cr, Mo), enhance toughness and strength (Ni), or improve corrosion resistance (Cu, Ni). The label is independent of graphite form; you can have alloyed grey or alloyed ductile irons depending on treatment. Step-by-Step Solution: Identify the presence of alloying elements added for property improvement.Relate additions to targeted properties (wear, heat, corrosion resistance).Apply standard nomenclature: such products are “alloy cast irons.”Therefore select option (d). Verification / Alternative check: Standards and handbooks classify irons with significant Ni, Cr, Mo, Cu as alloy cast irons (e.g., Ni-hard, high-Cr irons). Why Other Options Are Wrong: White iron refers to carbide-dominated microstructure, not necessarily alloyed. Nodular and malleable describe graphite morphology and heat treatment, not simply the presence of alloying elements. Grey iron only is too restrictive. Common Pitfalls: Confusing graphite shape control (via Mg treatment) with alloying; the two are orthogonal levers in cast iron design. Final Answer: alloy cast iron

Question 14

Induction hardening — control of case depth State whether the statement is correct: In induction hardening, the depth of hardening (case depth) is controlled by controlling the voltage applied to the induction coil.

Accepted Answer

Introduction / Context: Induction hardening produces a hardened surface layer by rapid, localized heating and quenching. Control of case depth is crucial for achieving desired wear resistance without compromising core toughness. Understanding the main control variables is a key practical skill in heat treatment engineering. Given Data / Assumptions: Induction system characterized by frequency, power (kW), coil geometry, and process time. Steel composition allows martensite formation upon quench. Voltage is one of several electrical parameters related to power delivery. Concept / Approach: Case depth in induction hardening is governed primarily by electromagnetic penetration depth (skin depth) and thermal diffusion. Skin depth δ ≈ k / √f (where f is frequency, k depends on material properties) shows that higher frequency reduces penetration, yielding shallower cases. Heating time and power density influence how far heat diffuses before quench. Voltage alone is not the primary, independent knob; it is part of the circuit achieving required power but does not directly set case depth like frequency and time do. Step-by-Step Solution: Relate case depth to skin depth: δ ∝ 1 / √f.Adjust process: choose frequency for target δ, then set power and time to reach austenitising temperature at that depth.Recognize voltage is secondary and coupled to coil design and load; it does not uniquely determine depth.Hence, the statement attributing depth control to voltage is incorrect. Verification / Alternative check: Process documentation specifies frequency bands (e.g., medium vs. high frequency) for desired case depths (e.g., 1–6 mm), confirming frequency/time control. Why Other Options Are Wrong: Options asserting correctness overlook the dominant roles of frequency and heating time; carbon content affects hardness potential, not electromagnetic penetration. The explanatory option (e) correctly states the real dependencies. Common Pitfalls: Using excessive power causing overheating instead of greater depth; attempting deep cases with very high frequency, which inherently limits penetration. Final Answer: Incorrect

Question 15

Alloy identification — what is bronze? Bronze is correctly defined as which of the following alloy systems?

Accepted Answer

Introduction / Context: Bronze and brass are two historically important copper alloys with distinct base systems. Correctly identifying these alloys is vital for selecting materials with appropriate mechanical properties, corrosion resistance, and casting behavior in engineering applications. Given Data / Assumptions: Bronze traditionally refers to copper–tin alloys. Brass refers to copper–zinc alloys. Modern marketing sometimes broadens “bronze” to include other copper-based alloys (e.g., aluminium bronze), but classical metallurgical usage is copper–tin. Concept / Approach: In standard metallurgical terminology, bronze = Cu–Sn, and brass = Cu–Zn. While there are variants like phosphor bronze (Cu–Sn–P) and aluminium bronze (Cu–Al), the foundational identification for “bronze” in basic materials questions remains copper and tin. This distinction correlates with differences in strength, wear resistance, and corrosion behavior. Step-by-Step Solution: Match “bronze” to copper–tin system.Exclude copper–zinc (brass) and multi-component mixtures that are not the canonical bronze.Therefore, select “Copper and tin”.Note that specialised bronzes may include minor additions (e.g., P) but remain Cu–Sn based. Verification / Alternative check: Property tables list typical bronzes (e.g., tin bronzes) with Sn ~ 3–12% for bearings and springs, confirming Cu–Sn as the system. Why Other Options Are Wrong: (a) identifies brass. (c) is a ternary alloy that could exist but is not the standard definition of bronze. (e) refers to aluminium bronze, a specific subclass; the general textbook definition is Cu–Sn. Common Pitfalls: Relying on trade names that stretch “bronze” to many copper alloys; for exams, stick to the classical definition. Final Answer: Copper and tin

Question 16

Ironmaking equipment — product of a blast furnace A blast furnace is primarily used to produce which product?

Accepted Answer

Introduction / Context: The blast furnace is the centerpiece of traditional integrated iron and steel works. It chemically reduces iron ore to molten iron using coke and fluxes under a hot blast of air. Knowing what product emerges directly from the blast furnace clarifies downstream processing steps in steelmaking flowsheets. Given Data / Assumptions: Burden: iron ore (e.g., sinter, pellets), coke, and limestone/dolomite. Hot blast air injected through tuyeres sustains combustion and reduction reactions. Molten products tapped: hot metal (pig iron) and slag. Concept / Approach: The blast furnace outputs hot metal commonly called pig iron when cast into “pigs.” It is high in carbon (about 3–4.5%) and contains Si, Mn, P, and S. This hot metal is subsequently refined in a basic oxygen furnace (BOF) or other steelmaking unit to become steel. Cast iron, wrought iron, and steel are not produced directly in the blast furnace; DRI is produced in shaft furnaces using reducing gases, not in a blast furnace. Step-by-Step Solution: Identify the unit operation: blast furnace reduction of ore to hot metal.Recognize the molten iron product: pig iron/hot metal.Exclude downstream products like steel and finished castings.Therefore, the correct answer is pig iron. Verification / Alternative check: Plant flow diagrams show blast furnace → basic oxygen furnace → continuous casting; DRI processes (Midrex, HYL) are separate direct-reduction routes. Why Other Options Are Wrong: Cast iron refers to final cast products made by remelting and casting; wrought iron is low-carbon iron produced by puddling (historic); steel requires refining the carbon-rich hot metal. Common Pitfalls: Conflating “cast iron” with “pig iron” because both are high in carbon; the former is a product class made in foundries, not the direct blast furnace output. Final Answer: Pig iron

Question 17

Metallurgy — Definition Check on Austenite Is the following statement correct for materials science terminology: “Austenite is a solid solution of carbon or iron carbide (cementite) in gamma-iron (γ-Fe)”?

Accepted Answer

Introduction / Context: Austenite is one of the fundamental phases in iron–carbon metallurgy. It forms the basis for many steel heat treatments such as hardening, normalising, and austempering. This question tests whether you know the precise definition of austenite and can distinguish it from other phases like cementite (iron carbide) and ferrite. Given Data / Assumptions: The statement claims austenite is a solid solution of “carbon or iron carbide” in gamma-iron. Gamma-iron (γ-Fe) is the face-centred cubic form of iron stable at elevated temperatures in steels. Carbon solubility in γ-Fe is finite but significant compared with α-Fe. Concept / Approach: The precise definition is: austenite is a solid solution of carbon in gamma-iron (γ-Fe). Cementite (Fe3C), commonly called iron carbide, is a separate intermetallic compound phase, not something that dissolves into γ-Fe as “iron carbide.” Thus, including “iron carbide” in the definition is incorrect. When Fe3C coexists with austenite, it appears as a separate phase in the microstructure (e.g., ledeburite or pearlite constituents), not as solute in γ-Fe. Step-by-Step Solution: Identify the matrix: γ-Fe (austenitic iron) has an FCC lattice.Recognise the solute: interstitial carbon atoms occupy octahedral sites → austenite is C in γ-Fe.Contrast with cementite: Fe3C is a distinct compound; it does not define austenite by dissolution.Conclude the statement is wrong because it adds “iron carbide” to the definition. Verification / Alternative check: Iron–carbon phase diagram shows austenite as a single phase field (γ) with carbon in solution; Fe3C appears as a separate phase line/field. Why Other Options Are Wrong: Correct/true variants: they contradict the standard definition.“True for hypo-/hyper-eutectoid steels only”: steel composition does not change the phase definition.“Only true at very high carbon”: still incorrect; Fe3C remains a separate phase. Common Pitfalls: Confusing “phase present with” versus “dissolved in.” Cementite may coexist with austenite, but it is not dissolved as “iron carbide.” Final Answer: Incorrect

Question 18

Cartridge Brass (≈70% Cu, 30% Zn) — Formability in Manufacturing Which processing routes are feasible for cartridge brass under typical room-temperature or moderately warm working conditions?

Accepted Answer

Introduction / Context: Cartridge brass (about 70% copper, 30% zinc) is renowned for its excellent cold-workability. It is routinely used for cartridge cases, radiators, condenser tubes, deep-drawn components, and flexible metal hose due to its combination of ductility, strength after work hardening, and good corrosion resistance. Given Data / Assumptions: All options describe standard metalworking operations: rolling, wire drawing, and tube forming. Material: cartridge brass with typical mill anneals between passes. Room-temperature or moderately warm forming is intended. Concept / Approach: Cartridge brass possesses high ductility and strain-hardening capacity, allowing substantial cold-reduction before requiring an anneal. It can be cold rolled to thin sheet and strip, drawn into wire with sequential reductions, and formed into seamless tubes by cupping-and-drawing or by tube-drawing processes. These are industrially standard for this alloy family. Step-by-Step Solution: Assess each process: rolling, wire drawing, and tube forming all require good ductility.Match with alloy capability: cartridge brass meets these requirements exceptionally well.Conclude that any one of the listed processes is feasible. Verification / Alternative check: Product catalogs show cartridge brass available as sheet/strip, wire, and tubes; cold-working is routine with intermediate anneals. Why Other Options Are Wrong: The specific single-process options (a–c) are each correct but incomplete.“Only after full annealing at 900° C” is misleading; while anneals are used between reductions, extreme high-temperature anneals are not a universal prerequisite for feasibility. Common Pitfalls: Confusing brass series; some brasses with higher zinc or lead vary in formability, but 70/30 cartridge brass is notably formable. Final Answer: any one of these

Question 19

Heat Treatment — Normalising of Hypo-Eutectoid Steel Is this statement accurate: “In the normalising process, a hypo-eutectoid steel is heated 30–50° C above the upper critical temperature (A3) and then cooled in still air”?

Accepted Answer

Introduction / Context: Normalising is a common heat treatment to refine grain size, homogenise microstructure, and improve mechanical properties prior to machining or service. It differs from full annealing mainly in the cooling stage and slightly in the austenitising temperature range. Given Data / Assumptions: Material: hypo-eutectoid steel (less than 0.8% carbon). Temperature: 30–50° C above the upper critical temperature A3. Cooling medium: still air (not furnace cooling). Concept / Approach: For hypo-eutectoid steels, normalising involves heating above A3 to fully convert ferrite plus pearlite into austenite. Cooling in still air leads to a finer pearlitic structure than full annealing, which uses slower furnace cooling and typically yields coarser pearlite and greater softness. The specified temperature margin (30–50° C above A3) is standard to ensure full austenitisation without excessive grain growth. Step-by-Step Solution: Identify critical temperatures: A1 (eutectoid), A3 (upper critical for hypo-eutectoid steels).Heat to A3 + 30–50° C to obtain homogeneous austenite.Cool in still air to produce refined pearlite/ferrite, improving strength and machinability.Therefore, the statement aligns with normalising practice. Verification / Alternative check: Shop procedures and textbooks specify air cooling for normalising and furnace cooling for full anneal; this confirms the distinction. Why Other Options Are Wrong: “False” or “furnace cooling” contradicts the definition of normalising.Oil quenching is not part of normalising and would produce martensite, not normalised structure.Eutectoid steels use A1 as reference; the prompt explicitly names hypo-eutectoid steels. Common Pitfalls: Confusing normalising with annealing or normalising with hardening; cooling medium differentiates these processes. Final Answer: True

Question 20

Surface Hardening — Fuel Source in Flame Hardening In conventional flame hardening of steel surfaces, is an oxy-acetylene flame commonly employed as the heat source before quenching?

Accepted Answer

Introduction / Context: Flame hardening is a surface-hardening method in which a high-temperature flame rapidly heats the surface of a steel component followed by immediate quenching. It produces a hard martensitic case while preserving a tougher core, similar in goal to induction hardening but with a different heating method. Given Data / Assumptions: Workpiece is a hardenable steel (adequate carbon content). Heat input is localised and rapid. Quench follows shortly after the surface reaches the austenitising temperature. Concept / Approach: An oxy-acetylene flame is widely used because it provides a concentrated, controllable heat source with high flame temperature, enabling quick austenitisation of the surface layer. After sufficient heating depth is achieved, water or polymer quench transforms the surface to martensite, yielding a hardened case while limiting distortion. Step-by-Step Solution: Select heating method: oxy-acetylene torch with appropriate nozzle/travel speed.Heat to austenitising temperature for the steel grade.Immediately quench the heated zone to form martensite and achieve surface hardness. Verification / Alternative check: Industrial practice and manuals list oxy-acetylene as the standard flame for flame hardening; induction hardening is a separate method using electromagnetic heating. Why Other Options Are Wrong: “No” contradicts common practice; torches are typical in flame hardening.Stainless-only or cast-iron-only restrictions are incorrect; various steels can be flame hardened if carbon content allows.Carburising is not required; flame hardening relies on transforming existing carbon steel, not adding carbon. Common Pitfalls: Confusing flame hardening with flame carburising; the latter adds carbon, while the former transforms by heating and quenching. Final Answer: Yes

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