Question 1

Cutting tool materials — common choice for high-speed machining Which of the following materials is most commonly used for manufacturing general-purpose cutting tools such as drills, taps, and end mills?

Accepted Answer

Introduction / Context: Tool materials must retain hardness and resist wear at elevated temperatures generated during cutting. High-speed steel, cemented carbides, and ceramics each serve niches. For many general-purpose tools, high-speed steel remains a versatile and economical choice. Given Data / Assumptions: General-purpose tools for steels and nonferrous alloys. Moderate cutting speeds and feeds as in conventional machine shops. Focus on common, widely used tool material categories. Concept / Approach: High-speed steel (HSS) combines hot hardness with toughness and ease of grinding. Alloying with tungsten or molybdenum, chromium, vanadium, and sometimes cobalt allows HSS to maintain cutting edges under heat. While carbides cut faster, HSS provides superior toughness and is easier to sharpen, making it ubiquitous for drills, taps, and milling cutters in varied conditions. Step-by-Step Solution: Identify the most versatile tool material for common shop tools.Compare candidates: plain alloy steels soften at cutting temperatures; stainless grades lack suitable hot hardness; HSS is designed for this role.Select high-speed steel as the best answer. Verification / Alternative check: Catalogs from tooling manufacturers show extensive HSS product lines for twist drills, taps, reamers, and end mills, confirming industrial prevalence. Why Other Options Are Wrong: Nickel steel, chrome steel, nickel-chrome steel: may be strong but lack the hot hardness and wear resistance necessary for cutting edges at speed. Martensitic stainless: offers corrosion resistance, not optimal hot hardness for cutting. Common Pitfalls: Assuming the hardest room-temperature alloy is best. Cutting performance depends on hot hardness and abrasion resistance at temperature, where HSS excels compared with ordinary alloy steels. Final Answer: high-speed steel

Question 2

Special steels — name of the 36% nickel iron alloy with ultra-low thermal expansion A steel alloy containing approximately 36% nickel is commonly known as:

Accepted Answer

Introduction / Context: Certain iron–nickel alloys exhibit near-zero thermal expansion over a wide temperature range. These are critical for precision instruments, clocks, measuring tapes, and bimetallic thermostats. The most famous composition is around 36% nickel. Given Data / Assumptions: Composition is about 36% Ni with balance iron and minor additions. Application requires dimensional stability under temperature changes. Concept / Approach: The 36% nickel iron alloy is known as Invar. Its exceptionally low coefficient of thermal expansion arises from magneto-volume effects that counter normal lattice expansion. This property differentiates it from stainless steels (which are corrosion-resistant), high-speed steels (tool steels), and generic heat-resisting steels. Step-by-Step Solution: Associate 36% Ni with the trade name: Invar.Recall characteristic: ultra-low thermal expansion, used in precision components.Select the correct option: invar. Verification / Alternative check: Materials handbooks list Invar 36 with linear thermal expansion roughly one tenth that of carbon steel over room-to-moderate temperature ranges. Why Other Options Are Wrong: Stainless steel: identified by chromium content for corrosion resistance, not low expansion. High-speed steel: tool steel class for cutting tools. Heat-resisting steel: a broad category, not the specific low-expansion alloy. Hadfield steel: high-manganese austenitic steel, unrelated to low expansion. Common Pitfalls: Confusing corrosion-resistant stainless grades with low-expansion alloys; they serve very different engineering purposes. Final Answer: invar

Question 3

Surface hardening of machine tool guideways — common industrial method Machine tool guide ways are usually surface-hardened by which of the following processes to provide a hard wear layer on a tough substrate?

Accepted Answer

Introduction / Context: Guideways on lathes, milling machines, and grinders must resist wear while maintaining geometric accuracy. A hard surface on a tough core is ideal. Several surface and through-hardening processes exist, but workshops traditionally select methods suited to large, long castings. Given Data / Assumptions: Components are large cast iron or steel beds and slides. Need localized hardening without major distortion. Economical shop-floor process preferred. Concept / Approach: Flame hardening uses an oxy-fuel flame to heat the surface rapidly followed by spray quenching, creating a martensitic case with minimal through-heating. It is well suited to cast iron guideways and large components. Induction hardening is also widely used, especially on steels with adequate carbon, but flame hardening remains a classical and common choice for machine tool beds because of equipment reach and simplicity for large sections. Step-by-Step Solution: Identify constraints: large size, need for localized treatment, minimal distortion.Match process: flame hardening offers flexible coverage on long guideways.Therefore, select flame hardening as the usual process. Verification / Alternative check: Manufacturing references note flame-hardened cast iron ways as a standard practice, with hardness often in the mid-50s HRC range on the surface. Why Other Options Are Wrong: Vacuum hardening: a through-hardening method for alloy steels, not ideal for massive beds. Martempering: quench technique to reduce distortion, not a localized surface process. Induction hardening: viable and used, but the conventional textbook answer for large guideways is flame hardening. Carburizing and quenching: diffusion case hardening suited to gears and low-carbon steels, not typical for cast iron guideways. Common Pitfalls: Assuming one method fits all; practical selection depends on workpiece size, material, equipment, and desired case depth. Final Answer: flame hardening

Question 4

Cast iron — key property package Which statement best describes common advantages of cast iron as an engineering material?

Accepted Answer

Introduction / Context: Cast iron is used extensively for machine beds, pipes, engine blocks, and cookware due to a unique combination of properties arising from its graphite-containing microstructure. Recognizing the complete property package explains its enduring popularity despite the availability of many modern alloys. Given Data / Assumptions: Grey and ductile cast irons considered as representative types. Properties evaluated for general engineering use. Concept / Approach: Graphite in cast iron provides excellent machinability by breaking chips and reducing tool wear. The castability is high because cast irons have relatively low pouring temperatures and good fluidity, enabling complex shapes and thin sections. The compressive strength is typically much higher than tensile strength, and is adequate for bases and columns where compressive loads dominate. Together these attributes make cast iron economical and functional. Step-by-Step Solution: Evaluate compressive strength: cast irons possess high compressive values relative to their tensile strength.Evaluate machinability: graphite acts as a solid lubricant and chip breaker, improving machinability.Evaluate casting characteristics: good fluidity and low shrinkage facilitate complex castings.Hence, all listed advantages apply. Verification / Alternative check: Property tables for grey iron show compressive strengths multiple times their tensile strengths, while foundry manuals cite its fluidity and mold-filling behavior. Machinability ratings commonly rank grey iron highly. Why Other Options Are Wrong: Picking any single attribute ignores the material's full set of advantages. Only high tensile strength is incorrect; cast iron is relatively weak in tension but excels in compression and machinability. Common Pitfalls: Assuming that poor tensile strength means poor overall performance; applications that emphasize compressive loads and damping benefit substantially from cast iron. Final Answer: all of these

Question 5

Cast-iron solidification terminology — composition at the eutectic State whether the statement is correct: “Iron–carbon alloys containing 4.3% carbon are known as hypo-eutectic cast irons.”

Accepted Answer

Introduction / Context: In the cast-iron range, the eutectic composition of the iron–carbon system is about 4.3% carbon. Understanding the terms hypo-eutectic, eutectic, and hyper-eutectic is critical for predicting primary phases and casting behavior. Given Data / Assumptions: Eutectic composition is approximately 4.3% carbon under idealized conditions. Hypo-eutectic means below the eutectic composition; hyper-eutectic means above. Terminology is independent of whether graphite or cementite forms subsequently. Concept / Approach: An alloy exactly at 4.3% carbon is eutectic, not hypo-eutectic. Hypo-eutectic cast irons have carbon less than 4.3% and will form primary austenite before the eutectic reaction during solidification. Hyper-eutectic cast irons have carbon greater than 4.3% and may form primary graphite in grey irons or primary cementite in white irons depending on conditions. Step-by-Step Solution: Identify reference composition: 4.3% C corresponds to eutectic.Compare statement: calls 4.3% C hypo-eutectic, which contradicts definition.Therefore, the statement is incorrect.Correct classification: 4.3% C is eutectic. Verification / Alternative check: Textbook Fe–C diagrams label the eutectic point near 4.3% C and 1147°C, confirming the terminology. Why Other Options Are Wrong: Correct or conditional options misapply the definitions of hypo and hyper relative to the eutectic composition. Cannot be determined is not applicable because the definition is precise. Common Pitfalls: Confusing eutectic (liquid to two solids) with eutectoid (solid to two solids at about 0.8% C in steels). The 4.3% value relates to the cast-iron eutectic. Final Answer: Incorrect

Question 6

Steel classification by carbon content — name at 0.8% C A plain carbon steel containing approximately 0.8% carbon is known as:

Accepted Answer

Introduction / Context: Steel nomenclature often references the eutectoid composition, near which austenite transforms into a lamellar mixture of ferrite and cementite upon slow cooling. Identifying this composition is central to understanding pearlite formation and heat treatment behavior. Given Data / Assumptions: Eutectoid composition in plain carbon steel is about 0.8% carbon (sometimes cited as 0.77% depending on source). Terminology: hypo-eutectoid < 0.8% C; eutectoid ≈ 0.8% C; hyper-eutectoid > 0.8% C. Concept / Approach: At the eutectoid composition, austenite decomposes isothermally at the eutectoid temperature to form 100% pearlite in equilibrium transformation. This composition is therefore called eutectoid steel. Below it, proeutectoid ferrite forms in addition to pearlite; above it, proeutectoid cementite forms. Step-by-Step Solution: Recognize 0.8% C as the eutectoid point in plain carbon steel.Assign the correct classification: eutectoid steel.Rule out hypo- and hyper-eutectoid by their definitions relative to 0.8% C. Verification / Alternative check: Standard Fe–C diagrams and materials texts consistently cite the eutectoid composition near 0.8% C, reinforcing the term eutectoid steel for that carbon level. Why Other Options Are Wrong: Hyper-eutectoid: refers to carbon content greater than 0.8% C. Hypo-eutectoid: refers to carbon content less than 0.8% C. Eutectic steel: eutectic applies to the cast-iron range near 4.3% C, not steels at 0.8% C. Common Pitfalls: Mixing up eutectic with eutectoid terminology; the former describes liquid to two solid transformation in cast irons, the latter describes solid-state transformation in steels. Final Answer: eutectoid steel

Question 7

Material property fundamentals: Ductility is the property of a solid engineering material due to which it __________ under tensile loading and plastic deformation without fracture.

Accepted Answer

Introduction / Context: Ductility is a foundational mechanical property used across design, manufacturing, and failure analysis. It indicates the extent of plastic deformation a material can undergo in tension before fracture. Recognising ductile behaviour is critical for processes such as wire drawing, extrusion, and forming, and for predicting how parts will behave under overload conditions. Given Data / Assumptions: We compare qualitative property definitions, not numerical values. Tension-dominated deformation is the context for ductility. Standard textbook definitions are assumed (e.g., elongation, reduction of area). Concept / Approach: Ductility describes the ability to sustain significant plastic strain in tension. Common indicators are percentage elongation and reduction of area measured on tensile specimens. Materials like low-carbon steel, copper, aluminium, and their alloys are typically ductile at room temperature, while cast iron and many ceramics are brittle. The process most directly enabled by ductility is wire drawing—successive tensile/plastic reductions of cross-section without cracking. Step-by-Step Solution: Recall the definition: ductility = ability to undergo plastic deformation under tension.Connect to manufacturing: the signature operation is drawing into wire.Evaluate options: the statement matching this definition is the ability to be drawn into wires.Therefore, choose “can be drawn into wires”. Verification / Alternative check: Metals with high elongation in tensile tests (e.g., annealed copper) are easily wire-drawn; brittle materials with minimal elongation cannot be safely drawn. Why Other Options Are Wrong: Breaks with little permanent distortion defines brittleness, the opposite of ductility. Rolled/hammered into sheets defines malleability (plastic flow in compression), not ductility. Resisting fracture under impact loads refers to toughness, not ductility. High elastic modulus means stiffness, an elastic (not plastic) property. Common Pitfalls: Confusing ductility with malleability, toughness, or stiffness; assuming any plasticity equals ductility without specifying tension versus compression modes. Final Answer: can be drawn into wires

Question 8

Materials for pipe fittings (elbow, tee, union): For pressure-tight pipe fittings that must be threaded and shock-resistant, which material is generally preferred?

Accepted Answer

Introduction / Context: Pipe fittings such as elbows, tees, and unions need adequate strength, good machinability for threading, and toughness to survive handling and service shocks. The choice of cast ferrous materials directly affects sealing reliability and resistance to cracking at threads and thin sections. Given Data / Assumptions: Fittings are commonly made by casting, followed by machining threads. Service involves internal pressure, tightening loads, and occasional impact. Economy and mass production are relevant. Concept / Approach: Malleable iron is produced by heat-treating white cast iron to convert carbides into temper carbon nodules, substantially improving ductility and toughness while retaining castability and machinability. This combination is ideal for threaded fittings. Pig iron is a raw ironmaking product, not a finished material for fittings. High carbon steel is overly hard/brittle for standard threaded casting routes, and spheroidal graphite (SG) iron, while tough, is more often used for larger ductile iron pipe and fittings; traditional small threaded fittings have long used malleable iron due to its balance of properties and cost. Step-by-Step Solution: Define requirements: castable, machinable, tough at threads.Match material family: malleable iron meets these needs.Eliminate non-ideal choices: pig iron (intermediate product), high carbon steel (poor thread toughness), grey iron (flake graphite → low tensile/impact strength).Conclude malleable iron is preferred. Verification / Alternative check: Industry standards and catalogues list “malleable iron fittings” for threaded services in plumbing, gas, and steam applications. Why Other Options Are Wrong: Pig iron is not a service material for fittings. SG iron is tough but less common for small threaded unions versus malleable iron; it favours push-on or flanged systems. High carbon steel is difficult to cast into complex, thin-threaded shapes economically. Grey iron has poor impact strength due to flake graphite. Common Pitfalls: Assuming any cast iron works for threaded fittings; ignoring the embrittling effect of flake graphite at stress concentrators like threads. Final Answer: Malleable iron

Question 9

Phosphor bronze properties: Which of the following accurately summarises the key engineering advantages of phosphor bronze (Cu–Sn–P)?

Accepted Answer

Introduction / Context: Phosphor bronze is a copper–tin alloy with a small phosphorus addition. It is widely used for springs, bearings, electrical contacts, and marine hardware. Knowing its property package helps designers select it over brass, aluminium bronze, or beryllium copper for specific applications. Given Data / Assumptions: Chemistry: copper base with several percent tin and trace phosphorus. Applications require wear resistance, corrosion resistance, and formability. Elasticity here refers to useful springiness (modulus and yield balance). Concept / Approach: Tin in bronze improves strength and corrosion resistance, especially in marine and industrial atmospheres. Phosphorus acts as a deoxidiser during melting and forms fine phosphides that improve wear resistance and fatigue properties. Many grades cold work well, enabling springs and thin-section components with good elastic recovery. Therefore, the most complete statement of its advantages is “all of these.” Step-by-Step Solution: Identify contributions: Sn → strength/corrosion; P → deoxidation/wear improvement.Relate to performance: good bearing life, spring behaviour, and formability.Compare to options: each listed property is valid.Select the comprehensive option “all of these”. Verification / Alternative check: Materials handbooks cite phosphor bronze for marine springs, bushings, and wear plates owing to the property bundle listed. Why Other Options Are Wrong: Choosing any single property underestimates the alloy’s balanced performance. “None of these” contradicts well-established characteristics. Common Pitfalls: Confusing phosphor bronze with brass (Cu–Zn) or aluminium bronze; overlooking that springiness depends on both modulus and yield strength, which are favourable in phosphor bronzes. Final Answer: all of these

Question 10

Critical transformations in steel: On heating plain carbon steel, the temperature at which the austenitic transformation is completed (change ends) is called the __________.

Accepted Answer

Introduction / Context: Heat treatment relies on the precise control of transformation temperatures. For steels, the critical points mark where austenite starts to form and where it is fully formed during heating. Recognising these temperatures ensures proper normalising, hardening, and annealing cycles. Given Data / Assumptions: Plain carbon steel behaviour is considered. Heating is slow enough to observe equilibrium critical points (Ac1, Ac3/Acm). Terminology: “change ends” refers to completion of the austenitisation during heating. Concept / Approach: The lower critical point (Ac1) is where austenite begins to form on heating. The upper critical point (Ac3 for hypoeutectoid or Acm for hypereutectoid) marks the completion of transformation to austenite—i.e., the end of change. Decalescence is the apparent temperature arrest during heating as transformation absorbs heat, while recalescence refers to heat evolution arrests on cooling. Thus, the correct term for completion on heating is the upper critical point. Step-by-Step Solution: Identify transformation start: Ac1 (lower critical point).Identify transformation end on heating: Ac3 or Acm (upper critical point).Relate terminology: “change ends” → upper critical point.Select “upper critical point”. Verification / Alternative check: TTT/phase diagrams and dilatometry show completion of austenitisation at Ac3/Acm, confirming the chosen term. Why Other Options Are Wrong: Lower critical is the start, not the end, on heating. Recalescence pertains to cooling; decalescence describes the phenomenon of heat absorption but is not the nomenclature for the end point. Martensite start is a cooling/quench transformation, unrelated to heating completion. Common Pitfalls: Interchanging decalescence/recalescence; forgetting that Ac1 is start and Ac3/Acm is completion on heating. Final Answer: upper critical point

Question 11

Polymer classification — thermosets: Thermosetting plastics are materials that, when formed under heat and pressure, result in a product that is __________.

Accepted Answer

Introduction / Context: Plastics are broadly divided into thermoplastics and thermosets. Understanding the difference is essential for selecting materials for electrical insulators, adhesives, coatings, and structural components where heat resistance and dimensional stability are required. Given Data / Assumptions: Thermosets undergo crosslinking during cure. Processing typically involves heat and pressure (or catalysts/UV). After cure, reheating does not soften the material. Concept / Approach: Thermosetting polymers cure via chemical reactions forming three-dimensional crosslinked networks. Once set, they become rigid and infusible; reheating does not cause flow, distinguishing them from thermoplastics, which soften upon reheating. Examples include phenolics, epoxies, and melamine-formaldehyde resins. Therefore, the correct description emphasises permanent hardening after forming under heat and pressure. Step-by-Step Solution: Identify thermoset behaviour: crosslinking → permanent set.Relate to processing: heat/pressure cure.Select the option that states “permanently hard product”.Reject options implying remeltability or generic usage statements. Verification / Alternative check: DSC/DMTA data show absence of melting transition after cure in thermosets; degradation occurs instead of melting upon reheating. Why Other Options Are Wrong: No chemical change/soften on heat describes thermoplastics. Flexibility/wear resistance may describe some thermoplastics; it is not defining for thermosets. Friction lining is an application, not a definition (and many linings use specific composites). Remeltability contradicts the essence of thermosets. Common Pitfalls: Assuming all plastics can be reshaped; ignoring cure schedules that determine final properties of thermosets. Final Answer: are formed into shape under heat and pressure and results in a permanently hard product

Question 12

Heat treatment aim — relieving internal stresses and improving machinability: Which heat treatment is chiefly used to relieve prior cold-work stresses in steel and to improve machinability without full phase refinement?

Accepted Answer

Introduction / Context: Cold working introduces high dislocation density and residual stresses, reducing ductility and complicating machining. Choosing the right subcritical heat treatment can restore workability economically without the longer cycles of full annealing. Given Data / Assumptions: Material: low to medium carbon steel after cold forming. Objective: relieve internal stresses, restore ductility, and improve machinability. Avoid unnecessary grain growth or long furnace times. Concept / Approach: Process annealing heats steel below the lower critical temperature (below Ac1) to allow recovery and partial recrystallisation, lowering hardness and relieving stresses caused by cold work. It is quicker and less transformative than full annealing, which heats above Ac1 to form austenite and then furnace cools for maximum softness and grain refinement. Normalising heats above Ac3 and air cools, usually increasing hardness relative to full anneal. Spheroidising targets high-carbon/tool steels to form spheroidised carbides for machining, not generally for low-carbon cold-worked parts. Step-by-Step Solution: Identify need: stress relief + better machinability for cold-worked steel.Match process: subcritical heating below Ac1 = process annealing.Eliminate others: full anneal is slower; normalising increases hardness; spheroidising is specialised for high-carbon steels.Choose process annealing. Verification / Alternative check: Shop practice: sheet/strip steels are routinely process-annealed between forming passes to restore ductility and machinability. Why Other Options Are Wrong: Normalising refines grains but typically raises hardness versus full anneal. Full annealing does increase machinability but is more time- and energy-intensive than needed for simple stress relief. Spheroidising is ideal for high-carbon steels to machine before hardening, not the general low-carbon case here. Common Pitfalls: Using full anneal when a faster, cheaper process anneal would suffice; confusing stress relief temperatures with austenitising ranges. Final Answer: process annealing

Question 13

Property identification: Evaluate the statement — “Malleability is the property of a material due to which it can be rolled or hammered into thin sheets.”

Accepted Answer

Introduction / Context: Precise property terminology avoids design mistakes. Malleability and ductility both describe plasticity, but in different loading modes. Malleability specifically refers to plastic deformation under compressive actions like rolling, hammering, or pressing into sheet or foil. Given Data / Assumptions: The statement associates malleability with forming thin sheets. Plastic deformation under compression is implied. Room-temperature behaviour is assumed unless otherwise stated. Concept / Approach: Malleability is the ability to withstand compressive plastic deformation without cracking, enabling processes like rolling sheet, coin striking, and foil making. Ductility, by contrast, concerns tensile plastic deformation (e.g., wire drawing). Materials such as gold, silver, and annealed copper show very high malleability and can be hammered into extremely thin leaf while remaining intact. Step-by-Step Solution: Clarify definitions: malleability → compression; ductility → tension.Connect to process: sheet/foil production relies on compressive deformation.The statement aligns perfectly with the accepted definition.Conclude that “Agree” is correct. Verification / Alternative check: Manufacturing practice confirms that metals known as malleable (e.g., Au, Ag, Cu, Al) are rolled/hammered into thin sections without cracking when properly annealed. Why Other Options Are Wrong: Limiting to cast irons is incorrect; many cast irons are not malleable in this sense. Calling it ductility confuses compression with tension behaviour. Restricting to cryogenic temperatures is unnecessary and often counterproductive (many metals lose ductility then). Common Pitfalls: Using ductility and malleability interchangeably; ignoring the loading mode that differentiates these properties. Final Answer: Agree

Question 14

Heat treatment vs machinability: Which heat treatment most commonly improves the machinability of steels but simultaneously lowers hardness and tensile strength?

Accepted Answer

Introduction / Context: Machining costs dominate many manufacturing operations. Selecting a heat treatment that softens the microstructure, reduces cutting forces, and improves surface finish can greatly improve throughput before final hardening or service. Given Data / Assumptions: We consider common carbon and low-alloy steels. Goal: better machinability even at the expense of hardness and tensile strength. Processes compared: normalising, full annealing, process annealing, spheroidising. Concept / Approach: Full annealing heats the steel above the upper critical temperature and then furnace cools to produce a coarse, soft pearlitic (sometimes ferrite–pearlite) structure with reduced dislocation density. This yields the lowest hardness among the listed options and therefore the best general improvement in machinability for many steels. Normalising air-cools, often yielding finer pearlite and higher hardness than full anneal. Process annealing is subcritical; it relieves stresses but may not soften as much as a full anneal. Spheroidising is excellent for machining high-carbon/tool steels by forming spheroidised carbides but is not the default for general steels. Step-by-Step Solution: Match objective: maximum softening and machinability → full anneal.Contrast with normalising: typically harder/finer structure → higher cutting forces.Contrast with process annealing: limited softening; mainly stress relief.Consider spheroidising: specialised for high-carbon tools, not broadest answer. Verification / Alternative check: Machining handbooks recommend full anneal before rough machining to reduce tool wear and power consumption in many carbon and low-alloy steels. Why Other Options Are Wrong: Normalising refines grains but does not maximise softness. Process annealing primarily restores ductility with modest hardness change. Spheroidising is best for high-carbon steels specifically, not general-purpose steels. Common Pitfalls: Assuming stress relief alone ensures machinability; overlooking that the softest condition usually machines best for general operations. Final Answer: full annealing

Question 15

Effect of silicon in cast iron: Which statement best describes silicon’s overall influence on cast iron microstructure and castability?

Accepted Answer

Introduction / Context: Silicon is the most important graphitising element in cast irons. By steering solidification towards graphite rather than cementite, silicon changes both mechanical behaviour and machinability, and it also enhances fluidity for thin-walled castings. Given Data / Assumptions: Typical grey iron silicon range about 1.5–3.0% depending on section and strength target. Cooling rates are ordinary for sand casting unless noted. We seek the most representative, single best statement. Concept / Approach: Silicon promotes graphite formation (graphitisation), which lowers hardness relative to white iron and improves machinability because graphite flakes act as chip breakers and solid lubricants. While silicon also helps fluidity, the most fundamental microstructural effect is the shift from carbide to graphite. Hence the best single description is that silicon makes cast iron softer and easier to machine. Options suggesting increased hardness or cementite promotion are contrary to silicon’s principal role. Step-by-Step Solution: Identify silicon’s metallurgical role: strong graphitiser in iron–carbon–silicon systems.Connect to properties: graphite → lower hardness, better machinability.Select the statement capturing this dominant effect.Choose “makes the iron soft and easily machinable”. Verification / Alternative check: Foundry practice raises Si to avoid chilled (white) structures and to enhance casting fill in thin sections, both consistent with softer, more machinable grey iron. Why Other Options Are Wrong: Increases hardness/brittleness: opposite of graphitisation’s typical effect. Makes iron white and hard: describes carbide-rich (low Si/high cooling) outcomes. Aids fusibility/fluidity: partially true but not as comprehensive as the correct microstructural effect. Promotes cementite: incorrect; silicon promotes graphite. Common Pitfalls: Overemphasising fluidity while ignoring the primary microstructural shift; confusing silicon’s effect with that of phosphorus or sulphur. Final Answer: makes the iron soft and easily machinable

Question 16

Allotropy of iron: Delta-iron (the high-temperature phase of iron just below the melting point) possesses which crystal structure?

Accepted Answer

Introduction / Context: Pure iron exhibits allotropy—different crystal structures at different temperatures. Understanding these phases is important for interpreting phase diagrams, heat treatment responses, and high-temperature properties relevant to steelmaking and casting operations. Given Data / Assumptions: Alpha-iron (ferrite) is stable at room temperature up to about 912°C with a BCC structure. Gamma-iron (austenite) is stable roughly from 912°C to 1394°C with an FCC structure. Delta-iron is stable from approximately 1394°C to 1539°C (melting point) with a BCC structure. Concept / Approach: The question states that delta-iron possesses a body-centred cubic space lattice. This matches the accepted allotropic sequence: BCC (alpha) → FCC (gamma) → BCC (delta) as temperature increases to the melting point. The electronic and packing differences between BCC and FCC influence diffusion rates, solubility of carbon, and transformation kinetics in steels. Step-by-Step Solution: Recall allotropy sequence with increasing temperature: alpha (BCC) → gamma (FCC) → delta (BCC).Identify delta-iron’s crystal structure: BCC.Compare with the statement: it claims BCC for delta-iron.Therefore, the statement is correct. Verification / Alternative check: Iron–carbon phase diagram labels the delta-ferrite region near the liquidus as BCC, consistent with the statement. Why Other Options Are Wrong: “Incorrect” would imply delta-iron is not BCC, which contradicts established metallurgical data. Other placeholders do not apply; the core truth is the BCC nature of delta-iron. Common Pitfalls: Confusing gamma (austenite, FCC) with delta; assuming the high-temperature phase must be FCC because gamma is. Final Answer: Correct

Question 17

Materials classification — malleability and ductility in solids Which class of solids is characteristically both malleable (can be hammered into thin sheets) and ductile (can be drawn into wires), as typically observed in engineering metals?

Accepted Answer

Introduction / Context: Malleability and ductility are hallmark mechanical properties used to distinguish classes of solids in materials science. Understanding which bonding type permits layers of atoms to slip without catastrophic fracture is essential for selecting materials for forming processes such as rolling, drawing, and stamping. Given Data / Assumptions: Malleability: ability to deform in compression into thin sheets without cracking. Ductility: ability to deform in tension into wires without breaking. Classes considered: metallic, ionic, covalent (network), and molecular solids. Concept / Approach: Metallic solids are held together by a non-directional metallic bond where valence electrons are delocalized, producing a 'sea of electrons'. This bonding allows atomic planes to slide under stress while maintaining cohesion, enabling extensive plastic deformation. In contrast, ionic solids have strong directional electrostatic attractions; shifting planes brings like charges adjacent, causing brittle failure. Covalent network solids (e.g., diamond, silica) have rigid, highly directional bonds that resist plastic flow, leading to brittleness. Molecular solids are bound by weak van der Waals forces; they are soft and often low-melting, not typically malleable or ductile in the engineering sense. Step-by-Step Solution: Identify the property pair (malleability + ductility) required for metalworking operations.Relate to bonding: non-directional metallic bonding accommodates slip systems.Evaluate alternatives: ionic and covalent networks fracture rather than plastically flow; molecular solids deform poorly and lack structural integrity.Therefore, the correct selection is metallic solids. Verification / Alternative check: Common metals (gold, silver, copper, aluminum, low-carbon steel) are routinely rolled into sheets and drawn into wires in industrial practice—direct empirical confirmation of malleability and ductility inherent to metallic bonding and crystal slip systems. Why Other Options Are Wrong: Ionic solids are brittle due to repulsive like-charge alignment upon slip; covalent network solids resist plasticity because bonds are highly directional; molecular solids lack strong bonding needed for structural ductility. Common Pitfalls: Confusing 'softness' with ductility—molecular solids may be soft but not ductile; also assuming all metals are equally ductile (some, like cast irons, are not metallic solids in the pure sense and are brittle due to graphite/carbide morphology). Final Answer: Metallic solids

Question 18

Empirical hardness–strength correlation for steels A carbon steel has a Brinell hardness number (BHN) of 100. Using the standard engineering correlation between BHN and ultimate tensile strength (UTS), the UTS is closest to which value (in N/mm²)?

Accepted Answer

Introduction / Context: In shop-floor quality control and quick estimates, engineers often convert hardness readings into approximate tensile strengths. For plain carbon and low-alloy steels, a widely used empirical rule links Brinell hardness number (BHN) to ultimate tensile strength (UTS). Given Data / Assumptions: Material: carbon steel. Measured hardness: BHN = 100. Use the common rule: UTS (MPa) ≈ k * BHN, with k around 3.45 for steels. Note: 1 MPa = 1 N/mm². Concept / Approach: The practical correlation is often stated as UTS (MPa) ≈ 3.45 * BHN for carbon/low-alloy steels in typical heat-treated or normalized conditions. While the constant can vary slightly (3.3–3.5) depending on condition, 3.45 is a standard mid-value for quick estimates. This allows rapid strength estimation without destructive tensile testing. Step-by-Step Solution: Write the rule: UTS ≈ 3.45 * BHN.Insert BHN = 100: UTS ≈ 3.45 * 100.Compute: UTS ≈ 345 MPa.Round to the nearest provided choice: 350 N/mm² (since 1 MPa = 1 N/mm²). Verification / Alternative check: Some handbooks use UTS (MPa) ≈ 3.5 * BHN; with BHN = 100, that gives 350 MPa—exactly matching the chosen option. The difference between 345 and 350 is within practical estimation tolerance for this rule of thumb. Why Other Options Are Wrong: 100 and 200 N/mm² are too low for BHN 100 steel; 700 and 1000 N/mm² correspond to much higher hardness levels (BHN ~ 200–300+). Choosing these would ignore the established proportionality. Common Pitfalls: Treating the correlation as exact across all alloys and heat treatments. It is an estimate for common steels; tool steels, austenitic stainless, and nonferrous alloys deviate from this factor. Final Answer: 350 N/mm²

Question 19

Application suitability of silicon (electrical) steel Evaluate the statement: “Nuts and bolts are made from silicon steel.” Decide whether it is correct, considering the typical uses and properties of silicon (electrical) steels.

Accepted Answer

Introduction / Context: Material selection for hardware such as nuts and bolts depends on tensile strength, toughness, and thread integrity. Silicon steels—often called electrical steels—are engineered for magnetic properties (low core loss, high permeability) rather than fastener performance. Given Data / Assumptions: Nuts and bolts typically require medium to high tensile strength and good toughness (e.g., low/medium carbon steels heat-treated to grades like 8.8, 10.9). Silicon steels contain 1–3.5% Si to tailor magnetic behavior and are supplied as thin laminations. No requirement for magnetic optimization exists in fasteners. Concept / Approach: Fastener steels prioritize strength classes and fatigue resistance. Common choices include boron steel, medium carbon alloy steels (e.g., 35CrMo, 4140), or low-carbon steels for low-strength hardware. In contrast, silicon electrical steels are relatively brittle in thick sections and are processed as sheets for transformer and motor cores. Their property set (magnetic, not mechanical) makes them unsuitable for threaded fasteners. Step-by-Step Solution: Identify required properties for nuts/bolts: high tensile strength, ductility, fatigue resistance.Identify silicon steel properties: magnetic optimization, lamination-grade sheets, not designed for high thread strength.Match application to material: mismatch between needs and silicon steel capabilities.Therefore, the statement is incorrect. Verification / Alternative check: Standards for fasteners (e.g., ISO property classes) list mechanical properties aligned with carbon/low-alloy steels and heat treatments, not electrical steels. Why Other Options Are Wrong: Qualifiers suggesting special fastener types or coatings do not solve the fundamental mismatch in mechanical properties; coatings manage corrosion, not core strength/ductility. Common Pitfalls: Assuming that the presence of silicon always improves strength; in electrical steels, silicon targets magnetics and can reduce formability in thick sections. Final Answer: Incorrect

Question 20

Constituents of hardened steel after quenching Identify the essential (dominant) microconstituent present in hardened steel immediately after quenching from the austenitic region.

Accepted Answer

Introduction / Context: Hardening of steel is achieved by heating into the austenite region and then quenching rapidly to transform austenite into a hard, supersaturated structure. Recognizing the resulting microconstituent is central to heat-treatment design and property control. Given Data / Assumptions: Steel is heated above the critical temperature to form austenite. Quenching is sufficiently rapid to avoid diffusion-controlled transformations. Carbon content is sufficient to form martensite upon quench. Concept / Approach: Under rapid cooling, the austenite-to-pearlite or bainite transformations (which require diffusion) are suppressed. Instead, a diffusionless shear transformation produces martensite: a supersaturated, body-centered tetragonal structure with very high hardness. Some retained austenite may remain depending on composition and quench severity, but the essential constituent responsible for hardness is martensite. Step-by-Step Solution: Heat steel to austenitize (single phase austenite).Quench rapidly to bypass the nose of the time–temperature–transformation curve.Form martensite by shear without carbon diffusion.Therefore, hardened steel contains predominantly martensite (with possible retained austenite). Verification / Alternative check: Hardness measurements (e.g., HRC 60+) and characteristic needle-like microstructure confirm martensite. Tempering reduces brittleness by controlled precipitation and stress relief, transforming martensite toward tempered structures. Why Other Options Are Wrong: Pearlite forms under slower cooling; austenite is the high-temperature phase not stable at room temperature; troostite and sorbite are tempered microstructures (historical terms) formed after tempering, not as-quenched hardening. Common Pitfalls: Confusing bainite or tempered martensite with as-quenched martensite; the question explicitly asks about hardened steel immediately after quenching. Final Answer: martensite

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