Question 1

Compare nylon 6 with nylon 6,6: which property is generally lower for nylon 6?

Accepted Answer

Introduction / Context: Nylon 6 (from caprolactam) and nylon 6,6 (from adipic acid and hexamethylene diamine) are closely related polyamides with similar mechanical profiles but different melting points and thermal stability. Knowing their differences guides fibre, film, and engineering plastic selection. Given Data / Assumptions: Nylon 6 has a lower melting temperature (≈220–225°C) than nylon 6,6 (≈260–265°C). Thermal-oxidative stability and heat deflection are generally higher for nylon 6,6. Other properties may be grade-dependent and close in value. Concept / Approach: The amide group density and crystallinity differences give nylon 6,6 a higher melting point and better high-temperature strength retention than nylon 6. Thus, “thermal stability” is the most consistent differentiator. Abrasion resistance, adhesion, and hardness depend on specific formulations and can overlap significantly between the two. Step-by-Step Solution: Identify robust differentiator: melting point and heat resistance.Relate to stability: nylon 6,6 > nylon 6 for thermal stability.Select “thermal stability” as lower for nylon 6. Verification / Alternative check: Datasheets for unfilled grades consistently show higher melting point and higher heat deflection temperature for nylon 6,6. Why Other Options Are Wrong: Abrasion/adhesion/hardness: not universally lower; depend strongly on additives, conditioning, and processing. Common Pitfalls: Assuming a single property trend covers all grades; fillers and moisture conditioning influence mechanical data significantly. Final Answer: Thermal stability

Question 2

Copolymers and homopolymers: which listed material is not a polymer of two monomers?

Accepted Answer

Introduction / Context: Polymers may be homopolymers (one monomer) or copolymers (two or more monomers). Identifying which materials are copolymers helps anticipate properties and applications. Given Data / Assumptions: Teflon is PTFE, produced from a single monomer (tetrafluoroethylene). Bakelite forms from phenol and formaldehyde (two monomers; condensation). SBR is a copolymer of styrene and butadiene. Concept / Approach: A homopolymer is built from repeating units of a single monomer; PTFE fits this definition. Both Bakelite and SBR involve two different monomers, classifying them as copolymers (though Bakelite forms a crosslinked network via condensation rather than addition). Step-by-Step Solution: Identify PTFE as homopolymer of tetrafluoroethylene → not a two-monomer polymer.Recognise Bakelite (phenol + formaldehyde) and SBR (styrene + butadiene) as two-monomer systems. Verification / Alternative check: Polymer reference charts list PTFE as a homopolymer with exceptional chemical resistance and low surface energy. Why Other Options Are Wrong: Bakelite and SBR both use two monomers; “None of these” therefore cannot be correct. Common Pitfalls: Assuming “network polymers” like Bakelite are not considered in the monomer count; despite crosslinking, two distinct monomers are used. Final Answer: Teflon

Question 3

Identify the polyolefin among the following polymers (choose the single correct polyolefin).

Accepted Answer

Introduction / Context: Polyolefins are polymers made from simple olefin monomers such as ethylene and propylene. They are ubiquitous due to low cost, chemical resistance, and versatile processing. Distinguishing a polyolefin from other vinyl or diene polymers is a foundational classification task. Given Data / Assumptions: Polypropylene (PP) is produced from propylene, a simple olefin → polyolefin. Polystyrene derives from styrene (vinyl aromatic), typically not classed as a polyolefin. Neoprene (polychloroprene) is a chlorinated diene elastomer, not a polyolefin. Concept / Approach: Polyolefins are saturated hydrocarbon backbones from alpha-olefin monomers (e.g., PE from ethylene, PP from propylene). Polystyrene contains a pendant phenyl group and distinct properties; it is categorised as a vinyl polymer, not a polyolefin. Neoprene contains chlorine and double-bond-derived structure, hence is an elastomer outside the polyolefin family. Step-by-Step Solution: Identify olefin-based polymer → polypropylene.Eliminate polystyrene (vinyl aromatic) and neoprene (chloroprene rubber).Select polypropylene as the sole polyolefin listed. Verification / Alternative check: Commodity polymer classifications list PE and PP as the principal polyolefins. Why Other Options Are Wrong: Polystyrene: not a polyolefin; different monomer family and properties.Neoprene: synthetic rubber; chlorinated diene polymer.None of these: incorrect because PP is indeed a polyolefin. Common Pitfalls: Generalising “vinyl polymers” and “polyolefins” as the same; they are overlapping but not identical categories. Final Answer: Polypropylene

Question 4

Polymer classification: which of the following materials is a homopolymer (formed from a single monomer species, without copolymerization)?

Accepted Answer

Introduction / Context: A homopolymer is produced by polymerizing only one type of monomer, while a copolymer results from two or more different monomers. Distinguishing these classes is foundational in polymer science because it predicts repeat-unit chemistry, crystallinity, processing window, and end-use properties. Given Data / Assumptions: The options include common industrial polymers and resins. We interpret “homopolymer” as the repeat unit originating from a single monomer type. Trade names correspond to well-known chemistries (e.g., Bakelite = phenol–formaldehyde). Concept / Approach: Neoprene is polychloroprene, obtained by addition polymerization of a single monomer, chloroprene (2-chloro-1,3-butadiene). Hence, it is a homopolymer. In contrast, Bakelite is a condensation polymer formed from phenol and formaldehyde (copolymer network). Nylon-66 is a condensation copolymer produced from hexamethylene diamine and adipic acid. Terylene (polyethylene terephthalate, PET) is a condensation copolymer of ethylene glycol and terephthalic acid. Step-by-Step Solution: Identify each polymer's monomer(s).Neoprene: single monomer → chloroprene → homopolymer.Bakelite: phenol + formaldehyde → copolymer network.Nylon-66: diamine + diacid → copolymer.Terylene (PET): glycol + acid → copolymer. Verification / Alternative check: Manufacturer data and standard textbooks list neoprene as polychloroprene with one monomer species in the chain backbone. Why Other Options Are Wrong: Bakelite: two monomers; crosslinked network.Nylon-66: two monomers (salt formation), not a homopolymer.Terylene/PET: two monomers, not a homopolymer. Common Pitfalls: Assuming trade names imply homopolymer; many common materials are actually copolymers or condensation networks. Final Answer: Neoprene

Question 5

Rubber compounding: which additive is most important for improving wear resistance and toughness in both natural rubber and SBR (styrene–butadiene rubber)?

Accepted Answer

Introduction / Context: Rubber compounds rarely use raw polymer alone. Fillers, plasticisers, antioxidants, curatives, and process aids are blended to achieve targeted properties. Among these, reinforcing fillers dramatically affect abrasion resistance, tensile strength, and cut growth behavior, which are critical in tyres, belts, and dynamic seals. Given Data / Assumptions: Base elastomers: natural rubber (NR) and SBR. Objective: improve wear (abrasion) performance and toughness. Typical compounding options are listed. Concept / Approach: Carbon black acts as a reinforcing filler by creating a strong polymer–filler interface and by forming a fine, percolated network in the matrix that resists crack initiation and propagation. It increases modulus, tensile strength, tear resistance, and abrasion resistance. Grades (N220, N330, etc.) are chosen based on surface area and structure to tune hysteresis and wear. Phosphorous, pine oil, and rosin are not primary wear-reinforcing agents in tyre-grade rubbers; they serve other niche roles. Step-by-Step Solution: Identify reinforcement need: high abrasion and toughness.Recognize carbon black as the standard reinforcing filler for NR and SBR.Select carbon black as the correct choice. Verification / Alternative check: Tyre tread formulations universally include carbon black or, in some modern cases, silica–silane systems as the main reinforcing phase. Why Other Options Are Wrong: Phosphorous: not a standard reinforcing filler in rubber.Pine oil, rosin: processing or tackifying aids, not primary wear improvers. Common Pitfalls: Confusing antioxidants or plasticisers with reinforcing fillers; overlooking that wear resistance is strongly tied to filler type and loading. Final Answer: Carbon black

Question 6

Manufacturing of plastic tubes and pipes: which moulding/forming process is most commonly employed for continuous hollow sections?

Accepted Answer

Introduction / Context: Thermoplastic processing routes differ by geometry and throughput. Long, continuous hollow profiles such as tubes, pipes, and corrugated conduits are key products in infrastructure and packaging, and their economics depend on steady, continuous production. Given Data / Assumptions: Part geometry: long, constant cross-section hollow parts. Materials: PE, PP, PVC, PVDF and similar thermoplastics. Need for continuous, high-throughput manufacturing. Concept / Approach: Extrusion forces molten polymer through an annular die with a calibrated mandrel or pin to create hollow sections. Downstream sizing tanks, vacuum calibration, haul-off, and cutting create finished pipes with tight OD/ID tolerances. Injection and transfer moulding make discrete parts in closed moulds; compression moulding suits thermosets and simple shapes. Therefore, extrusion is the standard method for pipes/tubes. Step-by-Step Solution: Match geometry (continuous hollow) to process capability.Identify extrusion as continuous shaping through a die.Confirm auxiliary steps: cooling, sizing, and cutting for final dimension control. Verification / Alternative check: Commercial pipe lines run single- or twin-screw extruders with vacuum calibration baths—industry standard. Why Other Options Are Wrong: Injection/transfer: discontinuous, closed-mould processes.Compression: mainly for thermosets or flat/simple shapes. Common Pitfalls: Assuming injection moulding due to popularity; process must match continuous profile geometry. Final Answer: Extrusion

Question 7

Rubber protection: antioxidants are added to guard against light, heat, and ozone degradation. Which compound class is a widely used rubber antioxidant?

Accepted Answer

Introduction / Context: Unsaturated elastomers degrade via oxidative chain scission, crosslink modification, and surface cracking (e.g., ozone cracking). Antioxidants and antiozonants protect rubber articles such as tyres, belts, and seals, improving service life and appearance. Given Data / Assumptions: The question targets chemical classes that function as antioxidants. Light, heat, and ozone are cited as stressors. We assume typical NR, SBR, and related elastomer systems. Concept / Approach: Alkylated diphenyl amines (and related p-phenylenediamines) are effective antioxidants/antiozonants that scavenge free radicals and intercept oxidative pathways, thereby delaying embrittlement and surface cracking. Carbon (elemental), magnesium (metal), and “Thiokol” (polysulfide rubber) are not used as antioxidants in standard compounding practice. Step-by-Step Solution: Recall common antioxidant families: amine-based and phenolic types.Match “alkylated diphenyl amine” with proven antioxidant function in rubber.Select it as the correct answer. Verification / Alternative check: Tyre compound data sheets list amine antioxidants/antiozonants (e.g., 6PPD) as standard ingredients. Why Other Options Are Wrong: Carbon and magnesium: not antioxidants in rubber compounding.Thiokol: itself an elastomer, not an antioxidant additive. Common Pitfalls: Confusing carbon black's reinforcement with antioxidant function; mixing base elastomers with additives. Final Answer: Alkylated diphenyl amine

Question 8

Ziegler–Natta polyethylene: the low-pressure Ziegler process most characteristically produces which type of polyethylene?

Accepted Answer

Introduction / Context: Polyethylene manufacturing follows distinct routes. LDPE is typically made via free-radical polymerization at very high pressures, producing highly branched chains. Ziegler–Natta catalysis, developed for low-pressure operation, yields linear chains with higher crystallinity, characteristic of HDPE. Given Data / Assumptions: Ziegler–Natta = transition-metal catalyst system (e.g., TiCl4 with organoaluminium cocatalyst). Operates at relatively low pressures and temperatures compared to LDPE processes. Target property: higher density through low branching and high crystallinity. Concept / Approach: The catalyst surface controls chain growth stereochemistry and suppresses long-chain branching, yielding linear polyethylene. This increases crystallinity and density, hence HDPE. Statements that the process uses no catalyst or very high pressure are incorrect; those describe neither Ziegler–Natta practice nor LDPE's classic high-pressure radical route. Step-by-Step Solution: Associate Ziegler–Natta with low-pressure catalysis.Recognize linear PE product → higher density.Select HDPE as the correct outcome. Verification / Alternative check: Process encyclopedias classify HDPE under Ziegler–Natta/catalytic routes, whereas LDPE is produced in tubular/autoclave reactors at 1000–3000 bar. Why Other Options Are Wrong: No catalyst: false—Ziegler–Natta requires catalysts.LDPE/high pressure: those describe the radical route, not Ziegler–Natta. Common Pitfalls: Assuming all polyethylene is made the same way; conflating product density with processing pressure. Final Answer: High-density polyethylene (HDPE)

Question 9

Concept check: identify the single WRONG statement about polymer fibres, rubber tackiness, melting behavior, and UV resistance (assume standard unstabilised materials).

Accepted Answer

Introduction / Context: Polymer exam questions often mix true and false facts about processing and properties. Here we must find the single incorrect statement by checking each claim against typical materials and conditions. Given Data / Assumptions: “Polymeric fibres” include commodity and technical fibres (e.g., PP, PAN, PET, nylon). “Uncured rubber” refers to raw, unvulcanised elastomer exhibiting tack. Thermoplastics may be amorphous or semicrystalline. Polyethylene (polythene) is considered without stabilisers or carbon black. Concept / Approach: (a) is wrong: many fibres are made by addition polymerisation (polypropylene fibres, polyacrylonitrile precursor fibres). (b) is correct: uncured rubber is tacky due to chain mobility and surface adhesion. (c) is broadly correct: many thermoplastics soften over a range; amorphous polymers lack a sharp melting point and exhibit glass transition; even semicrystalline grades show melting over a range. (d) is correct as worded: unstabilised PE degrades under UV, hence stabilisers or carbon black are commonly used. Step-by-Step Solution: Evaluate (a): demonstrably false → choose as the wrong statement.Confirm (b), (c), (d) as accurate under the given assumptions. Verification / Alternative check: Textile and fibre technology references list PP (addition polymer) as a staple fibre; UV resistance charts show PE needs stabilisation. Why Other Options Are Wrong (i.e., not the wrong statement): (b) Correct property of uncured rubber.(c) Consistent with thermal transitions in polymers.(d) Correctly notes poor UV resistance without additives. Common Pitfalls: Assuming “sharp melting” like small molecules; ignoring the role of crystallinity and molecular weight distribution. Final Answer: Polymeric fibres are never produced by addition polymerisation.

Question 10

True/false mix: choose the single WRONG statement about common polymers and rubbers (thermoplasticity, natural origin, and copolymer/homopolymer identities).

Accepted Answer

Introduction / Context: Accurate classification of common materials avoids design and processing mistakes. This question asks you to identify the single wrong statement in a list that mixes thermoplasticity, natural polymers, and specific polymer identities. Given Data / Assumptions: Thermoplastic vs thermoset definitions are standard. “Protein” refers to polypeptides from natural sources. Neoprene = polychloroprene (synthetic). SBR = styrene–butadiene rubber (copolymer). Concept / Approach: (a) True: polystyrene softens on heating and is recyclable as a thermoplastic. (b) True: proteins are natural polymers with peptide bonds (polyamides). (c) False: neoprene is a synthetic rubber produced by addition polymerisation of chloroprene; natural rubber is cis-1,4-polyisoprene from latex. (d) True as worded: polyethylene is a homopolymer (from ethylene); SBR is a copolymer of styrene and butadiene. Step-by-Step Solution: Check each statement against accepted definitions.Identify (c) as the only incorrect statement.Confirm (a), (b), (d) as correct. Verification / Alternative check: Materials datasheets and polymer encyclopedias list neoprene under synthetic rubbers and SBR explicitly as a copolymer. Why Other Options Are Not Wrong: (a) and (d) align with standard classifications.(b) proteins are indeed natural polymers. Common Pitfalls: Confusing “natural rubber” with any rubbery material; forgetting SBR's two-monomer composition. Final Answer: Neoprene is a natural rubber.

Question 11

Polymer film properties: given that cellulose acetate films have very high permeability (~5000 g per 100 m^2), which listed polymer typically exhibits the highest film elongation (around 100%)?

Accepted Answer

Introduction / Context: Film engineering balances permeability, strength, and extensibility. Elongation at break indicates ductility and toughness, influencing package integrity and drawdown during processing (e.g., blown film). Given Data / Assumptions: Target elongation magnitude: on the order of 100% or higher. Commodity films compared: polyethylene (PE), cellulose acetate/cellophane, and PTFE. Standard processing conditions and typical commercial grades are assumed. Concept / Approach: Polyethylene (particularly LDPE and LLDPE film grades) exhibits high elongation and toughness, routinely exceeding 100% elongation. Cellophane and cellulose acetate films are relatively stiff and brittle compared with PE and generally show lower elongation. PTFE films are creep-resistant and chemically inert but are not known for high elongation at break in typical thin-film applications. Step-by-Step Solution: Match high elongation requirement with film-grade PE behavior.Eliminate cellulose-based films and PTFE due to lower ductility.Select polythene as the correct answer. Verification / Alternative check: Datasheets for LDPE/LLDPE films report elongation at break between ~100% and 600% depending on formulation and orientation. Why Other Options Are Wrong: Cellophane/cellulose acetate: lower elongation; more brittle.PTFE: exceptional chemical resistance but not maximised for elongation at break. Common Pitfalls: Equating permeability traits with extensibility; these properties are largely independent. Final Answer: Polythene

Question 12

Market share insight: in which application area is the maximum consumption of polymers observed globally among the listed choices?

Accepted Answer

Introduction / Context: Understanding the demand landscape for polymers helps prioritise product development and recycling efforts. Volumes are dominated by high-throughput, short-life applications requiring lightweight, formable materials. Given Data / Assumptions: Choices include major but distinct application sectors. We consider global consumption trends for commodity plastics. Packaging encompasses films, bottles, containers, and laminates. Concept / Approach: Packaging consistently accounts for the largest share of polymer consumption because it leverages plastics’ low density, barrier tunability, transparency, toughness, and processability at scale. PE, PP, PET, and PS dominate packaging formats. Electrical insulation, toys, and coatings are important but smaller by volume compared with packaging. Step-by-Step Solution: Compare sector volumes: packaging > construction & electrical in many markets; toys are niche.Identify packaging as the single largest consumer of commodity polymers. Verification / Alternative check: Industry reports and market analyses list packaging as the top application by tonnage for PE, PP, PET, and multilayer films. Why Other Options Are Wrong: Electrical insulation, toys, coatings: significant but collectively smaller markets than packaging. Common Pitfalls: Confusing value with volume; electronics and coatings may have high value density but lower tonnage. Final Answer: Packaging

Question 13

Medical textiles: why is rayon considered superior to cotton for making gauze for wound treatment?

Accepted Answer

Introduction / Context: Gauze performance depends on absorbency, non-adherence to wound beds, and comfort. Rayon (viscose rayon) is a regenerated cellulose fibre engineered for high moisture management and softness, making it a common choice in medical dressings. Given Data / Assumptions: Rayon is regenerated cellulose; cotton is natural cellulose. Desired properties: high absorbency and low adhesion to wounds. We are comparing general, widely cited characteristics. Concept / Approach: Rayon fibres are manufactured to have a fine denier and high amorphous content that promotes capillary uptake. This yields very high water absorbency (often >90% of fibre weight). Rayon also forms soft, smooth fabrics that tend to cling less to wound surfaces compared with some cotton gauzes, reducing trauma during dressing changes. Being “synthetic” (option a) is not itself a performance reason; what matters are fibre morphology and sorption behavior. Step-by-Step Solution: Identify target properties: absorbency and non-adherence.Map rayon's regenerated-cellulose structure to high water uptake and soft hand.Conclude that both (b) and (c) justify the preference. Verification / Alternative check: Medical textile guidelines list viscose-rayon gauze as highly absorbent and low-adherent, often blended with cotton for cost and mechanical strength. Why Other Options Are Wrong: (a) Merely stating synthetic origin does not explain performance benefits.(b) and (c) together capture the functional advantages relevant to wound care. Common Pitfalls: Equating fibre origin with performance; overlooking surface finish, denier, and weave density in gauze design. Final Answer: Both (b) and (c).

Question 14

Polymer identification in fibres: polycaprolactam is commercially known as which nylon grade in fibre and engineering plastic applications?

Accepted Answer

Introduction / Context: Many examination questions ask you to map a chemical name to its widely used trade or generic polymer name. Polycaprolactam is one such case, intimately tied to the nylon family. Understanding these equivalences helps engineers choose the correct resin or fibre for performance, processing temperature, and chemical resistance. Given Data / Assumptions: Polycaprolactam is the chemical name given. Answer options include well-known polymer families used in textiles and packaging. We assume standard naming conventions in polymer science. Concept / Approach: Polycaprolactam is formed by ring-opening polymerisation of caprolactam to yield a linear polyamide. Within industry, this material is known as nylon 6 (also written as PA6). By contrast, nylon 6,6 is produced by condensation of hexamethylene diamine and adipic acid, orlon is polyacrylonitrile, and saran typically refers to polyvinylidene chloride or its copolymers. Step-by-Step Solution: Identify monomer → caprolactam.Polymerisation route → ring-opening to polycaprolactam.Industrial name mapping → polycaprolactam = nylon 6 (PA6).Eliminate lookalikes → nylon 6,6 (different monomer pair), orlon (PAN), saran (PVDC). Verification / Alternative check: Materials handbooks, fibre standards, and resin datasheets consistently list polycaprolactam as nylon 6. Typical melt temperature, crystallinity range, and moisture uptake values cross-reference between PA6 and polycaprolactam literature. Why Other Options Are Wrong: Orlon: polyacrylonitrile, not a polyamide.Nylon 6,6: condensation of two monomers; not polycaprolactam.Saran: PVDC, used for barrier films. Common Pitfalls: Confusing nylon 6 with nylon 6,6 because of similar names; ignoring that “6,6” indicates two six-carbon monomers, whereas nylon 6 comes from a single lactam. Final Answer: Nylon 6

Question 15

Phenol–formaldehyde resins: the condensation polymer made from phenol and formaldehyde is commonly called what?

Accepted Answer

Introduction / Context: Identifying polymer families by their synthesis routes is a core skill. Phenol–formaldehyde resins are classic thermosets prepared via condensation polymerisation, and they paved the way for modern plastics. This question checks if you can link the chemistry to its well-known commercial name. Given Data / Assumptions: Monomers: phenol and formaldehyde. Process: condensation polymerisation with crosslinking. We compare with unrelated polymer systems. Concept / Approach: Phenol and formaldehyde react to form resols/novolacs which cure to three-dimensional networks known commercially as Bakelite. Teflon (PTFE) is a fluoropolymer made by addition polymerisation of tetrafluoroethylene, polyester is a broad class often from diacids and diols, and nylon-66 is a polyamide from hexamethylene diamine and adipic acid. Step-by-Step Solution: Identify the given monomers → phenol + formaldehyde.Recognise condensation with crosslinking → thermoset.Map to trade name → Bakelite.Eliminate PTFE, polyester, nylon-66 as different chemistries. Verification / Alternative check: Standard polymer textbooks and historic accounts credit Bakelite as one of the first synthetic plastics, based on phenol–formaldehyde chemistry. Why Other Options Are Wrong: Teflon: fluoropolymer by addition; unrelated monomers.Polyester: diol + diacid/anhydride; not phenol–formaldehyde.Nylon-66: polyamide from diamine + diacid. Common Pitfalls: Assuming all condensation polymers are “polyesters” or “polyamides”; overlooking crosslinked resin behaviour of phenolics. Final Answer: Bakelite

Question 16

Natural polyamide fibres: which of the following is correctly classified as a natural polyamide fibre?

Accepted Answer

Introduction / Context: Polyamides can be synthetic (like nylons) or natural (like proteins). Natural fibres composed of proteins contain amide linkages in their backbone and are therefore polyamides by chemistry. The question asks you to select one natural polyamide fibre from the list. Given Data / Assumptions: Candidates include wool, silk, and cotton. We treat “natural polyamide” as a fibre whose polymer backbone comprises peptide (amide) bonds. One option must be chosen as correct. Concept / Approach: Wool is primarily keratin, a protein with peptide linkages, thus a natural polyamide fibre. Silk is also a protein (fibroin), and by strict chemistry, it too is a natural polyamide. However, many exam conventions select “wool” as the canonical answer in single-correct MCQs when only one response is permitted. Cotton is cellulose (a polysaccharide with glycosidic linkages), not a polyamide. Step-by-Step Solution: Classify wool → protein → peptide bonds → natural polyamide.Recognise silk is also protein, but select the standard single answer per exam convention.Eliminate cotton → cellulose (not polyamide). Verification / Alternative check: Textile chemistry references identify wool and silk as protein fibres; examination banks often key “wool” where only one choice is allowed. Why Other Options Are Wrong: Silk: chemically a protein, but MCQ expects a single answer; wool is the keyed choice in many banks.Cotton: polysaccharide; no amide bonds.None of these: incorrect because at least wool fits. Common Pitfalls: Marking both wool and silk when the interface allows only one correct choice; forgetting that cotton is carbohydrate-based. Final Answer: Wool

Question 17

Rubber technology: which substance is most commonly used as a vulcanisation agent for unsaturated rubbers?

Accepted Answer

Introduction / Context: Vulcanisation is the process of crosslinking rubber chains to improve elasticity, heat resistance, and mechanical strength. The chemical used for crosslinking—particularly in natural rubber and SBR—is central to performance and processing. Given Data / Assumptions: We consider traditional, widely-used systems for unsaturated rubbers like NR and SBR. Catalysts/activators/accelerators may be present, but the question asks for the primary vulcanising agent. Concept / Approach: Sulphur is the most common vulcanising agent for unsaturated rubbers, forming sulphidic crosslinks between polymer chains. Accelerators (e.g., sulfenamides, thiazoles) and activators (ZnO, stearic acid) control cure rate and crosslink structure. Bromine and platinum are not standard vulcanising agents for these rubbers; alumina is an inorganic filler, not a curative. Step-by-Step Solution: Identify material class → unsaturated rubber (NR/SBR).Recall standard cure → sulphur crosslinking with accelerator/activator system.Choose sulphur as the principal vulcanising agent. Verification / Alternative check: Tyre compounds and industrial rubber recipes universally cite sulphur cures or sulphur donors, with well-known cure kinetics and crosslink types (mono-, di-, polysulphidic). Why Other Options Are Wrong: Bromine: not a typical vulcanising agent for NR/SBR.Platinum: used as a catalyst in silicone (addition) cures, not for NR/SBR sulphur vulcanisation.Alumina: filler/reinforcement, not a cure agent. Common Pitfalls: Confusing silicone rubber cure (platinum-catalysed hydrosilylation) with sulphur vulcanisation of diene rubbers. Final Answer: Sulphur

Question 18

Thermal limits in service: most common plastics are generally considered safe up to a maximum continuous-use temperature of approximately how many °C?

Accepted Answer

Introduction / Context: Plastics soften, creep, or degrade when exposed to elevated temperatures. Designers therefore quote a maximum continuous-use temperature (CUT) or heat deflection temperature (HDT) limit for commodity thermoplastics. This question probes approximate, rule-of-thumb knowledge for general plastics—not high-performance engineering polymers. Given Data / Assumptions: “Most of the plastics” refers to commodity resins such as PE, PP, PVC, PS, ABS. Continuous-use temperature is meant, not brief spikes. Unfilled, unstabilised grades are implied. Concept / Approach: Commodity plastics typically serve well below 100 °C for continuous applications. HDT values for PE and PP are well below 100 °C (though PP can handle hot fill briefly). Engineering polymers like nylon, PBT, PC, or PPS can exceed 100–150 °C, while high-performance plastics (PEEK, PTFE, PI) go far beyond. The safe blanket answer for “most” common plastics is therefore around 100 °C. Step-by-Step Solution: Identify scope → commodity plastics.Relate CUT/HDT data → typically <= 100 °C for continuous duty.Select 100 °C as the approximate maximum for “most” plastics. Verification / Alternative check: Datasheets show PE and PVC CUTs in the 60–80 °C range, PP somewhat higher but still often <100 °C for continuous use; hence 100 °C is a sensible upper bound. Why Other Options Are Wrong: 150, 350, 450 °C: Typical only for engineering or high-performance polymers; not true for “most” plastics. Common Pitfalls: Confusing peak or short-term temperatures with continuous ratings; extrapolating from specialised materials to all plastics. Final Answer: 100

Question 19

Thermosets vs thermoplastics: which listed material is a thermosetting plastic (i.e., crosslinks irreversibly on curing)?

Accepted Answer

Introduction / Context: Thermosetting plastics form crosslinked networks during cure and cannot be remelted. Thermoplastics soften repeatedly on heating. Differentiating these classes is essential for specifying molding methods, recycling routes, and end-use temperature limits. Given Data / Assumptions: PVC, polythene (PE), and polystyrene are common thermoplastics. Bakelite is a classic phenolic thermoset. We assume standard, unmodified grades. Concept / Approach: Phenol–formaldehyde (Bakelite) cures irreversibly into a crosslinked network, exhibiting high heat resistance, good dimensional stability, and solvent resistance. PVC, PE, and PS are linear or lightly branched thermoplastics; they can be softened and reshaped by heat. Step-by-Step Solution: Map each polymer to its class (thermoplastic vs thermoset).Identify Bakelite as a thermoset.Eliminate PVC, PE, PS as thermoplastics. Verification / Alternative check: Processing methods: Bakelite is compression/transfer molded and cured; the others are extruded or injection molded without irreversible chemical crosslinking. Why Other Options Are Wrong: PVC/PE/PS are melt-processable repeatedly; they do not undergo thermoset curing by default. Common Pitfalls: Assuming additives or crosslinking agents turn all plastics into thermosets; base resins remain thermoplastic unless intentionally crosslinked (e.g., XLPE). Final Answer: Bakelite

Question 20

Rubber nomenclature: Buna-N is the common name for which elastomer family?

Accepted Answer

Introduction / Context: Trade and historical names for rubbers can be confusing. “Buna-N” appears widely in seals, gaskets, and fuel-system components. Being able to translate these names into chemical families is essential for compatibility and property selection. Given Data / Assumptions: Buna-N is widely used in oil-resistant applications. Options include several common elastomer families. We assume standard nomenclature used in industry. Concept / Approach: Buna-N stands for butadiene–acrylonitrile copolymer, known generically as nitrile rubber (NBR). Butyl rubber is isobutylene-isoprene copolymer, neoprene is polychloroprene, and thiokol refers to polysulfide rubbers. NBR’s acrylonitrile content imparts oil and fuel resistance, making it prevalent in automotive and industrial seals. Step-by-Step Solution: Decode Buna-N → butadiene + acrylonitrile → NBR.Match properties → oil/fuel resistance typical of nitrile rubber.Eliminate butyl, neoprene, and polysulfide as different chemistries. Verification / Alternative check: Seal catalogs and materials handbooks list Buna-N and NBR interchangeably, with grades defined by acrylonitrile percentage and cure system. Why Other Options Are Wrong: Butyl: gas impermeability focus; chemistry differs.Neoprene: polychloroprene; halogenated rubber with different compatibility.Thiokol: polysulfide; notable for solvent resistance and odor, not NBR. Common Pitfalls: Assuming “Buna” always means butyl; historically “Buna” was used broadly, but Buna-N specifically denotes NBR. Final Answer: Nitrile rubber

Polymer Technology Questions