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Question 1Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the Ellingham diagram for the formation of metal oxides, the standard Gibbs energy of formation of a metal oxide is plotted against temperature. What does the slope of a line for the reaction M + O2 → MO2 correspond to, given that the entropy change of the reaction is approximately constant over the temperature range?

  1. The negative of the entropy change of the reaction Correct answer.
  2. The heat capacity of the oxide
  3. The enthalpy change of the reaction
  4. The equilibrium constant of the reaction

Explanation

Ellingham plots graph ΔG° against T; since ΔG° = ΔH° − TΔS°, the slope equals −ΔS°. The entropy change is dominated by consumption of one mole of gaseous oxygen, so lines are roughly parallel with positive slope. The intercept at T = 0 gives ΔH°.

Question 2Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel is being solidified and the solid-liquid interface is planar and stable. According to the constitutional supercooling criterion, which condition promotes the breakdown of a planar interface into a cellular morphology?

  1. A high growth rate with a pure metal of zero solute content
  2. A high temperature gradient in the liquid and a low solute concentration
  3. A low growth rate with a partition coefficient equal to unity
  4. A low temperature gradient in the liquid and a high solute concentration Correct answer.

Explanation

Constitutional supercooling arises when the actual temperature gradient in the liquid is less than the slope of the liquidus at the interface. A steep solute buildup ahead of the interface (high concentration, low gradient) creates supercooled liquid and destabilizes the planar front. When the partition coefficient equals unity or solute is absent, no solute pileup occurs.

Question 3Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a blast furnace, the primary reducing agent for iron oxide in the upper stack is which species, and what is the overall role of coke in the lower zone?

  1. O2 reduces iron oxide in the stack, while coke provides structural support only
  2. CO2 reduces iron oxide in the stack, while coke provides the carbon for carburization only
  3. H2 reduces iron oxide in the stack, while coke acts only as a refractory lining
  4. CO reduces iron oxide in the stack, while coke provides heat and generates CO at the tuyeres Correct answer.

Explanation

In the upper stack, indirect reduction of iron oxides proceeds mainly by CO derived from the Boudouard reaction and coke combustion. In the lower zone, coke burns at the tuyeres to generate heat and CO and supports the burden. This indirect reduction dominates above roughly 800–900 °C.

Question 4Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A copper matte containing 40% Cu and 20% Fe is being converted. During the slag-blowing stage, which reaction describes removal of iron from the matte?

  1. FeS + Cu2O → FeO + 2Cu
  2. 2FeS + 3O2 → 2FeO + 2SO2, with FeO fluxed by silica Correct answer.
  3. Cu2S + O2 → 2Cu + SO2
  4. 2Cu + O2 → 2CuO

Explanation

In converting, the first (slag-forming) stage oxidizes iron sulfide to FeO with SO2 evolution. The FeO is fluxed by silica to form a fayalite-type slag. Copper sulfide remains in the matte until the second (blister) stage.

Question 5Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

The Gibbs phase rule for a system at constant pressure with C components and P phases is F = C − P + 1. For a binary eutectic system at the eutectic point, how many degrees of freedom remain?

  1. Two
  2. Three
  3. One
  4. Zero Correct answer.

Explanation

At the eutectic point, three phases (two solids and the liquid) coexist in a two-component system at fixed pressure. Substituting C = 2 and P = 3 into F = C − P + 1 gives F = 0. The eutectic temperature and composition are therefore both fixed.

Question 6Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A binary alloy with a face-centered cubic structure is being cold rolled. Which slip system is the most highly stressed when a tensile axis lies along the [001] direction?

  1. A {111} plane with slip direction [001]
  2. A {110} plane with slip direction [111]
  3. A {111} plane with slip direction [101] Correct answer.
  4. A {100} plane with slip direction [010]

Explanation

FCC metals slip on {111} planes along close-packed <110> directions. The resolved shear stress is maximized when the Schmid factor is high; the [101] direction lies in the {111} plane and gives a favorable orientation. Directions such as [001] are not close-packed and are not slip directions.

Question 7Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a binary isomorphous system, the lever rule is applied to a two-phase region at a given temperature. If the alloy composition is 40% B, the liquid composition is 30% B and the solid composition is 60% B, what is the fraction of solid?

  1. 1/2
  2. 1/3 Correct answer.
  3. 3/4
  4. 2/3

Explanation

The fraction of solid equals (C0 − CL)/(CS − CL) = (40 − 30)/(60 − 30) = 10/30 = 1/3. The lever rule uses the distance from the alloy composition to the liquid composition divided by the tie-line length.

Question 8Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the Hall–Héroult process, cryolite is added to alumina primarily for which purpose?

  1. To dissolve alumina and lower the operating temperature of the electrolyte Correct answer.
  2. To increase the electrical conductivity of the molten aluminum
  3. To act as a consumable anode material
  4. To oxidize impurities in the alumina feed

Explanation

Alumina melts at very high temperature, but dissolves in molten cryolite near 960 °C, making electrolysis practical. Cryolite also improves conductivity and allows the bath to operate at a lower temperature. The anode is carbon, not cryolite.

Question 9Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel specimen is tested in a Charpy impact machine and shows a ductile-to-brittle transition as temperature decreases. Which microstructural feature most strongly raises the transition temperature?

  1. A high volume fraction of retained austenite
  2. A fine grain size with a low carbon content
  3. A coarse grain size with large brittle second-phase particles Correct answer.
  4. A fully tempered martensitic structure with fine carbides

Explanation

Coarse grains and large brittle particles raise the ductile-to-brittle transition temperature because they reduce the ability to blunt cracks and increase stress concentration. Refining grain size lowers the transition temperature. Retained austenite and fine tempered carbides generally improve toughness.

Question 10Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a binary phase diagram, a peritectic reaction is characterized by which transformation?

  1. Liquid + liquid → solid on cooling
  2. Solid α → solid β + solid γ on cooling
  3. Liquid → solid α + solid β on cooling
  4. Liquid + solid α → solid β on cooling Correct answer.

Explanation

A peritectic reaction involves a liquid and a solid phase reacting to form a new solid phase on cooling. It differs from a eutectic, where a single liquid transforms into two solids. Peritectic points occur at compositions where the liquid and primary solid react.

Question 11Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

During the solidification of a pure metal, which of the following best explains the formation of a shrinkage cavity (pipe) in the riser?

  1. Thermal expansion of the mold walls during pouring
  2. Volume contraction on solidification combined with directional solidification toward the riser Correct answer.
  3. Gas evolution from dissolved hydrogen during freezing
  4. Solute rejection causing constitutional supercooling

Explanation

Most metals contract on solidification, and if solidification is directional toward the riser, the last liquid to freeze forms a pipe in the riser. This concentrates shrinkage defects in the feeder rather than the casting. Gas porosity and constitutional supercooling are separate phenomena.

Question 12Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In powder metallurgy, which pressing parameter most directly controls the green density of a compact made from a ductile metal powder?

  1. Sintering atmosphere
  2. Lubricant content
  3. Compaction pressure Correct answer.
  4. Particle shape distribution

Explanation

Green density increases with applied compaction pressure as particles deform and rearrange. Sintering affects final density after consolidation, not green density. Particle shape and lubricant influence densification but are secondary to pressure.

Question 13Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A welded joint in a low-alloy steel develops a hard, crack-susceptible heat-affected zone. Which measure most effectively reduces this problem?

  1. Using a higher carbon equivalent steel
  2. Increasing the cooling rate after welding
  3. Increasing the welding current without preheating
  4. Preheating the base metal before welding Correct answer.

Explanation

Preheating slows the cooling rate in the heat-affected zone, allowing softer transformation products to form and reducing hydrogen-assisted cracking. Higher carbon equivalent and faster cooling both increase hardness and cracking risk.

Question 14Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the context of metal extraction, what is the principal function of a flux such as limestone in a blast furnace?

  1. To combine with gangue to form a low-melting slag Correct answer.
  2. To reduce iron oxide directly to iron
  3. To act as a fuel in the raceway
  4. To provide carbon for carburization of iron

Explanation

Limestone decomposes to CaO, which combines with silica and alumina gangue to form a fluid slag that can be tapped and separated from hot metal. The slag also helps remove sulfur. Reduction and fuel roles belong to coke and CO.

Question 15Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A single crystal of a metal is deformed in tension, and slip occurs on a plane whose normal makes 45° with the tensile axis and in a direction that also makes 45° with the axis. What is the Schmid factor?

  1. 0.5 Correct answer.
  2. 0.25
  3. 0.707
  4. 1.0

Explanation

The Schmid factor is m = cos φ cos λ, where φ is the angle between the tensile axis and the slip-plane normal and λ is the angle between the axis and the slip direction. With φ = λ = 45°, m = 0.707 × 0.707 = 0.5. The maximum Schmid factor is 0.5.

Question 16Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

Which of the following best describes the role of a nucleating agent in the grain refinement of an aluminum casting?

  1. It promotes columnar grain growth
  2. It increases the freezing range of the alloy
  3. It provides heterogeneous nucleation sites that increase the number of grains Correct answer.
  4. It raises the pouring temperature

Explanation

Inoculants such as TiB2 particles act as substrates for heterogeneous nucleation, increasing nucleation density and producing finer equiaxed grains. A finer grain size improves strength and reduces hot tearing. Columnar growth is suppressed rather than promoted.

Question 17Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In electrochemical corrosion, a metal immersed in an electrolyte develops a mixed potential. Which condition corresponds to the corrosion potential of an actively corroding metal?

  1. The metal is fully passivated with no current flow
  2. The cathodic reaction rate is zero
  3. The anodic and cathodic currents are both zero
  4. The rate of anodic dissolution equals the rate of cathodic reduction Correct answer.

Explanation

At the corrosion potential, the net current is zero because the anodic and cathodic partial currents are equal and opposite. This mixed-potential concept underlies the Evans diagram. Passivation and zero reaction rates describe different states.

Question 18Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

Which of the following is the dominant strengthening mechanism in a precipitation-hardenable aluminum alloy after artificial aging?

  1. Solid-solution strengthening by copper atoms only
  2. Work hardening from prior cold rolling
  3. Interaction of dislocations with coherent or semi-coherent precipitates Correct answer.
  4. Grain-boundary sliding

Explanation

Aging produces fine precipitates that impede dislocation motion, giving peak hardness at an optimum precipitate size and spacing. Overaging coarsens precipitates and lowers strength. Solid-solution and work hardening contribute but are not the dominant mechanism after aging.

Question 19Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a basic oxygen furnace (BOF) steelmaking process, what is the primary purpose of blowing oxygen onto the molten pig iron?

  1. To reduce iron oxide back to iron
  2. To oxidize carbon and impurities and generate heat Correct answer.
  3. To add nitrogen to the steel
  4. To increase the sulfur content

Explanation

Oxygen blowing decarburizes the melt, oxidizes silicon, manganese, and phosphorus, and supplies heat through exothermic reactions. Lime is added to form slag for impurity removal. Nitrogen pickup and sulfur increase are undesirable.

Question 20Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A metal is subjected to a fatigue test with a fully reversed stress cycle. The S–N curve for a ferrous alloy shows a plateau at low stress. What does this plateau represent?

  1. The endurance limit below which the material has effectively infinite life Correct answer.
  2. The stress at which necking begins
  3. The ultimate tensile strength
  4. The yield strength of the material

Explanation

For many ferrous alloys, the S–N curve becomes horizontal at the endurance limit, meaning the material can endure a very large number of cycles without failure. Nonferrous alloys often lack a true endurance limit. Yield and tensile strength are monotonic properties.

Question 21Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a continuous casting process for steel, which parameter most directly controls the depth of the oscillation marks and the surface quality of the strand?

  1. Mold oscillation frequency and stroke combined with casting speed Correct answer.
  2. Ladle slag basicity
  3. Tundish refractory composition
  4. Secondary cooling water temperature only

Explanation

Oscillation marks form as the meniscus freezes against the mold during the upward stroke. Their depth depends on oscillation parameters and casting speed. Tundish and ladle variables affect cleanliness, not mark formation directly.

Question 22Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A binary alloy shows a eutectoid reaction in the solid state. Which transformation describes it?

  1. Two solid phases combine into one solid phase on heating
  2. A liquid and a solid combine to form a new solid phase
  3. One liquid phase decomposes into two solid phases on cooling
  4. One solid phase decomposes into two different solid phases on cooling Correct answer.

Explanation

A eutectoid reaction is a solid-state analogue of the eutectic: a single solid phase transforms into two distinct solid phases on cooling. In the Fe–C system, austenite transforms to ferrite and cementite at the eutectoid. The other options describe eutectic, peritectic, or inverse reactions.

Question 23Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In extractive metallurgy, the process of roasting is best described by which of the following?

  1. Melting a concentrate to separate slag from matte
  2. Heating a sulfide ore in air to oxidize it and remove sulfur as SO2 Correct answer.
  3. Dissolving an ore in acid to leach the metal
  4. Heating an ore in the absence of air to drive off moisture

Explanation

Roasting is an oxidation process in which sulfide ores are heated in air to convert sulfides to oxides and release sulfur dioxide. Calcination is heating without air to remove volatiles. Leaching and smelting describe different unit operations.

Question 24Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel component is to be surface hardened while retaining a tough core. Which treatment achieves this by diffusing carbon into the surface?

  1. Nitriding at low temperature
  2. Induction hardening of the surface
  3. Carburizing followed by quenching and tempering Correct answer.
  4. Annealing at high temperature

Explanation

Carburizing introduces carbon into the surface layer, which then transforms to martensite on quenching, producing a hard case with a tough core. Nitriding also hardens the surface but by nitrogen, and induction hardening does not add carbon. Annealing softens rather than hardens.

Question 25Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a binary phase diagram, the liquidus and solidus lines separate the single-phase liquid and solid regions. For an alloy that solidifies under equilibrium conditions, which statement is true about the composition of the first solid to form?

  1. It is richer in the lower-melting component than the liquid
  2. It is richer in the higher-melting component than the liquid from which it forms Correct answer.
  3. It is identical in composition to the liquid
  4. It is independent of the phase diagram

Explanation

During equilibrium solidification of an isomorphous system, the first solid to form is richer in the higher-melting component, consistent with the tie line connecting liquidus and solidus. The solid composition continuously shifts as temperature falls. The partition coefficient determines the degree of this enrichment.

Question 26Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the zone-refining purification of a metal, a molten zone is passed repeatedly along an impure bar. Which quantity primarily determines how effectively a solute is transported to the ends of the bar?

  1. The equilibrium distribution coefficient k of the solute between solid and liquid Correct answer.
  2. The vapour pressure of the host metal at the melting point
  3. The latent heat of fusion of the host metal
  4. The thermal conductivity of the crucible material

Explanation

Zone refining works because solute partitions unequally between solid and liquid, and this partition is quantified by k = Cs/Cl. Solutes with k far from unity are swept efficiently to the ends, while k close to one gives little separation. The other listed properties influence heat flow or evaporation but not the equilibrium partitioning that drives purification.

Question 27Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A blast furnace burden contains hematite (Fe2O3). Using the standard simplified reduction stoichiometry, how many moles of CO are required to reduce one mole of Fe2O3 completely to metallic iron?

  1. 1
  2. 2
  3. 4
  4. 3 Correct answer.

Explanation

The overall reduction may be written Fe2O3 + 3CO → 2Fe + 3CO2, so three moles of CO are consumed per mole of hematite. Each oxygen atom removed from the oxide is carried off as CO2, and hematite contains three oxygen atoms per iron unit. This stoichiometric ratio underlies coke-rate calculations in the blast furnace.

Question 28Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the cyanidation of a gold ore, finely divided gold is leached in an aerated alkaline cyanide solution. Which reaction best represents the dissolution of gold in this process?

  1. 2Au + 2OH- → Au2O + H2O + 2e-
  2. Au + 2H+ → Au2+ + H2
  3. Au + Cl- → AuCl + e-
  4. 4Au + 8CN- + O2 + 2H2O → 4[Au(CN)2]- + 4OH- Correct answer.

Explanation

Cyanidation is an electrochemical dissolution in which oxygen is the oxidant and cyanide is the complexing ligand, giving the aurous dicyano complex. The Elsner equation shown balances gold, cyanide, oxygen and hydroxyl. The alkaline condition prevents cyanide loss as HCN while the aurocyanide complex keeps the dissolved gold concentration low.

Question 29Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a green sand mould, which sand property most directly controls the mould's ability to deform and collapse as the casting contracts during cooling?

  1. Dry strength of the moulding sand
  2. Refractoriness of the silica grains
  3. Collapsibility of the moulding sand Correct answer.
  4. Permeability of the moulding sand

Explanation

Collapsibility is the property that lets the mould yield or break down as the solidifying and cooling casting shrinks, preventing hot tears and cracks. Refractoriness concerns resistance to high temperature, permeability concerns gas escape, and dry strength concerns mould integrity. Each of these serves a different function in the foundry.

Question 30Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A coarse-grained ingot is subjected to hot rolling above its recrystallisation temperature. Which microstructural change is expected during this deformation?

  1. A fully amorphous structure forms because of the high strain
  2. Grains elongate progressively without any new grain formation
  3. Precipitates coarsen until the alloy becomes single phase
  4. Simultaneous deformation and recrystallisation keep the structure equiaxed and refined Correct answer.

Explanation

In hot working above the recrystallisation temperature, strain hardening is continuously offset by dynamic recrystallisation, so the grains remain equiaxed and the coarse cast structure is refined. This is why hot rolling is used to break down ingot structure. Cold working, by contrast, elongates grains and accumulates dislocations.

Question 31Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a direct shear test of a soil, a normal stress of 100 kPa is applied and the sample fails at a shear stress of 70 kPa. If the cohesion intercept is 20 kPa, what is the angle of internal friction?

  1. About 27 degrees Correct answer.
  2. About 45 degrees
  3. About 22 degrees
  4. About 35 degrees

Explanation

The Mohr-Coulomb failure criterion gives tau = c + sigma tan(phi), so tan(phi) = (70 - 20)/100 = 0.5. The friction angle is therefore arctan(0.5), approximately 26.6 degrees. This calculation is the standard way of extracting phi from a shear box result.

Question 32Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel component is to be given a wear-resistant surface while retaining a tough, unhardened core. Which treatment is designed specifically for this purpose?

  1. Case carburising followed by quenching and low-temperature tempering Correct answer.
  2. Full through-hardening followed by tempering
  3. Solution treatment and artificial ageing
  4. Stress-relief annealing after welding

Explanation

Carburising enriches the surface in carbon so that, after quenching, only a shallow case transforms to hard martensite while the low-carbon core stays tough. Low-temperature tempering then relieves case brittleness without softening it. Through-hardening, by contrast, hardens the whole section.

Question 33Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a cupola furnace used for melting cast iron, what is the principal role of the limestone charged with coke and metal?

  1. It provides the main fuel for combustion
  2. It supplies carbon to the melt
  3. It deoxidises the molten iron
  4. It acts as a slag former that removes silica and ash Correct answer.

Explanation

Limestone (calcium carbonate) calcines in the cupola and the resulting lime combines with silica and ash to form a fluid slag that is tapped off. This slagging action protects the metal from excessive silica pickup and cleans the melt. Coke supplies fuel and carbon, not the limestone.

Question 34Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a binary eutectic system, an alloy of exactly eutectic composition is cooled slowly from the liquid. Which description of the resulting solid is correct?

  1. A single homogeneous solid solution of eutectic composition
  2. Primary crystals of one phase surrounded by a glassy matrix
  3. A two-phase mixture of the two terminal solid phases formed by coupled growth Correct answer.
  4. A single intermetallic compound with a fixed stoichiometry

Explanation

At the eutectic point the liquid simultaneously precipitates the two terminal solid phases, which grow cooperatively as a lamellar or rod-like aggregate. The product is therefore a two-phase mixture, not a single phase. This coupled growth is a defining feature of eutectic solidification.

Question 35Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the electrowinning of zinc from purified sulphate electrolyte, which anode reaction occurs during electrolysis?

  1. Reduction of hydrogen ions to hydrogen gas
  2. Oxidation of water to oxygen gas Correct answer.
  3. Oxidation of sulphate to sulphur trioxide
  4. Reduction of Zn2+ to zinc metal

Explanation

In zinc electrowinning the cathode reaction deposits zinc from sulphate solution while the anode is made of lead-silver alloy that catalyses oxygen evolution. The anode reaction is the oxidation of water to oxygen, which balances the cathodic reduction. Sulphate is not oxidised under these conditions.

Question 36Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A single crystal of a metal with a body-centred cubic lattice is deformed in tension. Which set of slip systems is most commonly activated in BCC metals at room temperature?

  1. {0001}<11-20>
  2. {110}<111> Correct answer.
  3. {111}<11-20>
  4. {111}<110>

Explanation

In BCC metals the most common slip plane is {110} with slip direction <111>, the direction of closest atomic packing. Other planes such as {112} and {123} may also operate, especially at low temperature or in certain orientations. The {111}<110> and basal systems belong to FCC and HCP metals respectively.

Question 37Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel is austempered by quenching from the austenitising temperature into a molten salt bath held at 350 degrees C and holding until transformation is complete. Which microstructure results?

  1. Ferrite plus spheroidised carbides
  2. Bainite Correct answer.
  3. Coarse pearlite
  4. Tempered martensite

Explanation

Austempering holds the steel isothermally in the bainite transformation range, between the pearlite and martensite-start temperatures, until the austenite has fully transformed to bainite. Because no martensite forms, no subsequent tempering is needed. The resulting bainite has good toughness at high hardness.

Question 38Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a flotation cell, a collector such as xanthate is added to a sulphide ore pulp. What is the function of the collector?

  1. It controls the pH of the pulp
  2. It depresses the gangue minerals selectively
  3. It stabilises the froth so bubbles do not burst
  4. It makes the target mineral surface hydrophobic so it attaches to bubbles Correct answer.

Explanation

Collectors adsorb selectively on the target mineral, rendering it hydrophobic so that air bubbles can attach and float it. Frothers stabilise the froth, depressants suppress unwanted minerals, and pH modifiers adjust pulp chemistry. Each reagent class has a distinct role in flotation.

Question 39Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A cylindrical riser is to be designed for a sand casting so that it feeds the casting before it solidifies. According to Chvorinov's rule, which riser geometry is preferred for a given volume?

  1. A thin flat plate
  2. A sphere Correct answer.
  3. A tall cylinder with a small diameter
  4. A short cylinder with a large diameter

Explanation

Chvorinov's rule states that solidification time is proportional to (volume/surface area) squared, so a shape with the largest modulus solidifies last and feeds the casting. For a given volume, a sphere has the smallest surface area and hence the largest modulus. Practical risers are cylindrical for moulding reasons but are made as compact as possible.

Question 40Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

During the solidification of a steel ingot, which condition favours the formation of centreline segregation and porosity?

  1. A very low superheat combined with vigorous electromagnetic stirring
  2. A wide freezing range with a shallow temperature gradient and poor feeding of the centre Correct answer.
  3. A high cooling rate that produces a fine equiaxed structure
  4. A steep temperature gradient with rapid directional cooling

Explanation

A wide freezing range means a long mushy zone in which solute-rich liquid is trapped and interdendritic feeding is difficult, so the centreline enriches in solute and shrinks. A shallow gradient widens this mushy zone further. Steep gradients and rapid cooling promote directional or fine equiaxed structures with less segregation.

Question 41Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A metal is being deformed at a temperature where recovery and recrystallisation occur readily. How does the true stress-strain curve differ from that obtained at room temperature?

  1. It reaches a steady-state flow stress after an initial rise Correct answer.
  2. It shows no strain hardening at all from the first increment of strain
  3. It shows a much higher flow stress that rises continuously
  4. It falls to zero because the metal becomes superplastic

Explanation

At high temperature, work hardening is balanced by dynamic recovery and recrystallisation, so the flow stress rises initially and then plateaus at a steady-state value. The room-temperature curve, by contrast, continues to rise as dislocations accumulate. Superplasticity requires fine grains and specific strain-rate conditions, not merely elevated temperature.

Question 42Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the Hall-Heroult cell, alumina is dissolved in molten cryolite at about 960 degrees C. Which statement correctly describes the role of the cryolite?

  1. It is consumed as the reductant for alumina
  2. It dissolves alumina and lowers the operating temperature relative to pure alumina Correct answer.
  3. It reacts with alumina to form aluminium metal directly
  4. It acts as a cathode coating that conducts current to the metal pad

Explanation

Alumina melts at over 2000 degrees C, so it is dissolved in molten cryolite, which provides a conducting solvent and lowers the working temperature to around 960 degrees C. Cryolite is not consumed in the net reaction and is not a reductant. The carbon anodes, not cryolite, provide the reducing carbon.

Question 43Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A steel shaft is to be induction hardened. Which combination of prior microstructure and heating condition gives the most uniform hardened case?

  1. Coarse pearlite heated slowly to just above the eutectoid temperature
  2. A spheroidised structure heated just below the eutectoid temperature
  3. Large primary carbides in a soft ferrite matrix heated briefly
  4. A fine, uniform ferrite-pearlite structure heated rapidly to well above the austenitising temperature Correct answer.

Explanation

Induction hardening relies on very rapid heating, so a fine uniform starting structure dissolves carbon quickly and gives a consistent austenite before quenching. Coarse or segregated structures need longer times and produce soft spots or retained carbides. Heating just below the eutectoid temperature would not austenitise the steel at all.

Question 44Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a powder metallurgy compact, the green strength is measured before sintering. Which factor most strongly increases green strength for a given powder?

  1. Increasing the compaction pressure Correct answer.
  2. Increasing the sintering temperature
  3. Increasing the particle size of the powder
  4. Adding a lubricant to reduce die-wall friction

Explanation

Green strength arises from mechanical interlocking and cold welding between particles, both of which increase as compaction pressure raises the density and contact area. Lubricants actually lower green strength by coating particle surfaces. Sintering temperature affects the sintered strength, not the green strength measured before sintering.

Question 45Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A stainless steel component in a chloride-containing environment suffers localised breakdown of its passive film and rapid pitting. Which alloying strategy most directly improves resistance to this attack?

  1. Increasing the sulphur content for machinability
  2. Increasing the carbon content
  3. Increasing the molybdenum content together with chromium Correct answer.
  4. Increasing the nickel content while reducing chromium

Explanation

Pitting resistance in chloride media depends on a stable passive film, which is promoted by chromium and further improved by molybdenum, often expressed through the pitting resistance equivalent number. Carbon and sulphur form inclusions that act as pit initiation sites. Reducing chromium while raising nickel undermines the passive film.

Question 46Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a continuous casting machine, the mould is oscillated with a stroke and frequency. What is the main purpose of this oscillation?

  1. To stir the liquid core and eliminate segregation entirely
  2. To increase the casting speed beyond the machine limit
  3. To cool the mould copper plate more rapidly
  4. To prevent the solidifying shell from sticking to the mould wall and to control oscillation-mark depth Correct answer.

Explanation

Mould oscillation, together with mould flux lubrication, prevents the shell from adhering to the copper mould and tearing. The stroke and frequency, related through the negative strip time, control the depth of oscillation marks. Oscillation does not eliminate segregation or set the cooling rate of the copper plate.

Question 47Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A binary alloy of composition C0 is cooled slowly in an isomorphous system. At a temperature in the two-phase region, the alloy consists of liquid of composition CL and solid of composition CS. If the alloy is 40 percent liquid by mass and CL - CS is 30 wt percent, what is the composition difference between the alloy and the solid?

  1. 12 wt percent Correct answer.
  2. 6 wt percent
  3. 18 wt percent
  4. 24 wt percent

Explanation

The lever rule gives the fraction of liquid as (C0 - CS)/(CL - CS). With a liquid fraction of 0.40 and a tie-line length of 30 wt percent, C0 - CS equals 0.40 times 30, which is 12 wt percent. This is the horizontal distance from the solid composition to the alloy composition.

Question 48Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In the electrolytic refining of copper, impure anodes are refined in a sulphate electrolyte. Where do the more noble impurities such as silver and gold report?

  1. They volatilise from the electrolyte as gases
  2. They form a slag layer on the electrolyte surface
  3. They collect as anode slime beneath the anode Correct answer.
  4. They dissolve and plate with the cathode copper

Explanation

During copper electrorefining, copper dissolves from the anode and plates on the cathode, while noble impurities that do not dissolve under the anode potential fall as anode slime. This slime is a valuable by-product from which gold and silver are recovered. Base-metal impurities such as nickel do dissolve and accumulate in the electrolyte.

Question 49Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

A cold-worked metal is annealed and the hardness is plotted against annealing time at a fixed temperature. The curve shows an initial plateau, then a sharp drop, then a second plateau. Which sequence of processes explains this shape?

  1. Recovery, then recrystallisation, then grain growth Correct answer.
  2. Recrystallisation, then recovery, then precipitation
  3. Precipitation, then grain growth, then recrystallisation
  4. Grain growth, then recovery, then recrystallisation

Explanation

During the initial plateau, recovery reduces internal stress with little change in hardness. The sharp drop corresponds to recrystallisation, which replaces the deformed grains with new strain-free ones. The final plateau reflects grain growth, which changes hardness only slowly.

Question 50Metallurgical Engineering (MT) - official paper syllabus domains not enumerated on the source page

In a basic oxygen furnace, the slag is designed to be highly basic and oxidising. Which consequence follows from this slag chemistry?

  1. Phosphorus is oxidised and held in the slag while carbon is removed as CO Correct answer.
  2. Carbon is oxidised only after all phosphorus has been removed from the slag
  3. Sulphur is reduced into the metal and retained there
  4. Silicon is retained in the steel to improve deoxidation

Explanation

An oxidising, highly basic slag oxidises phosphorus to phosphate and holds it in the slag, while carbon is removed as CO gas. Sulphur removal is also favoured by high basicity and oxidising conditions, so sulphur is not retained. Silicon is oxidised into the slag early in the blow rather than kept in the steel.