Physical Metallurgy — Part 2 MCQs with Explanations
Practice high-yield Metallurgical Engineering MCQs on Physical Metallurgy — Part 2 with worked academic solutions for PAEC, NESCOM, HIT, FPSC.
Questions
40
Difficulty Level
Exam-Level
Target Exams
PAEC, NESCOM, HIT...
Estimated Time
48 mins
Exam-Level PAEC NESCOM HIT FPSC
Verified for Commission Syllabus • Oct 14, 2026
40
Verified Practice Questions
Interactive test mode with detailed academic worked solutions
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Question 1 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the names in Group 1 with invarient reactions in binary phase diagram in Group 2 . Group 1 Group 2 P. Eutectic 1. S 1 = S 2 + S 3 Q. Eutectoid 2. L = S 1 + S 2 R. Peritectoid 3. L 1 = L 2 + S S. Monotectic 4. S 1 + S 2 = S 3
Correct Answer: Option A — P-2, Q-1, R-3, S-4
Detailed Academic Explanation:
P-2, Q-1, R-3, S-4 is the correct answer to “Match the names in Group 1 with invarient reactions in binary phase diagram in Group 2 . Group 1 Group 2 P. Eutectic 1. S 1 = S 2 + S 3 Q. Eutectoid 2. L = S 1 + S 2 R. Peritectoid 3. L 1 = L 2 + S S. Monotectic 4. S 1 + S 2 = S 3”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 2 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Assume that Sievert's law holds for solubility of hydrogen dissolved in a thin metal foil. If the partial pressure of hydrogen in contact with the foil is increased by a factor of 4, the solubility increases by a factor of
Correct Answer: Option B — 2
Detailed Academic Explanation:
2 is the correct answer to “Assume that Sievert's law holds for solubility of hydrogen dissolved in a thin metal foil. If the partial pressure of hydrogen in contact with the foil is increased by a factor of 4, the solubility increases by a factor of”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 3 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the heat treatment processes of steel in Group 1 with the microstructural features in Group 2 . Group 1 Group 2 P. Quenching 1. Bainite Q. Maraging 2. Martensite R. Tempering 3. Internietallic precipitates S. Austempering 4. Epsilon carbide
Correct Answer: Option C — P-2, Q-3, R-2, S-1
Detailed Academic Explanation:
P-2, Q-3, R-2, S-1 is the correct answer to “Match the heat treatment processes of steel in Group 1 with the microstructural features in Group 2 . Group 1 Group 2 P. Quenching 1. Bainite Q. Maraging 2. Martensite R. Tempering 3. Internietallic precipitates S. Austempering 4. Epsilon carbide”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 4 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The free energy change during melting of ice at 0°C is equal to
Correct Answer: Option B — 0
Detailed Academic Explanation:
0 is the correct answer to “The free energy change during melting of ice at 0°C is equal to”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 5 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
At which temperature, cementite changes ferromagnetic to paramagnetic character in iron carbon diagram
Correct Answer: Option D — 210°C
Detailed Academic Explanation:
210°C is the correct answer to “At which temperature, cementite changes ferromagnetic to paramagnetic character in iron carbon diagram”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 6 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The elastic strain energy of a unit length of an edge dislocation as compared to that of a screw dislocation is
Correct Answer: Option A — More
Detailed Academic Explanation:
More is the correct answer to “The elastic strain energy of a unit length of an edge dislocation as compared to that of a screw dislocation is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 7 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The peritectic reaction in binary systems is given by
Correct Answer: Option D — L + α = β
Detailed Academic Explanation:
L + α = β is the correct answer to “The peritectic reaction in binary systems is given by”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 8 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The critical nucleus size (in mm) when copper melt is under-cooled by 100K is Given data: Melting point: 1356 K Density: 8900 kg.m -3 Solid-liquid interfacial energy: 0.5 Jm -2 Latent heat of freezing: 13000 J. mol -1 Molar volume: 7 × 10 -6 m 3 mol -1
Correct Answer: Option A — 0.36
Detailed Academic Explanation:
0.36 is the correct answer to “The critical nucleus size (in mm) when copper melt is under-cooled by 100K is Given data: Melting point: 1356 K Density: 8900 kg.m -3 Solid-liquid interfacial energy: 0.5 Jm -2 Latent heat of freezing: 13000 J. mol -1 Molar volume: 7 × 10 -6 m 3 mol -1”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 9 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which of the following is true for martensitic transformation in steels?
Correct Answer: Option C — Symmetry of the product is different from the parent phase but the unit cell volume increases
Detailed Academic Explanation:
Symmetry of the product is different from the parent phase but the unit cell volume increases is the correct answer to “Which of the following is true for martensitic transformation in steels”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 10 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
If the interfacial energy is increase by 10%, the homogeneous nucleation barrier for a spherical particle increases by
Correct Answer: Option C — 33%
Detailed Academic Explanation:
33% is the correct answer to “If the interfacial energy is increase by 10%, the homogeneous nucleation barrier for a spherical particle increases by”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 11 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
A number of solid state phase transformations follow a sigmoidal pattern. In these eases, at any time the fraction transformed can be expressed as
Correct Answer: Option B — 1 + exp (-α2t)
Detailed Academic Explanation:
1 + exp (-α2t) is the correct answer to “A number of solid state phase transformations follow a sigmoidal pattern. In these eases, at any time the fraction transformed can be expressed as”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 12 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
From an alloy, two specimens are machined and tested separately under tension and compression. The engineering stress-strain curves as well as the true stress-true strain curves for tension and compression are plotted in the same diagram. Identify the correct statements: P. The engineering stress-strain curves in tension and compression are identical. Q. The true stress-true strain curve in compression is lower than the true stresstrue strain curve in tension due to Bauschinger effect. R. The true stress-true strain curve in tension is above the corresponding engineering stress-strain, curve. S. The true stress-true strain curve in tension and compression are identical.
Correct Answer: Option B — Q, R
Detailed Academic Explanation:
Q, R is the correct answer to “From an alloy, two specimens are machined and tested separately under tension and compression. The engineering stress-strain curves as well as the true stress-true strain curves for tension and compression are plotted in the same diagram. Identify the correct statements: P. The engineering stress-strain curves in tension and compression are identical. Q. The true stress-true strain curve in compression is lower than the true stresstrue strain curve in tension due to Bauschinger effect. R. The true stress-true strain curve in tension is above the corresponding engineering stress-strain, curve. S. The true stress-true strain curve in tension and compression are identical.”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 13 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which one of the following is a ferrite stabiliser in steels?
Correct Answer: Option C — Cr
Detailed Academic Explanation:
Cr is the correct answer to “Which one of the following is a ferrite stabiliser in steels”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 14 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
In Cu-Al phase diagram, the solubility of Al in Cu at room temperature is about 10% and that of Cu in Al is less than 1%. The Hume-Rothery rule that justifies this difference is
Correct Answer: Option A — size factor
Detailed Academic Explanation:
size factor is the correct answer to “In Cu-Al phase diagram, the solubility of Al in Cu at room temperature is about 10% and that of Cu in Al is less than 1%. The Hume-Rothery rule that justifies this difference is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 15 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The property of a material that cannot be significantly changed by heat treatment is
Correct Answer: Option D — Elastic modulus
Detailed Academic Explanation:
Elastic modulus is the correct answer to “The property of a material that cannot be significantly changed by heat treatment is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 16 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
A binary phase diagram of components P and Q displays an eutectoid reaction with terminal solid solutions a on the P rich side and β on the Q rich side. At the eutectoid temperature, the solubilities of Q in a and β are 5 and 90 wt% respectively. The densities of α and β phases are 9.5 and 2.49 gm/cm 3 respectively. At the eutectoid point, the alloy has α and β in the weight ratio 1 : 1. The eutectoid point occurs at composition . . . . . . . . wt % Q.
Correct Answer: Option B — 47.5
Detailed Academic Explanation:
47.5 is the correct answer to “A binary phase diagram of components P and Q displays an eutectoid reaction with terminal solid solutions a on the P rich side and β on the Q rich side. At the eutectoid temperature, the solubilities of Q in a and β are 5 and 90 wt% respectively. The densities of α and β phases are 9.5 and 2.49 gm/cm 3 respectively. At the eutectoid point, the alloy has α and β in the weight ratio 1 : 1. The eutectoid point occurs at composition . . . . . . . . wt % Q.”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 17 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the elements is Group I with their corresponding crystal structures in Group II . Group I Group II P. Tungsten 1. HCP Q. Nickel 2. Simple cubic R. Magnesium 3. BCC S. Polonium 4. FCC
Correct Answer: Option B — P-3, Q-4, R-1, S-2
Detailed Academic Explanation:
P-3, Q-4, R-1, S-2 is the correct answer to “Match the elements is Group I with their corresponding crystal structures in Group II . Group I Group II P. Tungsten 1. HCP Q. Nickel 2. Simple cubic R. Magnesium 3. BCC S. Polonium 4. FCC”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 18 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match each phase in Group 1 with a description in Group 2 . Group 1 Group 2 P. E-carbide 1. A three component eutectic of iron, iron carbide, iron phosphate found in cast iron Q. Sigma phase 2. An embrittling compound found in ferritic stainless steel R. E-ferrite 3. Obtained on tempering of hardened steels S. Steadite 4. Responsible for causing the weld deposit on austenitic stainless steels to be slightly magnetic
Correct Answer: Option D — P-3, Q-2, R-4, S-1
Detailed Academic Explanation:
P-3, Q-2, R-4, S-1 is the correct answer to “Match each phase in Group 1 with a description in Group 2 . Group 1 Group 2 P. E-carbide 1. A three component eutectic of iron, iron carbide, iron phosphate found in cast iron Q. Sigma phase 2. An embrittling compound found in ferritic stainless steel R. E-ferrite 3. Obtained on tempering of hardened steels S. Steadite 4. Responsible for causing the weld deposit on austenitic stainless steels to be slightly magnetic”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 19 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
A liquid phase sintered SiC-Ni composite has a solid-solid grain boundary energy (γ SiC-SiC ) of 0.80 J.M -2 and a solid-liquid (γ SiC-SiC ) interfacial energy of 0.45 J.m -2 . For a SiC grain size of 20 μm, the average interparticle (SiC-SiC) neck size (in μm) is:
Correct Answer: Option C — 9.16
Detailed Academic Explanation:
9.16 is the correct answer to “A liquid phase sintered SiC-Ni composite has a solid-solid grain boundary energy (γ SiC-SiC ) of 0.80 J.M -2 and a solid-liquid (γ SiC-SiC ) interfacial energy of 0.45 J.m -2 . For a SiC grain size of 20 μm, the average interparticle (SiC-SiC) neck size (in μm) is:”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 20 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
If the slip direction is at right angle to the tensile axis, then the value of critical resolve shear stress is
Correct Answer: Option A — 0
Detailed Academic Explanation:
0 is the correct answer to “If the slip direction is at right angle to the tensile axis, then the value of critical resolve shear stress is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 21 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Dislocation cross-slip is difficult in those materials which have
Correct Answer: Option D — low stacking fault energy
Detailed Academic Explanation:
low stacking fault energy is the correct answer to “Dislocation cross-slip is difficult in those materials which have”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 22 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The number of components is 2, the number of phases are 3 and degrees of freedom are 0. Then which of the following equilibrium is it?
Correct Answer: Option A — Invariant
Detailed Academic Explanation:
Invariant is the correct answer to “The number of components is 2, the number of phases are 3 and degrees of freedom are 0. Then which of the following equilibrium is it”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 23 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
In the hypothetical phase diagram, the melting point of each pure component is. 1000 K and the eutectic temperature is 800 K. The eutectic is located at the equi-atomic composition. The maximum solid solubility in α-phase is given by the mole fraction, Ne = 0.1. The freezing range (in K) of the alloy with composition, Ne = 0.1 is
Correct Answer: Option C — 160
Detailed Academic Explanation:
160 is the correct answer to “In the hypothetical phase diagram, the melting point of each pure component is. 1000 K and the eutectic temperature is 800 K. The eutectic is located at the equi-atomic composition. The maximum solid solubility in α-phase is given by the mole fraction, Ne = 0.1. The freezing range (in K) of the alloy with composition, Ne = 0.1 is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 24 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
With the increase in the degree of supercooling, the growth rate of a nucleus follows which one of the following trends?
Correct Answer: Option A — First increases and then decreases
Detailed Academic Explanation:
First increases and then decreases is the correct answer to “With the increase in the degree of supercooling, the growth rate of a nucleus follows which one of the following trends”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 25 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the heat treatment process in Group I with resultant microstructure Group II . Group I Group II P. Martempering 1. Coarse pearlite Q. Normalising 2. Fine pearlite R. Sub-critical annealing for long time 3. Tempered martensite S. Full annealing 4. Spheroidised cementite in the matrix of ferrite
Correct Answer: Option D — P-3, Q-2, R-4, S-1
Detailed Academic Explanation:
P-3, Q-2, R-4, S-1 is the correct answer to “Match the heat treatment process in Group I with resultant microstructure Group II . Group I Group II P. Martempering 1. Coarse pearlite Q. Normalising 2. Fine pearlite R. Sub-critical annealing for long time 3. Tempered martensite S. Full annealing 4. Spheroidised cementite in the matrix of ferrite”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 26 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the items in Group1 with those in Group2 . Group 1 Group 2 P. Fatigue 1. Cracking at extrusions and intrusions Q. Hall-petch equation 2. Dislocation pile up R. Creep 3. Slip and twinning S. Plastic deformation 4. Grain boundary sliding 5. Angle of grain boundary 6. Young's modulus
Correct Answer: Option C — P-1, Q-2, R-4, S-3
Detailed Academic Explanation:
P-1, Q-2, R-4, S-3 is the correct answer to “Match the items in Group1 with those in Group2 . Group 1 Group 2 P. Fatigue 1. Cracking at extrusions and intrusions Q. Hall-petch equation 2. Dislocation pile up R. Creep 3. Slip and twinning S. Plastic deformation 4. Grain boundary sliding 5. Angle of grain boundary 6. Young's modulus”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 27 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Determine the correctness or otherwise of the following Assertion (a) and Reason (r) . Assertion (a): Peak aging time for an Al-4 wt.% Cu alloy is indicated by a maximum in the hardness. Reason (r): The maximum volume fraction of the θ precipitates is formed at the peak aging time.
Correct Answer: Option A — a is true but r is false
Detailed Academic Explanation:
a is true but r is false is the correct answer to “Determine the correctness or otherwise of the following Assertion (a) and Reason (r) . Assertion (a): Peak aging time for an Al-4 wt.% Cu alloy is indicated by a maximum in the hardness. Reason (r): The maximum volume fraction of the θ precipitates is formed at the peak aging time.”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 28 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The stress required for Orowon dislocation bypass is 200 MPa in an alloy when the interprecipitate spacing is 500 nm. In the same alloy, if the inter-precipitate spacing is reduced to 200 nm, the stress required (in MPa) is
Correct Answer: Option C — 500
Detailed Academic Explanation:
500 is the correct answer to “The stress required for Orowon dislocation bypass is 200 MPa in an alloy when the interprecipitate spacing is 500 nm. In the same alloy, if the inter-precipitate spacing is reduced to 200 nm, the stress required (in MPa) is”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 29 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The self diffusion in FCC metals occurs by one of the following mechanisms.
Correct Answer: Option C — Vacancy
Detailed Academic Explanation:
Vacancy is the correct answer to “The self diffusion in FCC metals occurs by one of the following mechanisms.”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 30 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
A binary alloy system A-B shows the peritectic reaction, L + α (⟶β). An alloy of the peritectic composition when cooled under natural cooling conditions will show coring in
Correct Answer: Option B — β phase only
Detailed Academic Explanation:
β phase only is the correct answer to “A binary alloy system A-B shows the peritectic reaction, L + α (⟶β). An alloy of the peritectic composition when cooled under natural cooling conditions will show coring in”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 31 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
It is possible to form martensite
Correct Answer: Option D — only in iron-based alloys
Detailed Academic Explanation:
only in iron-based alloys is the correct answer to “It is possible to form martensite”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 32 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The following reactions are written such that the forward reaction occurs during cooling. Match the reactions to their names. Group-I Group-II P. Y = α + β 1. peritectoid Q. L+ α = β 2. Eutectoid R. L 1 = L 2 + α 3. Peritectic S. α + β = Y 4. Monotectic
Correct Answer: Option A — P-2, Q-3, R-4, S-1
Detailed Academic Explanation:
P-2, Q-3, R-4, S-1 is the correct answer to “The following reactions are written such that the forward reaction occurs during cooling. Match the reactions to their names. Group-I Group-II P. Y = α + β 1. peritectoid Q. L+ α = β 2. Eutectoid R. L 1 = L 2 + α 3. Peritectic S. α + β = Y 4. Monotectic”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 33 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the associated primary change in Group 2 with the corresponding phenomenon in Group 1 . Group 1 Group 2 P. Recovery 1. change in dislocation density Q. Recrystallisation 2. change in the shape of a phase R. Malleabilising 3. change in vacancy concentration S. Texture 4. change in grain orientation
Correct Answer: Option C — P-3, Q-1, R-2, S-4
Detailed Academic Explanation:
P-3, Q-1, R-2, S-4 is the correct answer to “Match the associated primary change in Group 2 with the corresponding phenomenon in Group 1 . Group 1 Group 2 P. Recovery 1. change in dislocation density Q. Recrystallisation 2. change in the shape of a phase R. Malleabilising 3. change in vacancy concentration S. Texture 4. change in grain orientation”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 34 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
An alloy of overall composition, X B = 0.7 was equilibrated at temperature T 1 . Microstructural analysis showed two phases, (α and β, and that the phase fraction of β was 0.75. Given that the equilibrium composition of β at Ti is 0.9 as shown in the phase diagram below, the maximum solubility of B in α (in mole fraction) at this temperature is . . . . . . . .
Correct Answer: Option B — 0.10
Detailed Academic Explanation:
0.10 is the correct answer to “An alloy of overall composition, X B = 0.7 was equilibrated at temperature T 1 . Microstructural analysis showed two phases, (α and β, and that the phase fraction of β was 0.75. Given that the equilibrium composition of β at Ti is 0.9 as shown in the phase diagram below, the maximum solubility of B in α (in mole fraction) at this temperature is . . . . . . . .”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 35 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
According to Griffith theory, which of the following are true for the fracture stress in a brittle material? P. Increases with increasing surface energy Q. Decreases with increasing elastic modulus R. Decreases with increasing length of pre-existing crack S. Decreases with increasing number of pre-existing cracks of same size
Correct Answer: Option B — Q, R, S
Detailed Academic Explanation:
Q, R, S is the correct answer to “According to Griffith theory, which of the following are true for the fracture stress in a brittle material? P. Increases with increasing surface energy Q. Decreases with increasing elastic modulus R. Decreases with increasing length of pre-existing crack S. Decreases with increasing number of pre-existing cracks of same size”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 36 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Match the following strengthening methods (Group I) in metallic alloys with typical mechanisms responsible (Group II) for them. Group I Group II P. Grain size strengthening 1. Orowan looping Q. Work hardening 2. Suzuki interaction R. Dispersion strengthening 3. Lomer-Cottrell-barrier S. Solid solution hardening 4. Dislocation pile up at grain boundaries
Correct Answer: Option C — P-4, Q-3, R-1, S-2
Detailed Academic Explanation:
P-4, Q-3, R-1, S-2 is the correct answer to “Match the following strengthening methods (Group I) in metallic alloys with typical mechanisms responsible (Group II) for them. Group I Group II P. Grain size strengthening 1. Orowan looping Q. Work hardening 2. Suzuki interaction R. Dispersion strengthening 3. Lomer-Cottrell-barrier S. Solid solution hardening 4. Dislocation pile up at grain boundaries”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 37 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which one of the following can avoid weld decay of austenitic stainless steels?
Correct Answer: Option A — Reducing carbon content
Detailed Academic Explanation:
Reducing carbon content is the correct answer to “Which one of the following can avoid weld decay of austenitic stainless steels”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 38 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Lath martensitic microstructure we can see in
Correct Answer: Option C — low or medium carbon steel
Detailed Academic Explanation:
low or medium carbon steel is the correct answer to “Lath martensitic microstructure we can see in”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 39 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Regarding recrystalisation, which one of the following statements is not correct?
Correct Answer: Option D — Finer the initial grain size, higher is the recrystallization temperature
Detailed Academic Explanation:
Finer the initial grain size, higher is the recrystallization temperature is the correct answer to “Regarding recrystalisation, which one of the following statements is not correct”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
Question 40 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which of the following manufacturing processes is not used for producing fine grained metals?
Correct Answer: Option B — Czochralski method
Detailed Academic Explanation:
Czochralski method is the correct answer to “Which of the following manufacturing processes is not used for producing fine grained metals”. This question is standard in competitive examination syllabi for Physical Metallurgy.
💡Key Conceptual Anchor:Physical Metallurgy
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💡Examiner's Advisory for Metallurgical Engineering
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.