SSRAO

Explanation:
Step 1: Crystal defects are classified by their dimensionality. Step 2: Zero-dimensional (point) defects include vacancies (missing atoms), interstitials (extra atoms in interstitial sites), and substitutional impurities. Step 3: Dislocations are one-dimensional (line) defects. Grain boundaries are two-dimensional (planar) defects. Stacking faults are also planar defects. Step 4: A vacancy is a lattice site from which an atom is missing — it is the simplest and most common point defect. Answer: C = Vacancy.

Explanation:
Step 1: Frenkel defect is a type of point defect in ionic crystals. Step 2: It occurs when an ion (usually the smaller cation) leaves its normal lattice site and occupies an interstitial position. Step 3: This creates a cation vacancy at the original site and a cation interstitial at the new site. Step 4: The crystal remains electrically neutral because the same ion is involved. Step 5: Schottky defect, by contrast, involves paired vacancies (cation + anion vacancies). Answer: C.

Explanation:
Step 1: Schottky defect is a type of point defect in ionic crystals. Step 2: It involves the removal of a pair of oppositely charged ions (one cation and one anion) from the interior of the crystal to the surface. Step 3: This creates a cation vacancy and an anion vacancy in equal numbers to maintain electrical neutrality. Step 4: Schottky defects are common in NaCl, KCl, and other alkali halides. Step 5: The crystal density decreases due to Schottky defects. Answer: B.

Explanation:
Step 1: A dislocation is a one-dimensional (line) defect in a crystal. Step 2: The Burgers vector (b) describes the magnitude and direction of lattice distortion caused by a dislocation. Step 3: For an edge dislocation: the Burgers vector is perpendicular to the dislocation line. The extra half-plane of atoms is inserted perpendicular to the slip direction. Step 4: For a screw dislocation: the Burgers vector is parallel to the dislocation line. Step 5: For a mixed dislocation: the Burgers vector is at an angle to the dislocation line. Answer: B.

Explanation:
Step 1: In a screw dislocation, the crystal lattice is distorted in a helical manner around the dislocation line. Step 2: The Burgers vector for a screw dislocation is parallel to the dislocation line. Step 3: There is no extra half-plane of atoms (that is characteristic of edge dislocations). Step 4: The slip plane for a screw dislocation is not uniquely defined — it can glide on any plane containing the dislocation line. Step 5: Screw dislocations are important in crystal growth (spiral growth mechanism). Answer: B.

Explanation:
Step 1: Dislocations are classified based on the relationship between the Burgers vector and the dislocation line. Step 2: Edge dislocation: Burgers vector ⊥ dislocation line. Step 3: Screw dislocation: Burgers vector ∥ dislocation line. Step 4: Mixed dislocation: Burgers vector at an angle (neither parallel nor perpendicular) to the dislocation line. It has both edge and screw components. Step 5: Most real dislocations in crystals are mixed dislocations. Answer: C.

Explanation:
Step 1: In FCC metals, the close-packed planes (111) stack in the sequence ABCABCABC... Step 2: A stacking fault occurs when this regular stacking sequence is interrupted. Step 3: For example, the sequence becomes ABCABABC... (intrinsic stacking fault) or ABCABACBC... (extrinsic stacking fault). Step 4: Stacking faults are bounded by partial dislocations (Shockley partials). Step 5: They are planar (2D) defects, not point or line defects. Answer: B.

Explanation:
Step 1: The elastic strain energy stored in a dislocation is concentrated in the region around the dislocation core. Step 2: For a screw dislocation, the energy per unit length = Gb²/(4π) × ln(R/r₀) where G is shear modulus, b is Burgers vector, R is outer radius, r₀ is core radius. Step 3: For an edge dislocation, the energy per unit length = Gb²/(4π(1-ν)) × ln(R/r₀). Step 4: As a rough approximation, the energy per unit length ≈ Gb²/2 (commonly used in materials science). Step 5: This shows that dislocations with smaller Burgers vectors have lower energy. Answer: B = Gb²/2.

Explanation:
Step 1: In ionic crystals, electrical conductivity occurs by ion migration (ionic conductivity). Step 2: Schottky defects create vacancies through which ions can migrate, increasing ionic conductivity. Step 3: Frenkel defects create interstitial ions that can move through the lattice, also increasing ionic conductivity. Step 4: Both types of defects provide pathways for ion movement, thereby increasing electrical conductivity. Step 5: This is why ionic conductors (like solid electrolytes) often have high defect concentrations. Answer: C.

Explanation:
Step 1: In FCC crystals, the slip system is {111}<110>. Step 2: The Burgers vector for a perfect dislocation in FCC is b = a/2<110> where a is the lattice parameter. Step 3: The magnitude of this Burgers vector = a/2 × √(1²+1²+0²) = a/2 × √2 = a/√2. Step 4: The notation a/2[110] means the Burgers vector has components (a/2, a/2, 0). Step 5: This is the most common Burgers vector in FCC metals. Answer: A = a/2[110].

Explanation:
Step 1: Crystal defects are classified by their dimensionality. Step 2: Point defects (0D): vacancies, interstitials, substitutional impurities. Step 3: Line defects (1D): dislocations (edge, screw, mixed). Step 4: Planar defects (2D): grain boundaries, twin boundaries, stacking faults, free surfaces. Step 5: Volume defects (3D): voids, precipitates, inclusions. Step 6: Grain boundaries separate regions of different crystallographic orientation and are 2D planar defects. Answer: C.

Explanation:
Step 1: The Frank-Read source is a mechanism by which a single dislocation segment can generate multiple dislocation loops. Step 2: A dislocation segment pinned at two points (by precipitates or other dislocations) bows out under applied stress. Step 3: The bowing segment eventually wraps around and reconnects, creating a complete dislocation loop and regenerating the original segment. Step 4: This process can repeat, generating many dislocation loops from a single source. Step 5: This explains why the dislocation density increases dramatically during plastic deformation (work hardening). Answer: B.

Explanation:
Step 1: Point defects are zero-dimensional defects involving individual atomic sites. Step 2: Vacancy: a missing atom at a lattice site — point defect ✓. Step 3: Interstitial atom: an extra atom in an interstitial position — point defect ✓. Step 4: Substitutional impurity: a foreign atom replacing a host atom — point defect ✓. Step 5: Dislocation: a one-dimensional (line) defect involving a row of atoms — NOT a point defect. Answer: D.

Explanation:
Step 1: For Schottky defects in an ionic crystal like NaCl, both cation and anion vacancies must be created in pairs. Step 2: The equilibrium concentration of Schottky defect pairs: n_s = N × exp(-Qs/2kT) where Qs is the energy to form one Schottky pair, N is the number of ion pairs, k is Boltzmann's constant, T is temperature. Step 3: The factor of 2 in the denominator arises because the energy Qs is shared between two vacancies (one cation + one anion). Step 4: This is different from the single vacancy case where n = N × exp(-Qv/kT). Answer: B = exp(-Qs/2kT).

Explanation:
Step 1: Slip occurs on the most densely packed planes in the most densely packed directions. Step 2: In FCC metals: Most densely packed planes = {111} (octahedral planes). Most densely packed directions = <110>. Step 3: Slip system = {111}<110>. There are 4 {111} planes × 3 <110> directions = 12 slip systems in FCC. Step 4: In BCC metals: Slip system = {110}<111> (48 possible slip systems). Step 5: In HCP metals: Slip system = {0001}<1120> (basal slip). Answer: B = {111}<110>.

Explanation:
Step 1: Schmid's law relates the applied tensile stress σ to the resolved shear stress τ on a slip system. Step 2: τ = σ cosφ cosλ where φ = angle between tensile axis and slip plane normal, λ = angle between tensile axis and slip direction. Step 3: The term cosφ cosλ is called the Schmid factor. Step 4: Slip begins when τ reaches the critical resolved shear stress (CRSS): τ_CRSS = σ_y cosφ cosλ. Step 5: Maximum Schmid factor = 0.5 when φ = λ = 45°. Answer: A.

Explanation:
Step 1: Low-angle grain boundaries (tilt boundaries) are formed by arrays of dislocations. Step 2: A tilt boundary consists of a wall of edge dislocations arranged vertically. Step 3: The misorientation angle θ between adjacent grains is related to the dislocation spacing D by: θ = b/D where b is the Burgers vector. Step 4: A twist boundary consists of a network of screw dislocations. Step 5: For small misorientation angles (θ < 15°), the boundary energy increases with θ. Answer: B = Edge dislocations.

Explanation:
Step 1: The Read-Shockley equation gives the energy of a low-angle grain boundary as a function of misorientation angle θ. Step 2: E = E₀θ(A - lnθ) where E₀ = Gb/(4π(1-ν)) and A is a constant related to the dislocation core energy. Step 3: This equation is valid for small angles (θ < 15°). Step 4: At small θ, the boundary energy increases with θ (more dislocations per unit area). Step 5: At large angles (high-angle boundaries), the Read-Shockley equation breaks down and the energy becomes approximately constant. Answer: A.

Explanation:
Step 1: F-centers (Farbe = color in German) are point defects that give color to otherwise transparent alkali halide crystals. Step 2: An F-center consists of an anion vacancy that has trapped an electron. Step 3: The trapped electron can absorb visible light photons and be excited to higher energy levels, giving the crystal a characteristic color. Step 4: For example, NaCl with F-centers appears yellow, KCl appears violet. Step 5: F-centers can be created by irradiation with X-rays or by heating in alkali metal vapor. Answer: C.

Explanation:
Step 1: Dislocation density ρ is defined as the total length of dislocation lines per unit volume (m/m³ = m⁻²). Step 2: In an annealed (well-annealed) metal: ρ ≈ 10¹⁰ to 10¹² m⁻². Step 3: In a lightly deformed metal: ρ ≈ 10¹² to 10¹³ m⁻². Step 4: In a heavily cold-worked metal: ρ ≈ 10¹⁴ to 10¹⁶ m⁻². Step 5: The increase in dislocation density during cold working is responsible for work hardening (strain hardening). Answer: D = 10¹⁶ m⁻².

Explanation:
Step 1: The Taylor hardening equation relates yield strength to dislocation density. Step 2: σ_y = σ₀ + αGb√ρ where σ₀ is the friction stress (lattice resistance), α is a constant (~0.5), G is shear modulus, b is Burgers vector, ρ is dislocation density. Step 3: This can be written as σ_y = σ₀ + k√ρ where k = αGb. Step 4: The √ρ dependence arises from the long-range stress fields of dislocations interacting with each other. Step 5: This equation explains why cold working (increasing ρ) increases yield strength. Answer: B.

Explanation:
Step 1: Cottrell atmosphere is a phenomenon in which solute atoms (interstitial or substitutional) segregate to the stress field around a dislocation. Step 2: The dislocation has a compressive stress field above the slip plane and a tensile stress field below. Step 3: Interstitial atoms (like carbon in iron) prefer the tensile region (below the slip plane) where they can relieve their misfit strain. Step 4: This segregation of solute atoms around the dislocation is called the Cottrell atmosphere. Step 5: The Cottrell atmosphere pins the dislocation, requiring extra stress to break free — this explains the upper and lower yield point phenomenon in mild steel. Answer: B.

Explanation:
Step 1: Mild steel (low carbon steel) exhibits a distinct upper yield point followed by a lower yield point in its stress-strain curve. Step 2: At the upper yield point, the applied stress is sufficient to break dislocations free from their Cottrell atmospheres (carbon and nitrogen interstitials). Step 3: Once free, the dislocations can move at a lower stress — this is the lower yield point. Step 4: The Lüders band (deformation band) propagates across the specimen at the lower yield stress. Step 5: This phenomenon is specific to BCC metals with interstitial solutes (C, N in iron). Answer: B.

Explanation:
Step 1: The Peierls-Nabarro stress is the minimum stress required to move a dislocation through a perfect crystal (lattice friction). Step 2: τ_PN = 2G exp(-2πw/b) where G is shear modulus, w is the dislocation width, b is the Burgers vector. Step 3: The dislocation width w = a/(1-ν) for edge dislocations, where a is the interplanar spacing. Step 4: Wide dislocations (large w) have lower Peierls stress — this is why FCC metals (wide dislocations) are more ductile than BCC metals. Step 5: The Peierls stress is the theoretical lower bound for the yield stress of a perfect crystal. Answer: A.

Explanation:
Step 1: Solid solutions are classified as substitutional or interstitial based on where the solute atoms reside. Step 2: Substitutional solid solution: solute atoms replace host atoms at lattice sites. Examples: Cu-Zn (brass), Cu-Ni, Au-Ag. Step 3: Interstitial solid solution: solute atoms occupy interstitial spaces between host atoms. This occurs when the solute atom is much smaller than the host atom. Step 4: Fe-C (steel): Carbon atoms (r = 0.077 nm) are much smaller than iron atoms (r = 0.124 nm) and occupy interstitial sites in the iron lattice. Step 5: The ratio r_solute/r_solvent < 0.59 for interstitial solid solutions. Answer: B = Fe-C.

Explanation:
Step 1: Hume-Rothery rules are empirical guidelines for predicting the extent of solid solubility between two metals. Step 2: The four Hume-Rothery rules are: (1) Atomic size factor: size difference < 15% for extensive solubility. (2) Crystal structure: same crystal structure for complete solubility. (3) Electronegativity: similar electronegativity (difference < 0.4). (4) Valence: same valence or higher valence solute in lower valence solvent. Step 3: If the atomic size difference > 15%, the lattice strain is too large and solubility is limited. Step 4: Cu-Ni system satisfies all Hume-Rothery rules and shows complete solid solubility. Answer: B = 15%.

Explanation:
Step 1: In FCC metals, the slip system is {111}<110>. Step 2: Number of {111} planes = 4 (namely (111), (1̄11), (11̄1), (111̄)). Step 3: Number of <110> directions per {111} plane = 3 (e.g., [110], [101], [011] and their negatives, but only 3 independent). Step 4: Total slip systems = 4 × 3 = 12. Step 5: These 12 slip systems provide FCC metals with excellent ductility (von Mises criterion requires 5 independent slip systems for polycrystalline ductility). Answer: B = 12.

Explanation:
Step 1: The Hall-Petch equation describes the relationship between yield strength and grain size. Step 2: σ_y = σ₀ + k_y/√d where σ₀ is the friction stress (lattice resistance to dislocation motion), k_y is the Hall-Petch slope (strengthening coefficient), d is the average grain diameter. Step 3: As grain size decreases (d decreases), yield strength increases. Step 4: This is because grain boundaries act as barriers to dislocation motion. Step 5: The Hall-Petch relationship breaks down at very small grain sizes (< ~10 nm) where the inverse Hall-Petch effect may occur. Answer: B.

Explanation:
Step 1: Crystal defects classified by dimensionality: 0D (point): vacancies, interstitials, substitutional impurities. 1D (line): dislocations. 2D (planar): grain boundaries, twin boundaries, stacking faults, free surfaces. 3D (volume): voids, precipitates, inclusions. Step 2: Vacancy = 0D point defect. Edge dislocation = 1D line defect. Twin boundary = 2D planar defect. Void = 3D volume defect. Step 3: A twin boundary is a special type of grain boundary where the crystal on one side is a mirror image of the crystal on the other side. Answer: C = Twin boundary.

Explanation:
Step 1: Self-diffusion in metals occurs primarily by the vacancy mechanism. Step 2: For an atom to diffuse, two steps are required: (1) A vacancy must form adjacent to the atom (requires energy Qf = vacancy formation energy). (2) The atom must jump into the vacancy (requires energy Qm = vacancy migration energy). Step 3: The total activation energy for self-diffusion: Q_SD = Qf + Qm. Step 4: This is why the activation energy for self-diffusion is higher than for impurity diffusion (where vacancies are already present). Step 5: Typical values: Qf ≈ 1 eV, Qm ≈ 0.5 eV, Q_SD ≈ 1.5 eV for metals. Answer: C.

Explanation:
Step 1: In FCC metals, a perfect dislocation with Burgers vector b = a/2[110] can dissociate into two Shockley partial dislocations. Step 2: a/2[110] → a/6[211] + a/6[121̄]. Step 3: Each partial dislocation has Burgers vector of the form a/6<112>. Step 4: The dissociation is energetically favorable because the energy of two partials (2 × (a/6)²|<112>|²) < energy of perfect dislocation ((a/2)²|<110>|²). Step 5: The region between the two partial dislocations is a stacking fault. Answer: B = a/6<112>.

Explanation:
Step 1: For Schottky defects in an ionic crystal, the equilibrium number of defect pairs is derived from thermodynamics. Step 2: The Gibbs free energy is minimized when: n = N exp(-Qs/2kT) where n = number of Schottky pairs, N = number of ion pairs, Qs = energy to form one Schottky pair, k = Boltzmann's constant, T = temperature. Step 3: The factor of 2 in the denominator arises because each Schottky pair consists of two vacancies (one cation + one anion), and the energy Qs is shared between them. Step 4: This is analogous to the Frenkel defect concentration: n = √(NN') exp(-Qf/2kT). Answer: B.

Explanation:
Step 1: Work hardening is the increase in strength and hardness of a metal due to plastic deformation. Step 2: During plastic deformation, the dislocation density increases from ~10¹⁰ m⁻² (annealed) to ~10¹⁶ m⁻² (heavily cold-worked). Step 3: As dislocation density increases, dislocations interact with each other and create obstacles to further dislocation motion. Step 4: These dislocation-dislocation interactions (jogs, tangles, Lomer-Cottrell locks) require higher stress to continue deformation. Step 5: The Taylor hardening equation: σ = σ₀ + αGb√ρ quantifies this effect. Answer: C.

Explanation:
Step 1: A Lomer-Cottrell lock is a type of sessile (immobile) dislocation formed by the reaction of two partial dislocations from different slip systems. Step 2: In FCC metals: a/6[121] + a/6[11̄2] → a/6[110]. Step 3: The resulting dislocation a/6[110] lies along the intersection of two {111} planes and cannot glide on either plane. Step 4: This creates a barrier to dislocation motion, contributing to work hardening. Step 5: Lomer-Cottrell locks are also called stair-rod dislocations because they lie along the edge of a stacking fault ribbon. Answer: B.

Explanation:
Step 1: In close-packed crystal structures, atoms are arranged in layers. Step 2: FCC stacking sequence: ABCABCABC... (three-layer repeat). Step 3: HCP stacking sequence: ABABAB... (two-layer repeat). Step 4: In HCP, the A and B layers alternate, with each B layer directly above the A layer gaps. Step 5: The difference in stacking sequence between FCC and HCP is the source of stacking faults in FCC metals (local HCP-like stacking). Answer: B = ABABAB.

Explanation:
Step 1: Cross-slip is the process by which a screw dislocation moves from one slip plane to another. Step 2: In FCC metals, screw dislocations dissociate into two Shockley partial dislocations separated by a stacking fault ribbon. Step 3: For cross-slip to occur, the partial dislocations must first recombine (constrict) to form a perfect dislocation. Step 4: In metals with low stacking fault energy (SFE), the partials are widely separated and difficult to recombine → cross-slip is difficult. Step 5: In metals with high SFE (e.g., Al, Ni), the partials are close together and easily recombine → cross-slip is easy. Answer: B = High stacking fault energy.

Explanation:
Step 1: Precipitation hardening strengthens metals by introducing fine precipitate particles that impede dislocation motion. Step 2: Two mechanisms exist: (1) Cutting mechanism: dislocations cut through coherent precipitates (small, soft precipitates). (2) Orowan mechanism: dislocations bow around incoherent precipitates (large, hard precipitates). Step 3: In the Orowan mechanism, the dislocation bows between precipitates, eventually bypassing them and leaving a dislocation loop around each precipitate. Step 4: The Orowan stress: τ = Gb/L where L is the inter-precipitate spacing. Step 5: As precipitates coarsen (L increases), the Orowan stress decreases (overaging). Answer: B.

Explanation:
Step 1: The Burgers circuit is a closed loop drawn around a dislocation in a crystal. Step 2: The circuit is drawn atom by atom in a clockwise direction (using the RH/FS convention). Step 3: In a perfect crystal, the same circuit would close perfectly. Step 4: In a crystal with a dislocation, the circuit fails to close — the closure failure vector is the Burgers vector b. Step 5: The Burgers vector characterizes the magnitude and direction of the lattice distortion caused by the dislocation. Answer: B.

Explanation:
Step 1: Carbon occupies octahedral interstitial sites in both FCC (austenite) and BCC (ferrite) iron. Step 2: In FCC iron: octahedral site radius = 0.414r_Fe = 0.414 × 0.124 = 0.0513 nm. Step 3: In BCC iron: octahedral site radius = 0.154r_Fe = 0.154 × 0.124 = 0.0191 nm. Step 4: Carbon atom radius = 0.077 nm. The FCC octahedral site is larger and can accommodate carbon with less strain. Step 5: Maximum solubility: γ-iron = 2.14 wt% C (at 1147°C), α-iron = 0.022 wt% C (at 727°C). Answer: A.

Explanation:
Step 1: The vacancy formation energy Qv is the energy required to remove an atom from the interior of a crystal to the surface. Step 2: Typical values for metals: Al: Qv ≈ 0.68 eV, Cu: Qv ≈ 1.28 eV, Fe: Qv ≈ 1.4 eV, W: Qv ≈ 3.6 eV. Step 3: The range 0.5-2.0 eV covers most common metals. Step 4: Higher melting point metals generally have higher vacancy formation energies. Step 5: The equilibrium vacancy concentration at room temperature is extremely small (< 10⁻¹⁰) for most metals. Answer: B = 0.5-2.0 eV.

Explanation:
Step 1: Twin boundaries are special types of planar defects where the crystal on one side is a mirror image of the crystal on the other side. Step 2: At a twin boundary, the atoms are in a special coincidence site lattice (CSL) relationship. Step 3: The twin boundary has a very low energy because the atoms on both sides of the boundary are in nearly correct positions. Step 4: Typical energies: Grain boundary energy ≈ 0.5-1.0 J/m². Twin boundary energy ≈ 0.01-0.1 J/m² (coherent twin). Step 5: This is why annealing twins are commonly observed in FCC metals (low energy configuration). Answer: B.

Explanation:
Step 1: In an FCC unit cell, there are 4 atoms per unit cell. Step 2: Number of tetrahedral voids in FCC unit cell = 8 (located at positions like (1/4, 1/4, 1/4)). Step 3: Number of octahedral voids in FCC unit cell = 4 (located at edge centers and body center). Step 4: Tetrahedral voids per atom = 8/4 = 2. Step 5: Octahedral voids per atom = 4/4 = 1. Answer: B = 2 tetrahedral voids per atom.

Explanation:
Step 1: The radius ratio r/R (solute/solvent) determines which interstitial site an atom can occupy. Step 2: For a triangular void (2D): r/R = 0.155. Step 3: For a tetrahedral void: r/R = 0.225. Step 4: For an octahedral void: r/R = 0.414. Step 5: For a cubic void: r/R = 0.732. Step 6: If r/R > 0.414 for an octahedral site, the atom is too large and will distort the lattice. Answer: C = 0.414.

Explanation:
Step 1: Volterra (1907) developed a mathematical model for dislocations based on elastic theory. Step 2: The Volterra model classifies dislocations based on the relative displacement of the two halves of a crystal: Edge dislocation: displacement perpendicular to the dislocation line. Screw dislocation: displacement parallel to the dislocation line. Mixed dislocation: displacement at an angle to the dislocation line. Step 3: Any dislocation can be decomposed into edge and screw components. Step 4: The Burgers vector uniquely characterizes the type of dislocation. Answer: B.

Explanation:
Step 1: Dislocation motion can occur by glide (conservative) or climb (non-conservative). Step 2: Glide: dislocation moves in its slip plane under applied shear stress — no diffusion required. Step 3: Climb: edge dislocation moves perpendicular to its slip plane — requires diffusion of vacancies or interstitials. Step 4: Positive climb: dislocation moves up (away from extra half-plane) by absorbing vacancies or emitting interstitials. Step 5: Negative climb: dislocation moves down by emitting vacancies or absorbing interstitials. Step 6: Climb is important at high temperatures where diffusion is significant. Answer: B.

Explanation:
Step 1: Stacking fault energy (SFE) is the energy per unit area of a stacking fault. Step 2: Metals with low SFE: Ag (~16 mJ/m²), Cu (~45 mJ/m²), brass (~10 mJ/m²). Step 3: Metals with high SFE: Al (~200 mJ/m²), Ni (~125 mJ/m²). Step 4: High SFE → narrow stacking fault ribbons → easy cross-slip → good recovery → less work hardening. Step 5: Low SFE → wide stacking fault ribbons → difficult cross-slip → more work hardening → better for deep drawing. Answer: B = ~200 mJ/m².

Explanation:
Step 1: When two dislocations intersect, steps are created on each dislocation. Step 2: A kink is a step in the dislocation line that lies within the slip plane — it can move by glide. Step 3: A jog is a step in the dislocation line that lies out of the slip plane — it cannot move by glide and requires climb (diffusion). Step 4: Jogs on edge dislocations can move by climb. Jogs on screw dislocations require climb and are strong obstacles to dislocation motion. Step 5: Jogs are important in work hardening because they impede dislocation motion. Answer: B.

Explanation:
Step 1: The Peach-Koehler equation gives the force per unit length on a dislocation due to an applied stress field. Step 2: F = (σ·b) × ξ where σ is the stress tensor, b is the Burgers vector, ξ is the unit vector along the dislocation line. Step 3: This force drives dislocation glide and climb. Step 4: For a simple case of a screw dislocation under shear stress τ: F = τb (force per unit length). Step 5: The Peach-Koehler force is the fundamental driving force for dislocation motion. Answer: A.

Explanation:
Step 1: In a BCC (Body-Centered Cubic) unit cell: Corner atoms: 8 corners × (1/8 atom per corner) = 1 atom. Body center atom: 1 × 1 = 1 atom. Step 2: Total atoms per unit cell = 1 + 1 = 2. Step 3: For FCC: 8 × (1/8) + 6 × (1/2) = 1 + 3 = 4 atoms per unit cell. Step 4: For SC: 8 × (1/8) = 1 atom per unit cell. Step 5: The number of atoms per unit cell determines the packing efficiency and the number of interstitial sites. Answer: B = 2.

Explanation:
Step 1: Diffusion rate depends on the activation energy Q: D = D₀ exp(-Q/RT). Step 2: Interstitial diffusion: small atoms (C, N, H, O, B) move through interstitial spaces. Activation energy is relatively low because no vacancy is needed. Step 3: Substitutional diffusion: atoms exchange with vacancies. Requires vacancy formation energy + migration energy → higher total activation energy. Step 4: Example: D_C in Fe at 900°C ≈ 10⁻¹¹ m²/s (interstitial). D_Fe in Fe at 900°C ≈ 10⁻²⁰ m²/s (substitutional). Step 5: Interstitial diffusion is ~10⁹ times faster than self-diffusion. Answer: B.

Explanation:
Step 1: Creep is time-dependent plastic deformation at elevated temperatures. Step 2: Nabarro-Herring creep (diffusional creep) occurs at high temperatures and low stresses. Step 3: Mechanism: vacancies diffuse from regions of high stress (grain boundaries under tension) to regions of low stress (grain boundaries under compression) through the grain interior. Step 4: This causes the grain to elongate in the tensile direction. Step 5: Creep rate: ε̇ = A σ D_v / (d² kT) where d is grain size, D_v is vacancy diffusivity. Answer: C.

Explanation:
Step 1: Coble creep is a type of diffusional creep that occurs at lower temperatures than Nabarro-Herring creep. Step 2: Mechanism: vacancies diffuse along grain boundaries (rather than through the grain interior). Step 3: Grain boundary diffusion is faster than lattice diffusion at lower temperatures. Step 4: Coble creep rate: ε̇ = A σ D_gb δ / (d³ kT) where D_gb is grain boundary diffusivity, δ is grain boundary width. Step 5: Coble creep has a stronger grain size dependence (d³) than Nabarro-Herring creep (d²). Answer: B.

Explanation:
Step 1: Schmid factor = cosφ cosλ where φ = angle between tensile axis and slip plane normal, λ = angle between tensile axis and slip direction. Step 2: Maximum Schmid factor occurs when d(cosφ cosλ)/dφ = 0 subject to the constraint φ + λ = 90° (for the most favorable geometry). Step 3: With φ + λ = 90°: Schmid factor = cosφ cos(90°-φ) = cosφ sinφ = (1/2)sin2φ. Step 4: Maximum when sin2φ = 1, i.e., 2φ = 90°, φ = 45°. Then λ = 45°. Step 5: Maximum Schmid factor = cos45° × cos45° = 0.5. Answer: C = 45°.

Explanation:
Step 1: The Burgers vector b of a perfect dislocation must be a lattice translation vector. Step 2: This ensures that the crystal is perfect (no stacking fault) on both sides of the dislocation. Step 3: If b is not a lattice translation vector, the dislocation is a partial dislocation and is bounded by a stacking fault. Step 4: For FCC: b = a/2<110> (lattice translation vector). For BCC: b = a/2<111> (lattice translation vector). Step 5: Partial dislocations (e.g., Shockley partials with b = a/6<112>) are not lattice translation vectors. Answer: A.

Explanation:
Step 1: The equilibrium vacancy concentration Nv/N = exp(-Qv/kT). Step 2: At the melting point, the thermal energy kTm is significant. Step 3: For most metals, Qv/kTm ≈ 8-10. Step 4: Nv/N = exp(-9) ≈ 1.2 × 10⁻⁴. Step 5: This means approximately 1 in 10,000 lattice sites is vacant at the melting point. This is why quenching from near the melting point can trap a significant number of vacancies. Answer: B = 10⁻⁴.

Explanation:
Step 1: Frank partial dislocations are sessile (immobile) dislocations in FCC metals. Step 2: They are formed by the collapse of a vacancy disk or the condensation of interstitials on a {111} plane. Step 3: The Burgers vector of a Frank partial = a/3<111> (perpendicular to the {111} plane). Step 4: Since b = a/3<111> is not in the {111} slip plane, the Frank partial cannot glide — it is sessile. Step 5: Frank partials bound intrinsic stacking faults (vacancy type) or extrinsic stacking faults (interstitial type). Answer: C = a/3<111>.

Explanation:
Step 1: The Portevin-Le Chatelier (PLC) effect is characterized by serrated (jerky) flow in the stress-strain curve. Step 2: It occurs due to dynamic strain aging: as dislocations move, they are repeatedly pinned by solute atoms (Cottrell atmospheres) and then break free. Step 3: This repeated pinning and unpinning causes the serrated appearance of the stress-strain curve. Step 4: PLC effect is observed in Al-Mg alloys, mild steel, and other alloy systems at intermediate temperatures and strain rates. Step 5: It is also called the jerky flow or serrated yielding phenomenon. Answer: B.

Explanation:
Step 1: In an FCC unit cell, there are 4 atoms per unit cell. Step 2: Octahedral voids in FCC: 12 edge centers × (1/4) + 1 body center × 1 = 3 + 1 = 4 octahedral voids per unit cell. Step 3: Octahedral voids per atom = 4/4 = 1. Step 4: Tetrahedral voids in FCC: 8 per unit cell → 8/4 = 2 per atom. Step 5: This is why the maximum solubility of carbon in FCC iron (austenite) is 2.14 wt% — there is 1 octahedral void per Fe atom. Answer: A = 1.

Explanation:
Step 1: Zener pinning describes the retardation of grain boundary migration by second-phase particles. Step 2: Each particle exerts a pinning force on the grain boundary. Step 3: The Zener limiting grain size: D = 4r/(3f) where r = particle radius, f = volume fraction of particles. Step 4: Smaller particles (smaller r) and higher volume fraction (larger f) give smaller limiting grain size. Step 5: This principle is used in grain refinement of steels and aluminum alloys by adding fine dispersoids. Answer: B = 4r/(3f).

Explanation:
Step 1: Ni₃Al (γ' phase) has an L1₂ ordered structure. Step 2: Dislocations in Ni₃Al are superdislocations with b = a<110>. Step 3: The Kear-Wilsdorf (KW) lock forms when a superdislocation cross-slips from the {111} plane to the {100} (cube) plane. Step 4: The cross-slipped segment on the {100} plane is sessile (cannot glide on {111}). Step 5: This creates a locked configuration that resists further dislocation motion, explaining the anomalous yield strength increase with temperature in Ni₃Al. Answer: B.

Explanation:
Step 1: The Bauschinger effect is the reduction in yield strength when the direction of loading is reversed after plastic deformation. Step 2: During forward loading, dislocations pile up against obstacles (grain boundaries, precipitates). Step 3: The pile-up creates a back stress that opposes further forward motion but aids reverse motion. Step 4: When the load is reversed, the back stress assists dislocation motion in the reverse direction, reducing the yield stress in that direction. Step 5: The Bauschinger effect is important in metal forming processes involving cyclic loading. Answer: B.

Explanation:
Step 1: The Suzuki effect (chemical interaction) involves the segregation of solute atoms to stacking faults. Step 2: Solute atoms that lower the stacking fault energy will preferentially segregate to the stacking fault region. Step 3: This segregation lowers the stacking fault energy further, making the stacking fault more stable. Step 4: The Suzuki effect pins the partial dislocations bounding the stacking fault, making cross-slip more difficult. Step 5: This is different from the Cottrell effect (segregation to dislocation stress field) and the Suzuki effect (segregation to stacking fault). Answer: A.

Explanation:
Step 1: In BCC metals, slip occurs on multiple planes: {110}, {112}, and {123} planes, all in the <111> direction. Step 2: {110}<111>: 6 planes × 2 directions = 12 slip systems. Step 3: {112}<111>: 12 planes × 1 direction = 12 slip systems. Step 4: {123}<111>: 24 planes × 1 direction = 24 slip systems. Step 5: Total = 12 + 12 + 24 = 48 slip systems. This is why BCC metals can deform on many planes (pencil glide). Answer: C = 48.

Explanation:
Step 1: The Snoek effect is an anelastic relaxation phenomenon in BCC metals containing interstitial solutes. Step 2: In BCC iron, carbon atoms occupy octahedral interstitial sites. There are three types of octahedral sites (x, y, z-type) that are not equivalent under uniaxial stress. Step 3: Under applied stress, carbon atoms preferentially jump to the sites that relieve the most strain energy. Step 4: This stress-induced ordering of interstitial atoms causes internal friction (damping) at a characteristic temperature and frequency. Step 5: The Snoek peak in iron occurs at ~40°C at 1 Hz. Answer: A.

Explanation:
Step 1: Griffith (1921) developed a criterion for brittle fracture based on energy balance. Step 2: A crack propagates when the strain energy released equals the surface energy created. Step 3: Critical stress: σ_c = √(2Eγ_s/πa) where E = Young's modulus, γ_s = surface energy, a = half crack length. Step 4: The critical stress is inversely proportional to √a (crack length). Step 5: Longer cracks require lower stress to propagate — this is why surface defects and notches are dangerous. Answer: B = Crack length.

Explanation:
Step 1: The Peierls-Nabarro stress: τ_PN = 2G exp(-2πw/b) where w = dislocation width ≈ a/(1-ν). Step 2: The dislocation width w is proportional to the interplanar spacing a. Step 3: Large interplanar spacing → large w → small τ_PN (easier to move). Step 4: Small Burgers vector → small b → large 2πw/b → small τ_PN. Step 5: Therefore, slip occurs most easily on planes with large interplanar spacing (most densely packed planes) in directions with small Burgers vector (most densely packed directions). Answer: B.

Explanation:
Step 1: In the Volterra model, a dislocation is created by making a cut in a cylinder and displacing the two surfaces. Step 2: If the displacement is perpendicular to the dislocation line (cut edge) → Edge dislocation. Step 3: If the displacement is parallel to the dislocation line → Screw dislocation. Step 4: The screw dislocation gets its name from the helical arrangement of atoms around the dislocation line (like a screw thread). Step 5: The Burgers vector of a screw dislocation is parallel to the dislocation line. Answer: B.

Explanation:
Step 1: The Zener-Hollomon parameter Z is used to describe hot deformation behavior. Step 2: Z = ε̇ exp(Q/RT) where ε̇ = strain rate, Q = activation energy for hot deformation, R = gas constant, T = absolute temperature. Step 3: Z combines the effects of strain rate and temperature into a single parameter. Step 4: High Z (high strain rate or low temperature) → more work hardening, less recovery. Step 5: The flow stress during hot deformation: σ = f(Z). Answer: A = Z = ε̇ exp(Q/RT).

Explanation:
Step 1: The Orowan mechanism involves dislocations bowing between dispersed particles. Step 2: The Orowan stress (simplified): τ = Gb/L where L is the inter-particle spacing. Step 3: The more accurate Ashby-Orowan equation: τ = Gb/(2πL) × ln(L/b) (for edge dislocations) or τ = Gb/(2πL) × ln(L/b) (for screw dislocations). Step 4: This accounts for the logarithmic dependence on the ratio L/b. Step 5: The Ashby-Orowan equation gives a better fit to experimental data than the simple Orowan equation. Answer: B.

Explanation:
Step 1: Nabarro-Herring creep involves diffusion of vacancies through the grain interior. Step 2: The creep rate: ε̇_NH = A × (σΩD_v)/(d²kT) where σ = applied stress, Ω = atomic volume, D_v = lattice diffusivity, d = grain size, k = Boltzmann's constant, T = temperature. Step 3: The creep rate is proportional to σ/d². Step 4: Smaller grain size → faster Nabarro-Herring creep (more grain boundary area for vacancy sources/sinks). Step 5: This is why fine-grained materials are not suitable for high-temperature applications. Answer: B = σ/d².

Explanation:
Step 1: The Cottrell-Bilby theory describes the kinetics of solute atom segregation to dislocations. Step 2: The number of solute atoms per unit length of dislocation that have segregated at time t: n(t) ∝ (D_s t)^(2/3) where D_s is the solute diffusivity. Step 3: The temperature dependence comes through D_s = D₀ exp(-Q/kT). Step 4: At a given time, the segregation is proportional to T⁻²/³ (through the temperature dependence of diffusivity). Step 5: This theory explains the time-dependent strain aging (static strain aging) in mild steel. Answer: C = T⁻²/³.

Explanation:
Step 1: The Hall-Petch equation: σ_y = σ₀ + k_y/√d. Step 2: σ₀ = friction stress (Peierls stress + solid solution strengthening). Step 3: k_y = Hall-Petch slope = stress intensity factor for grain boundary strengthening. Step 4: k_y is related to the stress required to activate dislocation sources in the adjacent grain (to propagate slip across the grain boundary). Step 5: Higher k_y means grain boundaries are more effective barriers to slip transmission. Answer: B.

Explanation:
Step 1: Sievert's law describes the solubility of diatomic gases (H₂, N₂, O₂) in liquid and solid metals. Step 2: The dissolution reaction: ½X₂(gas) → X(dissolved). Step 3: Equilibrium constant K = [X]/√p_X₂. Step 4: Therefore, [X] = K√p_X₂ — solubility is proportional to √p. Step 5: This is why hydrogen porosity in welds increases with the square root of hydrogen partial pressure. Answer: B = √p.

Explanation:
Step 1: The Orowan equation relates the macroscopic plastic strain rate to the microscopic dislocation parameters. Step 2: ε̇ = ρ_m × b × v where ρ_m = mobile dislocation density (m⁻²), b = Burgers vector (m), v = average dislocation velocity (m/s). Step 3: This equation shows that plastic strain rate increases with: more mobile dislocations, larger Burgers vector, or faster dislocation velocity. Step 4: The dislocation velocity v depends on the applied stress: v = (τ/τ₀)^m where m is the stress exponent. Step 5: The Orowan equation is fundamental to understanding plastic deformation kinetics. Answer: A = ε̇ = ρbv.

Explanation:
Step 1: Solid solution strengthening occurs when solute atoms impede dislocation motion. Step 2: The Friedel-Fleischer model for weak obstacles (dilute solid solutions): Δτ ∝ c^(1/2) × ε^(3/2) where c = solute concentration, ε = misfit parameter. Step 3: However, for the Labusch model (stronger interactions): Δτ ∝ c^(2/3). Step 4: The c^(2/3) dependence is observed experimentally for many solid solution systems. Step 5: The strengthening increases with solute concentration and atomic size misfit. Answer: C = c^(2/3).

Explanation:
Step 1: Different creep mechanisms dominate in different temperature-stress regimes (shown in deformation mechanism maps). Step 2: Dislocation glide: low temperature, high stress. Step 3: Dislocation creep (power-law creep): intermediate temperature, intermediate stress. Step 4: Nabarro-Herring creep (lattice diffusion): high temperature (T > 0.8Tm), low stress. Step 5: Coble creep (grain boundary diffusion): intermediate temperature (0.5-0.8Tm), low stress. Answer: B = High temperature and low stress.

Explanation:
Step 1: At a dislocation node (junction of three or more dislocations), the Burgers vectors must be conserved. Step 2: Using the convention that all dislocations point away from the node: Σb_i = 0 (Frank's rule). Step 3: This is analogous to Kirchhoff's current law at an electrical node. Step 4: Example: b₁ + b₂ + b₃ = 0 for three dislocations meeting at a node. Step 5: This conservation law constrains the possible dislocation reactions and determines which reactions are energetically favorable. Answer: A.

Explanation:
Step 1: The Debye-Waller factor (temperature factor) accounts for the reduction in X-ray diffraction intensity due to thermal vibrations of atoms. Step 2: I = I₀ exp(-2M) where M = 8π²<u²>sin²θ/λ² and <u²> is the mean square displacement of atoms. Step 3: At higher temperatures, atoms vibrate more → larger <u²> → smaller Debye-Waller factor → lower diffraction intensity. Step 4: Point defects also contribute to <u²> by creating local lattice distortions. Step 5: The Debye-Waller factor is important in quantitative X-ray diffraction analysis. Answer: B.

Explanation:
Step 1: Kröger-Vink notation is used to describe point defects in ionic crystals. Step 2: The notation is: Species_Site^charge where charge is: × = neutral, • = positive, ' = negative. Step 3: V_Na' means: V = vacancy, Na = sodium site, ' = one negative charge (relative to the perfect lattice). Step 4: A sodium vacancy has one negative charge because removing Na⁺ leaves a site with effective charge -1. Step 5: Similarly: V_Cl• = chlorine vacancy (effective charge +1). Answer: A = V_Na'.

Explanation:
Step 1: The Zener anisotropy factor A measures the degree of elastic anisotropy in cubic crystals. Step 2: A = 2C₄₄/(C₁₁-C₁₂) where C₁₁, C₁₂, C₄₄ are elastic constants. Step 3: For an isotropic material: A = 1 (C₄₄ = (C₁₁-C₁₂)/2). Step 4: A > 1: stiffer in <111> direction (e.g., Cu: A = 3.2). A < 1: stiffer in <100> direction (e.g., Mo: A = 0.77). Step 5: The anisotropy factor affects dislocation energies and elastic interactions. Answer: A.

Explanation:
Step 1: The Stokes-Einstein equation relates the diffusion coefficient of a spherical particle to its size and the viscosity of the medium. Step 2: D = kT/(6πηr) where k = Boltzmann's constant, T = temperature, η = dynamic viscosity, r = particle radius. Step 3: This equation applies to diffusion in liquids and is used to estimate diffusion coefficients of large molecules. Step 4: It shows that smaller particles diffuse faster (larger D for smaller r). Step 5: The equation is derived from Stokes' law for drag force on a sphere and Einstein's relation D = kT/f where f is the friction coefficient. Answer: A.

Explanation:
Step 1: The Gibbs-Thomson effect (also called the Kelvin effect) describes how the chemical potential of a curved surface differs from that of a flat surface. Step 2: For a spherical particle of radius r: μ(r) = μ(∞) + 2γΩ/r where γ = surface energy, Ω = molar volume. Step 3: Smaller particles have higher chemical potential → higher solubility. Step 4: This drives Ostwald ripening: small particles dissolve and large particles grow. Step 5: The Gibbs-Thomson effect is important in precipitation hardening and coarsening of precipitates. Answer: B.

Explanation:
Step 1: The Portevin-Le Chatelier (PLC) effect produces a characteristic serrated or jerky appearance in the stress-strain curve. Step 2: The serrations occur because dislocations are repeatedly pinned by solute atoms (dynamic strain aging) and then break free. Step 3: Each serration corresponds to: stress rises as dislocations are pinned → stress drops as dislocations break free → cycle repeats. Step 4: This is different from the upper/lower yield point (which occurs only once at the start of plastic deformation). Step 5: PLC effect is observed in Al-Mg alloys at room temperature and in mild steel at elevated temperatures. Answer: B.

Explanation:
Step 1: Nabarro-Herring creep: ε̇_NH ∝ σ D_v / (d² kT). Step 2: The d⁻² dependence arises because: the flux of vacancies ∝ 1/d (concentration gradient over grain diameter), and the strain rate ∝ flux/d (strain per unit length). Step 3: Coble creep: ε̇_Coble ∝ σ D_gb δ / (d³ kT) → d⁻³ dependence. Step 4: Smaller grains → faster diffusional creep (more grain boundary area). Step 5: This is why fine-grained materials creep faster at high temperatures. Answer: B = d⁻².

Explanation:
Step 1: When mild steel is loaded beyond the upper yield point, dislocations break free from their Cottrell atmospheres. Step 2: A band of plastic deformation (Lüders band) nucleates at a stress concentration and propagates across the specimen. Step 3: The Lüders band propagates at the lower yield stress (constant stress). Step 4: The stress-strain curve shows a plateau (Lüders strain) during which the band sweeps across the specimen. Step 5: After the Lüders band has swept the entire specimen, uniform work hardening begins. Answer: B = Lower yield stress.

Explanation:
Step 1: In BCC metals, the slip system is {110}<111> (and also {112}<111> and {123}<111>). Step 2: The Burgers vector for a perfect dislocation in BCC = a/2<111>. Step 3: The magnitude = a/2 × √(1²+1²+1²) = a√3/2. Step 4: For BCC iron (a = 0.2866 nm): |b| = 0.2866 × √3/2 = 0.248 nm. Step 5: The a/2<111> vector is the shortest lattice translation vector in BCC, giving the lowest dislocation energy. Answer: A = a/2[111].

Explanation:
Step 1: The Suzuki effect involves segregation of solute atoms to stacking faults. Step 2: The driving force is the difference in stacking fault energy (SFE) between the pure solvent and the alloy. Step 3: If the solute lowers the SFE, it will preferentially segregate to the stacking fault region (which has HCP-like stacking). Step 4: This is a chemical interaction — the solute has a lower energy in the HCP-like environment of the stacking fault. Step 5: The Suzuki segregation is distinct from the Cottrell segregation (elastic misfit) and occurs specifically at stacking faults. Answer: B.

Explanation:
Step 1: Ashby deformation mechanism maps (also called Frost-Ashby maps) display the dominant deformation mechanism as a function of normalized stress and temperature. Step 2: The axes are: x-axis = homologous temperature T/Tm, y-axis = normalized shear stress τ/G. Step 3: Different regions of the map correspond to: elastic deformation, dislocation glide, power-law creep, diffusional creep (Nabarro-Herring and Coble), and grain boundary sliding. Step 4: These maps are useful for selecting materials and conditions to avoid unwanted deformation mechanisms. Answer: A.

Explanation:
Step 1: In the Volterra model, a dislocation is created by making a cut in a cylinder and displacing the two surfaces. Step 2: If the displacement is perpendicular to the dislocation line (cut edge) → Edge dislocation. Step 3: The extra half-plane of atoms is inserted perpendicular to the slip direction. Step 4: The Burgers vector of an edge dislocation is perpendicular to the dislocation line. Step 5: Edge dislocations can only glide in their slip plane (defined by the dislocation line and Burgers vector). Answer: B = Edge dislocation.

Explanation:
Step 1: When a grain boundary intersects a spherical particle of radius r, the particle exerts a pinning force on the boundary. Step 2: The maximum pinning force per particle: F_max = πrγ where γ = grain boundary energy per unit area. Step 3: This force arises because the grain boundary area (and hence energy) is reduced when the boundary passes through the particle. Step 4: The total pinning pressure on the grain boundary: P = 3fγ/(2r) where f = volume fraction of particles. Step 5: Setting P equal to the driving pressure for grain growth gives the Zener limiting grain size D = 4r/(3f). Answer: A = πrγ.

Explanation:
Step 1: Frenkel defects involve an ion leaving its lattice site and occupying an interstitial position. Step 2: For an ion to fit in an interstitial site, it must be small enough to fit without excessive lattice distortion. Step 3: Cations are generally smaller than anions in ionic crystals. Step 4: When the cation is much smaller than the anion, the cation can easily fit into interstitial sites → Frenkel defects are common. Step 5: Examples: AgBr, AgCl (Ag⁺ is small and mobile), ZrO₂. Schottky defects are more common in NaCl, KCl where ions are of comparable size. Answer: B.

Explanation:
Step 1: The Peierls-Nabarro model treats the dislocation as a continuous distribution of infinitesimal dislocations. Step 2: The dislocation width w (half-width of the misfit distribution): w = a/(2(1-ν)) for edge dislocation, w = a/2 for screw dislocation. Step 3: Here a = interplanar spacing of the slip plane. Step 4: Wider dislocations (larger a) have lower Peierls stress → easier to move. Step 5: This explains why slip occurs on the most widely spaced planes (most densely packed planes). Answer: B = a (interplanar spacing).

Explanation:
Step 1: The Cottrell-Lomer (or Lomer-Cottrell) lock is a sessile dislocation formed by the reaction of two dislocations from different slip systems. Step 2: In FCC: a/2[110] + a/2[011] → a/2[101] (Lomer reaction). Step 3: The resulting dislocation a/2[101] lies along the intersection of two {111} planes. Step 4: This dislocation cannot glide on either {111} plane → it is sessile (locked). Step 5: Lomer-Cottrell locks are important barriers to dislocation motion and contribute to work hardening. Answer: B.

Explanation:
Step 1: Creep mechanisms are classified based on the dominant atomic process. Step 2: Power-law creep (dislocation creep): ε̇ ∝ σⁿ (n = 3-8), involves dislocation climb and glide. Step 3: Diffusional creep: involves vacancy diffusion, linear in stress (n = 1). Step 4: Nabarro-Herring creep: diffusion through grain interior → diffusional creep. Step 5: Coble creep: diffusion along grain boundaries → also diffusional creep. Both Nabarro-Herring and Coble creep are types of diffusional creep. Answer: B = Diffusional creep.

Explanation:
Step 1: The Burgers vector b is the fundamental characteristic of a dislocation. Step 2: It describes: (1) The magnitude of lattice distortion (|b| = magnitude of slip). (2) The direction of slip (direction of b). (3) The type of dislocation (edge: b ⊥ line; screw: b ∥ line). Step 3: The energy of a dislocation ∝ Gb² (proportional to |b|²). Step 4: The Burgers vector is conserved along a dislocation line and at dislocation nodes. Step 5: The Burgers vector is determined experimentally using the Burgers circuit in TEM. Answer: B.

Explanation:
Step 1: In BCC iron, there are three types of octahedral interstitial sites: x-type (along [100]), y-type (along [010]), z-type (along [001]). Step 2: In the absence of stress, all three types are equivalent and equally occupied. Step 3: Under uniaxial stress along [001], the z-type sites are expanded and have lower energy for carbon atoms. Step 4: Carbon atoms preferentially jump to z-type sites → stress-induced ordering (Snoek ordering). Step 5: This ordering causes anelastic strain and internal friction (Snoek peak). Answer: B.

Explanation:
Step 1: Griffith (1921) developed his fracture theory for perfectly brittle materials. Step 2: Assumptions: (1) Material is linearly elastic (no plastic deformation). (2) Crack is sharp (zero tip radius). (3) Crack propagates when strain energy release rate G = 2γ_s (surface energy). Step 3: The Griffith equation: σ_c = √(2Eγ_s/πa) applies to perfectly brittle materials like glass and ceramics. Step 4: For metals, Irwin and Orowan modified the theory to include plastic deformation at the crack tip: σ_c = √(2E(γ_s + γ_p)/πa). Step 5: Answer: B = Perfectly brittle (linear elastic).

Explanation:
Step 1: The Zener-Hollomon parameter Z = ε̇ exp(Q/RT) combines strain rate and temperature. Step 2: It is used to describe hot deformation (hot working) behavior of metals. Step 3: During hot deformation, dynamic recovery and recrystallization occur simultaneously with deformation. Step 4: The flow stress during hot deformation: σ = f(Z) — higher Z gives higher flow stress. Step 5: Z is used to predict microstructure (grain size, recrystallization fraction) after hot working. Answer: B = Hot deformation behavior.

Explanation:
Step 1: The Peierls stress (also called lattice friction stress or Peierls-Nabarro stress) is the minimum stress required to move a dislocation through a perfect crystal. Step 2: It arises from the periodic potential energy of the crystal lattice (Peierls potential). Step 3: τ_PN = 2G exp(-2πw/b) where w = dislocation width, b = Burgers vector. Step 4: The Peierls stress is the theoretical lower bound for the yield stress of a perfect crystal. Step 5: In practice, the yield stress is much lower due to thermal activation and pre-existing dislocations. Answer: B.

Explanation:
Step 1: Hume-Rothery rules for extensive solid solubility: (1) Atomic size factor: size difference < 15%. (2) Crystal structure: same crystal structure. (3) Electronegativity: similar values. (4) Valence: same or similar. Step 2: For complete solid solubility (like Cu-Ni), both metals must have the same crystal structure (both FCC). Step 3: If the crystal structures are different, the solubility is limited because the solute cannot maintain the host crystal structure at high concentrations. Step 4: Example: Cu (FCC) and Zn (HCP) have limited solubility (max ~35% Zn in Cu). Answer: B = Same crystal structure. Note: Option A says 'different' which is WRONG. Correct answer is B.

Explanation:
Step 1: Nabarro-Herring creep is a diffusional creep mechanism. Step 2: Under applied stress, there is a gradient in vacancy concentration between grain boundaries under tension (high vacancy concentration) and grain boundaries under compression (low vacancy concentration). Step 3: Vacancies diffuse through the grain interior from tensile boundaries to compressive boundaries. Step 4: Equivalently, atoms diffuse in the opposite direction (from compressive to tensile boundaries). Step 5: This causes the grain to elongate in the tensile direction. Answer: B = Vacancies through the grain interior.

Explanation:
Step 1: In FCC metals, a perfect dislocation b = a/2[110] can dissociate into two Shockley partial dislocations. Step 2: a/2[110] → a/6[211] + a/6[12̄1]. Step 3: Each Shockley partial has Burgers vector of the form a/6<112>. Step 4: The magnitude: |a/6<112>| = a/6 × √(1²+1²+2²) = a/6 × √6 = a/√6. Step 5: The region between the two Shockley partials is a stacking fault. The partials are repelled by their elastic interaction but attracted by the stacking fault energy. Answer: B = a/6<112>.

Explanation:
Step 1: The Zener anisotropy factor A = 2C₄₄/(C₁₁-C₁₂). Step 2: For an elastically isotropic material: C₄₄ = (C₁₁-C₁₂)/2 → A = 1. Step 3: A > 1: material is stiffer in <111> direction (e.g., Cu: A = 3.2, Fe: A = 2.4). Step 4: A < 1: material is stiffer in <100> direction (e.g., Mo: A = 0.77). Step 5: A = 1 means the elastic properties are the same in all directions (isotropic). Answer: B = Elastically isotropic.

Explanation:
Step 1: The Cottrell atmosphere requires solute atoms to diffuse to the dislocation. Step 2: At very low temperatures: diffusion is too slow → solute atoms cannot reach the dislocation → no Cottrell atmosphere. Step 3: At very high temperatures: diffusion is very fast → solute atoms can follow the dislocation as it moves → no pinning (solute drag instead). Step 4: At intermediate temperatures: diffusion is fast enough to form the atmosphere but slow enough that the dislocation cannot escape → maximum pinning effect. Step 5: This explains why the upper/lower yield point is most pronounced at intermediate temperatures in mild steel. Answer: C.

Explanation:
Step 1: Different creep mechanisms have different stress exponents n in the power law: ε̇ ∝ σⁿ. Step 2: Nabarro-Herring creep (lattice diffusion): n = 1 (linear in stress). Step 3: Coble creep (grain boundary diffusion): n = 1 (linear in stress). Step 4: Power-law (dislocation) creep: n = 3-8 (typically 4-5 for metals). Step 5: The linear stress dependence (n = 1) of diffusional creep arises because the vacancy flux is proportional to the stress-induced concentration gradient. Answer: A = n = 1.

Explanation:
Step 1: The Griffith equation for critical stress: σ_c = √(2Eγ_s/πa). Step 2: Rearranging for critical crack length: σ_c² = 2Eγ_s/(πa_c). Step 3: a_c = 2Eγ_s/(πσ²). Step 4: This gives the critical crack length for a given applied stress σ. Step 5: Cracks longer than a_c will propagate spontaneously; shorter cracks are stable. Answer: A = a_c = 2Eγ_s/(πσ²).

Explanation:
Step 1: The PLC effect requires dynamic strain aging — solute atoms must be mobile enough to pin dislocations during deformation. Step 2: In mild steel (Fe-C, Fe-N), the PLC effect occurs at temperatures where carbon and nitrogen atoms are sufficiently mobile. Step 3: Temperature range: approximately 100-300°C (blue brittleness range). Step 4: At room temperature, diffusion is too slow for dynamic strain aging in mild steel. Step 5: At very high temperatures, diffusion is too fast and solute atoms cannot pin dislocations. Answer: B = Room temperature to 300°C (approximately).

Explanation:
Step 1: A Frank partial dislocation has Burgers vector b = a/3<111>, which is perpendicular to the {111} plane. Step 2: For a dislocation to glide, its Burgers vector must lie in the slip plane. Step 3: Since b = a/3<111> is perpendicular to the {111} plane (not in the plane), the Frank partial cannot glide on the {111} plane. Step 4: Therefore, the Frank partial is sessile (immobile by glide). Step 5: Frank partials can only move by climb (requiring diffusion). Answer: B = Sessile dislocation.

Explanation:
Step 1: The Snoek relaxation is an anelastic relaxation due to stress-induced ordering of interstitial atoms (C, N) in BCC iron. Step 2: The relaxation time τ = τ₀ exp(Q/kT) where Q is the activation energy for interstitial diffusion. Step 3: The Snoek peak occurs when ωτ = 1 (ω = angular frequency = 2π × 1 Hz). Step 4: For carbon in α-iron: Q ≈ 0.87 eV, τ₀ ≈ 10⁻¹⁴ s. Step 5: The Snoek peak for C in Fe occurs at approximately 40°C at 1 Hz. For N in Fe, it occurs at about -30°C at 1 Hz. Answer: B = 40°C.

Explanation:
Step 1: Orowan equation: ε̇ = ρ_m × b × v. Step 2: For a given strain rate ε̇ and Burgers vector b: v = ε̇/(ρ_m × b). Step 3: If ρ_m doubles (2ρ_m): v_new = ε̇/(2ρ_m × b) = v/2. Step 4: The dislocation velocity halves when the dislocation density doubles (at constant strain rate). Step 5: This makes physical sense: more dislocations each need to move less to produce the same macroscopic strain rate. Answer: B = Halves.

Explanation:
Step 1: The Zener pinning pressure is the total force per unit area exerted by particles on a grain boundary. Step 2: Number of particles per unit area of grain boundary: n_A = 3f/(2πr²) (for random distribution). Step 3: Force per particle: F = πrγ. Step 4: Total pinning pressure: P = n_A × F = (3f/(2πr²)) × πrγ = 3fγ/(2r). Step 5: Setting P equal to the driving pressure for grain growth (2γ/D) gives the Zener limiting grain size D = 4r/(3f). Answer: A = P = 3fγ/(2r).

Explanation:
Step 1: Nabarro-Herring creep rate: ε̇_NH ∝ D_v/d². Step 2: Coble creep rate: ε̇_Coble ∝ D_gb δ/d³. Step 3: Ratio: ε̇_NH/ε̇_Coble ∝ D_v d/(D_gb δ). Step 4: At high temperatures: D_v/D_gb increases (lattice diffusion activates more strongly). At large grain sizes: d is large → ratio increases. Step 5: Therefore, Nabarro-Herring creep dominates at high temperatures and large grain sizes. Coble creep dominates at lower temperatures and smaller grain sizes. Answer: B.

Explanation:
Step 1: Frank's rule (Burgers vector conservation) states that at any dislocation node, the sum of Burgers vectors of all dislocations meeting at the node must be zero. Step 2: Convention: all dislocations are considered to point away from the node. Step 3: Σb_i = 0 (vector sum). Step 4: This is analogous to Kirchhoff's current law in electrical circuits. Step 5: This rule constrains possible dislocation reactions: a reaction b₁ → b₂ + b₃ is possible only if b₁ = b₂ + b₃ (Burgers vector conservation). Answer: B = Σb_i = 0.

Explanation:
Step 1: Peierls-Nabarro stress: τ_PN = 2G exp(-2πw/b). Step 2: The dislocation width w ∝ a (interplanar spacing of slip plane). Step 3: FCC slip plane {111}: large interplanar spacing → wide dislocations (large w). Step 4: BCC slip plane {110}: smaller interplanar spacing → narrower dislocations (smaller w). Step 5: Wider dislocations (larger w/b) → smaller τ_PN → easier to move → FCC metals are generally more ductile than BCC metals at low temperatures. Answer: B.

Explanation:
Step 1: The interaction energy between an interstitial atom and an edge dislocation depends on the position (r, θ) of the atom relative to the dislocation. Step 2: U = A sinθ/r where A = Gb(1+ν)/(3π(1-ν)) × ΔV, ΔV = volume misfit of the interstitial atom, r = distance from dislocation, θ = angle from slip plane. Step 3: The sinθ dependence means: maximum interaction below the slip plane (θ = -90°, tensile region) and minimum above (θ = +90°, compressive region). Step 4: Interstitial atoms (which expand the lattice) prefer the tensile region (below the slip plane). Step 5: This drives segregation of C and N to the tensile region of edge dislocations in iron. Answer: A.

Explanation:
Step 1: Nabarro-Herring creep involves diffusion of vacancies through the grain interior. Step 2: The creep rate: ε̇_NH = A σ D_v Ω/(d² kT) where A = numerical constant (~10-40), σ = applied stress, D_v = lattice (vacancy) diffusivity, Ω = atomic volume, d = grain size, k = Boltzmann's constant, T = temperature. Step 3: Coble creep: ε̇_Coble = A σ D_gb δ Ω/(d³ kT) where D_gb = grain boundary diffusivity, δ = grain boundary width. Step 4: Power-law creep: ε̇ = A σⁿ exp(-Q/RT). Answer: A.