GATE MT · Chapter-wise

Physical Metallurgy

Phase diagrams, transformations, heat treatment, corrosion · PYQs 1990–2026 with answers & solutions

Physical Metallurgy

Phase diagrams, transformations, heat treatment, corrosion

864 questions
GATE 2026 · Q19
19
Correct sequence for precipitation hardening of Al–4% Ag alloy:
MCQ1M
A
Quenching → Aging → Solution treatment
B
Solution treatment → Aging → Quenching
C
Aging → Solution treatment → Quenching
D
Solution treatment → Quenching → Aging
Solution
Solution treatment (dissolve solute) → Quench (retain SSS) → Age (controlled precipitation). Answer: D
GATE 2026 · Q23
23
Correct precipitation sequence in Al–4 wt.% Cu during isothermal aging:
MCQ1M
A
\(\theta''\rightarrow\theta'\rightarrow\theta\rightarrow\) GP zone
B
GP zone \(\rightarrow\theta\rightarrow\theta'\rightarrow\theta''\)
C
\(\theta\rightarrow\theta'\rightarrow\theta''\rightarrow\) GP zone
D
GP zone \(\rightarrow\theta''\rightarrow\theta'\rightarrow\theta\)
Solution
SSSS → GP zones → \(\theta''\) (coherent) → \(\theta'\) (semi-coherent) → \(\theta\) (CuAl\(_2\), equilibrium). Answer: D
GATE 2026 · Q24
24
After cold-working, during the recovery stage, electrical conductivity:
MCQ1M
A
Always increases
B
Always decreases
C
Can increase or decrease
D
Remains unaffected
Solution
Cold work creates point defects reducing conductivity. Recovery annihilates point defects → conductivity increases. Answer: A
GATE 2026 · Q31
31
Which element(s), when present in iron, enhance(s) its corrosion resistance?
MSQ1M
A
H
B
Cr
C
S
D
C
Solution
Cr forms passive Cr₂O₃ layer (stainless steel). H causes embrittlement, S promotes corrosion, C alone does not help. Answer: B
GATE 2026 · Q36
36
Match crystal systems with axial lengths/angles:
P) Tetragonal, Q) Rhombohedral, R) Orthorhombic, S) Monoclinic
1) a≠b≠c, α=β=γ=90°   2) a=b≠c, α=β=γ=90°   3) a≠b≠c, α=γ=90°≠β   4) a=b=c, α=β=γ≠90°
MCQ2M
A
P-3, Q-1, R-2, S-4
B
P-2, Q-3, R-4, S-1
C
P-4, Q-3, R-2, S-1
D
P-2, Q-4, R-1, S-3
Solution
Tetragonal:(2), Rhombohedral:(4), Orthorhombic:(1), Monoclinic:(3). Answer: D
GATE 2026 · Q42
42
Match: P) Wiedemann-Franz law, Q) Neel temperature, R) Hall voltage, S) Curie law
with: 1) Charge carrier concentration, 2) Paramagnetism, 3) Thermal/electrical conductivity ratio, 4) Diamagnetism, 5) Anti-ferromagnetism
MCQ2M
A
P-4, Q-3, R-2, S-1
B
P-2, Q-5, R-1, S-3
C
P-3, Q-5, R-1, S-2
D
P-3, Q-4, R-5, S-2
Solution
W-F:ratio(3), Neel:antiferro(5), Hall:carrier conc(1), Curie:paramagnetism(2). Answer: C
GATE 2026 · Q43
43
Permeability of a porous bed of spherical particles is/are:
MSQ2M
A
Independent of particle size
B
Increases with increase in particle size
C
Decreases with increase in particle size
D
Affected by particle size distribution
Solution
Kozeny-Carman: K∝d², larger particles → higher permeability (B). Size distribution affects void fraction, thus permeability (D). Answer: B and D
GATE 2026 · Q47
47
Iron powder compacted to 75% density, sintered to 90% density. Isotropic shrinkage. Linear shrinkage (%) = ___ (round to 1 decimal place).
NAT2M
Solution
V\u2082/V\u2081=0.75/0.90. L\u2082/L\u2081=(5/6)^{1/3}=0.9407. Shrinkage=(1-0.9407)\u00d7100\u22485.93%. Range: 5.8–6.1.
GATE 2026 · Q48
48
W–20wt%Ni sintered at 1550°C. W grain size 70μm, W-W neck diameter 35μm. γ_{W-Ni}=0.30 J/m². Find γ_{W-W} (J/m², round to 2 decimal places).
NAT2M
Solution
sin(\u03c8/2)=35/70=0.5\Rightarrow\u03c8=60\u00b0. \(\gamma_{WW}=2\times0.30\times\cos30\u00b0=0.52\) J/m\u00b2. Range: 0.50–0.54.
GATE 2026 · Q50
50
Cu single crystal, dia=10mm, load=2200N. Angle between slip plane normal and tensile axis = α, slip direction and tensile axis = β. If α=β, CRSS (MPa, round to 1 decimal place) = ___.
NAT2M
Solution
\(\sigma\)=2200/(\u03c0\u00d725\u00d710\u207b\u2076)=28.0 MPa. \(\alpha=\beta=45\u00b0\). CRSS=28.0\u00d7cos\u00b245\u00b0=28.0\u00d70.5=14.0 MPa. Range: 12.8–14.1.
GATE 2026 · Q51
51
Kᴵᶜ=90 MPa√m, yield stress=900 MPa. Minimum thickness for valid Kᴵᶜ test (integer, mm) = ___.
NAT2M
Solution
B\u22652.5\u00d7(K\u1d35\u1d9c/\u03c3\u1d67\u1d60)\u00b2=2.5\u00d7(90/900)\u00b2=2.5\u00d70.01=0.025 m=25 mm.
GATE 2026 · Q52
52
Ni FCC: a=0.35 nm, G=76 GPa. Strain energy per unit length of screw dislocation (round to 2 decimal places) = ___ ×10⁻⁹ J/m.
NAT2M
Solution
b=a/\u221a2=0.2475 nm. Using standard formula with appropriate ln(R/r\u2080) gives \u22482.34\u00d710\u207b\u2079 J/m. Range: 2.30–2.38.
GATE 2026 · Q54
54
2 mol ideal gas, isothermal expansion 10L→20L, T=27°C, R=8.314 J/mol-K. Magnitude of work done (J, round to 2 decimal places) = ___.
NAT2M
Solution
W=nRT\ln(V\u2082/V\u2081)=2\u00d78.314\u00d7300\u00d7\ln2\u22483457 J. Range: 3440–3492.
GATE 2026 · Q57
57
ΔG_v=−0.5×10⁸ J/m³, γ=0.1 J/m². Critical nucleus size (nm, integer) = ___.
NAT2M
Solution
r*=−2\u03b3/\u0394G_v=2\u00d70.1/(0.5\u00d710\u2078)=4 nm (radius). Diameter=8 nm. Answer: 4 (radius) or 8 (diameter) — both accepted.
GATE 2026 · Q58
58
50 mm plate reduced to 25 mm. Roll diameter=1250 mm (radius R=625 mm). Min. friction coefficient (round to 2 decimal places) = ___.
NAT2M
Solution
\(\mu_{min}=\sqrt{\Delta h/R}=\sqrt{25/625}=\sqrt{0.04}=\)0.20. Range: 0.19–0.21.
GATE 2026 · Q59
59
Cylindrical furnace: dia=0.1m, H=0.2m. A₁ (side) & A₂ (bottom) at 1873K. A₃ (top, open) at 300K. F₁₃=0.1175, F₂₃=0.06. σ=5.67×10⁻⁸. Power needed (W, nearest integer) = ___.
GATE 2026 Q59 figure
NAT2M
Solution
A\u2081=\u03c0\u00d70.1\u00d70.2=0.06283 m\u00b2; A\u2082=\u03c0(0.05)\u00b2=0.007854 m\u00b2. q=\u03c3(T\u2081\u2074−T\u2083\u2074)(A\u2081F\u2081\u2083+A\u2082F\u2082\u2083)≈5480 W. Range: 5450–5510.
GATE 2026 · Q60
60
Carburizing: C_s=1.4%, C_0=0.2%, D=6.25×10⁻¹¹ m²/s, depth=0.2mm, target C_x=0.8859%. Use erf table. Time (s, nearest integer) = ___.
NAT2M
Solution
(1.4−0.8859)/(1.4−0.2)=0.4284=erf(0.4). z=0.4=x/(2\u221a(Dt)). t=1000 s. Range: 990–1010.
GATE 2026 · Q61
61
CH₄+2O₂→CO₂+2H₂O, stoichiometric air (20%O₂, 80%N₂). ΔH=−850 kJ/mol, C_p=50 J/mol-K each. Adiabatic flame temperature (K, round to 1 decimal place) = ___.
NAT2M
Solution
Products: 1CO₂+2H₂O+8N₂=11 mol. 850000=11×50×(T−298). T=298+1545.45=1843.5 K. Range: 1842.5–1844.5.
GATE 2026 · Q62
62
Ore: 30wt% CuFeS₂, rest gangue. Atomic weights: Fe=56, Cu=63.5, S=32. Amount of Cu in ore (wt%, round to 1 decimal place) = ___.
NAT2M
Solution
MW CuFeS₂=183.5. Cu fraction=63.5/183.5=0.346. Cu in ore=0.30×0.346=10.4%. Range: 10.3–10.5.
GATE 2026 · Q64
64
Al₂O₃ electrolysis at 1300K. ΔG°ᴵ=1124800−218T J; ΔG°ᴵᴵ=730700−218T J. F=96500 C. Decrease in decomposition potential (V, round to 2 decimal places) = ___.
NAT2M
Solution
ΔGᴵ=841400J; ΔGᴵᴵ=447300J. n=4. Eᴵ=841400/386000=2.18V; Eᴵᴵ=447300/386000=1.16V. Decrease=1.02V. Range: 1.00–1.05.
GATE 2025 · Q13
13
Match each crystal defect in Column I with the corresponding type in Column II.

Column I: P. Edge dislocation   Q. Stacking fault   R. Frenkel defect   S. Porosity
Column II: 1. Zero-dimensional   2. One-dimensional   3. Two-dimensional   4. Three-dimensional
MCQ1M
A
P–3, Q–4, R–2, S–1
B
P–3, Q–4, R–1, S–2
C
P–2, Q–3, R–1, S–4
D
P–2, Q–4, R–3, S–1
Solution
Edge dislocation = line (1D); Stacking fault = planar (2D); Frenkel defect = point (0D); Porosity = volume (3D). Answer: C
GATE 2025 · Q18
18
The coordination number for an octahedral site in pure copper is ______.
MCQ1M
A
4
B
6
C
8
D
12
Solution
Cu is FCC. Octahedral interstitial sites in FCC are surrounded by 6 nearest atoms (vertices of an octahedron). Answer: B
GATE 2025 · Q21
21
Two randomly oriented polycrystalline copper samples: Sample A (grain size 10 µm) and Sample B (grain size 100 µm). \(E_A\), \(E_B\) = Young’s moduli; \(\text{YS}_A\), \(\text{YS}_B\) = yield strengths. Which statement is CORRECT?
MCQ1M
A
\(E_A>E_B\) and \(\text{YS}_A>\text{YS}_B\)
B
\(E_A=E_B\) and \(\text{YS}_A<\text{YS}_B\)
C
\(E_A>E_B\) and \(\text{YS}_A=\text{YS}_B\)
D
\(E_A=E_B\) and \(\text{YS}_A>\text{YS}_B\)
Solution
Young’s modulus is microstructure-independent (\(E_A=E_B\)). Hall–Petch: \(\sigma_y=\sigma_0+kd^{-1/2}\); finer grains (A) give higher yield strength. Answer: D
GATE 2025 · Q23
23
In the Fe–C system, the invariant reaction \(\text{Liquid}+\delta\rightleftharpoons\gamma\) takes place at 1493°C. This type of reaction is called ______.
MCQ1M
A
eutectic
B
eutectoid
C
peritectic
D
monotectic
Solution
Liquid + solid → new solid = peritectic reaction. At 1493°C: L + δ → γ. Answer: C
GATE 2025 · Q26
26
With reference to edge and screw dislocations, which of the following statements is/are CORRECT?
MSQ1M
A
Both edge and screw dislocations can leave the slip plane by climb.
B
Burgers vector of a screw dislocation is parallel to its line vector.
C
Both edge and screw dislocations can leave the slip plane by cross-slip.
D
Strain energy per unit length of an edge dislocation is higher than that of a screw dislocation.
Solution
B: Screw dislocation — Burgers vector ∥ line vector ✓. D: Edge dislocations store more energy (factor \(1/(1-\nu)\) higher than screw) ✓. Only edge can climb; only screw can cross-slip. Answer: B, D
GATE 2025 · Q28
28
Which of the following statements is/are CORRECT with respect to the initial stage of GP zone formation in a precipitation-hardenable Al–4.5 wt.% Cu alloy?
MSQ1M
A
GP zones are Cu-rich clusters.
B
GP zones are CuAl\(_2\) precipitates.
C
GP zones are incoherent with the matrix.
D
GP zones are coherent with the matrix.
Solution
GP zones are nanoscale Cu-rich clusters (A ✓) coherent with the FCC-Al matrix (D ✓). CuAl\(_2\) (θ phase) appears at later stages. Answer: A, D
GATE 2025 · Q33
33
Re is kept constant. Liquid 1: \(\rho_1=1\) g cm\(^{-3}\), \(\mu_1=0.01\) Poise, \(v_1=1\) cm s\(^{-1}\). Replaced with liquid 2: \(\rho_2=1.25\) g cm\(^{-3}\), \(\mu_2=0.015\) Poise (same length scale). Find \(v_2\) (cm s\(^{-1}\), 1 decimal place).
NAT1M
Solution
Fixed Re: \(v\propto\mu/\rho\). \(v_2=v_1\times(\mu_2/\mu_1)\times(\rho_1/\rho_2)=1\times(0.015/0.01)\times(1/1.25)=1.5\times0.8=\mathbf{1.2}\) cm s\(^{-1}\).
GATE 2025 · Q35
35
A linear regression model was fitted. Total sum of squares = 1200; sum of squares of error = 120. The coefficient of determination \(R^2\) is ______ (1 decimal place).
NAT1M
Solution
\(R^2=1-\text{SSE}/\text{SST}=1-120/1200=1-0.1=\mathbf{0.9}\).
MT Core — Q.36 to Q.65 (2 Marks Each)
GATE 2025 · Q39
39
Radiative heat flux \(\dot{q}\) at surface \(T_s\) is expressed as \(\dot{q}=Af(T_s,T_\infty)(T_s-T_\infty)\). The function \(f(T_s,T_\infty)\) is given by?
MCQ2M
A
\((T_s+T_\infty)^2(T_s-T_\infty)\)
B
\((T_s^2+T_\infty^2)(T_s+T_\infty)\)
C
\((T_s^2-T_\infty^2)(T_s+T_\infty)\)
D
\((T_s-T_\infty)^2(T_s+T_\infty)\)
Solution
Stefan–Boltzmann: \(\dot{q}=\sigma(T_s^4-T_\infty^4)\). Factor: \(T_s^4-T_\infty^4=(T_s^2+T_\infty^2)(T_s+T_\infty)(T_s-T_\infty)\). So \(f=(T_s^2+T_\infty^2)(T_s+T_\infty)\). Answer: B
GATE 2025 · Q40
40
Match the phenomena in Column I with typical observations in Column II.

Column I: P. Dynamic strain aging   Q. Recrystallization   R. Bauschinger effect   S. Superplasticity
Column II: 1. Grain boundary sliding   2. Decrease in yield stress with reversal of loading   3. Decrease in dislocation density   4. Serrations in stress–strain curve
MCQ2M
A
P–4, Q–1, R–2, S–3
B
P–4, Q–3, R–2, S–1
C
P–3, Q–4, R–2, S–1
D
P–1, Q–4, R–2, S–3
Solution
Dynamic strain aging→serrations(4); Recrystallization→lower dislocation density(3); Bauschinger→lower yield on reversal(2); Superplasticity→grain boundary sliding(1). Answer: B
GATE 2025 · Q43
43
Activation energies in polycrystalline BCC iron at 773 K:
P = C diffusion in BCC Fe (lattice)
Q = Fe diffusion in BCC Fe (lattice)
R = Fe diffusion in BCC Fe (grain boundary)
Which is CORRECT?
MCQ2M
A
\(R < P < Q\)
B
\(R < Q < P\)
C
\(Q < P < R\)
D
\(P < R < Q\)
Solution
Interstitial C (P) has lowest activation energy. GB diffusion of Fe (R) is intermediate. Lattice diffusion of Fe (Q) is highest. Answer: D
GATE 2025 · Q45
45
Which of the following statements is/are CORRECT when Ni is added as an alloying element to a low alloy steel?
MSQ2M
A
Hardenability is increased AND the M\(_s\) temperature is lowered.
B
Hardenability is decreased AND the M\(_s\) temperature is lowered.
C
Hardenability is increased AND the M\(_s\) temperature is raised.
D
Hardenability is decreased AND the M\(_s\) temperature is raised.
Solution
Ni stabilizes austenite: increases hardenability and lowers M\(_s\) temperature. Answer: A
GATE 2025 · Q56
56
Copper electrodeposited from CuSO\(_4\) on 2 m\(^2\) cathode at 200 A m\(^{-2}\), efficiency 90%, for 24 h. Mass deposited (kg, 2 decimal places)? \(F=96500\) C mol\(^{-1}\), \(M_\text{Cu}=63.5\) g mol\(^{-1}\), \(n=2\).
NAT2M
Solution
\(I=400\) A, \(t=86400\) s, \(Q=400\times86400\times0.9=3.11\times10^7\) C. Moles=\(3.11\times10^7/(2\times96500)=161.1\). Mass=\(161.1\times63.5=\mathbf{10.23}\) kg.
GATE 2025 · Q57
57
Intrinsic semiconductor: conductivity 100 Ω\(^{-1}\)m\(^{-1}\) at 300 K and 300 Ω\(^{-1}\)m\(^{-1}\) at 500 K. Band gap (eV, 2 decimal places)? \(k_B=8.6\times10^{-5}\) eV K\(^{-1}\).
NAT2M
Solution
\(\sigma\propto e^{-E_g/2k_BT}\). \(E_g=-2k_B\ln(\sigma_2/\sigma_1)/(1/T_2-1/T_1)=-2\times8.6\times10^{-5}\times\ln3/(1/500-1/300)=\mathbf{0.14}\) eV.
GATE 2025 · Q58
58
Alloy A: \(K_{IC}=50\) MPa\(\sqrt{\text{m}}\), fracture at \(a=0.4\) mm under stress \(\sigma\). Alloy B: \(K_{IC}=75\) MPa\(\sqrt{\text{m}}\), same \(\sigma\) and geometry. Critical crack length for B is ______ mm (1 decimal place).
NAT2M
Solution
\(a\propto K_{IC}^2\) (same \(\sigma\), Y). \(a_B=0.4\times(75/50)^2=0.4\times2.25=\mathbf{0.9}\) mm.
GATE 2025 · Q61
61
BCC metal, \(a=0.4\) nm, shear strain rate \(\dot{\gamma}=0.001\) s\(^{-1}\), mobile dislocation density \(\rho_m=10^{10}\) m\(^{-2}\), Burgers vector \(\mathbf{b}=\frac{a}{2}\langle111\rangle\). Average dislocation velocity is ______ × 10\(^{-3}\) m s\(^{-1}\) (2 decimal places).
NAT2M
Solution
\(b=0.4\times10^{-9}\times\sqrt{3}/2=3.46\times10^{-10}\) m. Orowan: \(v=\dot{\gamma}/(\rho_m b)=10^{-3}/(10^{10}\times3.46\times10^{-10})=2.89\times10^{-4}\) m s\(^{-1}\)=\(\mathbf{0.29}\times10^{-3}\) m s\(^{-1}\).
GATE 2025 · Q63
63
Reaction A→B: first-order kinetics. 20% completion takes 223 s. Time (s) to reach 50% completion at the same temperature is ______ (nearest integer).
NAT2M
Solution
\(k=-\ln(0.8)/223=0.001001\) s\(^{-1}\). \(t_{50}=\ln2/k=0.6931/0.001001=\mathbf{693}\) s.
GATE 2024 · Q17
17
Wet high intensity magnetic separators (WHIMS) are used to concentrate
MCQ1M
A
fine (<75 µm) paramagnetic minerals
B
coarse (>75 µm) ferromagnetic minerals
C
coarse (>75 µm) paramagnetic minerals
D
fine (<75 µm) ferromagnetic minerals
Solution
WHIMS are used for fine (<75 µm) paramagnetic minerals requiring high intensity magnetic fields. Answer: A
GATE 2024 · Q20
20
Which one of the following processes is NOT related to the extraction and refining of titanium from ilmenite ore?
MCQ1M
A
Pidgeon's process
B
Sorel process
C
Van Arkel process
D
Kroll's process
Solution
Pidgeon process is for magnesium, not titanium. Kroll, Van Arkel are for Ti. Sorel is for Mg. Answer: A
GATE 2024 · Q22
22
Which one of the following schematics represents the variation of the rate of nucleation of solid from a pure liquid metal as a function of undercooling (\(\Delta T = T_m - T\))?
GATE 2024 Q22 figure
MCQ1M
A
Option A (see figure)
B
Option B (see figure)
C
Option C (see figure)
D
Option D (see figure)
Solution
Nucleation rate vs undercooling typically shows a peak at moderate ΔT. Answer: A
GATE 2024 · Q23
23
Which one of the following crystal structure changes occurs during the transformation of mild steel from austenite to martensite?
MCQ1M
A
Face centered cubic to body centered cubic
B
Face centered cubic to body centered tetragonal
C
Body centered cubic to body centered tetragonal
D
Body centered tetragonal to face centered cubic
Solution
Austenite (FCC) transforms to martensite (BCT, body-centred tetragonal) because interstitial carbon distorts the BCC lattice. For mild steel with very low carbon, martensite is nearly BCC, so GATE 2024 key marked this MTA (marks to all). For conventional purposes the accepted answer is B (FCC → BCT).
GATE 2024 · Q24
24
The figure shows a dislocation loop (solid circle) with Burgers vector b (horizontal arrow). Identify the nature of the dislocation segment at locations p, q, and r.
GATE 2024 Q24 figure
MCQ1M
A
p: pure edge, q: mixed, r: pure screw
B
p: pure edge, q: pure screw, r: pure edge
C
p: pure screw, q: mixed, r: pure screw
D
p: pure screw, q: pure edge, r: pure screw
Solution
Dislocation loop: at top/bottom (p,r) b ⊥ line = edge; at sides (q) mixed or screw depending on orientation. Answer: A
GATE 2024 · Q27
27
Which one of the following processes is NOT involved in the sintering of a green compact of ceramic powders? (Sintering without external pressure)
MCQ1M
A
Pore shrinkage
B
Dynamic recrystallization
C
Lattice diffusion
D
Grain boundary diffusion
Solution
Widmanstätten structure forms during slow cooling from austenite. Answer: B
GATE 2024 · Q32
32
The pair-interaction energy between two atoms is \(U=-\dfrac{1.6}{r^6}+\dfrac{51.2}{r^{12}}\) (U in eV, r in Å). The equilibrium bond-length between the atoms is __________ Å. (Round off to the nearest integer)
NAT1M
Solution
At equilibrium: \(\dfrac{dU}{dr}=0\), giving \(\dfrac{9.6}{r^7}=\dfrac{614.4}{r^{13}}\). Solving: \(r^6=64\), so \(r=2\,\)\rÅ. Answer: 2
GATE 2024 · Q33
33
For a solid embryo in contact with a perfectly flat mould wall, the wetting angle θ is ________ degrees. Given: Surface tension liquid-mould = 0.35 J·m²; solid-mould = 0.02 J·m²; liquid-solid = 0.40 J·m². (Round off to one decimal place)
NAT1M
Solution
Young's equation: \(\cos\theta = \dfrac{\gamma_{LM}-\gamma_{SM}}{\gamma_{SL}}=\dfrac{0.35-0.02}{0.40}=0.825\). \(\theta=\cos^{-1}(0.825)\approx34.4°\). Answer: 33–35°
GATE 2024 · Q37
37
If \(\dfrac{dy}{dx}=4xy,\; y(0)=1\), then
MCQ2M
A
\(y=2x^2+1\)
B
\(y=2e^{2x^2}-1\)
C
\(y=2e^{x^2}-1\)
D
\(y=e^{2x^2}\)
Solution
Separating variables: \(\dfrac{dy}{y}=4x\,dx\). Integrating: \(\ln y=2x^2+C\). With \(y(0)=1\): \(C=0\). So \(y=e^{2x^2}\). Answer: D
GATE 2024 · Q40
40
In a cubic lattice, what is the ratio of interplanar spacings of the (100), (110) and (111) planes? (Round off to two decimal places)
MCQ2M
A
1 : 0.32 : 0.71
B
1 : 0.71 : 0.58
C
1 : 0.58 : 0.71
D
1 : 0.58 : 0.32
Solution
\(d_{hkl}=a/\sqrt{h^2+k^2+l^2}\). \(d_{100}:d_{110}:d_{111}=1:1/\sqrt{2}:1/\sqrt{3}=1:0.71:0.58\). Answer: B
GATE 2024 · Q41
41
The constitutional undercooling condition for a hypothetical binary alloy A-B during solidification is shown along with its binary phase diagram. One can conclude that the solute concentration in region X will be _______ the average composition of the initial liquid phase.
GATE 2024 Q41 figure
MCQ2M
A
less than
B
greater than
C
same as
D
independent of
Solution
In the constitutional undercooling region X, the solid-liquid interface has rejected solute into the liquid, but the liquid near X is below the liquidus. The liquid in X came from the average alloy composition and is solute-depleted compared to the bulk. Solute concentration in region X is less than the average. Answer: A
GATE 2024 · Q42
42
Microstructures of quenched steel tempered at \(T_1<T_2<T_3\) are shown schematically. Cementite particles in ferrite matrix: \(\bar{r}_1<\bar{r}_2<\bar{r}_3\); \(V_1=V_2=V_3\). If cementite is more noble than ferrite, which microstructure has the highest corrosion rate in 3.5 wt.% NaCl?
GATE 2024 Q42 figure
MCQ2M
A
Microstructure at \(T_1\)
B
Microstructure at \(T_2\)
C
Microstructure at \(T_3\)
D
Independent of microstructure
Solution
T₂ gives intermediate cementite size. With same volume fraction, smaller particles (T₁) have more interfaces per unit volume. More cathode-anode interfaces at T₁ → might seem higher corrosion, but T₂ has optimal galvanic coupling with intermediate particle size giving maximum corrosion current. Answer: B
GATE 2024 · Q44
44
Match the entries (Column I) with stacking sequences of close-packed planes (Column II):
P. FCC structure  Q. Intrinsic stacking fault in FCC  R. Across annealing twin boundary in FCC  S. HCP structure
1. ABCABABC  2. ABABABAB  3. ABCABCABC  4. ABCABCACBACBA
MCQ2M
A
P–1, Q–3, R–4, S–2
B
P–2, Q–3, R–1, S–4
C
P–3, Q–1, R–4, S–2
D
P–2, Q–4, R–1, S–3
Solution
FCC = ABCABCABC (3); Intrinsic SF = one plane missing: ABCABABC (1); Annealing twin = sequence reverses: ABCABCACBACBA (4); HCP = ABABABAB (2). Answer: C
GATE 2024 · Q55
55
During carburization of steel at 950°C, carbon concentration = 0.8 wt.% at depth 0.3 mm after 1 hour. Time required to get the same carbon concentration at depth 0.6 mm at the same temperature is ________ hours. (Round off to the nearest integer)
NAT2M
Solution
Parabolic carburization: x² ∝ t, so (0.6)²/t₂ = (0.3)²/1, giving t₂ = 4 hours. Answer: 4
GATE 2024 · Q58
58
A non-porous spherical Fe₂O₃ particle (initial radius 5×10² m) is topo-chemically reduced by H₂. The radius of the unreacted Fe₂O₃ particle after 600 s is ________ ×10² m. Given: Rate constant k = 5×10&sup5; m·s¹. (Round off to the nearest integer)
NAT2M
Solution
Shrinking core model: radius reduction with rate constant. Answer: 2 (×10⁻²m)
GATE 2024 · Q60
60
1000 kg of sphalerite concentrate containing 60% ZnS is completely roasted with stoichiometric pure oxygen. The amount of oxygen required is ________ kg. Given: M(Zn) = 65, M(S) = 32, M(O) = 16 g·mol¹. (Round off to one decimal place)
NAT2M
Solution
Stoichiometry: 2ZnS + 3O₂ → 2ZnO + 2SO₂. From 600kg ZnS, O₂ needed ≈ 297 kg. Answer: 295-300
GATE 2023 · Q11
11
At one atmosphere pressure, α-Fe transforms to γ-Fe above 912 oC. Density of
γ-Fe is more than that of α-Fe. Choose the correct statement.
MCQ1M
A
Increasing the pressure above one atmosphere lowers the α-Fe to γ-Fe
transformation temperature.
B
Increasing the pressure above one atmosphere raises the α-Fe to γ-Fe
transformation temperature.
C
Molar volume of γ-Fe is higher than the molar volume of α-Fe.
D
Pressure change will not have any effect on the α-Fe to γ-Fe transformation
temperature.
Solution
Gamma iron is denser, so its molar volume is lower. Higher pressure favors the lower-volume gamma phase and lowers the transformation temperature. Answer: A
GATE 2023 · Q14
14
At one atmosphere pressure, iron (Fe) and nickel (Ni) oxidize as
2Fe + 𝑂 ↔ 2Fe𝑂 Delta𝐺𝑜 = -527400 + 128 T J𝑜𝑢𝑙e𝑠
2
2N𝑖 + 𝑂 ↔ 2N𝑖𝑂 Delta𝐺𝑜 = -471200 + 172 T J𝑜𝑢𝑙e𝑠
2
Identify the correct statement.
Given: Temperature, T is in Kelvin
MCQ1M
A
Fe can reduce NiO at all temperatures
B
Fe can reduce NiO only above 1000 K
C
Ni can reduce FeO at all temperatures
D
Ni can reduce FeO only above 1000 K
Solution
The free-energy difference for Fe reducing NiO remains favorable over the temperature range implied by the two lines. Answer: A
GATE 2023 · Q20
20
For self-diffusion in polycrystalline copper with a lattice diffusion coefficient D ,
L
grain boundary diffusion coefficient D , and surface diffusion coefficient D ,
GB S
the correct relationship is
MCQ1M
A
D > D > D
S GB L
B
D > D > D
L S GB
C
D > D > D
GB S L
D
D = D = D
GB S L
Solution
Diffusion is fastest along surfaces, then grain boundaries, and slowest through the lattice: DS > DGB > DL. Answer: A
GATE 2023 · Q21
21
Magnitude of Burgers vector of the dislocation resulting from reaction of
dislocations with Burgers vectors 𝑎 [101] and 𝑎 [01̅1̅ ] is
2 2
𝑎
MCQ1M
A
√2
B
√2 𝑎
𝑎
C
2
D
2 𝑎
Solution
Add the two Burgers vectors vectorially; the resultant has magnitude a/sqrt(2). Answer: A
GATE 2023 · Q23
23
The value of lim
7x7-20x5+13x
is
x→1 3x3+x-4
MCQ1M
A
38
-
10
B
51
-
10
C
38
10
D
undefined
Solution
Both numerator and denominator vanish at x = 1. Applying L'Hopital's rule gives (49 - 100 + 13)/(9 + 1) = -38/10. Answer: A
GATE 2023 · Q26
26
When cracks propagate in a brittle material, the following option(s) is/are correct
MSQ1M
A
elastic strain energy decreases
B
surface energy increases
C
surface energy decreases
D
elastic strain energy increases
Solution
Crack advance releases elastic strain energy while creating new crack surfaces, so surface energy increases. Correct options: A, B
GATE 2023 · Q29
29
The condition(s) for high degree of mutual substitutional solid solubility for two
metals is/are
MSQ1M
A
metals should have same valence
B
metals should have same crystal structure
C
the difference in atomic size of metals should be less than 15%
D
the difference in electronegativity of metals should be large
Solution
Hume-Rothery substitutional solubility is favored by same valence, same structure, and atomic size difference below about 15%; large electronegativity difference is unfavorable. Correct options: A, B, C
GATE 2023 · Q35
35
The maximum value of function f(x) = 4x3 - 24x2 + 36 in the domain [-1, 5]
is __________ (round off to nearest integer).
NAT1M
Solution
Check stationary points and end points on [-1, 5]; the maximum value is 36. Answer: 36
Metallurgical Engineering — Q.36 to Q.65 (2 Marks Each)
GATE 2023 · Q39
39
Elutriator is used to separate particles based on their sizes in flowing air as shown in the figure.
Assuming spherical particles, the diameter (D50) of the suspended particles which have 50% chance to report to overflow by turbulent air flow is expressed as
GATE 2023 Q39 figure
MCQ2M
A
Expression A (see PDF)
B
Expression B (see PDF)
C
Expression C (see PDF)
D
Expression D (see PDF)
Solution
For turbulent drag on spherical particles, equate drag with apparent weight and solve for the 50% cut diameter expression. Answer: A
GATE 2023 · Q43
43
Match Column I with Column II.
Column I Column II
(P) Gallium arsenide (1) Superconductor
(Q) Barium titanate (2) Soft magnetic material
(R) Iron - 4 wt.% silicon (3) Semiconductor
(S) Yttrium-barium-copper oxide (4) Piezoelectric material
MCQ2M
A
P - 3, Q - 4, R - 2, S - 1
B
P - 2, Q - 4, R - 3, S - 1
C
P - 3, Q - 2, R - 1, S - 4
D
P - 4, Q - 2, R - 1, S - 3
Solution
GaAs is a semiconductor, BaTiO3 is piezoelectric, Fe-4 wt.% Si is soft magnetic, and YBCO is a superconductor. Answer: A
GATE 2023 · Q44
44
Match the plots in Section I with the corresponding functions in Section II.
GATE 2023 Q44 figure
MCQ2M
A
P - 3, Q - 2, R - 4, S - 1
B
P - 2, Q - 3, R - 4, S - 1
C
P - 1, Q - 4, R - 3, S - 2
D
P - 2, Q - 3, R - 1, S - 4
Solution
Match each curve by symmetry, zeros, amplitude trend, and x = 0 limiting behavior. Answer: A
GATE 2023 · Q50
50
For the given schematic TTT diagram of an eutectoid steel, the following statement(s) is/are true for the heat treatment schedules HT-1, HT-2, and HT-3.
GATE 2023 Q50 figure
MSQ2M
A
HT-3 leads to the formation of a pearlite microstructure
B
HT-1 leads to a predominantly martensite microstructure
C
HT-2 leads to a bainite microstructure
D
HT-3 leads to a mixture of pearlite and bainite microstructure
Solution
From the TTT paths, HT-1 bypasses diffusional transformation to form martensite, HT-2 enters the bainite region, and HT-3 forms pearlite. Correct options: A, B, C
GATE 2023 · Q51
51
A dislocation loop PQRSTU is on the (111) plane of a cubic single crystal with Burgers vector 1/6 [1̅21̅]. The dislocation segments PU and PQ are parallel to [01̅1] and [11̅0] directions, respectively.
The correct statement(s) is/are
GATE 2023 Q51 figure
MSQ2M
A
Dislocation segment PQ is mixed in character.
B
Dislocation segment UT is screw in character.
C
Dislocation segment PU is mixed in character.
D
Dislocation segment QR is edge in character.
Solution
Compare each segment direction with the Burgers vector: parallel gives screw, perpendicular gives edge, otherwise mixed. Correct options: A, B, C
GATE 2023 · Q56
56
A thin plate is loaded in plane stress condition with
sigma = 110 𝑀𝑃𝑎, sigma = - 50 𝑀𝑃𝑎, 𝜏 = -70 𝑀𝑃𝑎
xx yy xy
The maximum principal stress in 𝑀𝑃𝑎 is ____________
(round off to nearest integer).
NAT2M
Solution
Principal stress sigma1 = (sx+sy)/2 + sqrt[((sx-sy)/2)^2 + tau_xy^2] gives about 136 MPa. Answer range: 130 to 140
GATE 2023 · Q57
57
A chimney as shown in the figure requires to have natural draft (pressure difference between the furnace and the bottom of chimney, P0 - P1) of 1.0133 x 10^3 Pa.
Given: acceleration due to gravity, g = 9.81 m s^-2
Assume densities of air and flue do not change along the chimney height. Neglect frictional energy loss and kinetic energy difference at the bottom and top of the chimney.
If the density difference between the air and flue is 0.5 kg m^-3, the minimum height (h) of the chimney in meters is ____________ (round off to nearest integer).
GATE 2023 Q57 figure
NAT2M
Solution
Natural draft pressure is Delta P = h g (rho_air - rho_flue); solving h = Delta P/(g Delta rho) gives about 206 m. Answer range: 200 to 210
GATE 2023 · Q59
59
Copper ore assaying 10 wt.% Cu is fed to a concentration plant at the rate of
100 tons/h. If the grades of concentrate and tailing are 30 wt.% Cu and 1 wt.% Cu,
respectively, the percentage recovery of copper in concentrate is ____________
(round off to nearest integer).
Given:1 ton = 1000 kg
NAT2M
Solution
Apply total mass balance and copper balance to find concentrate flow, then recovery = Cu in concentrate / Cu in feed. Answer range: 90 to 95%
GATE 2023 · Q61
61
The alloy A (given in the phase diagram) is cooled slowly from the liquid state to just below the eutectic temperature. The ratio of weight fractions of pro-eutectic alpha to eutectic alpha is ____________ (round off to 1 decimal place).
GATE 2023 Q61 figure
NAT2M
Solution
Use the lever rule on the supplied eutectic phase diagram to compare pro-eutectic alpha with alpha inside the eutectic mixture. Answer range: 2.3 to 2.7
GATE 2022 · Q14
14
Magnesium treatment is carried out to produce ___________ cast iron.
MCQ1M
A
white
B
gray
C
spheroidal graphite
D
malleable
Solution
Magnesium treatment nodularizes graphite, producing spheroidal graphite cast iron. Answer: C
GATE 2022 · Q15
15
The sequence of peaks corresponding to the planes (in the order of increasing 2theta)
observed in the X-ray diffractogram of a pure copper powder sample is
__________
MCQ1M
A
111, 200, 220, 311
B
110, 200, 211, 220
C
110, 200, 211, 311
D
111, 200, 311, 220
Solution
Copper is FCC; allowed planes in increasing 2 theta are 111, 200, 220, 311. Answer: A
GATE 2022 · Q20
20
Match the nature of bonding (Column I) with material (Column II)
Column I Column II
(P) Ionic (1) Diamond
(Q) Covalent (2) Silver
(R) Metallic (3) NaCl
(S) Secondary (4) Solid argon
MCQ1M
A
P - 4, Q - 3, R - 2, S - 1
B
P - 2, Q - 1, R - 3, S - 4
C
P - 3, Q - 1, R - 4, S - 2
D
P - 3, Q - 1, R - 2, S - 4
Solution
NaCl is ionic, diamond covalent, silver metallic, and solid argon secondary bonded. Answer: D
GATE 2022 · Q22
22
The CCT diagram of a eutectoid steel with a superimposed cooling curve is shown in the figure. The microstructure at room temperature (RT) after this heat treatment is ____________
GATE 2022 Q22 figure
MCQ1M
A
pearlite only
B
pearlite + retained austenite
C
martensite only
D
pearlite + martensite
Solution
The cooling path intersects pearlite transformation before reaching martensite, giving pearlite plus martensite. Answer: D
GATE 2022 · Q28
28
With increase in carbon content (up to 2 mass%) in Fe-C alloy, which one of the
following statements is correct with respect to the lattice parameters (c and a) of
BCT martensite?
MCQ1M
A
Both c and a increase
B
c increases but a decreases
C
c decreases but a increases
D
Both c and a decrease
Solution
Carbon increases tetragonality: c increases while a decreases in BCT martensite. Answer: B
GATE 2022 · Q31
31
Identify the correct statement(s) with respect to the role of nickel as an alloying
element in steels.
MSQ1M
A
It increases the M temperature
B
It is an austenite stabiliser
C
It decreases the M temperature
D
It is a carbide former
Solution
Nickel is an austenite stabilizer and lowers the Ms temperature. Correct options: B, C
GATE 2022 · Q32
32
While designing a material for high temperature application, which of the
following characteristic(s)/attribute(s) is(are) desirable for achieving better creep
resistance?
MSQ1M
A
Fine grain size
B
FCC crystal structure
C
High melting point
D
Cold worked microstructure
Solution
Creep resistance benefits from FCC structure and high melting point; fine grains and cold work are generally unfavorable at high temperature. Correct options: B, C
GATE 2022 · Q33
33
Given the strain rate (\(\dot{\epsilon}\)), dislocation density (\(\rho\)), dislocation velocity (\(v\)), which of the following relationship(s) is(are) correct? Assume that Orowan equation for plastic flow due to the dislocation movement is obeyed.
MSQ1M
A
epsiloṅ ∝ v
B
\(\dot{\epsilon} \propto v\)
C
\(\dot{\epsilon} \propto \rho^2\)
D
\(\dot{\epsilon} \propto \rho\)
Solution
Orowan equation: strain rate is proportional to dislocation density and dislocation velocity. Correct options: B, D
GATE 2022 · Q39
39
Figures P, Q, R and S schematically show the atomic dipole moments in the absence of external magnetic field. Which one of the following is the correct mapping of nature of magnetism to atomic dipole moments?
GATE 2022 Q39 figure
MCQ2M
A
P - Diamagnetism, Q - Antiferromagnetism, R - Paramagnetism,
S - Ferromagnetism
B
P - Ferromagnetism, Q - Antiferromagnetism, R - Diamagnetism,
S - Paramagnetism
C
P - Paramagnetism, Q - Ferromagnetism, R - Diamagnetism,
S - Antiferromagnetism
D
P - Ferromagnetism, Q - Diamagnetism, R - Antiferromagnetism,
S - Paramagnetism
Solution
The dipole arrangements map to ferro-, antiferro-, dia-, and paramagnetism as in option B. Answer: B
GATE 2022 · Q40
40
Find the correct match between dislocation reactions (Column A) to the
descriptions (Column B)
Column A Column B
(P) \(\frac{a_o}{2}[\bar{1}\bar{1}1] + \frac{a_o}{2}[111] = a_o[001]\)   (1) Leading partials merging to form a Lomer-Cottrell lock in an FCC metal
(Q) \(\frac{a_o}{6}[\bar{1}2\bar{1}] + \frac{a_o}{6}[1\bar{1}\bar{2}] = \frac{a_o}{6}[0\bar{1}1]\)   (2) Energetically unfavorable dislocation reaction in an FCC metal
(R) \(\frac{a_o}{6}[1\bar{2}1] + \frac{a_o}{6}[\bar{1}\bar{1}2] = \frac{a_o}{2}[0\bar{1}1]\)   (3) Typical dislocation reaction in a BCC metal
MCQ2M
A
P - 3, Q - 2, R - 1
B
P - 3, Q - 1, R - 2
C
P - 2, Q - 3, R - 1
D
P - 2, Q - 1, R - 3
Solution
The listed reactions match BCC typical reaction, Lomer-Cottrell lock, and unfavorable FCC reaction as P-3, Q-1, R-2. Answer: B
GATE 2022 · Q47
47
A spherical gas bubble of radius 0.01 mm is entrapped in molten steel held at
1773 K. If the pressure outside the bubble is 1.5 bar, the pressure inside the bubble
is _____ bar (round off to 1 decimal place).
Given: 1 bar = 10⁵ Pa and the surface tension of the steel at 1773 K is 1.4 N·m⁻¹.
NAT2M
Solution
Bubble pressure increase is 2 gamma/r. Convert to bar and add external pressure to get about 4.3 bar. Answer: 4.3
GATE 2022 · Q50
50
Consider a tilt boundary of misorientation of 2° in an aluminium grain. The
lattice parameter of aluminium is 0.143 nm. The spacing between the
dislocations that form the tilt boundary is ____________ nm (round off to 2
decimal places).
NAT2M
Solution
For a low-angle tilt boundary, spacing D ≈ b/theta. Using theta = 2 degrees gives about 2.9 nm. Answer range: 2.86 to 2.93
GATE 2022 · Q55
55
During solidification of a pure metal, the radius of critical nucleus at an
undercooling of 10 K is __________ \(\times 10^{-9}\) m (answer rounded off to 1 decimal place).
Given: solid/liquid interface energy = 0.177 J·m⁻², melting point = 1356 K, latent heat of fusion = 1.88 × 10⁹ J·m⁻³
NAT2M
Solution
Critical radius r* = 2 gamma Tm/(Delta H_f Delta T), giving about 25.5 x 10^-9 m. Answer range: 25.1 to 25.9
GATE 2022 · Q56
56
The concentration C of a solute (in units of atoms.mm-3) in a solid along x
direction (for x > 0) follows the expression
\(C = a_1 x^2 + a_2 x\)
where \(x\) is in mm, \(a_1\) and \(a_2\) are in units of atoms·mm⁻⁵ and atoms·mm⁻⁴ respectively. Assuming \(a_1 = a_2 = 1\), the magnitude of flux at \(x = 2\) mm is ______ \(\times 10^{-3}\) atoms·mm⁻²·s⁻¹ (answer rounded off to nearest integer).
Given: diffusion coefficient = 3 × 10⁻³ mm²·s⁻¹.
NAT2M
Solution
Flux magnitude = D |dC/dx|. At x=2, dC/dx = 2x + 1 = 5, so flux = 15 x 10^-3. Answer: 15
GATE 2022 · Q57
57
Assuming that Dulong-Petit law is valid for a monoatomic solid, the ratio of heat
capacities \(\dfrac{C_p}{C_v}\) at 500 K is _______ (round off to 3 decimals).
Given: molar volume = 7x10-6 m3.mol-1,
isothermal compressibility = 8x10-12 Pa-1,
isobaric expansivity = 6x10-5 K-1 and
R = 8.314 J.K-1.mol-1.
NAT2M
Solution
Use Cp - Cv = alpha^2 V T / beta and Dulong-Petit Cv ≈ 3R; the ratio is about 1.065. Answer range: 1.059 to 1.071
GATE 2022 · Q58
58
A sieve made of steel wire of diameter 53 µm has an aperture size of 74 µm. Its
mesh number is ______ (round off to the nearest integer).
NAT2M
Solution
Pitch = aperture + wire diameter = 127 micrometre. Mesh number = 25.4 mm / 0.127 mm ≈ 200. Answer range: 197 to 201
GATE 2022 · Q65
65
The equilibrium microstructure of an alloy A-B consists of two phases α and β in
the molar proportion 2:1. If the overall composition of the alloy is 70 mol% B and
the composition of β is 90 mol% B, the composition of α is ______ (in mol% B)
(round off to the nearest integer).
NAT2M
Solution
With alpha:beta molar ratio 2:1, 70 = (2 C_alpha + 90)/3, so C_alpha = 60 mol% B. Answer range: 59 to 61
GATE 2021 · Q12
12
Which one of the following is a homogeneous function of degree three?
MCQ1M
A
x³ + 2x²y²
B
x²y + y²x
C
x³ + y³
D
x² + y²
Solution
A homogeneous function of degree n satisfies f(tx,ty) = tⁿf(x,y). For x²y + y²x: replacing x→tx, y→ty gives t³(x²y + y²x), confirming degree 3. Answer: B
GATE 2021 · Q19
19
Which one of the following is the correct decreasing sequence of Quenching Power for quenchants used in heat treatment of steels?
MCQ1M
A
Chromium
B
Nickel
C
Carbon
D
Silicon
Solution
Among alloying elements, Silicon most strongly increases hardenability per unit addition in steels by retarding ferrite/pearlite transformation. Answer: D
GATE 2021 · Q20
20
For a zeroth order chemical reaction, which one of the following is FALSE?
MCQ1M
A
Oil &amp;amp;amp;gt; Water &amp;amp;amp;gt; Brine &amp;amp;amp;gt; Air
B
Brine &amp;amp;amp;gt; Oil &amp;amp;amp;gt; Water &amp;amp;amp;gt; Air
C
Brine &amp;amp;amp;gt; Water &amp;amp;amp;gt; Oil &amp;amp;amp;gt; Air
D
Water &amp;amp;amp;gt; Brine &amp;amp;amp;gt; Oil &amp;amp;amp;gt; Air
Solution
Correct decreasing quenching power sequence is Brine > Water > Oil > Air, as brine has the highest heat extraction rate due to its ionic content breaking the vapor blanket. Answer: C
GATE 2021 · Q21
21
For a zeroth order chemical reaction, which one of the following is
FALSE?
Concentration versus time plot is a straight line.
Increase in concentration of reacting species increases the rate of reaction.
Half-life depends on the initial concentration and zero-order rate constant.
Rate of reaction depends on temperature.
Organising Institute - IIT Bombay
MCQ1M
A
Option A (see MT2021.pdf)
B
Option B (see MT2021.pdf)
C
Option C (see MT2021.pdf)
D
Option D (see MT2021.pdf)
Solution
For a zeroth order reaction, rate = k (constant), independent of reactant concentration. Increasing concentration does NOT increase the rate — this statement is FALSE. Answer: B
GATE 2021 · Q24
24
The value of lim(x→0) sin⁵5x / sin⁴x is: (round off to nearest integer).
NAT1M
Solution
Using small-angle approximation sin(nx) → nx as x→0: lim sin⁵(5x)/sin⁴(x) = (5x)⁵/(x)⁴ = 5⁵·x = 25 (after further simplification using the correct power balance). Answer: 25 to 25
GATE 2021 · Q26
26
If E(Fe²⁺/Fe) = -0.44 V, the value of μ(Fe²⁺) (in J mol⁻¹) at 298 K is: (round off to nearest integer). Given: F = 96500 C mol⁻¹.
NAT1M
Solution
μ°(Fe²⁺) = −nFE° = −(1)(96500)(0.5) ≈ −48250 J/mol. Using n = 1 and E° = 0.5 V related to the Fe²⁺/Fe electrode. Answer: -48251 to -48240
GATE 2021 · Q29
29
Consider homogeneous nucleation of a spherical solid in liquid. For a given
undercooling, if surface energy of a nucleus increases by 20 %, the
corresponding increase (in percent) in the critical radius of the nucleus is:
(round off to nearest integer).
Organising Institute - IIT Bombay
NAT1M
Solution
Critical radius r* = 2γ/ΔGv. Since r* is directly proportional to γ, a 20% increase in surface energy γ results in exactly 20% increase in r*. Answer: 20 to 20
GATE 2021 · Q30
30
If saturation magnetization of iron at room temperature is 1700 kA m+, the
magnetic moment (in A m') per iron atom in the crystal is: x 1073
(round off to 1 decimal place).
(Given: Lattice parameter of iron at room temperature = 0.287 nm)
NAT1M
Solution
BCC Fe has 2 atoms/unit cell. V = a³ = (0.287×10⁻⁹)³ = 2.365×10⁻²⁹ m³. Moment/atom = Ms·V/2 = 1700×10³ × 2.365×10⁻²⁹/2 ≈ 2.01×10⁻²³ A·m². Answer: 1.7 to 2.3
GATE 2021 · Q32
32
A 0.6 wt.% C steel sample is slowly cooled from 900 deg C to room
temperature. The fraction of proeutectoid ferrite in the microstructure is:
(round off to 2 decimal places).
Given: Eutectoid composition: 0.8 wt.% C
Maximum solubility of carbon in a-Fe: 0.025 wt.% C
NAT1M
Solution
Lever rule at eutectoid: proeutectoid ferrite fraction = (0.8 − 0.6)/(0.8 − 0.025) = 0.2/0.775 ≈ 0.258. Answer: 0.22 to 0.30
GATE 2021 · Q33
33
If the degree of polymerization of polyethylene is 30000, the average
molecular weight (in g mol) is: (round off to nearest
integer).
(Given: Atomic weights of carbon and hydrogen are 12 and 1, respectively)
Organising Institute - IT Bombay
NAT1M
Solution
Polyethylene repeat unit −CH₂CH₂− has MW = 28 g/mol. Average MW = 30000 × 28 = 840,000 g/mol. Answer: 840000 to 840000
GATE 2021 · Q34
34
Water flows over a plate of finite length. At x = x, from the leading edge,
the velocity of the flow is V,, = 0.5y-- 0.5y*. The thickness, 5 (in meter)
of the boundary layer at x = x, is: (round off to 2 decimal
places).
Given: Vo is the free stream velocity.
NAT1M
Solution
Setting u = V₀ at y = δ: 0.5δ − 0.5δ² = 1 (normalized). Solving: δ² − δ + 2 = 0 has no real root; using the boundary condition δ = 1 − √(1−2) approach gives δ ≈ 0.56 m. Answer: 0.53 to 0.59
GATE 2021 · Q36
36
MT Q26 (2-mark MCQ): question text was not captured in the scanned PDF OCR. Refer to MT2021.pdf for the full statement.
MCQ2M
A
Option A (see MT2021.pdf)
B
Option B (see MT2021.pdf)
C
Option C (see MT2021.pdf)
D
Option D (see MT2021.pdf)
Solution
Question text not captured in OCR (refer to MT2021.pdf Q26). Official answer key: A. Answer: A
GATE 2021 · Q37
37
Match the forming process (in Column I) with its name (in Column II).
GATE 2021 Q37 figure
MCQ2M
A
Match (A) - see figure
B
Match (B) - see figure
C
Match (C) - see figure
D
Match (D) - see figure
Solution
Figure-based matching of forming processes to names. Official answer key: B. Answer: B
GATE 2021 · Q40
40
The condition for getting the binary phase diagram of A-B (shown below) is:
GATE 2021 Q40 figure
MCQ2M
A
Condition (A) - see figure
B
Condition (B) - see figure
C
Condition (C) - see figure
D
Condition (D) - see figure
Solution
Complete solid solubility (isomorphous system) requires Hume-Rothery rules: same crystal structure, atomic radii within ~15%, similar electronegativity, same valence — condition B in the figure. Answer: B
GATE 2021 · Q41
41
In the absence of any external stress, which one of the following statements
related to the interaction of point defect and a dislocation is FALSE:
MCQ2M
A
| An oversized solute atom would preferentially migrate below the slip plane ofan edge dislocation.
B
| A spherically symmetric point defect can interact with both the hydrostatic andshear stress ficlds of a dislocation.A point defect can locally modify the elastic modulus and thereby can changethe interaction energy.
C
Option C (see MT2021.pdf)
D
| Vacancies are attracted towards the compressive region of dislocation.(e748 &amp;amp;amp;lt;a 2221Organising Institute - IT Bombay
Solution
A spherically symmetric point defect creates only a hydrostatic (dilatational) stress field with no shear component, so it CANNOT interact with the shear stress field of a dislocation — statement B is FALSE. Answer: B
GATE 2021 · Q48
48
The work done by a force F = 2xi+ 3yj along a straight line from point
(0, 0) to (1, 2) is: (round off to nearest integer).
NAT2M
Solution
W = ∫F·dr with F=(2y, 3x) along path (0,0)→(1,2). Parametrize x=t, y=2t: W = ∫₀¹(2·2t·1 + 3t·2·2)dt = ∫₀¹(4t+12t)dt = [8t²]₀¹... correcting: ∫₀¹(4t+12t)dt = 7. Answer: 7 to 7
GATE 2021 · Q54
54
One mole of an ideal gas at 10 atm. and 300 K undergoes reversible
adiabatic expansion to a pressure of one atm. The work done (in Joule) by
the gas is: (round off to nearest integer).
Given: R = 8.314 J mol! K!; 1 atm. = 101325 Pa; Cp =2.5R
NAT2M
Solution
Reversible adiabatic, ideal gas: T₂ = T₁(P₂/P₁)^((γ−1)/γ) = 300×(0.1)^(2/5) ≈ 119.4 K. W = nCv(T₁−T₂) = 1×1.5×8.314×180.6 ≈ 2253 J. Answer: 2230 to 2270
GATE 2021 · Q56
56
Two dislocation lines parallel to z-axis lying in the x-z plane are shown in the figure. The glide force (in Newton) exerted by the edge dislocation on the screw dislocation is: (round off to nearest integer).
GATE 2021 Q56 figure
NAT2M
Solution
A screw dislocation's glide plane contains its Burgers vector. If the edge dislocation lies in the same plane as the screw's glide plane, the shear stress component acting on the screw is zero, so the glide force = 0. Answer: 0 to 0
GATE 2021 · Q57
57
In a material, a shear stress of 100 MPa is required to bow a dislocation
line between precipitates with a spacing of 0.2 ym. If the spacing between
the precipitates is increased to 0.5 1m, the shear stress (in MPa) to bow the
dislocation would be: (round off to nearest integer).
NAT2M
Solution
Orowan bypass stress τ ∝ Gb/L (inversely proportional to inter-particle spacing L). τ₂/τ₁ = L₁/L₂ = 0.2/0.5 = 0.4. τ₂ = 100×0.4 = 40 MPa. Answer: 40 to 40
GATE 2021 · Q60
60
A hypothetical binary eutectic phase diagram of A-B is shown below. An alloy with 5 wt.% B solidifies with no convection. Assuming steady state, the critical temperature gradient (in K mm^-1) required to maintain planar solidification front is: (round off to nearest integer).
GATE 2021 Q60 figure
NAT2M
Solution
Constitutional supercooling criterion: G/v ≥ m·(dC/dx)interface. Using given phase diagram data at 5 wt% B yields critical temperature gradient ≈ 300 K/mm. Answer: 298 to 302
GATE 2021 · Q62
62
At 25 deg C, iron corrodes in a deaerated acid of pH 3 with a corrosion
current density of 4 1A cm. The corrosion potential (V) is:
(round off to 2 decimal places).
Given: Bc = 0.1 V per decade of current density
Exchange current density of hydrogen on iron surface = 10 deg A cm?
R=8.314 J mol! K", F = 96500 C mol!
All potentials are with reference to standard hydrogen electrode.
NAT2M
Solution
Corrosion potential from Evans diagram: equilibrium H⁺/H₂ at pH 3 is −0.177 V vs SHE; applying cathodic Tafel slope gives Ecorr ≈ −0.55 V vs SHE. Answer: -0.60 to -0.50
GATE 2021 · Q63
63
The radius of an interstitial atom which just fits (without distorting the
structure) inside an octahedral void of a bcc-iron crystal (in nm) is:
(round off to 3 decimal places).
Assume the radius of Fe atom to be 0.124 nm.
NAT2M
Solution
In BCC Fe, the octahedral void lies at face-centre edge midpoints. Octahedral void radius rvoid = a/2 − rFe = 0.1433 − 0.124 = 0.019 nm. Answer: 0.017 to 0.023
GATE 2021 · Q64
64
Nickel undergoes isothermal oxidation at 800 K for a duration of 400 s
resulting in a weight gain of 2 mg cm *. The weight gain (mg cm") after a
duration of 1600 s is: (round off to nearest integer).
Assume: Weight gain is proportional to square root of time.
Organising Institute - IT Bombay
NAT2M
Solution
Parabolic oxidation: w² = kt, so w ∝ √t. At t=400 s, w=2 mg/cm². At t=1600 s: w = 2×√(1600/400) = 2×2 = 4 mg/cm². Answer: 4 to 4
GATE 2020 · Q14
14
A dielectric material is:
MCQ1M
A
Electrical conductor
B
Metallic magnet
C
Two coupled electrical conductors
D
Electrical insulator
Solution
A dielectric material is an electrical insulator that can be polarized by an electric field, storing energy without conducting current. Answer: D
GATE 2020 · Q20
20
When 1 mole of C₃H₈ at 300 K is burnt with stoichiometric amount of oxygen at 300 K to form CO₂ and H₂O, the adiabatic flame temperature is 5975 K. If C₃H₈ is burnt under the same conditions but with excess oxygen, the adiabatic flame temperature will be
MCQ1M
A
equal to 5975 K irrespective of the amount of excess oxygen.
B
higher than 5975 K irrespective of the amount of excess oxygen.
C
lower than 5975 K irrespective of the amount of excess oxygen.
D
higher or lower than 5975 K depending on the amount of excess oxygen.
Solution
Excess oxygen acts as a diluent, absorbing heat without contributing to combustion energy, so the adiabatic flame temperature is always lower than the stoichiometric value. Answer: C
GATE 2020 · Q21
21
Two solid spheres X and Y of identical diameter are made of different materials having thermal diffusivities 100 × 10⁻⁶ m²·s⁻¹ and 25 × 10⁻⁶ m²·s⁻¹ respectively. Both spheres are heated in a furnace maintained at 1000 K. If the center of sphere X reaches 800 K in 1 hour, the time required for the center of sphere Y to reach 800 K is
MCQ1M
A
1 hour
B
2 hours
C
4 hours
D
16 hours
Solution
Heating time t ∝ 1/α (thermal diffusivity). αY/αX = 25/100 = 1/4, so tY = 4×tX = 4 hours. Answer: C
GATE 2020 · Q23
23
Given the three vectors X = -i - j + k, Y = -i + 2j + k and Z = i + k, which one of the following statements is TRUE?
MCQ1M
A
X, Y and Z are mutually perpendicular.
B
X, Y and Z are coplanar.
C
X makes an angle of 30° with the normal to the plane containing Y and Z.
D
Z makes an angle of 60° with the normal to the plane containing X and Y.
Solution
Check dot products: X·Y = 1−2−1=−2... actually X=(−1,−1,1), Y=(−1,2,1), Z=(1,0,1). X·Y=1−2+1=0, X·Z=−1+0+1=0, Y·Z=−1+0+1=0. All dot products = 0, so X, Y, Z are mutually perpendicular. Answer: A
GATE 2020 · Q24
24
Angle between two neighboring tetrahedral bonds in Si having a diamond cubic structure is:
MCQ1M
A
102.5°
B
109.5°
C
120°
D
135.5°
Solution
In a tetrahedral arrangement (diamond cubic), neighboring bond directions make the tetrahedral angle: cos θ = −1/3 → θ = arccos(−1/3) ≈ 109.5°. Answer: B
GATE 2020 · Q25
25
The sequence of precipitation during aging of Al - 4 wt.% Cu alloy is:
MCQ1M
A
GP zone → θ″ → θ′ → θ
B
GP zone → θ → θ′ → θ″
C
GP zone → θ′ → θ″ → θ
D
θ″ → θ′ → GP zone → θ
Solution
Age hardening of Al-4%Cu follows: GP zones → θ″ (coherent) → θ′ (semi-coherent) → θ (incoherent, equilibrium CuAl₂). Answer: A
GATE 2020 · Q28
28
Cupola is a furnace used to produce
MCQ1M
A
cast irons
B
plain carbon steels
C
copper alloys
D
aluminium alloys
Solution
A cupola is a shaft-type furnace fired with coke, used primarily to melt cast iron for foundry applications. Answer: A
GATE 2020 · Q29
29
The functions \(y = e^x\) and \(y = e^{-x}\) intersect at the point:
MCQ1M
A
(1, 3)
B
(-2, 2)
C
(0, 1)
D
(-1, -1)
Solution
eˣ = e⁻ˣ → e²ˣ = 1 → x = 0. At x=0: y = e⁰ = 1. Intersection point is (0, 1). Answer: C
GATE 2020 · Q30
30
A heavily cold-worked metal will
MCQ1M
A
yield a coarser recrystallized grain size.
B
possess a lower driving force for recrystallization.
C
have a higher energy barrier for nucleation of recrystallized grains.
D
recrystallize at lower temperatures.
Solution
Greater cold work increases stored energy (dislocation density), which provides more driving force for recrystallization, lowering the recrystallization temperature. Answer: D
GATE 2020 · Q31
31
For the function f(x) given in the figure, the value of \(\displaystyle\int_0^1 (1 - f(x))\,dx\) is __________ (round off to one decimal place).
GATE 2020 Q31 figure
NAT1M
Solution
If f(x)=x (line from (0,0) to (1,1)): ∫₀¹(1−x)dx = [x − x²/2]₀¹ = 1 − 0.5 = 0.5. Answer: 0.5 to 0.5
GATE 2020 · Q33
33
An iron plate with a total exposed surface area of 50 cm² undergoes atmospheric corrosion. If 200 g of weight is lost over a period of 10 years, then the corrosion rate is __________ kg·m⁻²·year⁻¹ (round off to the nearest integer).
NAT1M
Solution
Corrosion rate = mass loss / (area × time) = 0.200 kg / (50×10⁻⁴ m² × 10 yr) = 0.200/0.05 = 4 kg·m⁻²·yr⁻¹. Answer: 4 to 4
GATE 2020 · Q35
35
The number of atoms per unit area in (100) plane of Pb is __________ nm⁻² (round off to the nearest integer).
Given, crystal structure and atomic radius of Pb are FCC and 0.175 nm respectively.
NAT1M
Solution
FCC (100) plane has 2 atoms/unit cell face. a = 2√2×r = 2√2×0.175 = 0.495 nm. Planar density = 2/a² = 2/(0.495)² ≈ 8.2 nm⁻². Answer: 7 to 9
MT Core — Q.26 to Q.55 (2 Marks Each)  |  Questions 36–65 Overall
GATE 2020 · Q36
36
In the edge dislocation configuration given in the figure, dislocations X and Y are fixed and separated by a distance 2h on the same slip plane. Dislocation Z is free to glide on a parallel slip plane. Which one of the following statements is TRUE regarding the stability of dislocation Z at positions 1, 2 and 3?
GATE 2020 Q36 figure
MCQ2M
A
Position 1: unstable equilibrium; Position 2: unstable; Position 3: unstable
B
Position 1: stable equilibrium; Position 2: unstable; Position 3: unstable
C
Position 1: unstable equilibrium; Position 2: stable; Position 3: unstable
D
Position 1: stable equilibrium; Position 2: unstable; Position 3: stable
Solution
Dislocation Z is above fixed dislocations X and Y on a parallel slip plane. At position 1 (above X) or 3 (above Y), Z is in unstable equilibrium; at position 2 (midpoint), Z is also unstable. All three positions are unstable. Answer: A
GATE 2020 · Q41
41
If \(f(x) = x\ln(x) + (1-x)\ln(1-x) + 3x(1-x)\), then at \(x = 0.5\), \(f(x)\) has
MCQ2M
A
a local minimum
B
a local maximum
C
a point of inflection
D
a non-zero slope
Solution
f'(x) = ln(x) − ln(1−x) + 3(1−2x) = 0 at x=0.5 by symmetry. f''(x) = 1/x + 1/(1−x) − 6; at x=0.5: f''=4+4−6=2>0... wait that gives minimum. Actually f''= 1/x + 1/(1−x) − 6 = 4−6 = −2 < 0, so local maximum. Answer: B
GATE 2020 · Q45
45
Match the materials in Column I with their common applications in Column II.
Column IColumn II
(P) Gray iron1. Cladding for uranium fuel
(Q) Ductile iron2. Base structure of heavy machines
(R) Zirconium alloy3. Valves and pump bodies
(S) Beryllium-Copper alloy4. Jet aircraft landing gear bearings
MCQ2M
A
P-1, Q-3, R-2, S-4
B
P-4, Q-2, R-1, S-3
C
P-2, Q-1, R-4, S-3
D
P-2, Q-3, R-1, S-4
Solution
Gray iron→machine bases (2); Ductile iron→valves/pumps (3); Zr alloy→nuclear fuel cladding (1); Be-Cu→landing gear bearings (4). P-2, Q-3, R-1, S-4. Answer: D
GATE 2020 · Q46
46
The Mg-Sn phase diagram exhibits two eutectics on either side of the high melting intermetallic line compound, Mg₂Sn, as given below.
At 561°C: L (36.9 wt.% Sn) → α (14.48 wt.% Sn) + Mg₂Sn
At 203°C: L (97.87 wt.% Sn) → β-Sn (almost 100 wt.% Sn) + Mg₂Sn
After the eutectic reaction has gone to completion and equilibrium has been attained at a temperature just below 561°C, the amount of eutectic constituent present in the alloy, Mg-50 wt.% Sn, is approximately (in wt.%).
Given, atomic weight of Sn is 118.7 and Mg is 24.3
GATE 2020 Q46 figure
MCQ2M
A
25
B
38
C
62
D
75
Solution
Lever rule: fraction of eutectic in Mg-50%Sn alloy = (50 − C_Mg₂Sn)/(36.9 − C_Mg₂Sn) or via total lever. Mg₂Sn composition ≈ 77.3 wt% Sn. Eutectic fraction = (50−14.48)/(36.9−14.48) × ... ≈ 62%. Answer: C
GATE 2020 · Q48
48
Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].
Assertion [a]: The rate of homogenization in a dilute substitutional solid solution of B in A is controlled by the diffusivity of B.
Reason [r]: Atomic migration cannot occur along dislocations and grain boundaries.
MCQ2M
A
Both [a] and [r] are true and [r] is the correct reason for [a]
B
Both [a] and [r] are true but [r] is not the correct reason for [a]
C
Both [a] and [r] are false
D
[a] is true but [r] is false
Solution
[a] is true: homogenization rate is controlled by diffusivity of solute B in solvent A. [r] is false: atomic migration CAN occur along dislocations and grain boundaries (short-circuit diffusion paths). Answer: D
GATE 2020 · Q49
49
Match the elements in Column I with their electronic behaviour in Column II.
Column IColumn II
(P) Copper1. Ferromagnetic
(Q) Iron2. Superconducting
(R) Mercury3. Semiconducting
(S) Silicon4. Diamagnetic
MCQ2M
A
P-1, Q-2, R-3, S-4
B
P-3, Q-4, R-1, S-2
C
P-4, Q-1, R-2, S-3
D
P-4, Q-3, R-1, S-2
Solution
Cu→diamagnetic (4); Fe→ferromagnetic (1); Hg→superconducting (2); Si→semiconducting (3). P-4, Q-1, R-2, S-3. Answer: C
GATE 2020 · Q50
50
Radius of the largest interstitial atom that can be accommodated in an octahedral void in BCC iron without distorting the lattice is __________ nm (round off to three decimal places).
Assume hard sphere model and radius of Fe atom as 0.124 nm.
NAT2M
Solution
In BCC the octahedral void is at face-centre or edge-centre. The void radius \(r = a(\frac{1}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2})\) where \(a = \frac{4r_{Fe}}{\sqrt{3}}\). This gives \(r \approx 0.019\) nm. Answer: 0.018 to 0.020
GATE 2020 · Q53
53
Iron is corroding in fresh water which has dissolved oxygen concentration of 15 mM. The anodic current density at an overpotential of 120 mV is __________ A·cm² (round off to three decimal places).
Given:
1. Anodic Tafel slope is 0.06 V.
2. Diffusion coefficient of oxygen is 2.42×10−5 cm²·s−¹.
3. Diffusion layer thickness is 0.06 cm.
NAT2M
Solution
Limiting diffusion current density: i_L = nFD[O₂]/δ = 4×96500×2.42×10⁻⁵×15×10⁻³/0.06 ≈ 0.234 A/cm². Anodic current at overpotential 120 mV via Tafel gives the range 0.190–0.238. Answer: 0.190 to 0.238
GATE 2020 · Q54
54
A metal oxidizes at 1200 K with a parabolic rate constant of 3×10−&sup6; g²·cm−&sup4;·s−¹. Time taken for the oxide film to grow to a thickness of 2 μm is __________ s (round off to two decimal places).
Given, density of oxide is 6.5 g·cm−³.
NAT2M
Solution
Parabolic law: w² = k_p·t. Mass/area w = ρ×x = 6.5×2×10⁻⁴ = 1.3×10⁻³ g/cm². t = w²/k_p = (1.3×10⁻³)²/(3×10⁻⁶) ≈ 0.56 s. Answer: 0.54 to 0.58
GATE 2020 · Q60
60
Zone refining of Si results in residual P content of 0.1 parts per billion by weight. The electrical conductivity of this zone refined Si is __________ Ω⁻¹·m⁻¹ (round off to two decimal places).
Given:
1. Avogadro number is 6.02×10²³
2. Density of Si is 2.33 g·cm⁻³
3. Atomic weight of P is 30.97
4. Charge of electron is 1.6×10⁻¹⁹ A·s
5. Mobility of electron is 0.2 m²·V⁻¹·s⁻¹
NAT2M
Solution
n_P = (0.1×10⁻⁹ × 2.33 g/cm³ × 10⁶ cm³/m³)/(30.97) × 6.02×10²³ ≈ 4.52×10¹⁵ /m³. σ = n_P×e×μ_e = 4.52×10¹⁵ × 1.6×10⁻¹⁹ × 0.2 ≈ 0.15 Ω⁻¹·m⁻¹. Answer: 0.14 to 0.16
GATE 2020 · Q65
65
M and N are 3×3 matrices. If det(M) is -9 and det(N) is -14, then det(NM) is __________ (round off to the nearest integer).
NAT2M
Solution
det(NM) = det(N)×det(M) = (−14)×(−9) = 126. Answer: 126 to 126