GATE 2026 — Metallurgical Engineering (MT)

GATE 2026 Question Paper

All 65 questions · Answers & solutions · Organized by institute: IIT Guwahati

65
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General Aptitude — Q.1 to Q.5 (1 Mark Each)
1
‘The team ______ more than 300 runs in 20 overs __________ rains. However, some players needed to improve their batting skills.’
Choose the option with the correct sequence of words to fill the blanks.
MCQ1M
A
score; despite
B
scoring; instead of
C
scored; despite
D
scoring; in spite of
Solution
Past tense needed: “scored”; contrast conjunction: “despite rains.” Answer: C
2
If a positive real \(x\) satisfies \(\log_2 x + \log_{\sqrt{2}} x = 48\), then the value of \(x\) is
MCQ1M
A
\(2^{16}\)
B
\(4^{16}\)
C
\(2^{14}\)
D
\(4^{14}\)
Solution
Let \(t=\log_2 x\). Then \(\log_{\sqrt2}x=2t\). So \(3t=48\Rightarrow t=16\Rightarrow x=2^{16}\). Answer: A
3
The next figure (indicated by ‘?’) in the sequence is
GATE 2026 Q3 figure
MCQ1M
A
Option A (see figure)
B
Option B (see figure)
C
Option C (see figure)
D
Option D (see figure)
Solution
Tracking the circle and triangle positions through the grid sequence, the next figure matches Option A. Answer: A
4
‘All the mangoes in the basket are good.’
If the above statement is false, then which one of the following statements is necessarily true?
MCQ1M
A
All the mangoes in the basket are not good.
B
No mango in the basket is good.
C
In the basket, some of the mangoes are good and some are not good.
D
There exists at least one mango in the basket that is not good.
Solution
The logical negation of “All P are Q” is “There exists at least one P that is not Q.” Answer: D
5
Consider the following statements about four numbers:
(S1) The average of the four numbers is 25
(S2) Each number is at most 40
(S3) Each number is at least 20
Choose the option that is necessarily correct.
MCQ1M
A
(S1) and (S2) together imply (S3)
B
(S2) and (S3) together imply (S1)
C
(S1) and (S3) together imply (S2)
D
(S1) implies (S3)
Solution
If avg=25 (sum=100) and each ≥20, remaining sum after 4×20=20 spread over 4 gives each ≤40. So (S1)+(S3)↠(S2). Answer: C
General Aptitude — Q.6 to Q.10 (2 Marks Each)
6
Choose the correct sequence: “People are crowding around ___ pit into which ___ elephant has fallen... bewildered ___ miserable... look up ___ a vast, curiosity-stricken crowd.”
MCQ2M
A
an; a; at; and
B
a; an; and; at
C
and; a; an; at
D
at; a; an; and
Solution
a pit / an elephant / and (conjunction) / at (preposition). Answer: B
7
Five products P,Q,R,S,T. 250 items sold, avg price Rs.60. S=2T, R=3T, Q=4T. Prices: P=100, Q=50, R=40, S=60, T=60. Quantity of P sold?
MCQ2M
A
40
B
50
C
60
D
70
Solution
Let T=t. P+10t=250; 100P+500t=15000. Solving: t=20, P=50. Answer: B
8
String P length \(l\) (diameter), K = semicircle. Both shortened by \(x\); K becomes full circle with P as diameter. Value of \(x/l\):
GATE 2026 Q8 figure
MCQ2M
A
\(\pi\)
B
\(\dfrac{\pi-1}{2\pi}\)
C
\(\dfrac{\pi}{2(\pi-1)}\)
D
\(\dfrac{\pi}{\pi-1}\)
Solution
K=\(\pi l/2\) (semicircle). After: \(\pi l/2 - x = \pi(l-x)\). Solving: \(x/l = \pi/[2(\pi-1)]\). Answer: C
9
Meritorius, brother, son, daughter: seating in 2×2 grid. (i) daughter & brother same column; (ii) son diagonally across sibling of worst orator; (iii) best & worst same row. Who is best orator?
GATE 2026 Q9 figure
MCQ2M
A
Meritorius
B
Meritorius’ brother
C
Meritorius’ son
D
Meritorius’ daughter
Solution
Working through the constraints: Meritorius’ brother is the best orator. Answer: B
10
Which pattern (P, Q, R, S) generates the given figure?
GATE 2026 Q10 figure
MCQ2M
A
P
B
Q
C
R
D
S
Solution
Pattern Q tiles to generate the given figure. Answer: B
Metallurgical Engineering — Q.11 to Q.35 (1 Mark Each)
11
Given \(f(t)=e^{-at}\). The Laplace transform \(\mathcal{L}[f(t)]=F(s)\). Which is correct?
MCQ1M
A
\(F(s)=\frac{1}{s-a}\)
B
\(F(s)=\frac{s}{s^2-a^2}\)
C
\(F(s)=\frac{a}{s^2-a^2}\)
D
\(F(s)=\frac{1}{s+a}\)
Solution
Standard result: \(\mathcal{L}[e^{-at}]=\frac{1}{s+a}\). Answer: D
12
Correct pair of eigenvectors for \(\begin{bmatrix}1&2\\2&4\end{bmatrix}\)?
MCQ1M
A
\(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\end{bmatrix}\)
B
\(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\)
C
\(\begin{bmatrix}-1\\-2\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\)
D
\(\begin{bmatrix}-1\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\-1\end{bmatrix}\)
Solution
Eigenvalues: \(\lambda=0,5\). For \(\lambda=0\): \(v=[-2,1]^T\). For \(\lambda=5\): \(v=[1,2]^T\). Answer: A
13
Given \(w=f(ax+by)\), value of \(\left(b\frac{\partial w}{\partial x}-a\frac{\partial w}{\partial y}\right)\) is:
MCQ1M
A
\(-a\)
B
\(b\)
C
\(b-a\)
D
0
Solution
\(\frac{\partial w}{\partial x}=af'(u)\), \(\frac{\partial w}{\partial y}=bf'(u)\). Expression \(=baf'-abf'=0\). Answer: D
14
Which fusion welding technique results in least Heat-Affected Zone (HAZ)?
MCQ1M
A
Submerged Arc Welding (SAW)
B
Tungsten Inert Gas Welding (TIG)
C
Electron Beam Welding (EBW)
D
Oxy-Acetylene Welding (OAW)
Solution
EBW has extremely concentrated energy, very low heat input per unit length → smallest HAZ. Answer: C
15
During metallography, Nital is most commonly used for etching ____________.
MCQ1M
A
Bronze
B
Brass
C
Mild Steel
D
Aluminium
Solution
Nital (HNO₃ in ethanol) is the standard etchant for ferrous metals including mild steel. Answer: C
16
In a face-centered cubic metal, Shockley partial is:
MCQ1M
A
Perfect and mobile dislocation
B
Perfect and immobile dislocation
C
Imperfect and immobile dislocation
D
Imperfect and mobile dislocation
Solution
Shockley partial: b = a/6⟨112⟩ — NOT a lattice vector (imperfect), glissile on {111} plane (mobile). Answer: D
17
Deformation mechanism map is used for determining which property?
MCQ1M
A
Fatigue strength
B
Creep rate
C
Tensile strength
D
Impact toughness
Solution
Ashby deformation mechanism maps show dominant creep mechanisms and strain rates vs. T/T_m. Answer: B
18
Which dislocation dissociation reaction is feasible in FCC metals?
MCQ1M
A
\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{2}1]+\frac{a}{6}[\bar{1}\bar{1}2]\)
B
\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[112]+\frac{a}{6}[21\bar{1}]\)
C
\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{1}2]+\frac{a}{6}[\bar{1}\bar{2}\bar{1}]\)
D
\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{2}1]+\frac{a}{6}[2\bar{1}\bar{1}]\)
Solution
Check A: vectors add to \(\frac{a}{6}[0\bar{3}3]=\frac{a}{2}[0\bar{1}1]\) \(\checkmark\). Energy decreases: \(a^2/2 > a^2/6+a^2/6\) \(\checkmark\). Answer: A
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
20
Which one is NOT a state function?
MCQ1M
A
Enthalpy
B
Entropy
C
Work
D
Internal Energy
Solution
Work and heat are path functions. H, S, U are state functions. Answer: C
21
For a regular solution (\(\Delta H_{mix}\) = enthalpy of mixing, \(\Delta S_{mix}\) = entropy of mixing):
MCQ1M
A
Both \(\Delta H_{mix}\) and \(\Delta S_{mix}\) are finite
B
\(\Delta H_{mix}\) is zero, \(\Delta S_{mix}\) is finite
C
\(\Delta H_{mix}\) is finite, \(\Delta S_{mix}\) is zero
D
Both are zero
Solution
Regular solution: \(\Delta H_{mix}=\Omega x_Ax_B\neq0\) (finite), \(\Delta S_{mix}=-R\sum x_i\ln x_i\) (ideal/random mixing, finite). Answer: A
22
During spinodal decomposition, uphill diffusion occurs:
MCQ1M
A
From higher to lower concentration and from higher to lower chemical potential
B
From higher to lower concentration and from lower to higher chemical potential
C
From lower to higher concentration and from lower to higher chemical potential
D
From lower to higher concentration and from higher to lower chemical potential
Solution
Inside spinodal: higher concentration → lower chemical potential; atoms still flow high→low μ but that means low→high concentration. Answer: D
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
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
25
Red mud is generated in the production of:
MCQ1M
A
Aluminium
B
Iron
C
Titanium
D
Copper
Solution
Red mud (bauxite residue) is alkaline waste from the Bayer process for alumina extraction. Answer: A
26
Residual stress can be determined by which technique?
MCQ1M
A
X-ray Diffraction (XRD)
B
Tensile Testing
C
Thermo-Gravimetric Analysis (TGA)
D
Optical Microscopy
Solution
XRD measures d-spacing shifts to calculate residual stress via sin²ψ method. Answer: A
27
Sherwood number for convective mass transfer (laminar flow over flat plate) is a function of:
MCQ1M
A
Schmidt number and Reynolds number
B
Weber number and Reynolds number
C
Schmidt number and Weber number
D
Weber number and Prandtl number
Solution
Analogy: Nu=f(Re,Pr) → Sh=f(Re,Sc). Sc=ν/D replaces Pr=ν/α. Answer: A
28
Convective heat transfer coefficient is NOT dependent on:
MCQ1M
A
Solid-fluid interfacial area
B
Thermal conductivity of solid
C
Roughness of solid surface
D
Viscosity of fluid
Solution
h depends on fluid properties and flow/geometry — NOT on the thermal conductivity of the solid. Answer: B
29
Ergun equation is NOT applied in which unit operation?
MCQ1M
A
Blast Furnace
B
Sintering
C
Roasting
D
LD Converter
Solution
Ergun applies to packed beds. LD Converter uses liquid steel bath with O₂ lancing — not a packed bed. Answer: D
30
Here “A” is Helmholtz free energy and “G” is Gibbs free energy. Choose correct option(s).
MSQ1M
A
“A” provides criterion for equilibrium at constant T and P
B
“G” provides criterion for equilibrium at constant T and P
C
“A” provides criterion for equilibrium at constant T and V
D
“G” provides criterion for equilibrium at constant T and V
Solution
G (Gibbs): equilibrium at constant T,P. A (Helmholtz): equilibrium at constant T,V. Answer: B and C
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
32
Value of scalar triple product \(\vec{a}\cdot(\vec{b}\times\vec{c})\) (answer in integer).
\(\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}\), \(\vec{b}=\hat{i}+2\hat{j}-3\hat{k}\), \(\vec{c}=3\hat{i}+4\hat{j}-\hat{k}\)
NAT1M
Solution
Determinant = 2(2\(\cdot\)(-1)-(-3)\(\cdot\)4)+3((-1)-(-9))+4(4-6)=2(10)+3(8)+4(-2)=20+24-8=36
33
20 thermometers, 3 defective. Draw 2 without replacement. Probability (%) that none is defective (round to 2 decimal places) is _____ %.
NAT1M
Solution
P = (17/20)(16/19) = 272/380 = 71.58%. Answer range: 71.00 to 72.00
34
For binary A-B phase diagram at constant pressure, degree of freedom at point X (in two-phase L+S region) is (integer).
GATE 2026 Q34 figure
NAT1M
Solution
Gibbs phase rule at constant P: F=C-P+1=2-2+1=1.
35
Two parallel plates 2 mm apart, lower plate moves at 4 m/s, shear force 5 N/m². Viscosity (round to 2 decimal places) = _____ ×10⁻³ N·s/m².
NAT1M
Solution
\(\mu=\tau/(dv/dy)=5/2000=2.50\times10^{-3}\) N\u00b7s/m\u00b2. Answer range: 2.40 to 2.60
Metallurgical Engineering — Q.36 to Q.65 (2 Marks Each)
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
37
Match: P) Paris Law, Q) Schmid Factor, R) Larson-Miller Parameter, S) Portevin-Le Chatelier Effect
with 1) Creep, 2) Fatigue, 3) Dynamic Strain Aging, 4) Critical Resolved Shear Stress
MCQ2M
A
P-3, Q-1, R-2, S-4
B
P-2, Q-4, R-1, S-3
C
P-2, Q-4, R-3, S-1
D
P-1, Q-3, R-2, S-4
Solution
Paris Law:Fatigue(2), Schmid:CRSS(4), Larson-Miller:Creep(1), PLC:DSA(3). Answer: B
38
Match defects: P) Edge Cracking, Q) Flashline Cracking, R) Chevron Cracking, S) Cracked Core
with processes: 1) Casting, 2) Forging, 3) Rolling, 4) Extrusion
MCQ2M
A
P-3, Q-4, R-2, S-1
B
P-4, Q-3, R-1, S-2
C
P-4, Q-2, R-3, S-1
D
P-3, Q-2, R-4, S-1
Solution
Edge cracking:Rolling(3), Flashline:Forging(2), Chevron:Extrusion(4), Cracked core:Casting(1). Answer: D
39
Match NDT: P) Internal flaws in railroad wheel, Q) In-service crack monitoring, R) Inclusion in mild steel, S) Surface crack in Al alloy
with 1) Ultrasonic, 2) Radiography, 3) Dye Penetrant, 4) Acoustic Emission
MCQ2M
A
P-1, Q-4, R-2, S-3
B
P-3, Q-2, R-4, S-1
C
P-3, Q-2, R-1, S-4
D
P-1, Q-3, R-4, S-2
Solution
Internal:UT(1), In-service monitoring:AE(4), Inclusion:Radiography(2), Surface:Dye penetrant(3). Answer: A
40
Steady-state laminar flow: \(\eta\frac{1}{r}\frac{d}{dr}\!\left(r\frac{dv_z}{dr}\right)-\frac{dP}{dz}=0\)
GATE 2026 Q40 figureWhich statement is NOT correct?
MCQ2M
A
The fluid is Newtonian
B
Shear stress is maximum at the center (r=0)
C
Radial velocity is zero
D
There is no variation of v₂ in z-direction
Solution
For Hagen-Poiseuille flow, shear stress τ=η(dv/dr) is ZERO at r=0 and MAXIMUM at the wall. Answer: B
41
Match: P) COREX, Q) Bayer, R) Matte Smelting, S) MIDREX
with products: 1) DRI, 2) Pig Iron, 3) Alumina, 4) Copper
MCQ2M
A
P-4, Q-3, R-2, S-1
B
P-2, Q-3, R-4, S-1
C
P-1, Q-3, R-4, S-2
D
P-2, Q-4, R-3, S-1
Solution
COREX:Pig iron(2), Bayer:Alumina(3), Matte:Copper(4), MIDREX:DRI(1). Answer: B
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
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
44
PDF: \(f(x)=0.5\) for \(0
NAT2M
Solution
\(E[X^2]=\int_0^2 0.5x^2\,dx=4/3\). Var\(=4/3-1=1/3\approx\)0.33. Range: 0.32 to 0.34.
45
Trapezoidal rule, n=3: \(\int_0^{0.3}e^{-x^2}dx\) = ___ (round to 2 decimal places).
NAT2M
Solution
h=0.1; f(0)=1, f(0.1)=0.990, f(0.2)=0.9608, f(0.3)=0.9139. Trap=(0.05)(1+1.98+1.9216+0.9139)=0.291. Range: 0.27–0.31.
46
ODE \(10x^2y''-20xy'+22.4y=0\), solution \(y=c_1x^{m_1}+c_2x^{m_2}\). Value of \(m_1+m_2\) (integer) = ___.
NAT2M
Solution
Euler-Cauchy: \(10m(m-1)-20m+22.4=0\Rightarrow m^2-3m+2.24=0\). By Vi\u00e8ta: \(m_1+m_2=3\). Answer: 3
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.
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.
49
BCC metal, XRD: \(\lambda=0.154\) nm, 2\(\theta\)=60\u00b0 for {200} plane. Atomic radius (nm, round to 3 decimal places) = ___.
NAT2M
Solution
Bragg: d\(_{200}\)=\(\lambda/(2\sin30\u00b0)\)=0.154 nm. a=2d=0.308 nm. BCC: r=\(\frac{\sqrt3}{4}a=0.133\) nm. Range: 0.132–0.134.
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.
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.
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.
53
20g Au (MW=197) + 20g Ag (MW=108) ideal mixing. R=8.314 J/mol-K. Total entropy of mixing (J/K, round to 2 decimal places) = ___.
NAT2M
Solution
n\u2090\u1d64=0.1015, n\u2090\u1d58=0.1852, x\u2090\u1d64=0.354, x\u2090\u1d58=0.646. \(\Delta S_{mix}=-nR\sum x_i\ln x_i\approx\)1.55 J/K. Range: 1.50–1.60.
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.
55
C\(_p\)=20+5\u00d710\u207b\u00b3T J/mol-K. 2 mol heated 300K\u2192600K. Change in enthalpy (integer, J) = ___.
NAT2M
Solution
\(\Delta H=2\int_{300}^{600}(20+5\times10^{-3}T)dT=2\times6675=\)13350 J. Range: 13340–13360.
56
Ellingham: Reaction I (solid): ΔG°=(−338900−15.2T lnT+247T) J. Reaction II (liquid): ΔG°=(−390800−15.2T lnT+285.3T) J. Melting point (K, round to 1 decimal place) = ___.
NAT2M
Solution
At T_m, both equal: −338900+247T=−390800+285.3T ↠ 51900=38.3T ↠ T=1354.8 K. Range: 1353.9–1356.3.
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.
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.
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.
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.
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.
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.
63
Blast furnace hot metal: 4%C, 1.5%Si, rest Fe. Ore: 85%Fe₂O₃, 15% gangue. 2% Fe lost in slag. Ore needed per 1000 kg hot metal (kg, round to 1 decimal place) = ___.
NAT2M
Solution
Fe in hot metal=945 kg. Required Fe input=945/0.98=963.3 kg. Ore×0.85×(112/160)=963.3 ↠ ore=1618.9 kg. Range: 1615.5–1625.5.
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.
65
Scalar field \(\phi(x,y,z)=x^2-yz\). Magnitude of \(\nabla\phi\) at P(3,4,1) (round to 2 decimal places) = ___.
NAT2M
Solution
\(\nabla\phi=(2x,-z,-y)\). At P(3,4,1): (6,-1,-4). |\(\nabla\phi\)|=\(\sqrt{36+1+16}=\sqrt{53}\approx\)7.28. Range: 7.08–7.48.