All 65 questions · Answers & solutions · Organized by institute: IIT Guwahati
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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
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 \(x/l\):
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?
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?
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?
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
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). \(\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}\), \(\vec{b}=\hat{i}+2\hat{j}-3\hat{k}\), \(\vec{c}=3\hat{i}+4\hat{j}-\hat{k}\)
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) = ___.
\(\Delta H=2\int_{300}^{600}(20+5\times10^{-3}T)dT=2\times6675=\)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.
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) = ___.
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) = ___.