AskMeMetallurgy- Gate MT 2017 Question paper with solution

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GATE MT 2017 paper with solution
Asked: July 6, 2020 In: Powder MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q47.
GATE MT 2017 paper with solution gate mt previous year questions with solution liquid phase sintering
Best AnswerAskmemetallurgy Official ✓ Silver
o cial account of Askmemetallurgy
Added an answer on July 7, 2020 at 6:31 pm
Correct answer is option C. Given that γs-s = 0.52 J/m2 and γs-l= 0.30 J/m2 2γs-l
cos(Θ/2)=γs-s On simpli cation, (Θ/2)= 29.92 sin(Θ/2)=( x/2)/(D/2) x=20 μm
1 Answer 86 Views Answer
W-Ni compact is prepared by liquid phase sintering is 1500°C. If the size of the tungsten grains is
40µm and the interfacial tungsten- tungsten and tungsten-nickel energies are 0.52 and 0.30 Jm-2
respectively, the predicted average neck size (in µm) of sintered tungsten grain is: (Melting points
of tungsten and nickel are 3410°C and 1453°C, respectively) (A) 10 (B) 15 (C) 20 (D) 25
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Asked: July 6, 2020 In: Manufacturing TechnologyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q46.
cold rolling GATE MT 2017 paper with solution gate mt previous year questions with solution
A 250mm thick slab of nickel alloy is subjected to cold rolling using a roll of diameter 450mm. If the
angle of bite during rolling is 10°, the maximum possible reduction (in mm) during rolling is ____
(answer up to two decimal places)
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Correct answer is 6.97 we know that, μ=tan Θ given Θ= 100 Maximum possible
reduction= Δh=μ2R= (tan 10)2×225= 6.97 mm
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Asked: July 5, 2020 In: Foundry TechnologyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q45.
Chvorinov’s law GATE MT 2017 paper with solution gate mt previous year questions with solution
Correct answer is 11.78 Using Chvorinov’s rule, tf= k(V/A)2 For casting shape,
Volume= a3 = (5)3 Area= 6a2= 6(5)2 tc= (53)/(6(5)2) = 1.6min For riser shape,
(cylinder) d/k= 0.5 ⇒ h/d= 2; h=4r Volume= πr2h= πr2(4r)= 4πr3 Area= 2πr(r+h)=
2πr(4r+r)= 2 πr(5r)=10 πr2 tr= 3.2min On dividing both the soli Read More
Total time for solidification of a cubic casting of dimensions 5.0 cm×5.0 cm ×5.0cm is 1.6 min. A
cylindrical riser with diameter to height ratio 0.5 is required so that the time for solidification of riser
is 3.2 min. Applying Chvorinov’s law, the height of the riser (in cm) is _____ (answer up to two
decimal places) (Assume that exponent (n) in Chvorinov’s equation is 2)
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Asked: July 5, 2020 In: Mechanical MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q44.
GATE MT 2017 paper with solution gate mt previous year questions with solution
ratio of resilience to toughness
Correct answer is 0.125 Given that, σ=450×106 MPa and E= 4.5 ×109 MPa
Engineering strain in elastic limit= σ/E= 450×106 / 4.5 ×109 = 0.1 Plastic strain=
0.45- 0.1= 0.35 Resilience/toughness= Area of region of A/(Area of region of A+ Area
of region of B) = 1/2×0.1×450×106 /( 1/2(0.1×450×106 + 0.35×4 Read More
2 Answers 116 Views Answer
A perfectly elastic-plastic material has a yield stress of 450 MPa and fractures at a strain 0f 0.45.
The ratio of resilience to toughness for this material is _____ (answer up to three decimal places)
Given the young’s modulus E=4.5 GPa
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PollAsked: July 5, 2020
In: Physical Metallurgy & Heat treatment
Askmemetallurgy Official ✓ Silver
GATE MT 2017 Q43. A single crystal of a FCC metal is subjected to
a sufficiently large tensile stress along the [110] direction to
activate some of the slip systems. Which one of the following slip
systems will be activated:
❓ Participate in Poll, Choose Your Answer.
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(A) a/2[1-10][111]

GATE MT 2017 paper with solution gate mt previous year questions with solution slip systems
Correct answer is option B. Schmid's law(m)= cosΦcosλ Slip system is activated
when schmid factor is not equal to zero. (A) m= (1×1 + 1×1 + 0×0)/(√2×√2) × (1×1 +
1×1 + 1×1)/(√2×√3)= 0 (B) m= (1×0 + 1×1 + 0×1)/(√2×√2) × (1×1 + 1×1 + 0×1-)/(√2×√3)=
1/√6 (C) m= (1×0 + 1×1 + 1×0)/(√2×√2) × (1×1 + 1×1- + Read More
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(B) a/2[011][111-]
(C) a/2[011][11-1]
(D) a/2 [110][1-11-]
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Asked: July 5, 2020
GATE MT 2017 Q41.
GATE MT 2017 paper with solution gate mt previous year questions with solution heat treatment
During heat treatment of a cold worked metal, recrystallization is 20% complete after 100s. The
transformation (in %) in 400s is _____(answer up to two decimal places) (Assume Avrami
constant, n=2)
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Correct answer is 97.18 We know that, Avrami equation, γ= 1- exp(-atn) As per
question, γ=0.20, t= 100 sec γ=f, t= 400 sec Two cases arises in this question. Case 1:
0.20 = 1- exp(-a 100n) ln 0.8= -a( 100)2 Case 2: ln(1-f)= -a( 400)2 On dividing both the
cases: ln0.8/ln(1-f)=(100/400)2 f=0.9718 tr Read More
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GATE MT 2017 Q42. At low temperatures, two parallel edge
dislocations lying on parallel planes are shown in different
configurations below.
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(A) P-3,Q-2,R-4,S-1
(B) P-4,R-1,R-3,S-2
(C) P-1,Q-3,R-2,S-4

two parallel edge dislocations
Correct answer is option A.
Submit Results
(D) P-2,Q-4,R-1,S-3
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Asked: July 4, 2020 In: Mechanical MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q40.
fatigue loading GATE MT 2017 paper with solution gate mt previous year questions with solution
A steel component is subjected to fatigue loading : σ(maximum)= 200 MPa, σ(minimum)=0. The
component has an initial crack length of 1mm. Propagation of crack is governed by: da/dN= 10-12
(ΔK)3, where, the crack length a is in metres, N is the number of cycles and ΔK is in MPa m1/3.
The crack length of (in m) after one million cycles will be _____ (answer up to three decimal
places)
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Correct answer is 0.0125 Given, da/dN= 10-12 (ΔK)3 We know that, ΔK=σ√πamax
da/dN= 10-12 (σ√πamax)3 = ∫10-6 to af (a-3/2) da =10-12 ×(200)3 ×(3.14)3/2 ∫o to
106 dN af= 0.0125
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Asked: July 4, 2020
GATE MT 2017 Q39.
GATE MT 2017 paper with solution gate mt previous year questions with solution intrinsic semiconductor
Correct answer is option B. For intrinsic semiconductor σ=(μene + μnnn)ni ni=
σ/(μene + μnnn) = 10-6/(0.75+0.06)(1.6×10-19) = 7.7×1012/m3
For an intrinsic semiconductor, the room temperature electrical conductivity is 10-6 Ω-1m-1. If the
electron and hole mobilities are 0.75 and 0.06 m2V-1s-1 respectively, the intrinsic carrier
concentration (per m3) at room temperature is: (A) 1 × 1012 (B) 7 ×1012 (C) 3 ×1012 (D) 1.1
×1014
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GATE MT 2017 Q38. Arrange the magnetic moment of neighbouring
atoms in a one-dimensional lattice in Group 1 to the corresponding
magnetic material in Group 2.
GATE MT 2017 paper with solution gate mt previous year questions with solution magnetic moment
Correct answer is option B.
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(A) P-4,Q-1,R-3,S-2
(B) P-3,Q-4,R-1,S-2
(C) P-2,Q-4,R-1,S-3
(D) P-1,Q-2,R-3,S-4
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Asked: July 4, 2020
In: Thermodynamics & Rate Processes
GATE MT 2017 Q37.
GATE MT 2017 paper with solution gate mt previous year questions with solution predominance area
Correct answer is (-0.5). Consider line BC: 2NiO+ 2SO2 + O2 → 2NiSO4 K=
[NiSO4]2/([NiO]2[pSO2][pO2] K= 1/([pSO2][pO2]) Simplifying: log K= -2log pSO2-log
pO2 log pSO2 = -1/2(pO2)- 1/2( pO2) The above equation in the form of y=mx +c
Slope= -(1/2)= (-0.5)
Assuming the solid phases to be pure, the slope of line BC in the predominance area diagram
schematically shown below is _____ (answer up to two decimal places)
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Asked: July 4, 2020
GATE MT 2017 Q36.
GATE MT 2017 paper with solution gate mt previous year questions with solution homogenous nucleation
For homogenous nucleation of solid in liquid of a pure metal, the critical edge length (in nm) of a
cube shaped nucleus is ______. (answer up to two decimal places) Given surface energy γ=0.177
Jm-2 ; change in volume free energy ΔGv = -2.8 × 108 Jm-3)
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Correct answer is 2.52 For cubic nucleus: ΔG=6a2γs + a3ΔGv Critical length=
dΔG/da= 0= 12aγs +3a2 a= -4γs/ ΔGv = (4×0.1770)/(2.8 × 108)= 2.52 nm
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Asked: July 3, 2020
GATE MT 2017 Q35.
third law of thermodynamics
Correct answer is 0.94 As per question, we can write ΔS(IV)= ΔS(I) + ΔS(II) + ΔS(III)=0
36.8+ΔS(II)-37.8=0 ΔS(II)=0.94
Pure orthorhombic sulphur transforms to stable monoclinic sulphur above 368.5 K. Applying the
third law of thermodynamics, the value of entropy (in JK-1) of transformation at 368.5 K is _____
(answer up to two decimal places) Given: Entropy change associated with heating orthorhombic
sulphur from 0 K to 368.5K is 36.86 J K-1. Entropy change associated with cooling monoclinic
sulphur from 368.5 K to 0 K is -37.8 J K-1.
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GATE MT 2017 Q34. For each of the following crystallographic
system listed in column 1. Match the corresponding minimum
symmetry in column 2.
crystallographic system GATE MT 2017 paper with solution gate mt previous year questions with solution
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(A) P-3,Q-4,R-2,S-1
(B) P-4,Q-3,R-2,S-1
(C) P-1,Q-2,R-4,S-3
(D) P-4,Q-3,R-1,S-2

Correct answer is option D. Tetragonal= One- 3 fold rotation Cubic= 4- three fold
rotation Monoclinic= 1-two fold rotation Rhombohedral= 1-three fold rotation
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Asked: July 3, 2020 In: Extractive MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q33.
cathodic current e ciency electrowinning GATE MT 2017 paper with solution
gate mt previous year questions with solution
Correct answer is 11547 E ciency (η)= actual energy/total energy E×η×Z×I= V×t
where E= speci c energy consumption Z=M/nF (n = valency) E=VnF/ηM = 3.5
×2×96500/(0.9×65)= 11547
A zinc electrowinning cell is being operated at a current of 400 A, voltage of 3.5V, and a cathodic
current efficiency of 90%, the specific energy consumption (in kg KJ-1 zinc) is _____. (answer up
to two decimal places) Given Atomic weight of zinc =65
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GATE MT 2017 Q32. Pure metals A and B form two real binary solid
solutions α and β at a temperature T and pressure P. The free
energy versus composition plots for both the solutions are given
below. The condition for chemical equilibrium is:
GATE MT 2017 paper with solution gate mt previous year questions with solution real binary solid solutions
Correct answer is option C. At point A, μAα = μAβ At point B, μBα = μBβ
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(A) Mole fraction of A in α=mole fraction of A in βand mole fraction of B in α= mole fraction of
B in β
(B) Mole fraction of B in α= mole fraction of A in β and mole fraction of A in α = mole fraction
of B in β
(C) Activity of A in α= activity of A in β and activity of B in α=activity of B in β
(D) Activity of A in α= activity of B in β and activity of B in α= activity of A in β

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Asked: July 3, 2020
GATE MT 2017 Q31.
Correct answer is 2. Fe(OH)3 → Fe+3 +3(OH)- Kp= 10-39= [Fe+3][OH]-3 log(10-39)
=log(Fe+3)+3log(OH)- -39=-3+3log(OH)- pOH = 12 We know that, pH +pOH= 14 pH=2

A solution contains 10-3 M of Fe+3 at 25°C. The solubility of product of Fe(OH)3 is 10-39.
Assuming activity equals concentration, the maximum pH at which Fe+3 will precipitate as
Fe(OH)3 is _______. (answer up to two decimal places)
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Asked: July 2, 2020
GATE MT 2017 Q30.
A stoichiometric equation of CO and pure oxygen at 1atm and 25°C flows into a combustion
reactor. The molar flow rate of CO entering the reactor is 1kg-mol.h-1. The adiabatic flame
temperature (in K) for the combustion of CO with stoichiometric oxygen is ______. (answer up to
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adiabatic ame temperature GATE MT 2017 paper with solution
Correct answer is 6707.09 K ΔHf = ∫T0Tf CpdT Tf= adiabatic ame temperature Tf =
298 + 2822000/44 = 6707.09 K
two decimal places). Given ΔH0298 (CO-CO2) =-282000 KJ (kg-mol-CO)-1, Cp(CO2)= 44 KJ (kg-
mol K)-1
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Asked: July 2, 2020 In: Iron & Steel MakingAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q29.
GATE MT 2017 paper with solution gate mt previous year questions with solution Henrian activity coe cient
Correct answer is 0.0034. Given reaction: C+ O= CO T=1600+273=1873K log k=
1160/T + 2.003 = 1160/1873 +2.003 K= 419.109 k= PCO / [%O][%C] 419.109=1/[%O]
[0.7] [%O]= 0.0034
In primary steel making, dissolved Oxygen (O) reacts with carbon (c ) to produce CO(g), at 1 atm
pressure according to the reaction C+ O= CO. The equilibrium constant for this reaction is
logK=1160/T + 2.003 , where T is kelvin. Assuming the Henrian activity coefficient of both O and C
to be unity, the dissolved oxygen content (in wt%) of a plain carbon steel melt with 0.7% C at
1600°C is _____(answer up to four decimal places)
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PollAsked: July 2, 2020 In: Iron & Steel MakingAskmemetallurgy Official ✓ Silver
GATE MT 2017 28. A ladle containing molten steel is being
discharged. The relevant forces are listed in column 1. Match them
with their corresponding expressions in column 2. µ=viscosity, U=
characteristic velocity, L= characteristic length, g=acceleration due
to gravity, P=pressure.
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(A) P-4,Q-3,R-2, S-1
(B) P-1, Q-3, R-2, S-4
(C) P-2,Q-3,R-4,S-1
(D) P-4,Q-3,R-1,S-2

ladle containing molten steel
Correct answer is option A. Pressure force= PL2 Inertial force=ρU2L2 Gravity force=
ρgL3 Viscous force= μUL
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Asked: July 2, 2020
GATE MT 2017 Q27.
GATE MT 2017 paper with solution gate mt previous year questions with solution thermal equilibrium
Best AnswerNareshN Iron
Metallurgy and material technology in YSR Engineering college of yogi
vemana university
Added an answer on July 3, 2020 at 10:43 am
1 Reply
P+F=C+1 2+F=2+1 F=1
CaCo3(s) dissociates in a closed system according to the reaction : CaCO3(s) = CaO(s) + CO2(g)
Assuming the reaction is in thermal equilibrium, the degree(s) of freedom, F = _____.
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GATE MT 2017 Q26. The pourbaix plot of reaction Al3+ + 2H2O =
AlO2- + 4H+ in potential( E) versus pH diagram is :
GATE MT 2017 paper with solution gate mt previous year questions with solution pourbaix plot
Correct answer is option C. Given reaction: Al3+ + 2H2O = AlO2- + 4H+ No
involvement of electrons takes place in the above reaction. Thus, it is not a redox
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A
B
C
D

reaction. Potential is only dependent on pH.
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Asked: July 1, 2020 In: Iron & Steel MakingAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q25.
GATE MT 2017 paper with solution gate mt previous year questions with solution LD steelmaking
During the end blow period in LD steelmaking, the de-carburisation rate is expressed by the
equation : dc/dt= -(c-c*) are the instantaneous and equilibrium concentration of carbon in steel
respectively, in units of wt%. Given that c*= 0.04 wt % and c(t=0 min) =0.4 wt%, the concentration
of carbon in steel (in wt%) at t=1 min is ______ (answer up to three decimal places)
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Asked: July 1, 2020
GATE MT 2017 24.
GATE MT 2017 paper with solution gate mt previous year questions with solution radiation and convection
A continuous cast steel slab, 1m × 1m × 0.1m at 1298 K cools in air. The initial rate of heat loss (in
kW) from the top surface of slab by radiation and convection is _____ (answer up to two decimal
places) (Given (i) Ambient temperature= 298K (ii) emissivity of steel=0.8 (iii) convective heat
transfer coefficient = 4.6 W m-2K-1 (iv) Stefan-Boltzmann constant (σ)= 5.7 × 108 W m-2 K-4
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PollAsked: July 1, 2020 In: Testing Of MaterialAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q23. Dye penetrant test is based on the principle of:
dye penetrant test GATE MT 2017 paper with solution gate mt previous year questions with solution
Correct answer is option D. Dye penetrant inspection (DP), also called liquid
penetrate inspection (LPI) or penetrant testing (PT), is a widely applied and low-cost
inspection method used to check surface-breaking defects in all non-porous
materials (metals, plastics, or ceramics). LPI is used to de Read More
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A. Polarised sound waves in liquid
B. Magnetic domain
C. Absorption of X-rays
D. Capillary action
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Asked: July 1, 2020
GATE MT 2017 Q22.

stoichiometric metal oxidation reaction
Correct answer is option C. As per question, A(s) + O2(g) = AO2 (s) For Ellingham
diagram's slope = -ΔS0 Three cases arises in this question, Case 1: If T<T1 i.e. only
O2(g) is present slope= -(-ΔS0) = ΔS0 i.e change in entropy=(products-reactants)
Case 2: If T1<T<T2 i.e O2(g) and A (liquid Read More
T1 and T2 are melting points of pure metal A and pure stoichiometric oxide AO2, respectively and
T1<T2. The stoichiometric metal oxidation reaction A(s) + O2(g) = AO2 (s) is in equilibrium at 1
atm pressure at temperature less than T1. If the temperature increases, which schematic
represents the correct standard free energy change versus temperature plot?
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PollAsked: July 1, 2020 In: Manufacturing TechnologyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q21. Which one of the following manufacturing
techniques is used for making window glass?
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A. Investment casting
B. Patenting
C. Spray forming
D. Float-bath method

GATE MT 2017 paper with solution gate mt previous year questions with solution manufacturing techniques
Askmemetallurgy Official ✓ Silver o cial account of Askmemetallurgy
Correct answer is option D. A sheet of glass is made by oating molten glass on a
bed of molten metal, typically tin, although lead and other various low-melting
point alloys are also used. This method gives the sheet uniform thickness and very
at surfaces. Modern windows are made from oat gl Read More
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PollAsked: June 30, 2020
In: Manufacturing Technology
GATE MT 2017 Q19. Schematic diagram shows rolling of a slab. P
and Q are the points on the surface of the workpiece near entrance
and exit, respectively. With reference to the work piece, which one
of the following statements is TRUE?
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GATE MT 2017 paper with solution gate mt previous year questions with solution rolling of a slab
Correct answer is option C. Frictional force acts in the rolling direction to pull the
slab, whereas at exit point the frictional force acts in the opposite direction.
Submit Results
A. Frictional force is along rolling direction at both P and Q.
B. Frictional force is opposite to rolling direction at both P and Q.
C. Frictional force is along the rolling direction at P and opposite to rolling direction at Q.
D. Frictional force is opposite to rolling direction at P and along the rolling direction at Q.
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Asked: June 30, 2020 In: Mechanical MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q20.
GATE MT 2017 paper with solution gate mt previous year questions with solution power law behaviour
A material which shows power law behaviour σ- = 50 Ē0.3, is being wire drawn. The maximum
strain per pass in annealed condition (assume ideal work and efficiency ƞ=1) is _____. (answer up
to two decimal places)
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Asked: June 30, 2020 In: Testing Of MaterialAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q16.
bragg angle GATE MT 2017 paper with solution gate mt previous year questions with solution
The second peak in the powder X-ray diffraction pattern of a FCC metal occurs at a Bragg angle Ɵ
(in degrees)= _______ (answer up to two decimal places) (Given λCuKα= 0.154nm; lattice
parameter of metal = 0.36nm)
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GATE MT 2017 Q18.
GATE MT 2017 paper with solution gate mt previous year questions with solution Poisson’s ratio
A rod is elastically deformed by a uniaxial stress resulting in a strain of 0.02. If the Poisson’s ratio
is 0.3, the volumetric strain is ____ .(answer up to three decimal places)
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GATE MT 2017 17. Four alloys C1,C2, C3,C4 shown in the phase
diagram are poured at a temperature T1 in a mould. During
solidification, which one of these alloys is expected to have the
highest fluidity?

GATE MT 2017 paper with solution gate mt previous year questions with solution phase diagram
Best AnswerShetty Gold
Added an answer on July 1, 2020 at 1:35 am
1 Reply
Option C is correct. Because at C3 there is a eutectic point at which melting point is
low and has high uidity. The other compositions have a freezing range where
there are two di erent phases so uidity decreases.
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A. C1
B. C2
C. C3
D. C4
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GATE MT 2017 Q15.
fracture strength GATE MT 2017 paper with solution gate mt previous year questions with solution
A brittle material (young’s modulus= 60 GPa and surface energy =0.5 Jm-2 ) has a surface crack
of length 2 µm. the fracture strength (in MPa) of this material is _______ (answer up to two
decimal places)
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Asked: June 29, 2020
GATE MT 2017 Q14.
hypothetical microstructure
Spherical α-phase particles are depicted in the hypothetical microstructure section shown below.
Using the superimposed grid on the microstructure, the estimated volume fraction of α phase is
_______( answer up to three decimal places)
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GATE MT 2017 Q13.
GATE MT 2017 paper with solution gate mt previous year questions with solution unit cell
A cubic metal has density of 19000 kg m-3, lattice parameter of 0.4nm and atomic weight of 183.
The effective number of atoms in an unit cell of this metal is ______.
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PollAsked: June 29, 2020 In: Mechanical MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q12. Stress required to operate a Frank-Read
source of length L is approximately given by:
frank-read source GATE MT 2017 paper with solution gate mt previous year questions with solution
Added an answer on June 30, 2020 at 4:49 pm
Correct answer is option A. The relation between the stress to operate Frank-Read
source of length 'L' and burger vector 'b' : τ = Gb/L
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A. Gb/L
B. Gb^2/L
C. Gb^2/L^2
D. Gb^2/2L^2
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GATE MT 2017 Q11.
GATE MT 2017 paper with solution gate mt previous year questions with solution iron-carbon alloy
In an iron-carbon alloy containing 0.35 wt% C, the mass fraction of pearlite just below the eutectoid
temperature is _____. (answer up to two decimal places)
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GATE MT 2017 Q10.
GATE MT 2017 paper with solution gate mt previous year questions with solution tetrahedral void
For a FCC metal, radius of the largest sphere that can fir in tetrahedral void (in mm) is ______
(answer up to three decimal places). (Given lattice parameter =0.401 nm)
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GATE MT 2017 Q9. Both creep resistance and tensile strength of a
metal can be enhanced by the:
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A. Increase in the grain size
B. Decrease in the grain size
C. Addition of dispersoids

creep resistance and tensile strength GATE MT 2017 paper with solution
Best AnswerRohit.km Platinum Gate aspirant.
1 Reply
The correct option is C. We know that the tensile strength of materials can be
enhanced by several strengthening mechanisms like: Solid solution strengthening
Grain Boundary Strengthening Dispersion Strengthening Strain hardening and
some more processes. As per question if we see there are two mecha Read More
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D. Annealing
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Asked: June 28, 2020 In: Powder MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q7.
GATE MT 2017 paper with solution gate mt previous year questions with solution theoretical density
Tungsten powder is pressed at 15MPa to a green density of 55%. After sintering the compact
attains 86.5% of its theoretical density. Assuming uniform shrinkage, the linear shrinkage (in % ) is
________ (answer up to two decimal places)
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GATE MT 2017 Q8. Primary mechanisms of accommodating plastic
strain at low temperatures in crystalline metals are:
GATE MT 2017 paper with solution gate mt previous year questions with solution plastic strain
1 Reply
The correct option is A. Twinning as we know is an active mechanism at lower
temperatures. During twinning, the atoms mirror image themselves in a plane
forming a twin boundary and thus decrease the strain around it. This mechanism
needs lesser energy than slip but relieves lesser strain comparative Read More
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A. Twinning and dislocation- slip
B.Dislocation-climb and dislocation- slip
C. Dislocation-slip and di usion
D. Viscous ow and dislocation-slip
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Asked: June 28, 2020 In: Extractive MetallurgyAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q6.

GATE MT 2017 paper with solution gate mt previous year questions with solution sieve analysis
The sieve analysis of ground quartz particles is given in the table below: The cumulative mass
fraction of particles of size less than 1.68mm is ________. (answer up to two decimal places)
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GATE MT 2017 Q5. The sequence of precipitation to reach stable
equilibrium during ageing of Al-4.5 wt% Cu alloy is :
al-4.5 wt %cu alloy GATE MT 2017 paper with solution gate mt previous year questions with solution
The correct option is B. The Al-4.5Cu alloy as the composition lies near the solubility
limit of Cu in Al the solution results in precipitate formation when the temperature
is below the Solvus line. Here below Solvus two phases can exist α and CuAl2
intermediate compound. This ppt is a equilibrium o Read More
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(A) GP zone- θ'-θ''-θ
(B) GP zone- θ''-θ'-θ
(C) GP zone- θ-θ''-θ
(D) GP zone- θ''-θ-θ'

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GATE MT 2017 Q4
GATE MT 2017 paper with solution gate mt previous year questions with solution solubility of hydrogen
Correct answer is 4920 mm3/kg. Given reaction: H2= 2[H] K= [H]2/ pH2 [H]∝ √pH2
Two cases arises in this question. Case 1: 1.64× 104 ∝ 1 Case 2: [H2]∝ √0.09 On
dividing both the cases (i.e case1/case2) [H2]= 4920 mm3/kg
Hydrogen dissolves in Pd by the reaction H2=2[H]. At 3000C and pH2= 1 atm, the solubility of
hydrogen in Pd is 1.6× 104 mm3 (STP) per kg of Pd. At 3000C and pH2= 0.09 atm, the solubility of
hydrogen in Pd in mm3 (STP) per kg of Pd is _______. (answer upto one decimal place)
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Asked: June 27, 2020 In: Iron & Steel MakingAskmemetallurgy Official ✓ Silver
GATE MT 2017 Q3
The rate of dissolution of Al particles in liquid steel is proportional to concentration difference (ΔC).
ΔC is defined by (Given (i) Cb=bulked concentration of dissolved Al in liquid steel (ii) C*=saturation
concentration of Al in liquid steel at the given temperature (iii) Cm= Density of Al/Atomic weight of
Al (A) C*-Cb (B)Cb-Cm (C) C*-Cm (D) √C*Cm– Cb
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GATE MT 2017 paper with solution gate mt previous year questions with solution rate of dissolution
Added an answer on June 28, 2020 at 3:18 am
1 Reply
Option A is correct. Rate of transfer of Al = Ak m, Al [C*Al – Cb Al ] where A is
surface area, km, Al is given by Higbie's theory of penetration. So option A is
correct
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GATE MT 2017 Q2
electrochemical reaction GATE MT 2017 paper with solution gate mt previous year questions with solution
1 Reply
Option D is correct
For the electrochemical reaction, Cu2++Zn= Zn 2++Cu, the standard cell potential at 250C and 1
atm pressure is (Given E0(Cu2+/Cu) =0.337 V and E0(Zn2+/Zn) = -0.763V) ( A) -0.426 V (B)
0.426 V (C) 0.55 V (D) 1.1V
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GATE MT 2017 Q1. The pressure(P) versus volume (V) diagram
given below represents reversible isothermal curves at
temperatures T1,T2,T3. Consider one mole of ideal gas for all the
three isothermal processes, which one of the following is TRUE?
GATE MT 2017 paper with solution gate mt previous year questions with solution isothermal processes
Option B is correct By ideal gas equation PV=nRT PV∝ T from the graph we can say
T2 is higher temperature and then T3, and T1 has low temperature compare to
other. So, T2>T3>T1.
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(A) T1>T2>T3
(B) T2>T3>T1
(C) T3>T1>T2
(D) T2<T1<T3

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