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Askmemetallurgy

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  1. Asked: September 20, 2021In: Books & Notes

    Can anyone share Gate mt 2021 solved paper pl?

    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on September 21, 2021 at 7:36 pm
    This answer was edited.

    Pl find the solution here : https://askmemetallurgy.com/question-tag/gate-mt-2021-question-paper-with-solution/

    Pl find the solution here : https://askmemetallurgy.com/question-tag/gate-mt-2021-question-paper-with-solution/

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  2. Asked: June 25, 2021In: Career/ Jobs/ Exam

    What are the job opportunities after Diploma in metallurgy and in which companies

    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on June 25, 2021 at 10:54 am

    Below are the companies who recruit diploma metallurgy engineers: At Govt. / PSUs Various SAIL plant (Bokaro, Rourkela, Burnpur, Bhilai, Durgapur), Vizag steel plant, Defence Research development organisation, Bhabha Research Atomic center, Non-Teaching staff at various IITs & NITs, MIDHANI, OrdRead more

    Below are the companies who recruit diploma metallurgy engineers:

    At Govt. / PSUs

    Various SAIL plant (Bokaro, Rourkela, Burnpur, Bhilai, Durgapur), Vizag steel plant, Defence Research development organisation, Bhabha Research Atomic center, Non-Teaching staff at various IITs & NITs, MIDHANI, Ordinance Factory, SSC selection post (1 year experience basis usually), Nuclear Fuel complex. 

    For Private companies, follow this link https://askmemetallurgy.com/list-of-companies-recruiting-metallurgical-engineers/ 

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  4. Asked: February 4, 2021In: Career/ Jobs/ Exam

    Is there any jobs for Diploma in Metallurgical Engineering for girls?

    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on February 26, 2021 at 7:52 pm
    This answer was edited.

    Hi pragya, Please follow our Latest jobs page and LinkedIn page to get every Freshers and experienced based jobs in the field of Metallurgy. Our page: https://askmemetallurgy.com/latest-jobs/ Linkedin page: https://www.linkedin.com/company/askmemetallurgy/ Alternatively, you can request an activatioRead more

    Hi pragya,

    Please follow our Latest jobs page and LinkedIn page to get every Freshers and experienced based jobs in the field of Metallurgy.

    Our page:

    https://askmemetallurgy.com/latest-jobs/

    Linkedin page:

    https://www.linkedin.com/company/askmemetallurgy/

    Alternatively, you can request an activation for Askmemetallurgy Whatsapp job alert to get all these alert directly in your whatsapp.

    https://askmemetallurgy.com/askmemetallurgy-is-now-on-whatsapp/

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  5. Asked: March 17, 2020In: Thermodynamics & Rate Processes

    GATE MT 2012 Q48. Common Data Questions Common Data for Questions 48 and 49: A steel ball (density ρsteel = 7200 kg/m3 ) is placed in an upward moving liquid Al (density ρAl = 2360 kg/m3 , viscosity μAl = 110–3 Pa.s and Reynolds number = 5105 ). The force (F) exerted on the steel ball is expressed as F = f πR 2 (ρAlv 2 /2) where, f is friction factor (=0.2), v is the velocity of liquid Al and R is the radius of steel ball. The force exerted on the steel ball is

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on December 27, 2020 at 2:46 pm

    Correct answer is option A. Given that, Reynolds number = 5 × 105 We know that, Reynolds number, R = ρvD/μ On equating, 5×105 = 2360×v ×2R/ 10-3 On simplifying,  0.211 = v × 2R V = 0.211/2R Given that, force exerted F =fπR2ρAl V2 /2  On substituting the given values, F = 0.2×3.14×R2× 2360 ×( 0.211/2Read more

    Correct answer is option A.

    Given that, Reynolds number = 5 × 105

    We know that, Reynolds number, R = ρvD/μ

    On equating, 5×105 = 2360×v ×2R/ 10-3

    On simplifying,  0.211 = v × 2R

    V = 0.211/2R

    Given that, force exerted F =fπR2ρAl V2 /2 

    On substituting the given values, F = 0.2×3.14×R2× 2360 ×( 0.211/2R) 2/ 2

    = 8.25

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  6. Asked: March 15, 2020In: Thermodynamics & Rate Processes

    GATE MT 2012 Q7. A fluid is flowing with a velocity of 0.5 m/s on a plate moving with a velocity of 0.01 m/s in the same direction. The velocity at the interface of the fluid and plate is

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on December 27, 2020 at 2:22 pm

    Correct answer is option B. Given that, velocity of plate =0.01 m/s As the fluid is flowing in the same direction, the velocity at the interface of the fluid and plate is same i.e 0.01 m/s

    Correct answer is option B.

    Given that, velocity of plate =0.01 m/s

    As the fluid is flowing in the same direction, the velocity at the interface of the fluid and plate is same i.e 0.01 m/s

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  7. Asked: March 16, 2020In: Mineral Processing & Non-ferrous Metallurgy

    GATE MT 2012 Q37. Match the processes given in Group I with the corresponding metals in Group II

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 22, 2020 at 9:48 pm

    Correct answer is option B.  A method for removing oxygen from metal or metalloid oxides, by electrolysis in a fused salt, to produce elements or alloys, improved so as to prevent problems of cell contamination and poor power efficiency, particularly in scaled up electro-deoxidation processes. GoldRead more

    Correct answer is option B.

    •  A method for removing oxygen from metal or metalloid oxides, by electrolysis in a fused salt, to produce elements or alloys, improved so as to prevent problems of cell contamination and poor power efficiency, particularly in scaled up electro-deoxidation processes.
    • Gold cyanidation is a hydrometallurgical technique for extracting gold from low-grade ore by converting the gold to a water-soluble coordination complex.
    • Matte smelting is the most common way of smelting copper–iron sulfur concentrates. The primary purpose of matte smelting is to turn the sulfide minerals in solid copper concentrate into three products: molten matte, molten slag, and offgas.
    • A method for removing oxygen from metal or metalloid oxides, by electrolysis in a fused salt, to produce elements or alloys, improved so as to prevent problems of cell contamination and poor power efficiency, particularly in scaled up electro-deoxidation processes.
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  8. Asked: March 16, 2020

    GATE MT 2011 Q39. Match the heat treatment processes in group 1 with resultant microstructure of steel in group 2.

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 4:21 pm

    Correct answer is option D. Martempering is also known as stepped quenching or interrupted quenching. In this process, steel is heated above the upper critical point (above the transformation  range) and then quenched in a salt, oil, or lead bath kept at a temperature of 150-300 °C.  The workpiece iRead more

    Correct answer is option D.

    • Martempering is also known as stepped quenching or interrupted quenching. In this process, steel is heated above the upper critical point (above the transformation  range) and then quenched in a salt, oil, or lead bath kept at a temperature of 150-300 °C.  The workpiece is held at this temperature above martensite start (Ms) point until the temperature becomes uniform throughout the cross-section of workpiece. After that it is cooled in air or oil to room temperature. The steel is then tempered. In this process,  austenite is transformed to martensite by step quenching , at a rate fast enough to avoid the formation of ferrite, pearlite  or bainite.
    • Normalizing is a heat treatment used on steel so as to refine its crystal structure and produces a more uniform and desired grain size distribution. The final microstructure consists of fine pearlite.
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  9. Asked: March 16, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 Q40. In case of homogenous nucleation , the critical edge length of a cube shaped nucleus is (γ: energy per unit area of the interface between product and the parent phase Δg gibbs free energy per unit volume )

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 4:09 pm

    Correct answer is option A. ΔG= a3.Δg + 6a2γ dΔG/da = 0= 3a2 . Δg + 12 aγ a= -( 4γ/ Δg)

    Correct answer is option A.

    ΔG= a3.Δg + 6a2γ

    dΔG/da = 0= 3a2 . Δg + 12 aγ

    a= -( 4γ/ Δg)

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  10. Asked: March 17, 2020In: DT & NDT

    GATE MT 2011 Q46. Match those listed in Group 1 with the NDT methods listed in Group 2.

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 4:02 pm

    Correct answer is option A. Drawing is a metalworking process which uses tensile forces to stretch metal, glass, or plastic. As the metal is drawn (pulled), it stretches thinner, into a desired shape and thickness. Drawing is classified in two types: sheet metal drawing and wire, bar, and tube drawiRead more

    Correct answer is option A.

    • Drawing is a metalworking process which uses tensile forces to stretch metal, glass, or plastic. As the metal is drawn (pulled), it stretches thinner, into a desired shape and thickness. Drawing is classified in two types: sheet metal drawing and wire, bar, and tube drawing.
    • Crankshafts are the main rotating parts of an engine that are installed on a connecting rod and can convert the up and down movement into a circular movement of the connecting rod. Typically, in the crankshaft manufacturing process, a billet of suitable size is given a heat treatment till the required forging temperature, and then it is successively pounded or pressed into the required shape by pressing the billet between a pair of dies under high pressure.
    • Rolling is a metal forming process in which metal stock is passed through one or more pairs of rolls to reduce the thickness, to make the thickness uniform, and/or to impart a desired mechanical property. The final product is either sheet or plate.
    • Stretch forming is a metal forming process in which a piece of sheet metal is stretched and bent simultaneously over a die in order to form large contoured parts.
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  11. Asked: March 17, 2020In: Mechanical Metallurgy

    GATE MT 2011 Q55. The elastic strain energy per unit length of dislocation line in copper is:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:47 pm

    Correct answer is option D.

    Correct answer is option D.

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  12. Asked: March 17, 2020In: Mechanical Metallurgy

    GATE MT 2011 Q54. Shear modulus of copper is 45 GPa. Lattice parameter of copper is 3.61 A0. The magnitude of burgers vector in copper is:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:47 pm

    Correct answer is option A.

    Correct answer is option A.

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  13. Asked: March 17, 2020In: Iron & Steel Making

    GATE MT 2011 Q53. The CO/CO2 molar ratio in the top gas is:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:46 pm

    Correct answer is option C.

    Correct answer is option C.

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  14. Asked: March 17, 2020In: Iron & Steel Making

    Linked answer questions. Statements for linked answer questions 52 and 53: In an ideal blast furnace, the input and output are as follows: GATE MT 2011 Q52. The amount of oxygen in CO and CO2 leaving with the top gas:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:45 pm

    Correct answer is option C.

    Correct answer is option C.

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  15. Asked: March 17, 2020In: Physical Metallurgy & Heat treatment

    GATE MT 2011 Q51. At the eutectoid temperature, the ratio of α and β phases in the specimen observed under microscope is:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:44 pm

    Correct answer is option C.

    Correct answer is option C.

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  16. Asked: March 17, 2020In: Physical Metallurgy & Heat treatment

    Common data for Q50 and Q51. A binary phase diagram of components P and Q displays an eutectoid with terminal solid solutions α on the P rich side and β on the Q rich side. At the eutectoid temperature, the solubilities of Q in α and β are 5 and 90 wt% respectively. The densities of α and β are 9.5 and 2.49 g/cm3, respectively. GATE MT 2011 Q50. : At the eutectoid point, the alloys has α and β in the weight ratio 1:1. The eutectoid point occurs at composition:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:43 pm

    Correct answer is option B.

    Correct answer is option B.

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  17. Asked: March 17, 2020In: Metal Working (Forming)

    GATE MT 2011 Q47. What is the ideal extrusion pressure if the effective flow stress in compression is 250 MPa?

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:38 pm
    This answer was edited.

    Correct answer is option A. P= kA0Ln(A0/Af) P= 250 ln16 = 693.15 MPa    

    Correct answer is option A.

    P= kA0Ln(A0/Af)

    P= 250 ln16 = 693.15 MPa

     

     

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  18. Asked: March 16, 2020In: Physical Metallurgy & Heat treatment

    GATE MT 2011 Q45. Assertion ‘a’ : during casting of element grain refinement can be achieved by addition of certain allowing elements Reason ‘r’ : the addition of alloying element may result in the formation of deoxidation product or inter metallic compounds which may act as nucleation sites for grain refinement.

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:36 pm

    Correct answer is option B. Grain refinement, also known as inoculation, is the set of techniques used to implement grain boundary strengthening . One method for controlling grain size in aluminium alloys is by introducing particles to serve as nucleants, such as Al–5%Ti. Grains will grow via heteroRead more

    Correct answer is option B.

    Grain refinement, also known as inoculation, is the set of techniques used to implement grain boundary strengthening . One method for controlling grain size in aluminium alloys is by introducing particles to serve as nucleants, such as Al–5%Ti. Grains will grow via heterogenous nucleation ; that is, for a given degree of undercooling beneath the melting temperature, aluminium particles in the melt will nucleate on the surface of the added particles. Grains will grow in the form of dendrites growing radially away from the surface of the nucleant. Solute particles can then be added (called grain refiners) which limit the growth of dendrites, leading to grain refinement. Al-Ti-B alloys are the most common grain refiner for Al alloys.

     

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  19. Asked: March 16, 2020In: Physical Metallurgy & Heat treatment

    GATE MT 2011 Q43. Which one of the following reactions in fcc/bcc Crystal’s with lattice parameter a is energetically favourable?

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:28 pm

    Correct answer is option A. a/2 [1-10] + a/2 [01-1] → a/2 [1-01] a2/4(2)  + a2/4(2) →a2/4(2) b12 + b22 > b32  

    Correct answer is option A.

    a/2 [1–10] + a/2 [01–1] → a/2 [1–01]

    a2/4(2)  + a2/4(2) →a2/4(2)

    b12 + b22 > b32

     

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  20. Asked: March 16, 2020In: Physical Metallurgy & Heat treatment

    GATE MT 2011 Q41. For a cubic metal with lattice parameter 3.92A0, the first order diffraction peels from the x ray powder diffraction pattern taken with CuKx , radiation (λ=1.545 A0,) occur at 2 θ values of 39.7, 46.2, 67.5, 81.3. The crystal structure of metal is

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:23 pm

    Correct answer is option B. Given, a= 3.920 λ = 2dsin θ λ = 2(a/√h2+ k2 +l2) sinθ λ2/ 4a2 = sin 2θ/ (h2+ k2 +l2) (h2+ k2 +l2)= sin 2θ.  4a2 / λ2

    Correct answer is option B.

    Given, a= 3.920

    λ = 2dsin θ

    λ = 2(a/√h2+ k2 +l2) sinθ

    λ2/ 4a2 = sin 2θ/ (h2+ k2 +l2)

    (h2+ k2 +l2)= sin 2θ.  4a2 / λ2

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  21. Asked: March 15, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 38. A furnace wall consists of two layers. The middle layer of 450mm is made of light weight brick of thermal conductivity 1 w/m K and the outside layer of 900mm is made of thermal conductivity 2 w/m K. The hot face of the inside layer is temperature 1300 K and the cold face of the outer layer is at 400k. The temperature at the interface between the two layers is:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 3:09 pm

    Correct answer is option B. Q. = kA dT/dx 1× (1300-T) / 4T = 2× (T- 400)/ 900 On simplification, T= 850 K

    Correct answer is option B.

    Q. = kA dT/dx

    1× (1300-T) / 4T = 2× (T- 400)/ 900

    On simplification, T= 850 K

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  22. Asked: March 15, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 Q36. For the following reactions, the standard free energy change us given by at 1773 K as follows: If chromium oxide powder reduces by hydrogen in a fluidized bed, the minimum pH2 / pH2O ratio that has to be maintained at the exit of the reactor:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 2:53 pm

    Correct answer is  option D. Equation 1: 2/3 Cr2O3 = 4/3 Cr + O2  ΔG01 Equation 2: 2H2 + O2 = 2H2O  ΔG0 2 Equation 1 + Equation 2= 2/3 Cr2O3 + H2 = 4/3 Cr + 2H2O  ΔG0 3 ΔG0 3 = 150800 = -RT ln(pH2O/ pH2) (pH2/pH2O) = 166.5    

    Correct answer is  option D.

    Equation 1: 2/3 Cr2O3 = 4/3 Cr + O2  ΔG01

    Equation 2: 2H2 + O2 = 2H2O  ΔG0 2

    Equation 1 + Equation 2= 2/3 Cr2O3 + H2 = 4/3 Cr + 2H2O  ΔG0 3

    ΔG0 3 = 150800 = -RT ln(pH2O/ pH2)

    (pH2/pH2O) = 166.5

     

     

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  23. Asked: March 15, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 Q32. The molar free energy at 1300 K for the transformation of solid Cu to liquid Cu will be :

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 2:42 pm
    This answer was edited.

    Correct answer is option C. Equation 1:  2 Cu (s) + 1/2 O2  = Cu2 O    ΔG10 Equation 2: 2 Cu (l) + 1/2 O2  = Cu2 O    ΔG20 2 Cu (s)= 2 Cu (l)  ΔG3 = ΔG20 - ΔG10 ΔG3 = 1088 J For 1 mole of Cu = 544 J  

    Correct answer is option C.

    Equation 1:  2 Cu (s) + 1/2 O2  = Cu2 O    ΔG10

    Equation 2: 2 Cu (l) + 1/2 O2  = Cu2 O    ΔG20

    2 Cu (s)= 2 Cu (l)  ΔG3 = ΔG20 – ΔG10

    ΔG3 = 1088 J

    For 1 mole of Cu = 544 J

     

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  24. Asked: March 14, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 Q30. During fully developed laminar flow in a circular pipe, the velocity profile is a parabolic and symmetric around the axis. The velocity at the tube wall is zero. The ratio of the average to the maximum velocity is:

    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 2:31 pm

    Correct answer is option B. Circular pipe: ν = k(a2 - r2) νmax = ka2 (r=0) νavg = ∫∫ νr dr dθ / ∫∫ r dr dθ r=0 → a θ = 0→ 2r νavg  = ka2/2 = νmax/ 2  

    Correct answer is option B.

    Circular pipe: ν = k(a2 – r2)

    νmax = ka2 (r=0)

    νavg = ∫∫ νr dr dθ / ∫∫ r dr dθ

    r=0 → a

    θ = 0→ 2r

    νavg  = ka2/2 = νmax/ 2

     

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  25. Asked: March 14, 2020In: Thermodynamics & Rate Processes

    GATE MT 2011 Q27. For a reaction A-B, if the rate of change in concertation of A (Ca), can ne written as -dCA/dt =KC2A, then the change in concertation with time from initial of A, Ca0 is given by:

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 21, 2020 at 2:23 pm

    Correct answer is option A. ca0∫ca -dc/c2 = ∫kdt 1/cA- 1/c0 = kt

    Correct answer is option A.

    ca0∫ca -dc/c2 = ∫kdt

    1/cA– 1/c0 = kt

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  26. Asked: March 14, 2020In: Welding Technology

    GATE MT 2011 Q24. The nature of submerged arc welding flux with basicity index of 0.5 is

    Best Answer
    Askmemetallurgy Official official account of Askmemetallurgy
    Added an answer on November 20, 2020 at 7:25 pm

    Correct answer is option D. Acid fluxes have a basicity index of 0.5 to 0.8; neutral fluxes 0.8 to 1.2; basic fluxes 1.2 to 2.5 and highly basic fluxes 2.5 to 4.0. The basicity of a flux has a major effect on the weld metal properties, most importantly the notch toughness.

    Correct answer is option D.

    • Acid fluxes have a basicity index of 0.5 to 0.8; neutral fluxes 0.8 to 1.2; basic fluxes 1.2 to 2.5 and highly basic fluxes 2.5 to 4.0. The basicity of a flux has a major effect on the weld metal properties, most importantly the notch toughness.
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