Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous
  1. Hardness of steel greatly improves with








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    Cyaniding: Carbon & Nitrogen diffuses to surface and rapidly quenched in case hardening.

    Correct Option: B

    Cyaniding: Carbon & Nitrogen diffuses to surface and rapidly quenched in case hardening.

  1. Cold working of steel is defined as working








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    Cold working is a phenomenon in which material is deformed in recrystallization temperature.

    Correct Option: C

    Cold working is a phenomenon in which material is deformed in recrystallization temperature.

  1. During heat treatment of steel, the hardness of various structures in increasing order is








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    Spherodite, Coarse Pearlite, Fine Pearlite, Martensite

    Structure Rockwell Hardness
    Coarse Pearlite RC15
    Fine Pearlite RC25
    Martensite RC65

    Correct Option: D

    Spherodite, Coarse Pearlite, Fine Pearlite, Martensite

    Structure Rockwell Hardness
    Coarse Pearlite RC15
    Fine Pearlite RC25
    Martensite RC65

  1. When 1.0% plane carbon steel is slowly cooled from the molten state to 740°C, the resulting structure will contain








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    The resulting structure will contain Austenite & cementite.

    Correct Option: D

    The resulting structure will contain Austenite & cementite.

  1. The thickness of a sheet is reduced by rolling (without any change in width) using 600 mm diameter rolls. Neglect elastic deflection of the rolls and assume that the coefficient of friction at the roll-workpiece interface is 0.05. The sheet enters the rotating rolls unaided. If the initial sheet thickness is 2 mm, the minimum possible final thickness that can be produced by this process in a single pass is ____ mm (round off to two decimal places).








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    (∆h)max = μ² . R
    ⇒ hi – hf = μ² . R
    ⇒ 2 – hf = (0.05)2 × 300 ⇒ hf = 2 – 0.75
    ⇒ hf = 1.25 mm

    Correct Option: D

    (∆h)max = μ² . R
    ⇒ hi – hf = μ² . R
    ⇒ 2 – hf = (0.05)2 × 300 ⇒ hf = 2 – 0.75
    ⇒ hf = 1.25 mm