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. Match List-I (welding defects)with List-II (Causes) and select the correct answer using the codes given below the lists:
    List-I
    A. Spatter
    B. Distortion
    C. Slag inclusion
    D. Porosity
    List-II
    1. Damp electrodes
    2. Arc blow
    3. Improper cleaning in multi pass welding
    4. Poor joint selection








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    Spotter - Arc blow
    Distortion - Poor joint selection
    Slag Inclusion - Improper cleaning in multipass welding
    Porosity - Damp electrodes

    Correct Option: D

    Spotter - Arc blow
    Distortion - Poor joint selection
    Slag Inclusion - Improper cleaning in multipass welding
    Porosity - Damp electrodes

  1. The current in Amperes used in resistance spot welding of plain carbon steel sheets (1 to 3 mm thick) lies within the range








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    Very high current is used in air gap.

    Correct Option: D

    Very high current is used in air gap.

  1. Consider the following statements : The size of the heat affected zone (HAZ) will increase with
    1. Increased starting temperature
    2. Increased welding speed
    3. Increased thermal conductivity of the base metal
    4. Increase in base metal thickness
    Which of the statements given above are correct?








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    H A Z increases due to higher temperature & higher heat dissipation due to higher thermal conductivity

    Correct Option: B

    H A Z increases due to higher temperature & higher heat dissipation due to higher thermal conductivity

  1. A workpiece of 2000 mm length and 300 mm width was machined by a planning operation with the feed set at 0.3 mm/stroke. If the machine tool executes 10 double strokes/min, the planning time for a single pass will be








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    l = 2000 mm
    w = 300 mm
    f = 0.3 mm/ stroke
    No. of stroke = 2 × 10
    = 20 stroke/ min {double stroke}

    Total Machining time =
    2000
    		 = 100 min
    20

    Correct Option: B

    l = 2000 mm
    w = 300 mm
    f = 0.3 mm/ stroke
    No. of stroke = 2 × 10
    = 20 stroke/ min {double stroke}

    Total Machining time =
    2000
    		 = 100 min
    20

  1. In a cutting test with 0.3 mm flank wear as tool failure criterion, a tool life of 10 min was obtained at a cutting velocity of 20 m/min. Taking tool life exponent as 0.25, the tool life in minutes at 40 m/min of cutting velocity will be








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    n = 0.25 V1 = 20m /min,
    T1 = 10 min V2 = 40m/min, T2 =?
    V1Tn1 = V2Tn2 = C
    20 (T0)0.25 = 40 (T2)0.25
    T2 = 0.625 min

    Correct Option: B

    n = 0.25 V1 = 20m /min,
    T1 = 10 min V2 = 40m/min, T2 =?
    V1Tn1 = V2Tn2 = C
    20 (T0)0.25 = 40 (T2)0.25
    T2 = 0.625 min