Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

  1. A displacement sensor (a dial indicator) measures the lateral displacement of a mandrel mounted on the taper hole inside a drill spindle. The mandrel axis is an extension of the drill spindle taper hole axis and the protruding portion of the mandrel surface is perfectly cylindrical. Measurements are taken with the sensor placed at two positions P and Q as shown in the figure.
    The readings are recorded as Rx = maximum deflection minus minimum deflection, corresponding to sensor position at X, over one rotation.

    If Rp = RQ > o, which one the following would be consistent with the observation?









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    Equal deflect i on i n bot h sensor s means eccentricity parallel to the axis.

    Correct Option: C

    Equal deflect i on i n bot h sensor s means eccentricity parallel to the axis.


  1. A ring gauge is used to measure









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    Ring gauge is used to measure both outside diameter and roundness.

    Correct Option: C

    Ring gauge is used to measure both outside diameter and roundness.



  1. Match the following
    Feature to be inspected
    P. Pitch and Angle errors of screw thread
    Q. Flatness error of a surface
    R. Alignment error of a machine sideways
    S. Profile of a cam
    Instrument
    1. Auto Collimator
    2. Optical interferometer
    3. Dividing Head and Dial Gauge
    4. Spirit Level
    5. Sine bar
    6. Tool maker's Microscope









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    P-6, Q-4, R-1, S-3

    Correct Option: C

    P-6, Q-4, R-1, S-3


  1. Two slip gauges of 10 mm width measuring 1.000 mm and 1.002 mm are kept side by side in contact with each other lengthwise. An optical flat is kept resting on the slip gauges as shown in the figure. Mono-chromatic light of wavelength 0.0058928 mm is used in the inspection. The total number of straight fringes that can be observed on both slip gauges is









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    ∆h =
    N
    ×
    λ
    22

    ∆h = 1.002 – 1.00 = 0.002 mm
    λ = 0.0058928 mm
    N =
    4 × ∆h
    =
    4 × 0.002
    λ0.0058928

    N = 1.357 ≃ 2

    Correct Option: A

    ∆h =
    N
    ×
    λ
    22

    ∆h = 1.002 – 1.00 = 0.002 mm
    λ = 0.0058928 mm
    N =
    4 × ∆h
    =
    4 × 0.002
    λ0.0058928

    N = 1.357 ≃ 2



  1. A threaded nut of M 16, ISO metric type, having 2 mm pitch with a pitch diameter of 14.701 mm is to be checked for its pitch diameter using two or three number of balls or rollers of the following sizes









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    Metric Thread M16
    Pitch = 2 mm
    Pitch Dia = 14.701
    θ = 30°

    dw =
    P
    sec30° =
    2
    sec30° = 1.155 mm
    22

    Correct Option: B

    Metric Thread M16
    Pitch = 2 mm
    Pitch Dia = 14.701
    θ = 30°

    dw =
    P
    sec30° =
    2
    sec30° = 1.155 mm
    22