Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

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Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous
  1. In a CAD package, a point P(6,3,2) is projected along a vector Vrarr; (-2,1,-1). The projection of this point on X-Y plane will be








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    Let X (x, y, o) be the projection of point p (6, 3, 2) The unit vector of line joining point P and X should be same as Vˆ

    where D = √(6 − x)² + (3 − y)² + 2²
    on equating, Px and V, we get

    6 − x
    		 , =
    −2
    		 ,
    3 − y
    		 =
    1
    		,
    2
    		 =
    −1
    D√6D√6D√6

    So D = −2√6
    j = 3 −
    (−2√6)
    		= 5
    √6

    x = 6 −
    (−2 × −2√6)
    		= 2.
    √6

    z = 0

    Correct Option: D


    Let X (x, y, o) be the projection of point p (6, 3, 2) The unit vector of line joining point P and X should be same as Vˆ

    where D = √(6 − x)² + (3 − y)² + 2²
    on equating, Px and V, we get

    6 − x
    		 , =
    −2
    		 ,
    3 − y
    		 =
    1
    		,
    2
    		 =
    −1
    D√6D√6D√6

    So D = −2√6
    j = 3 −
    (−2√6)
    		= 5
    √6

    x = 6 −
    (−2 × −2√6)
    		= 2.
    √6

    z = 0

  1. To obtain dimension of 61.18 mm using slip gauges, the most appropriate combination is








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    Normally, slip gauge box set does not have slip of less than 1 mm thickness.

    Correct Option: A

    Normally, slip gauge box set does not have slip of less than 1 mm thickness.

  1. A 30 mm hole shaft assembly results in minimum and maximum clearances of 0.03 mm and 0.30 mm respectively. The hole has a unilateral tolerance with zero fundamental deviation. If the tolerance on the shaft is 0.08 mm, the maximum size of the hole is








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    Min. clearance
    = Holemin – Shaftmax = 0.03 .......(i)
    Max clearance
    = Holemin – Shaftmax = 0.30 ............(ii)
    Shaftmax – Shaftmin = 0.08 ...............(iii)
    Since, hole has unilateral tolerance,
    Hmin = 30 mm
    From eq. (i), we get
    Shaftmax = 30 – 0.03 = 29.97 min
    From eq. (iii), we get
    Shaftmin = 29.89 mm
    From eq. (ii), we get
    Holemax = 30.19 mm

    Correct Option: C

    Min. clearance
    = Holemin – Shaftmax = 0.03 .......(i)
    Max clearance
    = Holemin – Shaftmax = 0.30 ............(ii)
    Shaftmax – Shaftmin = 0.08 ...............(iii)
    Since, hole has unilateral tolerance,
    Hmin = 30 mm
    From eq. (i), we get
    Shaftmax = 30 – 0.03 = 29.97 min
    From eq. (iii), we get
    Shaftmin = 29.89 mm
    From eq. (ii), we get
    Holemax = 30.19 mm

  1. A shaft of diameter 20−0.10+0.20 mm and a hole of diameter 20+0.20 +0.10mm when assembled would yield.








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    NA

    Correct Option: C

    NA

  1. In an interference microscope, a groove produces a band distortion of 4 band spacing. If the wave length of the monochromatic light source is 0.5 microns, the groove depth is.... mm








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    Groove depth =
    h
    		 .
    λ
    		 =
    4
    		 ×
    0.5
    		 μm = 0.5μm
    2222

    Correct Option: A

    Groove depth =
    h
    		 .
    λ
    		 =
    4
    		 ×
    0.5
    		 μm = 0.5μm
    2222