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. In an engineering drawing one finds the designation of 20G7f8, the position of tolerance of the hole is indicated by








  1. View Hint View Answer Discuss in Forum

    Capital letter for Hole → G
    Small letter for shaft → f
    7 for → IT7
    8 for → IT8
    IT → Tolerance zone width

    Correct Option: A

    Capital letter for Hole → G
    Small letter for shaft → f
    7 for → IT7
    8 for → IT8
    IT → Tolerance zone width

  1. The hole 40+0.020  +0.000 and shaft, 40+0.010  +0.000 when assembled will result in








  1. View Hint View Answer Discuss in Forum

    Correct Option: C

Direction: A disc of 200 mm outer and 80 mm inner diameter is faced at a feed of 0.1 mm/rev with a depth of cut of 1 mm. The facing operation is undertaken at a constant cutting speed of 90 m/min in a CNC lathe. The main (tangential) cutting force is 200 N.

  1. Neglecting the contribution of the feed force towards cutting power, the specific cutting energy in J/mm3 is








  1. View Hint View Answer Discuss in Forum

    Specific cutting energy (SCE) =
    FC
    		 =
    200
    		 = 2 J / mm3
    d × f0.1 × 1 × 1000

    Correct Option: B

    Specific cutting energy (SCE) =
    FC
    		 =
    200
    		 = 2 J / mm3
    d × f0.1 × 1 × 1000

  1. Two cutting tools are being compared for a machining operation. The tool life equations are
    Carbide tool VT1.6 = 3000
    HSS tool VT0.6 = 200
    where V is the cutting speed in m/min and T is the tool life in min. The carbide tool will provide higher tool life if the cutting speed in m/min exceeds








  1. View Hint View Answer Discuss in Forum

    Carbide tool: VT1.6 = 3000
    HSS tool: VT 0.6 = 200

    T =
    3000
    		1 /1.6
    V

    T =
    200
    		1 /0.6
    V

    T =
    3000
    		1 /1.6 =
    200
    		1 /0.6
    VV

    V = 39.389 m/min

    Correct Option: B

    Carbide tool: VT1.6 = 3000
    HSS tool: VT 0.6 = 200

    T =
    3000
    		1 /1.6
    V

    T =
    200
    		1 /0.6
    V

    T =
    3000
    		1 /1.6 =
    200
    		1 /0.6
    VV

    V = 39.389 m/min

  1. Assuming approach and over-travel of the cutting tool to be zero, the machining time in min is








  1. View Hint View Answer Discuss in Forum

    t =
    π( d02 - d12 )
    		 =
    π[ (0.2)2 - (0.08)2 ]
    4.fr.V4 × 0.1 × 10-3 × 90

    time, t = 2.93 minutes

    Correct Option: A

    t =
    π( d02 - d12 )
    		 =
    π[ (0.2)2 - (0.08)2 ]
    4.fr.V4 × 0.1 × 10-3 × 90

    time, t = 2.93 minutes