Materials Science and Manufacturing Engineering Miscellaneous
- 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
- 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 2 2
∆h = 1.002 – 1.00 = 0.002 mm
λ = 0.0058928 mmN = 4 × ∆h = 4 × 0.002 λ 0.0058928
N = 1.357 ≃ 2Correct Option: A
∆h = N × λ 2 2
∆h = 1.002 – 1.00 = 0.002 mm
λ = 0.0058928 mmN = 4 × ∆h = 4 × 0.002 λ 0.0058928
N = 1.357 ≃ 2
- 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 2 2 Correct Option: B
Metric Thread M16
Pitch = 2 mm
Pitch Dia = 14.701
θ = 30°dw = P sec30° = 2 sec30° = 1.155 mm 2 2
- A grinding ratio of 200 implies that the
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Grinding ratio is defined as
= Volume of work material removed (Vm) Volume of wheel wear (Vw) 200 = Volume of work material removed (Vm) Volume of wheel wear (Vw)
Grinding wheel wears 0.005 times the volume of material removed.Correct Option: B
Grinding ratio is defined as
= Volume of work material removed (Vm) Volume of wheel wear (Vw) 200 = Volume of work material removed (Vm) Volume of wheel wear (Vw)
Grinding wheel wears 0.005 times the volume of material removed.
- Using the Taylor's tool life equation with exponent n = 0.5, if the cutting speed is reduced by 50%, the ratio of new tool life to original tool life is
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V2 = V1 2
V1T10.5 = V2T20.5V1T10.5 = V1 T20.5 2 ∴ T2 0.5 = 2 T1 ∴ T2 = 21/0.5 = 4 T1 Correct Option: A
V2 = V1 2
V1T10.5 = V2T20.5V1T10.5 = V1 T20.5 2 ∴ T2 0.5 = 2 T1 ∴ T2 = 21/0.5 = 4 T1