Materials Science and Manufacturing Engineering Miscellaneous Difficult Questions and Answers | Page - 8

Materials Science and Manufacturing Engineering Miscellaneous

In a wire-cut EDM process the necessary conditions that have to be met for making a successful cut are that

1. wire and sample are electrically nonconducting
1. wire and sample are electrically conducting
1. wire is electrically conducting and sample is electrically non-conducting
1. sample is electrically conducting and wire is electrically non-conducting

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In EDM or wirecut EDM, the work and tool must be electrically conductive otherwise the current passage will not takes place.

Correct Option: B

In EDM or wirecut EDM, the work and tool must be electrically conductive otherwise the current passage will not takes place.

Observation of a slip gauge on flatness interferometer produced fringe counts numbering 10 and 14 for two readings.The second reading is taken by rotating the setup by 180°. Assume that both faces of the slip gauge are flat and the wavelength of the radiation is 0.5086 μm. The parallelism error (in μm) between the two faces of the slip gauge is

1. 0.2543
1. 1.172
1. 0.5086
1. 0.800

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Parallelism
= n₂ - n₁ × λ = 14 - 10 × 0.5086
2 2 2 2

= 0.5086 µm.

Correct Option: C

Parallelism
= n₂ - n₁ × λ = 14 - 10 × 0.5086
2 2 2 2

= 0.5086 µm.

To measure the effective diameter of an external metric thread (included angle is 60°) of 3.5 mm pitch, a cylindrical standard of 30.5 mm diameter and two wires of 2 mm diameter each are used. The micrometer reading over the standard and over the wires are 16.532 mm and 15.398 mm respectively. The effective diameter (in mm) of the thread is

1. 33.336
1. 30.397
1. 29.366
1. 26.397

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NA

Correct Option: B

NA

A small bore is designated 25H7. The lower (minimum) and upper (maximum) limits of the . bore are 25.000 mm and 25.021 respectively. When the bore is designated as 25H8, then the upper limit is 25.033 mm. When the bore is designated as 25H6, then the upper limit of the bore in mm is

1. 25.001
1. 25.005
1. 25.009
1. 25.013

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i = 0.45³√D + 0.0011)
= 0.45 ³√23.24 + 0.001 (23.24)
i = 1.31mm
IT7 = 16i = 16 × 1.31 = 21 µm
IT6 = 10i = 10 × 1.31 = 13.1µm
upper limit for Hole 25H₆ = 250.013mm

Correct Option: D

i = 0.45³√D + 0.0011)
= 0.45 ³√23.24 + 0.001 (23.24)
i = 1.31mm
IT7 = 16i = 16 × 1.31 = 21 µm
IT6 = 10i = 10 × 1.31 = 13.1µm
upper limit for Hole 25H₆ = 250.013mm

The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit i (microns) = 0.45 (D^1/3) + 0.001 D. The unit of D is mm. Diameter step is 18 – 30. If the fundamental deviation for hole H is zero and IT8 = 26i, the maximum and minimum limits of dimension for a 25H8 hole (in mm) are

1. 24.984 and 24.967
1. 25.017 and 24.984
1. 25.033 and 25.00
1. 25.00, 24.967

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D = √18 &time; 30 = 23.24mm
i = 0.45³√D + 0.001D
= 0.45 ³√23.24 + 0.001 (23.24) = 1.307 mm
IT8 = 26i = 26 × 1.307 ≈ 33.98 34 m = 0.034mm
25H8 = 250.00 + 0.034mm.

Correct Option: C

D = √18 &time; 30 = 23.24mm
i = 0.45³√D + 0.001D
= 0.45 ³√23.24 + 0.001 (23.24) = 1.307 mm
IT8 = 26i = 26 × 1.307 ≈ 33.98 34 m = 0.034mm
25H8 = 250.00 + 0.034mm.