Question 1 |

An optical flat is used to measure the height difference between a reference slip
gauge A and a slip gauge B. Upon viewing via the optical flat using a
monochromatic light of wavelength 0.5 \mu m , 12 fringes were observed over a
length of 15 mm of gauge B. If the gauges are placed 45 mm apart, the height
difference of the gauges is ______ \mu m .
(Answer in integer)

5 | |

9 | |

10 | |

12 |

Question 1 Explanation:

\begin{aligned}
\lambda & =0.5 \mu \mathrm{m} \\
\mathrm{n} & =12 \text { per } 15 \mathrm{~mm} \\
\theta & =\frac{\lambda}{2 \times \frac{15}{12}}=\frac{0.5}{2 \times 15} \times 12 \\
\mathrm{~h} & =\theta \times 45=\frac{0.5}{2 \times 15} \times 12 \times 45 \\
& =9 \mu \mathrm{m}
\end{aligned}

Question 2 |

A coordinate measuring machine (CMM) is used to determine the distance between Surface
SP and Surface SQ of an approximately cuboidal shaped part. Surface SP is declared as the
datum as per the engineering drawing used for manufacturing this part. The CMM is used
to measure four points P1, P2, P3, P4 on Surface SP, and four points Q1, Q2, Q3, Q4 on
Surface SQ as shown. A regression procedure is used to fit the necessary planes.
The distance between the two fitted planes is _______ mm.
(Answer in integer)

4 | |

5 | |

6 | |

8 |

Question 2 Explanation:

Since y-coordinate of all four points P_{1}, P_{2}, P_{3} and \mathrm{P}_{4} is zero. Hence the regression plane fitted by points P_{1}, P_{2}, P_{3} and P_{4} is x-z plane.

The distance of any point (i.e. \mathrm{Q}_{1}, \mathrm{Q}_{2}, \mathrm{Q}_{3} and \mathrm{Q}_{4} ) from x-z plane is equal to y co-ordinate

Therefore,

The distance of Q_{1} from x-z plan =6 \mathrm{~mm}

The distance of Q_{2} from x-z plan =6 \mathrm{~mm}

The distance of Q_{3} from x-z plan =4 \mathrm{~mm}

The distance of Q_{4} from x-z plan =4 \mathrm{~mm}

The distance of best fitted plane by Q_{1}, Q_{2}, Q_{3} and Q_{4} from x-z plane is equal to average of 6,6,4 and 4 , which is equal to 5 \mathrm{~mm}.

Hence, the distance between two fitted planes is 5 \mathrm{~mm}.

The distance of any point (i.e. \mathrm{Q}_{1}, \mathrm{Q}_{2}, \mathrm{Q}_{3} and \mathrm{Q}_{4} ) from x-z plane is equal to y co-ordinate

Therefore,

The distance of Q_{1} from x-z plan =6 \mathrm{~mm}

The distance of Q_{2} from x-z plan =6 \mathrm{~mm}

The distance of Q_{3} from x-z plan =4 \mathrm{~mm}

The distance of Q_{4} from x-z plan =4 \mathrm{~mm}

The distance of best fitted plane by Q_{1}, Q_{2}, Q_{3} and Q_{4} from x-z plane is equal to average of 6,6,4 and 4 , which is equal to 5 \mathrm{~mm}.

Hence, the distance between two fitted planes is 5 \mathrm{~mm}.

Question 3 |

A part, produced in high volumes, is dimensioned as shown. The machining
process making this part is known to be statistically in control based on sampling
data. The sampling data shows that D1 follows a normal distribution with a mean
of 20 mm and a standard deviation of 0.3 mm, while D2 follows a normal
distribution with a mean of 35 mm and a standard deviation of 0.4 mm. An
inspection of dimension C is carried out in a sufficiently large number of parts.

To be considered under six-sigma process control, the upper limit of dimension C should be ______ mm.

(Rounded off to one decimal place)

To be considered under six-sigma process control, the upper limit of dimension C should be ______ mm.

(Rounded off to one decimal place)

12.4 | |

16.5 | |

18.6 | |

24.4 |

Question 3 Explanation:

D1 and D2 follow normal distribution.

So,

Upper limit =\mu +3\sigma

Lower limit =\mu -3\sigma

\bar{D1}=20mm, \bar{D2}=35mm, \sigma _{D2}=0.4mm,\sigma _{D1}=0.3mm

Mean value of (\bar{C}) =\bar{D2}-\bar{D1}=35-20-15

Standard Devialtion of C

\sigma _C^2=\sigma _{D2}^2 -\sigma _{D1}^2

\sigma _C=\sqrt{0.4^2-0.3^2}=0.2645

Upper Limit of C \bar{C}+6 \times \sigma _C=15+6 \times 0.2645=16.587

So,

Upper limit =\mu +3\sigma

Lower limit =\mu -3\sigma

\bar{D1}=20mm, \bar{D2}=35mm, \sigma _{D2}=0.4mm,\sigma _{D1}=0.3mm

Mean value of (\bar{C}) =\bar{D2}-\bar{D1}=35-20-15

Standard Devialtion of C

\sigma _C^2=\sigma _{D2}^2 -\sigma _{D1}^2

\sigma _C=\sqrt{0.4^2-0.3^2}=0.2645

Upper Limit of C \bar{C}+6 \times \sigma _C=15+6 \times 0.2645=16.587

Question 4 |

The best size wire is fitted in a groove of a metric
screw such that the wire touches the flanks of the
thread on the pitch line as shown in the figure. The
pitch (p) and included angle of the thread are 4 mm
and 60^{\circ} , respectively. The diameter of the best size
wire is ___________ mm (round off to 2 decimal
places).

1.12 | |

2.31 | |

4.25 | |

3.36 |

Question 4 Explanation:

Pitch (p) = 4 mm

Thread angle 2\alpha =60^{\circ}

Semi Thread Angle \alpha =30^{\circ}

Diameter of best size wire

\begin{aligned} d_w&=\frac{p}{2}\sec \alpha \\ d_w&=\frac{4}{2} \sec 30^{\circ}\\ d_w&=2.039 mm=2.31mm \end{aligned}

Thread angle 2\alpha =60^{\circ}

Semi Thread Angle \alpha =30^{\circ}

Diameter of best size wire

\begin{aligned} d_w&=\frac{p}{2}\sec \alpha \\ d_w&=\frac{4}{2} \sec 30^{\circ}\\ d_w&=2.039 mm=2.31mm \end{aligned}

Question 5 |

A shaft of diameter 25^{^{-0.04}}_{-0.07} mm is assembled in a
hole of diameter 25^{^{+0.02}}_{0.00} mm.

Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft-hole assembly.

Allowance and limit parameter (Column I)

P. Allowance

Q. Maximum clearance

R. Maximum material limit for hole

Quantitative value (Column II)

1. 0.09 mm

2. 24.96 mm

3. 0.04 mm

4. 25.0 mm

Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft-hole assembly.

Allowance and limit parameter (Column I)

P. Allowance

Q. Maximum clearance

R. Maximum material limit for hole

Quantitative value (Column II)

1. 0.09 mm

2. 24.96 mm

3. 0.04 mm

4. 25.0 mm

P-3, Q-1, R-4 | |

P-1, Q-3, R-2 | |

P-1, Q-3, R-4 | |

P-3, Q-1, R-2 |

Question 5 Explanation:

(1) Allowance = Lower limit of hole - upper limit of shaft

Allowance = 25.00 - 24.96 = 0.04 mm

(2) Maximum clearance C_{max} = Upper limit of hole - lower limit of shaft

C_{max} = 25.02 - 24.93 = 0.09 mm

(3) Maximum material limit for hole = minimum size of hole = 25.00

There are 5 questions to complete.

Please correct the answer to Q. 29. It’s (a) outside diameter but not roundness.

Thank You Aftaab Alam,

We have updated the answer.