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 solid part (see figure) of polymer material is to be fabricated by additive
manufacturing (AM) in square-shaped layers starting from the bottom of the part
working upwards. The nozzle diameter of the AM machine is a/10 mm and the
nozzle follows a linear serpentine path parallel to the sides of the square layers
with a feed rate of a/5 mm/min.

Ignore any tool path motions other than those involved in adding material, and any other delays between layers or the serpentine scan lines.

The time taken to fabricate this part is _______ minutes. (Answer in integer)

Ignore any tool path motions other than those involved in adding material, and any other delays between layers or the serpentine scan lines.

The time taken to fabricate this part is _______ minutes. (Answer in integer)

5000 | |

6000 | |

8000 | |

9000 |

Question 2 Explanation:

\begin{aligned}
V&=(3a)^2 \times 1.5a+ (2a)^2 \times a+(a)^2 \times 0.5a\\
&= 18a^3\\
T&=\frac{V}{\left ( \frac{a}{10} \right )^2 \times \left ( \frac{a}{5} \right )} \\
&=\frac{18a^3}{a^3} \times 100 \times 5=9000 min
\end{aligned}

Question 3 |

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 3 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 4 |

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 4 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 5 |

A steel sample with 1.5 wt. % carbon (no other alloying elements present) is
slowly cooled from 1100^{\circ}C to just below the eutectoid temperature (723^{\circ}C). A
part of the iron-cementite phase diagram is shown in the figure. The ratio of the
pro-eutectoid cementite content to the total cementite content in the
microstructure that develops just below the eutectoid temperature is ________.

(Rounded off to two decimal places)

(Rounded off to two decimal places)

0.54 | |

0.35 | |

0.85 | |

0.95 |

Question 5 Explanation:

Mass friction of proeutectoid cementite \mathrm{W}_{\mathrm{Fe}_{3}} C^{\prime}=\frac{1.5-0.8}{6.7-0.8}=0.1186

Mass fraction of total cementite

\mathrm{W}_{\mathrm{Fe}_{3}} \mathrm{C}=\frac{1.5-0.035}{6.7-0.035}=0.2197

So, \quad \frac{W_{F_{3}} C^{\prime}}{W_{F_{e 3}} C}=\frac{0.1186}{0.2197}=0.54

There are 5 questions to complete.

question number -12 is mostly to engineering mechanics than metrology, manufacturing