# Manufacturing Engineering

 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)

 A 5 B 9 C 10 D 12
GATE ME 2023      Metrology and Inspection
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)

 A 5000 B 6000 C 8000 D 9000
GATE ME 2023      Machining and Machine Tool Operation
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)

 A 4 B 5 C 6 D 8
GATE ME 2023      Metrology and Inspection
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 $5 \mathrm{~mm}$.
 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)

 A 12.4 B 16.5 C 18.6 D 24.4
GATE ME 2023      Metrology and Inspection
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 $\bar{C}+6 \times \sigma _C=15+6 \times 0.2645=16.587$
 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)

 A 0.54 B 0.35 C 0.85 D 0.95
GATE ME 2023      Engineering Materials
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.

### 1 thought on “Manufacturing Engineering”

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