Metrology and Inspection

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Question 1

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).

A	1.12
B	2.31
C	4.25
D	3.36

GATE ME 2022 SET-2 Manufacturing Engineering

Question 1 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}

Question 2

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

A	P-3, Q-1, R-4
B	P-1, Q-3, R-2
C	P-1, Q-3, R-4
D	P-3, Q-1, R-2

GATE ME 2022 SET-2 Manufacturing Engineering

Question 2 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

Question 3

The type of fit between a mating shaft of diameter

25.0^{\begin{matrix} +0.010\\ -0.010 \end{matrix}}

mm and a hole of diameter

25.015^{\begin{matrix} +0.015\\ -0.015 \end{matrix}}

mm is __________.

A	Clearance
B	Transition
C	Interference
D	Linear

GATE ME 2022 SET-1 Manufacturing Engineering

Question 3 Explanation:

If,

D_{hole} = 25.00 mm,

D_{shaft} = 25.01 mm

( Interference fit.)

D_{hole} = 25.03 mm,

D_{shaft} = 24.99 mm

(Clearance fit)
Some of the assemblies provide clearance fit and some provides interference fit.
Hence, It is transition fit

Question 4

A 76.2 mm gauge block is used under one end of a 254 mm sine bar with roll diameter of 25.4 mm. The height of gauge blocks required at the other end of the sine bar to measure an angle of

30^{\circ}

is __________mm (round off to two decimal places).

A	203.2
B	145.6
C	214.3
D	542.6

GATE ME 2021 SET-2 Manufacturing Engineering

Question 4 Explanation:

\begin{aligned} \sin 30^{\circ} &=\frac{h_{1}}{254} \\ \text{or}\qquad h_{1} &=254 \sin 30^{\circ}=127 \mathrm{~mm} \\ H_{2} &=76.2+127=203.2 \mathrm{~mm} \end{aligned}

\\ Case II:

\begin{aligned} \sin 30^{\circ} &=\frac{76.2-h_{1}}{254} \\ \text{or}\qquad h_{1} &=76.2-254 \sin 30^{\circ}\\ &=-50.8 \mathrm{~mm} \end{aligned}

Which is not feasible
Therefore final answer 203.2 mm.

Question 5

The allowance provided in between a hole and a shaft is calculated from the difference between

A	lower limit of the shaft and the upper limit of the hole
B	upper limit of the shaft and the upper limit of the hole
C	upper limit of the shaft and the lower limit of the hole
D	lower limit of the shaft and the lower limit of the hole

GATE ME 2021 SET-2 Manufacturing Engineering

Question 5 Explanation:

It is minimum clearance or maximum interference. It is the intentional difference between the basic dimensions of the mating parts. The allowance may be positive or negative.

Question 6

Two smooth identical spheres each of radius 125 mm and weight 100 N rest in a horizontal channel having vertical walls. The distance between vertical walls of the channel is 400 mm.

The reaction at the point of contact between two spheres is _______N (round off to one decimal place).

A	110
B	125
C	258
D	324

GATE ME 2021 SET-1 Manufacturing Engineering

Question 6 Explanation:

\begin{aligned} B C &=250^{2}-150^{2} \\ \cos \theta &=\frac{200}{250} \\ \theta &=36.869^{\circ} \end{aligned}

\begin{aligned} R_{2} \cos \theta &=100 \\ R_{2} &=\frac{100 \times 250}{200}=125 \mathrm{~N} \end{aligned}

Question 7

Consider the surface roughness profile as shown in the figure.

The center line average roughness (

R_a \text{ in }\mu m

) of the measured length (L) is

A	0
B	1
C	2
D	4

GATE ME 2021 SET-1 Manufacturing Engineering

Question 7 Explanation:

R_{G}=\frac{\sum_{i=1}^{n} y}{n}=\frac{4}{4}=1

Question 8

Two rollers of diameters

D_1

(in mm) and

D_2

(in mm) are used to measure the internal taper angle in the V-groove of a machined component. The heights

H_1

(in mm) and

H_2

(in mm) are measured by using a height gauge after inserting the rollers into the same V-groove as shown in the figure.

Which one of the following is the correct relationship to evaluate the angle

\alpha

as shown in the figure?

A	$\sin \alpha =\frac{(D_1-D_2)}{2(H_1-H_2)-(D_1-D_2)}$
B	$\cos \alpha =\frac{(D_1-D_2)}{2(H_1-H_2)-2(D_1-D_2)}$
C	$cosec\; \alpha =\frac{(H_1-H_2)-(D_1-D_2)}{2(D_1-D_2)}$
D	$\sin \alpha =\frac{(H_1-H_2)}{(D_1-D_2)}$

GATE ME 2020 SET-2 Manufacturing Engineering

Question 8 Explanation:

\begin{aligned} O B &=\left(H_{1}-\frac{D_{1}}{2}\right)-\left(H_{2}-\frac{D_{2}}{2}\right)\\ &=\left(H_{1}-H_{2}\right)-\left(\frac{D_{1}-D_{2}}{2}\right) \\ O A &=\frac{D_{1}-D_{2}}{2} \\ \sin \alpha &=\frac{\frac{\left(D_{1}-D_{2}\right)}{2}}{\left(H_{1}-H_{2}\right)-\frac{\left(D_{1}-D_{2}\right)}{2}} \\ \sin \alpha &=\frac{\left(D_{1}-D_{2}\right)}{2\left(H_{1}-H_{2}\right)-\left(D_{1}-D_{2}\right)} \end{aligned}

Question 9

The figure below shows a symbolic representation of the surface texture in a perpendicular lay orientation with indicative values (I through VI) marking the various specifications whose definitions are listed below.

P: Maximum Waviness Height (mm);
Q: Maximum Roughness Height (mm);
R: Minimum Roughness Height (mm);
S: Maximum Waviness Width (mm);
T: Maximum Roughness Width (mm);
U: Roughness Width (mm);