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Standard VII
History
Tippu Challenges The British Rulers
Tolerance of ...
Question
Tolerance of the dimension
x
in the two component assembly shown below in (all dimensions are in mm)
A
±
0.040
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B
±
0.045
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C
±
0.025
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D
±
0.030
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Solution
The correct option is
B
±
0.045
X
m
a
x
=
50.02
−
(
37.985
+
9.99
)
−
2.045
m
m
X
m
i
n
=
49.98
−
(
38.015
+
10.0
)
=
1.955
m
m
Tolerance
=
X
m
a
x
−
X
m
i
n
=
2.045
−
1.955
=
0.09
m
m
X
=
2
±
0.045
m
m
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3
Similar questions
Q.
In the assembly shown below, the part dimensions are :
L
1
=
22.0
±
0.01
mm
,
L
2
=
L
3
=
10.0
±
0.005
mm
,
Assuming the normal distributions of part dimensions, the dimension
L
4
in mm for assembly condition would be,
Diagram 1
Q.
In the assembly shown below, the part dimensions are:
L
1
=
22.0
±
0.01
m
m
L
2
=
L
3
=
10.0
±
0.005
m
m
Assuming the normal distribution of part dimensions, the dimension
L
4
in mm for assembly conditions would be:
Q.
GO and GO-NO plug gauges are designed for measuring a hole of nominal diameter 25 mm with a hole tolerance of
±
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m
. Considering 10% of work tolerance to be the gauge tolerance and no wear condition, the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system is
Q.
The following data are given for calculating limits of dimension and tolerances for a hole: Tolerance unit
i
=
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(
D
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/
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)
+
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.
The unit of D is mm. Diameter step is
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30.
If the fundamental deviation for hole H is zero and
IT8=26i
, the maximum and minimum limits of dimension for a
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Q.
A hole and mating shaft are to have a nominal assembly size of 70 mm. The assembly is to have maximum clearance of 0.2 mm and minimum clearance of 0.1 mm and hole tolerance is 1.5 times the shaft tolerance. The tolerance on hole, based on basic hole standard, is
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