In an interchangeable assembly, shafts of size 25.000 -0.0100⁺⁰∙⁰⁴⁰ mm mate with holes of size 25.000 -0.010⁺⁰∙⁰²⁰ mm. The maximum possible clearance in the assembly will be
A. 10 microns
B. 20 microns
C. 30 microns
D. 60 microns
Answer: Option C
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The correct answer is: 60 microns
Max shaft = 24.99 mm
Min shaft = 24.96 mm
Max Hole = 25.02 mm
Min Hole 25.01 mm
Min. clearance = Min. Hole - Max Shaft = 20 microns
Max. clearance = Max Hole - Minimum shaft = 60 microns
From the above que clearance is only 20microns how is it possible for 30microns its the wrong ans know
Maximum possible clearance = Hole max size - Shaft min size = 25.02 -24.99 = 0.03 (30 microns)
Here 25.02 = 25 + 0.02 (max tolerance value) &
24.99 = 25 - 0.01 (min tolerance value)
25.02-24.99= .03
This is interference fit. Max upper limit - Min lower limit gives clearance. So, 0.04-0.01 = 0.03 mm = 30 microns.
Please show the procedure
nice question.