Question

$\text{Holes of diameter}{25.0}_{+0.020}^{+0.040}mm$
$\text{are assembled interchangeably with}$

$\text{The pins of diameter}{25}_{-0.008}^{+0.005}mm\text{.}$

$\text{The minimum clearance in the assembly will be}$

• A. 0.048 mm
• B. 0.015 mm
• C. 0.005 mm
• D. 0.008 mm

Answer 1

The correct option is B 0.015 mm

$\text{Minimum clearances}=(\text{Hole}{)}_{\text{minimum}}-(\text{Shaft}{)}_{\text{maximum}}$

$=25.02-25.005=0.015mm$