Holes of diameter 25+0.020+0.040 mm are assembled interchangeably with the pins of diameter 25−0.008+0.005 mm. The minimum clearance in the assembly will be
0.048 mm
0.015 mm
0.005 mm
0.008 mm
Correct Option: B
Minimum clearance ⇒ minimum hole – maximum shaft = 25 + .020 – 25 + .005 = 0.015 mm
