In an interchangeable assembly,

$Shafts of size {25.000}_{- 0.0100}^{- 0.040} m m mate with$

$Holes of size {25.000}_{- 0.000}^{+ 0.020} m m .$

The maximum possible clearance in the assembly will be

30 microns

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 microns

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 microns

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 microns

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 60 microns
$Minimum shaft size = 25.000 - 0.04 = 24.96 m m$

$Maximum hole size = 25.000 + 0.02 = 25.02 m m$

$∴ Maximum possible clearance = Maximum size of hole - Minimum size of shaft$

$= 25.02 - 24.96$

$= 0.060 mm = 60 microns$

Suggest Corrections

Similar questions

{25.000}^{\begin{matrix} + 0.040 - 0.010 \end{matrix}} m m

{25.000}^{\begin{matrix} + 0.030 - 0.020 \end{matrix}} m m

Holes of diameter {25.0}_{+ 0.020}^{+ 0.040} m m

are assembled interchangeably with

The pins of diameter 25_{- 0.008}^{+ 0.005} m m .

The minimum clearance in the assembly will be

A shaft of {20.00}_{- 0.005}^{+ 0.015} m m diameter

A hole of {20.00}_{+ 0.025}^{- 0.010} m m diameter

Consider a hole - shaft interchangeable assembly as 40 H \begin{matrix} + 0.02 0.000 \end{matrix} j \begin{matrix} + 0.01 - 0.01 \end{matrix} .

The allowance (in microns) in the assembly is

Join BYJU'S Learning Program