$A hole 40 K \begin{matrix} + 0.004 - 0.001 \end{matrix} has to be inspected by using GO/NO GO gages.$
$The sizes of the GO and NO GO plug gages, respectively will be ( in mm)$

40.004, 39.996

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40.001, 40.004

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39.999, 40.004

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40.004, 39.999

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Solution

The correct option is C 39.999, 40.004
$GO gage size = maximum material limit of$

$hole = minimum size hole$

$= 40 - 0.001 = 39.999 mm$

$NO GO gage size = minimum material limit$

$of hole = maximum size hole$

$= 40 + 0.004 = 40.004 mm$

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A hole 40 K+0.004−0.001 has to be inspected by using GO/NO GO gages. The sizes of the GO and NO GO plug gages, respectively will be ( in mm)

The correct option is C 39.999, 40.004GO gage size=maximum material limit of hole=minimum size hole =40−0.001=39.999 mm NO GO gage size = minimum material limit of hole = maximum size hole =40+0.004=40.004 mm

$A hole 40 K \begin{matrix} + 0.004 - 0.001 \end{matrix} has to be inspected by using GO/NO GO gages.$
$The sizes of the GO and NO GO plug gages, respectively will be ( in mm)$

The correct option is C 39.999, 40.004
$GO gage size = maximum material limit of$

$hole = minimum size hole$

$= 40 - 0.001 = 39.999 mm$

$NO GO gage size = minimum material limit$

$of hole = maximum size hole$

$= 40 + 0.004 = 40.004 mm$