A circular hole of 30-0-060-0-030 mm diameter is electroplated with the plating thickness as 500-10-10 microns-Neglecting the gauge tolerance- the diameter in mm of the GO plug gauge is

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Standard XII

Chemistry

Magnetic Properties

A circular ho...

Question

$A circular hole of 30 \begin{matrix} + 0.060 + 0.030 \end{matrix} mm diameter is electroplated with the plating thickness as 500 \begin{matrix} + 10 - 10 \end{matrix} microns.$
$Neglecting the gauge tolerance, the diameter (in mm) of the GO plug gauge is$

29.01

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31.08

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29.52

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31.05

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Solution

The correct option is A 29.01
The size of GO gauge corresponds to the maximum material limit of hole.

$Maximum material limit of hole =lower$

$limit of hole =30.03 mm$

$Maximum material limit of plating =510$

$μ m = 0.510 mm$

$∴ Diameter of GO gauge$

$= 30.03 - 2 \times 0.510 = 29.01 mm$

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A circular hole of 30+0.060+0.030 mm diameter is electroplated with the plating thickness as 500+10−10 microns. Neglecting the gauge tolerance, the diameter (in mm) of the GO plug gauge is

The correct option is A 29.01The size of GO gauge corresponds to the maximum material limit of hole. Maximum material limit of hole =lower limit of hole =30.03 mm Maximum material limit of plating =510 μm=0.510 mm ∴ Diameter of GO gauge =30.03−2×0.510=29.01 mm

$A circular hole of 30 \begin{matrix} + 0.060 + 0.030 \end{matrix} mm diameter is electroplated with the plating thickness as 500 \begin{matrix} + 10 - 10 \end{matrix} microns.$
$Neglecting the gauge tolerance, the diameter (in mm) of the GO plug gauge is$

The correct option is A 29.01
The size of GO gauge corresponds to the maximum material limit of hole.

$Maximum material limit of hole =lower$

$limit of hole =30.03 mm$

$Maximum material limit of plating =510$

$μ m = 0.510 mm$

$∴ Diameter of GO gauge$

$= 30.03 - 2 \times 0.510 = 29.01 mm$