A cylindrical pin of 25-0-020-0-010mm diameter is electroplated- and the plating thickness is 2-0- 0-005mm- Neglecting the gauge tolerance- the diameter in mm- up to 3 decimal points accuracy of the GO ring gauge to inspect the plated pin is

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Standard X

Physics

Electric Current

A cylindrical...

Question

A cylindrical pin of $25_{+ 0.010}^{+ 0.020} m m$ diameter is electroplated, and the plating thickness is ${2.0}^{\pm 0.005} m m$ . Neglecting the gauge tolerance, the diameter (in mm, up to 3 decimal points accuracy) of the GO ring gauge to inspect the plated pin is
29.03

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Solution

The correct option is A 29.03
GO ring gauge will inspect maximum metal conditions i.e. UL of shaft i.e. Largest size after platting.

$UL after plating$

$= 25.02 + 2.005 + 2.005 m m$

$= 29.030 m m$

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A cylindrical pin of 25+0.020+0.010mm diameter is electroplated, and the plating thickness is 2.0±0.005mm. Neglecting the gauge tolerance, the diameter (in mm, up to 3 decimal points accuracy) of the GO ring gauge to inspect the plated pin is29.03

The correct option is A 29.03GO ring gauge will inspect maximum metal conditions i.e. UL of shaft i.e. Largest size after platting. UL after plating =25.02+2.005+2.005mm =29.030mm

A cylindrical pin of $25_{+ 0.010}^{+ 0.020} m m$ diameter is electroplated, and the plating thickness is ${2.0}^{\pm 0.005} m m$ . Neglecting the gauge tolerance, the diameter (in mm, up to 3 decimal points accuracy) of the GO ring gauge to inspect the plated pin is
29.03

The correct option is A 29.03
GO ring gauge will inspect maximum metal conditions i.e. UL of shaft i.e. Largest size after platting.

$UL after plating$

$= 25.02 + 2.005 + 2.005 m m$

$= 29.030 m m$