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Question

A screw gauge has50 divisions on its circular scale. The circular scale is4units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively:


A

Positive,0.1mm

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B

Positive,0.1μm

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C

Positive,10μm

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D

Negative,2μm

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Solution

The correct option is C

Positive,10μm


Step 1. Given data :

Displacement =0.5mm

Screw gauge divisions = 50

Step 2. Find the least count of the screw gauge:

LeastCount=pitchno.ofdivisions

=(0.550)mm=1×10-5m=10μm

And the nature of the zero error is positive because it is4 units ahead of pitch scale marking.

Hence, option (C) is correct.


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