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Example · Example 7

Q.A screw gauge has a pitch of 11 mm and a circular scale with 100100 divisions. Calculate its least count. A wire's diameter is then read as main scale =2= 2 mm, circular scale =45= 45 divisions, with a zero error of −2-2 divisions. Find the true diameter of the wire.

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Step 1 — least count.

Least count=PitchNumber of circular divisions=1 mm100=0.01 mm\text{Least count} = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{1\ \text{mm}}{100} = 0.01\ \text{mm}

Step 2 — observed reading.

Observed reading=Main scale reading+(Circular scale reading×Least count)\text{Observed reading} = \text{Main scale reading} + (\text{Circular scale reading}\times\text{Least count})

=2 mm+(45×0.01 mm)=2 mm+0.45textmm=2.45 mm= 2\ \text{mm} + (45\times0.01\ \text{mm}) = 2\ \text{mm} + 0.45\\text{mm} = 2.45\ \text{mm} …

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