Physics

The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge ?

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As we know,

$\text {Least Count} = \dfrac{\text{Pitch}}{\text {Total no. of div on circular head}} \\[0.5em]$

Given,

Pitch = 0.5 mm

Number of divisions on circular head = 100

Substituting the values in the formula above we get,

$\text {Least count} = \dfrac{0.5}{100} \\[0.5em] \text {Least count} = 0.005 \text { mm} = 0.0005 \text { cm} \\[0.5em]$

Hence, the least count of the screw gauge = 0.005 mm or 0.0005 cm.

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