The least count of a screw gauge can be decreased by

increasing the total number of divisions on the circular scale

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decreasing the total number of divisions on the circular scale

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breaking the screw gauge into two

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constructing the gauge with a mixture of iron and lead

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Solution

The correct option is A increasing the total number of divisions on the circular scale
For a screw gauge,
$L.C. = \frac{Pitch of the screw}{Circular scale divisions}$

Hence, the least count of a screw gauge can be decreased by decreasing the pitch or increasing the total number of divisions on the circular scale.

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