How can the least count of a screw gauge be increased ?

Increasing number of divisions on circular scale

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Increasing number of divisions on linear scale

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Increasing pitch of screw gauge

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All of the above

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Solution

The correct options are
A Increasing number of divisions on circular scale
B Increasing number of divisions on linear scale
least count =Value of one main scale division / Total number of divisions on vernier
So if we increase number of divisions then least count can be increased.

Option A and B both are correct.

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