Physics, asked by javedkhan6511, 1 year ago

A screw guage has a pitch of 1mm and 100 division on circular scale calulate and write it's least count

Answers

Answered by lajjachhatwani

1

Least count=pitch÷no.of divisions

So LC=1mm÷100div.

=0.01

Answer=0.01is the least count of the screwgauge......

Hope it helps

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