Answer:
Here,
no. of divisions on circular scale = 100
no. of
complete rotations =4
distance
moved = 2 mm.
\[Pitch=\frac{dis\tan
ce\,moved}{no.\,of\,complete\,rotations}=\frac{2mm}{4}=0.5mm\]
\[Least\,count=\,\frac{Pitch}{no.\,of\,divisions\,on\,circular\,scale}=\frac{0.5mm}{100}=0.0005mm\]
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