question_answer

When circular scale of a screw gauge carrying 100 divisions is given four complete rotations, the head of the screw moves through 2 mm. Calculate pitch and least count of the screw gauge.

Answer:

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