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

Explanation

Here, no of divisions on circular scale =100
no. of complete rotations = 4
distacne moved =2mm.
` Pitch = ("distance moved")/("no. of complete rotations") = (2mm)/(4)`
0.5mm
Least count =` ("Pitch")/("no. ofn divisions on circular scale")`
`=(0.5mm)/(100) = 0.005mm`