Question

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.

Answer 1

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`