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

5

Explanation

Here, no. of divisions on circular scale = 100
No. of complete rotations = 4
Distance moved = 2mm
Pitch `=("Distance moved")/("Number of complete rotations") = (2mm)/4`
= 0.5 mm
Least count = `("Pitch")/("Number of divisions on circular scale")`
`= (0.5mm)/(100) = 0.005 mm implies 5xx10^(-3)` mm.