The circular scale of a screw gauge has `200` divisions. When it is given `4` complete rotations, it moves through `2mm`. The least count of the screw gauge is

Options

A. `0.25 xx 10^(-2) cm`
B. `0.25 xx 10^(-3) cm`
C. 0.001 cm
D. 0.001 mm

Answer

(b)

Explanation

Pitch `= (2)/(4)mm = 0.5 mm`
`L.C = (pitch)/(200) = (0.5)/(200)mm = 0.25 xx 10^(-2) mm`
`=0.25 xx10^(-3) cm`