Consider a Vernier callipers in which each `1cm` on the main scale is divided into `8` equal divisions and a screw gauge with `100` divisions in its circular scale. In the Vernier callipers, `5` divisions of the Vernier scale coincide with `4` divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

Explanation

For Vernier calipers
`1MSD = (1)/(8) cm`
`5 VSD = 4MSD`
`1 VSD = (4)/(5) MSD = (4)/(5) MSD = (4)/(5) xx (1)/(8) = (1)/(10)cm`
`LC` of vernier calliper `= (1)/(8) cm - (1)/(10) cm = 0.025cm`
`(A) & (B)`
pitch of screw gauger `=2 xx (0.025) = 0.05 cm`
Leastcount of screw gague `= (0.05)/(100) cm = 0.005 mm`
`(C) & (D)` Least count of linear scale of screw gague `= 0.05 cm`
pitch `= 0.05 xx 2cm = 0.1 cm`
Leastcount of crew gauge `= (0.1)/(100) cm = 0.01 mm` .