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

Options

A. (a) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gayge is `0.01mm`
B. (b) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is `0.005mm`
C. ( c ) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is `0.01mm`
D. (d) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is `0.005mm`

Answer

B, C

Explanation

(b,c )
Vernier callipers
`1MSD = (1cm)/(8) = 0.125cm`
`5 VSD = 4MSD`
:. `5VSD = 4 xx (1)/(8)cm = 0.5cm`
:. `1 VSD = 0.1cm`
`= 1VSD = 0.1cm`
`L.C = 1MSD - 1VSD`
`=0.0125cm - 0.1cm`
`=0.025cm`
Screw gauge
One complete revolution = 2M.S.D`
If the pitch of screw gauge is twice the `L.C` of vernier callipers then pitch `=2 xx 0.025 =0.05 cm`.
`L.C` of screw Gauge
`= (pitch)/(Total no. of divisions of circular scale)`
`=(0.05)/(100)cm = 0.0005cm = 0.005mm`.
(b) is a correct option
Now if the least count of the linear scale of the screw gauge is twice the least count of vernier callipers then.
`L.C` of linear scale of screw gauge `= 2 xx 0.025 = 0.05 cm`.
Then pitch `= 2 xx 0.05 = 0.1cm`.
Then `L.C of screw gauge = (0.1)/(100)cm = 0.001cm = 0.01mm`.
(c) is a correct option.