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Consider a Vernier callipers in which ea...

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:

A

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is `0.01mm`

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

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

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`

Text Solution

Verified by Experts

The correct Answer is:
B, C
Vernier callipers
`1MSD=(1cm)/(8)=0.125cm`
`5VSD=4MSD`
`:. 1VSD=0.1cm`
`L.C=1MSD-1VSD`
`=0.125cmM-0.1cm`
Screw gauge
One complete revoltuion `=2M.S.D`
If the pitch of screw gauge is twice the `L.C` of vernier callipers then pitch `=2xx0.025=0.05cm`
`L.C` of screw Gauge
`=("pitch")/("Total no. of division of cicrularscale")`
`=(0.05)/(100)cm=0.005cm=0.005mm`
(`b`) is correct option
Now if the least count of the linear scale of the screw gauge is twice the least count of venier callipers then,
`L.C` of linear scale of screw gauge
`=2xx0.025=0.05cm`
Then pitch `=2xx0.05=0.1cm`
Then `L.C` of screw gauge
`=(0.1)/(100)cm=0.001cm=0.01mm`
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