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1 cm of main scale of a vernier calliper...

`1 cm` of main scale of a vernier callipers is divided into `10` divisions. The least count of the callipers is `0.05 cm`, then the varnier scale must have

A

`10` divisions

B

`20` divisions

C

`25` divisions

D

`50` divisions

Text Solution

Verified by Experts

The correct Answer is:
B
Let Vernier scale has `N` divisions.
Then: `(N)VSD=(10)MSD`
`1VSD=((10)/(N))MSD`
`LC=1MSD-1VSD`
`=(1-(10)/(N))MSD`
or `(0.025cm)=(1-(10)/(N))((1cm)/(10))`
Solving we get,
`N=20`
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