One main scale division of a Vernier calipers is 'a' cm and $n^{th}$ division of the Vernier scale coincides with $(n-1)^{th}$ division of the main scale. The least count of the calipers in mm is:
(A) $ \frac {10na}{(n-1)} $
(B) $ \frac {10a}{(n-1)}$
(C) $ \frac {10a}{n} $
(D) $ (\frac {n-1}{10n})a$
Solution
We have, 1 main scale division MSD = a
1 VSD $= \frac {(n-1)a}{n} $
Least Count = 1 MSD - 1 VSD $= a - \frac {(n-1)a}{n} $
So, least count $= \frac {a}{n}$ cm $= \frac {10a}{n}$ mm
Answer: (C)
