In a vernier calliper N divisions of vernier scale coincides with (N - 1) divisions of main scale (in which length of one division is 1mm). The least count of the instrument should be

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N-1

1/10N

1/N-1

answer is C.

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Detailed Solution

Least count= IMSD - 1 VSD

$\begin{array}{l} = 1 MSD - (\frac{N - 1}{N}) MSD \\ (∴ NVSD = (N - 1) MSD ∴ 1 VSD = \frac{N - 1}{N} MSD) \\ = \frac{1}{N} MSD = \frac{1}{N} \times \frac{1}{10} cm = \frac{1}{10 N} \end{array}$

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