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In Vernier calipers, m divisions of main scale coincide with (m + 1) divisions of vernier scale. If each division of main scale is d units, the least count of the instrument is

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$\frac{d}{m - 1}$

$\frac{d}{m + 1}$

$\frac{d}{m}$

$\frac{md}{m + 1}$

answer is B.

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Detailed Solution

Since m divisions of main scale is equivalent to (m + 1) divisions of vernier scale, one division of vernier scale is equivalent to $\frac{m}{m + 1}$ divisions of main scale.
So, L.C. = 1 division of M.S. - 1 division of V.S.

L.C = $d - \frac{m}{m + 1} d = \frac{1}{m + 1} d$

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