If n main scale divisions coincide with (n+1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into 6 equal parts, will be.

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$\frac{1}{6 (n + 1)} c m$

$\frac{5}{6 (n + 1)} cm$

$\frac{1}{2 n} cm$

$\frac{1}{5 n} cm$

answer is A.

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Detailed Solution

$n M S D = (n + 1) V S D$

$\begin{array}{r} 1 VSD = (\frac{n}{n + 1}) MSD \\ = (\frac{n}{n + 1}) \times \frac{1 cm}{6} = \frac{n}{6 (n + 1)} cm = \frac{n}{6 (n + 1)} cm \end{array}$

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