Question:
One main scale division of a vernier callipers is 'a' cm and \(n^{\text{th}}\) division of the vernier scale coincide with \((n - 1)^{\text{th}}\) division of the main scale. The least count of the callipers in mm is :
One main scale division of a vernier callipers is 'a' cm and \(n^{\text{th}}\) division of the vernier scale coincide with \((n - 1)^{\text{th}}\) division of the main scale. The least count of the callipers in mm is :
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Least count = 1 MSD - 1 VSD, with \(n\) VSD = \((n-1)\) MSD, then convert cm to mm.
Updated On: Nov 4, 2026
- \(\frac{10a}{n}\)
- \(\frac{10na}{(n - 1)}\)
- \(\left(\frac{n - 1}{10n}\right)a\)
- \(\frac{10a}{(n - 1)}\)
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The Correct Option is A
Solution and Explanation
Given \(n\) VSD = \((n - 1)\) MSD with 1 MSD = \(a\) cm.
1 VSD = \(\frac{(n - 1)}{n}a\) cm.
Least count = 1 MSD - 1 VSD = \(a - \frac{(n - 1)a}{n} = \frac{a}{n}\) cm.
In mm: \(\frac{a}{n}\times 10 = \frac{10a}{n}\) mm.
Answer: (1) \(\frac{10a}{n}\)
1 VSD = \(\frac{(n - 1)}{n}a\) cm.
Least count = 1 MSD - 1 VSD = \(a - \frac{(n - 1)a}{n} = \frac{a}{n}\) cm.
In mm: \(\frac{a}{n}\times 10 = \frac{10a}{n}\) mm.
Answer: (1) \(\frac{10a}{n}\)
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