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JEE Main 2024 (9 Apr, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2024 (9 Apr, Shift 1)

Q17JEE Main 2024 · 9 APR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

One main scale division of a vernier caliper is equal to $m$ units. If $n$ th division of main scale coincides with $(n + 1)$ th division of vernier scale, the least count of the vernier caliper is:

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Answer: (b)

$n\text{ MSD} = (n+1)\text{ VSD} \Rightarrow 1\text{ VSD} = \dfrac{nm}{n+1}$
LC $= m - \dfrac{nm}{n+1} = \dfrac{m}{n+1}$
So the answer is (b) $\dfrac{m}{n+1}$ .

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