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JEE Main 2022 (30 Jun, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (30 Jun, Shift 1)

Q31JEE Main 2022 · 30 JUN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

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 five equal parts, will be:

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Answer: (a)

1 MSD $= \dfrac{1\text{ cm}}{5} = 2$ mm
$n\text{ MSD} = (n+1)\text{ VSD} \Rightarrow$ LC $= \dfrac{1\text{ MSD}}{n+1} = \dfrac{2}{n+1}$ mm
So the answer is (a).

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