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JEE Main 2021 (16 Mar, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2021 (16 Mar, Shift 1)

Q41JEE Main 2021 · 16 MAR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n $-$ 1)th division of the main scale. The least count of the callipers in mm is:

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Answer: (a)

$n\text{ VSD} = (n-1)\text{ MSD} \Rightarrow 1\text{ VSD} = \dfrac{n-1}{n}a$
LC $= a - \dfrac{(n-1)a}{n} = \dfrac{a}{n}$ cm $= \dfrac{10a}{n}$ mm
So the answer is (a).

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