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NEET 2019 · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from NEET 2019

Q55NEET 2019Medium

Least Count, Vernier Callipers & Screw Gauge

The main scale of a vernier callipers has $n$ divisions/cm. $n$ divisions of the vernier scale coincide with $(n - 1)$ divisions of main scale. The least count of the vernier callipers is,

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Answer: (d)

1 MSD $= \dfrac{1}{n}$ cm; $n\text{ VSD} = (n-1)\text{ MSD} \Rightarrow 1\text{ VSD} = \dfrac{n-1}{n}$ MSD.
LC $= 1\text{ MSD} - 1\text{ VSD} = \dfrac{1}{n}\text{ MSD} = \dfrac{1}{n^2}$ cm
So the answer is (d) $\dfrac{1}{n^2}$ cm.

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