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

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

Q22JEE Main 2024 · 1 FEB · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:

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Answer: (a)

$10\text{ MSD} = 11\text{ VSD} \Rightarrow 1\text{ VSD} = \dfrac{10}{11}$ MSD
LC $= \left(1 - \dfrac{10}{11}\right)\times 5 = \dfrac{5}{11}$ units
So the answer is (a) $\dfrac{5}{11}$ .

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