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JEE Main 2019 (12 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2019 (12 Jan, Shift 1)

Q54JEE Main 2019 · 12 JAN · SHIFT 1Easy

Least Count, Vernier Callipers & Screw Gauge

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $\mu$ m diameter of a wire is:

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Answer: (c)

To measure $5\,\mu$ m the least count must be $5\,\mu\text{m} = 0.005$ mm.
$N = \dfrac{\text{pitch}}{\text{LC}} = \dfrac{1\text{ mm}}{0.005\text{ mm}} = 200$
So the answer is (c) 200.

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