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JEE Main 2020 (9 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2020 (9 Jan, Shift 1)

Q52JEE Main 2020 · 9 JAN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

If the screw on a screw-gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is:

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Answer: (d)

Pitch $= \dfrac{3\text{ mm}}{6} = 0.5\text{ mm}$
LC $= \dfrac{0.5}{50} = 0.01\text{ mm} = 0.001\text{ cm}$
So the answer is (d) $0.001\text{ cm}$ .

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