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JEE Main 2020 (3 Sep, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2020 (3 Sep, Shift 1)

Q49JEE Main 2020 · 3 SEP · SHIFT 1Easy

Least Count, Vernier Callipers & Screw Gauge

Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

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Answer: (b)

LC $= \dfrac{0.1\text{ cm}}{50} = 0.002$ cm
A correct reading must be a whole multiple of $0.002$ cm (last digit even).
Only $2.124$ cm $= 1062\times 0.002$ cm fits.
So the answer is (b) $2.124$ cm.

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