The pitch and the number of divisions, on the circular scale, for a… | JEE Main 2019 (9 Jan, Shift 2)

Q56JEE Main 2019 · 9 JAN · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

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Answer: (c)

LC $= \dfrac{0.5}{100} = 0.005\text{ mm}$ ; zero 3 divisions below mean line: zero error $= +3\times 0.005 = +0.015\text{ mm}$
Observed reading $= 5.5 + 48\times 0.005 = 5.74\text{ mm}$
Thickness $= 5.74 - 0.015 = 5.725\text{ mm}$ . So the answer is (c).

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Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2019 (9 Jan, Shift 2)

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