In a screw gauge, 5 complete rotations of the screw cause it to move… | JEE Main 2018 (15 Apr, Shift 1)

Q58JEE Main 2018 · 15 APR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25\text{ cm}$ There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

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Answer: (b)

Pitch $= \dfrac{0.25}{5} = 0.05\text{ cm}$ (one main scale division)
LC $= \dfrac{0.05}{100} = 0.0005\text{ cm}$
Thickness $= 4\times 0.05 + 30\times 0.0005 = 0.2 + 0.015 = 0.2150\text{ cm}$
So the answer is (b).

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Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2018 (15 Apr, Shift 1)

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