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Q.

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm and each rotation corresponds to one main scale division. 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, what is the thickness (in mm) of the wire?
(round off to nearest integer)

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Detailed Solution

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Least count $= \frac{0.25}{5 \times 100} cm = 5 \times 10^{- 4} cm$ ,

Reading $= 4 \times 0.05 cm + 30 \times 5 \times 10^{- 4} cm = 2.15 mm$

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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm and each rotation corresponds to one main scale division. 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, what is the thickness (in mm) of the wire? (round off to nearest integer)