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Question:

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 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:

0.3150cm

0.2150cm

0.0430cm

0.4300cm

Solution:

The correct option is B
In one rotation scale moves 0.25/5 = 0.05 cm
Least count = 0.05 × 1/100 cm
For 4 main scale division = 4 × 0.05 = 0.2 cm
For circular scale division = 30 × 0.05 × 1/100 = 0.015 cm
Thickness of wire = 0.2 + 0.015 = 0.2150 cm