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

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:

TestHub · Accepted Answer

Least count = Number of parts on vernier scale Value of 1 part on main scale ​ = 5 × 100 0.25 ​ cm = 5 × 1 0 − 4 cm Reading = 4 × 0.05 cm + 30 × 5 × 1 0 − 4 cm = ( 0.2 + 0.0150 ) cm = 0.2150 cm (Thickness of wire)

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