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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 . 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 error, the thickness of the wire is

Options

  1. A0 . 4300   cm
  2. B0 . 3150   cm
  3. C0 . 0430   cm
  4. D0 . 2150   cm

Correct answer

D. 0 . 2150   cm

Step-by-step solution

The Least count = 0.25 5 × 100   cm = 5 × 10 - 4   cm . Reading = 4 × 0.05   cm + 30 × 5 × 10 - 4   cm = 0.2 + 0.0150   cm = 0.2150   cm

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