The pitch of a screw gauge having 100 divisions on its circular scale is 1 mm. When the jaws are in contact, the zero of the circular scale lies 4 divisions below the reference line. With a wire held

Question

The pitch of a screw gauge having 100 divisions on its circular scale is 1 mm. When the jaws are in contact, the zero of the circular scale lies 4 divisions below the reference line. With a wire held between the jaws, the linear scale reads 5 mm and the 42nd division of the circular scale coincides with the reference line. Find the diameter of the wire.

Answer:

5.38

Solution:

LC $= \frac{1 mm}{100} = 0.01$ mm

Positive zero error $= 4 \times 0.01 = 0.04$ mm

Measured $= 5 + 42 \times 0.01 = 5.42$ mm

Actual diameter $= 5.42 - 0.04 = 5.38$ mm

TestHub · Accepted Answer

LC = 100 1 mm ​ = 0.01 mm Positive zero error = 4 × 0.01 = 0.04 mm Measured = 5 + 42 × 0.01 = 5.42 mm Actual diameter = 5.42 − 0.04 = 5.38 mm

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