The pitch of a screw gauge is `1 mm` and three are `100` divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies `6` divisions below the reference line. When a wire a placed between the jaws, `2` linear scale divisions are clearly visible while ` 62` divisions on circular scale coincide with the reference line. Determine the diameter of the wire.

Answer

B

Explanation

`LC = (1 mm)/(100) = 0.01 mm`
The instrument has a positive zero error,
`e = + n(LC) = +(6 xx 0.01) =0.06 mm`
Linear scale reading = `2 xx (1 mm) = 2 mm`
Circular scale reading
= `62 xx (0.01 mm) = 0.62 mm`
`:.` Measured reading
= `2 + 0.62 = 2.62 mm`
or true reading
= `2.62 - 0.06`
= `2.56 mm`.