Question

The pitch of a screw gauge is $0.5mm$ and there are $100$ divisions on its circular scale. The instruments reads $+2$ divisions when nothing is put in-between its jaws. In measuring the diameter of a wire, there are $8$ divisions on the main scale and $83^{rd}$ division coincides with the reference line. The diameter of the wire is $\frac{k}{100}$ (in $mm$ ), then find $k$ .

Answer 1

Least count $=\frac{0 . 5 mm}{100}$
Positive error $=2\times \frac{0 . 5 mm}{100}=\frac{1}{100}mm$
Scale reading $=8\times 0.5mm$ $+83\times \frac{0 . 5 mm}{100}-\frac{1}{100}mm$
$\Rightarrow 4.405mm\approx4.40mm$