In a screw gauge, there are 100 divisions on the circular scale and...

Least count of screw gauge is$= \dfrac{Pitch}{No . ofcirculardivisions}$

$= \dfrac{0 . 5mm}{100}$

Least count, $LC = 5 \times {10}^{- 3}mm$

The zero of circular scale lies $6$ divisions below the line of graduation. So, zero error is positive.

Positive zero error $= MSR + CSR ( LC )$

$= 0mm + 6\left(5 \times {10}^{- 3}mm\right)$

Diameter of wire, $d = MSR + CSR ( LC ) – Positivezeroerror$

$= 4 \times 0 . 5mm + 46\left(5 \times {10}^{- 3}\right) - 6\left(5 \times {10}^{- 3}\right)mm$

$= 2mm + 40 \times 5 \times {10}^{- 3}mm = 2 . 2mm$

Hence, diameter of wire is $220 \times {10}^{- 2}mm$.