In an experiment to find out the diameter of wire using screw...

From the data given in the question,

Pitch of the screw gauge$= 0 . 5mm$.

Total divisions on the circular scale$n = 50$.

Therefore, the least count of the screw gauge is $LC = \dfrac{\text{pitch}}{n} = \dfrac{0 . 5}{50} = 0 . 01mm$

Now, $MSR = 2 . 5mm$, $CSR = 45$

Diameter reading =$MSR + LC \times CSR -$ zero error

$⇒ d = 2 . 5 + 0 . 45 - \left(- 0 . 03\right)$

$⇒ d = 2 . 98mm$