In a screw gauge, there are $100$ divisions on the circular scale and the main scale moves by $0.5\ \text{mm}$ on a complete rotation of the circular scale. The zero of circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, $4$ linear scale divisions are clearly visible while $46^\text{th}$ division of the circular scale coincides with the reference line. The diameter of the wire is ______ $\times10^{-2}\ \text{mm}$.
Numerical value type. Enter your answer.
Answer: 220
Least count $=\dfrac{0.5}{100}=0.005\ \text{mm}$.
Zero of the circular scale below the reference line means a positive zero error: $+6\times0.005=0.03\ \text{mm}$.
Reading $=4\times0.5+46\times0.005=2.23\ \text{mm}$.
Diameter $=2.23-0.03=2.20\ \text{mm}=220\times10^{-2}\ \text{mm}$.
Solution by Sreeraj P, M.Sc Physics