In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions in circular scale and pitch of screw gauge is $0.1$ mm. When diameter of a sphere is measured, the reading of main scale is $5$ mm and $50^{\text{th}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is ______ mm.
Answer: (A) $5.045$
Least count $=\dfrac{\text{pitch}}{\text{divisions}}=\dfrac{0.1}{100}=0.001$ mm.
With the studs touching, the $5^{\text{th}}$ division is on the reference line: positive zero error $=+5\times0.001=+0.005$ mm.
Observed reading $=5+50\times0.001=5.050$ mm.
Corrected diameter $=5.050-0.005=5.045$ mm.
Solution by Sreeraj P, M.Sc Physics