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Q.
The circular divisions of shown screw gauge are $50$. It moves $0.5 \,mm$ on main scale in one rotation. The diameter of the ball is
ManipalManipal 2014
Solution:
Zero error $=5 \times \frac{0.5}{50}=0.05\, mm$
Actual measurement
$=2 \times 0.5\, mm +25 \times \frac{0.5}{50}-0.05\, mm$
$=1\, mm +0.25\, mm -0.05\, mm =1.20\, mm$