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
A) $2.25\ mm$
B) $2.20\ mm$
C) $1.20\ mm$
D) $1.25\ mm$
Explanation
Zero error $= 5 \times \frac{0.5}{50} = 0.05\ mm$ Actual measurement $= 2 \times 0.5\ mm + 25 \times \frac{0.5}{50} - 0.05\ mm$ $= 1\ mm + 0.25\ mm - 0.05\ mm = 1.20\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.