Q. Two full turns of the circular scale of a screw gauge cover a distance of $1\ mm$ on its main scale. The total number of divisions on the circular scale is $50$. Further, it is found that the screw gauge has a zero error of $- 0.03\ mm$. While measuring the diameter of a thin wire, a student notes the main scale reading of $3\ mm$ and the number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is
A) $3.73\ mm$
B) $3.67\ mm$
C) $3.38\ mm$
D) $3.32\ mm$
Explanation
Least count $= \frac{Pitch}{No.\ of\ div.\ in\ circular\ scale}$ $= \frac{0.5}{50} = 0.01\ mm$ Actual reading $= 0.01 \times 35 + 3 = 3.35\ mm$ Taking error into consideration $= 3.35 + 0.03 = 3.38\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.