Question

A screw gauge has $50$ divisions on its circular scale. The circular scale is $4 $ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively :

• A. Negative, $2\, \mu m$
• B. Positive, $10\, \mu m$
• C. Positive, $0.1\, \mu m$
• D. Positive, $0.1\, mm$

Answer 1

Answer: (B)

Least count of screw gauge
$=\frac{\text { Pitch }}{\text { no. of division on circular scale }}$
$=\frac{0.5}{50} mm =1 \times 10^{-5} m$
$=10 \mu m$
Zero error in positive