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 $0.5 \mathrm{~mm}$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively :
Correct Option: , 2
Solution:
Least count of screw gauge
$\frac{\text { Pitch }}{\text { sion on circular scale }}$
$=\frac{0.5}{50} \mathrm{~mm}=1 \times 10^{-5} \mathrm{~m}$
$=10 \mu \mathrm{m}$
Zero error in positive
Ans. (2)