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
ANegative, $2\, \mu m$
BPositive, $10\, \mu m$
CPositive, $0.1$ $\mu m$
DPositive, $0.1$ $mm$
JEE MAIN 2020, Medium
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BPositive, $10\, \mu m$
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
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