1.Units, Dimensions and Measurement
medium

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, $0.1$ $mm$

(JEE MAIN-2020)

Solution

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

Standard 11
Physics

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