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 :

Updated On: Jul 29, 2024
  • Negative, $2\, \mu m$
  • Positive, $10\, \mu m$
  • Positive, $0.1\, \mu m$
  • Positive, $0.1\, mm$
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The Correct Option is B

Solution and Explanation

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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The physical world includes the complications of the natural world around us. It is a type of analysis of the physical world around us to understand how it works. The fundamental forces that control nature are:

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