Question

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

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

Answer 1

The correct option is C

Positive,$10\mu m$

Step 1. Given data :

Displacement =$0.5mm$

Screw gauge divisions = $50$

Step 2. Find the least count of the screw gauge:

$LeastCount=\frac{pitch}{no.ofdivisions}$

$$\begin{gathered} =\left(\frac{0.5}{50}\right)mm \\ =1\times {10}^{-5}m \\ =10\mu m \end{gathered}$$

And the nature of the zero error is positive because it is$4$ units ahead of pitch scale marking.

Hence, option (C) is correct.