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Question 20

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 $$0.5\,\text{mm}$$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively:

$$\text{Pitch} = 0.5\text{ mm},\quad \text{Divisions} = 50$$

$$\text{LC} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm} = 10\ \mu\text{m}$$

Since the circular scale is 4 units ahead of the pitch scale marking prior to use, it registers a non-zero positive reading when the studs are in contact.

$$\text{Zero Error} = +4 \times \text{LC} = +40\ \mu\text{m}\quad \text{(Positive Error)}$$

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