A screw gauge has the least count of $0.01 ext{ mm}$ and there are $50$ divisions in its circular scale. The

Question

A screw gauge has the least count of $0.01	ext{ mm}$ and there are $50$ divisions in its circular scale. The pitch of the screw gauge is:

Accepted Answer

1. **Recall the formula:** The least count of a screw gauge is given by the formula: $	ext{Least Count} = \frac{	ext{Pitch}}{	ext{Number of divisions on the circular scale}}$.
2. **Rearrange for Pitch:** $	ext{Pitch} = 	ext{Least Count} 	imes 	ext{Number of divisions on the circular scale}$.
3. **Substitute the given values:** 
   $	ext{Least count} = 0.01	ext{ mm}$
   $	ext{Number of circular scale divisions} = 50$
   $	ext{Pitch} = 0.01	ext{ mm} 	imes 50 = 0.5	ext{ mm}$.

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