Question

15. A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correct up to 0.001 mm with its help? Ans. 500​

Answer 1

Answer:

$last \: count \: = \: \frac{pitch}{number \: of \: divisions}$

$0.001 = \frac{0.5}{n}$

$n = 500$

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Answer 2

Answer:

0.5÷0.001=500 divisions on the head