A screw gauge was found to have a pitch equal to 0.8 mm. What should be the number of divisions on its head so that you can read up to 0.001 cm using this?

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800

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Solution

The correct option is B
80

We know that, Least Count = $\frac{P i t c h}{N o . o f d i v i s i o n s o n h e a d s c a l e}$

Substituting the values, 0.001 cm = 0.01 mm = $\frac{0.8 m m}{N o . o f d i v i s i o n s o n h e a d s c a l e}$

Solving we get, No. of divisions on head scale = $\frac{0.8 m m}{0.01 m m}$ = 80

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A screw gauge was found to have a pitch equal to 0.8 mm. What should be the number of divisions on its head so that you can read up to 0.001 cm using this?

The correct option is B 80 We know that, Least Count = PitchNo. of divisions on head scale Substituting the values, 0.001 cm = 0.01 mm = 0.8mmNo. of divisions on head scale Solving we get, No. of divisions on head scale = 0.8mm0.01mm = 80