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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Solution

The correct option is B
80

We know that, $Least Count = \frac{Pitch}{No.of divisions on head scale} Substituting the values, 0.001 c m = 0.01 m m = \frac{0.8 m m}{Substituting the values,} On solving we get, No. of divisions on head scale = \frac{0.8 m m}{0.01 m m} = 80$

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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 scaleSubstituting the values,0.001 cm=0.01 mm=0.8mmSubstituting the values,On solving we get, No. of divisions on head scale=0.8mm0.01mm=80

The correct option is B
80

We know that, $Least Count = \frac{Pitch}{No.of divisions on head scale} Substituting the values, 0.001 c m = 0.01 m m = \frac{0.8 m m}{Substituting the values,} On solving we get, No. of divisions on head scale = \frac{0.8 m m}{0.01 m m} = 80$