A screw gauge has a pitch of 1 mm. What should be the number of divisions on its head so that the smallest measurement that can be made using this is 0.001 cm?

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Solution

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 = 0.01mm = $\frac{(1 m m)}{(N o . o f d i v i s i o n s o n h e a d s c a l e)}$

On solving we get, No. of divisions on head scale = $\frac{(1 m m)}{(0.01 m m)}$ = 100

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A screw gauge has a pitch of 1 mm. What should be the number of divisions on its head so that the smallest measurement that can be made using this is 0.001 cm?

We know that, Least Count = (Pitch)(No. of divisions on head scale) Substituting the values, 0.001 cm = 0.01mm = (1mm)(No. of divisions on head scale) On solving we get, No. of divisions on head scale = (1mm)(0.01mm) = 100