The head scale of a screw gauge has 100 divisions and the screw moves 4 mm ahead in four rotations of the circular head. The pitch and the least count are respectively

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Solution

We know that, Pitch = $\frac{(D i s t a n c e t r a v e l l e d o n t h e p i t c h s c a l e)}{(N o . o f r o t a t i o n s o f t h e h e a d s c a l e)}$ = $\frac{(4 m m)}{(4)}$ = 1mm

Also, Least Count = $\frac{(P i t c h)}{(N o . o f d i v i s i o n s o n t h e h e a d s c a l e)}$ = $\frac{(1 m m)}{(100)}$ = 0.01mm

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The head scale of a screw gauge has 100 divisions and the screw moves 4 mm ahead in four rotations of the circular head. The pitch and the least count are respectively

We know that, Pitch = (Distance travelled on the pitch scale)(No. of rotations of the head scale) = (4mm)(4) = 1mm Also, Least Count = (Pitch)(No. of divisions on the head scale) = (1mm)(100)= 0.01mm