Question

there are 50 divisions on a circular scale of screw gorge in these rotations of screw advances 1.5 millimetre on main scale . Find out pitch and least count of the screw gauge.

Answer 1

Answer:  

Explanation: no of divisions = 50

                      length in screw gauge on main scale = 1.5

                      pitch =?

                      least count = ?

pitch formula =   distance moved by scale/ no of rotations

                       =    1.5/50

                       =     0.03

least count formula = pitch / total no of divisions

                                  = 0.03/ 50

                                    = 0.006

   

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