Question

spindle advances 1 mm when the screw is turned
through two revolutions.
(i) What is the pitch of the screw gauge ?
(ii) What is the least count of the screw gauge?​

Answer 1

Answer:

Given,

Number of circular division=50 divisions

Linear distance moved in two complete rotation=1mm

Therefore,

(i)  Pitch of the screw gauge

pitch=  1/2*1mm

        =0.5 mm

(ii)  Least count of the screw gauge

Least count=     pitch ÷no of circular div              

                  =                0.5/50

                       =         5/10*5

                        =   0.01mm

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