A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two rotations. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the base line and 4th division of circular scale is in line with the base line. Find — (i) the pitch, (ii) the least count and (iii) the zero error, of the screw gauge.

(i) Pitch = Distance moved ahead in 1 revolution.

Given,

Total number of divisions on a circular scale = 50

Distance covered in two revolutions = 1 mm

∴, we get,

Pitch = 1/2 mm = 0.5 mm

Hence, pitch of the screw gauge = 0.5 mm

(ii) Pitch = 0.5 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

Least count = 0.5/50

Least count = 0.01 mm

Hence, the least count of the screw gauge = 0.01 mm

Zero error = Coinciding Division x Least Count

Coinciding Division = 4

L.C. = 0.01 mm

Substituting the values in the formula above we get,

Zero error = 4 x 0.01 = + 0.04 mm

Hence, zero error, of the screw gauge = + 0.04 mm