Question

13.
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.
Ans. (i) 0.5 mm (ii) 0.01 mm (iii) + 0.04 mm

Answer 1

Answer:

Given : 50 division on circular scale

(i) One complete rotation, or the distance travelled by the screw is the pitch

1 mm movement has 2 rotations.

∴ pitch = distance / rotation =

1

/

2

= 0.5 mm

(ii) Least count = Pitch/ total number of divisions on circular head

=0.5/50=0.01mm

(iii) Zero error = coinciding division × least count

=+4\times0.01mm=+0.04mm