Question

The pitch of a screw god is one millimetre what would be its least count if the number of circular divisions is 50

Answer 1

Explanation:

According to the question, the screw gauge has 50 divisions on its circular scale.

The Pitch is calculated as the screw completes one rotation completely, the distance travelled by the screw defines the pitch.

Thereby, from given data in question we have movement of 1 mm by 2 rotations

1. Pitch = distance / rotation

Pitch=\frac{1}{2}Pitch=

2

1

\bold{Pitch=0.5\ \mathrm{mm}}Pitch=0.5 mm

2. Least count is given by the ratio of pitch and the divisions present over the circular scale of screw gauge.

Least Count = pitch / number of division over circular scale

Least\ Count=\frac{0.5}{50}Least Count=

50

0.5

\bold{Least\ Count=0.01 \mathrm{mm}}Least Count=0.01mm

3. Zero Error of the screw gauge is calculated by the coinciding division of circular scale with zero and least count.

Zero Error = Coinciding division x Least Count

Zero\ Error=+4 \times 0.01\ \mathrm{mm}Zero Error=+4×0.01 mm

\bold{Zero\ Error=+0.04\ \mathrm{mm}}Zero Error=+0.04 mm