There are 50 divisions on the circular scale of a screw gauge. If the head of the screw is given 5 revolutions , the spindle advances by 5 mm. find the pitch and least count. When a reading is taken , the edge of the circular scale is found to be between 0.3 cm and 0.4 cm and 25th division of circular scale coincides with the base line.. find the reading. If the zero error is+ 0.004 cm, what will be the correct reading?
(What's the Main Scale Reading??? )
Answers
Answer:
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