a screw gauge has 50 divisions on its circular scale and its screw Moves 1 mm on turning it by 2 rotations when the flat end of the screw is in contact with the stud the zero of circular scale lies below the baseline and 4th division of circular scale is in line with the baseline. find
(1) the pitch
(2) the least count
(3) zero error of the screw gauge
Answers
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
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
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
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=1/2
pitch=0.5mm
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 of count=0.5/50
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