A screw gauge advances by 3mm in 6 rotations. There are 50 divisions on circular scale. Find least count of screw gauge
Answers
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=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
Zero error= +4×0.01 mm
Zero Error=+0.04 mm
Answer:
0.01 mm or 0.001 cm
Explanation:
Pitch : The translational motion of the screw is directly proportional to the total rotation of the head. The pitch of the instrument is the distance between two consecutive threads of the screw which is equal to the distance moved by the screw due to one complete rotation of the cap.
Least Count : In this case also, the minimum ( or least) measurement ( or count) of length is equal to one division on the head scale which is equal to the pitch divided by the total cap divisions.
Least Count (L.C.) = Pitch / Total no. of divisions
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.
Pitch = distance / rotation
Pitch = 3 / 6 = 0.5 mm
hence,
Least Count (L.C.) = Pitch / Total no. of divisions
= 0.5 mm / 50 = 1 / 100 mm
= 0.01 mm = 0.001 cm
Therefore, least count of screw ause will be 0.01 mm or 0.001 cm.