someone please help me in solving this...pls...
Answers
Answer:
Pitch of the screw gauge = 0.5 mm
Least count of the screw gauge = 0.01 mm
Error of the screw gauge = + 0.04 mm
The diameter of the needle = 0.70 mm
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
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
Total reading = reading on the sleeve + reading on the thimble
Reading on the sleeve = 0.5 mm
Thimble scale = 24 divisions
Reading on the thimble :
Given the least count as 0.01 mm it means :
1 division = 0.01 mm
The reading on the thimble scale is therefore = 24 × 0.01 =0.24 mm
Total reading = 0.5 mm + 0.24 mm
= 0.74 mm
However we have a zero error of + 0.04 mm.
This means the scale begins at 0.04 mm and not 0.
The correct reading is therefore :
= 0.74 mm - 0.04 mm = 0.70 mm
The correct diameter of the needle = 0.70 mm