In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is mm.
- 5.045
- 5.055
- 5.450
- 5.550
Answer
(1) 5.045
Solution
Least count = pitchnumber of circular divisions = 0.1100 = 0.001 mm
With the studs in contact the reference line is at the 5th division (above zero) ⇒ positive zero error:
Zero error = +5 0.001 = +0.005 mm
Observed reading = MSR + CSR LC = 5 + 50 0.001 = 5.050 mm
Corrected reading = observed reading - zero error
⇒ Diameter = 5.050 - 0.005 = 5.045 mm
Answer: (1)