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Question
A micrometre screw gauge has a positive zero error of 7 divisions, such that its main scale is marked in 1/2 mm and the circular scale has 100 divisions. The spindle of the screw advances by 1 division complete rotation.
If this screw gauge reading is 9 divisions on the main scale and 67 divisions on the circular scale for the diameter of a thin wire, calculate
- Pitch
- L.C.
- Observed diameter
- Corrected diameter
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Solution
(1) Pitch = `1/2` mm
= 0.5 mm
= 0.05 cm
(2) No. of circular scale divisions = 100
∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`
L.C. = `(0.05 "cm")/100`
= 0.0005 cm
(3) Main scale reading = 9 divisions = `9xx1/2` mm
= 4.5 mm
= 0.45 cm
Circular scale reading = 67 div.
∴ Observed diameter = M.S. reading + L.C. × C.S. reading
= 0.45 + 0.0005 × 67
= 0.45 + 0.0335
= 0.4835 cm
(4) Positive zero error = 7 divisions
Correction = −(Error × L.C.)
= −(7 × 0.0005) cm
= −0.0035 cm
∴ Corrected diameter = Observed diameter + Correction
= 0.4835 + (−0.0035)
= 0.4835 − 0.0035
= 0.4800 cm
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