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Question
A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on the main scale is 2 divisions and 79th circular scale division coincides with baseline.
If the number of divisions on the main scale is 10 to a centimetre and circular scale has 100 divisions, calculate
- pitch
- observed diameter
- least count
- corrected diameter.
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Solution
(1) The number of divisions on the main scale are 10 to a centimetre
⇒ Pitch = `"Unit"/"No. of divisions in unit"=(1 "cm")/10` = 0.1 cm
(2) No. of circular scale divisions = 100
∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`
= `0.1/100` cm
= 0.001 cm
(3) Main scale reading = 2 division
⇒ Main scale reading = 2 × Pitch = 0.2 cm
Circular scale reading = 79 division
∴ Observed diametre = M.S. reading + L.C. × C.S. reading
= 0.2 + 0.001 × 79
= 0.2 + 0.079 cm
= 0.279 cm
(4) Negative zero error = 7 division
∴ Correct = −(−7 × L.C.)
= −(−7 × 0.001) cm
= +0.007 cm
Correct diametre = Observed diametre + Correction
= 0.279 + 0.007
= 0.286 cm
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