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प्रश्न
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on the main scale is 3 divisions and the 24th circular scale division coincides with baseline.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
- pitch
- observed diameter.
- least count
- corrected diameter.
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उत्तर
(1) The number of divisions on the main scale are 20 to a centimetre
⇒ Pitch = `"Unit"/"No. of divisions in unit"=(1 "cm")/20` = 0.05 cm
(2) No. of circular scale divisions = 50
∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`
= `0.05/50` cm
L.C. = 0.001 cm
(3) Main scale reading = 3 division
⇒ Main scale reading = 3 × Pitch = 3 × 0.05 = 0.15 cm
Circular scale reading = 24 division
∴ Observed diametre = M.S. reading + L.C. × C.S. reading
= 0.15 + 0.001 × 24
= 0.15 + 0.024
= 0.174 cm
(4) Negative zero error = 8 division
Correction = −(−8 × L.C.)
= −(−8 × 0.001) cm
= +0.008 cm
Correct diametre = Observed diametre + Correction
= 0.174 + 0.008
= 0.182 cm
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