Question

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

1. pitch
2. observed diameter.
3. least count
4. corrected diameter.

Answer 1

(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