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Question
(a) A vernier scale has 20 divisions. It slides over the main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with the main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of the sphere.
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Solution
No. of divisions on vernier scale = 20
Pitch 0.5 mm
Least count = `"Pitch"/"No. of divisions on the vernier scale"`
L.C. = `0.5/20` mm
L.C. = 0.025 mm
L.C. = 0.0025 cm
(a) There is 38 number of main scale divisions on the left of the zero of the vernier scale.
⇒ Main scale reading = `38/2` mm = 1.9 cm
Vernier scale division coinciding with the main scale = 18th
Diameter of sphere = Main scale reading + L.C. × V.S.D.
= 1.9 + 0.0025 × 18
= 1.9 + 0.0450
= 1.945 cm
(b) Negative error = −0.04 cm
⇒ Correction = −(−0.04) = +0.04 cm
Corrected diameter of sphere = Observed reading + Correction
= 1.945 + (+0.04)
= 1.945 + 0.04
= 1.985 cm
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