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Question

a scale is calibrated into divisions of 0.5mm each . If 20divisions of this scale divided into 25 equal parts of a vernier , calculate the least count .

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Solution

1 M.S.D. = 0.5 mm ( main scale division)

25 V.S.D. = 20 M.S.D.

Thus,

1 V.S.D. = 20/25 M.S.D. = 0.8 M.S.D. = 0.8 x 0.5 mm = 0.4 mm

Now, Least Count of Vernier Caliper = 1 M.S.D. – 1.V.S.D. = 0.5 mm – 0.4 mm = 0.1 mm

Hence least count of vernier calliper is "0.1 mm

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