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Vernier Caliper

A vernier has...

Question

A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?

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Solution

Step 1, Given data
Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
Step 2, Finding the least count
We know,
L.C. = $\frac{V a l u e o f 1 m s d}{N u m b e r o f d i v i s i o n s o n v e r n i e r s c a l e}$
Putting all the values
$L C = \frac{1}{10} = 0.1$
Hence the least count is 0.1 mm or 0.01 cm

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