If the length of a vernier scale having 25 divisions correspond to 23 main scale divisions, then given that 1 M.S.D = 1 mm, the least count of the vernier calipers is :

80 mm

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0.04 mm

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0.08 cm

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None of the above

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Solution

The correct option is D None of the above
Least count of a vernier calipers is $L . C = 1 M . S . D - 1 V . S . D$ where $M . S . D =$ main scale division and $V . S . D =$ vernier scale division.
Here, $25 V . S . D = 23 M . S . D$ or $1 V . S . D = (23 / 25) M . S . D$
Thus, $L . C = 1 M . S . D - (23 / 25) M . S . D = (2 / 25) M . S . D = (2 / 25) (1 m m) = 0.08 m m$

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