In a Vernier Caliper 10 divisions of Vernier scale is equal to the...

Given that $1M . S . D = 1mm$

And $9M . S . D = 10V . S . D$

$⇒ 1V . S . D = 0 . 9M . S . D = 0 . 9mm$

L.C of vernier caliper $= 1 - 0 . 9 = 0 . 1mm = 0 . 01cm$

Zero error $= \left(10 - 4\right) \times 0 . 1mm = 0 . 6mm$ as at zero the, the zero of the vernier is shifted towards left zero error is negative. Therefore, zero error$= - 0 . 6mm$.

Reading $= M . S . R + V . S . R -$ Zero error

Reading$= 3 + 6 \times 0 . 01 - \left[- 0 . 06\right]$

Reading$= 3 + 0 . 06 + 0 . 06$

Reading$= 3 . 12cm$

The closest option is $3 . 10cm$