The smallest division on the main scale of a Vernier calipers is 0...

Given, $10VSD = 9MSD$

Least count of the vernier caliper,

$LC = MSD - VSD ⇒ LC = MSD - \dfrac{9MSD}{10} = \dfrac{MSD}{10} = \dfrac{0 . 1cm}{10} = 0 . 1mm ⇒ LC = 0 . 01cm$

Case I: When sphere is not present between jaws then,

main value = Zero error $= M . S . R + \left(L . C .\right) \times V . S . R$

Zero error $= \left[- 1 + 0.1 \times \left(10 - 6\right)\right]mm$

$= - 0.04cm$

Case II: When sphere is present between jaws then,

main value $= 3.1 + 0.01 \times 1$

$= 3.11cm$

So,

True value = Diameter of sphere $= 3.11 - \left(- 0.4\right)$

$= 3.15cm$