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

(i) As we know,

$\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier (n)}}$

Given,

Total number of divisions on vernier = 10

Value of one main scale division(x) = 1 mm

Substituting the values in the formula given above we get,

$\text{L.C.} = \dfrac{\text{1mm}}{\text{10}} \\[0.5em] \Rightarrow \text{L.C.} = 0.1 \text{mm} \\[0.5em] \Rightarrow \text{L.C.} = 0.01 \text{cm} \\[0.5em]$

Hence, least count of the vernier callipers is 0.01cm.