A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with a main scale division.

Find —

(i) the least count and

(ii) the zero error of the vernier callipers.

(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 (n) = 10

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

Substituting the values in the formula above, we get,

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

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

(ii) As we know,

Zero error = L.C. x Coinciding division (with sign)

and

Coinciding division = 3

L.C. = 0.01 cm

Substituting the values in the formula above we get,

$\text {Zero error} = \text {0.01 cm x 3} \\[0.5em] \Rightarrow \text {Zero error} = \text {0.03 cm} \\[0.5em]$

Hence, the zero error of the vernier callipers = + 0.03 cm.