The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line.

Find —

1. The least count, and
2. The diameter of the wire

(i) As we know,

$\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]$

Given,

Pitch = 1 mm

Number of divisions on circular head = 100

Substituting the values in the formula above we get,

$\text {Least count} = \dfrac{1}{100} \text{mm} \\[0.5em] \text {Least count} = 0.01 \text {mm} = 0.001 \text {cm}\\[0.5em]$

Hence, the least count of the screw gauge = 0.001 cm

(ii) As we know,

Diameter of the wire = main scale reading + circular scale reading     [Equation 1]

and

Circular scale reading = p x L.C.     [Equation 2]

p = 45

L.C. = 0.001 cm

Substituting the values in the Equation 2 we get,

Circular scale reading = 45 x 0.001 = 0.045

Hence, circular scale reading = 0.045 cm

Given, main scale reads 2 mm = 0.2 cm

Hence, main scale reading = 0.2 cm

Substituting the values in Equation 1 we get,

Diameter of the wire = 0.2 cm + 0.045 cm = 0.245 cm

Hence, the diameter of the wire is 0.245 cm.