The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In the measurement of the diameter of a wire, the main scale reads 2 mm and the 45th mark on the circular scale coincides with the baseline. Find — (1) The least count, and (2) The diameter of the wire

(i) Pitch = 1 mm

Number of divisions on circular head = 100

Substituting the values in the formula above we get,

Least count = 1/100 mm

$\text{Least count}=0.01\text{mm}=0.001\text{cm}$

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

(ii) Diameter of the wire = main scale reading + circular scale reading     Eq – 1

and

Circular scale reading = p × L.C.     Eq – 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.