Question

The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and 63rd division on the circular scale coincides with the reference line. The diameter of the wire is:

• A. 1.63 cm
• B. 0.163 cm
• C. 0.163 m
• D. 1.63 m

Answer 1

Answer: (B)

0.163 cm

Answer 2

Least count (LC) of screw gauge $= \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \, mm}{100} = 0.01 \, mm$
Main scale reading (MSR) = 1 mm
Circular scale reading (CSR) = 63 divisions
Final reading = MSR + CSR $\times$ LC = $1 \, mm + 63(0.01 \, mm) = 1.63 \, mm = 0.163 \, cm$