Question

In a screw gauge, 5 complete motations of the screw cause it to a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisiors and 30 circular scale divisions. Assuming negligible zero, the thickness of the wire is:

• A. $0.4300$ cm
• B. $0.2150$ cm
• C. $0.3150$ cm
• D. $0.0430$ cm

Answer 1

The correct option is B $0.2150$ cm

In one rotation, scale moves $\frac{0.25}{5}=0.05cm$

Least count $=0.05\times {10}^{-2}cm$

For $4$main scale division $=4\times 0.05=0.2cm$

For circular scale division $30\times 0.05\times {10}^{-2}=1.5\times {10}^{-2}cm$

Thickness of wire $0.2+0.015=0.2150cm$