Question

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

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

Answer 1

The correct option is D $0.2150cm$

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 divosion = $30\times 0.05\times {10}^{-2}=1.5\times {10}^{-2}cm$

Thickness of wire = $0.2+0.015=0.2150cm$