\(\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta\)

\(\frac{\Delta r / r}{\Delta l / I}=0.5\)

Substitute \(\Delta r / r=2 / 100\)

\(\frac{\Delta l}{l}=\frac{4}{100}\)

\(\therefore\% \text { increase }=\frac{\Delta l}{l} \times 100=4 \%\)

\(\because A \propto r^2\)

So \(\frac{\Delta A}{A}=\frac{2 \Delta r}{r}\)

\(\frac{4}{100}=2 \times \frac{\Delta r}{r}\)

\(\frac{2}{100}=\frac{\Delta r}{r}\)