b
\(Y =\frac{ T }{ A } \frac{ dx }{ dy }\)

\(m =20\,kg\)

\(A =0.4\,m^{2}\)

\(1=20\,m\)

let extension is \(dy\) in length \(dx\)

\(Y =\frac{\text { stress }}{\text { strain }}\)

\(Y =\frac{\frac{ T }{ A }}{\frac{ d }{ dx }}=\frac{ T }{ A } \cdot \frac{ dx }{ dy }\)

\(dy =\frac{ Tdx }{ AY }\)

Tension at a distance \(x\) from lower end \(=\frac{ mg }{\ell} x\)

So. \(\int_{0}^{\Delta l} dy =\int_{0}^{\ell} \frac{ mg }{\ell} x \frac{ dx }{ AY }\)

\(\Delta \ell=\frac{ mg }{\ell AY }\left[\frac{ x ^{2}}{2}\right]_{0}^{\ell}\)

\(\Delta \ell=\frac{ mg \ell}{2\,AY }\)

\(\Delta \ell=\frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}}\)

\(2500 \times 10^{-11}\)

\(\Delta \ell=25 \times 10^{-9}\)

\(= x \times 10^{-9}\)