By energy conservation
\(\frac{1}{2} \cdot \frac{ YA }{ L } \cdot x ^{2}=\frac{1}{2} mv ^{2}\)
\(\frac{0.5 \times 10^{9} \times 10^{-6} \times(0.04)^{2}}{0.1}=\frac{20}{1000} v ^{2}\)
\(\therefore \quad v ^{2}=400\)
\( v =20 m / s\)
[તારનો આડઇેદનું ક્ષેત્રણ $=0.005 \mathrm{~cm}^2 \gamma=2 \times 10^{11} \mathrm{Nm}^{-2}$ અને $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]