Young's modulus $$(Y)=\frac{\text { Stress }}{\text { Strain }}=\frac{4 m g l}{\pi d^{2} \Delta l}$$

$$\begin{gathered} \mathrm{Y}=\frac{4 m g l}{\pi d^{2} l} \\\\ =\frac{4 \times 1 \times 9.8 \times 2}{3.14 \times(0.4)^{2} \times(0.8) \times 10^{-6} \times 10^{-3}} \\\\ =2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \end{gathered} $$

Where, $$\quad$$ Mass $$(m)=1 \mathrm{~kg}$$

$$\begin{aligned}g & =9.8 \mathrm{~m} / \mathrm{s}^{2} \\\\ \text { Length of wire } & =2 \mathrm{~m} \\\\ \pi & =3.14 \\\\ \text { Diameter }(d) & =0.4 \mathrm{~mm} \times 10^{-3} \mathrm{~m} \end{aligned}$$

Extension in the length of wire

$$\begin{aligned}(\Delta l) & =0.8 \mathrm{~mm} \\\\ & =0.8 \times 10^{-3} \mathrm{~m} \\\\ \frac{\Delta y}{y} & =\frac{2 \Delta d}{d}+\frac{\Delta l}{l} \\\\ & =\frac{2 \times(0.01)}{0.4}+\frac{(0.05)}{0.8} \\\\ & =0.1125\end{aligned}$$

$$\begin{aligned}\Rightarrow \Delta \mathrm{Y} & =(\mathrm{Y}) \times 0.1125 \\\\ & =2 \times 10^{11} \times 0.1125 \\\\ & =0.225 \times 10^{11} \\\\ & \simeq 0.2 \times 10^{11} \\\\ \text { (b) }\left(2 \pm 0.2 \times 10^{11} \mathrm{~N/}\right. & \left.\mathrm{m}^{2}\right) \end{aligned}$$