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(-) Answer-strain energy release rate\begin{array}{l}G=p i \times \operatorname{strss}^{2} \times a / E \\a=\text { half crack length }=10 \mathrm{~cm}=0.1 \mathrm{~m} \\E=70 \times 10^{9} \mathrm{Nlm}^{2} \\G=200 \\\text { Sthess }=\sqrt{4.456 \times 10^{13}} \\=6675581.178 \mathrm{~N} / \mathrm{m}^{2}\end{array}\begin{aligned}\text { Stress } & =\sqrt{4.456 \times 10^{13}} \\& =6675581.178 \mathrm{~N} / \mathrm{m}^{2}\end{aligned}to find the load p\begin{array}{l}4 \sin g \\\sigma=\frac{P}{A} \\\text { area }=0.2 \times 0.5 \times 10^{-2} \mathrm{~m}^{2} \\6675581.178=\frac{P}{0.2 \times 0.5 \times 10^{-2}} \\P=6675.58 \mathrm{~N} \\P=6.6756 \mathrm{kN}\end{array}if you have any doubt please comment please upvote it will help me thanks. ...