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A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \mathrm{~mm}$. Then the elastic energy stored in the wire is
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$0.1 \mathrm{~J}$
Elastic energy per unit volume $=1 / 2 \times$ stress $\times$ strain
Elastic Energy $=1 / 2 x$ stress $x$ strain $x$ volume
$\begin{aligned} & =1 / 2 \times F / A \times(\Delta \mathrm{L} / \mathrm{L}) \times(\mathrm{AL}) \\ & =1 / 2 \times F \Delta \mathrm{L} \\ & =1 / 2 \times 200 \times 10^{-3}\end{aligned}$
Elastic Energy $=0.1 \mathrm{~J}$
Elastic Energy $=1 / 2 x$ stress $x$ strain $x$ volume
$\begin{aligned} & =1 / 2 \times F / A \times(\Delta \mathrm{L} / \mathrm{L}) \times(\mathrm{AL}) \\ & =1 / 2 \times F \Delta \mathrm{L} \\ & =1 / 2 \times 200 \times 10^{-3}\end{aligned}$
Elastic Energy $=0.1 \mathrm{~J}$
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