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A 5 metrelong wire is fixed to the ceiling. A weight of $10 \mathrm{~kg}$ is hung at the lower end and is 1 metre above the floor. The wire was elongated by $1 \mathrm{~mm}$. The energy stored in the wire due to stretching is
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0.05 joule
$W=\frac{1}{2} \times F \times l=\frac{1}{2} m g l$
$=\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{-1}=0.05 \mathrm{~J}$
$=\frac{1}{2} \times 10 \times 10 \times 1 \times 10^{-1}=0.05 \mathrm{~J}$
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