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# A wire suspended vertically from one of it's end is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1mm. Then the energy stored in the wire is 0.1 J 0.2 J 10 J 20 J

Arun
25763 Points
2 years ago

As we learnt in

Work Done in Stretching Wire / Elastic P.E. -

$=\frac{1}{2}\frac{YAl^{2}}{L}=\frac{1}{2}Fl$

- wherein

L - Length of wire

$l$ - increase in length

We know that

$E=\frac{1}{2}\times F\times \Delta l$

$=\frac{1}{2}\times 200\times 10^{-3}J=0.1\ J$