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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 is0.1 J 0.2 J 10 J 20 J

Abhishek Dogra , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 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

 

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