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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

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

Grade:11

1 Answers

Arun
25750 Points
5 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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