A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is

  • Option 1)

    0.2\: J

  • Option 2)

    10\: J

  • Option 3)

    20\: J

  • Option 4)

    0.1\: J

 

Answers (1)

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

Correct option is 4.


Option 1)

0.2\: J

This is an incorrect option.

Option 2)

10\: J

This is an incorrect option.

Option 3)

20\: J

This is an incorrect option.

Option 4)

0.1\: J

This is the correct option.

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