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# How to solve this problem- 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

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)

$0.1\: J$

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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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