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  • Solve this problem - Properties of Solids and Liquids - JEE Main-3

A  wire fixed at the upper end stretches by length l by applying a force F The work done in stretching is

  • Option 1)

    F/2l

  • Option 2)

    Fl

  • Option 3)

    2Fl

  • Option 4)

    Fl/2

 

Answers (1)

best_answer

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

 

 Work done by constant force

W=\frac{1}{2}\times stress\times strain \times volume

W=\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}.AL=\frac{Fl}{2}

W=\frac{Fl}{2}

Correct option is 4.

 


Option 1)

F/2l

This is an incorrect option.

Option 2)

Fl

This is an incorrect option.

Option 3)

2Fl

This is an incorrect option.

Option 4)

Fl/2

This is the correct option.

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perimeter

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