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  • A uniform elastic rod of cross section area A, neutral length L and Young's modulus Y is placed on a smooth horizontal surface.Now two horizontal forces (of magnitude 2F and 4F) directed along

A uniform elastic rod of cross section area A, neutral length L and Young's modulus Y is placed on a smooth horizontal surface.Now two horizontal forces (of magnitude 2F and 4F) directed along the length of rod and in opposite direction act at two of its end as shown. After the rod has acquired steady state. The extension of the rod will be

Option: 1

\frac{2FL}{AY}
 


Option: 2

\frac{4FL}{AY}

 

 


Option: 3

\frac{3FL}{AY}


Option: 4

\frac{3FL}{2AY}


Answers (1)

best_answer

Tension in rod of a distance x from right edge is

T= F(4-2\frac{x}{L})\; x= L

Total extension in rod = \int dL= \int \frac{T}{AY}dx

                                   x= 0

\Delta L= \frac{3FL}{AY}

Posted by

Ritika Harsh

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