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  • A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l.

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r is pulled by a force 2f. Find the increase in length of this wire.

Answers (1)

Let Y be Young's modulus of the material of the wire.

 Y= \frac{stress}{strain}, Stress=\frac{Force}{Area}=\frac{F}{A}

And  Strain= \frac{Change~ in~ Length}{Original~ length}

Given, For the first wire length = L

Applied force = f

Radius = r

Then, for the first wire,

Y= \frac{f}{\pi\cdot r^{2}} \div \frac{l}{L}= \frac{fL}{\pi.r^{2}.l}

l= \frac{fL}{\pi.r^{2}.Y}

Now, Find the change in length of the material for the second wire.

For the second wir, the length = 2L

Applied force = 2f

Radius = 2r

Therefore, the increased length of this wire can be calculated as 

l{}'= \frac{2f\times 2L}{\pi.(2r)^{2}.Y}= \frac{fL}{\pi.r^{2}.Y} ^{} 

As both the wires are made of the same material, their Young’s modulus is the same.

Hence, l=l{}'= \frac{fL}{\pi.r^{2}.Y} ^{}

Posted by

infoexpert24

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