Question

wire of length L and radius r fixed at one end and a force F applied to the other end produces an extension $ l. $ The extension produced in another wire of the same material of length 2L and radius 2r by a force 2F,is

• A. $ l $
• B. $ 2l $
• C. $ 4l $
• D. $ \frac{l}{2} $

Answer 1

Answer: (A)

When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Youngs modulus $ (Y) $ of the material of the body.
$ Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{l/A} $
where, F is force, A the area, $ l $ the change in length and L the original length.
$ \therefore $ $ Y=\frac{FL}{\pi {{r}^{2}}l} $ r being radius of the wire.
Given, $ {{r}_{2}}=2{{r}_{1}},{{L}_{2}}=2{{L}_{1}},{{F}_{2}}=2{{F}_{1}} $
Since, Youngs modulus is a property of material, we have
$ {{Y}_{1}}={{Y}_{2}} $
$ \therefore $ $ \frac{{{F}_{1}}{{L}_{1}}}{\pi _{1}^{2}{{l}_{1}}}=\frac{2{{F}_{1}}\times 2{{L}_{1}}}{\pi {{(2{{r}_{1}})}^{2}}{{l}_{2}}} $
$ \Rightarrow $ $ {{l}_{2}}={{l}_{1}}=l $
Hence, extension produced is-same as that in the other wire.