A) \[l\]
B) \[2l\]
C) \[4l\]
D) \[\frac{l}{2}\]
Correct Answer: A
Solution :
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.
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