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 Young's modulus (Y) of the material of the body. \[Y=\frac{stress}{strain}=\frac{F/A}{l/L}\] where, \[F\]is force, \[A\] the area, \[l\] the change in length and L the original length. \[Y=\frac{FL}{\pi {{r}^{2}}l}\] \[r\]being radius of the wire. Given,\[{{r}_{2}}=2{{r}_{1}},\,\,{{L}_{2}}=2{{L}_{1}},\,\,{{F}_{2}}=2{{F}_{1}}\] Since, Young's 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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