A) \[\frac{{{L}_{1}}+{{L}_{2}}}{2}\]
B) \[\sqrt{{{L}_{1}}{{L}_{2}}}\]
C) \[\frac{{{T}_{2}}{{L}_{1}}-{{T}_{1}}{{L}_{2}}}{{{T}_{2}}-{{T}_{1}}}\]
D) \[\frac{{{T}_{2}}{{L}_{1}}+{{T}_{1}}{{L}_{2}}}{{{T}_{2}}+{{T}_{1}}}\]
Correct Answer: C
Solution :
Let the initial length of the metal wire is L. The strain at tension \[{{T}_{1}}\]is \[\Delta {{L}_{1}}={{L}_{1}}-L\] The strain at tension \[{{T}_{2}}\] is\[\Delta {{L}_{2}}={{L}_{2}}-L\] Suppose, the Youngs modulus of the wire is Y, then \[\frac{\frac{{{T}_{1}}}{A}}{\frac{\Delta {{L}_{1}}}{L}}=\frac{\frac{{{T}_{2}}}{A}}{\frac{\Delta {{L}_{2}}}{L}}\] where, A is an cross-section of the wire. Assume to be same at all the situations. \[\Rightarrow \] \[\frac{{{T}_{1}}}{A}\times \frac{L}{\Delta {{L}_{1}}}=\frac{{{T}_{2}}}{A}\times \frac{L}{\Delta {{L}_{2}}}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{({{L}_{1}}-L)}=\frac{{{T}_{2}}}{({{L}_{2}}-L)}\] \[{{T}_{1}}({{L}_{2}}-L)={{T}_{2}}({{L}_{2}}-L);L=\frac{{{T}_{2}}{{L}_{1}}-{{T}_{1}}{{L}_{2}}}{{{T}_{2}}-{{T}_{1}}}\]
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