A) \[\frac{\sqrt{{{T}_{1}}{{T}_{2}}}}{{{L}_{1}}+{{L}_{2}}}\]
B) \[\frac{{{T}_{1}}{{L}_{1}}\,-\,\,{{T}_{2}}{{L}_{2}}}{{{T}_{1}}+{{T}_{2}}}\]
C) \[\frac{{{T}_{2}}{{L}_{1}}\,-\,\,{{T}_{1}}{{L}_{2}}}{{{T}_{2}}-{{T}_{1}}}\]
D) \[\frac{{{T}_{1}}{{L}_{1}}\,-\,\,{{T}_{2}}{{L}_{2}}}{{{T}_{1}}+{{T}_{2}}}\]
Correct Answer: C
Solution :
Let \[\operatorname{L} =original length of wire\] \[\therefore \,\,\,\,\,\,\,\,\,\Delta {{L}_{1}}={{L}_{1}}-L\] Similarly, change in length of second wire is \[\Delta {{L}_{2}}={{L}_{2}}-L\] Now, \[Y=\frac{{{T}_{1}}}{A}\times \frac{L}{\Delta {{L}_{1}}}=\frac{{{T}_{2}}}{A}\times \frac{L}{\Delta {{L}_{2}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{{{T}_{1}}}{\Delta {{L}_{1}}}\,=\frac{{{T}_{2}}}{\Delta {{L}_{2}}}\] \[\frac{{{T}_{1}}}{{{L}_{1}}-L}\,=\frac{{{T}_{2}}}{{{L}_{2}}-L}\] \[L=\frac{{{T}_{2}}{{L}_{1}}-{{T}_{1}}{{L}_{2}}}{{{T}_{2}}-{{T}_{1}}}\]
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