A) \[l\]
B) \[2l\]
C) \[l/2\]
D) \[4l\]
Correct Answer: C
Solution :
Youngs elastic coefficient \[Y=\frac{Fl}{\pi {{r}^{2}}\Delta l}\] Wires are of same material, so their Youngs coefficient will be same. \[\therefore \] \[\frac{{{F}_{1}}{{l}_{1}}}{\pi r_{1}^{2}\Delta {{l}_{1}}}=\frac{{{F}_{2}}{{l}_{2}}}{\pi r_{2}^{2}\Delta {{l}_{2}}}\] \[\frac{F\times L}{\pi {{r}^{2}}\times l}=\frac{F\times 2L}{\pi {{(2r)}^{2}}\times \Delta {{l}_{2}}}\] \[\Delta {{l}_{2}}=\frac{l}{2}\] Extension produced in the 2nd wire \[=\frac{l}{2}\]
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