A) \[9R\]
B) \[3R\]
C) \[R/2\]
D) \[R\]
Correct Answer: A
Solution :
Initial resistance \[{{R}_{1}}=R\] Initial length of the wave \[{{I}_{1}}=I\] Final length of the wave \[{{I}_{2}}=3\,I\] As the volume of the wire remains same after stretched \[\pi r_{1}^{2}I=\pi r_{2}^{2}\times 3\,I\] \[{{r}_{2}}=\frac{{{r}_{1}}}{\sqrt{3}}\] The resistance is given by \[R=\rho \times \frac{I}{A}=\rho \times \frac{1}{\pi {{r}^{2}}}\propto \frac{1}{{{r}^{2}}}\] Hence, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}=\frac{I}{3I}\times \frac{{{\left( \frac{{{r}_{1}}}{\sqrt{3}} \right)}^{3}}}{{{({{r}_{1}})}^{2}}}\] \[=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}\] \[{{R}_{2}}=9{{R}_{1}}=9R\]
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