A) \[6\,\Omega \]
B) \[2\,\Omega \]
C) \[8\,\Omega \]
D) \[4\,\Omega \]
Correct Answer: C
Solution :
If length is increased then from the formula \[R\propto {{l}^{2}},\] so \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)}^{2}}\] so \[{{R}_{2}}={{\left( \frac{2l}{l} \right)}^{2}}\times {{R}_{1}}=4\times 2=8\,\Omega \]
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