A) \[nR\]
B) \[{{n}^{2}}R\]
C) \[\frac{R}{n}\]
D) \[\frac{R}{{{n}^{2}}}\]
Correct Answer: B
Solution :
Given: \[{{l}_{1}}=l,{{l}_{2}}=nl\] As the volume of wire does not change after stretching so, \[{{l}_{1}}{{A}_{1}}={{l}_{2}}{{A}_{2}}\] \[\Rightarrow \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{l}{nl}=\frac{1}{n}\] The resistance of the wire is \[R=\rho \frac{l}{A}\] \[\Rightarrow \] \[R\propto \frac{l}{A}\] Hence, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}\] \[=\frac{1}{n}\times \frac{1}{n}=\frac{1}{{{n}^{2}}}\] Therefore, \[{{R}_{2}}={{n}^{2}}R\] \[(\because {{R}_{1}}=R)\]
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