A) 2R
B) 4R
C) 8R
D) R
Correct Answer: B
Solution :
The resistance of a wire of length\[l,\]area of cross-section A and specific resistance\[\rho \]is given by \[R=\rho \frac{l}{A}\] On stretching the wire, the volume remains the same. \[\therefore \] \[{{A}_{1}}{{l}_{1}}={{A}_{2}}{{l}_{2}}\] \[\Rightarrow \] \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{2l}{l}=2\] or \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{1}{2}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\rho {{l}_{1}}}{{{A}_{1}}}\times \frac{{{A}_{2}}}{\rho {{l}_{2}}}\] \[=\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)\left( \frac{{{A}_{2}}}{{{A}_{2}}} \right)=\left( \frac{l}{2l} \right)\left( \frac{1}{2} \right)\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[{{R}_{2}}=4{{R}_{1}}\] Given, \[{{R}_{1}}=R\] \[\therefore \] \[{{R}_{2}}=4R\]
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