A) \[\frac{R}{\sqrt{2}}\]
B) \[2R\]
C) \[\frac{R}{2}\]
D) \[\sqrt{2}R\]
Correct Answer: D
Solution :
Key Idea: Volume of wire does not change on stretching. Let initial length of wire be I and its becomes\[l\] on stretching. The initial cross-section of the wire is A, it becomes A after stretching. Initial volume is\[Al\], and final volume is \[Al\] Initial volume = Final volume \[Al=Al\] \[\Rightarrow \] \[\frac{A}{A}=\frac{l}{l}\] Given, \[\frac{l}{l}=n\] \[\therefore \] \[\frac{A}{A}=\frac{l}{l}=n\] Resistance of wire before stretching is \[R=\rho \frac{l}{A}\] ... (i) where \[\rho \] is specific resistance. Resistance becomes R after stretching \[R\rho \frac{l}{A}\] ... (i) From Eqs. (i) and (ii), we get \[\frac{R}{R}=\left( \frac{l}{l} \right)\,\,\left( \frac{A}{A} \right)=n\times n={{n}^{2}}\] \[\Rightarrow \] \[R={{n}^{2}}R\] Note: The resistance of a long and thin wire will be greater than the resistance of a small and thick wire of the same material.
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