A) \[4R\]
B) \[8R\]
C) \[16R\]
D) \[2R\]
Correct Answer: C
Solution :
Resistance of a wire of length l, area A and specific resistance \[\rho \] is given by \[R=\rho \frac{l}{A}\] Also, Volume (V) Length (l) Area where \[A=\pi {{r}^{2}}\](r is radius) When the wire is stretched its volume (V) remains constant. Hence, \[R=\frac{\rho V}{{{\pi }^{2}}{{r}^{4}}}\] ?.(i) When radius is halved \[R'=\frac{\rho V}{{{\pi }^{2}}{{\left( \frac{r}{2} \right)}^{4}}}\] ?..(ii) \[\therefore \] \[\frac{R'}{R}=\frac{16\rho V}{{{\pi }^{2}}{{r}^{4}}}\times \frac{{{\pi }^{2}}{{r}^{4}}}{\rho V}=16\] \[\Rightarrow \] \[R'=16R\] Hence, new resistance increases to sixteen times its original value.
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