A) 200 \[\Omega \] 400 \[\Omega \]
B) 100 \[\Omega \] 200 \[\Omega \]
C) 400 \[\Omega \] 160 \[\Omega \]
D) 400 \[\Omega \] 1600 \[\Omega \]
Correct Answer: D
Solution :
(i) Since, given wire of fixed mass is recanted to twice the length \[R\propto {{l}^{2}}\] .....(i) Let R' be the new resistance \[R'\propto l{{'}^{2}}\] ?...(2) Hence, \[\frac{R'}{R}={{\left( \frac{l'}{l} \right)}^{2}}\] or \[R'={{\left( \frac{{{l}_{1}}'}{l} \right)}^{2}}R\] Here, \[R=100\Omega \,\,l'=2l\] So, \[R'={{\left( \frac{2l}{l} \right)}^{2}}\times 100=400\Omega \] (ii) Now in terms of radius \[R\propto \frac{1}{{{r}^{4}}}\] Here, \[r'=\frac{r}{2}\] Hence, \[\frac{R'}{R}={{\left( \frac{r}{r'} \right)}^{4}}\] or \[R'={{\left( \frac{r}{r'} \right)}^{4}}\times R\] So, \[R'={{\left( \frac{\frac{r}{r}}{2} \right)}^{4}}R=16R\] \[=16\times 100=1600\Omega \]
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