A) \[16R\]
B) \[4R\]
C) \[2R\]
D) \[zero\]
Correct Answer: A
Solution :
Resistance of wire initially is \[R=\rho \frac{l}{A}\] or \[R=\rho \frac{l}{\pi {{r}^{2}}}\] Volume of wire remains same even after stretching it. Therefore, \[\pi {{r}^{2}}l=\pi r{{}^{2}}l\] where,\[r\]and\[l=\]initial radius and length of wire \[r\]and \[l=\]final radius and length of wire Since, \[r=\frac{r}{2},\]therefore \[\pi {{r}^{2}}l=\pi {{\left( \frac{r}{2} \right)}^{2}}l\Rightarrow l=4l\] Now, new resistance of wire is given by \[R=\rho \frac{l}{A}=\rho \frac{l}{\pi r{{}^{2}}}\] \[=\rho \cdot \frac{4l}{\pi \left( \frac{l}{2} \right)}=16\times \rho \cdot \frac{l}{\pi {{r}^{2}}}\] \[\Rightarrow \] \[R=16R\]
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