• # question_answer If a wire of resistance R is stretched to double its length, then the new resistance is: A)  $4R$               B)  $2R$C)  R                    D)  none of these

Initial Resistance ${{R}_{1}}=R$ Initial length     ${{l}_{1}}=l$                               Final lenght       ${{l}_{2}}=2l$ The volume of wire is given by $\pi r_{1}^{2}\times l$ .                Since the wire of volume remains same after stretch therefore,             $\pi r_{1}^{2}\times l=\pi r_{2}^{2}\times 2l$ $\Rightarrow$               ${{r}_{2}}=\frac{{{r}_{1}}}{{{r}_{2}}}$ The resistance of wire is given by $R=\rho \times \frac{l}{\Delta }=\rho \times \frac{l}{\pi {{r}^{2}}}$ $\Rightarrow$$R\propto \frac{l}{{{r}^{2}}}$ $\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}/r_{1}^{2}}{{{l}_{2}}/r_{2}^{2}}=\frac{l}{2l}=\frac{{{\left( {{r}_{1}}/\sqrt{2} \right)}^{2}}}{{{\left( {{r}_{1}} \right)}^{2}}}$ $\Rightarrow$               $\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$ $\Rightarrow$               ${{R}_{2}}=4{{R}_{1}}=4R$