A wire has a resistance 10 $\mathrm{\Omega }$. It is stretched by one-tenth of its original length. Then its resistance will be

Here volume remains constant. Thus
${\mathrm{\pi r}}_1^2\mathrm{l}={\mathrm{\pi r}}_2^2(11\mathrm{l}/10)$
or ${\mathrm{\pi r}}_2^2=\frac{10}{11}\left({\mathrm{\pi r}}_1^2\right)$
{ $\therefore$When wire is stretched by 1/10 of its original length, the new length of wire becomes (11 / / 10)}
Let the new resistance be R_2. Then
$\begin{array}{l}{\mathrm{R}}_2=\frac{\mathrm{\rho }\cdot \left(\frac{11}{10}\mathrm{l}\right)}{{\mathrm{\pi r}}_2^2}=\frac{(11/10)\mathrm{\rho l}}{(10/11){\mathrm{\pi r}}_1^2}\\ =\frac{(11/10)}{10/11)}\left[\frac{\mathrm{\rho l}}{{\mathrm{\pi r}}_1^2}\right]\\ =\frac{(11/10)}{(10/11)}\times 10=\frac{121}{10}=12\cdot 1\mathrm{\Omega }\end{array}$