Physics - Current Electricity Question with Solution | TestHub

$\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}$ Now, $1=1+\frac{1}{10}=\frac{111}{10}$ and therefore, $\mathrm{A}=\frac{10\mathrm{A}}{11}$ So ${\mathrm{R}}^{\prime }=\frac{\mathrm{p}\times \left(\frac{111}{10}\right)}{\frac{10\mathrm{A}}{11}}=\frac{\rho \mathrm{l}}{\mathrm{A}}\times \frac{(11{)}^2}{(10{)}^2}=1.21\mathrm{R}$ Now resistance becomes $1.21$ times of initial and specific resistance is the intrinsic property so remains same.