Physics - Current Electricity Question with Solution | TestHub

Let Initial length $=l_1$ Final length $=l_2$ Initial area $={\mathrm{A}}_1$ Final area $=A_1$ $\therefore$ Volume remains same $\therefore {\mathrm{A}}_1{\mathrm{l}}_1={\mathrm{A}}_2{\mathrm{l}}_2\Rightarrow {\mathrm{A}}_1{\mathrm{l}}_1={\mathrm{A}}_2\frac{{\mathrm{l}}_1}{4}$ $\Rightarrow 4{\mathrm{A}}_1={\mathrm{A}}_2$ Initial resistance, ${\mathrm{R}}_1=\frac{\rho l_1}{A_1}=160\Omega$ (given) Final resistance, ${\mathrm{R}}_2=\frac{{\rho \mathrm{l}}_2}{{\mathrm{A}}_2}$ $\therefore \frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=\frac{{\mathrm{l}}_2{\mathrm{A}}_1}{{\mathrm{A}}_2{\mathrm{l}}_1}=\frac{l_1}{4}\frac{{\mathrm{A}}_1}{4{\mathrm{A}}_1{l}_1}$ $\Rightarrow R_2=\frac{1}{16}{R}_1=\frac{1}{16}\times 160=10\Omega$